Theory of Structures II — Chapter 2: Force method

Asked 2 times
2078 Bhadra · 6 marks
2067 Asar · 5 marks

State and prove Castigliano's second theorem (theorem for determination of displacement in a structural system).

Answer

Statement

For a linearly elastic structure with supports that do not yield and no temperature effect, the partial derivative of the total strain energy $U$ with respect to any applied force $P_i$ (or couple) gives the displacement (or rotation) of the point of application in the direction of that force:

\Delta_i = \frac{\partial U}{\partial P_i}, \qquad \theta_i = \frac{\partial U}{\partial M_i}

Proof

Take an elastic body acted on by loads $P_1, P_2, \ldots, P_n$ producing displacements $\Delta_1, \Delta_2, \ldots, \Delta_n$ in the directions of the loads. The strain energy equals the work done by the loads:

U = \tfrac12 \left(P_1\Delta_1 + P_2\Delta_2 + \cdots + P_n\Delta_n\right)

Case 1. Apply all loads $P_1, \ldots, P_n$ gradually, then add a small extra load $dP_i$ at point $i$ . The total strain energy is

U + \frac{\partial U}{\partial P_i}\,dP_i

Case 2. Apply $dP_i$ first, then the loads $P_1, \ldots, P_n$ . The work done is the small energy of $dP_i$ alone, plus $U$ for the loads, plus the work of $dP_i$ moving through the displacement $\Delta_i$ that the loads produce at $i$ . Since $dP_i$ is already acting at full value, the last term is $dP_i\,\Delta_i$ . Neglecting the second-order term $\tfrac12 dP_i\,d\Delta_i$ , the total is

U + dP_i\,\Delta_i

Step 3. Strain energy depends only on the final state of loading, not on the order. Equate Case 1 and Case 2:

U + \frac{\partial U}{\partial P_i}\,dP_i = U + dP_i\,\Delta_i

\Delta_i = \frac{\partial U}{\partial P_i}

Hence the displacement in the direction of a load equals the partial derivative of the strain energy with respect to that load.

Use

For bending, $U = \int \dfrac{M^2}{2EI}\,dx$ , so

\Delta_i = \int \frac{M}{EI}\,\frac{\partial M}{\partial P_i}\,dx

At a point with no load, apply a dummy load $P$ , differentiate, then set $P = 0$ . For a redundant reaction $R$ at a rigid support, $\partial U/\partial R = 0$ ; if the support settles by $\delta$ , then $\partial U/\partial R = \delta$ .

Asked 2 times
2074 Chaitra · 4 marks
2070 Chaitra (old course) · 6 marks

Calculate the reaction at the prop of a propped cantilever with uniform distributed load throughout the span using Castigliano's theorem.

Answer

Take a propped cantilever of span $L$ , fixed at A and with a prop (roller) at B, carrying a UDL $w$ per unit length. Treat the prop reaction $R_B$ as the redundant.

 w kN/m
 vvvvvvvvvvvvvvvvvvv
 ||==================o
 A (fixed)           B (prop, R_B)
 |<------- L ------->|

Take $x$ from B towards A.

M_x = R_B\,x - \frac{w x^2}{2}, \qquad \frac{\partial M_x}{\partial R_B} = x

The prop is at the same level as the fixed end, so the deflection at B is zero:

\frac{\partial U}{\partial R_B} = \int_0^L \frac{M_x}{EI}\frac{\partial M_x}{\partial R_B}\,dx = 0

\int_0^L \left(R_B\,x^2 - \frac{w x^3}{2}\right)dx = 0

\frac{R_B L^3}{3} - \frac{w L^4}{8} = 0

R_B = \frac{3wL}{8}

Other reactions

R_A = wL - \frac{3wL}{8} = \frac{5wL}{8}

M_A = \frac{3wL}{8}\,L - \frac{wL^2}{2} = -\frac{wL^2}{8}

The fixed-end moment is $wL^2/8$ (hogging).

Answer: prop reaction $R_B = \dfrac{3wL}{8}$ upward.

Asked 2 times
2073 Shrawan · 6 marks
2069 Chaitra · 7 marks

Derive the expression of the three moment theorem for a continuous beam and explain its physical meaning.

Answer

Statement

For two adjacent spans $L_1$ and $L_2$ of a continuous beam between three supports A, B and C, the support moments are related by the three moment equation (Clapeyron's theorem):

M_A L_1 + 2M_B (L_1 + L_2) + M_C L_2 = -\left(\frac{6 a_1 \bar{x}_1}{L_1} + \frac{6 a_2 \bar{x}_2}{L_2}\right)

where $a_1, a_2$ are the areas of the free (simply supported) BM diagrams on spans 1 and 2, $\bar{x}_1$ is the distance of the centroid of $a_1$ from A, and $\bar{x}_2$ is the distance of the centroid of $a_2$ from C. If the supports settle (downward $\delta_A, \delta_B, \delta_C$ ), the term $+6EI\left[\dfrac{\delta_B-\delta_A}{L_1} + \dfrac{\delta_B-\delta_C}{L_2}\right]$ is added to the right side (EI constant).

Derivation (supports at the same level, EI constant)

Let the support moments be $M_A, M_B, M_C$ (sagging positive, so hogging moments are negative). The BM diagram on each span is the sum of the free BM diagram and a trapezoid of support moments.

Step 1: Slope at B in span AB. The deviation of A from the tangent at B equals the moment about A of the $M/EI$ diagram between A and B. The diagram is the free BM diagram (area $a_1$ , centroid $\bar{x}_1$ from A) plus two triangles from $M_A$ and $M_B$ . With the supports at the same level, this deviation equals $\theta_{B1}L_1$ :

\theta_{B1}L_1 = \frac{1}{EI}\left[a_1\bar{x}_1 + \frac{M_A L_1}{2}\cdot\frac{L_1}{3} + \frac{M_B L_1}{2}\cdot\frac{2L_1}{3}\right]

\theta_{B1} = \frac{1}{EI}\left[\frac{a_1\bar{x}_1}{L_1} + \frac{M_A L_1}{6} + \frac{M_B L_1}{3}\right]

Step 2: Slope at B in span BC. In the same way, taking moments about C (with $\bar{x}_2$ measured from C), the deviation of C from the tangent at B is $-\theta_{B2}L_2$ :

\theta_{B2} = -\frac{1}{EI}\left[\frac{a_2\bar{x}_2}{L_2} + \frac{M_C L_2}{6} + \frac{M_B L_2}{3}\right]

Step 3: Continuity. The beam is continuous over B, so the slope on the left of B equals the slope on the right: $\theta_{B1} = \theta_{B2}$ .

\frac{a_1\bar{x}_1}{L_1} + \frac{M_A L_1}{6} + \frac{M_B L_1}{3} = -\frac{a_2\bar{x}_2}{L_2} - \frac{M_C L_2}{6} - \frac{M_B L_2}{3}

Multiply by 6 and rearrange:

M_A L_1 + 2M_B (L_1 + L_2) + M_C L_2 = -\frac{6a_1\bar{x}_1}{L_1} - \frac{6a_2\bar{x}_2}{L_2}

Special results

UDL $w$ on a span $L$ : $\dfrac{6a\bar{x}}{L} = \dfrac{wL^3}{4}$ .
Point load $W$ at distance $a$ from the left support of a span $L$ , with $b = L - a$ : $\dfrac{6a\bar{x}}{L} = \dfrac{Wab(L+a)}{L}$ when $\bar{x}$ is measured from the left support, and $\dfrac{Wab(L+b)}{L}$ when measured from the right support.

Physical meaning

The equation expresses continuity of slope at the middle support B. The left side is the rotation at B produced by the unknown support moments, and the right side is the rotation produced by the loads on the two spans acting as simple beams. For a beam with $n$ supports it gives $n-2$ equations, the remaining equations come from end conditions (a fixed end adds an imaginary span of zero stiffness, a free overhang gives a known moment). Solving the equations gives the support moments, and the reactions and BM follow from statics.

2074 Chaitra · 10 marks

Determine forces in all members of the truss shown in figure below using the force method. AE for all members is constant. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 50 kN horizontal at C; A and D supported.]

Similar questions: Truss: force method, 100 kN and 50 kN (2072 Kartik)

Answer

Reading of the figure: A (0, 0) hinged and D (4, 0) roller, B (0, 3) and C (4, 3) at the top; members AB, CD (3 m), BC, AD (4 m) and diagonals AC, BD (5 m). The 50 kN load acts horizontally at C (towards the right). $AE$ is constant. $D_s = 6 + 3 - 8 = 1$ . Take the force in AC as the redundant $X$ .

Reactions

$\Sigma M_A = 0$ : $V_D(4) = 50(3) \Rightarrow V_D = 37.5$ kN ( $\uparrow$ ); $V_A = 37.5$ kN ( $\downarrow$ ); $H_A = 50$ kN ( $\leftarrow$ ).

Force tables (tension +; $S_0$ with AC removed; $n$ for $X=1$ )

Member	$L$ (m)	$S_0$	$n_1$	Final $S$
AB	3.000	+37.500	-0.600	+13.194
BC	4.000	+50.000	-0.800	+17.593
CD	3.000	0.000	-0.600	-24.306
AD	4.000	+50.000	-0.800	+17.593
AC	5.000	0.000	+1.000	+40.509
BD	5.000	-62.500	+1.000	-21.991

f_{11} = \frac{\sum n^2L}{AE} = \frac{17.280}{AE}, \qquad \Delta_{10} = \frac{\sum S_0nL}{AE} = \frac{-700.000}{AE}

Compatibility

f_{11}X + \Delta_{10} = 0 \ \Rightarrow\ X = -\frac{-700.000}{17.280} = 40.51\ \text{kN}

Final forces $S = S_0 + nX$

Member	Force (kN)	Nature
AB	+13.19	tension
BC	+17.59	tension
CD	-24.31	compression
AD	+17.59	tension
AC	+40.51	tension
BD	-21.99	compression

Answer: AC $= +40.51$ kN, BD $= -21.99$ kN, AB $= +13.19$ kN, BC $= +17.59$ kN, CD $= -24.31$ kN, AD $= +17.59$ kN (positive = tension).

2072 Kartik · 6 marks

Using Castigliano's theorem, determine the moment at the fixed support A of the propped cantilever beam loaded as shown in figure below. [Figure: beam AB, A fixed, B roller, span 6 m, UDL 5 kN/m; $E = 232\ \text{kN/mm}^2$ , $I = 112.5\times10^{6}\ \text{mm}^4$ .]

Similar questions: Castigliano: fixed-end moment, 15 kN/m, L = 5 m (2070 Asar)

Answer

Propped cantilever AB: A fixed, B a roller, span $L = 6$ m, UDL $w = 5$ kN/m. $E = 232$ kN/mm $^2$ and $I = 112.5\times10^6$ mm $^4$ give $EI = 232\times10^6\times1.125\times10^{-4} = 26100$ kNm $^2$ , but the reactions of a single-span propped cantilever do not depend on $EI$ .

Take the roller reaction $R_B$ as the redundant. With $x$ measured from B:

M = R_B\,x - \frac{wx^2}{2},\qquad \frac{\partial M}{\partial R_B} = x

The deflection at B is zero:

\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^6\left(R_Bx - 2.5x^2\right)x\,dx = 0

\frac{R_B(6)^3}{3} - \frac{2.5(6)^4}{4} = 0 \ \Rightarrow\ 72R_B = 810 \ \Rightarrow\ R_B = 11.25\ \text{kN}

(This is $3wL/8 = 3(5)(6)/8 = 11.25$ kN.)

Moment at the fixed support A

M_A = R_B(6) - \frac{5(6)^2}{2} = 67.5 - 90 = -22.5\ \text{kNm}

The fixed-end moment is 22.5 kNm (hogging, $= wL^2/8 = 5(36)/8$ ).

The vertical reaction at A is $30 - 11.25 = 18.75$ kN. The maximum sagging moment is $\dfrac{9wL^2}{128} = 12.66$ kNm at $x = \dfrac{3L}{8} = 2.25$ m from B.

Answer: $M_A = 22.5$ kNm (hogging); $R_B = 11.25$ kN, $R_A = 18.75$ kN.

2072 Kartik · 7 marks

Determine the forces in all members of the truss shown in figure below by using the force method. AE is constant for all members. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 100 kN vertical at C and 50 kN horizontal at D; A hinged, B roller.]

Similar questions: Truss: force method, 4 m x 3 m panel (2074 Chaitra)

Answer

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, C (4, 3) and D (0, 3) at the top; members AB, CD (4 m), BC, DA (3 m) and diagonals AC, BD (5 m). Loads: 100 kN downward at C and 50 kN horizontal at D (towards the right). $AE$ is constant. $D_s = 6 + 3 - 8 = 1$ . Take the force in AC as the redundant $X$ .

Reactions

$\Sigma M_A = 0$ : $V_B(4) = 100(4) - 50(3) \Rightarrow V_B = 137.50$ kN ( $\uparrow$ )
$V_A = 100 - 137.50 = -37.50$ kN ( $\uparrow$ )
$H_A = 50$ kN ( $\leftarrow$ )

Force tables (tension +)

Member	$L$ (m)	$S_0$	$n_1$	Final $S$
AB	4.000	+50.000	-0.800	+33.333
BC	3.000	-100.000	-0.600	-112.500
CD	4.000	0.000	-0.800	-16.667
DA	3.000	+37.500	-0.600	+25.000
AC	5.000	0.000	+1.000	+20.833
BD	5.000	-62.500	+1.000	-41.667

f_{11} = \frac{\sum n^2L}{AE} = \frac{17.280}{AE},\qquad \Delta_{10} = \frac{\sum S_0nL}{AE} = \frac{-360.000}{AE}

Compatibility

f_{11}X + \Delta_{10} = 0 \ \Rightarrow\ X = -\frac{-360.000}{17.280} = 20.83\ \text{kN}

Final forces $S = S_0 + nX$

Member	Force (kN)	Nature
AB	+33.33	tension
BC	-112.50	compression
CD	-16.67	compression
DA	+25.00	tension
AC	+20.83	tension
BD	-41.67	compression

Answer: AB $= +33.33$ , BC $= -112.50$ , CD $= -16.67$ , DA $= +25.00$ , AC $= +20.83$ , BD $= -41.67$ kN (positive = tension).

2070 Asar · 5 marks

Determine the moment at the fixed end of the propped cantilever beam shown in figure below using Castigliano's theorem. [Figure: beam, left end fixed, right end roller, span L = 5 m with UDL 15 kN/m.]

Similar questions: Castigliano: fixed moment, propped cantilever UDL (2072 Kartik)

Answer

Propped cantilever: fixed at the left end A, roller at B, span $L = 5$ m, UDL $w = 15$ kN/m. Take the roller reaction $R_B$ as the redundant and measure $x$ from B.

M = R_Bx - \frac{wx^2}{2} = R_Bx - 7.5x^2,\qquad \frac{\partial M}{\partial R_B} = x

The deflection at B is zero:

\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^5\left(R_Bx^2 - 7.5x^3\right)dx = 0

R_B\frac{5^3}{3} - 7.5\frac{5^4}{4} = 0 \ \Rightarrow\ 41.667\,R_B = 1171.875 \ \Rightarrow\ R_B = 28.125\ \text{kN} \ \left(= \frac{3wL}{8}\right)

Moment at the fixed end

M_A = R_B(5) - 7.5(5)^2 = 140.625 - 187.5 = -46.875\ \text{kNm}

The fixed-end moment is 46.875 kNm (hogging), equal to $\dfrac{wL^2}{8} = \dfrac{15(25)}{8}$ .

Check: $R_A = wL - R_B = 75 - 28.125 = 46.875$ kN.

Answer: $M_A = 46.875$ kNm (hogging); $R_B = 28.125$ kN and $R_A = 46.875$ kN.

2069 Chaitra · 10 marks

Use Castigliano's theorem to find the moment at point C of the propped cantilever beam loaded as shown in the figure below. Take EI to be constant. [Figure: beam A-B-D-C, A hinged, B support at 2 m, D at 4 m with 30 kNm moment, C fixed end at 6 m (2 m + 2 m + 2 m).]

Similar questions: Castigliano: fixed-end moment, 15 kN load (2066 Jestha)

Answer

Assumptions: A is a hinge at $x=0$ , B a roller at $x=2$ m, the 30 kNm couple acts clockwise at D ( $x=4$ m) and C is the fixed end at $x=6$ m. EI is constant. The beam has four vertical/rotational reactions ( $R_A,R_B,R_C,M_C$ ) and two equilibrium equations, so it is indeterminate to the 2nd degree. Take $R_A$ and $R_B$ (upward) as redundants. Castigliano's theorem for redundants that do not move: $\partial U/\partial R_A=0$ and $\partial U/\partial R_B=0$ .

Bending moments (x from A, sagging +)

Portion	M
A to B (0 to 2)	$R_Ax$
B to D (2 to 4)	$R_Ax+R_B(x-2)$
D to C (4 to 6)	$R_Ax+R_B(x-2)+30$

Compatibility equations

\frac{\partial U}{\partial R_A}=\frac{1}{EI}\int M\frac{\partial M}{\partial R_A}dx=0:\quad \int_0^2R_Ax\cdot x\,dx+\int_2^4[R_Ax+R_B(x-2)]\,x\,dx+\int_4^6[R_Ax+R_B(x-2)+30]\,x\,dx=0

which simplifies to

72R_A+37.333R_B+300=0

Similarly

\frac{\partial U}{\partial R_B}=0:\quad \int_2^4[R_Ax+R_B(x-2)](x-2)dx+\int_4^6[R_Ax+R_B(x-2)+30](x-2)dx=0

37.333R_A+21.333R_B+180=0

Solution

R_A=2.250\ \text{kN}\uparrow,\qquad R_B=-12.375\ \text{kN}\ (\text{negative: acts downward})

R_C=-(R_A+R_B)=10.125\ \text{kN}\uparrow

Moment at the fixed end C

M_C=R_A(6)+R_B(4)+30=13.500+(-49.500)+30=-6.00\ \text{kNm}

The negative sign shows hogging (tension at the top) at C.

Check: moments about C of all forces: $R_A(6)+R_B(4)+30+M_C$ -terms give $M_C=-6.00$ kNm ✓ (verified by an independent stiffness analysis).

Answer: $|M_C|=6.00$ kNm, hogging (for a clockwise 30 kNm couple); $R_A=2.25$ kN, $R_B=12.375$ kN (downward), $R_C=10.125$ kN.

2066 Jestha · 10 marks

Use Castigliano's theorem and find the moment at the fixed end of the propped cantilever loaded as shown below. EI is constant. [Figure: beam AB, A fixed, B roller; 15 kN at 2 m from A, B at 5 m (2 m + 3 m).]

Similar questions: Castigliano: moment at C of propped cantilever (2069 Chaitra)

Answer

Method: the propped cantilever is indeterminate to the 1st degree. Take the roller reaction $R_B$ (upward) as the redundant. B does not move, so by Castigliano's theorem $\partial U/\partial R_B=0$ .

Data: $L=5$ m, 15 kN at 2 m from the fixed end A (so 3 m from B), EI constant.

Bending moment (x measured from B towards A, sagging +)

$0\le x\le3$ : $M=R_Bx$ , $\;\partial M/\partial R_B=x$
$3\le x\le5$ : $M=R_Bx-15(x-3)$ , $\;\partial M/\partial R_B=x$

Condition $\partial U/\partial R_B=0$

\int_0^3R_Bx\cdot x\,dx+\int_3^5\left[R_Bx-15(x-3)\right]x\,dx=0

R_B\left[\frac{x^3}{3}\right]_0^5-15\int_3^5(x^2-3x)dx=0 \;\Rightarrow\; \frac{125}{3}R_B-15(8.667)=0 \;\Rightarrow\; R_B=\frac{130\times3}{125}=3.12\ \text{kN}\uparrow

Fixed-end moment

At A ( $x=5$ ):

M_A=R_B(5)-15(5-3)=15.60-30=-14.40\ \text{kNm}

The negative sign means hogging (tension at the top).

Check with the standard formula: $M_A=\dfrac{Pab(L+b)}{2L^2}=\dfrac{15\times2\times3\times(5+3)}{2\times25}=14.40$ kNm ✓. Also $R_A=15-3.12=11.88$ kN.

Answer: fixed-end moment $M_A=14.40$ kNm (hogging); $R_B=3.12$ kN, $R_A=11.88$ kN.

2079 Baishakh · 4 marks

Explain the physical significance of flexibility and stiffness matrices with suitable examples.

Answer

Flexibility matrix $[F]$

Each element $f_{ij}$ is the displacement at coordinate $i$ caused by a unit force at coordinate $j$ . The relation is $\{\Delta\} = [F]\{P\}$ . It measures how soft the structure is: a large $f_{ij}$ means a small force gives a large displacement. Column $j$ is the deflected shape produced by a unit load at $j$ . Diagonal terms are always positive, and the matrix is symmetric ( $f_{ij} = f_{ji}$ , Maxwell's theorem). It exists only for a stable, restrained structure and is the matrix used in the force method.

Stiffness matrix $[K]$

Each element $k_{ij}$ is the force needed at coordinate $i$ to give a unit displacement at coordinate $j$ while all other coordinates are held fixed. The relation is $\{P\} = [K]\{\Delta\}$ . It measures how stiff the structure is: column $j$ is the set of restraint forces that holds the structure in the shape of a unit displacement at $j$ . Diagonal terms are positive, and the matrix is symmetric. It is used in the displacement method.

The two are inverses: $[K] = [F]^{-1}$ .

Example: cantilever of length $L$

Coordinates: 1 = vertical deflection at the free end, 2 = rotation at the free end.

[F] = \begin{bmatrix} \dfrac{L^3}{3EI} & \dfrac{L^2}{2EI} \\[2mm] \dfrac{L^2}{2EI} & \dfrac{L}{EI} \end{bmatrix}, \qquad [K] = [F]^{-1} = \begin{bmatrix} \dfrac{12EI}{L^3} & -\dfrac{6EI}{L^2} \\[2mm] -\dfrac{6EI}{L^2} & \dfrac{4EI}{L} \end{bmatrix}

$f_{11} = L^3/3EI$ is the tip deflection due to a unit tip load.
$k_{22} = 4EI/L$ is the moment needed at the tip to give a unit rotation when the tip deflection is held zero.
Note $k_{11} = 12EI/L^3 \neq 1/f_{11} = 3EI/L^3$ . The stiffness is not the reciprocal of the flexibility for a matrix, because the stiffness holds the other coordinate fixed while the flexibility leaves it free.

2079 Baishakh · 6 marks

In the beam ABC, support B settles by $\lambda R_B$ units. Find the reaction $R_B$ using Castigliano's theorem. [Figure: beam ABC, AB = BC = l, EI constant, hinge supports at A and C, support B on a spring, UDL of $P/2l$ over the full length 2l.]

Answer

Beam ABC has $AB = BC = l$ , hinged ends A and C, a spring-type support at B, and UDL $w = \dfrac{P}{2l}$ over $2l$ (total load $P$ ). Support B settles $\lambda R_B$ , where $R_B$ is its upward reaction.

 w = P/2l
 vvvvvvvvvvvvvvvvvvvvvvvv
 o---------o---------o
 A         B(R_B)    C
 |<-- l -->|<-- l -->|

By symmetry $R_A = R_C = \dfrac{P - R_B}{2}$ .

For $0 \le x \le l$ from A (the other half is the same):

M = \frac{P - R_B}{2}\,x - \frac{P}{4l}\,x^2, \qquad \frac{\partial M}{\partial R_B} = -\frac{x}{2}

Castigliano's theorem for the upward redundant $R_B$ gives the upward displacement of B. B moves down by $\lambda R_B$ , so

\frac{\partial U}{\partial R_B} = -\lambda R_B

\frac{2}{EI}\int_0^l \left[\frac{P - R_B}{2}x - \frac{P}{4l}x^2\right]\left(-\frac{x}{2}\right)dx = -\lambda R_B

Evaluate the integral:

-\frac{1}{EI}\left[\frac{P - R_B}{2}\cdot\frac{l^3}{3} - \frac{P}{4l}\cdot\frac{l^4}{4}\right] = -\lambda R_B

\frac{(P - R_B)\,l^3}{6} - \frac{P l^3}{16} = EI\,\lambda R_B

\frac{5Pl^3}{48} - \frac{R_B l^3}{6} = EI\,\lambda\,R_B

R_B = \frac{5P}{8}\cdot\frac{1}{1 + \dfrac{6EI\lambda}{l^3}}

Answer: $R_B = \dfrac{5Pl^3}{8\,(l^3 + 6EI\lambda)}$ upward. For $\lambda = 0$ (rigid support), $R_B = \dfrac{5P}{8}$ (the continuous-beam value).

2079 Baishakh · 6 marks

Determine reactions at A using the force method. [Figure: L-shaped frame; horizontal member AB of length l carrying 10 kN/m UDL, vertical member BC of height l, EI constant; A hinged, C fixed.]

Answer

L-frame: A hinged (end of horizontal member AB, length $l$ , UDL $w = 10$ kN/m), B rigid corner, vertical member BC of height $l$ down to C fixed. EI is constant.

  w = 10 kN/m
 vvvvvvvvvvvvv
 A o----------B
               |
               | l
               |
              /// C (fixed)
 |<---- l ---->|

The frame is statically indeterminate to the 2nd degree ( $r = 2 + 3 = 5$ ). Release the hinge at A to get the cantilever C-B-A, and take $X_1$ = horizontal reaction at A and $X_2$ = vertical reaction at A as redundants.

Moment expressions

In AB, $x$ measured from A; in BC, $y$ measured down from B.

Member	$M_0$ (load)	$m_1$ (unit $X_1$ )	$m_2$ (unit $X_2$ )
AB	$-\dfrac{wx^2}{2}$	$0$	$x$
BC	$-\dfrac{wl^2}{2}$	$y$	$l$

Flexibility coefficients

\begin{aligned} f_{11} &= \frac{1}{EI}\int_0^l y^2\,dy = \frac{l^3}{3EI} \\ f_{12} &= \frac{1}{EI}\int_0^l y\,l\,dy = \frac{l^3}{2EI} \\ f_{22} &= \frac{1}{EI}\left[\int_0^l x^2\,dx + \int_0^l l^2\,dy\right] = \frac{4l^3}{3EI} \end{aligned}

Load displacements

\begin{aligned} \Delta_{10} &= \frac{1}{EI}\int_0^l y\left(-\frac{wl^2}{2}\right)dy = -\frac{wl^4}{4EI} \\ \Delta_{20} &= \frac{1}{EI}\left[\int_0^l x\left(-\frac{wx^2}{2}\right)dx + \int_0^l l\left(-\frac{wl^2}{2}\right)dy\right] = -\frac{5wl^4}{8EI} \end{aligned}

Compatibility (zero displacement at the hinge A)

\begin{aligned} \tfrac13 X_1 + \tfrac12 X_2 &= \frac{wl}{4} \\ \tfrac12 X_1 + \tfrac43 X_2 &= \frac{5wl}{8} \end{aligned}

Solving: $X_2 = \dfrac{3wl}{7}$ and $X_1 = \dfrac{3wl}{28}$ .

With $w = 10$ kN/m:

V_A = \frac{30\,l}{7} = 4.29\,l\ \text{kN}\ (\uparrow), \qquad H_A = \frac{15\,l}{14} = 1.07\,l\ \text{kN}\ (\rightarrow)

(with $l$ in metres). Check from the support at C: $V_C = 10l - 4.29l = 5.71\,l$ kN up, $H_C = 1.07\,l$ kN to the left, $M_C = 5l^2/14 = 0.357\,l^2$ kNm.

Answer: at A, $H_A = 3wl/28 = 1.07\,l$ kN (to the right) and $V_A = 3wl/7 = 4.29\,l$ kN upward.

2079 Baishakh · 10 marks

Find the BM at point D for the given two hinged parabolic arch. Take $I_x = I_c \sec\theta$ . [Figure: two-hinged parabolic arch AB, span 60 m (dimensions 30 m, 20 m, 10 m), rise 10 m; UDL 30 kN/m on the left portion up to crown C; 50 kN point load at D, 10 m from support B.]

Answer

A two-hinged arch has one redundant, the horizontal thrust $H$ . With $I_x = I_c\sec\theta$ we have $ds/I = dx/I_c$ , so the thrust is

H = \frac{\int M_0\, y\, dx}{\int y^2\, dx}

where $M_0$ is the bending moment in the equivalent simply supported beam of the same span.

Data and arch profile

Span $L = 60$ m, rise $h = 10$ m, A at the left support (origin). The UDL acts on $0 \le x \le 30$ m and the 50 kN load acts at $x = 50$ m (10 m from B).

y = \frac{4h\,x(L-x)}{L^2} = \frac{x(60-x)}{90}

Reactions of the simple beam

UDL resultant $= 30 \times 30 = 900$ kN at $x = 15$ m.

V_B = \frac{900(15) + 50(50)}{60} = 266.67\ \text{kN}, \qquad V_A = 950 - 266.67 = 683.33\ \text{kN}

Free bending moment $M_0$

Range (m)	$M_0$ (kNm)
$0 \le x \le 30$	$683.33x - 15x^2$
$30 \le x \le 50$	$683.33x - 900(x-15)$
$50 \le x \le 60$	$266.67(60-x)$

Integrals

\begin{aligned} \int_0^{60} y^2\,dx &= \frac{8h^2L}{15} = 3200 \\ \int_0^{30} M_0 y\,dx &= 1347500, \quad \int_{30}^{50} M_0 y\,dx = 855556, \quad \int_{50}^{60} M_0 y\,dx = 51852 \\ \int_0^{60} M_0 y\,dx &= 2254907 \end{aligned}

H = \frac{2254907}{3200} = 704.66\ \text{kN}

Bending moment at D ( $x = 50$ m)

\begin{aligned} y_D &= \frac{50(10)}{90} = 5.556\ \text{m} \\ M_0 &= V_B \times 10 = 2666.67\ \text{kNm} \\ M_D &= M_0 - H\,y_D = 2666.67 - 704.66(5.556) = -1248.10\ \text{kNm} \end{aligned}

Answer: $H = 704.66$ kN and the bending moment at D is $M_D = -1248.10$ kNm (hogging, tension on the top/extrados).

2079 Baishakh · 10 marks

Use the consistent deformation method to draw the BM diagram of the given frame. [Figure: frame with column AB (A fixed at the bottom, B at top, height 4 m) carrying 10 kN/m UDL along its height, beam BCD with BC = CD = 3 m, 20 kN downward at C, EI constant, D supported at the right end.]

Answer

Reading of the figure: A is fixed at the base of the column AB (4 m), the beam BCD has $BC = CD = 3$ m, the column carries a horizontal UDL of 10 kN/m (towards the right), a 20 kN load acts down at C, and D is on a roller (vertical support). EI is constant.

The frame is indeterminate to the first degree ( $r = 3 + 1 = 4$ ). Take the vertical reaction $R_D$ at D as the redundant.

        20 kN
          |
  B ------C------ D  (roller)
10 ->|     v     o
kN/m |
  A (fixed)

Step 1: Primary structure (D released), loads only

It is a cantilever fixed at A. Downward deflection of D under the loads is $\Delta_D$ .

Section	$M_0$ (kNm)	$m$ (unit load at D)
D	0	0
C	0	3.00
B (beam)	-60.00	6.00
B (column)	-60.00	6.00
mid AB	-80.00	6.00
A	-140.00	6.00

( $M_0$ is the moment of the loads on the cantilever (hogging, negative) and $m$ is the moment due to a unit upward force at D (sagging, positive), so their product is negative.)

For example, at A: $M_0 = -(10 \times 4 \times 2 + 20 \times 3) = -140$ kNm and $m = +6$ kNm per kN.

Step 2: Deformations by the unit-load method

\begin{aligned} \Delta_{D0} &= \int \frac{M_0\, m}{EI}\,dx = \frac{-2530}{EI}\ \ (\text{i.e. }2530/EI\ \text{downward}) \\ f_{DD} &= \int \frac{m^2}{EI}\,dx = \frac{216}{EI} \end{aligned}

(Column AB: $6^2 \times 4 = 144$ ; beam BC: $\frac{3}{3}(36 + 18 + 9) = 63$ ; beam CD: $\frac{3}{3}(9) = 9$ ; total 216.)

Step 3: Consistent deformation (deflection at D is zero)

\Delta_{D0} + f_{DD} R_D = 0 \ \Rightarrow\ R_D = \frac{2530}{216} = 11.713\ \text{kN}\ (\uparrow)

Step 4: Other reactions (statics)

$V_A = 20 - 11.713 = 8.287$ kN ( $\uparrow$ )
$H_A = 10 \times 4 = 40$ kN ( $\leftarrow$ )
$M_A$ from $\Sigma M_A = 0$ : $M_A = 40 \times 2 + 20 \times 3 - 11.713 \times 6 = 69.72$ kNm (anticlockwise reaction).

Step 5: Bending moments $M = M_0 + R_D\, m$

Point	BM (kNm) (+ve = tension on inner/under face)	Tension side
A	-69.72	outer (left) face of column
Mid AB	-9.72	outer face
B	10.28	inner (right) face of column / underside of beam
C	35.14	underside
D	0

The BM in the column is $M(y) = -69.72 + 40y - 5y^2$ (y from A), which changes sign at $y = 2.57$ m from A. In the beam, BM rises linearly from 10.28 kNm at B to 35.14 kNm under the 20 kN load, then falls linearly to zero at D.

Answer: $R_D = 11.71$ kN up; $M_A = 69.72$ kNm; peak sagging BM 35.14 kNm at C.

2078 Kartik · 6 marks

Find the reaction at support B using Castigliano's method. [Figure: beam ABC with 20 kN/m UDL over its full length; AB = 6 m, BC = 10 m; A hinged, B roller.]

Answer

Reading of the figure: a continuous beam ABC with $AB = 6$ m and $BC = 10$ m carrying 20 kN/m over the whole length, A hinged, B and C on rollers (a beam with only A and B supported would be determinate, so C is taken as a roller). EI is constant. Take $R_B$ as the redundant.

   20 kN/m
 vvvvvvvvvvvvvvvvvvvvvvvvvv
 A o--------o B-----------o C
 |<-- 6 -->|<---- 10 ---->|

Reactions in terms of $R_B$

Moments about C: $R_A(16) + R_B(10) = 20(16)(8)$

R_A = 160 - 0.625\,R_B

Bending moments (x from A)

\begin{aligned} 0 \le x \le 6:\quad M &= R_A x - 10x^2, & \frac{\partial M}{\partial R_B} &= -0.625x \\ 6 \le x \le 16:\quad M &= R_A x + R_B(x-6) - 10x^2, & \frac{\partial M}{\partial R_B} &= 0.375x - 6 \end{aligned}

Castigliano's theorem

The support B does not settle, so $\partial U/\partial R_B = 0$ :

\int_0^6 M\,\frac{\partial M}{\partial R_B}\,dx + \int_6^{16} M\,\frac{\partial M}{\partial R_B}\,dx = 0

Substituting $R_A$ and integrating gives

75.000\,R_B - 15800.00 = 0

R_B = \frac{15800.00}{75.000} = 210.67\ \text{kN}

Other reactions

R_A = 160 - 0.625(210.67) = 28.33\ \text{kN}, \qquad R_C = 320 - 210.67 - 28.33 = 81.00\ \text{kN}

Check: $M_B = 28.33(6) - 20(6)(3) = -190$ kNm, which agrees with the three moment equation $2M_B(16) = -\tfrac{20}{4}(6^3 + 10^3)$ .

Answer: $R_B = 210.67$ kN (upward).

2078 Kartik · 6 marks

Determine forces in all members of the truss shown in the figure if the member AC is too long by 10 mm. Take AE constant for all members. [Figure: rectangular truss ABCD, 4 m wide and 3 m high, with both diagonals; A hinged, D roller.]

Answer

Reading of the figure: A (0, 0) hinged, D (4, 0) roller at the base, B (0, 3) and C (4, 3) at the top, with both diagonals AC and BD. Bar AB and CD are 3 m, BC and AD are 4 m and AC, BD are 5 m. The truss is indeterminate to the 1st degree (internal); the supports are determinate. Take the force in AC as the redundant $X$ (tension +). Lack of fit $\lambda = +10$ mm $= 0.01$ m (member too long).

No external load acts, so the primary structure carries no force and the fabrication error alone produces the member forces.

Unit force system ( $X = 1$ in AC)

Member	$L$ (m)	$n$	$n^2L$
AB	3	$-0.6$	1.08
BC	4	$-0.8$	2.56
CD	3	$-0.6$	1.08
AD	4	$-0.8$	2.56
BD	5	$+1$	5.00
AC	5	$+1$	5.00
Total			17.28

f_{11} = \sum \frac{n^2 L}{AE} = \frac{17.28}{AE}

Compatibility

The member is too long, so after assembly it is compressed. The total elongation of the closed system is zero for the self-stressed state:

f_{11}X + \lambda = 0 \ \Rightarrow\ X = -\frac{0.01\,AE}{17.28} = -0.0005787\,AE

Member forces $S = n X$ (AE in kN)

Member	Force	Nature
AC	$-0.0005787\,AE$	compression
BD	$-0.0005787\,AE$	compression
AB, CD	$+0.0003472\,AE$	tension
BC, AD	$+0.0004630\,AE$	tension

Example: for $AE = 2\times10^5$ kN, AC = BD $= -115.74$ kN, AB = CD $= 69.44$ kN, BC = AD $= 92.59$ kN.

Check at joint A (horizontal): $S_{AD} + 0.8\,S_{AC} = (0.4630 - 0.8\times0.5787)\times10^{-3}AE = 0$ . There is no external load, so all reactions are zero.

Answer: AC = BD $= -5.787\times10^{-4}\,AE$ (compression); AB = CD $= +3.472\times10^{-4}\,AE$ ; BC = AD $= +4.630\times10^{-4}\,AE$ (tension).

2078 Kartik · 10 marks

Determine reactions at the hinged support of the frame loaded as shown in the figure using the force method. [Figure: frame ABCD, beam BC of span 10 m with I = 2I carrying 50 kN/m UDL; column AB 5 m long with 1.5I, A fixed; column CD 3 m long with I, D hinged.]

Answer

Reading of the figure: A (0, 0) fixed; B (0, 5) with column AB 5 m of stiffness $1.5EI$ ; beam BC of span 10 m with stiffness $2EI$ carrying 50 kN/m; column CD 3 m of stiffness $EI$ with D hinged. As the columns are of different lengths, D is taken 2 m above the level of A. Here $EI$ stands for $E\times I$ of the lightest member.

The frame has $r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at D and take the reactions at D as redundants: $X_1$ = horizontal ( $\rightarrow$ positive) and $X_2$ = vertical ( $\uparrow$ positive). The primary structure is a cantilever fixed at A.

  B ================ C   50 kN/m
  |                  |
  | 5 m              | 3 m
  |                  o D (hinge)
 A///

Moment diagrams (sign: + tension on the inner face of the frame)

$M_0$ (50 kN/m on the cantilever): $M_B = M_{A} = -50(10)^2/2 = -2500$ kNm, falling parabolically to 0 at C.
$m_1$ (unit horizontal force at D): CD varies 0 to 3, beam constant 3, column from 3 at B down to $-2$ at A.
$m_2$ (unit vertical force at D): beam varies 10 to 0 from B to C, column constant 10.

Flexibility coefficients and load terms

\begin{aligned} f_{11} &= \frac{61.778}{EI} \\ f_{12} &= \frac{91.667}{EI} \\ f_{22} &= \frac{500.000}{EI} \\ \Delta_{10} &= \frac{-16666.7}{EI} \\ \Delta_{20} &= \frac{-114583.3}{EI} \end{aligned}

Compatibility ( $\Delta_{D,x} = \Delta_{D,y} = 0$ )

\begin{aligned} 61.778\,X_1 + 91.667\,X_2 &= 16666.7 \\ 91.667\,X_1 + 500.000\,X_2 &= 114583.3 \end{aligned}

Solving:

X_1 = -96.51\ \text{kN},\qquad X_2 = 246.86\ \text{kN}

Final bending moments $M = M_0 + X_1 m_1 + X_2 m_2$ (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-2500.00	-2.00	10.00	161.62
B (column)	-2500.00	3.00	10.00	-320.93
B (beam)	-2500.00	3.00	10.00	-320.93
mid BC	-625.00	3.00	5.00	319.77
C (beam)	0.00	3.00	0.00	-289.53
C (column)	0.00	3.00	0.00	-289.53
D	0.00	0.00	0.00	0.00

The maximum sagging moment in BC is 319.87 kNm at 5.07 m from B.

Reactions

Hinge D: $H_D = 96.51$ kN ( $\leftarrow$ ), $V_D = 246.86$ kN ( $\uparrow$ )
Fixed support A: $H_A = 96.51$ kN ( $\rightarrow$ ), $V_A = 253.14$ kN ( $\uparrow$ ), $M_A = 161.62$ kNm (anticlockwise)

Check: $\Sigma V = 253.14 + 246.86 = 500.00$ kN $= 50 \times 10 = 500$ kN; $\Sigma H = 96.51 - 96.51 = 0$ .

Answer: at the hinged support D, $H_D = 96.51$ kN (towards the left) and $V_D = 246.86$ kN (upward).

2078 Kartik · 8 marks

A two hinge parabolic arch of span 60 m and central rise of 6 m is subjected to a concentrated load of 40 kN at the crown. Allowing for rib shortening and yielding of support B by 6 mm outwards horizontally, determine the horizontal thrust. Also draw the BMD. Take $I_0 = 6\times10^{3}\ \text{cm}^4$ [?], $A_c = 1000\ \text{cm}^2$ , $E = 10\ \text{kN/mm}^2$ and consider $I_x = I_0 \sec\alpha$ .

Answer

A two-hinged arch has one redundant, the horizontal thrust $H$ . Taking $I_x = I_0\sec\alpha$ , the compatibility equation with rib shortening and yielding of a support is

H = \frac{\displaystyle\int \frac{M_0\,y}{EI_0}\,dx - \delta}{\displaystyle\int \frac{y^2}{EI_0}\,dx + \int \frac{\cos^2\alpha}{EA_c}\,dx}

where $\delta$ is the outward yield of support B. (The small axial-force term of the free structure is neglected, as is usual.)

Data (converted to kN and m)

$L = 60$ m, $h = 6$ m, parabola $y = \dfrac{4h\,x(L-x)}{L^2} = \dfrac{x(60-x)}{150}$
$EI_0 = 10\,\text{kN/mm}^2 \times 6\times10^3\,\text{cm}^4 = 10^7 \times 6\times10^{-5} = 600$ kNm $^2$
$EA_c = 10^7 \times 0.1 = 10^6$ kN, $\delta = 6$ mm $= 0.006$ m
$V_A = V_B = 20$ kN; $M_0 = 20x$ for $0 \le x \le 30$ (peak 600 kNm at the crown)

Integrals

\begin{aligned} \int_0^{60} M_0 y\,dx &= 2\int_0^{30} 20x\cdot\frac{x(60-x)}{150}\,dx = 90000 \\ \int_0^{60} y^2\,dx &= \frac{8h^2L}{15} = 1152 \\ \int_0^{60} \cos^2\alpha\,dx &= 57.08 \end{aligned}

( $\tan\alpha = \dfrac{dy}{dx} = \dfrac{4h(L-2x)}{L^2}$ , so $\cos^2\alpha = \dfrac{1}{1+\tan^2\alpha}$ .)

\begin{aligned} \frac{\int M_0 y\,dx}{EI_0} &= \frac{90000}{600} = 150.000 \\ \frac{\int y^2\,dx}{EI_0} &= \frac{1152}{600} = 1.9200 \\ \frac{\int \cos^2\alpha\,dx}{EA_c} &= \frac{57.08}{10^6} = 5.708\times10^{-5} \end{aligned}

Horizontal thrust

H = \frac{150.000 - 0.006}{1.9200 + 5.708\times10^{-5}} = 78.12\ \text{kN}

For comparison: without rib shortening and yielding, $H = \dfrac{25PL}{128h} = 78.125$ kN. With the printed $I_0$ the arch is very flexible in bending, so rib shortening and a 6 mm yield change the thrust only slightly. (A stiffer rib, say $I_0 = 6\times10^6$ cm $^4$ , would give $H = 72.83$ kN.)

Bending moment $M = M_0 - H y$

$x$ (m)	$y$ (m)	$M_0$ (kNm)	$M$ (kNm)
0.0	0.000	0.0	0.00
7.5	2.625	150.0	-55.06
15.0	4.500	300.0	-51.54
22.5	5.625	450.0	10.58
30.0	6.000	600.0	131.28

The BMD is symmetric about the crown: zero at the hinges A and B, hogging (negative) about the quarter points, and sagging with a peak of 131.28 kNm under the load at the crown.

Answer: $H \approx 78.12$ kN; crown BM $= 131.28$ kNm (sagging), BM at $x = 15$ m $= -51.54$ kNm.

2078 Bhadra · 6 marks

Analyze the given loaded beam using Castigliano's theorem. The redundant should be considered as the support moment at A. EI constant. [Figure: beam ABCD, A fixed; 10 kNm moment applied at B (2 m from A); support at C, 2 m from B; 6 kN/m UDL on overhang CD of 1 m.]

Answer

Reading of the figure: A fixed; B is 2 m from A where a couple of 10 kNm (taken clockwise) is applied; C is a roller 2 m from B; CD is a 1 m overhang with 6 kN/m. EI is constant.

The fixity at A gives one redundant. Release the rotation at A and take the fixed-end moment $M_A$ (clockwise on the beam) as the redundant, so the beam becomes simply supported at A and C with an overhang. The condition is that the slope at A is zero:

\frac{\partial U}{\partial M_A} = \theta_A = 0

        10 kNm (cw)
 (fixed)  10 kNm cw   6 kN/m
 A o--------o--------C vvvv D
 |<-- 2 -->|<-- 2 -->|<- 1 ->|

Reaction at A in terms of $M_A$

Moments about C (the overhang load is 6 kN at 0.5 m from C):

4R_A + M_A + 10 + 6(0.5) = 0 \ \Rightarrow\ R_A = -\frac{M_A + 13}{4}

$\dfrac{\partial R_A}{\partial M_A} = -\dfrac{1}{4}$ (R_A is positive upward).

Bending moments (x from A, sagging positive)

\begin{aligned} 0 \le x \le 2:\quad M &= M_A + R_A x, & \frac{\partial M}{\partial M_A} &= 1 - \frac{x}{4} \\ 2 \le x \le 4:\quad M &= M_A + R_A x + 10, & \frac{\partial M}{\partial M_A} &= 1 - \frac{x}{4} \end{aligned}

The overhang CD does not depend on $M_A$ (its moment comes from the UDL alone), so it adds nothing to the integral.

Castigliano's theorem

\int_0^4 M\left(1-\frac{x}{4}\right)dx = 0

Evaluating each term with $M = M_A - \dfrac{(M_A+13)x}{4} + 10\,[x>2]$ :

\begin{aligned} M_A\int_0^4\left(1-\tfrac{x}{4}\right)^2dx &= \tfrac{4}{3}M_A \\ -\tfrac{13}{4}\int_0^4 x\left(1-\tfrac{x}{4}\right)dx &= -\tfrac{13}{4}\cdot\tfrac{8}{3} = -\tfrac{26}{3} \\ 10\int_2^4 \left(1-\tfrac{x}{4}\right)dx &= 10(0.5) = 5 \end{aligned}

\tfrac{4}{3}M_A - \tfrac{26}{3} + 5 = 0 \ \Rightarrow\ M_A = \tfrac{11}{4} = 2.75\ \text{kNm}

Other reactions

R_A = -\frac{2.75+13}{4} = -3.9375\ \text{kN}\ (\text{downward}), \qquad R_C = 6 - (-3.9375) = 9.9375\ \text{kN}\ (\uparrow)

Bending moments (kNm)

Point	BM
A	+2.75 (sagging)
B (left of couple)	-5.125
B (right of couple)	+4.875
C	-3.000
D	0

Answer: $M_A = 2.75$ kNm (clockwise, resisted by the wall), $R_A = 3.938$ kN downward, $R_C = 9.938$ kN upward.

2078 Bhadra · 10 marks

Analyse the portal frame ABCD shown in the figure below using the flexibility matrix method. Take $EI = 12000\ \text{kN/m}^2$ . [Figure: frame with column AB of height 4 m (2 m + 2 m) with 50 kN horizontal at mid-height, beam BC of 8 m carrying 25 kN/m UDL, column CD of height 3 m; A fixed, D supported.]

Answer

Reading of the figure: A (0, 0) fixed; column AB is 4 m high with a 50 kN horizontal load (towards the right) at mid-height E; beam BC is 8 m long with 25 kN/m; column CD is 3 m high, so D is 1 m above the level of A, and D is hinged. $EI = 12000$ kNm $^2$ is constant.

$r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at D and take the reactions at D as the redundants: $X_1$ = horizontal ( $\rightarrow$ ), $X_2$ = vertical ( $\uparrow$ ). The primary structure is a cantilever fixed at A. In the flexibility method the compatibility condition is $[F]\{X\} + \{\Delta_0\} = 0$ .

Moments in the primary structure (sign: + tension on the inner face)

Loads only ( $M_0$ ): at A, $-(50\times2 + 25\times8\times4) = -900$ ; at E, $-800$ ; at B, $-800$ ; the beam falls parabolically to 0 at C.
Unit $X_1$ ( $m_1$ ): $m_1 =$ lever arm of the unit horizontal force; at C and along BC it is $3$ , at B it is $3$ , at A it is $-1$ .
Unit $X_2$ ( $m_2$ ): at B it is $8$ , falling linearly to 0 at C; along AB it is $8$ .

Flexibility matrix and load vector

\begin{aligned} f_{11} &= \frac{90.333}{EI} \\ f_{12} &= \frac{128.000}{EI} \\ f_{22} &= \frac{426.667}{EI} \\ \Delta_{10} &= \frac{-9566.7}{EI} \\ \Delta_{20} &= \frac{-39200.0}{EI} \end{aligned}

In units of m/kN with $EI = 12000$ :

[F] = \begin{bmatrix} 0.00753 & 0.01067 \\ 0.01067 & 0.03556 \end{bmatrix},\qquad \{\Delta_0\} = \begin{Bmatrix} -0.79722 \\ -3.26667 \end{Bmatrix}\ \text{m}

Compatibility equations

\begin{aligned} 90.333\,X_1 + 128.000\,X_2 &= 9566.7 \\ 128.000\,X_1 + 426.667\,X_2 &= 39200.0 \end{aligned}

\{X\} = -[F]^{-1}\{\Delta_0\} \ \Rightarrow\ X_1 = -42.23\ \text{kN},\quad X_2 = 104.55\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-900.00	-1.00	8.00	-21.41
E	-800.00	1.00	8.00	-5.87
B	-800.00	3.00	8.00	-90.34
mid BC	-200.00	3.00	4.00	91.48
C	0.00	3.00	0.00	-126.70
D	0.00	0.00	0.00	0.00

Maximum sagging moment in BC: 91.89 kNm at 3.82 m from B.

Reactions

At D (hinge): $H_D = 42.23$ kN ( $\leftarrow$ ), $V_D = 104.55$ kN ( $\uparrow$ )
At A: $H_A = 7.77$ kN ( $\leftarrow$ ), $V_A = 95.45$ kN ( $\uparrow$ ), $M_A = 21.41$ kNm (anticlockwise)

Check: $\Sigma V = 200.00 = 25 \times 8 = 200$ kN; $\Sigma H = 50 - 7.77 - 42.23 = 0$ .

Answer: $X_1 = H_D = 42.23$ kN towards the left and $X_2 = V_D = 104.55$ kN upward.

2078 Bhadra · 6 marks

Compute the bar force in the member AB, and EF due to the decrease in temperature of 20°C in all vertical members only. Take $\alpha = 12\times10^{-6}/°C$ and $E = 2\times10^{5}\ \text{N/mm}^2$ , cross-sectional area of all members is 35 cm². [Figure: truss with bottom chord A-B-C-D (3 panels of 3 m), top joints F and E, height 3 m, with diagonals; A hinged, D roller.]

Answer

Reading of the figure: bottom chord A-B-C-D with three panels of 3 m (A hinged, D roller); top joints F above B and E above C at 3 m height; members AB, BC, CD, FE, AF, ED, the verticals BF and CE, and the two diagonals BE and CF. The truss has $m = 10$ , $j = 6$ , $r = 3$ , so $D_s = m + r - 2j = 1$ (internal). Take the force in CF as the redundant $X$ .

Data

$E = 2\times10^5$ N/mm $^2 = 2\times10^8$ kN/m $^2$ , $A = 35$ cm $^2 = 35\times10^{-4}$ m $^2$ , so $AE = 700000$ kN.
Vertical members BF and CE: $\Delta t = -20^\circ$ C, so each shortens by $e = \alpha\,\Delta t\,L = 12\times10^{-6}(-20)(3) = -7.20\times10^{-4}$ m $= -0.72$ mm.

There is no external load, so the primary structure carries no force. The temperature change alone causes the forces.

Force $n$ due to $X = 1$ in CF

Member	$L$ (m)	$n$
AB	3.000	0.0000
BC	3.000	-0.7071
CD	3.000	0.0000
FE	3.000	-0.7071
AF	4.243	0.0000
ED	4.243	0.0000
BF	3.000	-0.7071
CE	3.000	-0.7071
BE	4.243	+1.0000
CF	4.243	+1.0000

The outer members AB, CD, AF and ED carry zero force in this system.

Compatibility

\begin{aligned} f_{11} &= \sum\frac{n^2 L}{AE} = \frac{4(0.5)(3) + 2(1)(4.243)}{AE} = \frac{14.485}{AE} = 2.0693\times10^{-5}\ \text{m/kN} \\ \Delta_{1t} &= \sum n\,e = 2(-0.7071)(-0.00072) = 1.0182\times10^{-3}\ \text{m} \end{aligned}

f_{11}X + \Delta_{1t} = 0 \ \Rightarrow\ X = -\frac{1.0182\times10^{-3}}{2.0693\times10^{-5}} = -49.21\ \text{kN}

Member forces $S = nX$

Member	Force (kN)
AB	0.00
EF (FE)	34.79 (tension)
BC	34.79 (tension)
BF = CE	34.79 (tension)
BE = CF	-49.21 (compression)
AF, ED, CD	0

Answer: bar force in AB $= 0$ ; bar force in EF $= 34.79$ kN (tension).

2078 Bhadra · 10 marks

Draw shear force and bending moment diagram of the continuous beam using the three moment equation. [Figure: beam ABC, A fixed; 300 kN point load at 4 m from A; B support at 8 m; BC = 10 m with 50 kN/m UDL; C roller.]

Answer

Reading of the figure: A fixed, span AB = 8 m with a 300 kN load at 4 m from A, B an interior support, span BC = 10 m with 50 kN/m over the whole span, C a roller. EI is constant.

The three moment equation is $M_A L_1 + 2M_B(L_1+L_2) + M_C L_2 = -\dfrac{6a_1\bar x_1}{L_1} - \dfrac{6a_2\bar x_2}{L_2}$ with sagging moments positive. The fixed end A is treated by adding an imaginary span of zero length to its left, which makes the slope at A zero. At the roller C the moment is $M_C = 0$ .

Load terms

Span AB: free BM is a triangle with peak $\dfrac{300\times4\times4}{8} = 600$ kNm, area $a_1 = \tfrac12(8)(600) = 2400$ kNm $^2$ , $\bar x_1 = 4$ m. $\dfrac{6a_1\bar x_1}{L_1} = \dfrac{6(2400)(4)}{8} = 7200$ .
Span BC: UDL, $\dfrac{wL_2^3}{4} = \dfrac{50(10)^3}{4} = 12500$ .

Equations

At A (fixed end):

2M_A(8) + M_B(8) = -7200 \quad\Rightarrow\quad 16M_A + 8M_B = -7200

At B:

M_A(8) + 2M_B(8+10) + 0 = -7200 - 12500 \quad\Rightarrow\quad 8M_A + 36M_B = -19700

Solving:

M_B = -503.125\ \text{kNm},\qquad M_A = -198.4375\ \text{kNm}

Both are hogging.

Reactions

With sagging-positive moments, $M(x) = M_A + R_A x - 300\langle x-4\rangle$ in AB, so $M_B = M_A + 8R_A - 1200$ :

R_A = \frac{1200 + M_B - M_A}{8} = \frac{1200 + (-503.125) - (-198.4375)}{8} = 111.914\ \text{kN}

Shear just left of B $= 111.914 - 300 = -188.086$ kN. In BC, $M(x) = M_B + V_B' x - 25x^2$ with $M_C = 0$ at $x = 10$ :

V_B' = \frac{2500 - M_B}{10} = 300.3125\ \text{kN},\qquad R_B = 188.086 + 300.312 = 488.398\ \text{kN}

R_C = 500 - 300.3125 = 199.6875\ \text{kN}

Check: $R_A + R_B + R_C = 800.00$ kN $= 300 + 50(10) = 800$ kN.

Shear force (kN)

Section	SF
A	+111.91
E (just left / right of the 300 kN load)	+111.91 / -188.09
B (left / right)	-188.09 / +300.31
C	-199.69

SF is zero in BC at $x = 3.994$ m from C.

Bending moment (kNm)

Section	BM
A	-198.44 (hogging)
E (under the 300 kN load)	+249.22 (sagging)
B	-503.12 (hogging)
Maximum in BC at 3.99 m from C	+398.75 (sagging)
C	0

Points of contraflexure: 1.77 m from A; 1.33 m beyond E (towards B); and 2.01 m beyond B in BC.

Answer: $M_A = -198.44$ kNm and $M_B = -503.12$ kNm (hogging); $R_A = 111.91$ kN, $R_B = 488.40$ kN, $R_C = 199.69$ kN; peak sagging BM $= 398.75$ kNm in BC.

2076 Chaitra · 4 marks

Explain theorems on displacement with suitable illustration.

Answer

Displacements of an elastic structure can be found, and checked, with the following theorems.

1. Maxwell's reciprocal displacement theorem

In a linearly elastic structure, the deflection at point A due to a unit load at point B equals the deflection at B due to a unit load at A:

\delta_{AB} = \delta_{BA}

Illustration. Take a cantilever of length $L$ with A at mid-span and B at the free end.

Unit load at B: deflection at A $= \dfrac{5L^3}{48EI}$ .
Unit load at A: deflection at B $= \dfrac{(L/2)^2\left(3L - L/2\right)}{6EI} = \dfrac{5L^3}{48EI}$ .

The two are equal. The same holds for rotations and for a rotation-deflection pair (a unit couple at A and the rotation produced at B by a unit force).

2. Betti's law

The work done by a first system of loads $P_1$ through the displacements caused by a second system $P_2$ equals the work of $P_2$ through the displacements caused by $P_1$ : $\sum P_1\Delta_{12} = \sum P_2\Delta_{21}$ . Maxwell's theorem is the special case of two unit loads.

3. Castigliano's second theorem

The displacement at a point in the direction of a load equals the partial derivative of the strain energy with respect to that load: $\Delta = \dfrac{\partial U}{\partial P}$ . For a beam, $\Delta = \displaystyle\int \frac{M}{EI}\frac{\partial M}{\partial P}dx$ .

4. Unit load (virtual work) theorem

The displacement at a point is $\Delta = \displaystyle\int \frac{M\,m}{EI}\,dx$ , where $M$ is the moment due to the real loads and $m$ is the moment due to a unit load at the point and in the direction of the required displacement.

These theorems give the flexibility coefficients and are used to apply the compatibility condition in the force method.

2076 Chaitra · 6 marks

Find the reaction at support 'C' using Castigliano's theorem. [Figure: L-shaped frame; horizontal member BC of L m carrying W kN/m UDL, C on a roller; vertical member AB of L m, A fixed; EI constant.]

Answer

Reading of the figure: A (0, 0) is fixed at the bottom of the vertical member AB (length $L$ ), B is the corner, BC is the horizontal member (length $L$ ) carrying $W$ kN/m, and C is on a roller that gives a vertical reaction $R_C$ . EI is constant. The frame has one redundant, $R_C$ .

  B ================= C  (roller)
  |   W kN/m          o
  |
  | L
 A///   |<---- L ---->|

Bending moments (take $R_C$ upward)

Member CB ( $x$ from C): $M = R_C\,x - \dfrac{Wx^2}{2}$ , $\ \dfrac{\partial M}{\partial R_C} = x$
Member BA ( $y$ from B downward): the moment is constant along the column, $M = R_C L - \dfrac{WL^2}{2}$ , $\ \dfrac{\partial M}{\partial R_C} = L$

Castigliano's theorem

The roller does not move vertically, so $\dfrac{\partial U}{\partial R_C} = 0$ :

\frac{1}{EI}\left[\int_0^L \left(R_C x - \frac{Wx^2}{2}\right)x\,dx + \int_0^L \left(R_C L - \frac{WL^2}{2}\right)L\,dy\right] = 0

\frac{R_C L^3}{3} - \frac{WL^4}{8} + R_C L^3 - \frac{WL^4}{2} = 0

\frac{4}{3}R_C L^3 = \frac{5}{8}WL^4 \ \Rightarrow\ R_C = \frac{15\,WL}{32}

Other reactions

V_A = WL - \frac{15WL}{32} = \frac{17\,WL}{32}\ (\uparrow),\qquad H_A = 0,\qquad M_A = \left|\frac{15WL^2}{32} - \frac{WL^2}{2}\right| = \frac{WL^2}{32}

The moment in the column is $-WL^2/32$ (hogging at the corner side), and the maximum sagging moment in the beam occurs at $x = 15L/32$ from C.

Answer: $R_C = \dfrac{15\,W L}{32}$ (upward).

2076 Chaitra · 10 marks

Determine reactions at the hinged support using the force method when support D settles vertically downward by $200/EI$ . Take EI to be constant. [Figure: frame ABCD; beam BC (2EI) of 4 m with 10 kN/m UDL; column AB (EI) of height 4 m, A fixed; column CD (EI) of 4 m with D hinged and a further 2 m dimension marked below.]

Answer

Reading of the figure: A (0, 0) fixed, column AB 4 m ( $EI$ ), beam BC 4 m ( $2EI$ ) with 10 kN/m, column CD 4 m ( $EI$ ) with D hinged at the level of A (the extra 2 m dimension is not needed). Support D settles by $200/EI$ (downward), where $EI$ is the stiffness of the columns.

$D_s = 2$ . Release the hinge at D; redundants: $X_1$ = horizontal reaction ( $\rightarrow$ positive) and $X_2$ = vertical reaction ( $\uparrow$ positive).

Primary structure moments (sign: + tension on the inner face)

$M_0$ (UDL on the cantilever A-B-C): $-80$ kNm in the column (constant), at B; the beam falls parabolically from $-80$ at B to 0 at C.
$m_1$ (unit horizontal force at D): CD from 0 at D to 4 at C, beam constant 4, column rises from 4 at B to 0 at A.
$m_2$ (unit vertical force at D): zero in CD, falling linearly from 4 at B to 0 at C in the beam, and constant at 4 in the column.

Flexibility coefficients (in $1/EI$ units)

\begin{aligned} f_{11} &= \frac{74.667}{EI} \\ f_{12} &= \frac{48.000}{EI} \\ f_{22} &= \frac{74.667}{EI} \\ \Delta_{10} &= \frac{-853.3}{EI} \\ \Delta_{20} &= \frac{-1440.0}{EI} \end{aligned}

Compatibility with settlement

The displacement of D along $X_1$ is zero. The displacement along $X_2$ (upward) is $-200/EI$ because D moves down. The right side of each equation is $-\Delta_{i0} + \delta_i$ (second equation: $1440 - 200 = 1240$ ):

\begin{aligned} 74.667\,X_1 + 48.000\,X_2 &= 853.3 \\ 48.000\,X_1 + 74.667\,X_2 &= 1240.0 \end{aligned}

X_1 = 1.28\ \text{kN},\qquad X_2 = 15.78\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-80.00	0.00	4.00	-16.87
B (column)	-80.00	4.00	4.00	-11.74
mid BC	-20.00	4.00	2.00	16.70
C (beam)	0.00	4.00	0.00	5.13
D	0.00	0.00	0.00	0.00

Reactions

At D: $H_D = 1.28$ kN ( $\rightarrow$ ), $V_D = 15.78$ kN ( $\uparrow$ )
At A: $H_A = 1.28$ kN ( $\leftarrow$ ), $V_A = 24.22$ kN ( $\uparrow$ ), $M_A = 16.87$ kNm (anticlockwise)

Check: $V_A + V_D = 40.00$ kN $= 10\times4 = 40$ kN.

Answer: reactions at the hinged support D: horizontal $1.28$ kN (to the right) and vertical $15.78$ kN (upward).

2076 Chaitra · 10 marks

Find the member forces of the given loaded truss for the given external loadings and due to rise in temperature of all diagonal members by 20°C. Take $AE = 5000$ kN for all members and coefficient of thermal expansion $2.06\times10^{-6}/°C$ . Additionally, vertical members are 5 mm too long. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 30 kN horizontal at B and 50 kN downward at C; A hinged, D roller.]

Answer

Reading of the figure: A (0, 0) hinged and D (4, 0) roller at the base, B (0, 3) and C (4, 3) at the top, members AB, CD (verticals, 3 m), BC, AD (horizontals, 4 m) and diagonals AC, BD (5 m). Loads: 30 kN horizontal at B (towards C) and 50 kN downward at C. The truss has $D_s = 6 + 3 - 8 = 1$ (internal). Take the force in AC as the redundant $X$ (tension +).

Effects to be included

Diagonals AC and BD: temperature rise 20 $^\circ$ C, so $e = \alpha\,\Delta t\,L = 2.06\times10^{-6}\times20\times5 = 2.06\times10^{-4}$ m (elongation).
Verticals AB and CD: 5 mm too long, so $e = +0.005$ m.
$AE = 5000$ kN for all members.

Reactions (statics)

$\Sigma M_A = 0$ : $V_D(4) = 50(4) + 30(3)$ , so $V_D = 72.50$ kN ( $\uparrow$ ). $\Sigma F_y$ : $V_A = 50 - 72.50 = -22.50$ kN, i.e. 22.50 kN downward. $\Sigma F_x$ : $H_A = 30$ kN acting towards the left.

Primary structure (AC removed) and unit force system

$S_0$ = force due to loads; $n$ = force due to $X = 1$ in AC. Tension is positive.

Member	$L$ (m)	$S_0$ (kN)	$n$	$S_0 nL$	$n^2L$	$n\,e$ (mm)
AB	3	+22.500	-0.60	-40.500	1.080	-3.0000
BC	4	0.000	-0.80	0.000	2.560	0.0000
CD	3	-50.000	-0.60	+90.000	1.080	-3.0000
AD	4	+30.000	-0.80	-96.000	2.560	0.0000
AC	5	0.000	+1.00	0.000	5.000	+0.2060
BD	5	-37.500	+1.00	-187.500	5.000	+0.2060
Total				-234.000	17.280	-5.5880

Compatibility

\frac{\sum S_0 nL}{AE} + X\frac{\sum n^2L}{AE} + \sum n\,e = 0

\frac{-234.000}{5000} + X\,\frac{17.280}{5000} - 0.005588 = 0

X = \frac{-(-234.000 + 5000 \times -0.005588)}{17.280} = 15.159\ \text{kN}

(The thermal and lack-of-fit terms enter only through $\sum n\,e$ , which is $-0.005588$ m.)

Final forces $S = S_0 + nX$ (kN)

Member	Force	Nature
AB	+13.40	tension
BC	-12.13	compression
CD	-59.10	compression
AD	+17.87	tension
AC	+15.16	tension
BD	-22.34	compression

Check at joint C (vertical): $-S_{CD} - 0.6\,S_{AC} = 50.00$ kN, equal to the 50 kN load.

Answer: AB $= +13.40$ kN, BC $= -12.13$ kN, CD $= -59.10$ kN, AD $= +17.87$ kN, AC $= +15.16$ kN, BD $= -22.34$ kN (positive = tension). Reactions: $H_A = 30$ kN (left), $V_A = 22.50$ kN (downward), $V_D = 72.50$ kN (upward).

2076 Asoj · 6 marks

Enunciate Betti's law and Maxwell's reciprocal theorem and explain their uses.

Answer

Betti's law

Statement. In a linearly elastic structure acted on by two separate systems of loads, the work done by the forces of the first system through the displacements produced by the second system equals the work done by the forces of the second system through the displacements produced by the first system.

If system 1 has loads $P_i$ giving displacements $\Delta_{i1}$ and system 2 has loads $Q_j$ giving displacements $\Delta_{j2}$ at the same points, then

\sum P_i\,\Delta_{i2} = \sum Q_j\,\Delta_{j1}

Proof (outline). Apply $P$ first, then $Q$ : the total work is $U_P + U_Q + \sum P\,\Delta_{P(Q)}$ . Apply $Q$ first, then $P$ : the total work is $U_Q + U_P + \sum Q\,\Delta_{Q(P)}$ . Strain energy depends only on the final loads, not on the order, so the two cross terms are equal.

Maxwell's reciprocal theorem

Statement. The displacement at point A in the direction of a unit load applied at B equals the displacement at B in the direction of a unit load applied at A:

\delta_{AB} = \delta_{BA}, \qquad \text{or in general } f_{ij} = f_{ji}

It follows from Betti's law by taking each system as a single unit load. It holds for linear displacements, rotations and mixed pairs (force and couple).

Example. For a simply supported beam of span $L$ , the deflection at the centre due to a unit load at $L/4$ equals the deflection at $L/4$ due to a unit load at the centre, $\dfrac{11L^3}{768EI}$ .

Uses

It makes the flexibility matrix $[F]$ and the stiffness matrix $[K]$ symmetric, which halves the work of finding the coefficients in the force method.
It checks the computed flexibility coefficients ( $f_{12}$ against $f_{21}$ ).
It gives influence lines of deflection: the deflected shape under a unit load at B is the influence line for deflection at A, which is the basis of Muller-Breslau's principle.
It lets the deflection at a point be found from the deflection curve of another loading, which is easier to compute.

2076 Asoj · 10 marks

Compute the bar forces in the members BG, HC, and CF of the following loaded truss structure. AE = constant. [Figure: Pratt-type truss with top joints B, C, D, bottom joints A, H, G, F, E; 4 panels of 6.0 m (total 24 m), height 6 m, X-bracing in the panels; 8 kN downward at H; A hinged, E roller.]

Answer

Reading of the figure: bottom joints A (0, 0), H (6, 0), G (12, 0), F (18, 0), E (24, 0); top joints B (6, 6), C (12, 6), D (18, 6); end posts AB and DE; verticals BH, CG, DF; chords AH, HG, GF, FE, BC, CD; X-bracing in the two middle panels (diagonals BG, CH and CF, DG). A is hinged, E is a roller, 8 kN acts downward at H. $AE$ is constant.

$m = 15$ , $j = 8$ , $r = 3$ , so $D_s = m + r - 2j = 2$ (internal). Choose the redundants $X_1$ = force in BG and $X_2$ = force in CF (tension +). The primary structure keeps CH and DG as the single diagonals of the two panels.

Reactions

$V_E = 8\times6/24 = 2$ kN, $V_A = 6$ kN, $H_A = 0$ .

Force tables (tension +; $S_0$ = loads on the primary, $n_1$ , $n_2$ = unit redundants)

Member	$L$ (m)	$S_0$	$n_1$	$n_2$	Final $S$
AH	6.000	+6.000	0.000	0.000	+6.000
HG	6.000	+4.000	-0.707	0.000	+4.278
GF	6.000	+2.000	0.000	-0.707	+2.971
FE	6.000	+2.000	0.000	0.000	+2.000
BC	6.000	-6.000	-0.707	0.000	-5.722
CD	6.000	-4.000	0.000	-0.707	-3.029
AB	8.485	-8.485	0.000	0.000	-8.485
DE	8.485	-2.828	0.000	0.000	-2.828
BH	6.000	+6.000	-0.707	0.000	+6.278
CG	6.000	-2.000	-0.707	-0.707	-0.751
DF	6.000	0.000	0.000	-0.707	+0.971
BG	8.485	0.000	+1.000	0.000	-0.393
CH	8.485	+2.828	+1.000	0.000	+2.435
CF	8.485	0.000	0.000	+1.000	-1.373
DG	8.485	+2.828	0.000	+1.000	+1.455

Flexibility coefficients (in $1/AE$ )

\begin{aligned} f_{11} &= \sum n_1^2 L = 28.971, \quad f_{22} = \sum n_2^2L = 28.971, \quad f_{12} = \sum n_1n_2L = 3.000 \\ \Delta_{10} &= \sum n_1S_0L = 15.515, \quad \Delta_{20} = \sum n_2S_0L = 40.971 \end{aligned}

Compatibility equations

\begin{aligned} 28.971\,X_1 + 3.000\,X_2 &= -15.515 \\ 3.000\,X_1 + 28.971\,X_2 &= -40.971 \end{aligned}

X_1 = S_{BG} = -0.393\ \text{kN},\qquad X_2 = S_{CF} = -1.373\ \text{kN}

Member forces required

BG: $S = X_1 = -0.393$ kN (compression)
CF: $S = X_2 = -1.373$ kN (compression)
HC (CH): $S = S_0 + n_1X_1 + n_2X_2 = 2.828 + (1)(-0.393) + 0 = 2.435$ kN (tension)

Answer: BG $= -0.393$ kN, HC $= 2.435$ kN, CF $= -1.373$ kN (positive = tension).

2076 Asoj · 10 marks

Determine the moment at the fixed support and the rotation at the roller support of a propped cantilever beam of span 10 m loaded with a uniformly distributed load of 30 kN/m on its whole span and a point load of 50 kN at the centre using Castigliano's theorem.

Answer

The beam AB (span 10 m) is fixed at A and has a roller at B. Loads: 30 kN/m over the whole span and 50 kN at the centre. EI is constant. Take the roller reaction $R_B$ as the redundant.

        50 kN
          v      30 kN/m
 vvvvvvvvvvvvvvvvvvvvvvvvvv
 ||==========+===========o
 A (fixed)  5 m          B (roller, R_B)
 |<------------ 10 m ---->|

Bending moment (x measured from B)

M = R_B\,x - 15x^2 - 50\langle x-5\rangle, \qquad \frac{\partial M}{\partial R_B} = x

Redundant from Castigliano

The deflection at B is zero:

\frac{\partial U}{\partial R_B} = \frac{1}{EI}\int_0^{10} M\,x\,dx = 0

\int_0^{10}\left(R_B x^2 - 15x^3\right)dx - 50\int_5^{10}(x-5)x\,dx = 0

\frac{1000}{3}R_B - 37500 - 5208.33 = 0 \ \Rightarrow\ R_B = 128.125\ \text{kN}

(Check: UDL alone gives $3wL/8 = 112.5$ kN and the central load gives $5P/16 = 15.625$ kN; their sum is 128.125 kN.)

Moment at the fixed support

V_A = 350 - 128.125 = 221.875\ \text{kN}

M_A = R_B(10) - 30\cdot\frac{10^2}{2} - 50(5) = 128.125(10) - 1500 - 250 = -468.75\ \text{kNm}

The fixed-end moment is 468.75 kNm (hogging).

Rotation at the roller B

Apply a dummy couple $M'$ at B. Then $\partial M/\partial M' = 1$ at the sections (the extra term caused by the change of $R_B$ integrates to zero because $\int M\,x\,dx = 0$ ), so

\theta_B = \frac{\partial U}{\partial M'}\bigg|_{M'=0} = \frac{1}{EI}\int_0^{10} M\,dx

\int_0^{10} M\,dx = R_B\frac{10^2}{2} - 15\frac{10^3}{3} - 50\frac{5^2}{2} = 128.125(50) - 5000 - 625 = 781.25

\theta_B = \frac{781.25}{EI}\ \text{rad (anticlockwise)}

Answer: fixed-end moment $M_A = 468.75$ kNm (hogging); rotation at the roller $\theta_B = \dfrac{781.25}{EI}$ rad.

2076 Asoj · 6 marks

Write down the compatibility equation for a two hinged parabolic arch due to external loads, variation in temperature, rib shortening and yielding of supports.

Answer

A two-hinged arch is indeterminate to the first degree. Take the horizontal thrust $H$ as the redundant, release the hinge at B (roller), and write that the net horizontal movement of B is equal to the yield of the support.

          C
       .-----.
     /         \
   A o           o B  <-- H (redundant)
   |<----- L ----->|

Displacement of B in the released structure (outward positive)

External loads: $\Delta_{10} = \displaystyle\int \frac{M_0\,y}{EI}\,ds$ ( $M_0$ = BM of the equivalent simple beam, $y$ = ordinate of the arch axis). The axial-force term is small and is usually neglected.
Temperature rise $t$ : $\Delta_{1t} = \alpha\,t\,L$ (a fall in temperature gives a negative value).
Redundant $H$ (unit horizontal force at B gives $m = -y$ and $N = -\cos\theta$ ):

\Delta_{1H} = -H\,f_{11}, \qquad f_{11} = \int \frac{y^2}{EI}\,ds + \int \frac{\cos^2\theta}{AE}\,ds

The first term of $f_{11}$ is bending, and the second is the rib shortening due to the axial thrust $H\cos\theta$ .

Yielding of support: B moves outward by $\delta$ (positive outward).

Compatibility equation

The final horizontal movement of B must equal the support yield $\delta$ :

\Delta_{10} + \alpha\,t\,L - H\,f_{11} = \delta

\boxed{H = \frac{\displaystyle\int \frac{M_0\,y}{EI}\,ds + \alpha\,t\,L - \delta}{\displaystyle\int \frac{y^2}{EI}\,ds + \int \frac{\cos^2\theta}{AE}\,ds}}

Remarks

For $I = I_c\sec\theta$ the bending integrals become $\dfrac{1}{EI_c}\int M_0\,y\,dx$ and $\dfrac{1}{EI_c}\int y^2\,dx$ , and with $A = A_c\sec\theta$ the rib shortening term is $\dfrac{1}{EA_c}\int\cos^2\theta\,dx$ .
A temperature rise or an inward movement of B increases $H$ . An outward yield and rib shortening reduce $H$ (rib shortening acts like an elastic spring in the denominator).
After $H$ is found, the moment at any section is $M = M_0 - Hy$ , the normal thrust is $N = H\cos\theta + V_0\sin\theta$ and the radial shear is $V_r = V_0\cos\theta - H\sin\theta$ .

2075 Chaitra · 10 marks

Determine reactions at the hinged support in the frame shown in figure below using the force method. [Figure: frame with beam of 10 m (1.5I) carrying 30 kN/m UDL, right end on a hinged support with redundants 1 (vertical) and 2 (horizontal); column of 2I and height 6 m, fixed at the base; 50 kN horizontal load at 2 m below the beam.]

Answer

Reading of the figure: column AB is 6 m high (stiffness $2I$ ) and fixed at A; the beam BC is 10 m long ( $1.5I$ ) with 30 kN/m; the beam end C is on a hinged support at the level of the beam; a 50 kN horizontal load (towards the right) acts on the column 2 m below the beam, at E (4 m above A). Here $I$ stands for $EI$ units: the member stiffnesses are $2EI$ and $1.5EI$ .

$r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at C: $X_1$ = vertical reaction at C ( $\uparrow$ ), $X_2$ = horizontal reaction at C ( $\rightarrow$ ). The primary structure is a cantilever fixed at A.

 B ================= C (hinge)
 |    30 kN/m          X1 up, X2 horizontal
 |
 E ->  50 kN
 |
 A (fixed)

Moments in the primary structure (+ tension on the inner face)

$M_0$ : at A, $-(50\times4 + 30\times10\times5) = -1700$ ; at E and B, $-1500$ (the column between E and B carries only the beam load); falling parabolically to 0 at C.
$m_1$ (unit vertical at C): $10$ in the column (from B down to A) and falling linearly from 10 at B to 0 at C.
$m_2$ (unit horizontal at C): zero in the beam; in the column it grows linearly from 0 at B to $-6$ at A ( $-2$ at E).

Flexibility coefficients (in $1/EI$ units)

\begin{aligned} f_{11} &= \frac{522.222}{EI} \\ f_{12} &= \frac{-90.000}{EI} \\ f_{22} &= \frac{36.000}{EI} \\ \Delta_{10} &= \frac{-72000.0}{EI} \\ \Delta_{20} &= \frac{14433.3}{EI} \end{aligned}

Compatibility

\begin{aligned} 522.222\,X_1 - 90.000\,X_2 &= 72000.0 \\ -90.000\,X_1 + 36.000\,X_2 &= -14433.3 \end{aligned}

X_1 = V_C = 120.84\ \text{kN},\qquad X_2 = H_C = -98.82\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-1700.00	10.00	-6.00	101.35
E (50 kN level)	-1500.00	10.00	-2.00	-93.94
B (column)	-1500.00	10.00	0.00	-291.59
mid BC	-375.00	5.00	0.00	229.21
C	0.00	0.00	0.00	0.00

Maximum sagging moment in the beam: 243.38 kNm at 5.97 m from B.

Reactions

C (hinge): $V_C = 120.84$ kN ( $\uparrow$ ), $H_C = 98.82$ kN (towards the left)
A (fixed): $V_A = 179.16$ kN ( $\uparrow$ ), $H_A = 48.82$ kN (towards the right), $M_A = 101.35$ kNm (anticlockwise)

Check: $V_A + V_C = 300.00$ kN $= 30\times10 = 300$ kN; $\Sigma H$ : $48.82 + 50 - 98.82 = 0$ .

Answer: at the hinged support C, $V_C = 120.84$ kN upward and $H_C = 98.82$ kN towards the left.

2075 Chaitra · 10 marks

Determine the support reaction at support 'C' using Castigliano's theorem. EI = constant throughout. [Figure: beam ABCD, A fixed, a moment applied at B, 2 m from A; support C at 2 m from B; 6 kN/m UDL on CD of 1 m overhang.]

Answer

Reading of the figure: A fixed; a couple of 10 kNm (taken clockwise) is applied at B, 2 m from A; C is a roller support 2 m from B; the overhang CD (1 m) carries 6 kN/m. EI is constant. The beam is indeterminate to the first degree: take $R_C$ (upward) as the redundant.

          10 kNm (cw)        6 kN/m
 A ||======B===========C vvvvvv D
 fixed   |<- 2 ->|<- 2 ->|<- 1 ->|

Bending moments (s measured from the free end D; sagging positive)

Range	$M$	$\partial M/\partial R_C$
DC, $0 \le s \le 1$	$-3s^2$	0
CB, $1 \le s \le 3$	$-6(s-0.5) + R_C(s-1)$	$s-1$
BA, $3 \le s \le 5$	$-6(s-0.5) + R_C(s-1) - 10$	$s-1$

(The clockwise couple at B, lying on the D-side of any section between B and A, reduces the sagging moment by 10 kNm.)

Castigliano's theorem

C does not settle, so $\dfrac{\partial U}{\partial R_C} = 0$ :

\int_1^3 \left[-6(s-0.5) + R_C(s-1)\right](s-1)\,ds + \int_3^5 \left[-6(s-0.5) + R_C(s-1) - 10\right](s-1)\,ds = 0

Term by term over $1 \le s \le 5$ :

\begin{aligned} R_C\int_1^5 (s-1)^2\,ds &= \tfrac{64}{3}R_C \\ -6\int_1^5 (s-0.5)(s-1)\,ds &= -152 \\ -10\int_3^5 (s-1)\,ds &= -60 \end{aligned}

\tfrac{64}{3}R_C - 152 - 60 = 0 \ \Rightarrow\ R_C = \frac{212\times3}{64} = 9.9375\ \text{kN}\ (\uparrow)

Other reactions

Vertical: $R_A = 6 - 9.9375 = -3.9375$ kN, that is 3.9375 kN downward at A.
Fixed-end moment: $M_A = -6(4.5) + 9.9375(4) - 10 = 2.75$ kNm (sagging side, clockwise reaction on the beam).

Bending moment values (kNm)

Point	BM
A	+2.750
B (left of couple)	-5.125
B (right of couple)	+4.875
C	-3.000
D	0

Answer: the reaction at the roller C is $R_C = 9.9375$ kN upward ( $159/16$ kN).

2075 Chaitra · 6 marks

Find the bending moment at a given section x-x of the following loaded two hinged parabolic arch due to the given loading. Take $EI_c = 10000\ \text{kNm}^2$ . [Figure: two-hinged parabolic arch with horizontal dimensions 2 m, 6 m, 8 m and 2 m as marked, rise 3 m; 60 kN/m UDL on the left part; 85 kN point load on the right part; section x-x on the left part near the support.]

Answer

Reading of the figure: span $L = 18$ m, rise $h = 3$ m; A is at the left support. The dimensions 2, 6, 8 and 2 m are read as: section x-x at 2 m from A; the UDL of 60 kN/m covers the first 8 m (2 m + 6 m); the 85 kN load acts 16 m from A (8 m after the UDL, 2 m before B). The arch is taken as $I = I_c\sec\theta$ (so $EI_c = 10000$ kNm $^2$ cancels in $H$ ).

Arch profile

y = \frac{4h\,x(L-x)}{L^2} = \frac{x(18-x)}{27}

Reactions of the equivalent simple beam

UDL resultant $= 60\times8 = 480$ kN at 4 m.

V_B = \frac{480(4) + 85(16)}{18} = 182.222\ \text{kN},\qquad V_A = 565 - 182.222 = 382.778\ \text{kN}

Free bending moment $M_0$ (kNm)

Range	$M_0$
$0 \le x \le 8$	$382.778x - 30x^2$
$8 \le x \le 16$	$382.778x - 480(x-4)$
$16 \le x \le 18$	$182.222(18-x)$

Horizontal thrust

\int_0^{18} y^2\,dx = \frac{8h^2L}{15} = 86.4

\int M_0\,y\,dx = 15836.2 + 15802.5 + 297.0 = 31935.6

H = \frac{\int M_0 y\,dx}{\int y^2\,dx} = \frac{31935.6}{86.4} = 369.63\ \text{kN}

Bending moment at section x-x ( $x = 2$ m)

\begin{aligned} y &= \frac{2(16)}{27} = 1.1852\ \text{m} \\ M_0 &= 382.778(2) - 30(2)^2 = 645.56\ \text{kNm} \\ M_{xx} &= M_0 - Hy = 645.56 - 369.63(1.1852) = 207.48\ \text{kNm} \end{aligned}

Answer: $H = 369.63$ kN and the bending moment at x-x is $M = 207.48$ kNm (sagging).

2075 Asoj · 10 marks

Determine the horizontal and vertical reaction at the hinged support and also draw BMD using the force method. [Figure: portal frame; beam of 2EI, span 5 m, with 30 kN/m UDL; columns EI of height 4 m; left base fixed, right base hinged; 100 kN horizontal at the top-left joint.]

Answer

Reading of the figure: portal frame ABCD, columns AB and CD are 4 m high ( $EI$ ), the beam BC is 5 m long ( $2EI$ ) with 30 kN/m, a 100 kN horizontal load acts at B (towards C), A is fixed and D is hinged.

$r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at D and take $X_1$ = horizontal reaction ( $\rightarrow$ ) and $X_2$ = vertical reaction ( $\uparrow$ ) at D.

 100 kN
  ->  B ============ C   30 kN/m
      |              |
      | 4 m          | 4 m
      |              |
     ///A            o D (hinge)
      |<---- 5 m --->|

Primary structure (cantilever fixed at A)

$M_0$ (+ tension on the inner face): at B, $-30\times5\times2.5 = -375$ kNm (the 100 kN load acts at the joint B); at A, $-(375 + 100\times4) = -775$ kNm; the beam falls parabolically from $-375$ at B to 0 at C; column CD has no moment.
$m_1$ (unit horizontal force at D): CD from 0 at D to 4 at C, beam constant 4, column from 4 at B down to 0 at A.
$m_2$ (unit vertical force at D): beam from 5 at B to 0 at C, column AB constant 5.

Flexibility coefficients (in $1/EI$ units)

\begin{aligned} f_{11} &= \frac{82.667}{EI} \\ f_{12} &= \frac{65.000}{EI} \\ f_{22} &= \frac{120.833}{EI} \\ \Delta_{10} &= \frac{-5316.7}{EI} \\ \Delta_{20} &= \frac{-12671.9}{EI} \end{aligned}

Compatibility ( $\Delta_{Dx} = \Delta_{Dy} = 0$ )

\begin{aligned} 82.667\,X_1 + 65.000\,X_2 &= 5316.7 \\ 65.000\,X_1 + 120.833\,X_2 &= 12671.9 \end{aligned}

X_1 = H_D = -31.44\ \text{kN},\qquad X_2 = V_D = 121.79\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-775.00	0.00	5.00	-166.07
B (column)	-375.00	4.00	5.00	108.15
mid BC	-93.75	4.00	2.50	84.94
C (beam)	0.00	4.00	0.00	-125.78
D	0.00	0.00	0.00	0.00

The maximum sagging moment in the beam is 121.42 kNm at 0.94 m from B.

Reactions

At D (hinge): $H_D = 31.44$ kN (towards the left), $V_D = 121.79$ kN ( $\uparrow$ )
At A (fixed): $H_A = 68.56$ kN (towards the left), $V_A = 28.21$ kN ( $\uparrow$ ), $M_A = 166.07$ kNm (anticlockwise)

Check: $\Sigma H$ : $100 - 68.56 - 31.44 = 0$ ; $\Sigma V$ : $28.21 + 121.79 = 30\times5 = 150$ kN.

Answer: at the hinged support D, the horizontal reaction is $31.44$ kN (towards the left) and the vertical reaction is $121.79$ kN (upward). End moments: $M_A = 166.07$ kNm, $M_B = 108.15$ kNm, $M_C = -125.78$ kNm (signs as in the table).

2075 Asoj · 6 marks

Determine reaction at support B of the beam shown in figure below by Castigliano's method. [Figure: beam A-B-C, A fixed, B roller at 4 m from A, overhang BC = 2 m with 150 kN downward at C.]

Answer

Beam ABC: A fixed, B a roller 4 m from A, overhang BC = 2 m with 150 kN downward at C. EI is constant. The beam has one redundant; take the roller reaction $R_B$ (upward).

 ||==============o=======+ 150 kN
 A (fixed)       B       C
 |<----- 4 ----->|<- 2 ->|

Bending moments (x measured from C towards A, sagging positive)

\begin{aligned} CB\ (0 \le x \le 2):\quad & M = -150x, & \frac{\partial M}{\partial R_B} &= 0 \\ BA\ (2 \le x \le 6):\quad & M = -150x + R_B(x-2), & \frac{\partial M}{\partial R_B} &= x-2 \end{aligned}

Castigliano's theorem

The deflection at B is zero, so $\dfrac{\partial U}{\partial R_B} = 0$ . Let $s = x-2$ (from B):

\int_0^4 \left[R_B\,s - 150(s+2)\right]s\,ds = 0

\frac{R_B (4)^3}{3} - 150\left(\frac{4^3}{3} + 2\cdot\frac{4^2}{2}\right) = 0

21.333\,R_B - 150(21.333 + 16) = 0 \ \Rightarrow\ R_B = \frac{5600}{21.333} = 262.5\ \text{kN}\ (\uparrow)

Other reactions (check)

V_A = 150 - 262.5 = -112.5\ \text{kN (i.e. 112.5 kN downward)}

M_A = R_B(4) - 150(6) = 1050 - 900 = 150\ \text{kNm}

Bending moment

In the overhang $M = -150x$ , so $M_B = -300$ kNm (hogging). In AB, with $s$ from B, $M = R_B s - 150(s+2) = -300 + 112.5\,s$ , which is zero at $s = 2.667$ m from B and equals $+150$ kNm (sagging) at A ( $s = 4$ ).

Answer: $R_B = 262.5$ kN (upward); also $V_A = 112.5$ kN (downward) and $M_A = 150$ kNm.

2075 Asoj · 6 marks

A portal frame of span 6 m and height 5 m is hinged supported at both ends. The beam of the frame carries a uniformly distributed gravity load of intensity 50 kN/m. Use the force method to solve the frame considering the flexural stiffness EI to be constant. Determine the reactions at both supports.

Answer

Portal frame: span $L = 6$ m, height $h = 5$ m, both bases hinged, EI constant, beam load $w = 50$ kN/m. A hinged portal has $r = 4$ , so $D_s = 1$ . Take the horizontal reaction $H$ at D as the redundant (D is released into a roller).

   50 kN/m
 B vvvvvvvvvvvv C
 |              |
 | 5 m          | 5 m
 |              |
 A o            o D  <- H
 |<---- 6 m --->|

Primary structure (D on a roller)

Reactions: $V_A = V_D = 50\times6/2 = 150$ kN.

$M_0$ : columns 0; beam $M_0 = 150x - 25x^2$ (maximum $225$ kNm at mid-span).
$m$ (unit horizontal force at D): column CD: $m = y$ (from D); beam: $m = 5$ ; column AB: $m = y$ (from A).

Flexibility coefficient and load term

\begin{aligned} f_{11} &= \frac{1}{EI}\left[2\int_0^5 y^2dy + 5^2(6)\right] = \frac{1}{EI}\left[83.33 + 150\right] = \frac{233.33}{EI} \\ \Delta_{10} &= \frac{1}{EI}\int_0^6 M_0\,(5)\,dx = \frac{5}{EI}\cdot\frac{wL^3}{12} = \frac{5\times900}{EI} = \frac{4500}{EI} \end{aligned}

(The loads push D outwards by $\Delta_{10}$ in the primary structure, and the inward thrust $H$ brings it back by $f_{11}H$ .)

Compatibility (D does not move horizontally)

\Delta_{10} - f_{11}H = 0 \ \Rightarrow\ H = \frac{4500}{233.33} = 19.286\ \text{kN} \left(= \frac{135}{7}\right)

The horizontal reaction acts inwards (towards the frame) at each support.

Final results

Support	Horizontal (inwards)	Vertical (up)
A	19.286 kN	150 kN
D	19.286 kN	150 kN

Bending moments: $M_A = M_D = 0$ ; $M_B = M_C = -H h = -19.286\times5 = -96.43$ kNm (tension on the outside); mid-span $M = 225 - 96.43 = 128.57$ kNm (sagging).

Answer: at each hinge, $V = 150$ kN (upward) and $H = 19.286$ kN (towards the other support).

2075 Asoj · 8 marks

Determine the forces in all members of the truss shown below, using the force method. Take $EA = 10^{5}$ kN. [Figure: truss ABDC, 4 m wide and 3 m high, A hinged, B roller, diagonal CB; 50 kN horizontal at D.]

Answer

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, D (4, 3) and C (0, 3) at the top; members AB, BD, DC, CA and the two diagonals CB and AD (a single diagonal would leave the truss determinate, so both are taken). The 50 kN load acts horizontally at D towards the right. $D_s = 6 + 3 - 8 = 1$ . Take the force in AD as the redundant $X$ . $EA = 10^5$ kN is the same for all members, so it cancels from the forces.

Reactions

$\Sigma M_A = 0$ : $V_B(4) = 50(3) \Rightarrow V_B = 37.5$ kN ( $\uparrow$ ); $V_A = 37.5$ kN ( $\downarrow$ ); $H_A = 50$ kN ( $\leftarrow$ ).

Tables (tension +)

Member	$L$ (m)	$S_0$	$n_1$	Final $S$
AB	4.000	+50.000	-0.800	+17.593
BD	3.000	0.000	-0.600	-24.306
DC	4.000	+50.000	-0.800	+17.593
CA	3.000	+37.500	-0.600	+13.194
CB	5.000	-62.500	+1.000	-21.991
AD	5.000	0.000	+1.000	+40.509

f_{11} = \sum\frac{n^2L}{AE} = \frac{17.280}{AE} = 1.728\times10^{-4}\ \text{m/kN}, \qquad \Delta_{10} = \sum\frac{S_0nL}{AE} = \frac{-700.000}{AE}

Compatibility

\Delta_{10} + f_{11}X = 0 \ \Rightarrow\ X = -\frac{-700.000}{17.280} = 40.51\ \text{kN}

Final forces $S = S_0 + nX$

Member	Force (kN)	Nature
AB	+17.59	tension
BD	-24.31	compression
DC	+17.59	tension
CA	+13.19	tension
CB	-21.99	compression
AD	+40.51	tension

Answer: AD $= 40.51$ kN (T); AB $= +17.59$ kN, BD $= -24.31$ kN, DC $= +17.59$ kN, CA $= +13.19$ kN, CB $= -21.99$ kN (positive = tension).

2074 Chaitra · 10 marks

Analyze the structure given below using the force method. Draw shear force and bending moment diagrams. [Figure: frame a-d-b-c; horizontal member a-b of 24 ft (12 ft + 12 ft, 2I) with 120 kip downward load at d (mid-span); vertical member b-c of 18 ft (3I); a hinged, c hinged.]

Answer

Reading of the figure: beam a-b is 24 ft (12 ft + 12 ft) with stiffness $2EI$ , carrying 120 kip at d (mid-span); column b-c is 18 ft high with stiffness $3EI$ ; a and c are hinged. Units: kip and ft.

$r = 4$ , so $D_s = 1$ . Take the horizontal reaction $H_a$ at a as the redundant (a becomes a roller).

        120 k
   a o----d----b
   ^           |
   H_a         | 18 ft
               |
               o c
   |<- 12 ->|<- 12 ->|

Primary structure (a on a roller)

Reactions: $V_a = V_c = 60$ kip.

Section	$M_0$ (kip-ft)	$m$ (unit $H_a$ )
a	0	0
d	720	-9
b	0	-18
c	0	0

( $M_0 = 60\times12 = 720$ at d. A unit horizontal force at a causes a vertical reaction pair of $18/24 = 0.75$ , so $m_d = -0.75\times12 = -9$ and $m_b = -18$ (hogging).)

Coefficients (unit-load method; beam stiffness $2EI$ , column stiffness $3EI$ )

\begin{aligned} f_{11} &= \frac{1}{2EI}\left[\frac{12}{3}(81) + \frac{12}{3}(81 + 162 + 324)\right] + \frac{1}{3EI}\left[\frac{18}{3}(324)\right] = \frac{162 + 1134 + 648}{EI} = \frac{1944}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[\frac{12}{3}(720)(-9) + \frac{12}{6}\{2(720)(-9) + 720(-18)\}\right] = \frac{-12960 - 25920}{EI} = \frac{-38880}{EI} \end{aligned}

Compatibility (a does not move horizontally)

f_{11}H_a + \Delta_{10} = 0 \ \Rightarrow\ H_a = \frac{38880}{1944} = 20.00\ \text{kip}

Reactions

a: $H_a = 20.00$ kip ( $\rightarrow$ ), $V_a = 45.00$ kip ( $\uparrow$ )
c: $H_c = 20.00$ kip ( $\leftarrow$ ), $V_c = 75.00$ kip ( $\uparrow$ )

(The unit force on the primary structure gives a vertical reaction at a of $0.75$ per unit $H$ , so $V_a = 60 - 0.75\times20 = 45$ kip and $V_c = 60 + 15 = 75$ kip.)

Bending moment $M = M_0 + H_a m$ (kip-ft)

Point	BM	Tension side
a	0
d	540.0	bottom
b	-360.0	outside (top of beam, outer face of column)
c	0

Shear force (kip)

Member	SF
a-d	+45.0
d-b	-75.0
b-c	+20.0 (constant)

Axial force (kip)

Beam: $-20.0$ (compression); column: $-75.0$ (compression).

Answer: $H_a = H_c = 20.00$ kip; $V_a = 45.00$ kip, $V_c = 75.00$ kip; $M_d = 540.0$ kip-ft (sagging), $M_b = -360.0$ kip-ft (hogging).

2074 Chaitra · 6 marks

Derive the three moment equation and use it to solve a single span fixed beam with uniform distributed load throughout the span.

Answer

Derivation

Consider two adjacent spans AB ( $L_1$ ) and BC ( $L_2$ ) of a continuous beam with supports at the same level and constant $EI$ . Let $M_A, M_B, M_C$ be the support moments (sagging positive). On each span the BM diagram is the free (simple beam) diagram plus a trapezoid of support moments. Let $a_1, a_2$ be the areas of the free BM diagrams, $\bar x_1$ the distance of the centroid of $a_1$ from A, and $\bar x_2$ the distance of the centroid of $a_2$ from C.

Slope at B from span AB (moment of the $M/EI$ area about A equals the deviation of A from the tangent at B, which is $\theta_BL_1$ for level supports):

\theta_{B1} = \frac{1}{EI}\left[\frac{a_1\bar x_1}{L_1} + \frac{M_AL_1}{6} + \frac{M_BL_1}{3}\right]

Slope at B from span BC (taking moments about C):

\theta_{B2} = -\frac{1}{EI}\left[\frac{a_2\bar x_2}{L_2} + \frac{M_CL_2}{6} + \frac{M_BL_2}{3}\right]

The beam is continuous over B, so $\theta_{B1} = \theta_{B2}$ . Multiplying by $6$ gives the three moment equation:

M_AL_1 + 2M_B(L_1+L_2) + M_CL_2 = -\frac{6a_1\bar x_1}{L_1} - \frac{6a_2\bar x_2}{L_2}

Application: fixed beam AB with a UDL $w$ over the span $L$

A fixed end is treated as a continuous span of zero length beyond the end. Add an imaginary span $A'A$ of length $L_0 = 0$ at A and $BB'$ of length $L_0 = 0$ at B; the end moments $M_A$ and $M_B$ are the unknowns.

For the span AB: $a = \dfrac23\cdot L\cdot\dfrac{wL^2}{8} = \dfrac{wL^3}{12}$ , $\bar x = \dfrac L2$ , so $\dfrac{6a\bar x}{L} = \dfrac{wL^3}{4}$ .

Equation at A (spans $A'A$ and $AB$ , the first has zero length):

2M_AL + M_BL = -\frac{wL^3}{4}

Equation at B (spans $AB$ and $BB'$ ):

M_AL + 2M_BL = -\frac{wL^3}{4}

Subtracting gives $M_A = M_B$ . Then $3M_AL = -\dfrac{wL^3}{4}$ , so

M_A = M_B = -\frac{wL^2}{12}

Results

Reactions: $R_A = R_B = \dfrac{wL}{2}$ (symmetry).
Mid-span moment: $\dfrac{wL^2}{8} - \dfrac{wL^2}{12} = +\dfrac{wL^2}{24}$ (sagging).
Points of contraflexure at $x = \dfrac{L}{2}\left(1 \pm \dfrac{1}{\sqrt3}\right)$ , that is $0.211L$ from each end.

For example, with $w = 20$ kN/m and $L = 6$ m: $M_A = M_B = -60$ kNm, $R_A = R_B = 60$ kN, mid-span $+30$ kNm.

2074 Chaitra · 4 marks

Explain with example how the bending moment diagram is drawn for a statically indeterminate portal frame which undergoes settlement of one support.

Answer

Settlement of a support in an indeterminate frame produces bending moments even without any external load, because the frame is forced to fit the new support position. The BMD is drawn from the redundant reactions caused by the settlement.

Procedure

Find $D_s$ and choose the redundants (for example the reactions of the settling support). Remove them to get the primary structure.
The primary structure moves as a rigid body when the support settles, so $M_0 = 0$ and $\Delta_{i0} = 0$ .
Find the flexibility coefficients $f_{ij}$ from unit-load moment diagrams $m_i$ .
Write compatibility: the displacement at each redundant must equal the prescribed settlement $\delta_i$ (measured positive in the direction of $X_i$ ): $\sum_j f_{ij}X_j = \delta_i$ .
Solve for $X_j$ and compute $M = \sum X_j m_j$ . Since there are no member loads, the BM varies linearly between the joints, so the BMD is a set of straight lines joining the joint moments. Draw each ordinate on the tension side.

Note that the moments are proportional to $EI\,\delta/L^2$ , so a stiffer frame develops larger moments for the same settlement.

Example

Portal frame ABCD, $h = L = 4$ m, $EI = 20000$ kNm $^2$ constant, A fixed, D hinged. D settles vertically by $\delta = 10$ mm.

Redundants: $X_1$ = $H_D$ ( $\rightarrow$ ), $X_2$ = $V_D$ ( $\uparrow$ ). The unit moment diagrams are: $m_1$ = 0 at A rising to 4 at B, 4 along BC, falling to 0 at D; $m_2$ = 4 along AB, falling from 4 at B to 0 at C.

f_{11} = \frac{106.67}{EI},\quad f_{12} = \frac{64}{EI},\quad f_{22} = \frac{85.33}{EI}

\begin{aligned} 106.67X_1 + 64X_2 &= 0 \\ 64X_1 + 85.33X_2 &= -0.01\times20000 = -200 \end{aligned}

X_1 = H_D = 2.56\ \text{kN}\ (\rightarrow),\qquad X_2 = V_D = 4.26\ \text{kN}\ (\text{downward, since D is pulled down})

Bending moments (kNm; + tension on the inner face)

Point	BM
A	-17.05
B	-6.82
C	10.23
D	0

BM is linear in every member; it is hogging (outer tension) in the left column and at B, and sagging at C (inner tension).

2074 Asoj · 6 marks

Determine the moment at the fixed support of the following loaded beam using Castigliano's theorem. Take EI constant. [Figure: beam ABC, A fixed, B roller at 5 m from A, overhang BC = 2 m with a 50 kNm moment applied at C.]

Answer

Reading of the figure: A fixed; the roller B is 5 m from A; the overhang BC (2 m) carries a couple of 50 kNm at C, taken as clockwise. EI is constant. The beam has one redundant; take the roller reaction $R_B$ (upward).

 ||==================o=========  <- 50 kNm (cw) at C
 A (fixed)           B        C
 |<------ 5 m ------>|<- 2 ->|

Bending moments (x from C for the overhang, $s$ from B for AB)

Overhang BC: $M = -50$ kNm (constant, hogging), independent of $R_B$ .
Span AB: $M = R_B\,s - 50$ , $\ \dfrac{\partial M}{\partial R_B} = s$

Castigliano's theorem ( $\delta_B = 0$ )

\int_0^5 (R_B\,s - 50)\,s\,ds = 0 \ \Rightarrow\ \frac{R_B (5)^3}{3} - 50\frac{(5)^2}{2} = 0

41.667\,R_B = 625 \ \Rightarrow\ R_B = 15\ \text{kN}\ (\uparrow)

Moment at the fixed support

At A ( $s = 5$ ):

M_A = R_B(5) - 50 = 15(5) - 50 = +25\ \text{kNm}

The reaction moment at A is 25 kNm, which is half the applied couple (the carry-over factor of a propped cantilever is 1/2). It acts in the clockwise sense on the beam. The vertical reaction at A is $15$ kN downward.

Check: $\Sigma V = 0$ : $R_A + R_B = 0$ so $R_A = -15$ kN.

Answer: moment at the fixed support $M_A = 25$ kNm (clockwise); $R_B = 15$ kN upward, $R_A = 15$ kN downward. For an anticlockwise couple the results have the same magnitudes with all senses reversed.

2074 Asoj · 10 marks

A portal frame of span 4 m and height 4 m is fixed at both supports. The beam of the frame carries a uniformly distributed gravity load of intensity 30 kN/m. Use the force method to solve the frame considering the cross-sectional stiffness (EI) to be constant. Draw bending moment, shear force and normal thrust diagrams for the frame.

Answer

Portal frame: span 4 m, height 4 m, both bases fixed, EI constant, UDL $w = 30$ kN/m on the beam BC. $r = 6$ , so $D_s = 3$ .

Release the fixed support at D and take $X_1 = H_D$ ( $\rightarrow$ ), $X_2 = V_D$ ( $\uparrow$ ) and $X_3 = M_D$ (anticlockwise) as redundants. The primary structure is a cantilever fixed at A.

   30 kN/m
 B vvvvvvvvvv C
 |            |
 | 4 m        | 4 m
 |            |
A///        ///D   (both fixed)
 |<-- 4 m -->|

Unit and load moment diagrams (+ tension on the inside)

Section	$M_0$	$m_1$	$m_2$	$m_3$
A	-240	0	4	1
B	-240	4	4	1
mid BC	-60	4	2	1
C	0	4	0	1
D	0	0	0	1

( $M_0 = -30\times4\times2 = -240$ kNm at A and B, falling parabolically to 0 at C.)

Flexibility coefficients (in $1/EI$ units)

[f] = \begin{bmatrix} 106.67 & 64 & 32 \\ 64 & 85.33 & 24 \\ 32 & 24 & 12 \end{bmatrix},\qquad \{\Delta_0\} = \begin{Bmatrix} -3200 \\ -4800 \\ -1280 \end{Bmatrix}

Compatibility ( $[f]\{X\} = -\{\Delta_0\}$ )

\begin{aligned} 106.67X_1 + 64X_2 + 32X_3 &= 3200 \\ 64X_1 + 85.33X_2 + 24X_3 &= 4800 \\ 32X_1 + 24X_2 + 12X_3 &= 1280 \end{aligned}

X_1 = H_D = -10\ \text{kN},\qquad X_2 = V_D = 60\ \text{kN},\qquad X_3 = M_D = 13.33\ \text{kNm}

(The symmetric structure and load give $V_A = V_D = 60$ kN and $|H| = 10$ kN inwards at both supports.)

Reactions

Support	$H$ (inwards)	$V$ (up)	Moment
A	10 kN	60 kN	13.33 kNm
D	10 kN	60 kN	13.33 kNm

Bending moment diagram (kNm, + tension inside)

Point	BM
A, D	+13.33 (inner face tension)
B, C	-26.67 (outer face tension)
Mid-span of BC	+33.33 (bottom tension)

The BM in each column is linear, $M = 13.33 - 10y$ , zero at $y = 1.33$ m above the base. In the beam $M = -26.67 + 60x - 15x^2$ , zero at $0.51$ m and $3.49$ m from B, with the maximum $+33.33$ kNm at mid-span.

Shear force diagram (kN)

Columns: constant $10$ kN (horizontal), acting inwards at the base.
Beam: $+60$ kN at B, falling linearly through zero at mid-span to $-60$ kN at C.

Normal thrust diagram (kN)

Columns: $-60$ (compression) throughout.
Beam: $-10$ (compression) throughout.

Answer: $H = 10$ kN, $V = 60$ kN and $M = 13.33$ kNm at each base; $M_B = M_C = -26.67$ kNm; mid-span $M = +33.33$ kNm.

2074 Asoj · 12 marks

Analyse the truss shown in figure below using the "Force Method". Take the cross-sectional stiffness EA of the members to be constant. [Figure: truss ABCDE; bottom chord A-D-E (4 m + 4 m), top joints B and C at 3 m height, with diagonals; A hinged, E roller; 60 kN downward at D.]

Answer

Reading of the figure: A (0, 0) hinged, E (8, 0) roller; D (4, 0) on the bottom chord; top joints B (2, 3) and C (6, 3); members AB, BC, CE (top and sides), AD, DE (bottom chord), BD, CD and the two crossing diagonals AC and BE. The 60 kN load acts downward at D. $EA$ is constant. (The member arrangement is read from the description; the method is the same for any arrangement.)

$m = 9$ , $j = 5$ , $r = 3$ , so $D_s = m + r - 2j = 2$ . Choose $X_1$ = force in AC and $X_2$ = force in BE (tension +).

Reactions

By symmetry $V_A = V_E = 30$ kN and $H_A = 0$ .

Force tables (tension +)

Member	$L$ (m)	$S_0$	$n_1$	$n_2$	Final $S$
AB	3.606	-36.056	-0.537	0.000	-30.796
BC	4.000	-40.000	-0.596	-0.596	-28.331
CE	3.606	-36.056	0.000	-0.537	-30.796
AD	4.000	+20.000	-0.596	0.000	+25.834
DE	4.000	+20.000	0.000	-0.596	+25.834
BD	3.606	+36.056	+0.537	-0.537	+36.056
CD	3.606	+36.056	-0.537	+0.537	+36.056
AC	6.708	0.000	+1.000	0.000	-9.785
BE	6.708	0.000	0.000	+1.000	-9.785

Flexibility coefficients (in $1/AE$ units)

\begin{aligned} f_{11} &= f_{22} = 12.677, \qquad f_{12} = f_{21} = -0.661 \\ \Delta_{10} &= \sum S_0n_1L = 117.576, \qquad \Delta_{20} = \sum S_0n_2L = 117.576 \end{aligned}

Compatibility

\begin{aligned} 12.677\,X_1 + (-0.661)\,X_2 &= -117.576 \\ -0.661\,X_1 + 12.677\,X_2 &= -117.576 \end{aligned}

X_1 = S_{AC} = -9.78\ \text{kN},\qquad X_2 = S_{BE} = -9.78\ \text{kN}

Final member forces $S = S_0 + n_1X_1 + n_2X_2$

Member	Force (kN)	Nature
AB	-30.80	compression
BC	-28.33	compression
CE	-30.80	compression
AD	+25.83	tension
DE	+25.83	tension
BD	+36.06	tension
CD	+36.06	tension
AC	-9.78	compression
BE	-9.78	compression

Answer: AB $= CE = -30.80$ kN, BC $= -28.33$ kN, AD $= DE = +25.83$ kN, BD $= CD = +36.06$ kN, AC $= BE = -9.78$ kN (positive = tension).

2073 Shrawan · 10 marks

Determine slope at A and deflection at D of the beam shown in figure below using Castigliano's theorem. [Figure: beam A-B-C-D with 30 kN at B (4 m from A), 10 kN at D; BC = 6 m, CD = 2 m overhang; A hinged, C roller.]

Answer

Beam ABCD: A hinged, C roller (AC = 10 m), loads 30 kN at B (4 m from A) and 10 kN at the end D of the 2 m overhang CD. EI is constant. Apply a dummy couple $M'$ (clockwise) at A, and treat the load at D as a variable $P$ (equal to 10 kN at the end).

        30 kN                  10 kN
          v                      v
 A o------B---------------------C o--D
 hinge                         roller
 |<- 4 ->|<------- 6 ------->|<- 2 ->|

Reactions (with $M'$ and $P$ )

Moments about A:

R_C = \frac{30(4) + 12P + M'}{10},\qquad R_A = 30 + P - R_C

At $M' = 0$ , $P = 10$ : $R_C = 24$ kN, $R_A = 16$ kN.

Bending moments ( $x$ from A for AC; $s$ from D for the overhang)

\begin{aligned} AB\ (0 \le x \le 4):\ & M = M' + R_Ax, \\ BC\ (4 \le x \le 10):\ & M = M' + R_Ax - 30(x-4), \\ DC\ (0 \le s \le 2):\ & M = -Ps \end{aligned}

so $R_A = 30 + P - R_C = 18 - 0.2P - 0.1M'$ , with $\dfrac{\partial R_A}{\partial M'} = -0.1$ and $\dfrac{\partial R_A}{\partial P} = -0.2$ .

Range	$\partial M/\partial M'$	$\partial M/\partial P$
AB, BC	$1 - 0.1x$	$-0.2x$
DC	0	$-s$

At $M' = 0$ and $P = 10$ : $R_A = 16$ kN, $M = 16x$ in AB, $M = 120 - 14x$ in BC, $M = -10s$ in the overhang.

Slope at A

EI\,\theta_A = \int_0^4 16x\,(1-0.1x)\,dx + \int_4^{10}(120-14x)(1-0.1x)\,dx = 93.87 + 64.80

\theta_A = \frac{158.67}{EI}\ \text{rad (clockwise, the positive sense of } M'\text{)}

Deflection at D

EI\,\delta_D = \int_0^4 16x(-0.2x)\,dx + \int_4^{10}(120-14x)(-0.2x)\,dx + \int_0^2(-10s)(-s)\,ds

EI\,\delta_D = -68.27 - 134.40 + 26.67 = -176.00

The negative sign shows that D moves opposite to the 10 kN load.

\delta_D = \frac{176}{EI}\ \text{upward}

(The 30 kN load at B tilts the beam at C so that the overhang rises; this outweighs the downward bending of the overhang under its own 10 kN load, which is only $160/EI$ .)

Answer: slope at A $= \dfrac{158.67}{EI}$ rad (clockwise); deflection at D $= \dfrac{176}{EI}$ (upward).

2073 Shrawan · 1 mark

Define and explain the term primary structure.

Answer

The primary structure (also called the released or basic determinate structure) is the statically determinate and stable structure obtained from an indeterminate structure by removing the redundant forces, that is, redundant reactions or internal forces such as a support, a member, or a moment restraint. The removed redundants are then applied as unknown loads and found from compatibility conditions.

A structure with $D_s$ degrees of indeterminacy needs $D_s$ releases. The primary structure must remain stable, and it is not unique. Example: for a propped cantilever, the primary structure may be the cantilever (prop removed) or a simply supported beam (fixed-end moment released). In the force method every analysis starts by choosing it, finding $M_0$ for the loads and $m_i$ for unit redundants.

2073 Shrawan · 1 mark

Define and explain the term redundant force.

Answer

A redundant force (or redundant) is a reaction or internal force that is not needed for the static equilibrium of a structure, so it cannot be found from the equations of equilibrium alone. The number of redundants equals the degree of static indeterminacy $D_s$ .

The redundants are chosen and released to form the primary (determinate) structure, are treated as unknown external loads $X_1, X_2, \ldots$ , and are solved from the compatibility equations $\sum f_{ij}X_j + \Delta_{i0} = 0$ . Example: the prop reaction $R_B$ of a propped cantilever, or the horizontal thrust $H$ of a two-hinged arch.

2073 Shrawan · 1 mark

Define and explain the term flexibility coefficient.

Answer

The flexibility coefficient $f_{ij}$ is the displacement at coordinate $i$ (in the direction of the redundant $X_i$ ) caused by a unit force acting at coordinate $j$ , with all other redundants and loads absent.

f_{ij} = \int \frac{m_i\,m_j}{EI}\,dx

for flexural members (for trusses, $f_{ij} = \sum n_in_jL/AE$ ). Its unit is m/kN for a force coordinate or rad/kNm for a moment coordinate. The coefficients form the flexibility matrix $[F]$ , which is symmetric ( $f_{ij} = f_{ji}$ , Maxwell). Example: for a cantilever of length $L$ , the tip deflection due to a unit tip load is $f_{11} = L^3/3EI$ .

2073 Shrawan · 12 marks

Determine the forces in all members of the truss shown in figure below using the force method. AE is constant for all members. [Figure: truss with joints A, B (left), C, D (top) and E, F (bottom); panels 3 m + 3 m wide and 4 m high, double diagonals; 50 kN downward at C and 20 kN horizontal at D; A hinged, F roller.]

Answer

Reading of the figure: bottom joints A (0, 0), E (3, 0), F (6, 0); top joints B (0, 4), C (3, 4), D (6, 4). Members: chords AE, EF, BC, CD; verticals AB, CE, DF; and two crossing diagonals in each panel (BE, AC and CF, ED). A is hinged, F is a roller. Loads: 50 kN downward at C and 20 kN horizontal at D (towards the right). $AE$ is constant.

$m = 11$ , $j = 6$ , $r = 3$ , so $D_s = 11 + 3 - 12 = 2$ . Take $X_1$ = force in AC and $X_2$ = force in CF (tension +).

Reactions

$\Sigma M_A = 0$ : $V_F(6) = 50(3) + 20(4) \Rightarrow V_F = 38.333$ kN ( $\uparrow$ ); $V_A = 50 - 38.333 = 11.667$ kN ( $\uparrow$ ); $H_A = 20$ kN ( $\leftarrow$ ).

Force tables (tension +)

Member	$L$ (m)	$S_0$	$n_1$	$n_2$	Final $S$
AE	3.000	+20.000	-0.600	0.000	+26.046
EF	3.000	0.000	0.000	-0.600	+17.785
BC	3.000	-8.750	-0.600	0.000	-2.704
CD	3.000	-8.750	0.000	-0.600	+9.035
AB	4.000	-11.667	-0.800	0.000	-3.606
CE	4.000	-50.000	-0.800	-0.800	-18.226
DF	4.000	-38.333	0.000	-0.800	-14.620
BE	5.000	+14.583	+1.000	0.000	+4.507
AC	5.000	0.000	+1.000	0.000	-10.076
CF	5.000	0.000	0.000	+1.000	-29.641
ED	5.000	+47.917	0.000	+1.000	+18.275

Flexibility coefficients (in $1/AE$ units)

\begin{aligned} f_{11} &= \sum n_1^2L = 17.280, \quad f_{22} = \sum n_2^2L = 17.280, \quad f_{12} = \sum n_1n_2L = 2.560 \\ \Delta_{10} &= \sum S_0n_1L = 250.000, \quad \Delta_{20} = \sum S_0n_2L = 538.000 \end{aligned}

Compatibility

\begin{aligned} 17.280\,X_1 + 2.560\,X_2 &= -(250.000) \\ 2.560\,X_1 + 17.280\,X_2 &= -(538.000) \end{aligned}

X_1 = S_{AC} = -10.08\ \text{kN},\qquad X_2 = S_{CF} = -29.64\ \text{kN}

Final member forces $S = S_0 + n_1X_1 + n_2X_2$

Member	Force (kN)	Nature
AE	+26.05	tension
EF	+17.78	tension
BC	-2.70	compression
CD	+9.03	tension
AB	-3.61	compression
CE	-18.23	compression
DF	-14.62	compression
BE	+4.51	tension
AC	-10.08	compression
CF	-29.64	compression
ED	+18.28	tension

Answer: AE $= +26.05$ , EF $= +17.78$ , BC $= -2.70$ , CD $= +9.03$ , AB $= -3.61$ , CE $= -18.23$ , DF $= -14.62$ , BE $= +4.51$ , AC $= -10.08$ , CF $= -29.64$ , ED $= +18.28$ kN (positive = tension).

2073 Shrawan · 7 marks

Using the flexibility matrix method, determine the reactions at support D of the frame loaded as shown in figure below. Also draw SFD and BMD. Take EI = constant. [Figure: frame ABCD; A fixed, column AB 5 m high with 50 kN horizontal at B; beam BC with 30 kNm moment at 2 m from B, total span 5 m; column CD 3 m high, D on roller/hinge.]

Answer

Reading of the figure: A (0, 0) fixed; column AB is 5 m high with 50 kN horizontal at B (towards the right); beam BC is 5 m long with a couple of 30 kNm (taken clockwise) applied at F, 2 m from B; column CD is 3 m high (D is 2 m above the level of A) and D is hinged. EI is constant.

$D_s = 3 + 2 - 3 = 2$ . Release the hinge at D: $X_1$ = horizontal reaction ( $\rightarrow$ ), $X_2$ = vertical reaction ( $\uparrow$ ).

Flexibility matrix and load vector (flexibility method)

The primary structure is the cantilever fixed at A. The loads (50 kN at B and the couple at F) give $M_0$ , and unit forces at D give $m_1$ , $m_2$ (the table of moments below lists their values; + means tension on the inner face).

\begin{aligned} f_{11} &= \frac{65.667}{EI} \\ f_{12} &= \frac{50.000}{EI} \\ f_{22} &= \frac{166.667}{EI} \\ \Delta_{10} &= \frac{-46.7}{EI} \\ \Delta_{20} &= \frac{-4115.0}{EI} \end{aligned}

$[F]\{X\} = -\{\Delta_0\}$ :

\begin{aligned} 65.667\,X_1 + 50.000\,X_2 &= 46.7 \\ 50.000\,X_1 + 166.667\,X_2 &= 4115.0 \end{aligned}

X_1 = H_D = -23.44\ \text{kN},\qquad X_2 = V_D = 31.72\ \text{kN}

Final bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-280.00	-2.00	5.00	-74.50
B (column)	-30.00	3.00	5.00	58.28
F (left of couple)	-30.00	3.00	3.00	-5.16
F (right of couple)	0.00	3.00	3.00	24.84
C	0.00	3.00	0.00	-70.33
D	0.00	0.00	0.00	0.00

Shear force (kN)

Member	SF
AB	+26.56
BF, FC (beam)	-31.72
CD	+23.44

Reactions

D (hinge): $H_D = 23.44$ kN (towards the left), $V_D = 31.72$ kN ( $\uparrow$ )
A (fixed): $H_A = 26.56$ kN (towards the left), $V_A = 31.72$ kN (downward), $M_A = 74.50$ kNm

Check: $\Sigma H$ : $50 - 26.56 - 23.44 = 0$ ; $\Sigma V$ : $-31.72 + 31.72 = 0$ (no vertical load).

The BMD is linear in each member, with a jump of 30 kNm at F from the couple.

Answer: at D, $H_D = 23.44$ kN (towards the left) and $V_D = 31.72$ kN (upward).

2072 Chaitra · 12 marks

Using Castigliano's theorem, find the deflection at point B of the beam shown in figure below. Take constant EI through the length. [Figure: beam AC, A fixed, C roller; span L/2 + L/2 with point load P at B (mid-span).]

Answer

Beam AC of span $L$ : A fixed, C a roller, point load $P$ at B (mid-span). EI is constant. The beam is indeterminate to the 1st degree. First find the redundant $R_C$ and then the deflection at B.

          P
          v
 ||=======B=======o
 A (fixed)        C (roller, R_C)
 |<-- L/2 -->|<-- L/2 -->|

Step 1: Bending moments (x measured from C)

\begin{aligned} CB\ (0 \le x \le L/2):\quad & M = R_C\,x \\ BA\ (L/2 \le x \le L):\quad & M = R_C\,x - P\left(x - \frac{L}{2}\right) \end{aligned}

Step 2: Redundant reaction

The deflection at C is zero, so $\dfrac{\partial U}{\partial R_C} = 0$ :

\int_0^{L/2} R_C\,x^2\,dx + \int_{L/2}^{L}\left[R_C\,x - P\left(x-\frac L2\right)\right]x\,dx = 0

R_C\frac{L^3}{3} - P\left[\frac{x^3}{3} - \frac{Lx^2}{4}\right]_{L/2}^{L} = 0

R_C\frac{L^3}{3} - P\,\frac{5L^3}{48} = 0 \ \Rightarrow\ R_C = \frac{5P}{16}

(The bracket equals $\left(\dfrac{L^3}{3} - \dfrac{L^3}{4}\right) - \left(\dfrac{L^3}{24} - \dfrac{L^3}{16}\right) = \dfrac{L^3}{12} + \dfrac{L^3}{48} = \dfrac{5L^3}{48}$ .)

Step 3: Deflection at B by Castigliano's theorem

The load $P$ acts at B, so

\delta_B = \frac{\partial U}{\partial P} = \frac{1}{EI}\int M\,\frac{\partial M}{\partial P}\,dx

Because $\partial U/\partial R_C = 0$ , the dependence of $R_C$ on $P$ does not add anything, so only the explicit $P$ in $M$ is differentiated: $\dfrac{\partial M}{\partial P} = -\left(x - \dfrac L2\right)$ for $L/2 \le x \le L$ and $0$ for the part CB.

With $R_C = 5P/16$ :

\delta_B = \frac{1}{EI}\int_{L/2}^{L}\left[\frac{5P}{16}x - P\left(x - \frac L2\right)\right]\left[-\left(x-\frac L2\right)\right]dx

Let $u = x - L/2$ (from 0 to $L/2$ ), so $x = u + L/2$ :

\delta_B = \frac{P}{EI}\int_0^{L/2}\left[u - \frac{5}{16}\left(u + \frac L2\right)\right]u\,du = \frac{P}{EI}\int_0^{L/2}\left[\frac{11}{16}u^2 - \frac{5L}{32}u\right]du

\delta_B = \frac{P}{EI}\left[\frac{11}{48}\cdot\frac{L^3}{8} - \frac{5L}{64}\cdot\frac{L^2}{4}\right] = \frac{P L^3}{EI}\left(\frac{11}{384} - \frac{5}{256}\right)

\delta_B = \frac{PL^3}{EI}\cdot\frac{22 - 15}{768} = \frac{7PL^3}{768\,EI}

Answer: the deflection at B is $\delta_B = \dfrac{7PL^3}{768\,EI}$ downward; $R_C = \dfrac{5P}{16}$ .

2072 Chaitra · 4 marks

State and prove Maxwell's Reciprocal theorem.

Answer

Statement

In a linearly elastic structure, the deflection at point A due to a unit load at point B equals the deflection at B due to a unit load at A:

\delta_{AB} = \delta_{BA}

Here $\delta_{ij}$ is the displacement at $i$ caused by a unit load at $j$ .

Proof

Let loads $P_A$ and $P_B$ act at A and B. Let $\delta_{AA}, \delta_{BA}$ be the deflections at A and B caused by a unit load at A, and $\delta_{AB}, \delta_{BB}$ those caused by a unit load at B.

Case 1: apply $P_A$ first, then $P_B$ .

$P_A$ gradually applied: work $= \tfrac12 P_A(P_A\delta_{AA})$ .
$P_B$ then applied: its own work $= \tfrac12 P_B(P_B\delta_{BB})$ . While $P_B$ is applied, $P_A$ (already at full value) moves through the extra displacement $P_B\delta_{AB}$ at A and does work $P_A(P_B\delta_{AB})$ .

U_1 = \tfrac12 P_A^2\delta_{AA} + \tfrac12 P_B^2\delta_{BB} + P_AP_B\,\delta_{AB}

Case 2: apply $P_B$ first, then $P_A$ . In the same way,

U_2 = \tfrac12 P_B^2\delta_{BB} + \tfrac12 P_A^2\delta_{AA} + P_BP_A\,\delta_{BA}

Strain energy depends only on the final loads, not on the order of loading, so $U_1 = U_2$ . Hence

P_AP_B\,\delta_{AB} = P_AP_B\,\delta_{BA} \ \Rightarrow\ \delta_{AB} = \delta_{BA}

Example

A cantilever of length $L$ : the tip deflection due to a unit load at mid-span is $\dfrac{5L^3}{48EI}$ , which equals the mid-span deflection due to a unit load at the tip. The theorem also holds for rotations and for a force-couple pair, so the flexibility matrix is symmetric: $f_{ij} = f_{ji}$ .

2072 Chaitra · 12 marks

Determine the bar forces and reactions that develop in the statically indeterminate truss shown in figure below. [Figure: rectangular truss ABCD, 6.5 m wide and 5 m high, with both diagonals; A hinged, B roller; 400 kN horizontal at C. Cross-sectional area: member BD = 20 cm², other members = 15 cm²; Young's modulus = $240\times10^{6}\ \text{kN/m}^2$ .]

Answer

Reading of the figure: A (0, 0) hinged and B (6.5, 0) roller at the base, C (6.5, 5) and D (0, 5) at the top; members AB, CD (6.5 m), BC, DA (5 m), diagonals AC and BD (8.20 m). The 400 kN load acts horizontally at C (towards the right). $E = 240\times10^6$ kN/m $^2$ . Areas: BD $= 20$ cm $^2$ , all others $15$ cm $^2$ . So $EA_{BD} = 480000$ kN and $EA = 360000$ kN for the other members.

$D_s = 6 + 3 - 8 = 1$ . Take the force in BD as the redundant $X$ (tension +).

Reactions

$\Sigma M_A = 0$ : $V_B(6.5) = 400(5) \Rightarrow V_B = 307.69$ kN ( $\uparrow$ ); $V_A = 307.69$ kN ( $\downarrow$ ); $H_A = 400$ kN ( $\leftarrow$ ).

Force table (tension +; $S_0$ with BD removed)

Member	$L$ (m)	$EA$ (kN)	$S_0$ (kN)	$n$	$S_0nL/EA\ (\times10^{-3})$	$n^2L/EA\ (\times10^{-3})$
AB	6.500	360000	0.00	-0.7926	+0.0000	0.0113
BC	5.000	360000	-307.69	-0.6097	+2.6056	0.0052
CD	6.500	360000	0.00	-0.7926	+0.0000	0.0113
DA	5.000	360000	0.00	-0.6097	+0.0000	0.0052
AC	8.201	360000	+504.65	+1.0000	+11.4957	0.0228
BD	8.201	480000	+0.00	+1.0000	+0.0000	0.0171

\Delta_{10} = \sum\frac{S_0nL}{EA} = 0.014101\ \text{m},\qquad f_{11} = \sum\frac{n^2L}{EA} = 7.2877\times10^{-5}\ \text{m/kN}

Compatibility

\Delta_{10} + f_{11}X = 0 \ \Rightarrow\ X = -\frac{0.014101}{7.2877\times10^{-5}} = -193.49\ \text{kN}

Final forces $S = S_0 + nX$

Member	Force (kN)	Nature
AB	+153.37	tension
BC	-189.72	compression
CD	+153.37	tension
DA	+117.98	tension
AC	+311.16	tension
BD	-193.49	compression

Reactions

$H_A = 400$ kN ( $\leftarrow$ ), $V_A = 307.69$ kN ( $\downarrow$ ), $V_B = 307.69$ kN ( $\uparrow$ ).

Answer: AB $= +153.37$ , BC $= -189.72$ , CD $= +153.37$ , DA $= +117.98$ , AC $= +311.16$ , BD $= -193.49$ kN (positive = tension).

2072 Chaitra · 13 marks

Determine the reactions at support E and A and draw bending moment diagram of the frame shown in figure below by using the flexibility matrix method (force method). [Figure: frame; column AB (2EI) of 10 m with A fixed, 20 kN horizontal at B, with a 7 m dimension marked; beam BC-D (2EI) of 10 m carrying 75 kN/m UDL; column DE (EI) of 5 m, E hinged.]

Answer

Reading of the figure: A (0, 0) is fixed; column AB is 10 m high ( $2EI$ ) with a 20 kN horizontal load (towards the right) at 7 m above A; beam BD is 10 m long ( $2EI$ ) with 75 kN/m over its whole length; column DE is 5 m long ( $EI$ ), so E is 5 m above the level of A, and E is hinged.

$D_s = 3 + 2 - 3 = 2$ . Release the hinge at E: $X_1$ = horizontal reaction ( $\rightarrow$ ), $X_2$ = vertical reaction ( $\uparrow$ ). The primary structure is a cantilever fixed at A.

 B ================= D    75 kN/m
 |                   |
 |  20 kN ->         | 5 m
 |                   E (hinge)
 | 10 m
 A (fixed)

Primary-structure moments (+ tension on the inner face)

$M_0$ : at B, $-75\times10\times5 = -3750$ kNm; at the 20 kN level, $-3750$ ; at A, $-(3750 + 20\times7) = -3890$ kNm; the beam falls parabolically from $-3750$ at B to 0 at D.
$m_1$ (unit horizontal force at E): DE from 0 at E to 5 at D; beam constant 5; column from 5 at B to $-5$ at A.
$m_2$ (unit vertical force at E): beam from 10 at B to 0 at D; column constant 10.

Flexibility matrix $[F]$ and load vector $\{\Delta_0\}$ (in $1/EI$ units)

\begin{aligned} f_{11} &= \frac{208.333}{EI} \\ f_{12} &= \frac{125.000}{EI} \\ f_{22} &= \frac{666.667}{EI} \\ \Delta_{10} &= \frac{-30596.7}{EI} \\ \Delta_{20} &= \frac{-236825.0}{EI} \end{aligned}

Compatibility: $[F]\{X\} = -\{\Delta_0\}$

\begin{aligned} 208.333\,X_1 + 125.000\,X_2 &= 30596.7 \\ 125.000\,X_1 + 666.667\,X_2 &= 236825.0 \end{aligned}

X_1 = H_E = -74.68\ \text{kN},\qquad X_2 = V_E = 369.24\ \text{kN}

Final bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-3890.00	-5.00	10.00	175.80
Q (20 kN level)	-3750.00	2.00	10.00	-206.96
B (column)	-3750.00	5.00	10.00	-431.00
mid BD	-937.50	5.00	5.00	535.30
D (beam)	0.00	5.00	0.00	-373.40
E	0.00	0.00	0.00	0.00

Maximum sagging moment in the beam: 535.52 kNm at 5.08 m from B.

Reactions

E (hinge): $H_E = 74.68$ kN (towards the left), $V_E = 369.24$ kN ( $\uparrow$ )
A (fixed): $H_A = 54.68$ kN (towards the right), $V_A = 380.76$ kN ( $\uparrow$ ), $M_A = 175.80$ kNm (anticlockwise)

Check: $\Sigma H$ : $20 + 54.68 - 74.68 = 0$ ; $\Sigma V$ : $380.76 + 369.24 = 75\times10 = 750$ kN.

Answer: at E, $H_E = 74.68$ kN (towards the left) and $V_E = 369.24$ kN (upward); at A, $H_A = 54.68$ kN, $V_A = 380.76$ kN and $M_A = 175.80$ kNm.

2072 Kartik · 10 marks

Analyse the frame shown in figure below by using the force method and draw bending moment diagram. [Figure: frame; column (2I) of height 7 m, fixed at the base; 80 kN horizontal at the 3 m level; beam (I) of 10 m carrying 60 kN/m UDL, right end on a roller.]

Answer

Reading of the figure: column AB is 7 m high ( $2EI$ ) and fixed at A; the beam BC is 10 m long ( $EI$ ) with 60 kN/m; C is a roller (vertical support); a horizontal load of 80 kN (towards the right) acts on the column at 3 m above A.

$r = 3 + 1 = 4$ , so $D_s = 1$ . Take the vertical reaction $R_C$ at the roller as the redundant.

 B =================== C  (roller)
 |    60 kN/m
 |
 | 80 kN ->
 |
 A (fixed)

Primary structure (cantilever fixed at A)

Section	$M_0$ (kNm)	$m$ (unit $R_C$ )
A	-3240	10
80 kN level (3 m up)	-3000	10
B	-3000	10
mid BC	-750	5
C	0	0

( $M_0$ at B $= -60\times10\times5 = -3000$ ; at A it is $-3000 - 80\times3 = -3240$ .)

Flexibility coefficient and load term

\begin{aligned} f_{11} &= \frac{1}{2EI}\left[10^2(7)\right] + \frac{1}{EI}\left[\frac{10^2\times10}{3}\right] = \frac{350 + 333.33}{EI} = \frac{683.33}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[-10\,\frac{3(3240+3000)}{2} - 10(3000)(4)\right] + \frac{1}{EI}\left[-30\int_0^{10}(10-x)^3dx\right] \\ &= \frac{-46800 - 60000 - 75000}{EI} = \frac{-181800}{EI} \end{aligned}

Compatibility ( $\delta_C = 0$ )

f_{11}R_C + \Delta_{10} = 0 \ \Rightarrow\ R_C = \frac{181800}{683.33} = 266.05\ \text{kN}\ (\uparrow)

Final bending moments $M = M_0 + R_C\,m$ (kNm)

Section	BM	Tension side
A	-579.51	outer face of the column
80 kN level	-339.51	outer face
B	-339.51	outer (top of the beam)
Maximum in BC at 5.57 m from B	589.85	bottom of the beam
C	0

Reactions

C: $R_C = 266.05$ kN ( $\uparrow$ )
A: $V_A = 333.95$ kN ( $\uparrow$ ), $H_A = 80$ kN ( $\leftarrow$ ), $M_A = 579.51$ kNm (anticlockwise)

Check: $V_A + R_C = 600.00$ kN $= 60\times10 = 600$ kN.

Answer: $R_C = 266.05$ kN; $M_A = 579.51$ kNm; $M_B = -339.51$ kNm (hogging); maximum sagging moment $589.85$ kNm in the beam.

2072 Kartik · 6 marks

List the differences between force and displacement methods. Draw a neat sketch of a system and explain.

Answer

The force (flexibility) method takes the redundant forces as unknowns and satisfies compatibility of displacements. The displacement (stiffness) method takes the joint displacements as unknowns and satisfies equilibrium of the joints.

Point	Force method	Displacement method
Unknowns	Redundant forces or moments ( $D_s$ )	Joint displacements ( $D_k$ )
Governing condition	Compatibility of displacements	Equilibrium of joints
Matrix used	Flexibility matrix $[F]$	Stiffness matrix $[K]$
Equations	$[F]\{X\} = -\{\Delta_0\}$	$[K]\{\Delta\} = \{P\}$
Primary structure	Statically determinate (redundants removed)	Kinematically determinate (all joints fixed)
Best for	Structures with few redundants (low $D_s$ ), trusses, arches	Structures with few joint displacements (low $D_k$ ), multi-storey frames, continuous beams
Suitable for computer	Less systematic, since the primary structure is not unique	Very systematic, so it is the basis of computer programs
Effect of settlement/temperature	Enters through the load vector $\Delta_0$	Enters through fixed-end forces

Illustration: portal frame with fixed bases

        B ______ C
        |        |
        |        |
       A///    ///D     (both fixed)

For the portal frame with fixed bases, $D_s = 3$ and $D_k = 3$ (rotations of B and C, one sway) if axial deformation is neglected. If one base is replaced by a hinge, $D_s = 2$ , but $D_k$ stays 3 (or 2 with the modified stiffness $3EI/L$ ). A two-span continuous beam on rigid supports has $D_s = 1$ (the middle reaction) and $D_k = 1$ (rotation at the middle support, if the ends are hinged).

Force method steps: remove the redundants, find $M_0$ and $m_i$ , find $f_{ij}$ and $\Delta_{i0}$ , solve $[F]\{X\} = -\{\Delta_0\}$ , then superpose.

Displacement method steps: fix all joints, apply fictitious restraints, find the fixed-end forces, find the stiffness coefficients, solve $[K]\{\Delta\} = \{P\}$ for the joint displacements, then find the member end moments.

Choose the method that gives fewer unknowns: a frame with many redundant members but few joints suits the displacement method, whereas an arch or a truss with one or two redundants suits the force method.

2072 Kartik · 7 marks

Determine the support moments and draw bending moment diagram of the continuous beam shown in figure below by using the three moment equation. [Figure: beam 1-2-3, 1 fixed; 100 kN at 4 m from 1, 2 at 8 m (I); span 2-3 = 10 m (2I) with 40 kN/m UDL; overhang of 2 m with 20 kN at the end.]

Answer

Reading of the figure: the beam is fixed at 1; span 1-2 = 8 m with $I$ and a 100 kN load at 4 m from 1; span 2-3 = 10 m with $2I$ and 40 kN/m; the overhang 3-4 is 2 m with 20 kN at the free end. Supports 2 and 3 are on rollers. $E$ is constant.

Known moment at 3

The overhang is statically determinate: $M_3 = -20\times2 = -40$ kNm (hogging).

Three moment equation with different $I$

M_1\frac{L_1}{I_1} + 2M_2\left(\frac{L_1}{I_1}+\frac{L_2}{I_2}\right) + M_3\frac{L_2}{I_2} = -\frac{6a_1\bar x_1}{I_1L_1} - \frac{6a_2\bar x_2}{I_2L_2}

Load terms:

Span 1-2: free BM peak $= \dfrac{100\times4\times4}{8} = 200$ , $a_1 = \tfrac12(8)(200) = 800$ , $\bar x_1 = 4$ : $\dfrac{6a_1\bar x_1}{L_1} = 2400$ .
Span 2-3: $\dfrac{wL_2^3}{4} = \dfrac{40(10)^3}{4} = 10000$ .

Equation at 1 (fixed end): $2M_1\dfrac{8}{I} + M_2\dfrac{8}{I} = -\dfrac{2400}{I}$

2M_1 + M_2 = -300

Equation at 2: multiply by $I$ : $8M_1 + 2M_2\left(8 + \dfrac{10}{2}\right) + M_3\dfrac{10}{2} = -2400 - \dfrac{10000}{2}$

8M_1 + 26M_2 + 5(-40) = -7400 \ \Rightarrow\ 8M_1 + 26M_2 = -7200

Substituting $M_1 = \dfrac{-300 - M_2}{2}$ : $-1200 - 4M_2 + 26M_2 = -7200$ , so

M_2 = -272.727\ \text{kNm},\qquad M_1 = -13.636\ \text{kNm}

( $M_1$ is a small hogging moment at the fixed end.)

Reactions

With sagging-positive moments, in span 1-2: $M_2 = M_1 + 8R_1 - 100(4)$ , so

R_1 = \frac{400 + M_2 - M_1}{8} = \frac{400 + (-272.727) - (-13.636)}{8} = 17.614\ \text{kN}

In span 2-3 (taking $x$ from 2): $M_3 = M_2 + 10V_{2R} - 40\frac{10^2}{2}$ , so

V_{2R} = \frac{2000 + M_3 - M_2}{10} = \frac{2000 - 40 - (-272.727)}{10} = 223.273\ \text{kN}

R_2 = (100 - 17.614) + 223.273 = 305.659\ \text{kN},\qquad R_3 = 400 - 223.273 + 20 = 196.727\ \text{kN}

Check: $R_1 + R_2 + R_3 = 520.00$ kN $= 100 + 400 + 20$ .

Bending moment (kNm)

Section	BM
1 (fixed end)	-13.64
Under the 100 kN load	56.82
2	-272.73
Maximum in span 2-3 (5.58 m from 2)	350.41
3	-40.00
4 (free end)	0

Answer: $M_1 = -13.636$ kNm, $M_2 = -272.727$ kNm, $M_3 = -40$ kNm.

2071 Chaitra · 6 marks

Determine the moment at the fixed support of the propped cantilever beam using Castigliano's method. [Figure: beam AB, A fixed, B roller, span 12 m with 100 kN at 4 m from A and 50 kN at 8 m from A.]

Answer

Propped cantilever AB: A fixed, B a roller, span 12 m, loads 100 kN at 4 m and 50 kN at 8 m from A. EI is constant. Take the roller reaction $R_B$ as the redundant. Measure $x$ from B towards A.

         50 kN      100 kN
           v          v
 B o-------+----------+---------|| A (fixed)
 |<- 4 ->|<--- 4 --->|<--- 4 --->|

(Positions from B: 50 kN at 4 m, 100 kN at 8 m.)

Bending moments

\begin{aligned} 0 \le x \le 4:\quad & M = R_Bx, & \frac{\partial M}{\partial R_B} &= x \\ 4 \le x \le 8:\quad & M = R_Bx - 50(x-4), & \frac{\partial M}{\partial R_B} &= x \\ 8 \le x \le 12:\quad & M = R_Bx - 50(x-4) - 100(x-8), & \frac{\partial M}{\partial R_B} &= x \end{aligned}

Castigliano's theorem ( $\delta_B = 0$ )

\int_0^{12}R_Bx^2\,dx - 50\int_4^{12}(x-4)x\,dx - 100\int_8^{12}(x-8)x\,dx = 0

Evaluate each term:

\begin{aligned} R_B\frac{12^3}{3} &= 576\,R_B \\ 50\int_4^{12}(x^2 - 4x)\,dx &= 50\left[\frac{x^3}{3} - 2x^2\right]_4^{12} = 50\,(576 - 288 - 21.333 + 32) = 14933.3 \\ 100\int_8^{12}(x^2 - 8x)\,dx &= 100\left[\frac{x^3}{3} - 4x^2\right]_8^{12} = 100\,(0 - (170.667 - 256)) = 8533.3 \end{aligned}

576R_B = 14933.3 + 8533.3 = 23466.7 \ \Rightarrow\ R_B = 40.741\ \text{kN}

Moment at the fixed support

M_A = R_B(12) - 50(8) - 100(4) = 40.741(12) - 800 = -311.11\ \text{kNm}

The fixed-end moment is 311.11 kNm (hogging). The vertical reaction is $R_A = 150 - 40.741 = 109.259$ kN.

Answer: $M_A = 311.11$ kNm (hogging); $R_B = 40.741$ kN.

2071 Chaitra · 10 marks

Generate the flexibility matrix to determine the reactions at support D for the frame loaded as shown in the figure below. Also determine the reactions at support D and draw the bending moment diagram. Show all the steps. [Figure: frame; column AE-B (A fixed) with 4 m + 4 m... marked dimensions 2 m, 4 m, 4 m on the left; 30 kN horizontal at E; beam BFC with 4 m + 2 m; column CD with D hinged, 4 m high.]

Answer

Reading of the figure (dimensions as assumed): column ABE with A fixed at the base, E 4 m above A (30 kN horizontal, towards the right) and B 8 m above A; beam BFC with $BF = 4$ m and $FC = 2$ m; column CD, 4 m high, so D is 4 m above the level of A and is hinged. EI is constant.

$r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at D and take $X_1$ = horizontal reaction ( $\rightarrow$ ) and $X_2$ = vertical reaction ( $\uparrow$ ). The primary structure is the cantilever fixed at A.

Step 1: Coordinates and unit-load moments

$M_0$ is due to the 30 kN load only: $M_0 = -30\times4 = -120$ kNm at A, zero above E. The unit moments $m_1$ and $m_2$ are those caused by a unit horizontal force and a unit vertical force at D (see the table below, + tension on the inner face).

Step 2: Flexibility matrix $[F]$ and load vector $\{\Delta_0\}$

\begin{aligned} f_{11} &= \frac{160.000}{EI} \\ f_{12} &= \frac{72.000}{EI} \\ f_{22} &= \frac{360.000}{EI} \\ \Delta_{10} &= \frac{640.0}{EI} \\ \Delta_{20} &= \frac{-1440.0}{EI} \end{aligned}

[F] = \frac{1}{EI}\begin{bmatrix} 160.00 & 72.00 \\ 72.00 & 360.00 \end{bmatrix},\qquad \{\Delta_0\} = \frac{1}{EI}\begin{Bmatrix} 640.0 \\ -1440.0 \end{Bmatrix}

Step 3: Compatibility, $[F]\{X\} = -\{\Delta_0\}$

\begin{aligned} 160.000\,X_1 + 72.000\,X_2 &= -640.0 \\ 72.000\,X_1 + 360.000\,X_2 &= 1440.0 \end{aligned}

X_1 = H_D = -6.37\ \text{kN},\qquad X_2 = V_D = 5.27\ \text{kN}

Step 4: Bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-120.00	-4.00	6.00	-62.86
E	0.00	0.00	6.00	31.65
B (column top)	0.00	4.00	6.00	6.15
F	0.00	4.00	2.00	-14.95
C	0.00	4.00	0.00	-25.49
D	0.00	0.00	0.00	0.00

Step 5: Shear force (kN)

Member	SF
AE	+23.63
EB	-6.37
BF, FC	-5.27
CD	+6.37

Reactions

D: $H_D = 6.37$ kN (towards the left), $V_D = 5.27$ kN ( $\uparrow$ )
A: $H_A = 23.63$ kN (towards the left), $V_A = 5.27$ kN ( $\downarrow$ ), $M_A = 62.86$ kNm

Check: $\Sigma H$ : $30 - 23.63 - 6.37 = 0$ and $\Sigma V$ : $-5.27 + 5.27 = 0$ .

Answer: reactions at D: $H_D = 6.37$ kN (towards the left) and $V_D = 5.27$ kN (upward).

2071 Chaitra · 6 marks

List the properties of stiffness and flexibility matrices for a given system. Draw a neat sketch of a system and explain.

Answer

A system is described by $n$ coordinates (points and directions of force or displacement). The flexibility matrix $[F]$ gives displacements from forces, $\{\Delta\} = [F]\{P\}$ , and the stiffness matrix $[K]$ gives forces from displacements, $\{P\} = [K]\{\Delta\}$ .

Properties of the flexibility matrix $[F]$

It is a square matrix of order $n$ (number of coordinates).
It is symmetric, $f_{ij} = f_{ji}$ (Maxwell's reciprocal theorem).
The diagonal terms $f_{ii}$ are always positive (a force does positive work through its own displacement).
Off-diagonal terms may be positive, negative or zero.
It exists only for a stable, adequately supported system; a mechanism has no flexibility matrix.
It is positive definite, and $[F] = [K]^{-1}$ .
Column $j$ is the deformed shape caused by a unit force at coordinate $j$ .

Properties of the stiffness matrix $[K]$

It is a square, symmetric matrix of order $n$ : $k_{ij} = k_{ji}$ .
The diagonal terms $k_{ii}$ are always positive.
Off-diagonal terms may be positive, negative or zero.
It is positive definite when the rigid-body motion is prevented. If the system is free, $[K]$ is singular (it has zero determinant, because rigid-body motions need no force).
Column $j$ is the set of forces needed to hold the system in the shape of a unit displacement at $j$ (all others zero).
$[K] = [F]^{-1}$ , and for any coordinate $k_{ii} \ge 1/f_{ii}$ .

Illustration: cantilever of length $L$ with two coordinates

Coordinate 1 is the vertical deflection at the free end and coordinate 2 is the rotation at the free end.

  ||==========================o  -> (1) deflection
  fixed                       |  -> (2) rotation

[F] = \begin{bmatrix} \dfrac{L^3}{3EI} & \dfrac{L^2}{2EI} \\[2mm] \dfrac{L^2}{2EI} & \dfrac{L}{EI} \end{bmatrix},\qquad [K] = [F]^{-1} = \begin{bmatrix} \dfrac{12EI}{L^3} & -\dfrac{6EI}{L^2} \\[2mm] -\dfrac{6EI}{L^2} & \dfrac{4EI}{L} \end{bmatrix}

Both are symmetric with positive diagonals. Also $k_{11} = 12EI/L^3 > 1/f_{11} = 3EI/L^3$ , because $k_{11}$ holds the rotation at zero while $f_{11}$ leaves it free.

2071 Chaitra · 15 marks

Using the consistent deformation method analyse the frame shown in figure and draw bending moment, shear force and normal thrust diagram. [Figure: frame; left column 2EI with 10 kN/m UDL horizontally, height 3 m + 1 m...; beam EI of 2 m + 2 m with 15 kN downward at mid-span; right column 2EI.]

Answer

Reading of the figure (as assumed): portal frame of height 4 m and span 4 m. The left column AB ( $2EI$ ) is fixed at A and carries a horizontal UDL of 10 kN/m over its lower 3 m (towards the right); the beam BC ( $EI$ , 2 m + 2 m) carries 15 kN downward at mid-span F; the right column CD ( $2EI$ ) is hinged at D.

$r = 3 + 2 = 5$ , so $D_s = 2$ . By the consistent deformation method, release the hinge at D and take the reactions $X_1$ (horizontal, $\rightarrow$ ) and $X_2$ (vertical, $\uparrow$ ) at D as redundants. The displacements of D in the primary structure must be cancelled by the redundants, that is, the total displacement at D along each redundant is zero.

  B ----F---- C
  |   15 kN   |
 10->         | 4 m
 kN/m         |
  |           o D
  A (fixed)

Consistent deformation equations

\begin{aligned} \Delta_{10} + f_{11}X_1 + f_{12}X_2 &= 0 \\ \Delta_{20} + f_{21}X_1 + f_{22}X_2 &= 0 \end{aligned}

where $\Delta_{i0} = \int \dfrac{M_0\,m_i}{EI}dx$ (displacements of D in the primary structure) and $f_{ij} = \int\dfrac{m_im_j}{EI}dx$ .

\begin{aligned} f_{11} &= \frac{85.333}{EI} \\ f_{12} &= \frac{48.000}{EI} \\ f_{22} &= \frac{53.333}{EI} \\ \Delta_{10} &= \frac{-256.9}{EI} \\ \Delta_{20} &= \frac{-430.0}{EI} \end{aligned}

\begin{aligned} 85.333\,X_1 + 48.000\,X_2 &= 256.9 \\ 48.000\,X_1 + 53.333\,X_2 &= 430.0 \end{aligned}

X_1 = H_D = -3.09\ \text{kN},\qquad X_2 = V_D = 10.84\ \text{kN}

Final bending moments $M = M_0 + X_1m_1 + X_2m_2$ (kNm, + tension on the inside)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-75.00	0.00	4.00	-31.63
Q (top of UDL)	-30.00	3.00	4.00	4.10
B	-30.00	4.00	4.00	1.01
F (load)	0.00	4.00	2.00	9.33
C	0.00	4.00	0.00	-12.35
D	0.00	0.00	0.00	0.00

Shear force (kN)

Member	At lower end	At upper end
AQ (UDL zone)	+26.91	-3.09
QB	-3.09	-3.09
BF	+4.16	+4.16
FC	-10.84	-10.84
CD	+3.09	+3.09

(Column shears are horizontal forces; positive means towards the right on the left face.)

Normal thrust (kN, compression negative)

Member	Axial force
AB (left column)	-4.16
BC (beam)	-3.09
CD (right column)	-10.84

Reactions

D (hinge): $H_D = 3.09$ kN (towards the left), $V_D = 10.84$ kN ( $\uparrow$ )
A (fixed): $H_A = 26.91$ kN (towards the left), $V_A = 4.16$ kN ( $\uparrow$ ), $M_A = 31.63$ kNm (anticlockwise)

Check: $\Sigma H$ : $30 - 26.91 - 3.09 = 0$ (UDL resultant $= 10\times3 = 30$ kN); $\Sigma V$ : $4.16 + 10.84 = 15$ kN.

Answer: $H_D = 3.09$ kN, $V_D = 10.84$ kN; $H_A = 26.91$ kN, $V_A = 4.16$ kN, $M_A = 31.63$ kNm. The BMD, SFD and thrust values are tabulated above.

2071 Shrawan · 10 marks

Determine the reaction at B of the propped cantilever beam shown in figure below using Castigliano's theorem. Also draw the bending moment diagram. [Figure: beam AB, A fixed, B roller, 50 kN point load at 4 m from A, span 10 m (4 m + 6 m).]

Answer

Propped cantilever AB: A fixed, B a roller, span 10 m, 50 kN at 4 m from A (6 m from B). EI is constant. Take $R_B$ as the redundant and measure $x$ from B.

              50 kN
                v
 B o-----------+---------|| A (fixed)
 |<--- 6 m --->|<- 4 m ->|

Bending moments

\begin{aligned} 0 \le x \le 6:\quad & M = R_Bx, & \frac{\partial M}{\partial R_B} &= x \\ 6 \le x \le 10:\quad & M = R_Bx - 50(x-6), & \frac{\partial M}{\partial R_B} &= x \end{aligned}

Castigliano's theorem ( $\delta_B = 0$ )

\int_0^{10}R_Bx^2\,dx - 50\int_6^{10}(x-6)\,x\,dx = 0

\frac{1000}{3}R_B - 50\left[\frac{x^3}{3} - 3x^2\right]_6^{10} = 0

333.33\,R_B - 50\,(33.33 + 36) = 0 \ \Rightarrow\ R_B = \frac{3466.7}{333.33} = 10.40\ \text{kN}\ (\uparrow)

Other reactions

R_A = 50 - 10.40 = 39.60\ \text{kN}\ (\uparrow)

M_A = R_B(10) - 50(4) = 10.40(10) - 200 = -96.00\ \text{kNm}

The fixed-end moment is 96.00 kNm (hogging).

Bending moment diagram (kNm)

Section	BM
B	0
Under the load (6 m from B)	62.40 (sagging)
A (fixed end)	-96.00 (hogging)

BM is linear in each portion: from 0 at B to $+62.40$ under the load, then falling to $-96.00$ at A. It changes sign at $x = \dfrac{300}{39.6} = 7.58$ m from B (2.42 m from A).

        +62.4
        /\
  B ___/  \___
              \_______ x = 7.58 m (zero)
                       \
                        -96 at A

Answer: $R_B = 10.40$ kN (upward); $R_A = 39.60$ kN; $M_A = 96.00$ kNm (hogging).

2071 Shrawan · 3+7 marks

Explain why the flexibility method is called a Force Method. Using the force method determine the reactions in the continuous beam shown in figure below, if support B settles 18 mm and support C settles 12 mm. Given EI is constant, $E = 232\ \text{kN/mm}^2$ and $I = 112.5\times10^{6}\ \text{mm}^4$ . [Figure: beam ABC, all supports; AB = BC = 4.8 m.]

Answer

Why the flexibility method is a force method

In the flexibility method the unknowns are redundant forces (reactions or internal forces). The redundants are applied as loads on a statically determinate primary structure, and their values are found from the compatibility equations $\sum f_{ij}X_j + \Delta_{i0} = \delta_i$ , in which the flexibility coefficients $f_{ij}$ convert forces into displacements. Since the primary unknowns are forces, the method is called the force method (the displacements are found afterwards).

Continuous beam with settlements

Reading of the figure: beam ABC on three supports, A a hinge and B, C rollers, $AB = BC = 4.8$ m. $E = 232$ kN/mm $^2 = 232\times10^6$ kN/m $^2$ and $I = 112.5\times10^6$ mm $^4 = 1.125\times10^{-4}$ m $^4$ , so

EI = 232\times10^6\times1.125\times10^{-4} = 26100\ \text{kNm}^2

There is no load, so the reactions arise only from the settlements: B settles 18 mm and C settles 12 mm.

$D_s = 1$ . Remove the support B and take $R_B$ (upward) as the redundant. The primary structure is a simply supported beam AC of span $9.6$ m.

Displacement of B in the primary structure

C settles by 12 mm, so the straight primary beam moves as a rigid body, and the point B moves down by half of this:

\Delta_{10} = 0.5\times12 = 6\ \text{mm downward} = -0.006\ \text{m (in the direction of } R_B)

Flexibility coefficient

Deflection at mid-span of a simply supported beam due to a unit load there:

f_{11} = \frac{L^3}{48EI} = \frac{9.6^3}{48\times26100} = 7.062\times10^{-4}\ \text{m/kN}

Compatibility

In the actual beam B is displaced downward by 18 mm, that is, $-0.018$ m in the direction of $R_B$ :

\Delta_{10} + f_{11}R_B = -0.018 \ \Rightarrow\ R_B = \frac{-0.018 + 0.006}{7.062\times10^{-4}} = -16.99\ \text{kN}

The negative sign means that the reaction at B acts downward (the support pulls the beam down).

Other reactions

R_A = R_C = -\frac{R_B}{2} = 8.50\ \text{kN}\ (\uparrow)

Check: $R_A + R_B + R_C = 0$ .

Bending moment

$M_B = R_A\times4.8 = 40.78$ kNm (sagging, tension at the bottom); $M_A = M_C = 0$ . The SF is $+8.50$ kN in AB and $-8.50$ kN in BC.

Answer: $R_A = R_C = 8.50$ kN (upward) and $R_B = -16.99$ kN (i.e. 16.99 kN downward).

2071 Shrawan · 15 marks

Explain the physical meaning of the compatibility condition and derive the equation for it. A portal frame with hinged supports is subjected to a temperature variation as shown in figure below. Determine flexibility coefficients and calculate the redundant force with the help of the compatibility equation. Take $\alpha = 11\times10^{-6}/°C$ , $E = 5000\sqrt{f_{ck}}$ , $f_{ck} = 20$ MPa and constant flexural rigidity. [Figure: portal frame ABCD with hinged supports at A and D, height 6 m, span 3 m; temperature $t_1 = 20°C$ on the outer face and $t_2 = 10°C$ on the inner face of the beam; member cross-section 0.3 m x 0.6 m.]

Answer

Physical meaning of the compatibility condition

A statically indeterminate structure has more supports or members than are needed for equilibrium. When the redundants are removed, the primary structure deforms freely, and its displacements at the released points generally contradict the real conditions of the structure (for example, a fixed or hinged support cannot move). The compatibility condition states that the redundants must be of such a size that the displacements of the real structure at the releases agree with the real support conditions (usually zero, or equal to the settlement). It ensures that the deformed shape is continuous and fits the supports.

Derivation (one redundant $X_1$ )

Let the primary structure carry the loads and the unknown $X_1$ . By superposition, the displacement at the release point in the direction of $X_1$ is

\Delta_1 = \Delta_{10} + f_{11}X_1

where $\Delta_{10}$ = displacement of the primary structure due to the loads (and temperature, settlement of other supports, lack of fit) and $f_{11}$ = displacement due to a unit $X_1$ . Compatibility requires $\Delta_1$ to equal the actual displacement $\delta_1$ of the support (zero for a rigid support):

\Delta_{10} + f_{11}X_1 = \delta_1 \quad\Rightarrow\quad X_1 = \frac{\delta_1 - \Delta_{10}}{f_{11}}

For $n$ redundants: $\{\Delta_0\} + [F]\{X\} = \{\delta\}$ , with $f_{ij} = \int \dfrac{m_im_j}{EI}dx$ and $\Delta_{i0} = \int \dfrac{M_0m_i}{EI}dx$ . For a temperature change, the load term is

\Delta_{it} = \int m_i\,\frac{\alpha\,(t_1 - t_2)}{d}\,dx + \int n_i\,\alpha\,t_m\,dx,\qquad t_m = \frac{t_1+t_2}{2}

where $d$ is the depth, $m_i$ the unit BM and $n_i$ the unit axial force.

Portal frame with temperature variation

Reading of the figure: hinged portal ABCD, height $h = 6$ m, span $L = 3$ m, EI constant. The beam is warmer on the outer (top) face by $t_1 = 20^\circ$ C against $t_2 = 10^\circ$ C on the inner face. Section $0.3\times0.6$ m ( $d = 0.6$ m). Temperatures are changes from the original state.

$D_s = 4 - 3 = 1$ . Release D horizontally and take the thrust $H$ at D (positive outward) as the redundant.

   B ____________ C   t1 = 20 (outer)
   |              |   t2 = 10 (inner)
   |              |
   | 6 m          | 6 m
   |              |
  A o            o D <- H
   |<--- 3 m --->|

Elastic constants

E = 5000\sqrt{f_{ck}} = 5000\sqrt{20} = 22360.7\ \text{MPa} = 22360680\ \text{kN/m}^2

I = \frac{0.3\times0.6^3}{12} = 0.0054\ \text{m}^4,\qquad EI = 120748\ \text{kNm}^2

Flexibility coefficient

A unit outward force at D gives $m = y$ in each column (from the hinge) and $m = 6$ in the beam:

f_{11} = \frac{1}{EI}\left[2\cdot\frac{6^3}{3} + 6^2(3)\right] = \frac{144 + 108}{EI} = \frac{252}{EI} = 2.087\times10^{-3}\ \text{m/kN}

Displacement of D due to temperature (primary structure)

Mean temperature of the beam $t_m = (20+10)/2 = 15^\circ$ C; temperature difference $t_1 - t_2 = 10^\circ$ C.

Axial expansion of the beam (moves D outward): $\alpha\,t_m\,L = 11\times10^{-6}\times15\times3 = 4.950\times10^{-4}$ m
Curvature $\kappa = \dfrac{\alpha(t_1-t_2)}{d} = \dfrac{11\times10^{-6}\times10}{0.6} = 1.833\times10^{-4}$ per m. The beam bends concave downward (hot face on top), which moves D inward: $-\,\kappa\,(h)(L) = -1.833\times10^{-4}\times6\times3 = -3.300\times10^{-3}$ m

\Delta_{1t} = 4.950\times10^{-4} + (-3.300\times10^{-3}) = -2.805\times10^{-3}\ \text{m}\ (\text{inward})

Compatibility ( $\delta_1 = 0$ )

\Delta_{1t} + f_{11}H = 0 \ \Rightarrow\ H = \frac{2.805\times10^{-3}}{2.087\times10^{-3}} = 1.344\ \text{kN}

The support reaction at D is 1.344 kN acting outward (away from the frame) and an equal and opposite reaction acts at A. The vertical reactions are zero.

Bending moments

$M_B = M_C = H\,h = 1.344\times6 = 8.06$ kNm, with tension on the inside (sagging in the beam). The columns have a linear BM from 0 at the hinge to 8.06 kNm at the joint.

Answer: flexibility coefficient $f_{11} = 2.087\times10^{-3}$ m/kN, thermal displacement $\Delta_{1t} = -2.805\times10^{-3}$ m, redundant $H = 1.344$ kN (outward at each hinge), corner moment $8.06$ kNm.

2070 Chaitra (old course) · 8 marks

Compute the reactions and draw shear force and bending moment diagram for the frame shown in figure below. Use the consistent deformation method. [Figure: frame ABC; horizontal member AB (2I) of 6 m (3 m + 3 m) with 12 kN at mid-span, A hinged; vertical member BC (I) of 4 m with C hinged.]

Answer

Reading of the figure: the horizontal member AB is 6 m long ( $2I$ ) with a 12 kN load at mid-span D (3 m from A), A hinged; the vertical member BC is 4 m long ( $I$ ) with C hinged at its foot. $E$ is constant.

$r = 4$ , so $D_s = 1$ . Take the horizontal reaction $H_A$ at A as the redundant. Release A horizontally (roller): the primary structure is a simply supported frame.

        12 kN
          v
  A o-----D-----B
  H_A->         |
                | 4 m
                o C
  |<-- 3 -->|<-- 3 -->|

Primary structure (A on a roller)

$V_A = V_C = 6$ kN. A unit horizontal force at A is balanced by a horizontal reaction at C and a vertical couple: $V = 4/6 = 0.667$ kN.

Section	$M_0$ (kNm)	$m$ (unit $H_A$ )
A	0	0
D	18	-2.0
B	0	-4.0
C	0	0

(With $m$ negative at D and B because a unit force at A produces hogging in AB: $m_D = -0.667\times3 = -2$ , $m_B = -4$ .)

Deformations

\begin{aligned} f_{11} &= \frac{1}{2EI}\left[\frac{3}{3}(4) + \frac{3}{3}(4 + 8 + 16)\right] + \frac{1}{EI}\left[\frac{4}{3}(16)\right] = \frac{16 + 21.333}{EI} = \frac{37.33}{EI} \\ \Delta_{10} &= \frac{1}{2EI}\left[\frac{3}{3}(18)(-2) \cdot 1 + \frac{3}{6}\left\{2(18)(-2) + 18(-4)\right\}\right] = \frac{-54.00}{EI} \end{aligned}

Consistent deformation (A does not move horizontally)

\Delta_{10} + f_{11}H_A = 0 \ \Rightarrow\ H_A = \frac{54.00}{37.33} = 1.446\ \text{kN}\ (\rightarrow)

Reactions

A: $H_A = 1.446$ kN ( $\rightarrow$ ), $V_A = 5.036$ kN ( $\uparrow$ )
C: $H_C = 1.446$ kN ( $\leftarrow$ ), $V_C = 6.964$ kN ( $\uparrow$ )

Bending moment $M = M_0 + H_A\,m$ (kNm)

Point	BM	Tension side
A	0
D	15.11	bottom
B	-5.79	top of beam / outside of corner
C	0

Shear force and axial force (kN)

Member	SF	Axial force
AD	+5.04	-1.446 (compression)
DB	-6.96	-1.446 (compression)
BC	+1.45	-6.964 (compression)

Answer: $H_A = H_C = 1.446$ kN; $V_A = 5.036$ kN and $V_C = 6.964$ kN; $M_D = 15.11$ kNm (sagging), $M_B = -5.79$ kNm (hogging).

2070 Chaitra (old course) · 2+2 marks

Explain compatibility conditions. Also describe Maxwell's reciprocal theorem.

Answer

Compatibility conditions

Compatibility conditions are the geometric requirements that the deformed structure must satisfy: displacements must be continuous inside members and at the joints, and must agree with the support conditions (for example zero deflection at a rigid support, zero rotation at a fixed end, equal rotation of connected members at a rigid joint). They are independent of the material and loading.

In an indeterminate structure, equilibrium alone cannot give all the forces. The extra equations come from compatibility. In the force method, the redundants $X_j$ are found from

\Delta_{i0} + \sum_j f_{ij}X_j = \delta_i

Example: for a propped cantilever with prop reaction $R_B$ , the compatibility condition is that the deflection at the prop is zero, $\Delta_{B0} - f_{BB}R_B = 0$ , so $R_B = \dfrac{wL^4/8EI}{L^3/3EI} = \dfrac{3wL}{8}$ .

Maxwell's reciprocal theorem

In a linearly elastic structure, the displacement at A due to a unit load at B equals the displacement at B due to a unit load at A:

\delta_{AB} = \delta_{BA}, \qquad f_{ij} = f_{ji}

It follows from Betti's law (or from the independence of strain energy from the order of loading). It makes the flexibility and stiffness matrices symmetric, which reduces the number of coefficients to be found in the force method and gives a check on the computation. Example: for a cantilever of length $L$ , the deflection at mid-span due to a unit tip load is $\dfrac{5L^3}{48EI}$ , which equals the tip deflection due to a unit load at mid-span.

2070 Chaitra (old course) · 16 marks

In the two hinged parabolic arch shown below, find the values of bending moment, normal thrust and radial shear at section D due to the given loading and due to yielding of support B by 10 mm. Take $EI_c = 100\times10^{6}\ \text{kNm}^2$ , $I = I_c \sec\theta$ . Also draw bending moment diagram. [Figure: two-hinged parabolic arch AB, span 120 m (60 m + 60 m), rise 20 m at crown C; section D at 30 m from A; 10 kN/m UDL on the right half CB.]

Answer

A two-hinged arch has one redundant, the horizontal thrust $H$ . With $I = I_c\sec\theta$ the compatibility equation (support B yields outward by $\delta$ ) is

H = \frac{\displaystyle\frac{1}{EI_c}\int M_0\,y\,dx - \delta}{\displaystyle\frac{1}{EI_c}\int y^2\,dx} = H_{load} - \frac{\delta\,EI_c}{\int y^2\,dx}

Data

Span $L = 120$ m, rise $h = 20$ m, $EI_c = 100\times10^6$ kNm $^2$ , $\delta = 10$ mm $= 0.01$ m; UDL $10$ kN/m on CB (60 m to 120 m).

y = \frac{4h\,x(L-x)}{L^2} = \frac{x(120-x)}{180},\qquad \tan\theta = \frac{dy}{dx} = \frac{120 - 2x}{180}

Simple-beam reactions and $M_0$

UDL resultant $= 10\times60 = 600$ kN at 90 m: $V_B = \dfrac{600(90)}{120} = 450$ kN, $V_A = 150$ kN.

$0 \le x \le 60$ : $M_0 = 150\,x$
$60 \le x \le 120$ : $M_0 = 150\,x - 5(x-60)^2$

Thrust

\int_0^{120} y^2\,dx = \frac{8h^2L}{15} = 25600,\qquad \int_0^{120} M_0\,y\,dx = 11520000

H_{load} = \frac{11520000}{25600} = 450.00\ \text{kN}

H_{yield} = -\frac{\delta\,EI_c}{\int y^2dx} = -\frac{0.01\times100\times10^6}{25600} = -39.06\ \text{kN}

H = 450.00 + (-39.06) = 410.94\ \text{kN}

Section D ( $x = 30$ m)

\begin{aligned} y_D &= \frac{30(90)}{180} = 15\ \text{m},\qquad \tan\theta_D = \frac{120-60}{180} = \frac13,\ \ \theta_D = 18.43^\circ \\ M_0 &= V_A(30) = 4500\ \text{kNm},\qquad V_0 = V_A = 150\ \text{kN} \end{aligned}

Bending moment

M_D = M_0 - Hy_D = 4500 - 410.94(15) = -1664.06\ \text{kNm}

(Due to the load alone: $-2250.00$ kNm; the yield adds $+585.94$ kNm.)

Normal thrust

N_D = H\cos\theta + V_0\sin\theta = 410.94(0.9487) + 150(0.3162) = 437.28\ \text{kN}

Radial shear

Q_D = V_0\cos\theta - H\sin\theta = 150(0.9487) - 410.94(0.3162) = 12.35\ \text{kN}

Bending moment diagram $M = M_0 - Hy$ (kNm)

$x$ (m)	$y$ (m)	$M_0$	$M$
0	0.00	0.0	0.0
15	8.75	2250.0	-1345.7
30	15.00	4500.0	-1664.1
45	18.75	6750.0	-955.1
60	20.00	9000.0	781.2
75	18.75	10125.0	2419.9
90	15.00	9000.0	2835.9
105	8.75	5625.0	2029.3
120	0.00	0.0	0.0

The BMD is hogging for $0 < x < 54.3$ m (largest hogging about $-1683$ kNm near $x = 27$ m) and sagging for the rest, with the largest sagging moment about $2852$ kNm near $x = 88$ m, under the loaded half. It is zero at both hinges.

Answer: at D, $M = -1664.06$ kNm (hogging), $N = 437.28$ kN (compression), radial shear $Q = 12.35$ kN; $H = 410.94$ kN.

2070 Chaitra (old course) · 2+2 marks

Define the terms flexibility and stiffness.

Answer

Flexibility

Flexibility is the displacement produced by a unit force. The flexibility coefficient $f_{ij}$ is the displacement at coordinate $i$ due to a unit force at coordinate $j$ (all other coordinates free of load). It is a measure of how easily a structure deforms. The flexibility matrix $[F]$ gives $\{\Delta\} = [F]\{P\}$ . Its unit is m/kN (or rad/kNm). Example: for a cantilever of length $L$ with a tip load, $f_{11} = \dfrac{L^3}{3EI}$ .

Stiffness

Stiffness is the force required to produce a unit displacement. The stiffness coefficient $k_{ij}$ is the force needed at coordinate $i$ to produce a unit displacement at coordinate $j$ while all other coordinates are held fixed. The stiffness matrix $[K]$ gives $\{P\} = [K]\{\Delta\}$ . Its unit is kN/m (or kNm/rad). Example: a fixed-end beam of length $L$ requires a force $\dfrac{12EI}{L^3}$ to give a unit deflection at one end (the other end fixed against rotation). The two matrices are inverses of each other, $[K] = [F]^{-1}$ .

2070 Chaitra · 10 marks

Use Castigliano's theorem to determine forces induced in each member of the square truss loaded as shown below. [Figure: square truss ABCD, 4 m x 4 m, with both diagonals, AE constant; 120 kN downward at D and 90 kN horizontal (toward the left) at D; A hinged, B roller.]

Answer

Reading of the figure: A (0, 0) hinged, B (4, 0) roller, C (4, 4) and D (0, 4); members AB, BC, CD, DA (4 m) and diagonals AC, BD (5.657 m); 120 kN downward and 90 kN horizontal (towards the left) at D. $AE$ is constant. $D_s = 6 + 3 - 8 = 1$ . Take the force in AC as the redundant $X$ .

Reactions

$\Sigma M_A = 0$ : $V_B(4) + 90(4) = 0$ , hence $V_B = -90.00$ kN (downward 90 kN). $V_A = 120 + 90 = 210.00$ kN ( $\uparrow$ ), $H_A = 90$ kN ( $\rightarrow$ ).

Member forces in terms of $X$ ( $S = S_0 + nX$ )

Member	$L$ (m)	$S_0$	$n_1$	Final $S$
AB	4.000	-90.000	-0.707	-32.574
BC	4.000	0.000	-0.707	+57.426
CD	4.000	0.000	-0.707	+57.426
DA	4.000	-210.000	-0.707	-152.574
AC	5.657	0.000	+1.000	-81.213
BD	5.657	+127.279	+1.000	+46.066

Castigliano's theorem

The strain energy of the truss is $U = \sum \dfrac{S^2L}{2AE}$ . The relative displacement of the cut ends of AC must be zero, so

\frac{\partial U}{\partial X} = \sum \frac{S\,n\,L}{AE} = \frac{1}{AE}\left[\sum S_0nL + X\sum n^2L\right] = 0

\sum S_0nL = 1568.528,\qquad \sum n^2L = 19.314

X = S_{AC} = -\frac{1568.528}{19.314} = -81.21\ \text{kN}

Forces in all members

Member	Force (kN)	Nature
AB	-32.57	compression
BC	+57.43	tension
CD	+57.43	tension
DA	-152.57	compression
AC	-81.21	compression
BD	+46.07	tension

Answer: AB $= -32.57$ , BC $= +57.43$ , CD $= +57.43$ , DA $= -152.57$ , AC $= -81.21$ , BD $= +46.07$ kN (positive = tension).

2070 Chaitra · 15 marks

Draw shear force and bending moment diagrams for the frame given below. Use the force method. [Figure: frame; beam B-C-D with 5 kN/m UDL on BC (5 m) and a 2 m overhang CD with 20 kN at D; 20 kN horizontal at B; columns EI: AB (A fixed) of height 4 m, middle column CE hinged at E, right side with a 2 m dimension marked.]

Answer

Reading of the figure: left column AB (4 m, A fixed); beam B-C-D with BC = 5 m carrying 5 kN/m and a 2 m overhang CD with 20 kN at D; a middle column CE (4 m) hinged at E, below C; 20 kN horizontal at B (towards the right). $EI$ is constant for all members.

$r = 3 + 2 = 5$ , so $D_s = 2$ . Release the hinge at E and take $X_1$ = horizontal reaction ( $\rightarrow$ ), $X_2$ = vertical reaction ( $\uparrow$ ) at E as redundants. The primary structure is fixed at A.

 20 kN->  B ============ C ====== D   5 kN/m on BC
          |             |        v 20 kN
          | 4 m         | 4 m
          |             o E (hinge)
          A (fixed)
          |<--- 5 m --->|<- 2 ->|

Unit and load moments (+ tension on the inner/lower face)

$M_0$ is the moment in the primary structure under the real loads (20 kN at B, 5 kN/m on BC, 20 kN at D), with the column CE unstressed; $m_1$ and $m_2$ are the moments due to unit forces at E. Their values at the key sections are given in the table below.

\begin{aligned} f_{11} &= \frac{122.667}{EI} \\ f_{12} &= \frac{90.000}{EI} \\ f_{22} &= \frac{141.667}{EI} \\ \Delta_{10} &= \frac{-4050.0}{EI} \\ \Delta_{20} &= \frac{-6574.0}{EI} \end{aligned}

\begin{aligned} 122.667\,X_1 + 90.000\,X_2 &= 4050.0 \\ 90.000\,X_1 + 141.667\,X_2 &= 6574.0 \end{aligned}

X_1 = H_E = -1.93\ \text{kN},\qquad X_2 = V_E = 47.63\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-282.50	0.00	5.00	-44.35
B (column top)	-202.50	4.00	5.00	27.93
mid BC	-105.63	4.00	2.50	5.73
C (beam, left)	-40.00	4.00	0.00	-47.72
C (overhang)	-40.00	0.00	0.00	-40.00
D	0.00	0.00	0.00	0.00
C (column)	0.00	4.00	0.00	-7.72
E	0.00	0.00	0.00	0.00

At the joint C the member moments balance: -47.72 = -40.00 + (-7.72).

Shear force (kN)

Member	SF
AB	+18.07
BC (at B / at C)	-2.63 / -27.63
CD (overhang)	+20.00
CE	+1.93

Reactions

E (hinge): $H_E = 1.93$ kN (towards the left), $V_E = 47.63$ kN ( $\uparrow$ )
A (fixed): $H_A = 18.07$ kN (towards the left), $V_A = 2.63$ kN (downward), $M_A = 44.35$ kNm

Check: $\Sigma V$ : $-2.63 + 47.63 = 5\times5 + 20 = 45$ kN; $\Sigma H$ : $20 - 18.07 - 1.93 = 0$ .

In BC the moment falls from $+27.93$ kNm at B through zero at 2.86 m from B to -47.72 kNm at C.

Answer: BMD ordinates: $M_A = 44.35$ kNm, $M_B = 27.93$ kNm, $M_C = -47.72$ kNm (beam side), $M_D = 0$ and the BM in BC changes sign at 2.86 m from B; the SFD values are in the table above.

2070 Chaitra · 10 marks

Determine the horizontal reaction in the two hinged parabolic arch shown in figure below. Also determine the bending moment at C. ( $I = I_c \sec\theta$ ). [Figure: two-hinged parabolic arch AB, span 60 m (30 m + 30 m), rise 5 m at crown C; UDL 50 kN/m over the right half CB.]

Answer

A two-hinged arch has one redundant, the horizontal reaction (thrust) $H$ . With $I = I_c\sec\theta$ , $ds/I = dx/I_c$ , so

H = \frac{\int M_0\,y\,dx}{\int y^2\,dx}

Data and profile

Span $L = 60$ m, rise $h = 5$ m, UDL 50 kN/m on the right half CB ( $30 \le x \le 60$ , with A at the origin).

y = \frac{4h\,x(L-x)}{L^2} = \frac{x(60-x)}{180}

Reactions of the simple beam

UDL resultant $= 50\times30 = 1500$ kN at $x = 45$ m.

V_B = \frac{1500(45)}{60} = 1125\ \text{kN},\qquad V_A = 1500 - 1125 = 375\ \text{kN}

Free bending moment

$0 \le x \le 30$ : $M_0 = 375\,x$
$30 \le x \le 60$ : $M_0 = 375\,x - 25(x-30)^2$

Integrals

\begin{aligned} \int_0^{60}y^2dx &= \frac{8h^2L}{15} = \frac{8(25)(60)}{15} = 800 \\ \int_0^{30}M_0y\,dx &= 703125, \qquad \int_{30}^{60}M_0y\,dx = 1096875 \\ \int_0^{60}M_0y\,dx &= 1800000 \end{aligned}

Horizontal reaction

H = \frac{1800000}{800} = 2250.0\ \text{kN}

Bending moment at the crown C ( $x = 30$ m)

\begin{aligned} M_0 &= V_A(30) = 375(30) = 11250\ \text{kNm} \\ y_C &= h = 5\ \text{m} \\ M_C &= M_0 - H\,y_C = 11250 - 2250.0(5) = 0.0\ \text{kNm} \end{aligned}

The crown moment is zero for this loading, which is a known property of a parabolic two-hinged arch carrying a UDL over half the span.

Answer: $H = 2250.0$ kN; $M_C = 0.0$ kNm.

2070 Asar · 4 marks

Define force method and primary structure.

Answer

Force method

The force method (flexibility or compatibility method) is a method of analysing statically indeterminate structures in which the redundant forces are taken as the unknowns. The redundants are removed to get a determinate structure, the loads and the unit redundants are applied to it, and the redundants are found from compatibility equations that restore the real displacement conditions at the releases:

\Delta_{i0} + \sum_j f_{ij}X_j = \delta_i

Here $\Delta_{i0}$ is the displacement due to the loads, $f_{ij}$ the flexibility coefficient and $\delta_i$ the actual displacement of the support (usually zero). After finding the $X_j$ , the final forces follow by superposition, $M = M_0 + \sum X_jm_j$ . The number of unknowns equals the degree of static indeterminacy $D_s$ .

Primary structure

The primary structure is the statically determinate, stable structure obtained by removing the redundants (supports, members or moment restraints) from the indeterminate structure. It is not unique, but it must stay stable.

Example: a propped cantilever (fixed at A, roller at B) has $D_s = 1$ . If the roller reaction $R_B$ is removed, the primary structure is a cantilever; if the fixed-end moment $M_A$ is released, it is a simply supported beam.

2070 Asar · 6 marks

Generate the flexibility matrix for the coordinates shown in figure below. [Figure: beam of two spans of 10 m each, EI constant, left end fixed with coordinate 1 (rotation at the fixed end), interior support with coordinate 2 (rotation), right end roller.]

Answer

Reading of the figure: a beam of two spans AB and BC, each $L = 10$ m, with constant $EI$ ; A is a fixed end, B an interior support and C a roller. Coordinate 1 is the rotation at the fixed end A, and coordinate 2 is the rotation at the interior support B. To define flexibility coefficients, the restraints against these rotations are released: A becomes a hinge and the beam is cut into two simply supported spans by a hinge over B, so the primary structure is made of the two simply supported spans AB and BC. A unit moment at coordinate 1 acts at A and a unit moment at coordinate 2 acts at B (on both spans).

 (1) A o------------o B (2)------------o C
        |<-- 10 -->|     |<-- 10 -->|

Flexibility coefficients

The flexibility coefficient $f_{ij}$ is the rotation at coordinate $i$ due to a unit moment at coordinate $j$ .

Unit moment at coordinate 1 (A): it acts on span AB only. For a simply supported span of length $L$ with a unit moment at one end, the rotation at that end is $\dfrac{L}{3EI}$ and the rotation at the far end is $\dfrac{L}{6EI}$ :

f_{11} = \frac{L}{3EI} = \frac{10}{3EI} = \frac{3.333}{EI},\qquad f_{21} = \frac{L}{6EI} = \frac{10}{6EI} = \frac{1.667}{EI}

Unit moment at coordinate 2 (B): it acts on both spans, so each span turns by $\dfrac{L}{3EI}$ at B, and span AB rotates $\dfrac{L}{6EI}$ at A:

f_{22} = \frac{L}{3EI} + \frac{L}{3EI} = \frac{20}{3EI} = \frac{6.667}{EI},\qquad f_{12} = \frac{L}{6EI} = \frac{1.667}{EI}

( $f_{12} = f_{21}$ , as required by Maxwell's theorem.)

Flexibility matrix

[F] = \frac{1}{EI}\begin{bmatrix} 3.333 & 1.667 \\ 1.667 & 6.667 \end{bmatrix} = \frac{L}{6EI}\begin{bmatrix} 2 & 1 \\ 1 & 4 \end{bmatrix}

The matrix is symmetric with positive diagonal terms. Its inverse is the stiffness matrix for the same coordinates:

[K] = [F]^{-1} = \frac{6EI}{7L}\begin{bmatrix} 4 & -1 \\ -1 & 2 \end{bmatrix}

2070 Asar · 10 marks

Determine horizontal and vertical reactions at support D of the frame shown in figure below using the force method. [Figure: portal frame ABCD; beam BC (2EI) of 5 m with 30 kN/m UDL; columns AB and CD (EI) 4 m high; 110 kN horizontal at B; A fixed, D hinged.]

Answer

Reading of the figure: portal frame ABCD, columns AB and CD are 4 m high ( $EI$ ), the beam BC is 5 m long ( $2EI$ ) with 30 kN/m, a 110 kN horizontal load acts at B (towards C), A is fixed and D is hinged.

$D_s = 3 + 2 - 3 = 2$ . Release the hinge at D and take $X_1$ = horizontal reaction ( $\rightarrow$ ) and $X_2$ = vertical reaction ( $\uparrow$ ). The primary structure is the cantilever fixed at A.

 110 kN->  B ============ C   30 kN/m
           |              |
           | 4 m          | 4 m
           |              |
          ///A            o D (hinge)

Moment diagrams in the primary structure (+ tension on the inner face)

$M_0$ : in the beam, $-375$ kNm at B (UDL resultant $150\times2.5$ ) falling parabolically to 0 at C; in the column AB it varies linearly from $-375$ at B to $-(375 + 110\times4) = -815$ kNm at A.
$m_1$ (unit horizontal force at D): CD from 0 to 4, beam constant 4, column AB from 4 at B down to 0 at A.
$m_2$ (unit vertical force at D): beam from 5 at B to 0 at C, column AB constant 5.

Flexibility coefficients (in $1/EI$ units)

\begin{aligned} f_{11} &= \frac{82.667}{EI} \\ f_{12} &= \frac{65.000}{EI} \\ f_{22} &= \frac{120.833}{EI} \\ \Delta_{10} &= \frac{-5423.3}{EI} \\ \Delta_{20} &= \frac{-13071.9}{EI} \end{aligned}

Compatibility

\begin{aligned} 82.667\,X_1 + 65.000\,X_2 &= 5423.3 \\ 65.000\,X_1 + 120.833\,X_2 &= 13071.9 \end{aligned}

X_1 = H_D = -33.72\ \text{kN},\qquad X_2 = V_D = 126.32\ \text{kN}

Final bending moments (kNm)

Section	$M_0$	$m_1$	$m_2$	$M=M_0+\sum X m$
A	-815.00	0.00	5.00	-183.40
B (column)	-375.00	4.00	5.00	121.72
mid BC	-93.75	4.00	2.50	87.17
C (beam)	0.00	4.00	0.00	-134.88
D	0.00	0.00	0.00	0.00

Maximum sagging moment in BC: 131.07 kNm at 0.79 m from B.

Reactions

D (hinge): $H_D = 33.72$ kN (towards the left), $V_D = 126.32$ kN ( $\uparrow$ )
A (fixed): $H_A = 76.28$ kN (towards the left), $V_A = 23.68$ kN ( $\uparrow$ ), $M_A = 183.40$ kNm (anticlockwise)

Check: $\Sigma H$ : $110 - 76.28 - 33.72 = 0$ ; $\Sigma V$ : $23.68 + 126.32 = 30\times5 = 150$ kN.

Answer: at the hinged support D, the horizontal reaction is $33.72$ kN (towards the left) and the vertical reaction is $126.32$ kN (upward).

2070 Asar

Analyse the continuous beam shown in figure below by using the three moment theorem. Draw shear force and bending moment diagram. [Figure: beam ABCD, A fixed; 100 kN at 4 m from A; B at 8 m; BC = 10 m with 60 kN/m UDL; overhang CD = 2 m with 20 kN at D. Printed as the alternative (OR) to the force-method frame question above.]

Answer

Reading of the figure: A fixed; span AB = 8 m with 100 kN at 4 m from A; span BC = 10 m with 60 kN/m; overhang CD = 2 m with 20 kN at D. B and C are rollers and $EI$ is constant.

Moment at C (overhang)

$M_C = -20\times2 = -40$ kNm (hogging).

Load terms

Span AB: free BM peak $\dfrac{100\times4\times4}{8} = 200$ kNm, $a_1 = \tfrac12(8)(200) = 800$ , $\bar x_1 = 4$ m: $\dfrac{6a_1\bar x_1}{L_1} = \dfrac{6(800)(4)}{8} = 2400$
Span BC: $\dfrac{wL_2^3}{4} = \dfrac{60(10)^3}{4} = 15000$

Three moment equations

At A (fixed end): $2M_A(8) + M_B(8) = -2400$

2M_A + M_B = -300

At B: $M_A(8) + 2M_B(8+10) + M_C(10) = -2400 - 15000$

8M_A + 36M_B + 10(-40) = -17400 \ \Rightarrow\ 8M_A + 36M_B = -17000

Substituting $M_A = \dfrac{-300 - M_B}{2}$ : $-1200 - 4M_B + 36M_B = -17000$ , so $32M_B = -15800$ :

M_B = -493.750\ \text{kNm},\qquad M_A = 96.875\ \text{kNm}

( $M_A$ is positive: the fixed end is sagging for this loading.)

Reactions and shear

Span AB ( $M_B = M_A + 8R_A - 400$ ):

R_A = \frac{400 + M_B - M_A}{8} = \frac{400 - 493.75 - 96.875}{8} = -23.828\ \text{kN}

(negative: A is pulled down). Shear just left of B $= -23.828 - 100 = -123.828$ kN.

Span BC ( $M_C = M_B + 10V_{BR} - 60\frac{10^2}{2}$ ):

V_{BR} = \frac{3000 + M_C - M_B}{10} = \frac{3000 - 40 + 493.75}{10} = 345.375\ \text{kN}

R_B = 123.828 + 345.375 = 469.203\ \text{kN},\qquad R_C = 600 - 345.375 + 20 = 274.625\ \text{kN}

Check: $R_A + R_B + R_C = 720.00$ kN $= 100 + 600 + 20 = 720$ kN.

Shear force (kN)

Section	SF
A to load point E	-23.83
E to B	-123.83
B to C (at B / at C)	345.375 / -254.63
C to D (overhang)	+20.00

The SF is zero in BC at 5.76 m from B.

Bending moment (kNm)

Section	BM
A	96.87
E (under 100 kN)	1.56
B	-493.75
Maximum in BC (5.76 m from B)	500.28
C	-40.00
D	0

Answer: $M_A = 96.87$ kNm, $M_B = -493.75$ kNm, $M_C = -40$ kNm; $R_A = -23.828$ kN, $R_B = 469.203$ kN, $R_C = 274.625$ kN; maximum sagging moment in BC $= 500.28$ kNm.

2070 Asar · 5 marks

Determine the force in member BF of the redundant truss shown in figure below. Cross section areas of each member in cm² are given in the figure within brackets (values 15, 20, 25 and 30). [Figure: truss ABCDEF with 4 m height; bottom panels 3 m, 4 m and 3 m; loads 30 kN and 50 kN downward at bottom joints; A hinged, D roller; member areas as printed in the figure.]

Answer

The truss has one degree of redundancy ( $m + r - 2j = 10 + 3 - 12 = 1$ ). Take the force $X$ in the diagonal BF as the redundant and solve it by the force method (unit-load / strain energy).

Assumed data (the figure is not fully legible): joints A, B, C, D on the bottom chord at 0, 3, 7, 10 m; E, F on top, 4 m above B and C; loads 30 kN at B and 50 kN at C; A hinged, D roller; both diagonals BF and CE present in the middle panel. Areas: bottom chord 20 cm², top chord 25 cm², verticals 15 cm², all diagonals 30 cm². E is constant, so it cancels.

Step 1: Released (determinate) truss, loads only

Remove BF. Reactions: $R_D = (30\times3 + 50\times7)/10 = 44$ kN, $R_A = 36$ kN, $H_A = 0$ . Solving by joints gives the forces $S_0$ in the table.

Step 2: Unit load on the released truss

Apply a pair of unit tensile forces along BF and find the forces $S_1$ (BF itself has $S_1 = +1$ ).

Step 3: Compatibility

\sum S_0 S_1 \frac{L}{A} + X\sum S_1^2\frac{L}{A} = 0 \quad\Rightarrow\quad X = -\frac{\sum S_0S_1L/A}{\sum S_1^2L/A}

Member	L (m)	A (cm²)	$S_0$ (kN)	$S_1$	$S_0S_1L/A$	$S_1^2L/A$
AB	3.000	20	27.00	0.000	0.000	0.0000
BC	4.000	20	27.00	-0.707	-3.818	0.1000
CD	3.000	20	33.00	0.000	0.000	0.0000
EF	4.000	25	-33.00	-0.707	3.734	0.0800
AE	5.000	30	-45.00	0.000	0.000	0.0000
BE	4.000	15	30.00	-0.707	-5.657	0.1333
CF	4.000	15	44.00	-0.707	-8.297	0.1333
DF	5.000	30	-55.00	0.000	0.000	0.0000
CE	5.657	30	8.49	1.000	1.600	0.1886
BF	5.657	30	0.00	1.000	0.000	0.1886
Sum					-12.438	0.8238

X = -\frac{(-12.438)}{0.8238} = 15.10\ \text{kN}

Final member forces ( $S = S_0 + X S_1$ , tension +)

AB = 27.00, BC = 16.32, CD = 33.00, EF = -43.68, AE = -45.00, BE = 19.32, CF = 33.32, DF = -55.00, CE = 23.58, BF = 15.10 kN.

Answer: force in BF = 15.10 kN (tension) for the assumed areas. (Checked by a full direct-stiffness analysis of the 10-member truss.)

2070 Asar

Draw the bending moment (BM) diagram for the two hinged parabolic arch shown, $I = I_c \sec\theta$ . Calculate the BM value at an interval of 10 m. [Figure: two-hinged parabolic arch AB, span 60 m (20 m + 40 m), rise 8 m at crown C; 80 kN vertical load at D, 20 m from A. Printed as the alternative (OR) to the truss question above.]

Answer

A two-hinged arch is statically indeterminate to the first degree. The horizontal thrust $H$ is the redundant, found from the condition that the horizontal movement of the hinge B is zero:

H=\frac{\int M_0\,y\,\dfrac{ds}{EI}}{\int y^2\,\dfrac{ds}{EI}}

For $I=I_c\sec\theta$ we have $ds/I = dx/I_c$ ( $ds=dx\sec\theta$ ), so $H=\int M_0 y\,dx\,/\int y^2dx$ .

Data

Span $L=60$ m, rise $h=8$ m, load $P=80$ kN at $x=20$ m from A. Parabola:

y=\frac{4h\,x(L-x)}{L^2}=\frac{4\times8\,x(60-x)}{3600}

Reactions and simple-beam moment $M_0$

V_A=\frac{80\times40}{60}=53.33\ \text{kN},\qquad V_B=80-53.33=26.67\ \text{kN}

$M_0 = 53.33x$ for $x\le20$ and $M_0=26.67(60-x)$ for $x\ge20$ .

Horizontal thrust

\int_0^{60} M_0\,y\,dx=208592.6\ \text{kN m}^3,\qquad \int_0^{60} y^2dx=\frac{8h^2L}{15}=2048.0\ \text{m}^3

H=\frac{208592.6}{2048.0}=101.85\ \text{kN}

(Check with the standard result $H=\dfrac{5PL\,\alpha(1-2\alpha^2+\alpha^3)}{8h}$ with $\alpha=1/3$ : same value.)

Bending moment $M=M_0-Hy$ at 10 m intervals

x (m)	y (m)	$M_0$ (kNm)	$Hy$ (kNm)	$M$ (kNm)
0	0.00	0.00	0.00	0.00
10	4.44	533.33	452.67	80.66
20	7.11	1066.67	724.28	342.39
30	8.00	800.00	814.81	-14.81
40	7.11	533.33	724.28	-190.95
50	4.44	266.67	452.67	-186.01
60	0.00	0.00	0.00	0.00

Sign: positive = sagging (tension on the lower side).

The BM is zero at both hinges, reaches its maximum +342.4 kNm under the load, changes sign at about x = 29.5 m (where $M=(60-x)(26.667-0.9053x)=0$ ) and is hogging beyond that, with a numerical maximum of about -211 kNm near x = 44.7 m. The values at 10 m intervals plot the BM diagram: sagging on the left of x = 29.5 m and hogging on the right.

Answer: $H=101.85$ kN; BM at 0, 10, 20, 30, 40, 50, 60 m = 0.00, 80.66, 342.39, -14.81, -190.95, -186.01, 0.00 kNm.

2069 Asar · 4 marks

Using Castigliano's second theorem, determine the slope at A of the beam shown in figure below. EI is constant. [Figure: beam A-B-C, A hinged, B roller, AB = 5 m (2.5 m + 2.5 m) with 50 kN at mid-span; overhang BC = 2 m with 10 kN at C.]

Answer

Castigliano's second theorem: the rotation at a point equals the partial derivative of the strain energy with respect to a couple applied there:

\theta_A=\frac{\partial U}{\partial M_0}=\int \frac{M}{EI}\frac{\partial M}{\partial M_0}\,dx

Since no couple acts at A, apply a fictitious clockwise couple $M_0$ at A, write $M$ in terms of $M_0$ , differentiate, then put $M_0=0$ .

Reactions (with $M_0$ )

Taking moments about B (AB = 5 m, BC = 2 m):

5R_A + M_0 = 50(2.5) - 10(2) = 105 \;\Rightarrow\; R_A = 21-\frac{M_0}{5}\ \text{kN},\qquad R_B = 39+\frac{M_0}{5}\ \text{kN}

(Check: $R_A+R_B=60=50+10$ .)

Bending moments (x from A, sagging +)

Portion A to mid-span ( $0\le x\le2.5$ ): $M=R_Ax+M_0=\left(21-\frac{M_0}{5}\right)x+M_0$ , so $\dfrac{\partial M}{\partial M_0}=1-\dfrac{x}{5}$
Mid-span to B ( $2.5\le x\le5$ ): $M=R_Ax+M_0-50(x-2.5)$ , same derivative $1-\dfrac{x}{5}$
Overhang BC: $M=-10x'$ (x' from C), independent of $M_0$ , so it contributes nothing.

Slope at A (put $M_0=0$ )

EI\,\theta_A=\int_0^{2.5}21x\left(1-\frac{x}{5}\right)dx+\int_{2.5}^{5}\big[21x-50(x-2.5)\big]\left(1-\frac{x}{5}\right)dx

= 175/4 + 425/24 = 1475/24 = 61.46\ \text{kN m}^2

\theta_A=\frac{61.46}{EI}\ \text{rad (clockwise, same sense as }M_0)

Check: $50\times5^2/16=78.13$ (central load) minus $20\times5/6=16.67$ (hogging moment from the overhang) = 61.46. ✓

Answer: $\theta_A = 61.46/EI$ (kNm² units), clockwise.

2069 Asar · 2+3 marks

Define flexibility and stiffness. What are the properties of the flexibility matrix?

Answer

Flexibility

The flexibility $f_{ij}$ is the displacement at coordinate i caused by a unit force acting at coordinate j (all other coordinates carry no force). For a beam fixed at one end, the tip deflection due to a unit tip load is $f=L^3/3EI$ . Its unit is m/kN (displacement per unit force). The flexibility matrix $[F]$ collects all $f_{ij}$ , so that

\{\Delta\}=[F]\{P\}

Stiffness

The stiffness $k_{ij}$ is the force required at coordinate i to produce a unit displacement at coordinate j while all other coordinates are held fixed (zero displacement). For the same cantilever the tip stiffness is $k=3EI/L^3$ . Its unit is kN/m. The stiffness matrix $[K]$ gives

\{P\}=[K]\{\Delta\}

For a stable structure with the same coordinates, $[K]=[F]^{-1}$ .

Properties of the flexibility matrix

Square: its order equals the number of coordinates (redundants) chosen.
Symmetric: $f_{ij}=f_{ji}$ (Maxwell's reciprocal theorem).
Diagonal terms are positive: $f_{ii}>0$ , because a force always moves its own point in its own direction (work done is positive).
Positive definite: the strain energy $U=\tfrac12\{P\}^T[F]\{P\}$ is always positive, so the determinant is positive.
Inverse is the stiffness matrix: $[F]=[K]^{-1}$ .
Depends on the chosen released structure and coordinates, but the final answer does not.
Depends only on geometry and material (L, E, I, A), not on the loads. Off-diagonal terms may be positive, negative or zero.

2069 Asar · 10 marks

Use the force method (flexibility matrix) to find the reactions at supports B and C of the beam shown in figure below and also draw shear force and bending moment diagrams. [Figure: beam ABC, A fixed; 200 kN at 5 m from A; B support at 10 m; 100 kN at 5 m beyond B; C at 20 m total; spans 5 m, 5 m, 5 m, 5 m.]

Answer

Assumptions (figure values read as given): A is fixed at $x=0$ ; 200 kN acts at $x=5$ m; B is a roller support at $x=10$ m; 100 kN acts at $x=15$ m; C is a roller support at $x=20$ m. EI is constant. The beam is indeterminate to the second degree. Choose the reactions $R_B$ (coordinate 1) and $R_C$ (coordinate 2) as redundants, acting upward. The released structure is the cantilever fixed at A.

Step 1: Flexibility matrix of the cantilever (L = 20 m)

For a unit load at distance $a$ from the fixed end, the deflection at $x$ is $x^2(3a-x)/6EI$ for $x\le a$ and $a^2(3x-a)/6EI$ for $x\ge a$ .

f_{11}=\frac{10^3}{3EI}=\frac{333.33}{EI},\quad f_{22}=\frac{20^3}{3EI}=\frac{2666.67}{EI},\quad f_{12}=f_{21}=\frac{10^2(3\times20-10)}{6EI}=\frac{833.33}{EI}

Step 2: Deflections due to the loads (downward) on the released beam

At B (x = 10): the 200 kN load gives $\dfrac{200\times5^2(3\times10-5)}{6EI}=\dfrac{20833.3}{EI}$ and the 100 kN load (at 15 m, beyond B) gives $\dfrac{100\times10^2(3\times15-10)}{6EI}=\dfrac{58333.3}{EI}$ . At C (x = 20): $\dfrac{200\times5^2(60-5)}{6EI}=\dfrac{45833.3}{EI}$ and $\dfrac{100\times15^2(60-15)}{6EI}=\dfrac{168750}{EI}$ .

\Delta_1=\frac{79166.7}{EI},\qquad \Delta_2=\frac{214583.3}{EI}

Step 3: Compatibility (deflection at B and C must be zero)

\begin{bmatrix}333.33&833.33\\833.33&2666.67\end{bmatrix}\begin{Bmatrix}R_B\\R_C\end{Bmatrix}=\begin{Bmatrix}79166.7\\214583.3\end{Bmatrix}

R_B=166.07\ \text{kN}\uparrow,\qquad R_C=28.57\ \text{kN}\uparrow

Step 4: Remaining reactions

R_A=200+100-166.07-28.57=105.36\ \text{kN}\uparrow,\qquad M_A=-200(5)-100(15)+166.07(10)+28.57(20)\;\Rightarrow\; M_A=-267.86\ \text{kNm (hogging)}

Check: moments about A: $105.36\times0 +166.07\times10+28.57\times20-200\times5-100\times15=267.86$ = $|M_A|$ ✓.

Shear force (kN)

Section	SF
A (right)	+105.36
5 m (left / right of 200 kN)	+105.36 / -94.64
B (left / right)	-94.64 / +71.43
15 m (left / right of 100 kN)	+71.43 / -28.57
C (left)	-28.57

Bending moment (kNm, sagging +)

Section	BM
A	-267.86
5 m (under 200 kN)	+258.93
B (10 m)	-214.29
15 m (under 100 kN)	+142.86
C	0

SF changes sign at the 200 kN load (BM max +258.93), at B (hogging peak -214.29 where SF jumps) and under the 100 kN load. BM is zero at C.

Answer: $R_B=166.07$ kN, $R_C=28.57$ kN (both upward); also $R_A=105.36$ kN, $M_A=267.86$ kNm (hogging).

2069 Asar · 5 marks

A two hinged symmetrical parabolic arch of secant variation cross section having span 50 m and rise 8 m is loaded with a uniformly distributed load of 12 kN/m extending from the left hand support to the centre of the arch as shown in figure below. Determine the horizontal reaction at the support.

Answer

For a two-hinged arch the thrust $H$ makes the horizontal displacement of the support zero. With $I=I_c\sec\theta$ (so $ds/I=dx/I_c$ ):

H=\frac{\int_0^L M_0\,y\,dx}{\int_0^L y^2\,dx}

Data

$L=50$ m, $h=8$ m, $w=12$ kN/m on the left half (0 to 25 m). Parabola: $y=\dfrac{4hx(L-x)}{L^2}=\dfrac{32x(50-x)}{2500}$ .

Reactions and $M_0$

Total load $=12\times25=300$ kN acting at 12.5 m from A.

V_A=\frac{300\times(50-12.5)}{50}=225\ \text{kN},\qquad V_B=75\ \text{kN}

For $0\le x\le25$ : $M_0=225x-6x^2$
For $25\le x\le50$ : $M_0=75(50-x)$

Integrals

\int_0^{50}M_0\,y\,dx=400000\ \text{kN m}^3,\qquad \int_0^{50}y^2dx=\frac{8h^2L}{15}=\frac{8\times64\times50}{15}=1706.67\ \text{m}^3

H=\frac{400000}{1706.67}=234.375\ \text{kN}

Check: by symmetry a UDL on the half span gives half of the full-span thrust, $\tfrac12\cdot\dfrac{wL^2}{8h}=\tfrac12\times\dfrac{12\times2500}{64}=234.375$ kN. ✓

Answer: Horizontal reaction $H=234.37$ kN at each support (inward). Vertical reactions $V_A=225$ kN and $V_B=75$ kN.

2069 Asar · 10 marks

Generate the flexibility matrix for the coordinates shown in figure below and use this to determine the reactions at support D. Take EI is constant for all members. [Figure: portal frame ABCD, height 5 m, span 4 m; A fixed, D hinged; 2 kN/m UDL on the left column AB (printed near B) and a 5 kN horizontal load at C; coordinates 1 (vertical) and 2 (horizontal) at D.]

Answer

Assumptions: A is fixed, D is hinged. AB = CD = 5 m high, BC = 4 m, EI constant. The 2 kN/m UDL acts horizontally (to the right) on AB and the 5 kN load acts to the right at C. Axial deformations are ignored. The hinge D gives two unknown reactions, so the frame is indeterminate to the 2nd degree. The redundants are the coordinates at D: 1 = vertical reaction (up +) and 2 = horizontal reaction (right +).

Step 1: Released structure

Remove the hinge at D. The frame A-B-C-D becomes a cantilever fixed at A with a free end D.

Step 2: Moment diagrams (positive = tension on the inside face)

Case	on AB (A to B)	on BC (B to C)	on CD (C to D)
Unit vertical force at D ( $m_1$ )	4 (constant)	4 at B to 0 at C	0
Unit horizontal force at D ( $m_2$ )	0 at A to 5 at B	5 (constant)	5 at C to 0 at D
Given loads ( $M_L$ )	-50 at A to 0 at B (parabolic)	0	0

For the loaded case, $M_L=-\left[(5-y)^2+5(5-y)\right]$ on AB (y up from A): the UDL gives $2(5-y)^2/2$ and the 5 kN load at C gives $5(5-y)$ , both causing tension on the outside.

Step 3: Flexibility matrix ( $f_{ij}=\sum\int m_im_j\,dx/EI$ )

f_{11}=\frac{1}{EI}\left[4^2(5)+\frac{4^2(4)}{3}\right]=\frac{101.33}{EI},\quad f_{22}=\frac{1}{EI}\left[\frac{5^3}{3}+5^2(4)+\frac{5^3}{3}\right]=\frac{183.33}{EI}

f_{12}=f_{21}=\frac{1}{EI}\left[4\cdot\frac{5^2}{2}+5\cdot\frac{4\times4}{2}\right]=\frac{90.00}{EI}

[F]=\frac{1}{EI}\begin{bmatrix}101.33&90.00\\90.00&183.33\end{bmatrix}

Step 4: Displacements of D in the released structure due to the loads

\Delta_{1L}=\frac{1}{EI}\int m_1M_L\,dx=-416.67/EI,\qquad \Delta_{2L}=\frac{1}{EI}\int m_2M_L\,dx=-156.25/EI

(negative: D moves down and to the left in the released frame.)

Step 5: Compatibility (D cannot move)

[F]\begin{Bmatrix}R_1\\R_2\end{Bmatrix}=-\begin{Bmatrix}\Delta_{1L}\\\Delta_{2L}\end{Bmatrix} \;\Rightarrow\; \begin{Bmatrix}R_1\\R_2\end{Bmatrix}=\begin{Bmatrix}5.95\\-2.07\end{Bmatrix}\ \text{kN}

Reactions at A (equilibrium)

$H_A=-(2\times5+5)-(-2.07)=-12.93$ kN (to the left), $V_A=-5.95$ kN, $M_A=26.21$ kNm. Check $\sum F_x$ : loads $15$ kN to the right, reactions -12.93 and -2.07 kN, total $=0$ ✓.

Final BM (inside tension +): $M_A=-26.21$ , $M_B=13.45$ , $M_C=-10.34$ , $M_D=0$ kNm.

Answer: reactions at D: vertical $R_1=5.95$ kN upward and horizontal $R_2=2.07$ kN directed to the left.

2069 Chaitra · 18 marks

Find out the member forces in the truss shown in figure below using the force method. The axial rigidity of all vertical and horizontal members is EA and that for all inclined members is 2EA. [Figure: two-panel tall truss ABCDEF, 3 m wide, two panels of 4 m height each, with X-bracing; 50 kN horizontal at C and 50 kN horizontal at B; A hinged, D roller.]

Answer

Assumptions (figure not fully clear): joints A(0,0), B(0,4), C(0,8) on the left and D(3,0), E(3,4), F(3,8) on the right (3 m wide, two panels of 4 m). Members: verticals AB, BC, DE, EF and horizontals BE, CF (axial rigidity EA); diagonals AE, BD, BF, CE (axial rigidity 2EA, lengths 5 m). Both 50 kN loads act horizontally to the right at B and C. A is a hinge and D a roller (vertical reaction only). There is no member joining A and D (the supports hold them).

Degree of indeterminacy

$m=10$ , $r=3$ , $j=6$ : $D_s=m+r-2j=10+3-12=1$ . Take the force $X$ in the diagonal CE as the redundant.

Reactions

$\sum M_A=0$ : $R_D\times3=50\times8+50\times4\Rightarrow R_D=200$ kN (up), $R_{Ay}=200$ kN (down), $H_A=100$ kN (to the left).

Forces $S_0$ (CE removed, loads only) and $S_1$ (unit tension in CE)

Solved joint by joint. Compatibility: $\sum S_0S_1\dfrac{L}{AE}+X\sum S_1^2\dfrac{L}{AE}=0$ (L/AE in units of 1/EA, EA/2EA already allowed for).

Member	L (m)	AE	$S_0$ (kN)	$S_1$	$S_0S_1L/AE$	$S_1^2L/AE$	Final $S$ (kN)
AB	4.000	EA	66.67	0.000	0.000	0.0000	66.67
BC	4.000	EA	0.00	-0.800	0.000	2.5600	45.06
DE	4.000	EA	-200.00	0.000	0.000	0.0000	-200.00
EF	4.000	EA	-66.67	-0.800	213.333	2.5600	-21.61
BE	3.000	EA	-100.00	-0.600	180.000	1.0800	-66.21
CF	3.000	EA	-50.00	-0.600	90.000	1.0800	-16.21
AE	5.000	2EA	166.67	0.000	0.000	0.0000	166.67
BD	5.000	2EA	0.00	0.000	0.000	0.0000	0.00
BF	5.000	2EA	83.33	1.000	208.333	2.5000	27.01
CE	5.000	2EA	0.00	1.000	0.000	2.5000	-56.32
Sum					691.667	12.2800

X=-\frac{691.667}{12.2800}=-56.32\ \text{kN}

Final member forces ( $S=S_0+XS_1$ ; + tension, - compression)

Verticals: AB = 66.67, BC = 45.06, DE = -200.00, EF = -21.61 kN. Horizontals: BE = -66.21, CF = -16.21 kN. Diagonals: AE = 166.67, BD = 0.00, BF = 27.01, CE = -56.32 kN.

As a check, the whole truss was also analysed by the direct stiffness method and gives identical forces.

Answer: CE = -56.32 kN (compression), AE = 166.67 kN, BF = 27.01 kN, BD = 0.00 kN, verticals and horizontals as tabulated.

2068 Chaitra · 8 marks

Use the force method (flexibility matrix) to solve the truss as shown in figure below. [Figure: square truss 4 m x 4 m with both diagonals; 100 kN downward and 50 kN horizontal at the top-left joint; bottom-left hinged, bottom-right roller.]

Answer

Assumptions: square truss ABCD, 4 m x 4 m: A bottom-left (hinge), B bottom-right (roller, vertical reaction only), C top-right, D top-left. 100 kN acts downward and 50 kN horizontally (to the right) at D. Both diagonals AC and BD are present; EA is the same for all members.

Degree of indeterminacy

$m=6$ , $r=3$ , $j=4$ : $D_s=6+3-8=1$ . Choose the diagonal BD as redundant $X_1$ (coordinate 1). The flexibility matrix is $1\times1$ : $[F]=[f_{11}]$ .

Reactions (statically determinate)

Taking moments about A (the 50 kN force acts 4 m above A, the 100 kN force passes through A's vertical line):

R_B\times4=50\times4\ \Rightarrow\ R_B=50\ \text{kN}\uparrow

$\sum F_x=0$ : $H_A=50$ kN to the left. $\sum F_y=0$ : $V_A=100-50=50$ kN upward.

Member forces by joint method

$S_0$ : forces in the released truss (BD removed) due to the loads.
$S_1$ : forces due to a unit tension pair applied along BD.

Member	L (m)	$S_0$ (kN)	$S_1$	$S_0S_1L$	$S_1^2L$	Final $S$ (kN)
AB	4.000	0.00	-0.707	0.00	2.000	35.36
BC	4.000	-50.00	-0.707	141.42	2.000	-14.64
CD	4.000	-50.00	-0.707	141.42	2.000	-14.64
DA	4.000	-100.00	-0.707	282.84	2.000	-64.64
AC	5.657	70.71	1.000	400.00	5.657	20.71
BD	5.657	0.00	1.000	0.00	5.657	-50.00
Sum				965.69	19.314

Flexibility and compatibility

f_{11}=\frac{\sum S_1^2L}{EA}=\frac{19.314}{EA},\qquad \Delta_{10}=\frac{\sum S_0S_1L}{EA}=\frac{965.69}{EA}

f_{11}X_1+\Delta_{10}=0\ \Rightarrow\ X_1=-\frac{965.69}{19.314}=-50.00\ \text{kN}

The diagonal BD carries 50.00 kN (compression).

Final forces ( $S=S_0+X_1S_1$ )

AB = 35.36, BC = -14.64, CD = -14.64, DA = -64.64, AC = 20.71, BD = -50.00 kN (+ tension, - compression).

Answer: BD = -50.00 kN, AC = 20.71 kN, AB = 35.36 kN, BC = -14.64 kN, CD = -14.64 kN, DA = -64.64 kN.

2068 Chaitra · 5 marks

For the beam as shown, determine the slope at support B. Use Castigliano's second theorem. Take EI = constant. [Figure: beam AB of length l, A fixed, B hinged support with an applied moment M at B.]

Answer

Castigliano's second theorem: $\theta=\partial U/\partial M$ . The beam is a propped cantilever (A fixed, B on a roller/hinge support) with a clockwise couple $M$ at B. The vertical reaction $R_B$ is the redundant, found from $\partial U/\partial R_B=0$ because the support does not settle.

Take $x$ measured from B towards A (B at $x=0$ , A at $x=l$ ), $R_B$ upward.

Bending moment (sagging +)

M_x = M + R_Bx,\qquad \frac{\partial M_x}{\partial R_B}=x,\qquad \frac{\partial M_x}{\partial M}=1

Step 1: find $R_B$

\frac{\partial U}{\partial R_B}=\frac{1}{EI}\int_0^l(M+R_Bx)\,x\,dx=0 \;\Rightarrow\; \frac{Ml^2}{2}+\frac{R_Bl^3}{3}=0 \;\Rightarrow\; R_B=-\frac{3M}{2l}

(i.e. $3M/2l$ acting downward.)

Step 2: slope at B

\theta_B=\frac{\partial U}{\partial M}=\frac{1}{EI}\int_0^l(M+R_Bx)(1)\,dx=\frac{1}{EI}\left[Ml+R_B\frac{l^2}{2}\right]

\theta_B=\frac{1}{EI}\left[Ml-\frac{3M}{2l}\cdot\frac{l^2}{2}\right]=\frac{1}{EI}\left[Ml-\frac{3Ml}{4}\right]=\frac{Ml}{4EI}

Answer: $\theta_B=\dfrac{Ml}{4EI}$ , in the same (clockwise) sense as the applied moment. The fixed far end stiffens the beam, so the slope is smaller than the simply-supported value $Ml/3EI$ .

2068 Chaitra · 8 marks

Using Castigliano's second theorem, determine the vertical deflection at the 50 kN load in the beam shown in figure below. [Figure: cantilever beam, fixed at the left end; 100 kN at 3 m from the fixed end; 50 kN at the free end, 3 m further; $E = 211000\ \text{N/mm}^2$ , $I = 46875\times10^{4}\ \text{mm}^4$ .]

Answer

Castigliano's second theorem: $\delta=\partial U/\partial P$ . Call the 50 kN load $P_2$ at the free end ( $x=6$ m from the fixed end) and the 100 kN load $P_1$ at $x=3$ m. Treat both as symbols, differentiate with respect to $P_2$ , then substitute the numbers.

Bending moments (x from the fixed end A, hogging taken as negative)

$0\le x\le3$ : $M=-[P_1(3-x)+P_2(6-x)]$ , $\;\dfrac{\partial M}{\partial P_2}=-(6-x)$
$3\le x\le6$ : $M=-P_2(6-x)$ , $\;\dfrac{\partial M}{\partial P_2}=-(6-x)$

Deflection under the 50 kN load

\delta_2=\frac{1}{EI}\left[\int_0^3[100(3-x)+50(6-x)](6-x)\,dx+\int_3^6 50(6-x)^2dx\right]

\int_0^3[100(3-x)+50(6-x)](6-x)dx=5400\ \text{kN m}^3,\qquad \int_3^650(6-x)^2dx=450\ \text{kN m}^3

\delta_2=\frac{5850}{EI}=\frac{5850}{EI}\ \text{kN m}^3

Flexural rigidity

EI=211000\ \text{N/mm}^2\times46875\times10^4\ \text{mm}^4=9.8906\times10^{13}\ \text{N mm}^2=98906.25\ \text{kN m}^2

\delta_2=\frac{5850}{98906.25}=59.15\ \text{mm}\ \text{(downward)}

Check with standard results: tip deflection due to 100 kN at 3 m is $Pa^2(3L-a)/6EI=2250/EI$ and due to 50 kN at the tip is $PL^3/3EI=3600/EI$ ; total $5850/EI$ ✓.

Answer: vertical deflection under the 50 kN load = 59.15 mm downward.

2068 Chaitra · 8 marks

Use the force method (flexibility matrix) to solve the frame as shown in figure below. [Figure: portal frame; left column EI of 6 m, hinged at the base A; 48 kNm moment at the top-left joint; beam EI of 4 m (1 m + 3 m); right column EI of 4 m, fixed base.]

Answer

Assumptions: A(0,0) is hinged at the base of the 6 m left column AB; the 4 m beam BC is horizontal at the top of AB; the 4 m right column CD is fixed at D, whose base is 2 m above A. The 48 kNm moment acts clockwise at joint B. EI is the same for all members. Axial deformation is ignored.

Degree of indeterminacy

Reactions: hinge A (2) + fixed D (3) = 5; equations = 3. So the frame is indeterminate to the 2nd degree. Redundants: the reactions at the hinge A, coordinate 1 = horizontal $X_1$ (right +) and coordinate 2 = vertical $X_2$ (up +). Released structure: cantilever fixed at D with A free.

Moment diagrams of the released structure (positive = inside tension)

Case	AB (A to B)	BC (B to C)	CD (C to D)
Load: 48 kNm at B ( $M_L$ )	0	48	48
Unit horizontal force at A ( $m_1$ )	0 to -6	-6	-6 to -2
Unit vertical force at A ( $m_2$ )	0	0 to 4	4

(In case 1 the moment equals the height of the section above A; in case 2 it equals the horizontal distance from A.)

Flexibility coefficients ( $\int m_im_j\,dx/EI$ )

f_{11}=\frac{1}{EI}\left[\frac{6^3}{3}+6^2(4)+\frac{6^3-2^3}{3}\right]=\frac{285.33}{EI},\quad f_{22}=\frac{1}{EI}\left[\frac{4^3}{3}+4^2(4)\right]=\frac{85.33}{EI},\quad f_{12}=f_{21}=\frac{-112.00}{EI}

\Delta_{1L}=\frac{1}{EI}[(-6)(48)(4)+(-1)(48)(16)]=\frac{-1920}{EI},\qquad \Delta_{2L}=\frac{1}{EI}\left[\frac{4}{2}(48)(4)+4(48)(4)\right]=\frac{1152}{EI}

Compatibility (A cannot move)

\begin{bmatrix}285.33&-112.00\\-112.00&85.33\end{bmatrix}\begin{Bmatrix}X_1\\X_2\end{Bmatrix}=-\begin{Bmatrix}-1920\\1152\end{Bmatrix} \;\Rightarrow\; X_1=2.95\ \text{kN},\quad X_2=-9.63\ \text{kN}

So the hinge A develops a horizontal reaction of 2.95 kN to the right and a vertical reaction of 9.63 kN downward.

Reactions at D

$H_D=-2.95$ kN (left), $V_D=9.63$ kN (up), $M_D=3.59$ kNm (anticlockwise). Check, moments about A: $V_D(4)+|H_D|(2)+M_D=48.00$ kNm, equal to the applied 48 kNm ✓.

Final bending moments ( $M=M_L+X_1m_1+X_2m_2$ , inside tension +)

Joint	Value (kNm)
A	0
B (on AB)	-17.70
B (on BC)	30.30
C	-8.21
D (on CD)	3.59

Joint B: 30.30 - (-17.70) = 48 kNm, equal to the applied moment ✓. Negative values mean tension on the outside face. The BM varies linearly in every member (no loads between joints).

Answer: $X_1=H_A=2.95$ kN and $X_2=V_A=9.63$ kN (downward); $M_B=-17.70$ kNm in AB and $30.30$ kNm in BC, $M_C=-8.21$ kNm, $M_D=3.59$ kNm.

2068 Baishakh · 10 marks

Find support reactions of the given loaded beam using Castigliano's theorem. [Figure: two-span continuous beam, spans L and L, UDL w per metre on the left span; A hinged, B roller, C roller.]

Answer

Method: the two-span beam has 3 vertical reactions and 2 equilibrium equations, so it is indeterminate to the 1st degree. Take $R_B$ (upward) as the redundant. Since B does not settle, Castigliano's theorem gives $\partial U/\partial R_B=0$ .

Reactions in terms of $R_B$

$\sum M_C=0$ with the UDL of total load $wL$ acting at $L/2$ from A (that is $3L/2$ from C):

R_A(2L)+R_B(L)=wL\cdot\frac{3L}{2}\ \Rightarrow\ R_A=\frac{3wL}{4}-\frac{R_B}{2}

Bending moments (x from A)

$0\le x\le L$ : $M=R_Ax-\dfrac{wx^2}{2}$ , $\;\dfrac{\partial M}{\partial R_B}=-\dfrac{x}{2}$
$L\le x\le2L$ : $M=R_Ax-wL\left(x-\dfrac{L}{2}\right)+R_B(x-L)$ , $\;\dfrac{\partial M}{\partial R_B}=-\dfrac{x}{2}+(x-L)=\dfrac{x}{2}-L$

Condition $\partial U/\partial R_B=0$

\int_0^L M_1\left(-\frac{x}{2}\right)dx+\int_L^{2L}M_2\left(\frac{x}{2}-L\right)dx=0

Evaluating the integrals (with $R_A$ substituted) gives

\frac{L^{3} \left(- 5 L w + 8 R_{B}\right)}{48}=0\ \Rightarrow\ R_B=\frac{5wL}{8}

Other reactions

R_A=\frac{3wL}{4}-\frac{5wL}{16}=\frac{7wL}{16}\ \uparrow,\qquad R_C=wL-R_A-R_B=-\frac{wL}{16}

$R_C$ is negative, so the support at C pulls the beam downward with $wL/16$ .

Check: $R_A+R_B+R_C=\frac{7+10-1}{16}wL=wL$ ✓.

Answer: $R_A=\dfrac{7wL}{16}$ (up), $R_B=\dfrac{5wL}{8}$ (up), $R_C=\dfrac{wL}{16}$ (down).

2068 Baishakh · 12+8 marks

Compute the bar forces in all members due to: (i) given load and (ii) temperature rise by 30°C in the upper chord. Take $E = 2\times10^{5}\ \text{N/mm}^2$ , $\alpha = 10.8\times10^{-6}/°C$ . Take area of all members to be 30 cm². [Figure: truss with top joints under 9 kN and 18 kN downward loads; three panels of 4 m (total 12 m), height 3 m, diagonals; left support hinged, two roller supports at the lower chord.]

Answer

Assumptions (the figure is not clear): bottom joints A, B, C, D at 0, 4, 8, 12 m; top joints E and F, 3 m above B and C. Members: bottom chord AB, BC, CD; top chord EF; verticals BE, CF; end diagonals AE, DF and one middle diagonal BF (9 members). 9 kN acts downward at E and 18 kN downward at F. A is a hinge; the two rollers are at C and D (vertical reactions only). $EA=2\times10^5\ \text{N/mm}^2\times3000\ \text{mm}^2=6\times10^8$ N $=6\times10^5$ kN for every bar.

Degree of indeterminacy

$m=9$ , $r=4$ , $j=6$ : $D_s=9+4-12=1$ (external). Take the roller reaction at C, $X$ (upward), as the redundant. Primary structure: remove the roller at C, leaving a truss hinged at A and on a roller at D.

(i) Bar forces due to the loads

$S_0$ = forces in the primary truss under the loads; $S_1$ = forces for a unit upward force at C.

Bar	L (m)	$S_0$ (kN)	$S_1$	$S_0S_1L$	$S_1^2L$	Final $S$ (kN)
AB	4.000	16.00	-0.444	-28.44	0.790	7.10
BC	4.000	20.00	-0.889	-71.11	3.160	2.20
CD	4.000	20.00	-0.889	-71.11	3.160	2.20
EF	4.000	-16.00	0.444	-28.44	0.790	-7.10
AE	5.000	-20.00	0.556	-55.56	1.543	-8.87
BE	3.000	3.00	-0.333	-3.00	0.333	-3.68
CF	3.000	0.00	-1.000	0.00	3.000	-20.03
DF	5.000	-25.00	1.111	-138.89	6.173	-2.75
BF	5.000	-5.00	0.556	-13.89	1.543	6.13
Sum				-410.44	20.494

Vertical displacement of C in the primary truss (positive upward) $\Delta_C=\dfrac{\sum S_0S_1L}{EA}=\dfrac{-410.44}{6\times10^5}=-0.684$ mm, so C would move 0.684 mm downward without the roller. Compatibility (C cannot move):

X=-\frac{\sum S_0S_1L}{\sum S_1^2L}=-\frac{-410.44}{20.494}=20.03\ \text{kN (up)}

Final bar forces: AB = 7.10, BC = 2.20, CD = 2.20, EF = -7.10, AE = -8.87, BE = -3.68, CF = -20.03, DF = -2.75, BF = 6.13 kN (+ tension, - compression).

(ii) Temperature rise of 30°C in the upper chord (EF)

Free elongation of EF: $e=\alpha\,\Delta T\,L=10.8\times10^{-6}\times30\times4=1.296$ mm. The primary truss expands freely and no forces develop in it ( $S_0=0$ ); only the displacement at C changes. The compatibility equation with the temperature term is

\frac{X\sum S_1^2L}{EA}+S_1^{EF}\,e=0\Rightarrow X=-\frac{EA\,S_1^{EF}\,e}{\sum S_1^2L}=-16.86\ \text{kN}

(with $S_1^{EF}=0.444$ ). Bar forces $S=XS_1$ :

AB = 7.49, BC = 14.99, CD = 14.99, EF = -7.49, AE = -9.37, BE = 5.62, CF = 16.86, DF = -18.74, BF = -9.37 kN (+ tension, - compression).

Answer: (i) AB = 7.10, BC = 2.20, CD = 2.20, EF = -7.10, AE = -8.87, BE = -3.68, CF = -20.03, DF = -2.75, BF = 6.13 kN. (ii) AB = 7.49, BC = 14.99, CD = 14.99, EF = -7.49, AE = -9.37, BE = 5.62, CF = 16.86, DF = -18.74, BF = -9.37 kN. Both were checked with a full stiffness analysis.

2067 Asar · 15 marks

Use the consistent deformation method to solve the frame and draw bending moment, shear force and normal thrust diagrams. [Figure: frame; left column (2I) of 6 m, fixed base, carrying 1 kN/m UDL horizontally; beam (2I) of 6 m with 3 kN at 3 m from the left; right column (I) of 3 m, fixed base.]

Answer

Assumptions: A(0,0) and D are fixed bases. Column AB (2I) is 6 m high, beam BC (2I) is 6 m and column CD (I) is 3 m high, so D is 3 m above the level of A. The 1 kN/m UDL acts horizontally to the right on AB; the 3 kN load acts downward at 3 m from B on BC. Axial deformation is ignored. Positive bending moment = tension on the inside of the frame.

Degree of indeterminacy

Reactions: 3 + 3 = 6; equations = 3, so $D_s=3$ . Release the fixed support D. The redundants are $X_1=H_D$ (right +), $X_2=V_D$ (up +) and $X_3=M_D$ (anticlockwise +). The primary structure is a cantilever fixed at A.

Moments in the primary structure (kNm; $m_1,m_2,m_3$ for unit $X_1,X_2,X_3$ )

Section	$m_1$ (H=1)	$m_2$ (V=1)	$m_3$ (M=1)	$M_L$
A (AB)	-3.00	6.00	1.00	-27.00
B (AB)	3.00	6.00	1.00	-9.00
B (BC)	3.00	6.00	1.00	-9.00
C (BC)	3.00	0.00	1.00	0.00
C (CD)	3.00	0.00	1.00	0.00
D (CD)	0.00	0.00	1.00	0.00

Flexibility matrix and load displacements ( $\int m_im_j\,ds/EI$ )

[F]=\frac{1}{EI}\begin{bmatrix}45.00&27.00&13.50\\27.00&144.00&27.00\\13.50&27.00&9.00\end{bmatrix},\qquad \{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}6.75\\-303.75\\-51.75\end{Bmatrix}

(For example $f_{33}=\sum \dfrac{L}{EI}=\dfrac{6}{2}+\dfrac{6}{2}+\dfrac{3}{1}=9$ .)

Consistent deformation (D is fixed: all three displacements are zero)

[F]\{X\}=-\{\Delta_L\}\;\Rightarrow\; X_1=-2.404\ \text{kN},\quad X_2=1.842\ \text{kN},\quad X_3=3.831\ \text{kNm}

So $H_D=2.40$ kN to the left, $V_D=1.84$ kN upward, $M_D=3.83$ kNm anticlockwise.

Reactions at A (equilibrium)

$H_A=6-2.40=3.60$ kN (left), $V_A=3-1.84=1.16$ kN (up), $M_A=4.90$ kNm (anticlockwise). Check: $\sum F_x=6-3.60-2.40=0$ ✓, $\sum F_y=3-1.16-1.84=0$ ✓.

Bending moment diagram ( $M=M_L+\sum X_im_i$ , kNm; + inside tension)

Section	BM
A	-4.90
Mid-height of AB (3 m)	1.38
B	-1.33
Under 3 kN load (BC)	2.14
C	-3.38
D	3.83

AB bends in a parabola (UDL); BC is two straight lines with a peak under the load; CD is a straight line with a change of sign.

Shear force and normal thrust

Member	Axial N (+ tension)	Shear at start	Shear at end
AB	-1.16	3.60	-2.40
BC	-2.40	1.16	-1.84
CD	-1.84	2.40	2.40

Shear sign: positive when the force on the start end of the member acts along its local y-axis (90° anticlockwise from the direction A to B, B to C, C to D). Axial: negative = compression. Normal thrust in AB, BC, CD is therefore 1.16, 2.40 and 1.84 kN compression. The shear in AB goes from 3.60 kN at A to -2.40 kN at B (linear, due to the UDL), and in BC it is 1.16 kN before the 3 kN load and -1.84 kN after it.

Answer: $H_D=2.40$ kN, $V_D=1.84$ kN, $M_D=3.83$ kNm; $M_A=-4.90$ , $M_B=-1.33$ , $M_C=-3.38$ kNm (inside tension +).

2067 Asar · 5 marks

What is the consistent deformation method? Derive the formula.

Answer

Meaning

The consistent deformation method (also called the method of consistent displacements or the force/flexibility method) analyses a statically indeterminate structure by:

choosing a suitable number of reactions or internal forces equal to the degree of indeterminacy as redundants,
removing them to get a stable, statically determinate primary (released) structure,
finding the displacement of the released structure at each redundant due to the loads, and
writing that the redundant forces must restore the displacements to the value that actually exists in the real structure (zero at a rigid support, equal to the settlement if the support settles). These are the compatibility, or consistent-deformation, conditions.

Derivation (one redundant)

Let a beam be indeterminate to the first degree, with the support reaction $R$ at B as redundant.

Released structure under the loads. Remove the support at B. The loads cause a deflection $\Delta_{B0}$ at B (call it $\Delta_{10}$ , positive in the direction of $R$ ).
Released structure under a unit load at B in the direction of $R$ . The deflection at B is the flexibility coefficient $f_{11}$ (deflection per unit force). If the force is $R$ , the deflection is $f_{11}R$ (linear elasticity, superposition).
Superpose. The actual deflection at B is

\Delta_B=\Delta_{10}+f_{11}R

Compatibility. In the actual structure the support prevents deflection (or allows a known settlement $\delta_B$ ):

\Delta_{10}+f_{11}R=\delta_B\quad(\delta_B=0\text{ for an unyielding support})

R=\frac{\delta_B-\Delta_{10}}{f_{11}}

where $f_{11}=\int\dfrac{m_1^2}{EI}dx$ and $\Delta_{10}=\int\dfrac{m_1M_0}{EI}dx$ , with $m_1$ the moment due to unit load at B, and $M_0$ the moment due to the loads in the released structure.

Several redundants

For $n$ redundants $X_1,\ldots,X_n$ :

\Delta_{i0}+\sum_{j=1}^{n}f_{ij}X_j=\delta_i\quad(i=1,\ldots,n)\qquad\Longrightarrow\qquad [F]\{X\}=\{\delta\}-\{\Delta_0\}

$f_{ij}=f_{ji}$ (Maxwell). After finding $X$ , the final moment is $M=M_0+\sum X_jm_j$ , and the remaining reactions follow from equilibrium.

Example

Propped cantilever of span $L$ with UDL $w$ , redundant $R_B$ at the roller: $\Delta_{10}=-\dfrac{wL^4}{8EI}$ , $f_{11}=\dfrac{L^3}{3EI}$ , so $R_B=\dfrac{3wL}{8}$ .

2066 Jestha · 20 marks

Determine forces in bars BC and BF of the truss shown below, if all inclined members are found to be 2 mm too long and all vertical members are subjected to a decrease in temperature of 15°C. Area of cross-section of all members is 40 cm². Take $E = 2\times10^{5}\ \text{N/mm}^2$ , $\alpha = 10.8\times10^{-6}/°C$ . [Figure: truss with joints A, B, C on the bottom chord (A hinged, B roller, C roller; 5 m + 5 m), D, E, F on the top chord, height 3 m, with vertical and inclined members.]

Answer

Assumptions (figure read as follows): bottom joints A(0,0), B(5,0), C(10,0); top joints D(0,3), E(5,3), F(10,3). Members: bottom chord AB, BC; top chord DE, EF; verticals AD, BE, CF; inclined members DB and BF. A is a hinge; B and C are rollers (vertical reactions only). All areas 40 cm², $E=2\times10^5$ N/mm², so $EA=8\times10^5$ kN. The inclined members DB and BF are 2 mm too long; the verticals AD, BE, CF are cooled by 15°C ( $\alpha=10.8\times10^{-6}$ /°C).

Degree of indeterminacy

$m=9$ , $r=4$ , $j=6$ : $D_s=9+4-12=1$ . The truss is internally determinate, so the lack of fit and the temperature change only produce forces because of the extra support. Take the vertical reaction at B as the redundant $X$ .

Free (initial) elongations $e_0$

Inclined bars too long: $e_0=+2$ mm (DB and BF).
Verticals cooled: $e_0=-\alpha\,\Delta T\,L=-10.8\times10^{-6}\times15\times3000=-0.486$ mm.

Unit load on the primary truss

The primary truss (B roller removed) is loaded by a unit upward force at B, giving $S_1$ . Since no external load acts, $S_0=0$ and only the initial strains cause displacement of B.

Bar	L (m)	$S_1$	$e_0$ (mm)	$S_1e_0$ (mm)	$S_1^2L$ (m)	Final $S$ (kN)
AB	5.000	0.000	0.000	0.0000	0.000	0.00
BC	5.000	0.000	0.000	0.0000	0.000	0.00
DE	5.000	0.833	0.000	0.0000	3.472	149.83
EF	5.000	0.833	0.000	0.0000	3.472	149.83
AD	3.000	0.500	-0.486	-0.2430	0.750	89.90
BE	3.000	0.000	-0.486	0.0000	0.000	0.00
CF	3.000	0.500	-0.486	-0.2430	0.750	89.90
DB	5.831	-0.972	2.000	-1.9437	5.507	-174.73
BF	5.831	-0.972	2.000	-1.9437	5.507	-174.73

\Delta_{B}=\sum S_1e_0=-3.8873+(-0.4860)=-4.3733\ \text{mm}\quad(\text{negative: B moves 4.373 mm downward in the primary truss})

f_{11}=\frac{\sum S_1^2L}{EA}=\frac{19.458}{8\times10^5}=0.02432\ \text{mm/kN}

Compatibility ( $B$ cannot move): $\Delta_B+f_{11}X=0$

X=-\frac{-4.3733}{0.02432}=179.80\ \text{kN}

Bar forces $S=XS_1$ (initial strains do not produce forces in the determinate part):

Answer: force in BC = 0.00 kN and force in BF = -174.73 kN (+ tension, - compression). Other bars: AB = 0.00, DE = 149.83, EF = 149.83, AD = 89.90, BE = 0.00, CF = 89.90, DB = -174.73 kN.

2066 Bhadra · 10 marks

Compute the maximum central vertical deflection for a simply supported beam of span L loaded with a uniformly distributed load of w/unit length, EI is constant. Use Castigliano's theorem.

Answer

Castigliano's second theorem: the deflection at a point equals $\partial U/\partial P$ , where $P$ is a load at that point. There is no point load at mid-span, so apply a fictitious load $P$ downward at the centre and set $P=0$ after differentiating.

Reactions

R_A=R_B=\frac{wL}{2}+\frac{P}{2}

Bending moment (x from A; the beam is symmetrical, so work with half and double)

For $0\le x\le L/2$ :

M=\left(\frac{wL}{2}+\frac{P}{2}\right)x-\frac{wx^2}{2},\qquad \frac{\partial M}{\partial P}=\frac{x}{2}

Deflection at the centre

\delta_C=\frac{\partial U}{\partial P}=2\int_0^{L/2}\frac{M}{EI}\frac{\partial M}{\partial P}dx\Big|_{P=0} =\frac{2}{EI}\int_0^{L/2}\left(\frac{wLx}{2}-\frac{wx^2}{2}\right)\frac{x}{2}\,dx

=\frac{w}{2EI}\int_0^{L/2}\left(Lx^2-x^3\right)dx=\frac{w}{2EI}\left[\frac{Lx^3}{3}-\frac{x^4}{4}\right]_0^{L/2} =\frac{w}{2EI}\left[\frac{L^4}{24}-\frac{L^4}{64}\right]

=\frac{w}{2EI}\cdot\frac{(8-3)L^4}{192}=\frac{5wL^4}{384EI}

Answer: maximum central deflection $\delta_C=\dfrac{5wL^4}{384\,EI}$ (downward).

2066 Bhadra · 10 marks

Use the consistent deformation method to draw bending moment diagram of the chair-frame loaded with a couple as shown. Take $E = 2\times10^{4}$ MPa, $\ell = 3$ m, $M = 50$ kNm and $I = 4.5\times10^{8}\ \text{mm}^4$ . Also draw shear force and normal thrust diagrams corresponding to the bending moment diagram. [Figure: chair-shaped frame; fixed top support, vertical member (2I) of height $\ell$ , horizontal member (I) of length $\ell/3$ with the couple M applied at its end, then vertical member (2I) of height $\ell$ and horizontal span $2\ell/3$ down to a fixed base.]

Answer

Assumptions (the figure is only described): the chair frame is A(0,6) - B(0,3) - C(1,3) - D(1,0) - E(3,0) with $\ell=3$ m: AB vertical $2I$ (3 m), BC horizontal $I$ ( $\ell/3=1$ m), CD vertical $2I$ (3 m) and DE horizontal $I$ ( $2\ell/3=2$ m). A (top) and E (base) are fixed. The couple $M=50$ kNm acts clockwise at joint C. Axial deformation is neglected. $E=2\times10^4$ MPa $=2\times10^7$ kN/m², $I=4.5\times10^8$ mm⁴ $=4.5\times10^{-4}$ m⁴, so $EI=9000$ kNm² and $2EI=18000$ kNm². Since only the relative stiffnesses matter for the redundants, EI is a common factor.

Degree of indeterminacy

$D_s=6-3=3$ . Release E and take $X_1=H_E$ (right +), $X_2=V_E$ (up +), $X_3=M_E$ (anticlockwise +) as redundants. The primary structure is the cantilever A-B-C-D-E fixed at A.

Moments in the primary structure (kNm)

Section	$m_1$ (H=1)	$m_2$ (V=1)	$m_3$ (M=1)	$M_L$
A (AB)	6.00	3.00	1.00	-50.00
B (AB)	3.00	3.00	1.00	-50.00
B (BC)	3.00	3.00	1.00	-50.00
C (BC)	3.00	2.00	1.00	-50.00
C (CD)	3.00	2.00	1.00	0.00
D (CD)	0.00	2.00	1.00	0.00
D (DE)	0.00	2.00	1.00	0.00
E (DE)	0.00	0.00	1.00	0.00

The couple at C acts on the part between the support A and C, so in the primary structure it produces a constant moment of 50 kNm on AB and BC (the sign is negative here) and no moment on CD and DE. Sign: positive = tension on the right-hand side when travelling A, B, C, D, E.

Flexibility matrix ( $\int m_im_j\,ds/EI$ , shown in units of 1/EI where $EI$ is the stiffness of $I$ )

[F]=\frac{1}{EI}\begin{bmatrix}45.00&32.25&12.00\\32.25&28.50&12.00\\12.00&12.00&6.00\end{bmatrix},\qquad \{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}-487.50\\-350.00\\-125.00\end{Bmatrix}

Consistent deformation (E is fixed)

[F]\{X\}=-\{\Delta_L\}\Rightarrow X_1=9.220\ \text{kN},\quad X_2=5.319\ \text{kN},\quad X_3=-8.245\ \text{kNm}

Reactions at A

$H_A=-9.220$ kN, $V_A=-5.319$ kN, $M_A=-13.032$ kNm (anticlockwise +). Check: $\sum H=0.000$ , $\sum V=-0.000$ ✓; moment equilibrium holds with the applied 50 kNm couple.

Bending moment diagram (kNm, sign as above)

Section	BM
A	13.03
B	-14.63
C (left, on BC)	-19.95
C (right, on CD)	30.05
D	2.39
E	-8.24

The jump at C (50.00 kNm) equals the applied couple. Every member has a straight-line BM because there are no loads along the members.

Shear force and normal thrust diagrams

Member	Axial N (+ tension)	Shear at start	Shear at end
AB	-5.32	-9.22	-9.22
BC	9.22	-5.32	-5.32
CD	-5.32	-9.22	-9.22
DE	9.22	-5.32	-5.32

Shear and axial force are constant along each member. Axial: negative = compression. In AB the thrust is 5.32 kN, in BC 9.22 kN, in CD 5.32 kN and in DE 9.22 kN (sign as tabulated). The shear in each member equals the constant slope of its BM diagram divided by the member length.

Answer: $H_E=9.22$ kN, $V_E=5.32$ kN, $M_E=-8.24$ kNm; $M_A=13.03$ kNm, $M_C=(-19.95, 30.05)$ kNm.

2066 Bhadra · 10 marks

A rectangular horizontal truss of span 12 m and height 9 m is with two diagonals and is supported by two hinges fixed at the base. A horizontal force of magnitude 100 kN is acting toward the truss at the left top joint. The diagonal connecting the loaded joint was manufactured 2 cm shorter than the assigned length. Calculate the forces induced in every member assuming Young's modulus and cross-sectional areas of every member to be $2\times10^{5}$ MPa and 1000 mm² respectively.

Answer

Setup: joints A(0,0) and B(12,0) at the base (both hinges), C(12,9) and D(0,9) at the top. Members: verticals AD, BC; top chord CD; diagonals AC and BD (bottom chord AB is not a member between the two hinges, or if present it carries no force because A and B cannot move). 100 kN acts horizontally to the right at D (the loaded joint). The diagonal BD joining the loaded joint is 20 mm short. $E=2\times10^5$ MPa, $A=1000$ mm², so $EA=2\times10^5$ kN and lengths: AD = BC = 9 m, CD = 12 m, AC = BD = 15 m.

Degree of indeterminacy

$m=5$ , $r=4$ , $j=4$ : $D_s=5+4-8=1$ . Take the horizontal reaction at B, $X=H_B$ (to the right +), as the redundant. In the primary truss B is a roller on the base line.

Forces in the primary truss

$S_0$ = bar forces due to the 100 kN load (the lack of fit is treated separately). $S_1$ = bar forces due to a unit force at B acting to the right.

Bar	L (m)	$S_0$ (kN)	$S_1$	$S_0S_1L$	$S_1^2L$	Final $S$ (kN)
BC	9.000	-75.00	-0.7500	506.250	5.0625	-85.33
CD	12.000	-100.00	-1.0000	1200.000	12.0000	-113.77
DA	9.000	0.00	-0.7500	0.000	5.0625	-10.33
AC	15.000	125.00	1.2500	2343.750	23.4375	142.21
BD	15.000	0.00	1.2500	0.000	23.4375	17.21
Sum				4050.00	69.000

Lack of fit term

The bar BD is shorter by 20 mm, i.e. its free elongation is $e_0=-0.02$ m. It contributes $S_1e_0=1.2500\times(-0.02)=-0.02500$ m to the displacement at B.

Compatibility ( $H_B$ gives no horizontal movement at B)

\frac{\sum S_0S_1L}{EA}+\sum S_1e_0+X\frac{\sum S_1^2L}{EA}=0

\frac{4050.000}{2\times10^5}+(-0.02500)+X\frac{69.0000}{2\times10^5}=0\;\Rightarrow\; X=H_B=13.77\ \text{kN}

(positive: $H_B$ acts to the right, in the same direction as the 100 kN load.)

Bar forces ( $S=S_0+XS_1$ , + tension, - compression)

BC = -85.33, CD = -113.77, DA = -10.33, AC = 142.21, BD = 17.21 kN.

Reactions: $H_B=13.77$ kN (right), $H_A=-113.77$ kN, i.e. 113.77 kN to the left; vertical reactions $V_A=-75.00$ kN and $V_B=75.00$ kN (couple of the 100 kN load). The result was checked with a full stiffness analysis.

Answer: BC = -85.33, CD = -113.77, DA = -10.33, AC = 142.21, BD = 17.21 kN (+ tension, - compression); AB carries zero.

2065 Shrawan · 20 marks

Use the consistent deformation method to analyze the bent frame shown in the figure below. Draw Axial Force Diagram, Shear Force Diagram, and Bending Moment Diagram for the shown system, if support A settles down by 10 mm, shifts towards left by 10 mm and rotates clockwise by 0.002 radians. [Figure: bent frame; column AC (3EI) of 4 m (2 m x 2 m) with 30 kN horizontal at mid-height, A fixed; beam CB (5EI) of 5 m carrying 20 kN/m UDL, B fixed.]

Answer

Assumptions: A(0,0), C(0,4), B(5,4). Column AC (3EI) is 4 m high with the 30 kN horizontal load (to the right) at mid-height; beam CB (5EI) is 5 m with 20 kN/m downward; B is fixed. Support A moves: 10 mm down, 10 mm to the left and rotates 0.002 rad clockwise. The value of EI is not given in the question, so the answer is derived in terms of EI (kNm² units, E in kN/m², I in m⁴), and a numerical example with $EI=20\,000$ kNm² is added. Axial deformations are ignored.

Step 1: Degree of indeterminacy and primary structure

Reactions = 3 + 3 = 6, equilibrium = 3, so $D_s=3$ . Release B. Redundants: $X_1=H_B$ (right +), $X_2=V_B$ (up +), $X_3=M_B$ (anticlockwise +). Primary structure: cantilever B-C-A fixed at A (which moves with the support).

Step 2: Displacement of B in the primary structure

(a) Due to the loads (using the moment diagrams of the primary structure): the 20 kN/m UDL on CB and the 30 kN force on AC give

\{\Delta_L\}=\frac{1}{EI}\begin{Bmatrix}733.33\\-2079.17\\-436.67\end{Bmatrix}

(b) Due to the movement of A. The primary structure moves as a rigid body with A: for A displaced $(u_A,v_A)=(-0.010,-0.010)$ m and rotated $\theta_A=-0.002$ rad (anticlockwise +), the point B at $(5,4)$ from A moves by

u_B=u_A-\theta_A\,y=-0.010+0.002\times4=-0.002\ \text{m},\quad v_B=v_A+\theta_A\,x=-0.010-0.002\times5=-0.020\ \text{m},\quad \theta_B=-0.002\ \text{rad}

\{\Delta_S\}=\{-0.002,\;-0.020,\;-0.002\}^T\ (\text{m, m, rad})

Step 3: Flexibility matrix of the primary structure (unit forces at B)

[F]=\frac{1}{EI}\begin{bmatrix}7.111&-13.333&-2.667\\-13.333&41.667&9.167\\-2.667&9.167&2.333\end{bmatrix}

(e.g. $f_{11}=\dfrac{1}{3EI}\displaystyle\int_0^4(4-y)^2dy=\dfrac{7.111}{EI}$ ; $f_{33}=\dfrac{4}{3EI}+\dfrac{5}{5EI}=\dfrac{2.333}{EI}$ .)

Step 4: Consistent deformation (B is fixed)

[F]\{X\}+\{\Delta_L\}+\{\Delta_S\}=\{0\}\quad\Rightarrow\quad \{X\}=\{X_L\}+EI\{x_S\}

\{X_L\}=\begin{Bmatrix}-19.286\ \text{kN}\\54.571\ \text{kN}\\-49.286\ \text{kNm}\end{Bmatrix},\qquad \{x_S\}=\begin{Bmatrix}0.003978\\0.004157\\-0.010929\end{Bmatrix}\ (\text{per unit }EI)

So $H_B=-19.29+0.00398\,EI$ kN, $V_B=54.57+0.00416\,EI$ kN and $M_B=-49.29-0.01093\,EI$ kNm. The settlement effects grow in proportion to EI.

Step 5: Results for the numerical example ( $EI=20\,000$ kNm²)

$H_B=60.27$ kN, $V_B=137.71$ kN, $M_B=-267.86$ kNm.

Bending moments (sagging/inside tension +, kNm):

Section	Loads only	Support movement (per unit EI)	Total, EI = 20 000
M at A	-9.29	-0.006054	-130.36
M at mid-height of AC	12.14	0.001902	50.18
M at C (column side)	-26.43	0.009857	170.71
M at C (beam side)	-26.43	0.009857	170.71
M at mid-span of CB	24.64	-0.000536	13.93
M at B	-49.29	-0.010929	-267.86

Shear and axial forces (EI = 20 000; axial: + tension):

Member	Axial N (+ tension)	Shear at start	Shear at end
AC	37.71	90.27	60.27
CB	60.27	-37.71	-137.71

The BMD is drawn from the table: AC has a straight-line moment with a kink at the 30 kN load, CB is parabolic (UDL) between the C and B values.

Answer: redundants at B: $H_B=-19.29+0.00398EI$ kN, $V_B=54.57+0.00416EI$ kN, $M_B=-49.29-0.01093EI$ kNm; for EI = 20 000 kNm²: $H_B=60.3$ kN, $V_B=137.7$ kN, $M_B=-267.9$ kNm.

2065 Shrawan · 8 marks

For the parabolic two hinged arch loaded symmetrically with concentrated loads as shown in the figure, determine the horizontal reaction. Use secant variation of moment of inertia. [Figure: two-hinged parabolic arch AB, span 8 x 10 m = 80 m, rise 5 m; vertical loads 4 kN, 10 kN, 8 kN, 10 kN, 8 kN, 10 kN, 4 kN symmetric about the crown.]

Answer

Method: a two-hinged arch is indeterminate to the first degree; the horizontal thrust $H$ is the redundant. Using the condition that the horizontal displacement of the support is zero (strain energy, bending only):

H=\frac{\int M_0\,y\,\dfrac{ds}{EI}}{\int y^2\,\dfrac{ds}{EI}}

With the secant variation $I=I_c\sec\theta$ , $ds/I=dx/I_c$ , so

H=\frac{\int_0^{L}M_0\,y\,dx}{\int_0^{L}y^2\,dx}

Data

$L=80$ m, rise $h=5$ m, parabolic axis $y=\dfrac{4hx(L-x)}{L^2}=\dfrac{20\,x(80-x)}{6400}$ . Loads (kN) at 10 m spacing: 4, 10, 8, 10, 8, 10, 4 at $x=10,20,\ldots,70$ m (total 54 kN).

Reactions

By symmetry $V_A=V_B=27$ kN.

Simple-beam moment $M_0$ and ordinate $y$

x (m)	y (m)	$M_0$ (kNm)	$M=M_0-Hy$ (kNm)
0	0.000	0.0	0.00
10	2.188	270.0	-18.26
20	3.750	500.0	5.83
30	4.688	630.0	12.29
40	5.000	680.0	21.11
50	4.688	630.0	12.29
60	3.750	500.0	5.83
70	2.188	270.0	-18.26
80	0.000	0.0	0.00

$M_0$ is linear between the loads, for example $M_0(10)=27\times10=270$ , $M_0(20)=27\times20-4\times10=500$ , $M_0(30)=27\times30-4\times20-10\times10=630$ , $M_0(40)=680$ .

Integrals

\int_0^{80}M_0\,y\,dx=140562.5\ \text{kN m}^3,\qquad \int_0^{80}y^2dx=\frac{8h^2L}{15}=\frac{8\times25\times80}{15}=1066.67\ \text{m}^3

H=\frac{140562.5}{1066.67}=131.78\ \text{kN}

The bending moment at the crown is $M_C=680-131.78\times5=21.11$ kNm (small compared with $M_0$ , which shows how effectively the thrust reduces the moments).

Answer: horizontal reaction $H=131.78$ kN at each support (inward); $V_A=V_B=27$ kN.

Past questions and answers

1.State and prove Castigliano's second theorem (theorem for determination of displacement in a structural system).

Statement

Proof

Use

2.Calculate the reaction at the prop of a propped cantilever with uniform distributed load throughout the span using Castigliano's theorem.

Other reactions

3.Derive the expression of the three moment theorem for a continuous beam and explain its physical meaning.

Statement

Derivation (supports at the same level, EI constant)

Special results

Physical meaning

4.Determine forces in all members of the truss shown in figure below using the force method. AE for all members is constant. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 50 kN horizontal at C; A and D supported.]

Reactions

Force tables (tension +; S0S_0S0​ with AC removed; nnn for X=1X=1X=1)

Compatibility

Final forces S=S0+nXS = S_0 + nXS=S0​+nX

Moment at the fixed support A

6.Determine the forces in all members of the truss shown in figure below by using the force method. AE is constant for all members. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 100 kN vertical at C and 50 kN horizontal at D; A hinged, B roller.]

Reactions

Force tables (tension +)

Compatibility

Final forces S=S0+nXS = S_0 + nXS=S0​+nX

7.Determine the moment at the fixed end of the propped cantilever beam shown in figure below using Castigliano's theorem. [Figure: beam, left end fixed, right end roller, span L = 5 m with UDL 15 kN/m.]

Moment at the fixed end

8.Use Castigliano's theorem to find the moment at point C of the propped cantilever beam loaded as shown in the figure below. Take EI to be constant. [Figure: beam A-B-D-C, A hinged, B support at 2 m, D at 4 m with 30 kNm moment, C fixed end at 6 m (2 m + 2 m + 2 m).]

Bending moments (x from A, sagging +)

Compatibility equations

Solution

Moment at the fixed end C

9.Use Castigliano's theorem and find the moment at the fixed end of the propped cantilever loaded as shown below. EI is constant. [Figure: beam AB, A fixed, B roller; 15 kN at 2 m from A, B at 5 m (2 m + 3 m).]

Bending moment (x measured from B towards A, sagging +)

Condition ∂U/∂RB=0\partial U/\partial R_B=0∂U/∂RB​=0

Fixed-end moment

10.Explain the physical significance of flexibility and stiffness matrices with suitable examples.

Flexibility matrix [F][F][F]

Stiffness matrix [K][K][K]

Example: cantilever of length LLL

11.In the beam ABC, support B settles by λRB\lambda R_BλRB​ units. Find the reaction RBR_BRB​ using Castigliano's theorem. [Figure: beam ABC, AB = BC = l, EI constant, hinge supports at A and C, support B on a spring, UDL of P/2lP/2lP/2l over the full length 2l.]

12.Determine reactions at A using the force method. [Figure: L-shaped frame; horizontal member AB of length l carrying 10 kN/m UDL, vertical member BC of height l, EI constant; A hinged, C fixed.]

Moment expressions

Flexibility coefficients

Load displacements

Compatibility (zero displacement at the hinge A)

Data and arch profile

Reactions of the simple beam

Free bending moment M0M_0M0​

Integrals

Bending moment at D (x=50x = 50x=50 m)

14.Use the consistent deformation method to draw the BM diagram of the given frame. [Figure: frame with column AB (A fixed at the bottom, B at top, height 4 m) carrying 10 kN/m UDL along its height, beam BCD with BC = CD = 3 m, 20 kN downward at C, EI constant, D supported at the right end.]

Step 1: Primary structure (D released), loads only

Step 2: Deformations by the unit-load method

Step 3: Consistent deformation (deflection at D is zero)

Step 4: Other reactions (statics)

Step 5: Bending moments M=M0+RD mM = M_0 + R_D\, mM=M0​+RD​m

15.Find the reaction at support B using Castigliano's method. [Figure: beam ABC with 20 kN/m UDL over its full length; AB = 6 m, BC = 10 m; A hinged, B roller.]

Reactions in terms of RBR_BRB​

Bending moments (x from A)

Castigliano's theorem

Other reactions

16.Determine forces in all members of the truss shown in the figure if the member AC is too long by 10 mm. Take AE constant for all members. [Figure: rectangular truss ABCD, 4 m wide and 3 m high, with both diagonals; A hinged, D roller.]

Unit force system (X=1X = 1X=1 in AC)

Compatibility

Member forces S=nXS = n XS=nX (AE in kN)

17.Determine reactions at the hinged support of the frame loaded as shown in the figure using the force method. [Figure: frame ABCD, beam BC of span 10 m with I = 2I carrying 50 kN/m UDL; column AB 5 m long with 1.5I, A fixed; column CD 3 m long with I, D hinged.]

Moment diagrams (sign: + tension on the inner face of the frame)

Flexibility coefficients and load terms

Compatibility (ΔD,x=ΔD,y=0\Delta_{D,x} = \Delta_{D,y} = 0ΔD,x​=ΔD,y​=0)

Final bending moments M=M0+X1m1+X2m2M = M_0 + X_1 m_1 + X_2 m_2M=M0​+X1​m1​+X2​m2​ (kNm)

Reactions

Data (converted to kN and m)

Integrals

Horizontal thrust

Bending moment M=M0−HyM = M_0 - H yM=M0​−Hy

19.Analyze the given loaded beam using Castigliano's theorem. The redundant should be considered as the support moment at A. EI constant. [Figure: beam ABCD, A fixed; 10 kNm moment applied at B (2 m from A); support at C, 2 m from B; 6 kN/m UDL on overhang CD of 1 m.]

Reaction at A in terms of MAM_AMA​

Bending moments (x from A, sagging positive)

Castigliano's theorem

Other reactions

Bending moments (kNm)

State and prove Castigliano's second theorem (theorem for determination of displacement in a structural system).

Calculate the reaction at the prop of a propped cantilever with uniform distributed load throughout the span using Castigliano's theorem.

Derive the expression of the three moment theorem for a continuous beam and explain its physical meaning.

Determine forces in all members of the truss shown in figure below using the force method. AE for all members is constant. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 50 kN horizontal at C; A and D supported.]

Force tables (tension +; $S_0$ with AC removed; $n$ for $X=1$ )

Final forces $S = S_0 + nX$

Determine the forces in all members of the truss shown in figure below by using the force method. AE is constant for all members. [Figure: rectangular truss ABCD with both diagonals, 4 m wide and 3 m high; 100 kN vertical at C and 50 kN horizontal at D; A hinged, B roller.]

Final forces $S = S_0 + nX$

Determine the moment at the fixed end of the propped cantilever beam shown in figure below using Castigliano's theorem. [Figure: beam, left end fixed, right end roller, span L = 5 m with UDL 15 kN/m.]

Use Castigliano's theorem to find the moment at point C of the propped cantilever beam loaded as shown in the figure below. Take EI to be constant. [Figure: beam A-B-D-C, A hinged, B support at 2 m, D at 4 m with 30 kNm moment, C fixed end at 6 m (2 m + 2 m + 2 m).]

Use Castigliano's theorem and find the moment at the fixed end of the propped cantilever loaded as shown below. EI is constant. [Figure: beam AB, A fixed, B roller; 15 kN at 2 m from A, B at 5 m (2 m + 3 m).]

Condition $\partial U/\partial R_B=0$

Explain the physical significance of flexibility and stiffness matrices with suitable examples.

Flexibility matrix $[F]$

Stiffness matrix $[K]$

Example: cantilever of length $L$

In the beam ABC, support B settles by $\lambda R_B$ units. Find the reaction $R_B$ using Castigliano's theorem. [Figure: beam ABC, AB = BC = l, EI constant, hinge supports at A and C, support B on a spring, UDL of $P/2l$ over the full length 2l.]

Determine reactions at A using the force method. [Figure: L-shaped frame; horizontal member AB of length l carrying 10 kN/m UDL, vertical member BC of height l, EI constant; A hinged, C fixed.]

Free bending moment $M_0$

Bending moment at D ( $x = 50$ m)

Use the consistent deformation method to draw the BM diagram of the given frame. [Figure: frame with column AB (A fixed at the bottom, B at top, height 4 m) carrying 10 kN/m UDL along its height, beam BCD with BC = CD = 3 m, 20 kN downward at C, EI constant, D supported at the right end.]

Step 5: Bending moments $M = M_0 + R_D\, m$

Find the reaction at support B using Castigliano's method. [Figure: beam ABC with 20 kN/m UDL over its full length; AB = 6 m, BC = 10 m; A hinged, B roller.]

Reactions in terms of $R_B$

Determine forces in all members of the truss shown in the figure if the member AC is too long by 10 mm. Take AE constant for all members. [Figure: rectangular truss ABCD, 4 m wide and 3 m high, with both diagonals; A hinged, D roller.]

Unit force system ( $X = 1$ in AC)

Member forces $S = n X$ (AE in kN)

Determine reactions at the hinged support of the frame loaded as shown in the figure using the force method. [Figure: frame ABCD, beam BC of span 10 m with I = 2I carrying 50 kN/m UDL; column AB 5 m long with 1.5I, A fixed; column CD 3 m long with I, D hinged.]

Compatibility ( $\Delta_{D,x} = \Delta_{D,y} = 0$ )

Final bending moments $M = M_0 + X_1 m_1 + X_2 m_2$ (kNm)

Bending moment $M = M_0 - H y$

Analyze the given loaded beam using Castigliano's theorem. The redundant should be considered as the support moment at A. EI constant. [Figure: beam ABCD, A fixed; 10 kNm moment applied at B (2 m from A); support at C, 2 m from B; 6 kN/m UDL on overhang CD of 1 m.]

Reaction at A in terms of $M_A$

Force $n$ due to $X = 1$ in CF

Member forces $S = nX$

Draw shear force and bending moment diagram of the continuous beam using the three moment equation. [Figure: beam ABC, A fixed; 300 kN point load at 4 m from A; B support at 8 m; BC = 10 m with 50 kN/m UDL; C roller.]

Explain theorems on displacement with suitable illustration.

Find the reaction at support 'C' using Castigliano's theorem. [Figure: L-shaped frame; horizontal member BC of L m carrying W kN/m UDL, C on a roller; vertical member AB of L m, A fixed; EI constant.]

Bending moments (take $R_C$ upward)

Flexibility coefficients (in $1/EI$ units)

Final forces $S = S_0 + nX$ (kN)

Enunciate Betti's law and Maxwell's reciprocal theorem and explain their uses.

Force tables (tension +; $S_0$ = loads on the primary, $n_1$ , $n_2$ = unit redundants)

Flexibility coefficients (in $1/AE$ )

Determine the moment at the fixed support and the rotation at the roller support of a propped cantilever beam of span 10 m loaded with a uniformly distributed load of 30 kN/m on its whole span and a point load of 50 kN at the centre using Castigliano's theorem.

Write down the compatibility equation for a two hinged parabolic arch due to external loads, variation in temperature, rib shortening and yielding of supports.

Flexibility coefficients (in $1/EI$ units)

Determine the support reaction at support 'C' using Castigliano's theorem. EI = constant throughout. [Figure: beam ABCD, A fixed, a moment applied at B, 2 m from A; support C at 2 m from B; 6 kN/m UDL on CD of 1 m overhang.]