Analysis by the Virtual Work Method

The virtual work (unit load) method finds the displacement of a chosen point by applying a unit virtual load at that point, in the direction of the required displacement, on the same structure.

Steps

1. Analyse the structure under the real loads and find the internal forces: $M$ (bending moment), $N$, $V$ (for trusses, the member forces $F$).
2. Remove the real loads and apply a unit load (or unit couple for rotation) at the point and in the direction of the required displacement.
3. Analyse the structure under this unit load and find the virtual internal forces $m$ (and $n$ or $k$ for a truss).
4. Write the virtual work equation: external virtual work = internal virtual work.
5. Integrate (or sum) over all members to get the displacement.

For beams and frames (bending only):

$$1\cdot\Delta = \int \frac{M\,m}{EI}\,dx$$

For trusses:

$$1\cdot\Delta = \sum \frac{F\,k\,L}{AE}$$

6. A positive result means the displacement is in the direction of the unit load; a negative result means the opposite.

Example

Find the deflection at the free end of a cantilever of length $L$ with a point load $P$ at that end.

• Real moment: $M = -Px$ (x from the free end).
• Apply a unit load at the free end: $m = -1\cdot x$.
• Then

$$\Delta = \int_0^L \frac{(-Px)(-x)}{EI}dx = \frac{PL^3}{3EI}$$

The sign is positive, so the deflection is downward, in the direction of the unit load.

Other effects (temperature, lack of fit, support settlement) are included by using the virtual forces with the corresponding deformations: $\Delta = \sum k\,\alpha\,\Delta T\,L$ for temperature and $\sum k\,e$ for lack of fit.

Given data and assumptions

A(0,0) hinge, D(4,0), B(8,0) roller; top joints E(2, 3.464) and C(6, 3.464); all members 4 m long (60$^\circ$). Loads: 15 kN at E and 30 kN at C, downward. $E = 200\times10^3$ N/mm$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Member area is not given; it is assumed the same for all members, $A = 1000$ mm$^2 = 10^{-3}$ m$^2$ (the loading part scales as $1/A$; the other parts do not depend on $A$). A unit downward load is applied at D.

Reactions: moments about A: $B_y\times8 = 15\times2 + 30\times6 = 210 \Rightarrow B_y = 26.25$ kN, $A_y = 18.75$ kN.

Member forces (tension +)

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{180.000}{10^{-3}\times2\times10^8} = 0.900\ \text{mm}\ (\downarrow)$$

(ii) DE and DC each 5 mm too long

$$\Delta_2 = \sum k\,e = (0.577)(5) + (0.577)(5) = 5.774\ \text{mm}\ (\downarrow)$$

(iii) Temperature rise of 20$^\circ$C in CD

$$\Delta L = 12\times10^{-6}\times20\times4000 = 0.96\ \text{mm},\qquad \Delta_3 = k\,\Delta L = 0.577\times0.96 = 0.554\ \text{mm}\ (\downarrow)$$

Total

$$\Delta_D = 0.900 + 5.774 + 0.554 = 7.228\ \text{mm}$$

Answer: vertical deflection of D = 7.23 mm downward (for $A = 1000$ mm$^2$; the load part scales as $1/A$).

Given data

A(0,0) hinge, E(4,0), B(8,0) roller; top joints C(2, 3.464) and D(6, 3.464); all members 4 m (60$^\circ$). 20 kN down at C. $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $A = 1000$ mm$^2 = 10^{-3}$ m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Loads: unit downward load at E.

Reactions (real): $\sum M_B = 0$: $A_y\times8 = 20\times6 \Rightarrow A_y = 15$ kN, $B_y = 5$ kN.

Member forces (tension +)

(i) Effect of the 20 kN load shown in the figure

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{80.000}{10^{-3}\times2\times10^8} = 0.400\ \text{mm}\ (\downarrow)$$

(ii) Temperature rise of 20$^\circ$C in CD (top chord)

$$\Delta L_{CD} = 12\times10^{-6}\times20\times4000 = 0.96\ \text{mm},\qquad \Delta_2 = k_{CD}\,\Delta L = (-0.577)(0.96) = -0.554\ \text{mm}\ (\uparrow)$$

(iii) CE 5 mm too long

$$\Delta_3 = k_{CE}\,e = (0.577)(5) = 2.887\ \text{mm}\ (\downarrow)$$

Total

$$\Delta_E = 0.400 - 0.554 + 2.887 = 2.732\ \text{mm}$$

Answer: vertical deflection of E = 2.73 mm downward (including the 20 kN load; temperature and lack of fit alone give 2.33 mm downward).

Real work method

The real work method equates the external work done by the actual loads (applied gradually) to the strain energy stored in the structure:

$$W_e = U \;\Rightarrow\; \tfrac12 P\,\Delta = \int \frac{M^2}{2EI}\,dx$$

The work is "real" because the loads move through their own actual displacements. Example: a cantilever with end load $P$: $\tfrac12P\Delta = \dfrac{P^2L^3}{6EI}$, so $\Delta = \dfrac{PL^3}{3EI}$.

 P |    /|
   |  /  |  Real work = area = 1/2 P*D
   |/____|
   +-----+--- Displacement
        D

Virtual work method

In the virtual work method, an imaginary (virtual) force system, usually a unit load at the point where the displacement is wanted, is applied. The external virtual work of the virtual forces moving through the real displacements equals the internal virtual work of the virtual internal forces on the real deformations:

$$1\cdot\Delta = \int \frac{M\,m}{EI}\,dx \qquad\text{(beams, frames)}$$ $$1\cdot\Delta = \sum \frac{F\,k\,L}{AE} \qquad\text{(trusses)}$$

Here the virtual load is constant while the real displacement occurs, so the virtual work is $1\times\Delta$ (no factor of $\tfrac12$).

 Real loads          Virtual load
 P1  P2              1 (at the point)
 v   v               v
 ______________      ______________
 -> M, F             -> m, k

Limitations of the real work method

1. Only one load can act on the structure; with several loads the energy equation has several unknown displacements.
2. It gives the displacement only at the point of application of the load, and only in the direction of the load.
3. It cannot give the displacement of a point where no load acts, or a rotation, unless a fictitious load is added.
4. It cannot account for temperature change, lack of fit or support settlement.
5. It is applicable only to linearly elastic structures.

Given data

A(0,0) hinge, E(5,0), B(10,0) roller; top joints D(2.5, 4.330) and C(7.5, 4.330); all members 5 m long (equilateral triangles). 80 kN downward at C. $E = 200\times10^3$ N/mm$^2 = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Areas: horizontal members AE, EB, DC = 2000 mm$^2$; inclined members = 1000 mm$^2$. Apply a unit load (down) at E.

Reactions (moments about A): $B_y\times10 = 80\times7.5 \Rightarrow B_y = 60$ kN, $A_y = 20$ kN. For the unit load at E: $A_y = B_y = 0.5$.

Member forces (tension +)

(i) Loading

Because the areas differ, use $\sum kFL/A$ with $A$ in each member:

$$\Delta_1 = \frac{1}{E}\sum\frac{kFL}{A} = 1.667\ \text{mm}\ (\downarrow)$$

(ii) Temperature fall of 20$^\circ$C in CD

$$\Delta L_{CD} = 12\times10^{-6}\times(-20)\times5000 = -1.2\ \text{mm},\qquad \Delta_2 = k\,\Delta L = (-0.577)(-1.2) = 0.693\ \text{mm}\ (\downarrow)$$

(iii) CE 8 mm too long

$$\Delta_3 = k_{CE}\,e = (0.577)(8) = 4.619\ \text{mm}\ (\downarrow)$$

Total

$$\Delta_E = 1.667 + 0.693 + 4.619 = 6.978\ \text{mm}$$

Answer: vertical deflection of joint E = 6.98 mm downward.

Given data and assumed geometry

From the figure: top chord joints D(0,6), E(5,6), F(10,6), G(15,6), H(20,6); bottom joints A(5,0), B(10,0), C(15,0). Members: top chord DE, EF, FG, GH; bottom chord AB, BC; verticals EA, FB, GC; diagonals DA, EB, GB, HC (13 members, 8 joints, 3 reactions: determinate). A hinge, C roller. Loads: 50 kN down at F, 30 kN up and 20 kN to the right at H. $A = 1500$ mm$^2 = 1.5\times10^{-3}$ m$^2$, $E = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Unit vertical load (down) at B.

Member forces (tension +); $F$ real, $k$ unit load at B

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{578.073}{1.5\times10^{-3}\times2\times10^8} = 1.927\ \text{mm}\ (\downarrow)$$

(ii) Temperature rise of 20$^\circ$C in the top chord

$$\Delta_2 = \sum k\,\alpha\,\Delta T\,L = (-0.417)(12\times10^{-6})(20)(5000)\times2 = -1.000\ \text{mm}$$

(only EF and FG have non-zero $k$).

(iii) Vertical members 10 mm too long

$$\Delta_3 = \sum k\,e = (-0.5)(10) + (0)(10) + (-0.5)(10) = -10.000\ \text{mm}$$

Total

$$\Delta_B = 1.927 - 1.000 - 10.000 = -9.073\ \text{mm}$$

Answer: vertical deflection of joint B = 9.07 mm upward (net). Loading gives 1.93 mm down, temperature 1.00 mm up and lack of fit 10.00 mm up.

Given data

A(0,0) hinge, E(2,0), B(4,0) roller; top joints C(1, 1.732) and D(3, 1.732); all members 2 m (60$^\circ$). 100 kN down at E. $A = 1000$ mm$^2 = 10^{-3}$ m$^2$, $E = 2\times10^5$ N/mm$^2 = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Unit vertical load (down) at E.

Member forces (tension +)

Reactions: $A_y = B_y = 50$ kN.

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{366.667}{10^{-3}\times2\times10^8} = 1.833\ \text{mm}\ (\downarrow)$$

(ii) CE and DE 8 mm too long

$$\Delta_2 = (0.577)(8) + (0.577)(8) = 9.238\ \text{mm}\ (\downarrow)$$

(iii) Temperature of CD decreased by 15$^\circ$C

$$\Delta L_{CD} = 12\times10^{-6}\times(-15)\times2000 = -0.36\ \text{mm},\qquad \Delta_3 = k_{CD}\Delta L = (-0.577)(-0.36) = 0.208\ \text{mm}\ (\downarrow)$$

Total

$$\Delta_E = 1.833 + 9.238 + 0.208 = 11.279\ \text{mm}$$

Answer: vertical deflection of E = 11.28 mm downward.

Given data and assumed geometry

Top chord joints C(0,4), D(3,4), E(6,4), F(9,4), G(12,4); bottom joints A(3,0), H(6,0), B(9,0). Members: top chord CD, DE, EF, FG; bottom chord AH, HB; verticals DA, EH, FB; diagonals CA, DH, FH, GB. A hinge, B roller. 20 kN down at E. $A = 1500$ mm$^2 = 1.5\times10^{-3}$ m$^2$, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Unit vertical load (down) at H.

Member forces (tension +)

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{135.000}{1.5\times10^{-3}\times2\times10^8} = 0.450\ \text{mm}\ (\downarrow)$$

(ii) Top chord temperature rise 20$^\circ$C

$$\Delta_2 = \sum k\alpha\Delta T L = 2\times(-0.375)(12\times10^{-6})(20)(3000) = -0.540\ \text{mm}$$

(DE and EF have $k = -0.375$; CD and FG have $k = 0$.)

(iii) All vertical members 10 mm too long

$$\Delta_3 = (-0.5)(10) + (0)(10) + (-0.5)(10) = -10.000\ \text{mm}$$

Total

$$\Delta_H = 0.450 - 0.540 - 10.000 = -10.090\ \text{mm}$$

Answer: vertical deflection of joint H = 10.09 mm upward (net).

Principle of virtual work (for deformable bodies): If a deformable structure in equilibrium under a real load system is given a virtual displacement field (compatible, small, and imaginary), the external virtual work done by the real forces equals the internal virtual work done by the real internal stresses on the virtual strains. Using the dual form (force virtual work), virtual loads are applied and real displacements occur.

Derivation (unit load form)

Consider a structure under real loads $P_1, P_2,\dots$ producing real internal forces and real deformations. To find the displacement $\Delta$ at a point, apply a virtual unit load at that point in the required direction. Let this virtual load produce internal forces: moment $m$, axial force $n$, shear $v$.

1. External virtual work. The unit load, kept constant, moves through the real displacement $\Delta$:

$$W_e = 1\times\Delta$$

2. Internal virtual work. In an element of length $dx$, the virtual internal forces move through the real deformations of the element. The real deformations are

$$d\theta = \frac{M\,dx}{EI},\qquad d\delta_{axial} = \frac{N\,dx}{AE},\qquad d\gamma = \frac{\alpha V\,dx}{GA}$$

so the internal virtual work of the element is

$$dW_i = m\,d\theta + n\,d\delta_{axial} + v\,d\gamma = \frac{mM}{EI}dx + \frac{nN}{AE}dx + \frac{\alpha vV}{GA}dx$$

3. Equate $W_e = W_i$ (the virtual force system is in equilibrium, and the real deformations are compatible):

$$1\cdot\Delta = \int \frac{mM}{EI}dx + \int\frac{nN}{AE}dx + \int\frac{\alpha vV}{GA}dx$$

For beams and frames, only the first term is generally kept. For trusses, $\Delta = \sum \dfrac{kFL}{AE}$.

The relation holds for any material behaviour, as long as the real deformations are small and compatible, and the virtual forces satisfy equilibrium.

Given data

Overhanging beam: span $AB = 5$ m (A hinge, B roller), overhang $BC = 2$ m, load $P = 20$ kN at C. Section 100 mm $\times$ 120 mm ($b = 0.1$ m, $d = 0.12$ m). $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $G = 80$ kN/mm$^2 = 8\times10^7$ kN/m$^2$.

        A            B      C
        ^------------^------+  20 kN
        |    5 m     | 2 m  v

$$I = \frac{0.1\times0.12^3}{12} = 1.44\times10^{-5}\ \text{m}^4,\quad EI = 2880\ \text{kNm}^2,\quad A = 0.012\ \text{m}^2$$

Reactions

$$R_A\times5 = 20\times2 \Rightarrow R_A = 8\ \text{kN (down)},\qquad R_B = 28\ \text{kN (up)}$$

Bending moments and shear forces (magnitudes)

Strain energy due to bending

$$U_b = \int_0^5\frac{(8x)^2}{2EI}dx + \int_0^2\frac{(20s)^2}{2EI}dx = \frac{1}{2EI}\left[\frac{64\times125}{3} + \frac{400\times8}{3}\right] = \frac{3733.33}{2\times2880}$$ $$U_b = 0.6481\ \text{kNm} = 648.1\ \text{J}$$

Strain energy due to shear

For a rectangle the form factor is $\alpha = 1.2$:

$$U_s = \frac{\alpha}{2GA}\left[8^2\times5 + 20^2\times2\right] = \frac{1.2\times1120}{2\times8\times10^7\times0.012} = 7.0\times10^{-4}\ \text{kNm} = 0.70\ \text{J}$$

Deflection at C (real work method)

External work = total strain energy:

$$\tfrac12 P\Delta_C = U_b + U_s = 0.6481 + 0.0007 = 0.6488\ \text{kNm}$$ $$\Delta_C = \frac{2\times0.6488}{20} = 0.06488\ \text{m}$$

Answer: $U_b = 0.648$ kNm (648 J), $U_s = 0.0007$ kNm (0.70 J), and $\Delta_C = 64.9$ mm downward (bending alone gives 64.81 mm; shear adds only 0.07 mm). Shear energy is about 0.1% of the bending energy.

Strain energy is the energy stored in a deformed elastic body, equal to the work done by the external loads when they are applied gradually: $U = \tfrac12 P\Delta$. It is recovered on unloading within the elastic limit.

Real work method for beam deflection

The method uses the principle of conservation of energy: the external work done by the load equals the internal strain energy of the beam.

1. Find the bending moment $M$ along the beam due to the single load $P$.
2. Strain energy: $U = \displaystyle\int \frac{M^2}{2EI}\,dx$ (shear energy is usually neglected).
3. External work: $W = \tfrac12 P\,\Delta$, where $\Delta$ is the deflection at the load point.
4. Equate $W = U$ and solve for $\Delta$.

Example. Cantilever of length $L$ with a load $P$ at the free end.

$$M = Px,\qquad U = \int_0^L \frac{P^2x^2}{2EI}dx = \frac{P^2L^3}{6EI}$$ $$\tfrac12 P\Delta = \frac{P^2L^3}{6EI} \;\Rightarrow\; \Delta = \frac{PL^3}{3EI}$$

The method gives the deflection only under a single load, at its point of application, in the direction of the load.

Given data and method

$A = 600$ mm$^2 = 6\times10^{-4}$ m$^2$, $E = 200$ GPa $= 2\times10^8$ kN/m$^2$, $\alpha = 19\times10^{-6}$/$^\circ$C. Joints: A(0,0), B(3,0), C(6,0), D(0,4), E(3,4), F(6,4); members AB, BC, DE, EF, AD, BE, CF, DB, BF. A hinge, C roller; 80 kN down at D and F. Apply a unit vertical load at B (downward) and use the virtual work method. Tension is positive.

Support reactions

By symmetry $A_y = C_y = 80$ kN. The 80 kN loads act directly over the supports, so they go straight down the verticals AD and CF ($F = -80$ kN, compression); all other members have zero force.

Forces in members ($F$ = real loading, $k$ = unit load at B)

(Real force $F$ is non-zero only in AD and CF.)

(i) External loading

$$\Delta_{B1} = \sum\frac{kFL}{AE} = \frac{320.00}{6\times10^{-4}\times2\times10^8} = 2.667\ \text{mm}\ (\downarrow)$$

(ii) Temperature fall of 15$^\circ$C in the inclined members DB and BF

Change in length of each: $\alpha\,\Delta T\,L = 19\times10^{-6}\times(-15)\times5 = -1.425$ mm.

$$\Delta_{B2} = \sum k\,\alpha\Delta T L = 2\times(0.625)(-1.425) = -1.781\ \text{mm}$$

(negative: joint B moves up.)

(iii) Lack of fit: vertical members 8 mm too long

$$\Delta_{B3} = \sum k\,e = (-0.5)(8) + (0)(8) + (-0.5)(8) = -8.000\ \text{mm}$$

(BE has $k = 0$.)

Total

$$\Delta_B = 2.667 - 1.781 - 8.000 = -7.115\ \text{mm}$$

Answer: (i) 2.67 mm down, (ii) 1.78 mm up, (iii) 8.00 mm up. The net vertical deflection of B is 7.11 mm upward.

Given data

Portal frame (EI constant): A(0,0) hinge, B(0,2), C(0,4), D(4,4), E(4,2), F(4,0) roller. 20 kN horizontal at B (to the right); UDL 10 kN/m on CD (down). Required: horizontal deflection at E using the unit load method.

   C ============== D
   |   10 kN/m      |
   |                |
 20->B              E  <- unit load (virtual)
   |                |
   A (hinge)        F (roller)

Real loading: reactions

$\sum F_x = 0$: $A_x = 20$ kN (left). $\sum M_A = 0$: $F_y\times4 = 20\times2 + 40\times2 \Rightarrow F_y = 30$ kN; $A_y = 40 - 30 = 10$ kN.

Virtual load: unit horizontal load at E (to the right)

$A_x = 1$ (left). Moments about A: $F_y\times4 = 1\times2$, so $F_y = 0.5$ (up) and $A_y = 0.5$ (down).

Bending moments ($s$ from the first end of each member, in the order A-B-C-D-E-F)

(The right column has no moment because the roller gives no horizontal force.)

Deflection

$$\Delta_{Eh} = \frac{1}{EI}\sum\int Mm\,ds = \frac{53.33 + 240 + 426.67}{EI} = \frac{720}{EI}$$

Answer: horizontal deflection of E $= \dfrac{720}{EI}$ kNm$^3$/EI, towards the right (direction of the unit load).

Given data

Frame with $A(0,0)$ hinge, $B(0,4)$ (15 kN to the right), $C(0,5)$, $D(5,5)$, $E(5,2)$ roller. Left column $AC = 5$ m ($EI$), beam $CD = 5$ m ($2EI$) with UDL 25 kN/m, right column $DE = 3$ m ($EI$). Required: horizontal displacement and rotation at the roller E.

Real loading reactions

$$\sum M_A = 0:\; E_y\times5 = 15\times4 + 125\times2.5 \Rightarrow E_y = 74.5\ \text{kN},\quad A_y = 125 - 74.5 = 50.5\ \text{kN},\quad A_x = 15\ \text{kN (left)}$$

Virtual cases

• Case 1: unit horizontal load at E (to the right): $A_x = 1$, $E_y = 0.4$ (up), $A_y = 0.4$ (down).
• Case 2: unit couple (anticlockwise) at E: $E_y = 0.2$ (down), $A_y = 0.2$ (up), $A_x = 0$.

Bending moments ($s$ along the member, A to B to C to D to E)

Integrals ($\int Mm\,ds/EI_{member}$, in units of $1/EI$)

Results

$$\Delta_{Eh} = \frac{1435.83}{EI}\ \text{(to the right)},\qquad \theta_E = \frac{90.10}{EI}\ \text{(anticlockwise)}$$

Answer: horizontal displacement of E $= 1435.8/EI$ (m, with $EI$ in kNm$^2$) to the right; rotation at E $= 90.10/EI$ rad anticlockwise. Taking AB as 4 m, BC as 1 m (as dimensioned in the figure), the right column as 3 m and E at the roller level 2 m above A.

Given data and assumptions

Square panel truss: A(0,0), D(5,0), B(0,5), C(5,5); members AB, BC, CD, AD and diagonal BD. A is a hinge and D a roller. Loads at C: 50 kN downward and 100 kN horizontal to the right (away from the frame). $AE$ constant. Required: horizontal displacement of C.

Step 1: Real forces (tension +)

Reactions: $\sum M_A = 0$: $D_y\times5 = 50\times5 + 100\times5 \Rightarrow D_y = 150$ kN (up); $A_y = -100$ kN (down); $A_x = -100$ kN.

Joint C: CD carries the 50 kN vertical load: $F_{CD} = -50$ kN. The horizontal 100 kN is taken by BC: $F_{BC} = +100$ kN. Joint B: $F_{BD}\cos45^\circ = -F_{BC}\Rightarrow F_{BD} = -141.42$ kN; $F_{AB} = -F_{BD}\sin45^\circ = +100$ kN. Joint A / D: $F_{AD} = +100$ kN.

Step 2: Virtual forces (unit horizontal load at C to the right)

$k_{BC} = 1$, $k_{BD} = -\sqrt2 = -1.414$, $k_{AB} = 1$, $k_{AD} = 1$, $k_{CD} = 0$.

Step 3: Table

$$\Delta_{Ch} = \sum\frac{kFL}{AE} = \frac{2914.2}{AE}$$

Answer: horizontal displacement of C $= \dfrac{2914.2\ \text{kNm}}{AE}$, towards the right (direction of the unit load).

Given data

Frame fixed at the base A (0,0). Left column AB: 4 m high, $2EI$, with 50 kN horizontal (to the right) at 2 m above the base. Beam BC: 6 m, $1.5EI$, UDL 20 kN/m (down). Right column CD: 3 m, $EI$, hanging down from C to the free end D, where a 30 kN horizontal load acts towards the left. Required: horizontal deflection at the free end D by virtual work.

    B =========== C        UDL 20 kN/m on BC
    |             |
 50->| (2 m)      | 3 m
    |             D <- 30 kN (free end)
    A (fixed)

Virtual load

Unit horizontal load at D, to the left (same direction as the 30 kN). Distances: $s$ measured from D along DC, from C along CB, and from B along BA. Since the structure is cantilevered from A, no reactions are needed to find $M$ and $m$ (moments are taken on the free-end side).

The column is split at the 50 kN load, where the expression for $M$ changes.

Integrals

$$\Delta_{Dh} = \frac{3633.33}{EI}$$

Answer: horizontal deflection of the free end $= 3633.3/EI$ (kNm$^3$/EI), towards the left (direction of the unit load and the 30 kN force). Assumed: 50 kN to the right and 30 kN to the left.

Given data

Portal frame: A(0,0) hinge, B(0,3), C(3,3), D(3,0) roller. Beam BC: $I$, 3 m, with 100 kN at 1 m from B. Columns AB and CD: $2I$, 3 m. UDL 30 kN/m on AB acting to the right, and 40 kN horizontal at B to the right. Required: horizontal deflection of the roller D by virtual work.

Real loading: reactions

Horizontal: $A_x = 40 + 30\times3 = 130$ kN (to the left).

Moments about A (clockwise positive): the horizontal loads give $40\times3 + 90\times1.5 = 255$ kNm clockwise; the 100 kN load gives $100\times1 = 100$ kNm clockwise; the roller reaction $D_y\times3$ acts anticlockwise.

$$D_y = \frac{255 + 100}{3} = 118.33\ \text{kN (up)},\qquad A_y = 100 - 118.33 = -18.33\ \text{kN (i.e. 18.33 kN down)}$$

Virtual load

Unit horizontal load at D (to the right). The hinge A resists it with $A_x = 1$; the vertical reactions are zero. Hence $m = s$ in AB, $m = 3$ in BC and $m = 3 - s$ in CD (s measured from the start of each member).

Moments and integrals

Check: at C, $M = 355 - 118.33\times3 = 0$, as it must be at the roller column (no moment in CD).

$$\Delta_{Dh} = \frac{433.13 + 737.50 + 710.00}{EI} = \frac{1880.6}{EI}$$

Answer: horizontal deflection of the roller D $= 1880.6/EI$ (kNm$^3$/EI), to the right, where $EI$ is the stiffness of the beam BC (columns $2EI$).

Given data and assumptions

Frame fixed at A: left column AB 6 m, top member BC 5 m, right column CD hanging 3 m (D free). Temperature: $+20^\circ$C on the outer face and $+10^\circ$C on the inner face of all members (assumed uniform for each member). Depth $d = 0.20$ m, $\alpha = 12\times10^{-6}$/$^\circ$C.

Principle

In each member, the temperature varies linearly across the depth:

• Mean temperature rise $T_m = \dfrac{20 + 10}{2} = 15^\circ$C, which causes an axial strain $\varepsilon = \alpha T_m = 1.8\times10^{-4}$ (elongation of $\alpha T_m\,ds$ for each element).
• Temperature difference $\Delta T = 20 - 10 = 10^\circ$C causes a curvature $\kappa = \dfrac{\alpha\,\Delta T}{d} = \dfrac{12\times10^{-6}\times10}{0.2} = 6\times10^{-4}$ per m, with the hotter (outer) fibres longer.

By the virtual work (unit load) method, with virtual axial force $n$ and moment $m$ from a unit load at D:

$$\Delta = \int n\,\alpha T_m\,ds + \int m\,\frac{\alpha\,\Delta T}{d}\,ds$$

(Equivalent geometric summation: each element's curvature rotates the rest of the frame about it, and each element's axial elongation moves D along the member.)

Evaluation (outer face hotter, in every member)

Mean-temperature (axial) effects on D: AB (6 m) lifts D by $1.8\times10^{-4}\times6000 = +1.08$ mm; BC (5 m) moves D to the right by $0.90$ mm; CD (3 m) lowers D by $0.54$ mm. Net: $\Delta_h = +0.90$ mm, $\Delta_v = +0.54$ mm.

Curvature effects (each element rotates the rest of the frame about itself by $\kappa\,ds$): integrating over AB, BC and CD gives $\Delta_h = -11.70$ mm and $\Delta_v = -25.50$ mm (left and down).

Total:

$$\Delta_h = 0.90 - 11.70 = -10.80\ \text{mm},\qquad \Delta_v = 0.54 - 25.50 = -24.96\ \text{mm}$$

Answer: horizontal deflection of D $= 10.8$ mm towards the left; vertical deflection $= 24.96$ mm downward. (Results assume the 20$^\circ$C face is the outside of the frame for every member.)

Given data and assumed geometry

Wall supports at C (0,0) and D (0,8) (both pinned); bottom chord C-B-A with $CB = BA = 8$ m (B at (8,0), A at (16,0)); E is the top middle joint at (8,4) on the sloping top chord D-E-A. Members: CB, BA, DE, EA, EC, EB. Loads: 8 kN at B and 4 kN at A (downward). $A = 1500$ mm$^2 = 1.5\times10^{-3}$ m$^2$, $E = 200$ kN/mm$^2 = 2\times10^8$ kN/m$^2$, $\alpha = 12\times10^{-6}$/$^\circ$C. Unit vertical load (down) at E.

Lengths: $DE = EA = EC = \sqrt{8^2+4^2} = 8.944$ m, $EB = 4$ m.

Member forces (tension +)

(i) Loading

$$\Delta_1 = \frac{\sum kFL}{AE} = \frac{268.328}{1.5\times10^{-3}\times2\times10^8} = 0.894\ \text{mm}\ (\downarrow)$$

(ii) Temperature rise of 30$^\circ$C in the top chord (DE and EA)

$$\Delta_2 = \sum k\alpha\Delta T L = (1.118)(12\times10^{-6})(30)(8.944) = 3.600\ \text{mm}\ (\downarrow)$$

(iii) AE and EC fabricated 5 mm too long

$$\Delta_3 = \sum k\,e = (0)(5) + (-1.118)(5) = -5.590\ \text{mm}\ (\text{up})$$

Total

$$\Delta_E = 0.894 + 3.600 - 5.590 = -1.096\ \text{mm}$$

Answer: vertical deflection of E $= 1.10$ mm upward (net). Individual effects: load 0.89 mm down, temperature 3.60 mm down, lack of fit 5.59 mm up.

Given data

Frame: A(0,0) hinge, B(0,3), C(0,6), D(4,6), E(4,2) roller. Left column AC: $EI$ (3 m + 3 m); beam CD: $2EI$, 4 m, UDL 20 kN/m; right column DE: $EI$, 4 m. 10 kN horizontal at B (to the right). Required: horizontal deflection of E.

Real loading: reactions

$$\sum M_A = 0:\; E_y\times4 = 10\times3 + 80\times2 \Rightarrow E_y = 47.5\ \text{kN},\quad A_y = 80 - 47.5 = 32.5\ \text{kN},\quad A_x = 10\ \text{kN (left)}$$

(The 10 kN load gives 30 kNm clockwise about A, the UDL resultant 80 kN at $x = 2$ m gives 160 kNm.)

Virtual load: unit horizontal load at E (to the right)

$A_x = 1$ (left); moments about A: $E_y\times4 = 1\times2$, so $E_y = 0.5$ (up), $A_y = 0.5$ (down).

Moments and integrals ($s$ from the start of each member, A-B-C-D-E)

$$\Delta_{Eh} = \frac{90 + 405 + 426.67}{EI} = \frac{921.67}{EI}$$

Answer: horizontal deflection of E $= 921.7/EI$ (kNm$^3$/EI), to the right.