Displacement method

The slope-deflection equations express the end moments of a member in terms of the end rotations, the chord rotation and the loads on the member. They are derived here for a prismatic member AB of length $L$ and flexural rigidity $EI$ (the same derivation applies to every span of a continuous beam).

Symbols and sign convention

• $M_{AB}, M_{BA}$: end moments at A and B, clockwise on the member end = positive
• $\theta_A,\theta_B$: rotations of the tangents at A and B, clockwise positive
• $\Delta$: relative transverse displacement of B with respect to A (e.g. support settlement); $\psi=\Delta/L$ is the chord rotation, positive clockwise (B moves down relative to A)
• $FEM_{AB}, FEM_{BA}$: fixed-end moments due to the loads on the span (clockwise +)
• Assumptions: linear elastic material, small deformations, axial and shear deformations neglected

Derivation (conjugate beam / moment-area)

Treat the member as simply supported at A and B carrying the end moments $M_{AB}$, $M_{BA}$ and the span loads. The end moments are then the sum of the effect of the loads on a fixed-ended member and the effect of the end displacements.

1. Rotation due to end moments only. For end moments $M_{AB}$ at A and $M_{BA}$ at B (clockwise), the slopes of the tangent relative to the chord are, by the moment-area theorem,

$$\theta_A-\psi=\frac{L}{6EI}\left(2M_{AB}-M_{BA}\right),\qquad \theta_B-\psi=\frac{L}{6EI}\left(2M_{BA}-M_{AB}\right)$$

(the end moments act with opposite sense on the beam, so one is sagging and the other hogging; this is the origin of the minus signs).

2. Solve for the moments. Multiply the first equation by 2 and add the second:

$$2(\theta_A-\psi)+(\theta_B-\psi)=\frac{L}{6EI}\left(3M_{AB}\right)\;\Rightarrow\;M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-3\psi\right)$$

In the same way

$$M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-3\psi\right)$$

3. Add the effect of the loads on the span. With the span loaded, the member is first held fixed at both ends; this gives the fixed-end moments $FEM_{AB}$ and $FEM_{BA}$. The end rotations and the chord rotation are then applied. By superposition:

$$\boxed{M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-\frac{3\Delta}{L}\right)+FEM_{AB}}$$ $$\boxed{M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-\frac{3\Delta}{L}\right)+FEM_{BA}}$$

Standard fixed-end moments (clockwise +)

Special cases

• Far end hinged or roller ($M_{BA}=0$): $M_{AB}=\dfrac{3EI}{L}\left(\theta_A-\psi\right)+FEM_{AB}-\dfrac{FEM_{BA}}{2}$
• Fixed end: $\theta=0$ for that end. Rigid support without settlement: $\psi=0$.

Use in a continuous beam

Write the two equations for every span, express each end moment in terms of the unknown joint rotations, and apply the joint equilibrium conditions $\sum M=0$ at each support (including any overhang moment). Solve the simultaneous equations for the rotations, then back-substitute to get the final end moments. For frames with sway, one more equation (the shear / storey equilibrium) is added for each independent sway.

For a prismatic beam of span $L$ and rigidity $EI$, fixed against rotation and translation at both ends, one end is then given a rotation or a settlement. Use the slope-deflection equations (clockwise end moments positive; no load on the span, so FEM = 0):

$$M_{AB}=\frac{2EI}{L}\left(2\theta_A+\theta_B-\frac{3\Delta}{L}\right),\qquad M_{BA}=\frac{2EI}{L}\left(2\theta_B+\theta_A-\frac{3\Delta}{L}\right)$$

(a) End A rotates by $\theta$ (clockwise), no settlement

Put $\theta_A=\theta$, $\theta_B=0$, $\Delta=0$:

$$M_{AB}=\frac{4EI\theta}{L},\qquad M_{BA}=\frac{2EI\theta}{L}$$

Both are clockwise on the member ends. The moment at the far end is half of that at the rotated end (carry-over factor 1/2).

Reactions. Taking moments about B for the whole member (clockwise positive), with $V_A$ the upward force on the member at A:

$$M_{AB}+M_{BA}+V_AL=0\;\Rightarrow\;V_A=-\frac{6EI\theta}{L^2}$$

So at A the reaction is $6EI\theta/L^2$ downward and at B it is $6EI\theta/L^2$ upward (the two forces form a couple that balances the end moments).

  A: M = 4EI.theta/L (clockwise)    B: M = 2EI.theta/L (clockwise)
  A: V = 6EI.theta/L^2 down         B: V = 6EI.theta/L^2 up

(b) End B settles down by $\Delta$, no rotation

Put $\theta_A=\theta_B=0$ and $\Delta$ downward (positive chord rotation $\psi=\Delta/L$):

$$M_{AB}=M_{BA}=-\frac{6EI\Delta}{L^2}$$

Both are anticlockwise, which means hogging at A and sagging at B.

Reactions. From moments about B:

$$M_{AB}+M_{BA}+V_AL=0\;\Rightarrow\;V_A=\frac{12EI\Delta}{L^3}\ (\text{upward}),\qquad V_B=\frac{12EI\Delta}{L^3}\ (\text{downward})$$

The settlement pulls the end B down, so the support must exert a downward force at B and an upward force at A.

Summary

These are exactly the stiffness coefficients $4EI/L$, $2EI/L$, $6EI/L^2$ and $12EI/L^3$ used in the stiffness (displacement) method.

Principle

The moment distribution method (Hardy Cross, 1930) is an iterative form of the slope-deflection method for continuous beams and frames without having to solve simultaneous equations. The idea is:

1. Lock every joint against rotation. Each loaded member then behaves as a fixed-ended beam and develops fixed-end moments (FEM). At a joint these are generally unequal, so there is an unbalanced moment = $\sum$FEM of the members meeting at the joint.
2. Release one joint at a time. It rotates until the joint is in equilibrium. The balancing moment (equal to the unbalanced moment with reversed sign) is distributed to the members in proportion to their stiffness.
3. A part of each distributed moment is carried over to the far end of the member (if the far end is fixed against rotation).
4. The carried-over moments disturb the neighbouring joints, so the balancing and carry-over are repeated until the carry-over moments are negligibly small.
5. The final end moment = FEM + all distributed moments + all carried-over moments.

Equilibrium of every joint and slopes compatibility of the members are therefore satisfied when the iteration converges.

Distribution factor (DF)

The fraction of the unbalanced moment taken by a member at a joint:

$$DF_{jk}=\frac{K_{jk}}{\sum K},\qquad K=\frac{4EI}{L}\ (\text{far end fixed}),\quad K=\frac{3EI}{L}\ (\text{far end hinged/roller})$$

$\sum DF=1$ at every joint. For a fixed support $DF=0$ (it takes no rotation); for a simple end support with no member beyond it, $DF=1$.

Carry-over factor (COF)

A moment $M$ applied at the near end of a member whose far end is fixed produces $M/2$ at the far end, so $COF=\tfrac12$. It is 0 if the far end is hinged or roller (no moment can develop there).

Example

Beam ABC: AB = 6 m, BC = 4 m, EI constant; A is fixed, B and C are simple supports; AB carries 10 kN/m. Since the far end C is hinged, BC is stiffened by the factor 3/4.

Distribution factors at B: $K_{BA}=\dfrac{EI}{6}=0.1667EI$, $K_{BC}=\dfrac{3}{4}\cdot\dfrac{EI}{4}=0.1875EI$ (relative values). $DF_{BA}=0.471$, $DF_{BC}=0.529$.

Fixed-end moments: $FEM_{AB}=-\dfrac{wL^2}{12}=-\dfrac{10\times6^2}{12}=-30$ kNm, $FEM_{BA}=+30$ kNm; BC has no load. (C is hinged: $FEM_{BC}=FEM_{CB}=0$.)

Distribution:

• Unbalanced moment at B = $+30$ kNm; balancing moment = $-30$ kNm.
• $BA$ takes $-30\times0.471=-14.12$; $BC$ takes $-30\times0.529=-15.88$.
• Carry-over to A: $\tfrac12\times(-14.12)=-7.06$. No carry-over to C (hinged).
• Joint B is now balanced and A, C need no more balancing (A is fixed), so the process ends in one cycle.

Joint check: $M_{BA}+M_{BC}=0.00$ ✓. The final moments equal the exact slope-deflection result, hence the method gives exact answers for this beam (several cycles are needed only when there are more joints to be balanced).

Assumptions: the 10 kN/m UDL acts downward on BC, the 40 kN load at B acts to the right and the 50 kNm moment at C is clockwise. EI is constant, axial deformation is neglected, A and D are fixed.

Degrees of freedom

$\theta_B,\theta_C$ (joint rotations, clockwise +) and $\Delta$ (horizontal sway of beam BC, right +). Since BC is axially rigid, B and C move equally, so $\psi_{AB}=\psi_{CD}=\Delta/4$ and $\psi_{BC}=0$.

Member relation: $M_{ij}=\dfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)+FEM_{ij}$ (clockwise end moments +).

Fixed-end moments

$FEM_{BC}=-\dfrac{wL^2}{12}=-\dfrac{10\times8^2}{12}=-53.33$ kNm, $FEM_{CB}=+53.33$ kNm. The columns carry no load between joints, so their FEMs are zero.

Stiffness equations for the members

• $M_{AB}=\frac{2EI}{4}\left(\theta_{B} - (0.7500)\Delta_1\right)$
• $M_{BA}=\frac{2EI}{4}\left(2\theta_{B} - (0.7500)\Delta_1\right)$
• $M_{BC}=\frac{2EI}{8}\left(2\theta_{B} + \theta_{C}\right) - 53.33$
• $M_{CB}=\frac{2EI}{8}\left(2\theta_{C} + \theta_{B}\right) + 53.33$
• $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

Equilibrium conditions

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=50$ (the applied clockwise moment)
• Storey shear: the column shears resist the 40 kN load, $\dfrac{M_{AB}+M_{BA}}{4}+\dfrac{M_{CD}+M_{DC}}{4}+40=0$

Substituting the member equations gives $[K]\{d\}=\{F\}$:

$$EI\begin{bmatrix}1.5000 & 0.2500 & -0.3750 \\ 0.2500 & 1.5000 & -0.3750 \\ -0.3750 & -0.3750 & 0.3750\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}53.333 \\ -3.333 \\ 40.000\end{Bmatrix}$$

(First two rows: joint B and C. Third row: sway equation. The load vector holds the applied loads minus the fixed-end effects.)

Solution

$$EI\theta_{B}=87.6667,\quad EI\theta_{C}=42.3333,\quad EI\Delta_{1}=236.6667$$

Final end moments (kNm, clockwise on the member end +)

Checks:

• Joint B: (-1.08) + (1.08) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (96.42) + (-46.42) = 50.00 kNm (applied clockwise moment 50.00) ✓
• Storey shear: $\dfrac{(-46.00)+(-114.00)}{4}+40=0.00$ ✓

Bending moment diagram ordinates (sagging/inside tension +, kNm)

Reactions: A: $H=-11.50$, $V=27.81$, $M=44.92$; D: $H=-28.50$, $V=52.19$, $M=67.58$ (M anticlockwise +).

Answer: $M_{AB}=-44.92$, $M_{BA}=-1.08$, $M_{BC}=1.08$, $M_{CB}=96.42$, $M_{CD}=-46.42$, $M_{DC}=-67.58$ kNm; sway $EI\Delta=236.67$.

Assumptions: the figure shows AB = 2 m + 4 m = 6 m with the 40 kN load 2 m from A, BC = 10 m, CD = 2 m overhang with 30 kN at D. A is taken as a fixed end; B and C are simple supports. Stiffness: AB = 1.5EI, BC = 2EI, CD = EI.

Step 1: Overhang CD

Moment at C due to the overhang: $M_C=30\times2=60$ kNm (hogging). This is a known moment acting on joint C.

Step 2: Fixed-end moments (clockwise +)

• AB: $FEM_{AB}=-\dfrac{40\times2\times4^2}{6^2}=-35.56$, $FEM_{BA}=+\dfrac{40\times2^2\times4}{6^2}=17.78$ kNm
• BC: $FEM_{BC}=-\dfrac{10\times10^2}{12}=-83.33$, $FEM_{CB}=+83.33$ kNm

No support settles ($\psi=0$) and $\theta_A=0$.

Step 3: Slope-deflection equations

• $M_{AB}=\frac{2(1.5EI)}{6}\left(\theta_{B}\right) - 35.56$
• $M_{BA}=\frac{2(1.5EI)}{6}\left(2\theta_{B}\right) + 17.78$
• $M_{BC}=\frac{2(2EI)}{10}\left(2\theta_{B} + \theta_{C}\right) - 83.33$
• $M_{CB}=\frac{2(2EI)}{10}\left(2\theta_{C} + \theta_{B}\right) + 83.33$

Step 4: Joint equilibrium

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=0$ with $M_{CD}=-60$ kNm (the overhang moment), i.e. $M_{CB}=60$

$$EI\begin{bmatrix}1.8000 & 0.4000 \\ 0.4000 & 0.8000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}65.556 \\ -23.333\end{Bmatrix}$$ $$EI\theta_{B}=48.2639,\quad EI\theta_{C}=-53.2986$$

Step 5: Final end moments (kNm)

Bending moment diagram

Reactions: A: $H=0.00$, $V=17.56$, $M=11.42$; B: $H=0.00$, $V=73.04$, $M=0.00$; C: $H=0.00$, $V=79.40$, $M=0.00$.

Answer: $M_{AB}=-11.42$, $M_{BA}=66.04$, $M_{BC}=-66.04$, $M_{CB}=60.00$ kNm (A fixed); $M_C=60$ kNm hogging; span BC sagging maximum about 62.00 kNm.

Assumptions: AB = 2 m + 4 m = 6 m (A fixed) carries a clockwise couple of 50 kNm at 2 m from A; BC = 5 m is a free-ended overhang with 30 kN/m; the column BD = 6 m is fixed at D. Relative stiffness: AB = 1.5I, BC = 2I, BD = I (E constant). No sway occurs because the fixed support at A prevents the horizontal movement of the girder.

Step 1: Overhang BC

The overhang is statically determinate: $M_{BC}=\dfrac{30\times5^2}{2}=375$ kNm (hogging). It is a fixed moment on joint B and takes no part in the distribution (DF = 0).

Step 2: Stiffness and distribution factors (only joint B is free)

• Joint B: relative stiffness $\Sigma k$ = 0.4167; BA: $k=EI/L$ = 0.2500, DF = 0.6000; BC: overhang, DF = 0 (its moment is a known fixed value); BD: $k=EI/L$ = 0.1667, DF = 0.4000

Step 3: Fixed-end moments (clockwise +)

For a couple $M$ at distance $a$ from A: $FEM_{AB}=\dfrac{Mb(2a-b)}{L^2}$, $FEM_{BA}=\dfrac{Ma(2b-a)}{L^2}$ with $a=2$, $b=4$, $L=6$: $FEM_{AB}=\dfrac{50\times4\times(4-4)}{36}=0$, $FEM_{BA}=\dfrac{50\times2\times(8-2)}{36}=16.67$ kNm. Overhang moment on B: $-375$ kNm. Column BD has no load.

Step 4: Moment distribution

Unbalanced moment at B = $16.67-375=-358.33$ kNm, so the balancing moment is $+358.33$ kNm, shared 0.6 : 0.4. Half of each share is carried over to the far fixed ends.

Step 5: Bending moment diagram (kNm)

• Joint B: $231.67+143.33-375=0$ ✓
• Column BD: top moment 143.33, base moment 71.67 (carry-over of one half)
• Overhang: 375 kNm at B (statics)

BM sign: positive = tension on the right-hand side when travelling A to B to C and B to D (sagging for the horizontal beam). The overhang BC has a parabolic diagram from 0 at C to 375 kNm (hogging) at B.

Answer: $M_{AB}=107.50$, $M_{BA}=231.67$, $M_{BC}=-375.00$, $M_{BD}=143.33$, $M_{DB}=71.67$ kNm (clockwise end moments +).

Assumptions (figure partly unclear): A(0,0), B(4,0), C(9,0) lie on one horizontal girder: AB = 4 m (3EI), BC = 5 m (2EI) with 8 kN/m downward. Column BE (2EI) is 4 m long and carries the 20 kN horizontal load (to the right) at its mid-height; E is a hinge. Column CD (EI) is 7 m long and fixed at D. A is fixed. The 20 kNm moment at B is clockwise. Because A is fixed, the girder cannot sway, so the unknowns are only the joint rotations $\theta_B,\theta_C$ and the hinge rotation $\theta_E$.

Fixed-end moments (clockwise +)

• BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{8\times5^2}{12}=\mp16.67$ kNm
• BE (point load 20 kN at mid-height): $\mp\dfrac{PL}{8}=\mp\dfrac{20\times4}{8}=\mp10$ kNm (for the load acting to the right on this downward member: $FEM_{BE}=10.00$, $FEM_{EB}=-10.00$ kNm)
• AB and CD: no loads, so zero.

Slope-deflection (stiffness) equations

• $M_{AB}=\frac{2(3EI)}{4}\left(\theta_{B}\right)$
• $M_{BA}=\frac{2(3EI)}{4}\left(2\theta_{B}\right)$
• $M_{BC}=\frac{2(2EI)}{5}\left(2\theta_{B} + \theta_{C}\right) - 16.67$
• $M_{CB}=\frac{2(2EI)}{5}\left(2\theta_{C} + \theta_{B}\right) + 16.67$
• $M_{BE}=\frac{2(2EI)}{4}\left(2\theta_{B} + \theta_{E}\right) + 10.00$
• $M_{EB}=\frac{2(2EI)}{4}\left(2\theta_{E} + \theta_{B}\right) - 10.00$
• $M_{CD}=\frac{2EI}{7}\left(2\theta_{C}\right)$
• $M_{DC}=\frac{2EI}{7}\left(\theta_{C}\right)$

Equilibrium

• Joint B: $M_{BA}+M_{BC}+M_{BE}=20$ (applied clockwise moment)
• Joint C: $M_{CB}+M_{CD}=0$
• Hinge E: $M_{EB}=0$

$$EI\begin{bmatrix}6.6000 & 0.8000 & 1.0000 \\ 0.8000 & 2.1714 & 0.0000 \\ 1.0000 & 0.0000 & 2.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \theta_{E}\end{Bmatrix}=\begin{Bmatrix}26.667 \\ -16.667 \\ 10.000\end{Bmatrix}$$ $$EI\theta_{B}=4.7900,\quad EI\theta_{C}=-9.4402,\quad EI\theta_{E}=2.6050$$

Final end moments (kNm, clockwise on the member end +)

Checks:

• Joint B: (14.37) + (-16.55) + (22.18) = 20.00 kNm (applied clockwise moment 20.00) ✓
• Joint C: (5.39) + (-5.39) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint E: (0.00) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Hinge: $M_{EB}=0.00$ ✓

Bending moment ordinates (sagging/inside + according to member direction):

Reactions: A: $H=-14.39$, $V=-5.39$, $M=-7.18$; E: $H=-4.45$, $V=27.62$, $M=0.00$; D: $H=-1.16$, $V=17.77$, $M=2.70$.

Answer: $M_{AB}=7.18$, $M_{BA}=14.37$, $M_{BC}=-16.55$, $M_{CB}=5.39$, $M_{BE}=22.18$, $M_{EB}=0$, $M_{CD}=-5.39$, $M_{DC}=-2.70$ kNm.

Assumptions: the girder A-B-C-D-E has A fixed, a roller at D and rigid column connections at B (BG, 6 m, 2I) and C (CH, 6 m, I), both fixed at their bases. Spans AB = BC = CD = 6 m, overhang DE = 2 m with 60 kN at E; 10 kN/m UDL on BC and CD (B to D). Relative I: AB = I, BC = 1.5I, CD = 2I. The 80 kN horizontal load at B acts to the right.

Sway check

The girder is held horizontally by the fixed support at A, so it cannot sway. The 80 kN load is carried as axial force in the girder to support A and causes no bending moment. The analysis is therefore a non-sway moment distribution with joints B, C and D free to rotate.

Step 1: Overhang

$M_D=60\times2=120$ kNm (hogging), a fixed moment on joint D.

Step 2: Fixed-end moments (clockwise +)

For the UDL spans: $\mp wL^2/12=\mp10\times36/12=\mp30$ kNm on BC and CD.

• AB: $FEM_{AB}=0.00$, $FEM_{BA}=0.00$ kNm
• BC: $FEM_{BC}=-30.00$, $FEM_{CB}=30.00$ kNm
• CD: $FEM_{CD}=-30.00$, $FEM_{DC}=30.00$ kNm
• DE: $FEM_{DE}=-120.00$, $FEM_{ED}=0.00$ kNm
• BG: $FEM_{BG}=0.00$, $FEM_{GB}=0.00$ kNm
• CH: $FEM_{CH}=0.00$, $FEM_{HC}=0.00$ kNm

Step 3: Distribution factors ($k=EI/L$)

• Joint B: relative stiffness $\Sigma k$ = 0.7500; BA: $k=EI/L$ = 0.1667, DF = 0.2222; BC: $k=EI/L$ = 0.2500, DF = 0.3333; BG: $k=EI/L$ = 0.3333, DF = 0.4444
• Joint C: relative stiffness $\Sigma k$ = 0.7500; CB: $k=EI/L$ = 0.2500, DF = 0.3333; CD: $k=EI/L$ = 0.3333, DF = 0.4444; CH: $k=EI/L$ = 0.1667, DF = 0.2222
• Joint D: relative stiffness $\Sigma k$ = 0.3333; DC: $k=EI/L$ = 0.3333, DF = 1.0000; DE: overhang, DF = 0 (its moment is a known fixed value)

Step 4: Moment distribution (kNm)

Step 5: Final end moments and BMD

Joint B: $8.82+(-26.45)+(17.63)=0.00$ ✓; Joint C: $17.26+(-4.35)+(-12.90)=0.00$ ✓; Joint D: $120.00+(-120.00)=0.00$ ✓.

BM sign: positive = sagging for the girder, tension on the right-hand side when travelling along the member as listed.

Reactions: A: $H=-81.18$, $V=-2.20$, $M=-4.41$; G: $H=4.41$, $V=33.74$, $M=-8.82$; H: $H=-3.23$, $V=39.19$, $M=6.45$; D: $H=0.00$, $V=109.27$, $M=0.00$.

Answer: $M_{AB}=4.41$, $M_{BA}=8.82$, $M_{BC}=-26.45$, $M_{CB}=17.26$, $M_{CD}=-4.35$, $M_{DC}=120.00$, $M_{DE}=-120.00$ kNm; columns: $M_{BG}=17.63$, $M_{GB}=8.82$, $M_{CH}=-12.90$, $M_{HC}=-6.45$ kNm.

Assumptions: A(0,0) and D are fixed; the girder BC (2I, 8 m) is horizontal at the top, AB = 6 m (1.5I) and CD = 4 m (I), so D is 2 m above A. The UDL of 20 kN/m acts downward on BC and the 10 kN load at C acts to the right. Axial deformations are ignored. Take $EI$ as the stiffness of I.

Unknown displacements

$\theta_B$, $\theta_C$ and the sway $\Delta$ of the girder. Chord rotations: $\psi_{AB}=\Delta/6$, $\psi_{CD}=\Delta/4$, $\psi_{BC}=0$.

Fixed-end moments

$FEM_{BC}=-\dfrac{20\times8^2}{12}=-106.67$ kNm, $FEM_{CB}=+106.67$ kNm. Columns carry no load.

Member stiffness equations

• $M_{AB}=\frac{2(1.5EI)}{6}\left(\theta_{B} - (0.5000)\Delta_1\right)$
• $M_{BA}=\frac{2(1.5EI)}{6}\left(2\theta_{B} - (0.5000)\Delta_1\right)$
• $M_{BC}=\frac{2(2EI)}{8}\left(2\theta_{B} + \theta_{C}\right) - 106.67$
• $M_{CB}=\frac{2(2EI)}{8}\left(2\theta_{C} + \theta_{B}\right) + 106.67$
• $M_{CD}=\frac{2EI}{4}\left(2\theta_{C} - (0.7500)\Delta_1\right)$
• $M_{DC}=\frac{2EI}{4}\left(\theta_{C} - (0.7500)\Delta_1\right)$

Equilibrium equations

• Joint B: $M_{BA}+M_{BC}=0$
• Joint C: $M_{CB}+M_{CD}=0$
• Storey shear: $\dfrac{M_{AB}+M_{BA}}{6}+\dfrac{M_{CD}+M_{DC}}{4}+10=0$

Combining gives the stiffness matrix equation:

$$EI\begin{bmatrix}2.0000 & 0.5000 & -0.2500 \\ 0.5000 & 2.0000 & -0.3750 \\ -0.2500 & -0.3750 & 0.2708\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C} \\ \Delta_{1}\end{Bmatrix}=\begin{Bmatrix}106.667 \\ -106.667 \\ 10.000\end{Bmatrix}$$

Typical terms: $K_{11}=\dfrac{4(1.5EI)}{6}+\dfrac{4(2EI)}{8}=2EI$, $K_{12}=\dfrac{2(2EI)}{8}=0.5EI$, $K_{13}=-\dfrac{6(1.5EI)}{6^2}=-0.25EI$ (sign convention of the rows above, positive sway to the right).

Solution

$$EI\theta_{B}=71.6049,\quad EI\theta_{C}=-70.1235,\quad EI\Delta_{1}=5.9259$$

Final end moments (kNm, clockwise on the member end +)

Checks:

• Joint B: (70.12) + (-70.12) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Joint C: (72.35) + (-72.35) = 0.00 kNm (applied clockwise moment 0.00) ✓
• Storey shear: $\dfrac{104.44}{6}+\dfrac{-109.63}{4}+10=0.00$ ✓

Reactions: A: $H=17.41$, $V=79.72$, $M=-34.32$; D: $H=-27.41$, $V=80.28$, $M=37.28$.

Answer: $M_{AB}=34.32$, $M_{BA}=70.12$, $M_{BC}=-70.12$, $M_{CB}=72.35$, $M_{CD}=-72.35$, $M_{DC}=-37.28$ kNm.

Assumptions: the girder A-B-C is horizontal with A fixed and C on a roller; AB = 4 m (2EI) with the 60 kN load at mid-span; BC = 4 m (EI) with 10 kN/m downward; the column BD (EI, 4 m) hangs below B and is fixed at D. The fixed support A prevents sway, so only $\theta_B$ and $\theta_C$ are unknown.

Fixed-end moments (clockwise +)

• AB: $\mp\dfrac{PL}{8}=\mp\dfrac{60\times4}{8}=\mp30$ kNm
• BC: $\mp\dfrac{wL^2}{12}=\mp\dfrac{10\times4^2}{12}=\mp13.33$ kNm
• BD: no load

Stiffness (slope-deflection) equations

• $M_{AB}=\frac{2(2EI)}{4}\left(\theta_{B}\right) - 30.00$
• $M_{BA}=\frac{2(2EI)}{4}\left(2\theta_{B}\right) + 30.00$
• $M_{BC}=\frac{2EI}{4}\left(2\theta_{B} + \theta_{C}\right) - 13.33$
• $M_{CB}=\frac{2EI}{4}\left(2\theta_{C} + \theta_{B}\right) + 13.33$
• $M_{BD}=\frac{2EI}{4}\left(2\theta_{B}\right)$
• $M_{DB}=\frac{2EI}{4}\left(\theta_{B}\right)$

Equilibrium

• Joint B: $M_{BA}+M_{BC}+M_{BD}=0$
• Roller C: $M_{CB}=0$

Writing these in terms of $\theta_B,\theta_C$ gives the stiffness matrix equation:

$$EI\begin{bmatrix}4.0000 & 0.5000 \\ 0.5000 & 1.0000\end{bmatrix}\begin{Bmatrix}\theta_{B} \\ \theta_{C}\end{Bmatrix}=\begin{Bmatrix}-16.667 \\ -13.333\end{Bmatrix}$$ $$EI\theta_{B}=-2.6667,\quad EI\theta_{C}=-12.0000$$

Final end moments (kNm, clockwise on the member end +)

Checks: Joint B: $24.67+(-22.00)+(-2.67)=0.00$ ✓; Roller C: $M_{CB}=0.00$ ✓.

Bending moment diagram ordinates (sagging + for the girder; column: tension on the right when going down, kNm)

Reactions: A: $H=1.00$, $V=32.00$, $M=32.67$; D: $H=-1.00$, $V=53.50$, $M=1.33$; C: $H=0.00$, $V=14.50$, $M=0.00$.

Answer: $M_{AB}=-32.67$, $M_{BA}=24.67$, $M_{BC}=-22.00$, $M_{CB}=0$, $M_{BD}=-2.67$, $M_{DB}=-1.33$ kNm.

Assumptions: the girder F-E-D-C has a 1.5 m cantilever FE (3EI) with 3 kN at F, a 4 m span ED (3EI) with 15 kN/m, and a 1.5 m span DC (3EI) ending on a roller at C. Columns EA (2EI) and DB (EI) are each 4 m high and fixed at their bases. The roller at C allows horizontal movement, so the frame can sway; the moment distribution is done in two cases (sway prevented, then sway induced).

Step 1: Overhang FE

$M_{EF}=3\times1.5=4.5$ kNm (hogging), fixed moment on joint E.

Step 2: Distribution factors

• Joint E: relative stiffness $\Sigma k$ = 1.2500; EF: overhang, DF = 0 (its moment is a known fixed value); ED: $k=EI/L$ = 0.7500, DF = 0.6000; EA: $k=EI/L$ = 0.5000, DF = 0.4000
• Joint D: relative stiffness $\Sigma k$ = 2.5000; DE: $k=EI/L$ = 0.7500, DF = 0.3000; DC: $k=EI/L$ = 1.5000 (far end hinged: 3/4 factor), DF = 0.6000; DB: $k=EI/L$ = 0.2500, DF = 0.1000 (Roller end C: member DC is stiffened by 3/4 because its far end is hinged and the end moment at C is zero.)

Step 3: Case I, sway prevented by an imaginary horizontal support at E

Fixed-end moments from the loads: $\mp15\times4^2/12=\mp20$ kNm on ED.

Reaction of the imaginary support (from the column shears): $R_1=-1.94$ kN (acting to the left).

Step 4: Case II, sway $\Delta_0$ without external loads

Give the girder a trial sway to the right and choose its size so that the fixed-end moment in column EA is $-100$ kNm at both ends. Since $FEM=-6EI\Delta/L^2$ is proportional to $EI/L^2$ and both columns have the same height and the same $\Delta$, the column DB (EI) gets half of this: $-50$ kNm at both ends. All other members have zero FEM.

Reaction of the imaginary support: $R_2=59.69$ kN.

Step 5: Correction for sway

The imaginary support does not exist, so the restraint force must vanish: $R_1+k\,R_2=0$

$$k=-\frac{R_1}{R_2}=-\frac{-1.94}{59.69}=0.0325$$

Final moments $M=M_I+kM_{II}$ (kNm):

Support moments: $M_{AE}=1.26$, $M_{BD}=-2.88$ kNm. Joint E check: $M_{ED}+M_{EA}+M_{EF}=0.00$ ✓. Joint D check: $M_{DE}+M_{DB}+M_{DC}=0.00$ ✓. $M_{CD}=0$.

Bending moment ordinates (kNm)

Answer: $M_{EA}=5.76$, $M_{AE}=1.26$, $M_{ED}=-10.26$, $M_{DE}=19.22$, $M_{DB}=-4.14$, $M_{BD}=-2.88$, $M_{DC}=-15.08$ kNm (clockwise end moments +); $M_{EF}=4.50$ kNm.