Moment Distribution · Problem 4 of 5All problems

Moment Distribution Solved ProblemMoment distribution frame: three-member joint

Problem. Three members meet at joint B of the frame. A and C are fixed, and D is pinned. Column A–B is 4 m and column B–C is 3 m, both with I; beam B–D is 6 m with 2I. A uniform load of 12 kN/m acts on the beam. Axial deformation is neglected. Use moment distribution to find the end moments, and draw the shear and moment diagrams.

Frame with three members at joint B: fixed at A and C, pinned at D, 4 m and 3 m columns with I, 6 m beam with 2I, 12 kN/m uniform load on the beam — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: 6.08 kN
$M_{A}$: −8.10 kN·m
$C_{x}$: −10.80 kN
$M_{C}$: −10.80 kN·m
$D_{x}$: 4.73 kN
$D_{y}$: 29.70 kN
$A_{y} + C_{y}$: 42.30 kN

Solution

Moment distribution for frames (end moments clockwise + on the member)

At each joint free to rotate (columns and beams together), moments are distributed with $DF_{ij} = k_{ij}/\textstyle\sum k$ ; the carry-over factor is $\times\tfrac12$ . The frame is assumed to have no sidesway ( $EA \approx \infty$ , ψ = 0). Sign convention (as in Hibbeler): a member end moment is positive when it acts clockwise on the member (counterclockwise on the joint). Kassimali's counterclockwise-positive values have the opposite sign.

FEM table: fixed-end moments

#	Loading	$(\mathrm{FEM})_i$	$(\mathrm{FEM})_j$
1	Uniform load (full span) w over the whole span	$-\dfrac{wL^2}{12}$	$+\dfrac{wL^2}{12}$
2	Point load at midspan P at x = L/2	$-\dfrac{PL}{8}$	$+\dfrac{PL}{8}$
3	Point load (general) P at a from end i, b from end j	$-\dfrac{P\,a\,b^2}{L^2}$	$+\dfrac{P\,a^2 b}{L^2}$
4	Triangular load (0 → w) 0 at end i, w at end j	$-\dfrac{wL^2}{30}$	$+\dfrac{wL^2}{20}$
5	Partial uniform load w from x = a to x = b	$-w\Big[\tfrac{x^2}{2}-\tfrac{2x^3}{3L}+\tfrac{x^4}{4L^2}\Big]_a^b$	$+w\Big[\tfrac{x^3}{3L}-\tfrac{x^4}{4L^2}\Big]_a^b$
6	Couple moment M₀ (counterclockwise +) at a from end i, ξ = a/L	$M_0\big(1-4\xi+3\xi^2\big)$	$M_0\big(3\xi^2-2\xi\big)$
7	Support settlement (chord rotation) ψ = (Δⱼ − Δᵢ)/L, settlement down +	$-\dfrac{6EI}{L}\psi$	$-\dfrac{6EI}{L}\psi$
8	Pinned far end (modified FEM) if end j is a pin	$(\mathrm{FEM})'_{i} = (\mathrm{FEM})_i - \tfrac{1}{2}(\mathrm{FEM})_j$	$M_{j} = 0$

Member fixed at both ends. Sign convention: an end moment is positive when it acts clockwise on the member (counterclockwise on the joint), as in Hibbeler. Use row 8 for a pinned end. The values match the engine's load integration exactly.

1Fixed-end moments and released ends

Each member is first treated as fixed at both ends to get its fixed-end moments. At a pinned support with a single member (e.g. a pinned column base) the moment is zero, so that end is released at the start:

All joints locked: fixed-end moments at the ends of the loaded members (kN·m)

(\mathrm{FEM})'_{\text{near}} = (\mathrm{FEM})_{\text{near}} - \tfrac{1}{2}\,(\mathrm{FEM})_{\text{far}}

FEM table, row 8

Member BD (L = 6.00 m): FEM table, row 1

(\mathrm{FEM})_{BD} = -36\ \text{kN·m},\qquad (\mathrm{FEM})_{DB} = 36\ \text{kN·m}

\Rightarrow\ (\mathrm{FEM})'_{BD} = -54,\quad M_{DB} = 0

D is pinned and released from the start: −36 − 36/2 = −54. That's why nothing is carried to D later.

2Distribution factors (joints free to rotate)

At each joint free to rotate, add up the member stiffnesses $k = \tfrac{4EI}{L}$ ( $\tfrac{3EI}{L}$ if the far end is pinned, 0 for an overhang):

DF_{ij} = \frac{k_{ij}}{\sum k}

Joint B: $DF_{BA} = 0.300$ $DF_{BC} = 0.400$ $DF_{BD} = 0.300$

Stiffnesses: A–B 4EI/4 = 1, B–C 4EI/3 = 1.333, B–D 3(2EI)/6 = 1. The total is 3.333, so the factors are 0.3, 0.4, 0.3. Count all three members.

3Moment distribution table

The FEM row lists the modified fixed-end moments. Balance each joint (distribute the negative of the unbalanced moment in proportion to DF), then carry over $\times\tfrac12$ to the far ends. Repeat until the carry-over moments are negligible.

	A	B			C
	$M_{AB}$	$M_{BA}$	$M_{BC}$	$M_{BD}$	$M_{CB}$
DF	—	0.300	0.400	0.300	—
FEM	0	0	0	−54.00	0
Dist. 1		16.20	21.60	16.20
CO 1	8.10				10.80
ΣM	8.10	16.20	21.60	−37.80	10.80

B is the only joint that rotates. +54 is distributed and half goes to the fixed column ends: 8.1 and 10.8. Nothing goes to D, so one cycle finishes the table.

4Result: M diagram and check

M_{AB} = 8.10\ \text{kN·m}

M_{BA} = 16.20\ \text{kN·m}

M_{BC} = 21.60\ \text{kN·m}

M_{BD} = -37.80\ \text{kN·m}

M_{CB} = 10.80\ \text{kN·m}

✓ The results match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 2.3 × 10⁻⁶ kN·m).

5Member equilibrium check (free-body diagrams)

Each member is cut free: the end moments come from the table and the end shears from member equilibrium ( $\sum M = 0,\ \sum F = 0$ ). The arcs and arrows are the end moments and forces acting on the member (kN, kN·m); each box is in equilibrium by itself.

6Support reactions and equilibrium check

Because axial deformation is neglected, the end moments cannot tell how axial force in a straight member chain between two supports is shared between them. The results table gives only the sum of those reactions; the other reactions and the M and V diagrams are not affected.

Find V_D by hand: (12 × 6 × 3 − 37.8) / 6 = 29.7. The remaining 42.3 kN goes down the column line; the moments can't split it between A and C.

Bending moment diagram of the frame solved by moment distribution: 16.20, 21.60, and 37.80 kN·m at joint B — Figure 2. Bending moment (kN·m).

Shear diagram of the frame solved by moment distribution: 6.08 in the lower column, 10.80 in the upper column, and 42.30 kN at the B end of the beam — Figure 3. Shear force (kN).

Results

$A_{x}$: 6.08 kN
$M_{A}$: −8.10 kN·m
$C_{x}$: −10.80 kN
$M_{C}$: −10.80 kN·m
$D_{x}$: 4.73 kN
$D_{y}$: 29.70 kN
$A_{y} + C_{y}$: 42.30 kN

Matches the independent stiffness solution.

Common mistakes

Two columns and a beam meet at joint B, and all three enter the stiffness total.
The largest factor, 0.4, belongs to the stiffest member, the short upper column. Reversed factors still total 1, so the error is hard to see.
When 3EI/L is used at a pinned end, no moment is carried over to that end. If you carry over, the stiffness must be 4EI/L.

Same problem, another method

Slope deflection

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