Moment Distribution · Problem 5 of 5All problems

Moment Distribution Solved ProblemMoment distribution frame with overhang

Problem. The frame is fixed at A, pinned at C, and free at D. Column A–B is 4 m with I, beam B–C is 6 m with 2I, and overhang C–D is 2 m. A 10 kN/m uniform load acts on the beam, and a 20 kN point load acts at D. Axial deformation is neglected. Use moment distribution to find the end moments and support reactions, and draw the shear and moment diagrams.

Frame with an overhang: fixed at A, pinned at C, 4 m column with I, 6 m beam with 2I, 2 m overhang; 10 kN/m uniform load on the beam, 20 kN point load at the tip — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: 4.69 kN
$A_{y}$: 25.42 kN
$M_{A}$: −6.25 kN·m
$C_{x}$: −4.69 kN
$C_{y}$: 54.58 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.

Overhang moments

Overhangs are statically determinate. Each applies a known cantilever moment to the joint it is attached to and takes no part in the distribution (its stiffness is 0):

M^{\text{overhang}}_{C} = -40\ \text{kN·m}

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

(\mathrm{FEM})_{BC} = -30\ \text{kN·m},\qquad (\mathrm{FEM})_{CB} = 30\ \text{kN·m}

Statics gives the overhang moment: 20 × 2 = 40. It isn't an unknown in the distribution; it enters the balance at C as a known value.

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: $DF_{BA} = 0.429$ $DF_{BC} = 0.571$

Joint C: $DF_{CB} = 1.000$

Here C–B starts held: k = 4(2EI)/6 = 1.333, and A–B is 4EI/4 = 1. At B, 0.429 and 0.571. At C one member takes everything, factor 1.

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
	$M_{AB}$	$M_{BA}$	$M_{BC}$	$M_{CB}$
DF	—	0.429	0.571	1.000
FEM	0	0	−30.00	30.00
Dist. 1		12.86	17.14	10.00
CO 1	6.43		5.00	8.57
Dist. 2		−2.14	−2.86	−8.57
CO 2	−1.07		−4.29	−1.43
Dist. 3		1.84	2.45	1.43
CO 3	0.92		0.71	1.22
Dist. 4		−0.31	−0.41	−1.22
CO 4	−0.15		−0.61	−0.20
Dist. 5		0.26	0.35	0.20
CO 5	0.13		0.10	0.17
Dist. 6		−0.04	−0.06	−0.17
CO 6	−0.02		−0.09	−0.03
Dist. 7		0.04	0.05	0.03
CO 7	0.02		0.01	0.02
Dist. 8		−0.01	−0.01	−0.02
CO 8			−0.01
Dist. 9		0.01	0.01
ΣM	6.25	12.50	−12.50	40.00

C is free to rotate, so it carries to B every cycle and the table runs long. Shortcut: release C from the start, FEM −30 + 5 = −25, and 3EI/L gives 0.5 and 0.5 at B. One cycle gives +12.5, the same result.

4Result: M diagram and check

M_{AB} = 6.25\ \text{kN·m}

M_{BA} = 12.50\ \text{kN·m}

M_{BC} = -12.50\ \text{kN·m}

M_{CB} = 40\ \text{kN·m}

✓ The results match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 1.6 × 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

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	4.69	25.42	-6.25
C	-4.69	54.58	0

Equilibrium check (ΣFx, ΣFy, and ΣM about the origin):

ΣFx0.00 (reactions) + 0.00 (loads) = 0.00 kN ✓

ΣFy80.00 (reactions) − 80.00 (loads) = 0.00 kN ✓

ΣM₀340.00 (reactions) − 340.00 (loads) = 0.00 kN·m ✓

Check $V_C$ by hand: (10 × 6 × 3 − 12.5 + 40) / 6 + 20 = 54.58. Total: 25.42 + 54.58 = 60 + 20.

Bending moment diagram of the frame with an overhang: 6.25 at A, 12.50 at B, and 40.00 kN·m at C — Figure 2. Bending moment (kN·m).

Shear diagram of the frame with an overhang: 4.69 in the column, 25.42 at the B end of the beam, and 20.00 kN in the overhang — Figure 3. Shear force (kN).

Results

$A_{x}$: 4.69 kN
$A_{y}$: 25.42 kN
$M_{A}$: −6.25 kN·m
$C_{x}$: −4.69 kN
$C_{y}$: 54.58 kN

Matches the independent stiffness solution.

Common mistakes

An overhang has zero stiffness and takes no part in distribution. Statics gives its moment.
When 3EI/L is used at a pinned end, nothing is carried over to it. This table uses 4EI/L, so moments are carried over to C.
C is a pin support, but the moment at the end of the beam isn't zero, because the overhang applies 40 kN·m there.

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