Moment Distribution · Problem 1 of 5All problems

Moment Distribution Solved ProblemMoment distribution three-span beam, unequal stiffness

Problem. The three-span beam is fixed at A and D and has rollers at B and C. The spans are 5, 6, and 4 m, with moments of inertia I, 2I, and I. A 12 kN/m uniform load acts on the first span, and a 60 kN point load acts at the middle of the last span. Use moment distribution to find the end moments and support reactions, and draw the shear and moment diagrams.

Beam with 5, 6, and 4 m spans: fixed at A and D, rollers at B and C, middle span 2I; 12 kN/m uniform load on the first span, 60 kN point load at the middle of the last span — Figure 1. The beam, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 34.15 kN
$M_{A}$: 31.91 kN·m
$B_{y}$: 25.68 kN
$C_{y}$: 23.50 kN
$D_{y}$: 36.67 kN
$M_{D}$: −38.90 kN·m

Solution

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

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. Each span is first treated as fixed at both ends to get its fixed-end moments. At a pin or roller end support the moment must be zero, so that end is released at the start and half of its moment is carried over to the other end (near = the end that stays fixed; far = the released end):

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 AB (L = 5.00 m): FEM table, row 1

(\mathrm{FEM})_{AB} = -25\ \text{kN·m},\qquad (\mathrm{FEM})_{BA} = 25\ \text{kN·m}

Member BC (L = 6.00 m):

(\mathrm{FEM})_{BC} = 0\ \text{kN·m},\qquad (\mathrm{FEM})_{CB} = 0\ \text{kN·m}

Member CD (L = 4.00 m): FEM table, row 2

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

Uniform load: wL²/12 = 12 × 25 / 12 = 25. Point load: PL/8 = 60 × 4 / 8 = 30. The middle span has no load, so its FEM is zero.

2Distribution factors (interior joints)

At each interior joint, find the member stiffnesses $k = \tfrac{4EI}{L}$ ( $\tfrac{3EI}{L}$ if the far end is pinned); the distribution factor is

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

Joint B: $DF_{BA} = 0.375$ $DF_{BC} = 0.625$

Joint C: $DF_{CB} = 0.571$ $DF_{CD} = 0.429$

Put the 2I in the stiffness: at B, 4EI/5 = 0.8 and 4(2EI)/6 = 1.333, a split of 0.375 and 0.625. At C, 1.333 and 1, a split of 0.571 and 0.429.

3Moment distribution table

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

	A	B		C		D
	$M_{AB}$	$M_{BA}$	$M_{BC}$	$M_{CB}$	$M_{CD}$	$M_{DC}$
DF	—	0.375	0.625	0.571	0.429	—
FEM	−25.00	25.00	0	0	−30.00	30.00
Dist. 1		−9.38	−15.63	17.14	12.86
CO 1	−4.69		8.57	−7.81		6.43
Dist. 2		−3.21	−5.36	4.46	3.35
CO 2	−1.61		2.23	−2.68		1.67
Dist. 3		−0.84	−1.40	1.53	1.15
CO 3	−0.42		0.77	−0.70		0.57
Dist. 4		−0.29	−0.48	0.40	0.30
CO 4	−0.14		0.20	−0.24		0.15
Dist. 5		−0.07	−0.12	0.14	0.10
CO 5	−0.04		0.07	−0.06		0.05
Dist. 6		−0.03	−0.04	0.04	0.03
CO 6	−0.01		0.02	−0.02		0.01
Dist. 7		−0.01	−0.01	0.01	0.01
CO 7			0.01	−0.01
ΣM	−31.91	11.18	−11.18	12.21	−12.21	38.90

A and D are fixed: they don't distribute, they only collect what's carried over. Only B and C balance, in turn.

4Results and check

M_{AB} = -31.91\ \text{kN·m}

M_{BA} = 11.18\ \text{kN·m}

M_{BC} = -11.18\ \text{kN·m}

M_{CB} = 12.21\ \text{kN·m}

M_{CD} = -12.21\ \text{kN·m}

M_{DC} = 38.90\ \text{kN·m}

✓ The results match the matrix (direct stiffness) solution (largest difference 8.0 × 10⁻⁷ kN·m).

5Member equilibrium check (free-body diagrams)

Each member is cut free: the end moments come from above 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	Ry (kN)	M (kN·m, ccw +)
A	34.15	31.91
B	25.68	0
C	23.50	0
D	36.67	-38.90

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

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

ΣFy120.00 (reactions) − 120.00 (loads) = 0.00 kN ✓

ΣM₀930.00 (reactions) − 930.00 (loads) = 0.00 kN·m ✓

Reactions come from the end moments. $V_A$ : (12 × 5 × 2.5 + 31.91 − 11.18) / 5 = 34.15. Total: 34.15 + 25.68 + 23.50 + 36.67 = 60 + 60.

Bending moment diagram of the three-span beam: −31.91 at A, −11.18 at B, −12.21 at C, and −38.90 kN·m at D — Figure 2. Bending moment (kN·m).

Shear diagram of the three-span beam: 34.15 at A and −0.17 kN in the middle span — Figure 3. Shear force (kN).

Results

$A_{y}$: 34.15 kN
$M_{A}$: 31.91 kN·m
$B_{y}$: 25.68 kN
$C_{y}$: 23.50 kN
$D_{y}$: 36.67 kN
$M_{D}$: −38.90 kN·m

Matches the independent stiffness solution.

Common mistakes

Even with distribution factors on the wrong members the joint total is still 1, so the error is easy to miss. The larger factor belongs to the stiffer member.
The 2I of the middle span must go into its stiffness: k = 4(2EI)/6.
Fixed ends don't distribute. A and D only receive carried-over moments.

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