Moment Distribution · Problem 2 of 5All problems

Moment Distribution Method ExampleTwo-span beam, 3EI/L at the pinned ends

Problem. The two-span beam has a pin at A and rollers at B and C. A uniform load of 12 kN/m acts over its full length. EI is constant. Use moment distribution to find the end moments and support reactions, and draw the shear and moment diagrams.

Continuous beam with 4 m and 6 m spans: pin at A, rollers at B and C, 12 kN/m uniform load over the full length — Figure 1. The beam, its load, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 13.50 kN
$B_{y}$: 77.50 kN
$C_{y}$: 29.00 kN

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 = 4.00 m): FEM table, row 1

(\mathrm{FEM})_{AB} = -16\ \text{kN·m},\qquad (\mathrm{FEM})_{BA} = 16\ \text{kN·m}

\Rightarrow\ (\mathrm{FEM})'_{BA} = 24,\quad M_{AB} = 0

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

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

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

Released end supports: A, C (the moment at these ends is zero, and the member stiffness is taken as $3EI/L$ ).

Releasing A sends half of −16 to B: 16 + 8 = 24. From C: −36 − 18 = −54.

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: $DF_{BA} = 0.600$ $DF_{BC} = 0.400$

3EI/4 and 3EI/6 are in the ratio 0.6 to 0.4. The shorter member is stiffer, so it takes the larger share.

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.

	B
	$M_{BA}$	$M_{BC}$
DF	0.600	0.400
FEM	24.00	−54.00
Dist. 1	18.00	12.00
ΣM	42.00	−42.00

The unbalance is 24 − 54 = −30, so +30 is distributed. There's one joint and both ends were released first, so nothing carries over. Starting with 4EI/L, A and C would be balanced too, with carry-overs, and a few cycles would reach the same 42.

4Results and check

M_{BA} = 42\ \text{kN·m}

M_{BC} = -42\ \text{kN·m}

✓ The results match the matrix (direct stiffness) solution (largest difference 7.1 × 10⁻¹⁵ kN·m).

No external moment acts at B, so $M_{BA} = -M_{BC}$ must hold.

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	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	0	13.50	0
B	0	77.50	0
C	0	29.00	0

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 ✓

ΣM₀600.00 (reactions) − 600.00 (loads) = 0.00 kN·m ✓

Bending moment diagram of the beam solved by moment distribution: −42.00 at support B and a maximum of 35.04 kN·m in span — Figure 2. Bending moment (kN·m).

Shear diagram of the beam solved by moment distribution: 13.50 at A and 43.00 kN just right of B — Figure 3. Shear force (kN).

Results

$A_{y}$: 13.50 kN
$B_{y}$: 77.50 kN
$C_{y}$: 29.00 kN

Matches the independent stiffness solution.

Common mistakes

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.
Even with reversed distribution factors the joint total is still 1, so the error is easy to miss. The larger factor, 0.6, belongs to the short member BA.
Once the reactions are found, check vertical equilibrium: 13.5 + 77.5 + 29 = 12 × 10.

Same problem, another method

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