Slope Deflection · Problem 2 of 7All problems

Slope Deflection Method ExampleSlope deflection support settlement, two-span beam

Problem. The two-span beam has a pin at A and rollers at B and C. A uniform load of 10 kN/m acts over its full length, and support B settles 15 mm. EI = 30,000 kN·m². Use the slope-deflection method to find the end moments and support reactions, and draw the shear and moment diagrams.

Continuous beam with two 6 m spans: pin at A, rollers at B and C, 10 kN/m uniform load, support B settling 15 mm — Figure 1. The beam, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 28.75 kN
$B_{y}$: 62.50 kN
$C_{y}$: 28.75 kN

Solution

1Unknowns (joint rotations)

Slope-deflection is a displacement method: the unknowns are the rotations of the joints that are free to rotate. The rotation is zero at a fixed support. There is no sidesway; the chord rotation ψ caused by the support settlement is not an unknown but a known value (step 2). Sign convention (as in Hibbeler): end moments, joint rotations θ, and chord rotations ψ are positive clockwise, with end moments taken as acting on the member (counterclockwise on the joint). Kassimali's counterclockwise-positive values have the opposite sign.

Unknowns: $\theta_{A}$ $\theta_{B}$ $\theta_{C}$

EI is the same in every span, so it is factored out and the unknowns are solved for as $EI\theta$ (kN·m²).

2Fixed-end moments

All joints locked: fixed-end moments at the ends of the loaded members (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.

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

FEM table, row 1

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

FEM table, row 1

A support settles, so the span rotates by $\psi = \dfrac{\Delta_j - \Delta_i}{L}$ , and the term $-3\psi$ is added to the slope-deflection equation $M_{ij}=\tfrac{2EI}{L}(2\theta_i+\theta_j-3\psi)+(\mathrm{FEM})_{ij}$ . FEM table, row 7

\psi_{AB} = 2.500\times 10^{-3}\ \text{rad},\qquad -\tfrac{6EI}{L}\psi_{AB} = -75\ \text{kN·m}

\psi_{BC} = -2.500\times 10^{-3}\ \text{rad},\qquad -\tfrac{6EI}{L}\psi_{BC} = 75\ \text{kN·m}

$\psi$ isn't an unknown: 0.015 / 6 = 0.0025 rad. Its sign flips between the members, because A–B goes down to the right and B–C comes back up.

3Slope-deflection equations

For each member end, $M_{ij} = \tfrac{2EI}{L}(2\theta_i + \theta_j - 3\psi) + (\mathrm{FEM})_{ij}$ :

M_{AB} = \frac{2EI}{L}\left(2\theta_{A} + \theta_{B} - 3\psi_{AB}\right) + (\mathrm{FEM})_{AB}

\Rightarrow\; M_{AB} = 0.667\,EI\theta_{A} + 0.333\,EI\theta_{B} - 105

M_{BA} = \frac{2EI}{L}\left(2\theta_{B} + \theta_{A} - 3\psi_{AB}\right) + (\mathrm{FEM})_{BA}

\Rightarrow\; M_{BA} = 0.667\,EI\theta_{B} + 0.333\,EI\theta_{A} - 45

M_{BC} = \frac{2EI}{L}\left(2\theta_{B} + \theta_{C} - 3\psi_{BC}\right) + (\mathrm{FEM})_{BC}

\Rightarrow\; M_{BC} = 0.667\,EI\theta_{B} + 0.333\,EI\theta_{C} + 45

M_{CB} = \frac{2EI}{L}\left(2\theta_{C} + \theta_{B} - 3\psi_{BC}\right) + (\mathrm{FEM})_{CB}

\Rightarrow\; M_{CB} = 0.667\,EI\theta_{C} + 0.333\,EI\theta_{B} + 105

4Joint equilibrium equations

At each free joint the end moments sum to zero (when no external moment acts there): $\sum M = 0$ . This gives a system of linear equations in the unknowns:

\text{(A)}\quad 0.667\,EI\theta_{A} + 0.333\,EI\theta_{B} = 105

\text{(B)}\quad 0.333\,EI\theta_{A} + 1.333\,EI\theta_{B} + 0.333\,EI\theta_{C} = 0

\text{(C)}\quad 0.333\,EI\theta_{B} + 0.667\,EI\theta_{C} = -105

5Solution (rotations)

EI\theta_{A} = 157.50\ \text{kN·m}^2

EI\theta_{B} = 0\ \text{kN·m}^2

EI\theta_{C} = -157.50\ \text{kN·m}^2

The structure and load are symmetric, so $\theta_B = 0$ and $\theta_A = -\theta_C$ . Use that to check your solution.

6End moments and check

Substitute the rotations back into the slope-deflection equations:

M_{AB} = 0\ \text{kN·m}

M_{BA} = 7.50\ \text{kN·m}

M_{BC} = -7.50\ \text{kN·m}

M_{CB} = 0\ \text{kN·m}

✓ The results match the matrix (direct stiffness) solution (largest difference 1.8 × 10⁻¹⁴ kN·m).

Check $M_{BA}$ by hand: 0.333 × 157.5 − 45 = 7.5.

7Member 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.

8Support reactions and equilibrium check

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	0	28.75	0
B	0	62.50	0
C	0	28.75	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 ✓

ΣM₀720.00 (reactions) − 720.00 (loads) = 0.00 kN·m ✓

Bending moment diagram of the settling beam solved by slope deflection: −7.50 at B and a maximum of 41.33 kN·m in span — Figure 2. Bending moment (kN·m).

Shear diagram of the settling beam solved by slope deflection: 28.75 at A and 31.25 kN just right of B — Figure 3. Shear force (kN).

Results

$A_{y}$: 28.75 kN
$B_{y}$: 62.50 kN
$C_{y}$: 28.75 kN

Matches the independent stiffness solution.

Common mistakes

If the sign of $\psi$ is reversed, the result can still satisfy statics, so an equilibrium check won't catch the error. An open-access textbook makes this mistake in four examples.
The fixed-end moments from load and the settlement terms are computed separately and added. Here −30 and +30 come from the load, −75 and +75 from settlement.
Once the reactions are found, check vertical equilibrium: 28.75 + 62.5 + 28.75 = 10 × 12.

Same problem, another method

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