Slope Deflection · Problem 1 of 7All problems

Slope Deflection Method ExampleTwo-span continuous beam

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 the slope-deflection method 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

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. With no sidesway and no support settlement, $\psi = 0$ . 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).

A and C are pinned, so they rotate too. That gives three unknowns and three equations.

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

FEM table, row 1

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

FEM table, row 1

3Slope-deflection equations

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

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

\Rightarrow\; M_{AB} = 1.000\,EI\theta_{A} + 0.500\,EI\theta_{B} - 16

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

\Rightarrow\; M_{BA} = 1.000\,EI\theta_{B} + 0.500\,EI\theta_{A} + 16

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

\Rightarrow\; M_{BC} = 0.667\,EI\theta_{B} + 0.333\,EI\theta_{C} - 36

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

\Rightarrow\; M_{CB} = 0.667\,EI\theta_{C} + 0.333\,EI\theta_{B} + 36

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 1.000\,EI\theta_{A} + 0.500\,EI\theta_{B} = 16

\text{(B)}\quad 0.500\,EI\theta_{A} + 1.667\,EI\theta_{B} + 0.333\,EI\theta_{C} = 20

\text{(C)}\quad 0.333\,EI\theta_{B} + 0.667\,EI\theta_{C} = -36

A shortcut: treat A and C as pinned ends from the start and $\tfrac{3EI}{4} + \tfrac{3EI}{6}$ leaves one unknown: 1.25 $EI\theta_B$ = 54 − 24, so $EI\theta_B = 24$ . The result is the same.

5Solution (rotations)

EI\theta_{A} = 4\ \text{kN·m}^2

EI\theta_{B} = 24\ \text{kN·m}^2

EI\theta_{C} = -66\ \text{kN·m}^2

6End moments and check

Substitute the rotations back into the slope-deflection equations:

M_{AB} = 0\ \text{kN·m}

M_{BA} = 42\ \text{kN·m}

M_{BC} = -42\ \text{kN·m}

M_{CB} = 0\ \text{kN·m}

✓ The results match the matrix (direct stiffness) solution (largest difference 8.9 × 10⁻¹⁵ kN·m).

First check: $M_{AB}$ and $M_{CB}$ must be zero, because A and C are pins. At B, $M_{BA} + M_{BC} = 0$ .

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	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 slope deflection: −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 slope deflection: 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, the fixed-end moments must also be adjusted for the pinned end. Here 24 and 54 are used instead of 16 and 36.
If $\theta_A$ and $\theta_C$ are unknowns, joints A and C need equilibrium equations too. Without them there are more unknowns than equations.
Once the reactions are found, check vertical equilibrium: 13.5 + 77.5 + 29 = 12 × 10.

Same problem, another method

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