Slope Deflection · Problem 3 of 7All problems

Slope Deflection Method ExampleSlope deflection triangular load, three-span beam

Problem. The three-span beam has a pin at A and rollers at B, C, and D. All spans are 6 m. The outer spans carry a triangular load that is zero at the outer support and 12 kN/m at the inner support, and the middle span carries a 12 kN/m uniform load. 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 three 6 m spans: triangular load from 0 to 12 kN/m on the outer spans, 12 kN/m uniform load on the middle span — Figure 1. The beam, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: 6.48 kN
$B_{y}$: 65.52 kN
$C_{y}$: 65.52 kN
$D_{y}$: 6.48 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: $\theta_{A}$ $\theta_{B}$ $\theta_{C}$ $\theta_{D}$

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} = -14.40\ \text{kN·m},\qquad (\mathrm{FEM})_{BA} = 21.60\ \text{kN·m}

FEM table, row 4

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

FEM table, row 1

(\mathrm{FEM})_{CD} = -21.60\ \text{kN·m},\qquad (\mathrm{FEM})_{DC} = 14.40\ \text{kN·m}

FEM table, rows 1 + 4 (superposition)

A triangle's two ends aren't equal: wL²/20 = 21.6 at the heavy end, wL²/30 = 14.4 at the light end. The middle span has wL²/12 = 36.

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} = 0.667\,EI\theta_{A} + 0.333\,EI\theta_{B} - 14.40

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

\Rightarrow\; M_{BA} = 0.667\,EI\theta_{B} + 0.333\,EI\theta_{A} + 21.60

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

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

\Rightarrow\; M_{CD} = 0.667\,EI\theta_{C} + 0.333\,EI\theta_{D} - 21.60

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

\Rightarrow\; M_{DC} = 0.667\,EI\theta_{D} + 0.333\,EI\theta_{C} + 14.40

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} = 14.40

\text{(B)}\quad 0.333\,EI\theta_{A} + 1.333\,EI\theta_{B} + 0.333\,EI\theta_{C} = 14.40

\text{(C)}\quad 0.333\,EI\theta_{B} + 1.333\,EI\theta_{C} + 0.333\,EI\theta_{D} = -14.40

\text{(D)}\quad 0.333\,EI\theta_{C} + 0.667\,EI\theta_{D} = -14.40

5Solution (rotations)

EI\theta_{A} = 17.28\ \text{kN·m}^2

EI\theta_{B} = 8.64\ \text{kN·m}^2

EI\theta_{C} = -8.64\ \text{kN·m}^2

EI\theta_{D} = -17.28\ \text{kN·m}^2

The structure is symmetric: $\theta_D = -\theta_A$ and $\theta_C = -\theta_B$ . The results must follow that.

6End moments and check

Substitute the rotations back into the slope-deflection equations:

M_{AB} = 0\ \text{kN·m}

M_{BA} = 33.12\ \text{kN·m}

M_{BC} = -33.12\ \text{kN·m}

M_{CB} = 33.12\ \text{kN·m}

M_{CD} = -33.12\ \text{kN·m}

M_{DC} = 0\ \text{kN·m}

✓ The results match the matrix (direct stiffness) solution (largest difference 1.8 × 10⁻¹⁵ kN·m).

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	6.48	0
B	0	65.52	0
C	0	65.52	0
D	0	6.48	0

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

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

ΣFy144.00 (reactions) − 144.00 (loads) = 0.00 kN ✓

ΣM₀1296.00 (reactions) − 1296.00 (loads) = 0.00 kN·m ✓

Check $V_A$ by hand: (36 × 2 − 33.12) / 6 = 6.48. In the triangle, V = 0 gives $x^2 = 6.48$ , so x = 2.55 m and the span moment is 11.

Bending moment diagram of the beam with triangular loads: −33.12 at supports B and C and 20.88 kN·m in the middle span — Figure 2. Bending moment (kN·m).

Shear diagram of the beam with triangular loads: 6.48 at A and 36.00 kN just right of B — Figure 3. Shear force (kN).

Results

$A_{y}$: 6.48 kN
$B_{y}$: 65.52 kN
$C_{y}$: 65.52 kN
$D_{y}$: 6.48 kN

Matches the independent stiffness solution.

Common mistakes

Under a triangular load the largest span moment isn't at midspan. Its location comes from the zero-shear section, which takes a quadratic equation.
For a triangular load, wL²/20 belongs at the heavy end and wL²/30 at the light end. The two shouldn't be swapped.
Once the reactions are found, check vertical equilibrium: 2 × (6.48 + 65.52) = 2 × 36 + 72.

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