Slope Deflection · Problem 4 of 7All problems

Slope Deflection Method ExampleSlope deflection frame example: three-member joint

Problem. Three members meet at joint B of the frame. A and C are fixed, and D is pinned. Column A–B is 4 m and column B–C is 3 m, both with I; beam B–D is 6 m with 2I. A uniform load of 12 kN/m acts on the beam. Axial deformation is neglected. Use the slope-deflection method to find the end moments, and draw the shear and moment diagrams.

Frame with three members at joint B: fixed at A and C, pinned at D, 4 m and 3 m columns with I, 6 m beam with 2I, 12 kN/m uniform load on the beam — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: 6.08 kN
$M_{A}$: −8.10 kN·m
$C_{x}$: −10.80 kN
$M_{C}$: −10.80 kN·m
$D_{x}$: 4.73 kN
$D_{y}$: 29.70 kN
$A_{y} + C_{y}$: 42.30 kN

Solution

1Unknowns and assumptions

Slope-deflection is a displacement method: the unknowns are the rotations of the joints that are free to rotate. This frame has no sidesway, so $\psi = 0$ for every member; axial deformation is neglected ( $EA \approx \infty$ , the usual hand-method assumption). 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.

Fixed (θ = 0): $\theta_{A} = 0$ $\theta_{C} = 0$
Unknowns: $\theta_{B}$ $\theta_{D}$

The members have different EI; each member carries its ratio $k = EI/EI_0$ (table below), and the unknowns are solved for as $EI_0\theta$ .

There's no sway: A, C, and D hold the frame horizontally and the members don't change length. Only B and D rotate.

2Fixed-end moments

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.

All joints locked: fixed-end moments at the ends of the loaded members (kN·m)

Member	L (m)	EI/EI₀	(FEM)ᵢⱼ (kN·m)	(FEM)ⱼᵢ (kN·m)
A–B	4.00	1.00	0	0
B–C	3.00	1.00	0	0
B–D	6.00	2.00	-36.00	36.00

B–D: FEM table, row 1

3Slope-deflection equations

For each member end, $M_{ij} = \tfrac{2EI}{L}\left(2\theta_i + \theta_j\right) + (\mathrm{FEM})_{ij}$ (ψ = 0):

M_{AB} = 0.500\,EI\theta_{B} + 0

M_{BA} = 1.000\,EI\theta_{B} + 0

M_{BC} = 1.333\,EI\theta_{B} + 0

M_{CB} = 0.667\,EI\theta_{B} + 0

M_{BD} = 1.333\,EI\theta_{B} + 0.667\,EI\theta_{D} - 36

M_{DB} = 1.333\,EI\theta_{D} + 0.667\,EI\theta_{B} + 36

4Joint equilibrium equations

At each joint free to rotate, the end moments balance the external moment: $\sum M_{\text{end}} + M_{\text{ext}} = 0$ . This gives a system of linear equations in the unknowns:

\text{(B)}\quad 3.667\,EI\theta_{B} + 0.667\,EI\theta_{D} = 36

\text{(D)}\quad 0.667\,EI\theta_{B} + 1.333\,EI\theta_{D} = -36

Three members add up at B: the coefficient of $\theta_B$ is 1 + 1.333 + 1.333 = 3.667. Skip one and the equation still solves, but the result is wrong.

5Solution (rotations)

EI_0\theta_{B} = 16.20\ \text{kN·m}^2

EI_0\theta_{D} = -35.10\ \text{kN·m}^2

6End moments and check

Substitute the rotations back into the slope-deflection equations:

M_{AB} = 8.10\ \text{kN·m}

M_{BA} = 16.20\ \text{kN·m}

M_{BC} = 21.60\ \text{kN·m}

M_{CB} = 10.80\ \text{kN·m}

M_{BD} = -37.80\ \text{kN·m}

M_{DB} = 0\ \text{kN·m}

✓ The results match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 2.3 × 10⁻⁶ kN·m).

First check: $M_{DB} = 0$ , because D is pinned. At B, 16.2 + 21.6 − 37.8 = 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

Because axial deformation is neglected, the end moments cannot tell how axial force in a straight member chain between two supports is shared between them. The results table gives only the sum of those reactions; the other reactions and the M and V diagrams are not affected.

Find V_D by hand: (12 × 6 × 3 − 37.8) / 6 = 29.7. The remaining 72 − 29.7 = 42.3 kN goes down the column line.

Bending moment diagram of the frame: 16.20, 21.60, and 37.80 kN·m at joint B — Figure 2. Bending moment (kN·m).

Shear diagram of the frame: 6.08 in the lower column, 10.80 in the upper column, and 42.30 kN at the B end of the beam — Figure 3. Shear force (kN).

Results

$A_{x}$: 6.08 kN
$M_{A}$: −8.10 kN·m
$C_{x}$: −10.80 kN
$M_{C}$: −10.80 kN·m
$D_{x}$: 4.73 kN
$D_{y}$: 29.70 kN
$A_{y} + C_{y}$: 42.30 kN

Matches the independent stiffness solution.

Common mistakes

Three members meet at joint B, and the end moments of all three enter its equilibrium equation.
D is pinned and can't be treated as fixed. It has its own equation, and its end moment must come out zero.
With EA ≈ ∞ the total vertical load on the columns is known ( $V_A + V_C = 42.3$ kN), but the end moments can't give each column's share.

Same problem, another method

Moment distribution

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