Slope Deflection · Problem 5 of 7All problems

Slope Deflection Method ExampleSlope deflection multi-span frame

Problem. The frame is pinned at A and fixed at D, E, and F. The beam spans are 6 m, and the columns under B and C are 4 m. A 40 kN point load acts at the middle of B–C, and a 10 kN/m uniform load acts on C–D. EI is constant and axial deformation is neglected. Use the slope-deflection method to find the end moments, and draw the shear and moment diagrams.

Three-span frame: pinned at A, fixed at D, E, and F, 6 m spans, 4 m columns, 40 kN point load at the middle of B–C, 10 kN/m uniform load on C–D — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{y}$: −1.18 kN
$D_{y}$: 29.66 kN
$M_{D}$: −29.33 kN·m
$E_{x}$: 5.31 kN
$E_{y}$: 19.16 kN
$M_{E}$: −7.08 kN·m
$F_{x}$: −0.76 kN
$F_{y}$: 52.36 kN
$M_{F}$: 1.01 kN·m
$A_{x} + D_{x}$: −4.55 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): $\theta_{D} = 0$ $\theta_{E} = 0$ $\theta_{F} = 0$
Unknowns: $\theta_{A}$ $\theta_{B}$ $\theta_{C}$

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

A is pinned, so it rotates too: $\theta_A$ , $\theta_B$ , $\theta_C$ . An exam sometimes asks only for these equations.

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)	(FEM)ᵢⱼ (kN·m)	(FEM)ⱼᵢ (kN·m)
A–B	6.00	0	0
B–C	6.00	-30.00	30.00
C–D	6.00	-30.00	30.00
E–B	4.00	0	0
F–C	4.00	0	0

B–C: FEM table, row 2 C–D: FEM table, row 1

For B–C, PL/8 = 40 × 6 / 8 = 30. For C–D, wL²/12 = 10 × 36 / 12 = 30. Both are 30, so the fixed-end moments cancel at C.

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} = 0.667\,EI\theta_{A} + 0.333\,EI\theta_{B} + 0

M_{BA} = 0.667\,EI\theta_{B} + 0.333\,EI\theta_{A} + 0

M_{BC} = 0.667\,EI\theta_{B} + 0.333\,EI\theta_{C} - 30

M_{CB} = 0.667\,EI\theta_{C} + 0.333\,EI\theta_{B} + 30

M_{CD} = 0.667\,EI\theta_{C} - 30

M_{DC} = 0.333\,EI\theta_{C} + 30

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

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

M_{FC} = 0.500\,EI\theta_{C} + 0

M_{CF} = 1.000\,EI\theta_{C} + 0

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{(A)}\quad 0.667\,EI\theta_{A} + 0.333\,EI\theta_{B} = 0

\text{(B)}\quad 0.333\,EI\theta_{A} + 2.333\,EI\theta_{B} + 0.333\,EI\theta_{C} = 30

\text{(C)}\quad 0.333\,EI\theta_{B} + 2.333\,EI\theta_{C} = 0

Coefficient of $\theta_B$ : 0.667 + 0.667 + 1 = 2.333. The column gives 4EI/4 = 1 and the beams 4EI/6 = 0.667. C's right-hand side is 0, because 30 − 30 = 0.

5Solution (rotations)

EI\theta_{A} = -7.08\ \text{kN·m}^2

EI\theta_{B} = 14.16\ \text{kN·m}^2

EI\theta_{C} = -2.02\ \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} = 7.08\ \text{kN·m}

M_{BC} = -21.24\ \text{kN·m}

M_{CB} = 33.37\ \text{kN·m}

M_{CD} = -31.35\ \text{kN·m}

M_{DC} = 29.33\ \text{kN·m}

M_{EB} = 7.08\ \text{kN·m}

M_{BE} = 14.16\ \text{kN·m}

M_{FC} = -1.01\ \text{kN·m}

M_{CF} = -2.02\ \text{kN·m}

✓ The results match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 3.6 × 10⁻⁶ kN·m).

Check: at B, 7.08 − 21.24 + 14.16 = 0. At the column end, $M_{EB} = M_{BE}/2$ , because E is fixed.

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.

Bending moment diagram of the multi-span frame: 21.24 at the B end of B–C, 33.37 at its C end, and 29.33 kN·m at D — Figure 2. Bending moment (kN·m).

Shear diagram of the multi-span frame: 17.98 at the B end of B–C and 30.34 kN at the C end of C–D — Figure 3. Shear force (kN).

Results

$A_{y}$: −1.18 kN
$D_{y}$: 29.66 kN
$M_{D}$: −29.33 kN·m
$E_{x}$: 5.31 kN
$E_{y}$: 19.16 kN
$M_{E}$: −7.08 kN·m
$F_{x}$: −0.76 kN
$F_{y}$: 52.36 kN
$M_{F}$: 1.01 kN·m
$A_{x} + D_{x}$: −4.55 kN

Matches the independent stiffness solution.

Common mistakes

Three members meet at each of joints B and C. The column end moment must be included in the equilibrium equation.
A is pinned, so its rotation can't be taken as zero. $\theta_A$ is an unknown with its own equation.
A load's fixed-end moments have opposite signs at the two ends of a member. Writing them with the same sign breaks joint equilibrium.

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