Slope Deflection · Problem 7 of 7All problems

Slope Deflection Method ExampleSlope deflection unequal columns, overhang and sidesway

Problem. The frame is fixed at A and E. Column A–B is 4 m with 2I, column E–C is 6 m with 2I, and B–C and C–D are 4 m with I; D is free. Column A–B carries a lateral triangular load that is 12 kN/m at A and zero at B, and overhang C–D carries a 10 kN/m uniform load. Axial deformation is neglected. Use the slope-deflection method to find the end moments and support reactions, and draw the shear and moment diagrams.

Frame with unequal columns: fixed at A and E, 4 m and 6 m columns with 2I, 4 m beam, 4 m overhang; lateral triangular load from 12 kN/m to zero on the column, 10 kN/m uniform load on the overhang — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: −32.24 kN
$A_{y}$: −16.22 kN
$M_{A}$: 41.92 kN·m
$E_{x}$: 8.24 kN
$E_{y}$: 56.22 kN
$M_{E}$: −11.28 kN·m

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 sways: for each story, a translation unknown $\Delta$ and a story shear equation are added; in the columns the chord rotation is $\psi = \Delta/h$ . 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): $\theta_{A} = 0$ $\theta_{E} = 0$
Unknowns: $\theta_{B}$ $\theta_{C}$ $\Delta_{1}$

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

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	2.00	-9.60	6.40
B–C	4.00	1.00	0	0
C–D (overhang)	4.00	1.00	—	—
E–C	6.00	2.00	0	0

The end moment of an overhang (cantilevered member) is known from statics; it is not an unknown in the equations and enters the equilibrium of its joint as a known value.

Chord rotations of the columns (from story drift):

$\psi_{AB} = \Delta_{1}/4$ $\psi_{EC} = \Delta_{1}/6$

The columns have different heights, so the ψ values differ: $\psi_{AB} = \Delta_1/4$ , $\psi_{EC} = \Delta_1/6$ . Triangular load FEM: wL²/20 = 9.6 and wL²/30 = 6.4. Overhang: 10 × 4 × 2 = 80.

3Slope-deflection equations

For each member end, $M_{ij} = \tfrac{2EI}{L}\left(2\theta_i + \theta_j - 3\psi\right) + (\mathrm{FEM})_{ij}$ (ψ = Δ/h in the columns):

M_{AB} = 1.000\,EI\theta_{B} - 0.750\,EI\Delta_{1} - 9.60

M_{BA} = 2.000\,EI\theta_{B} - 0.750\,EI\Delta_{1} + 6.40

M_{BC} = 1.000\,EI\theta_{B} + 0.500\,EI\theta_{C} + 0

M_{CB} = 1.000\,EI\theta_{C} + 0.500\,EI\theta_{B} + 0

M_{CD} = -80\ \text{kN·m}\quad\text{(overhang: known from statics)}

M_{EC} = 0.667\,EI\theta_{C} - 0.333\,EI\Delta_{1} + 0

M_{CE} = 1.333\,EI\theta_{C} - 0.333\,EI\Delta_{1} + 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{(B)}\quad 3.000\,EI\theta_{B} + 0.500\,EI\theta_{C} - 0.750\,EI\Delta_{1} = -6.40

\text{(C)}\quad 0.500\,EI\theta_{B} + 2.333\,EI\theta_{C} - 0.333\,EI\Delta_{1} = 80

(Known moment from the overhang at joint C: -80.00 kN·m, moved to the right-hand side.)

The 80 on C's right-hand side comes from the overhang: a known moment, not an unknown. No equation is written for the free end.

5Story shear (ΣFx) equations

Joint equilibrium cannot determine the sidesway, so a horizontal shear equation is written for a cut through each story (the column shears at the cut resist the lateral loads above it). In terms of the unknowns:

\text{(Story 1)}\quad -0.750\,EI\theta_{B} - 0.333\,EI\theta_{C} + 0.486\,EI\Delta_{1} = 7.20

Right-hand side 7.20 kN: fixed-end force that the load on member A–B transfers above the cut, 7.20 kN.

The 7.2 on the right-hand side is the fixed-end force the triangular load gives the top of the column: 3wL/20 = 3 × 12 × 4 / 20. The total lateral load is 24 kN, and 16.8 of it goes straight to A.

6Solution (rotations and translations)

EI_0\theta_{B} = 2.87\ \text{kN·m}^2

EI_0\theta_{C} = 40.37\ \text{kN·m}^2

EI_0\Delta_{1} = 46.92\ \text{kN·m}^3\;\;(\Delta_{1} = 1.56\ \text{mm})

7End moments and check

Substitute the rotations back into the slope-deflection equations:

M_{AB} = -41.92\ \text{kN·m}

M_{BA} = -23.06\ \text{kN·m}

M_{BC} = 23.06\ \text{kN·m}

M_{CB} = 41.81\ \text{kN·m}

M_{CD} = -80\ \text{kN·m}

M_{EC} = 11.28\ \text{kN·m}

M_{CE} = 38.19\ \text{kN·m}

✓ The results match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 5.1 × 10⁻⁶ kN·m).

8Member 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.

Story shear check: the sum of the column end shears must resist the lateral load above the cut:

Σ column shears = 24.00 kN; lateral load above the cut = 24.00 kN ✓

9Support reactions and equilibrium check

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	-32.24	-16.22	41.92
E	8.24	56.22	-11.28

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

ΣFx-24.00 (reactions) + 24.00 (loads) = 0.00 kN ✓

ΣFy40.00 (reactions) − 40.00 (loads) = 0.00 kN ✓

ΣM₀320.00 (reactions) − 320.00 (loads) = 0.00 kN·m ✓

$V_A$ is negative: A is in tension. The overhang load turns the frame about C. Vertical total: −16.22 + 56.22 = 10 × 4.

Bending moment diagram of the frame with unequal columns: 41.92 at A, 80.00 at the C end of the overhang, and 11.28 kN·m at E — Figure 2. Bending moment (kN·m).

Shear diagram of the frame with unequal columns: 32.24 at A, 8.24 in column E–C, and 40.00 kN at the C end of the overhang — Figure 3. Shear force (kN).

Results

$A_{x}$: −32.24 kN
$A_{y}$: −16.22 kN
$M_{A}$: 41.92 kN·m
$E_{x}$: 8.24 kN
$E_{y}$: 56.22 kN
$M_{E}$: −11.28 kN·m

Matches the independent stiffness solution.

Common mistakes

The sway is the same in both columns, but ψ = Δ/h depends on column height. Here it is Δ/4 and Δ/6.
If the sign of $\psi$ or of a fixed-end moment is reversed, the result can still satisfy statics, so an equilibrium check won't catch the error.
The overhang moment isn't an unknown. Statics gives 80 kN·m, which goes directly into the equilibrium equation at C.

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