Slope Deflection · Problem 6 of 7All problems

Slope Deflection Method ExampleSlope deflection frame with sidesway: portal

Problem. The portal frame is fixed at A and D. The columns are 4 m and beam B–C is 6 m. A uniform load of 10 kN/m acts on the beam, and a 30 kN lateral load acts to the right at joint B. EI is constant and axial deformation is neglected. Use the slope-deflection method to find the end moments and support reactions, and draw the shear and moment diagrams.

Portal frame: fixed at A and D, 4 m columns, 6 m beam, 10 kN/m uniform load on the beam, 30 kN lateral load at joint B — Figure 1. The frame, its loads, and its dimensions.

Try it yourself before reading the solution. Show the results

$A_{x}$: −6.56 kN
$A_{y}$: 22.00 kN
$M_{A}$: 24.75 kN·m
$D_{x}$: −23.44 kN
$D_{y}$: 38.00 kN
$M_{D}$: 47.25 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): $\theta_{A} = 0$ $\theta_{D} = 0$
Unknowns: $\theta_{B}$ $\theta_{C}$ $\Delta_{1}$

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

There's a lateral load and nothing holds the column tops, so the story sways. Three unknowns: $\theta_B$ , $\theta_C$ , $\Delta_1$ .

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	4.00	0	0
B–C	6.00	-30.00	30.00
D–C	4.00	0	0

B–C: FEM table, row 1

Chord rotations of the columns (from story drift):

$\psi_{AB} = \Delta_{1}/4$ $\psi_{DC} = \Delta_{1}/4$

The columns have equal height, so both have $\psi = \Delta_1/4$ . The beam doesn't rotate under sway: $\psi = 0$ .

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} = 0.500\,EI\theta_{B} - 0.375\,EI\Delta_{1} + 0

M_{BA} = 1.000\,EI\theta_{B} - 0.375\,EI\Delta_{1} + 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_{DC} = 0.500\,EI\theta_{C} - 0.375\,EI\Delta_{1} + 0

M_{CD} = 1.000\,EI\theta_{C} - 0.375\,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 1.667\,EI\theta_{B} + 0.333\,EI\theta_{C} - 0.375\,EI\Delta_{1} = 30

\text{(C)}\quad 0.333\,EI\theta_{B} + 1.667\,EI\theta_{C} - 0.375\,EI\Delta_{1} = -30

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 -0.375\,EI\theta_{B} - 0.375\,EI\theta_{C} + 0.375\,EI\Delta_{1} = 30

Right-hand side 30.00 kN: lateral load at joint B, 30.00 kN.

The third equation doesn't come from a joint: the column shears must carry the 30 kN. A column's shear is $-(M_{ij} + M_{ji})/h$ .

6Solution (rotations and translations)

EI\theta_{B} = 46.50\ \text{kN·m}^2

EI\theta_{C} = 1.50\ \text{kN·m}^2

EI\Delta_{1} = 128\ \text{kN·m}^3\;\;(\Delta_{1} = 4.27\ \text{mm})

7End moments and check

Substitute the rotations back into the slope-deflection equations:

M_{AB} = -24.75\ \text{kN·m}

M_{BA} = -1.50\ \text{kN·m}

M_{BC} = 1.50\ \text{kN·m}

M_{CB} = 46.50\ \text{kN·m}

M_{DC} = -47.25\ \text{kN·m}

M_{CD} = -46.50\ \text{kN·m}

✓ The results match the method's own reference, the axially rigid (EA ≈ ∞) matrix solution (largest difference 2.1 × 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 = 30.00 kN; lateral load above the cut = 30.00 kN ✓

Find the column shears by hand: (24.75 + 1.5) / 4 = 6.56 and (47.25 + 46.5) / 4 = 23.44. They total 30, the lateral load.

9Support reactions and equilibrium check

Equilibrium check: loads and support reactions together

Support	Rx (kN)	Ry (kN)	M (kN·m, ccw +)
A	-6.56	22.00	24.75
D	-23.44	38.00	47.25

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

ΣFx-30.00 (reactions) + 30.00 (loads) = 0.00 kN ✓

ΣFy60.00 (reactions) − 60.00 (loads) = 0.00 kN ✓

ΣM₀300.00 (reactions) − 300.00 (loads) = 0.00 kN·m ✓

Bending moment diagram of the swaying portal frame: 24.75 at A, 1.50 at B, 46.50 at C, and 47.25 kN·m at D — Figure 2. Bending moment (kN·m).

Shear diagram of the swaying portal frame: 6.56 in the left column and 23.44 kN in the right column — Figure 3. Shear force (kN).

Results

$A_{x}$: −6.56 kN
$A_{y}$: 22.00 kN
$M_{A}$: 24.75 kN·m
$D_{x}$: −23.44 kN
$D_{y}$: 38.00 kN
$M_{D}$: 47.25 kN·m

Matches the independent stiffness solution.

Common mistakes

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. An open-access textbook makes this mistake in four examples.
Joint equations can't determine the sway. Without the story shear equation, the system is one equation short.
All lateral loads above the cut go on the right-hand side of the story shear equation.

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