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Slope-Deflection Method for
Analysis of Statically
Indeterminate Structures
Slope-Deflection Method
Principle of the Slope-Deflection Method
The displacement method of analysis, which includes the slope-deflection method, consists of writing equations that satisfy equilibrium and the force- displacement requirements. This provides a direct solution for unknown displacements. Once the displacements have been obtained, the member forces and support reactions can be found from compatibility equations.
The key steps in the displacement method are:
Label all supports and joints (nodes), identify degrees of freedom (i.e. displacement and/or rotation) at these nodes, and identify the spans between nodes. Write slope-deflection equations for each span that relate moment and displacement at each node of the span. Write equilibrium equations for each unknown degree of freedom (at a node) of the structure and solve for the unknown nodal displacements. Use the slope-deflection equations to determine the member forces.
Slope-Deflection Equations
The slope-deflection equations relate the unknown rotations and translations to the loads applied on a structure. The equations are derived by considering the moments at each node due to:
Angular displacement of the near end (θA) Angular displacement of the far end (θB) Rotation of the chord (ψ = Δ/L) Applied loads
The general slope-deflection equations for a span are:
Near end moment: M_N = (2EI/L)θ_A + (EI/L)θ_B + (6EI/L^2)Δ + (FEM)_N
Far end moment: M_F = (EI/L)θ_A + (2EI/L)θ_B - (6EI/L^2)Δ + (FEM)_F
Where: - M_N, M_F = internal moments at the near and far ends of the span
- θ_A, θ_B = angular displacements/rotations at the near and far ends - Δ = relative displacement between the ends - (FEM)_N, (FEM)_F = fixed-end moments at the near and far ends - E = Young's modulus, I = moment of inertia, L = span length
Example 1: Beams
Consider the following simply supported beam with a concentrated load at midspan.
Step 1: Identify the degrees of freedom and spans. - There are 3 nodal rotations: θ_A = 0, θ_B, θ_C = 0 - There are 2 spans: AB and BC - There is 1 unknown deflection
Step 2: Determine the fixed-end moments. - For span AB: FEM_AB = 0, FEM_BA = 0 - For span BC: FEM_BC = -30(6)/20 = -7.2 kNm, FEM_CB = 30(6)/20 = 10.8 kNm
Step 2b: Write the slope-deflection equations. Span AB: M_AB = (2EI/L)θ_A
- (EI/L)θ_B + (6EI/L^2)Δ + 0 M_BA = (EI/L)θ_A + (2EI/L)θ_B - (6EI/L^2)Δ
- 0
Span BC: M_BC = (2EI/L)θ_B + (EI/L)θ_C + (6EI/L^2)Δ + (-7.2) M_CB = (EI/L)θ_B + (2EI/L)θ_C - (6EI/L^2)Δ + 10.
Step 3: Write the equilibrium equations at node B. Using the positive sign convention for bending moments at the joint as anti-clockwise: ΣM_B = M_AB + M_BC = 0 (2EI/L)θ_B + (EI/L)θ_B + (6EI/L^2)Δ + (-7.2) = 0 3.5EIθ_B + 6EIΔ/L = 7.
Solving for the unknown displacement Δ: Δ = 6.17 mm
Step 4: Determine the member forces. Span AB: M_A = (2EI/L)θ_A + (EI/ L)θ_B + (6EI/L^2)Δ + 0 = 1.54 kNm M_B = (EI/L)θ_A + (2EI/L)θ_B - (6EI/ L^2)Δ + 0 = 3.09 kNm
Span BC: M_B = (2EI/L)θ_B + (EI/L)θ_C + (6EI/L^2)Δ + (-7.2) = -3.09 kNm M_C = (EI/L)θ_B + (2EI/L)θ_C - (6EI/L^2)Δ + 10.8 = -12.86 kNm
The shear force and bending moment diagrams can then be plotted based on these results.
Slope-Deflection Method: Beams
Example 1: Beams
Step 5(a): Determine Support Reactions
Use the free body diagram and equilibrium equations of each segment to determine the support reactions.
The support reactions are: - A: 0.58 kN, 1.54 kNm - B: 4.95 kN, 3.09 kNm - C: 13.63 kN, -3.09 kNm, 12.86 kNm
Span CD: MCD = 0.444EI/L * θC - 0.00889EI/L * Δ = -182 kNm MDC = 0.889EI/L * θC - 0.00889EI/L * Δ = 151 kNm
Determine Support Reactions
Use the free body diagram and equilibrium equations of each segment to determine the support reactions.
The support reactions are: - A: 66.7 kN, 73.5 kNm - B: 121.1 kN, 112 kNm - C: 43.8 kN, -151 kNm - D: 74 kN, 182 kNm
Shear Force and Bending Moment Diagrams
Draw the shear force and bending moment diagrams for the beam.
Shear Force Diagram: - 66.7 kN, -77.3 kN, 43.8 kN, -74 kN
Bending Moment Diagram: - -182 kNm, -73.5 kNm, -112 kNm, 151 kNm
Use a software to check whether the results are correct.
Exercise 1: Beams
Draw the shear force and bending moment diagrams for the given beam, and find the maximum bending moment value along the beam. Assume that EI is constant.
Shear Force Diagram: - 15 kN, 127.5 kN, -112.5 kN
Bending Moment Diagram: - -135 kNm, 68.2 kNm, -90 kNm
The maximum bending moment value along the beam is 135 kNm.
Exercise 2: Beams
Determine the maximum bending moment along the beam if support B settles by 80 mm.
Bending Moment Diagram: - -24 kNm, 3 kNm, -3 kNm
The maximum bending moment along the beam is 24 kNm.
Exercise 3: Beams
Determine the support reactions and draw the bending moment diagram for the given beam.
Support Reactions: - A: 116.7 kN - B: 116.7 kN - C: 33.3 kN
Bending Moment Diagram: - 100 kNm, 200 kNm, 300 kNm, 500 kNm
Exercise 4: Beams
Determine the support reactions and draw the bending moment diagram for the given beam.
Support Reactions: - A: 82.0 kN - B: 187.1 kN - C: 115.9 kN - D: 67.5 kNm - E: 122.1 kNm
Bending Moment Diagram: - 73.9 kNm, 126 kNm, 38.2 kNm
Slope-Deflection Method: Planar Frames
What is Sidesway?
Sidesway is the horizontal translation at the nodes of a frame when it is subject to loads. Sidesway is typically caused by horizontal loads, but could also be caused by vertical loads. Sidesway is negligible if there are sufficient constraints against horizontal translation at the nodes.
Example 3: Frames without Sidesway
Step 1: Identify Degrees of Freedom and Spans
Identify the degrees of freedom (i.e., displacement and/or rotation at nodes) and the spans between nodes.
4 nodal rotations: θA = 0, θB, θC, and θD = 0 3 spans: AB, BC, and CD Two unknown deflections
Determine Fixed-End Moments
Span BC: FEM = -80 kNm FEM = 80 kNm
Spans AB and CD: FEM = 0
Step 2(b): Write Slope-Deflection Equations
Span AB:
M_AB = (2EI/L_AB^2)[(θ_A - θ_B) + (3/L_AB)v_A] M_BA = (2EI/ L_AB^2)[(θ_B - θ_A) + (3/L_AB)v_B]
Span BC:
M_BC = (2EI/L_BC^2)[-(θ_B - θ_C) + (3/L_BC)v_B] M_CB = (2EI/ L_BC^2)[-(θ_C - θ_B) + (3/L_BC)v_C]
Bending Moment Diagram: - Pin support at A and D - Rigid joint at B and C - Applied loads: - 88.75 kN at B - 61.25 kN at C - 7.8 kN at A - 42.2 kN at D - Maximum bending moment: - 62.5 kNm at B - 68.75 kNm at C
Exercise 7: Frames without Sidesway
Determine the support reactions and draw the bending moment diagram for this frame.
Bending Moment Diagram: - Pin support at A, B, C, and D - Applied loads: - 14.4 kNm at A - 17 kNm at B - 11.4 kNm at C - 22.7 kNm at D - Support reactions: - 56.8 kN at A - 38.2 kN at B - 6.8 kN at C - 6.8 kN at D - Maximum bending moment: - 60.2 kNm at B - 26.1 kNm at C
Exercise 8: Frames without Sidesway
Determine the support reactions and draw the bending moment diagram for this frame if supports A and D settle by 17 mm and 25 mm respectively.
Bending Moment Diagram: - Rigid support at B and E - Pin support at A, C, and D - Applied loads: - 234 kN at B - 132 kN at E - Support reactions: - 63 kN at A - 53 kN at D - 68 kN at C - Maximum bending moment: - 398 kNm at B - 150 kNm at C - 128 kNm at D
Slope-Deflection Method: Planar Frames
What is Sidesway?
Sidesway is the horizontal translation at the nodes of a frame when it is subject to loads. It is typically caused by horizontal loads, but could also be caused by vertical loads.
Planar frame without inclined members: - No sidesway
Planar frame with inclined members: - Sidesway is possible
Example 5: Frames with Sidesway
Draw the bending moment diagram for this frame. Assume that EI is constant.
Step 1: Identify degrees of freedom and spans - 4 nodal rotations: θA = 0, θB, θC, and θD = 0 - 2 nodal translations: B = C = Δ - 3 spans: AB, BC, and CD - 3 unknown deflections
Step 2(a): Determine fixed-end moments - Span AB: FEM_AB = 0, FEM_BA = 0 - Span BC: FEM_BC = 0, FEM_CB = 0 - Span CD: FEM_CD = 0, FEM_DC = 0
Step 2(b): Determine chord rotations - Span AB: θAB = Δ/4, θBA = -Δ/4 - Span BC: θBC = 0, θCB = 0 - Span CD: θCD = Δ/6, θDC = -Δ/
Step 3(a): Write equilibrium equations at nodes - Node B: 9Δ/5 - 2Δ/5 - 3Δ/ = 0 - Node C: 4Δ/5 - 2Δ/5 - 2Δ/3 = 0
Step 3(b): Write equilibrium equations for the structure - ∑Fx = 0: 3Δ/8 - Δ/
Step 3(c): Solve unknown deflection - Solve the system of equations to find Δ = 1 mm
Step 4: Determine member forces - Span AB: MAB = -347 kNm, MBA = - kNm - Span BC: MBC = 225 kNm, MCB = 158 kNm - Span CD: MCD = - kNm, MDC = -183 kNm
Determine support reactions: - A: 143 kN - B: 143 kN - C: 76.6 kN - D: 57 kN
Draw the bending moment diagram.
Example 6: Frames with Sidesway
Draw the bending moment diagram for this frame. Assume that EI is constant, while joint C is pin-connected.
5 nodal rotations: θA = 0, θB, θCB, θCD, and θD = 0 2 nodal displacements: B = C = Δ 3 spans: AB, BC, and CD
The steps to solve this example are similar to the previous one, but with the additional consideration of the pin-connected joint at C.
4 Unknown Deflections
Step 1: Identify Degrees of Freedom and Spans
Identify the degrees of freedom (i.e., displacement and/or rotation at nodes) and the spans between nodes.
2 independent rotations at pin joint C
Step 2(b): Determine Cord Rotations
Span AB: - $$\theta_{AB} = \frac{\Delta}{4}$$
Span BC: - $$\theta_{BC} = \frac{0}{3}$$
Span CD: - $$\theta_{CD} = \frac{\Delta}{4}$$
Step 2: Write Slope-Deflection Equations
Span AB: - $$M_{AB} = \frac{2EI}{L}\left(2\theta_A + \theta_B\right) + \frac{3EI\Delta}{4L}$$ - $$M_{BA} = \frac{2EI}{L}\left(\theta_A + 2\theta_B\right) + \frac{3EI\Delta}{4L}$$
Example 7: Frames with Sidesway
Step 1: Identify Degrees of Freedom and Spans
The frame has 3 spans: AB, BC, and CD. There are 4 nodal rotations: $ \theta_A = 0$, $\theta_B$, $\theta_C$, and $\theta_D = 0$.
Step 2(b): Determine Cord Rotations
Span AB: - $$\theta_{AB} = \frac{\Delta}{3}$$
Span BC: - $$\theta_{BC} = \frac{\Delta}{3.6}$$
Span CD: - $$\theta_{CD} = \frac{\Delta}{6}$$
Step 2(b): Determine Cord Rotations (Continued)
Expressing the rotations in terms of the horizontal displacement $\Delta$: - $$\theta_B = 1.155\Delta$$ - $$\theta_{AB} = 0.385\Delta$$ - $$ \theta_{BC} = -0.160\Delta$$ - $$\theta_{CD} = 0.167\Delta$$
Step 2(c): Write Slope-Deflection Equations
Span AB: - $$M_{AB} = \frac{2EI}{3}\left(2(0) + 3(0.385\Delta)\right) = 0.667EI\Delta$$ - $$M_{BA} = \frac{2EI}{3}\left(3(0.385\Delta) + 2(0)\right) = 1.333EI\Delta$$
Span BC: - $$M_{BC} = \frac{2EI}{3.6}\left(2(-0.160\Delta) + 3(0)\right) + 32.4 = -0.556EI\Delta + 32.4$$ - $$M_{CB} = \frac{2EI} {3.6}\left(3(-0.160\Delta) + 2(0)\right) + 32.4 = 1.111EI\Delta + 32.4$$
Span CD: - $$M_{CD} = \frac{2EI}{6}\left(2(0) + 3(0.167\Delta)\right) = 0.667EI\Delta$$ - $$M_{DC} = \frac{2EI}{6}\left(3(0.167\Delta) + 2(0)\right) = 0.333EI\Delta$$
Step 3(a): Write Equilibrium Equations at Nodes
Node B: - $$M_{BA} + M_{BC} = 0$$ - $$1.333EI\Delta - 0.556EI\Delta + 32.4 = 0$$ - $$2.444EI\Delta = 32.4$$
Node C: - $$M_{CB} + M_{CD} = 0$$ - $$1.111EI\Delta - 0.667EI\Delta + 32.4 = 0$$ - $$0.556EI\Delta = 32.4$$
Step 3(b): Write Equilibrium Equations for the Structure
Equilibrium of moments at point O: - $$10.2(6.235) - 12.235(3) - 108(1.8) = 0$$ - $$2.4(0.667EI\Delta) - 3.4(1.333EI\Delta) - 2.04(0.667EI\Delta) + 1.04(0.333EI\Delta) - 108(1.8) = 0$$ - $$6.133EI\Delta = 194.4$$
Bending Moment Diagram
Step 3(c): Solve Unknown Deflection
From node B: - $$2.444EI\Delta = 32.4$$ - $$\Delta = 35.5$$
From node C: - $$0.556EI\Delta = 32.4$$ - $$\Delta = -33.3$$
From equilibrium of moments at point O: - $$6.133EI\Delta = 194.4$$ - $$ \Delta = 71.4$$
Step 4: Determine Member Forces
Span AB: - $$M_{AB} = 0.667EI(35.5) = 31.3 \text{ kNm}$$ - $$M_{BA} = 1.333EI(35.5) = 7.7 \text{ kNm}$$
Span BC: - $$M_{BC} = 1.111EI(35.5) - 0.556EI(-33.3) - 0.267EI(71.4) + 32.4 = 7.6 \text{ kNm}$$ - $$M_{CB} = 0.556EI(35.5) - 1.111EI(-33.3) - 0.267EI(71.4) + 32.4 = 34.2 \text{ kNm}$$
Span CD: - $$M_{CD} = 0.667EI(-33.3) - 0.167EI(71.4) = 34.1 \text{ kNm} $$ - $$M_{DC} = 0.333EI(-33.3) - 0.167EI(71.4) = 23.0 \text{ kNm}$$
Step 5: Bending Moment Diagram
The bending moment diagram for the frame is shown below:
Structural Analysis of Frames with Sidesway
Exercise 9: Frames with Sidesway
The given frame has a pin support at the bottom and rigid joints at the top. The objective is to determine the maximum bending moment in the frame, assuming that the flexural rigidity (EI) is constant.
The frame is subjected to the following loads: - 34.7 kN at joint A - 13.3 kN at joint B - 28.8 kN at joint C - 28.8 kN at joint D
The bending moment diagram for this frame is provided, showing the following maximum bending moments: - 80 kNm at joints A and D - 64 kNm at joints B and C - 75 kNm at the midspan of the top beam
Exercise 10: Frames with Sidesway
This frame has fixed supports at A and D, with rigid joints at B and C. The Young's modulus (E) is 200 GPa.
The frame is subjected to the following loads: - 176 kN at joint A - 58 kN at joint B - 184 kN at joint C - 58 kN at joint D
Bending Moment Diagram
The slope-deflection method is a powerful tool for analyzing statically indeterminate structures, such as beams and planar frames. It involves writing slope-deflection equations for each span and imposing equilibrium conditions at the nodes to solve for the unknown deflections and member forces.
The main difference between sway and non-sway frames is the need to consider the equilibrium of forces for the entire structure in the case of sway frames, in addition to the equilibrium of moments at the nodes.
