Slope Deflection Examples

Slope Deflection Examples - Slope Deflection Examples Fixed...

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Slope Deflection Examples: Fixed End Moments For a member AB with a length L and any given load the fixed end moments are given by: ( ) 2 2 2 AB B FEM g g L = A ( ) 2 2 2 BA B A FEM g g L = Where: g B and g A are the moments of the bending moment diagrammes of the statically determinate beam about B and A respectively. Example: Determine the fixed end moments of a beam with a point load. Simply supported beam with bending moment diagramme. Centroid in accordance with standard tables. ( ) 2 2 2 AB B FEM g g L = A 2 2 2 2 3 2 3 AB L Wab b L L Wab a L FEM L L L + + = 2 2 2 2 3 AB Wab b L a L FEM L + = 2 2 2 3 AB Wab b a b a FEM L + + = 2 2 2 AB Wab FEM L = In a similar way FEM BA may be calculated. Slope-deflection Page 1 of 20 7/23/2003
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Use of slope-deflection equations: Example 1: Determine the bending moment diagramme of the following statically indeterminate beam. The unknowns are as follows: θ A , θ B , θ C = 0, ψ AB = 0, ψ BC = 0 We require two equations to solve the two unknown rotations: 0 0 A AB M M = = ( ) 2 2 3 AB A B AB EI M F L θ θ ψ = + + AB EM ( ) 2 1 2 4 8 AB A B EI M θ θ = + + 0 4 0,5 5 AB A B M EI EI θ θ = + + 0 A M = EI θ A + 0,5 EI θ B + 5 = 0 (1) 0 0 B BA BC M M M = + = ( ) 2 2 3 BA A B AB BA EI M F L θ θ ψ = + + EM ( ) 2 1 2 4 8 BA A B EI M θ θ = + 0 4 0,5 1 5 BA A B M EI EI θ θ = + ( ) 2 2 3 BC B C BC BC EI M F L θ θ ψ = + + EM ( ) 2 2 5 2 6 1 BC B EI M θ = + 6 2 0,66667 15 BC B M θ = + M BA + M BC = 0 0,5 EI θ A + 1,66667 EI θ B + 10 = 0 (2) Solve the unknowns: θ A = - 2,35294/EI θ B = - 5,29412/EI Calculate the values of the moments: M BA = 0,5 x –2,35294 – 5,29412 – 5 = -11,471 kN.m Slope-deflection Page 2 of 20 7/23/2003
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M BC = 0,6667 x – 5,29412 + 15 = + 11,471 kN.m ( ) 2 2 3 CB B C BC CB EI M F L θ θ ψ = + + EM ( ) 2 2 5 6 1 CB B EI M θ = + 6 2 M CB = 0,3333 x –5,29412 – 15 = -16,675 kN.m Draw the bending moment diagramme. The Modified Slope-Deflection Equation with a Hinge at A: We would like to eliminate θ A from the equation as we know that M AB = 0. ( ) 2 2 3 AB A B AB EI M F L θ θ ψ = + + AB EM Solve for θ A . 3 2 2 2 2 AB B A FEM L EI θ ψ θ = − + AB ( ) 2 2 3 BA A B AB BA EI M F L θ θ ψ = + + EM Replace θ A in this equation. 3 2 1 2 3 2 2 2 B AB BA B AB BA AB EI M F L θ ψ θ ψ = + + EM FEM ( ) 3 1 2 BA B AB BA AB EI M FEM L θ ψ = + FEM This equation may be used to reduce the number of unknown rotations. Solve the previous problem using the modified slope-deflection equation. Slope-deflection Page 3 of 20 7/23/2003
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The unknowns are as follows: θ A use the modified slope-deflection equation, θ B , θ C = 0, ψ AB = 0, ψ BC = 0 The total number of unknowns is reduced to θ B 0 0 B BA BC M M M = + = ( ) 3 1 2 BA B AB BA AB EI M FEM L θ ψ = + FEM ( ) 3 10 4 1 10 4 4 8 2 BA B EI M θ = 8 0,75 7,5 BA A M EI θ =
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