Explanation Concept Used: Write the expression for the net force balance in the vertical direction of the beam. ∑ F Y = 0 ........ (I) Here, ∑ F Y is the sum of all the vertical forces. Write the expression for the net force balance in the horizontal direction in the beam. ∑ F X = 0 ........ (II) Here, ∑ F X is the sum of all the horizontal forces. Write the expression for the net moment about end A of the beam. ∑ M A = 0 .......... (III) Here, ∑ M A is the sum of all the moments taken about the end A . Write the expression for internal moment due to displacement and loading. M N = 2 E I L ( 2 θ N + θ F − ψ ) + ( F E M ) N ...... (IV) Here, the internal moment is M N , the flexural rigidity is E I , the span rotation is ψ , the deflection is θ and the fixed end moment is ( F E M ) N . Calculation: Calculate the fixed end moments. The fixed end moments will be zero at every end because there is no transverse external loading. Calculate fixed end moments of member AD. F E M A D = − w l 2 20 Here, the load is w and the length is l . Substitute 30 kN / m for w and 3 m for l . F E M A D = − 30 kN / m × ( 3 m ) 2 20 = − 270 20 = − 13.5 kN ⋅ m Calculate the fixed end moments of member DA. F E M D A = w l 2 30 Substitute 30 kN / m for w and 3 m for l . F E M D A = 30 kN / m × ( 3 m ) 2 30 = 270 30 = 9 kN ⋅ m In all other members the fixed end moments will be zero because there is no external loading on these members. ( F E M ) D C = 0 ( F E M ) C D = 0 ( F E M ) C B = 0 ( F E M ) B C = 0 Write the value of angular rotation as shown below. ψ A D = ψ ψ D A = ψ ψ B C = ψ ψ C B = ψ ψ C D = 0 ψ D C = 0 The value of slope at A and B will be zero as A and B are fixed ends. Write the slope deflection equation for AD from Equation (IV). M A D = 2 E I L A D ( 2 θ A + θ D − 3 ψ A D ) + ( F E M ) A D Substitute 3 m for L A D , 0 for θ A and − 13.5 kN ⋅ m for ( F E M ) A D . M A D = 2 E I 3 m ( 2 × 0 + θ D − 3 ψ ) − 13.5 kN ⋅ m = ( 2 E I θ D 3 − 2 E I ψ − 13.5 ) ...... (V). Write the slope deflection equation for DA from Equation (IV). M D A = 2 E I L D A ( 2 θ D + θ A − 3 ψ D A ) + ( F E M ) D A Substitute 3 m for L D A , 0 for θ A and 9 kN ⋅ m for ( F E M ) D A . M D A = 2 E I 3 m ( 2 θ D + 0 − 3 ψ ) + 9 kN ⋅ m = ( 4 E I θ D 3 − 2 E I ψ + 9 kN ⋅ m ) ...... (VI). Write the slope deflection equation for CD from Equation (IV). M C D = 2 E I L C D ( 2 θ C + θ D − 3 ψ C D ) + ( F E M ) C D Substitute 3 m for L C D , 0 for ( F E M ) C D and ψ C D . M C D = 2 E I 3 m ( 2 θ C + θ D − 3 × 0 ) + 0 = ( 4 E I θ C + 2 E I θ D 3 ) ...... (VII). Write the slope deflection equation for DC from Equation (IV). M D C = 2 E I L D C ( 2 θ D + θ C − 3 ψ D C ) + ( F E M ) D C Substitute 3 m for L D C , 0 for ( F E M ) D C and ψ D C . M D C = 2 E I 3 m ( 2 θ D + θ C − 3 × 0 ) + 0 = ( 2 E I θ C + 4 E I θ D 3 ) ...... (VIII). Write the slope deflection equation for BC from Equation (IV). M B C = 2 E I L B C ( 2 θ B + θ C − 3 ψ B C ) + ( F E M ) B C Substitute 3 m for L B C , 0 for ( F E M ) B C , and ψ for ψ B C . M B C = 2 E I 3 m ( 2 θ B + θ C − 3 ψ ) + 0 = ( 2 E I θ C 3 − 2 E I ψ ) ...... (IX). Write the slope deflection equation for CB from Equation (IV). M C B = 2 E I L C B ( 2 θ C + θ B − 3 ψ C B ) + ( F E M ) C B Substitute 3 m for L C B , 0 for ( F E M ) C B , and ψ for ψ C B . M C B = 2 E I 3 m ( 2 θ C + θ B − 3 ψ ) + 0 = ( 4 E I θ C 3 − 2 E I ψ ) ...... (X). Write the joint equilibrium Equation for joint D. M D A + M D C = 0 Substitute ( 4 E I θ D 3 − 2 E I ψ + 9 kN ⋅ m ) for M D A and ( 2 E I θ C + 4 E I θ D 3 ) for M D C . ( 4 E I θ D 3 − 2 E I ψ + 9 kN ⋅ m ) + ( 2 E I θ C + 4 E I θ D 3 ) = 0 ( 4 E I θ D 3 − 2 E I ψ + 9 + 2 E I θ C 3 + 4 E I θ D 3 ) = 0 8 E I θ D 3 + 2 E I θ C 3 − 2 E I ψ = − 9

The moments at joints A, B, C, and D, then draw the moment diagram for each member of the frame.

Structural Analysis (10th Edition)

10th Edition

ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 10, Problem 10.22P

To determine

The moments at joints A, B, C, and D, then draw the moment diagram for each member of the frame.

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