Explanation Calculation: The following figure shows the frame ABC: Figure (1) Consider the reaction C y at joint C as redundant. The following figure shows the deflection of frame using principle of superposition. Figure (2) Write the compatibility Equation from the deflected shape of the frame. 0 = Δ C + C y f C C ...... (I) Here, the displacement at joint C due to real loads is Δ C , the vertical reaction at joint C is C y and the flexibility coefficient at joint C is f C C . Write the expression of virtual work method to calculate the displacement at joint C due to real loads. Δ C = ∫ 0 L M m E I d x Here, the moment in the frame caused by real loads is Δ C , the virtual moment in the frame caused by the external virtual unit load is m and E I is the flexural rigidity. The following figure shows the free body diagram of the frame. Figure (3) Write the expression to calculate the vertical forces in the beam. ∑ F y = 0 Here, the forces in y-direction is F y . A y = 8 kN / m × 9 m A y = 72 kN Write the expression to calculate the horizontal forces in the beam. ∑ F x = 0 Here, the forces in y-direction is F x . A x = 20 kN Write the expression for the moment about A. ∑ M A = 0 . Here, the moment about A is M A . ( 20 kN × 6 m ) + ( 72 kN × 9 m ) − ( 8 kN/m × 9 m × 9 m 2 ) − ( 20 kN × 3 m ) − M A = 0 M A = 120 + 648 − 324 − 60 M A = 384 kN-m Consider a section X at a distance of x 1 from A. Write the expression for the moment about x. ∑ M x = 0 M 1 = ( A y × x 1 ) − ( 8 × x 1 × x 1 2 ) − M A ...... (II) Substitute 72 kN for A y and 384 kN-m for M A in Equation (II). M 1 = ( 72 × x 1 ) − ( 8 × x 1 × x 1 2 ) − 384 = ( 72 x 1 − 4 x 1 2 − 384 ) kN-m Calculate the moment M 2 ∑ M 2 = 0 M 2 = 0 Write the expression to calculate the deflection at C. Δ C = ∫ 0 L M m E I d x = ∫ 0 L 1 M 1 m 1 E I d x 1 + ∫ 0 L 2 M 2 m 2 E I d x 2 ...... (III) Here, the deflection at C is Δ C . Substitute 9 m for L 1 , 6 m for L 2 , ( 72 x 1 − 4 x 1 2 − 384 ) kN-m for M 1 , 0 for M 2 , ( 9 − x 1 ) for m 1 and 0 for m 2 in Equation (III). Δ C = ∫ 0 9 ( 72 x 1 − 4 x 1 2 − 384 ) ( 9 − x 1 ) E I A B d x 1 + ∫ 0 6 ( 0 ) ( 0 ) E I B C d x 2 = ∫ 0 9 ( 648 x 1 − 72 x 1 2 − 36 x 1 2 + 4 x 1 3 − 3456 + 384 x 1 ) E I A B d x 1 = 1 E I A B ∫ 0 9 4 x 1 3 − 108 x 1 2 + 1032 x 1 − 3456 d x 1 = 1 E I A B ( 6561 − 26244 + 41796 − 31104 ) Δ C = 8991 E I A B Write the expression to calculate the flexibility coefficient at joint C

10th Edition

ISBN: 9780134610672

Author: Russell C. Hibbeler

Publisher: PEARSON

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Question

Chapter 9, Problem 9.16P

To determine

The reactions at the supports of the frame.

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Chapter 9, Problem 9.15P

Chapter 9, Problem 9.17P

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For the truss shown in Fig 2, determine the nodal displacement and member forces for all elements of the truss. Assume for each member A = 0.0015 m2 and E = 200 GPa please show all working, relevant FBD's and use ID's indicated in the diagram, if using an alpha numerical in equations please indicate where it is being applied to in the truss.

Determine the global stiffness matrix of the beam shown in Fig. 3. Assume supports at 1 and 3 are rollers and the support at 2 is a pinned support. Indicate the degrees of freedom in all the stiffness matrices. EI is constant, w=60kN/m, L1=1.25m and L2=3.45m please explain how the code numbers for global matrix are determined in detail

= 20 kips = 20 kips B w₁ = 2 kips/ft 20" 12'- 7760 12" 6"

Chapter 9 Solutions

Structural Analysis (10th Edition)

Ch. 9 - Determine the reactions at the suooorts, then draw...Ch. 9 - Prob. 9.2P Ch. 9 - Prob. 9.3P Ch. 9 - Prob. 9.4P Ch. 9 - Prob. 9.5P Ch. 9 - Prob. 9.6P Ch. 9 - Prob. 9.7P Ch. 9 - Prob. 9.8P Ch. 9 - Prob. 9.9P Ch. 9 - Prob. 9.10P

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