Explanation Given: The beams AB and BC are supported by a parabolic cable. Concept Used: Write the expression for maximum tension in the cable. T max = F H 2 + ( w ∘ L ) 2 ...... (I). Here, the uniform loading is w ∘ length of the cable is L and the horizontal reaction at D and E is F H . Write the expression for horizontal component of cable tension. F H = w ∘ L 2 2 h ...... (II). Here, sag in the cable is h . Write the expression for force in each hanger. T = w ∘ × d ...... (III). Here, spacing between the hangers is d The tension forces in bars connected to cables are equal as the sag in cable is parabolic. Calculation: The free body diagram of the beam ABC is shown below. Figure (1) Calculate the moment about point A. ∑ M A = 0 E y × ( 16 m ) + C y × ( 16 m ) = ( 80 kN ) × ( 4 m ) + ( 40 kN ) × ( 12 m ) ( 16 m ) ( E y + C y ) = 320 kN-m + 480 kN-m E y + C y = ( 800 kN-m 16 m ) E y + C y = 50 kN ...... (IV). The free body diagram of the portion BC is shown below. Figure (2) Calculate the moment about point B. ∑ M B = 0 E y × ( 8 m ) + C y × ( 8 m ) = ( 40 kN ) × ( 4 m ) + F H × ( 2 m ) ( 8 m ) × ( E y + C y ) = ( 160 kN-m ) + 2 F H E y + C y = 20 kN + 0.25 F H ...... (V). Substitute Equation (V) in Equation (IV). 20 kN + 0.25 F H = 50 kN F H = 120 kN Calculate the uniform loading in the cable. Substitute 120 kN for F H , 2 m for h and 8 m for L in the Equation (II). 120 kN = w ∘ ( 8 m ) 2 2 × ( 2 m ) w ∘ = 120 kN × 2 × ( 2 m ) ( 8 m ) 2 w ∘ = 480 kN-m 64 m 2 w ∘ = 7.5 kN/m Calculate maximum tension in the cable at points D and E. Substitute 7.5 kN/m for w ∘ , 120 kN for F H , and 8 m for L in the equation (I). T max = ( 120 kN ) 2 + ( 7.5 kN/m × 8 m ) 2 = ( 120 kN ) 2 + ( 60 kN ) 2 = 14400 kN 2 + 3600 kN 2 = 134.164 kN The tension at point D and E is 134.164 kN . Calculate the force in each hanger. Substitute 7.5 kN/m for w ∘ , and 2 m for L in the equation (III). T = 7.5 kN/m × 2 m =15 kN The free body diagram of the portion AC is shown below. Figure (3) Calculate the moment about point A. ∑ M A = 0 [ T × ( 2 m ) + T × ( 4 m ) + T × ( 6 m ) + T × ( 8 m ) + T × ( 10 m ) + T × ( 12 m ) + T × ( 14 m ) + C y × ( 16 m ) ] = [ ( 80 kN ) × ( 4 m ) + ( 40 kN ) × ( 12 m ) ] 56 T + 16 C y = ( 320 kN-m ) + ( 480 kN-m ) 56 T + 16 C y = 800 kN-m ...... (VI) Substitute 15 kN for T in the equation (VI). ( 56 m ) ( 15 kN ) + 16 C y = 800 kN-m 840 kN-m + 16 C y = 800 kN-m 16 C y = 800 kN-m − 840 kN-m C y = − 2.5 kN Calculate the forces in the y-direction. ∑ F y = 0 A y + C y + 7 T = 80 kN+40 kN A y + C y + 7 T = 120 kN ...... (VII). Substitute 15 kN for T and ( − 2.5 kN ) for C y in the equation (VII). A y − 2.5 kN + 7 ( 15 kN ) = 120 kN A y − 2.5 kN + 105 kN = 120 kN A y = 120 kN + 2.5 kN − 105 kN A y = 17.5 kN The free body diagram of section X-X between CI at a distance x from end C is shown below. Figure (4) Calculate the shear force. V x = − C y ...... (VI). Substitute 0 m for x for point C in the equation (VI). V C = − ( − 2.5 kN ) = 2.5 kN Substitute 2 m for x for point I in the equation (VI). V C − I = 2.5 kN Substitute 0 m for x for point C in the equation (VII). M C = ( − 2.5 kN ) × ( 0 m ) = 0 kN-m Substitute 2 m for x for point I in the equation (VII). M I = ( − 2.5 kN ) × ( 2 m ) = 5 kN-m The free body diagram of X-X section between HI at a distance x from end C is shown below. Figure (5) Calculate shear force V x = − C y − T ...... (VIII). Substitute 2 m for x for point I in the equation (VIII). V I = − ( − 2.5 kN ) − 15 kN = − 12.5 kN Substitute 4 m for x for point H in the equation (VIII). V H = − ( − 2.5 kN ) − 15 kN = − 12.5 kN Calculate bending moment M x = C y × x + T × ( x − 2 ) ...... (IX). Substitute 2 m for x for point I in the equation (IX). M I = ( − 2.5 kN ) × ( 2 m ) + 15 kN × ( 2 m − 2 ) = − 5 kN-m Substitute 4 m for x for point H in the equation (IX). M H = ( − 2.5 kN ) × ( 4 m ) + 15 kN × ( 4 m − 2 ) = 20 kN-m The free body diagram of X-X section between HG at a distance x from end C is shown below. Figure (6) Calculate shear force V H = − ( − 2.5 kN ) − 2 × ( 15 kN ) + 40 kN = 12.5 kN Substitute 6 m for x for point G in the equation (X). V G = − ( − 2.5 kN ) − 2 × ( 15 kN ) + 40 kN = 12.5 kN Calculate bending moment. M x = C y × x + T × ( x − 2 ) + T × ( x − 4 ) ...... (XI). Substitute 4 m for x for point H in the equation (XI). M H = ( − 2.5 kN ) × ( 4 m ) + 15 kN × ( 4 m − 2 ) + 15 kN × ( 4 m − 4 ) = 20 kN-m Substitute 6 m for x for point G in the equation (XI). M G = ( − 2.5 kN ) × ( 6 m ) + 15 kN × ( 6 m − 2 ) + 15 kN × ( 6 m − 4 ) − ( 40 kN ) × ( 2 m ) = − 5 kN-m The free body diagram of X-X section between GB at a distance x from end C is shown below. Figure (7) Calculate shear force V x = − C y − 3 T + 40 ...... (XII). Substitute 6 m for x for point G in the equation (XII). V G = − ( − 2.5 kN ) − 3 × ( 15 kN ) + 40 kN = − 2.5 kN Substitute 8 m for x for point B in the equation (XII). V B = − ( − 2.5 kN ) − 3 × ( 15 kN ) + 40 kN = − 2.5 kN Calculate bending moment. M x = C y × x + T × ( x − 2 ) + T × ( x − 4 ) + T × ( x − 6 ) − ( 40 kN ) × ( 2 m ) ...... (XIII). Substitute 6 m for x for point G in the equation (XIII). M G = ( ( − 2.5 kN ) × ( 6 m ) + 15 kN × ( 6 m − 2 ) + 15 kN × ( 6 m − 4 ) + 15 kN × ( 6 m − 6 ) − ( 40 kN ) × ( 2 m ) ) = − 5 kN-m Substitute 8 m for x for point B in the equation (XIII). M B = ( ( − 2.5 kN ) × ( 8 m ) + 15 kN × ( 8 m − 2 ) + 15 kN × ( 8 m − 4 ) + 15 kN × ( 8 m − 6 ) − ( 40 kN ) × ( 2 m ) ) = 0 kN-m The free body diagram of X-X section between BF at a distance x from end C is shown below

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition)

10th Edition

ISBN: 9780134702216

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 5, Problem 5.19P

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The shear force diagram for the member AC.

The bending moment diagram for the member AC.

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Chapter 5, Problem 5.18P

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Chapter 5 Solutions

Structural Analysis, Student Value Edition Plus Mastering Engineering With Pearson Etext -- Access Card Package (10th Edition)

Ch. 5 - Prob. 5.1P Ch. 5 - Prob. 5.2P Ch. 5 - Prob. 5.3P Ch. 5 - Prob. 5.4P Ch. 5 - Prob. 5.5P Ch. 5 - Prob. 5.6P Ch. 5 - Prob. 5.7P Ch. 5 - Prob. 5.8P Ch. 5 - Prob. 5.9P Ch. 5 - Prob. 5.10P

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The shear force diagram for the member AC. The bending moment diagram for the member AC.

The shear force diagram for the member AC. The bending moment diagram for the member AC.

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Chapter 5 Solutions

The shear force diagram for the member AC.

The bending moment diagram for the member AC.