Explanation Calculation: The following figure shows free body diagram of the frame. Figure-(1) Write the equilibrium equation for horizontal reaction of frame shown in Figure-(1). Σ F X = 0 − H A + ( 12 k ) sin θ = 0 H A = ( 12 k ) sin θ ...... (I). Here, horizontal reaction at A is H A . Calculate the angle of member AB shown in Figure-(1). sin θ = 4 5 cos θ = 3 5 Calculate the length of member AB using the Pythagoras theorem. A B = ( 8 ft ) 2 + ( 6 ft ) 2 = 64 ft 2 + 36 ft 2 = 100 ft 2 A B = 10 ft Substitute 4 5 for sin θ in Equation (I). H A = ( 12 k ) × ( 4 5 ) H A = 9.6 k Write the equilibrium equation for vertical reaction of frame shown in Figure-(1). Σ F Y = 0 V A + V D − ( 12 k ) cos θ − ( 2 k / ft ) × ( 12 ft ) = 0 V A + V D = ( 12 k ) cos θ + 24 k ...... (II). Here, vertical reactions at A and D are V A and V D . Substitute 3 5 for cos θ in Equation (II). V A + V D = ( 12 k ) × ( 3 5 ) + ( 24 k ) = ( 7.2 k ) + ( 24 k ) = 31.2 k ...... (III). Write the equilibrium equation for the sum of moment about point A. Σ M A = 0 − ( 12 k × 5 ft ) − ( 2 k / ft × 12 ft ) × ( 6 ft + 12 2 ft ) + V D × 24 ft = 0 24 V D − 6 0 k ⋅ ft − 288 k ⋅ ft = 0 24 V D = 348 k ⋅ ft V D = 14.5 k Substitute 14.5 k for V D in Equation (III). V A + 14.5 k = 31.2 k V A = 31.2 k − 14.5 k V A = 16.7 k Consider the section x-x at distance x from point A for member AB as shown in the figure below. Figure-(2) Calculate the shear force and moment in member AB at section x-x. Calculate the normal force in member AB. Consider the force along the axis of member AB. N − ( 9.6 k ) cos θ + ( 16.7 k ) sin θ = 0 ...... (IV). Substitute 3 5 for cos θ and 4 5 for sin θ in Equation (IV). N − ( 9.6 × 3 5 k ) + ( 16.7 × 4 5 k ) = 0 N − 5.76 k + 13.36 k = 0 N + 7.6 k = 0 N = − 7.6 k Calculate the shear force at section x-x. Consider the forces perpendicular to the axis of member AB as shown in Figure-(2). Σ V x = 0 V − ( 9.6 k ) sin θ − ( 16.7 k ) cos θ = 0 ...... (V). Here, shear force at section x-x is V . Substitute 3 5 for cos θ and 4 5 for sin θ in Equation (V). V − ( 9.6 × 4 5 k ) − ( 16.7 × 3 5 k ) = 0 V − 7.68 k − 10.02 k = 0 V − 17.7 k = 0 V = 17.7 k Write the expression for moment about point A. Σ M A = 0 M − ( 9.6 sin θ + 16.7 cos θ ) x k ⋅ ft = 0 ...... (VI). Substitute 3 5 for cos θ and 4 5 for sin θ in Equation (VI). M − ( 9.6 × 4 5 + 16.7 × 3 5 ) x k ⋅ ft = 0 M − ( 7.68 + 10.02 ) x = 0 M − 17.7 x = 0 M = 17.7 x ...... (VII). Here, moment at x-x is M . Moment at x = 0 ft Substitute 0 ft for x in Equation (VII). M A = ( 17.7 × 0 ) k ⋅ ft M A = 0 k ⋅ ft Here, moment at A is M A . Moment at x = 5 ft Substitute 5 ft for x in Equation (VII). M = ( 17.7 × 5 ) k ⋅ ft M = 88.5 k ⋅ ft Here, moment at mid of the AB is M . Consider the section x-x at distance x from point A for member AB as shown in figure below. Figure-(3) Calculate shear force at section x-x. Consider force perpendicular to the axis of member AB as shown in Figure-(3) Σ V x = 0 V − 9.6 sin θ − 16.7 cos θ + 12 k = 0 ...... (VIII). Substitute 3 5 for cos θ and 4 5 for sin θ in Equation (V). V − ( 9.6 × 4 5 k ) − ( 16.7 × 3 5 k ) + 12 k = 0 V − 7.68 k − 10.02 k + 12 k = 0 V − 5.7 k = 0 V = 5.7 k Here, shear force at section x-x is V . Write the expression for sum of moment about x-x for Figure-(3)

10th Edition

ISBN: 8220106821596

Author: HIBBELER

Publisher: PEARSON

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Question

Chapter 4, Problem 4.48P

To determine

The shear and moment diagram for each member of the frame.

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Chapter 4, Problem 4.47P

Chapter 4, Problem 4.49P

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P.3.3 Oil of sp.gr. 0.9 flows through a vertical pipe (upwards). Two points A and B one above the other 40 cm apart in a pipe are connected by a U-tube carrying mercury. If the difference of pressure between A and B is 0.2 kg/cm², 1- Find the reading of the manometer. 2- If the oil flows through a horizontal pipe, find the reading in manometer for the same difference in pressure between A and B. Ans. 1- R= 0.12913 m, 2- R = 0.1575 m,

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Chapter 4 Solutions

EBK STRUCTURAL ANALYSIS

Ch. 4 - Prob. 4.1P Ch. 4 - Prob. 4.2P Ch. 4 - Prob. 4.3P Ch. 4 - Prob. 4.4P Ch. 4 - Prob. 4.5P Ch. 4 - Prob. 4.6P Ch. 4 - Prob. 4.7P Ch. 4 - Prob. 4.8P Ch. 4 - Prob. 4.9P Ch. 4 - Prob. 4.10P

Ch. 4 - Prob. 4.11P Ch. 4 - Prob. 4.12P Ch. 4 - Prob. 4.13P Ch. 4 - Prob. 4.14P Ch. 4 - Prob. 4.15P Ch. 4 - Prob. 4.16P Ch. 4 - Prob. 4.17P Ch. 4 - Prob. 4.18P Ch. 4 - Prob. 4.19P Ch. 4 - Prob. 4.20P Ch. 4 - Prob. 4.21P Ch. 4 - Prob. 4.22P Ch. 4 - Prob. 4.23P Ch. 4 - Prob. 4.24P Ch. 4 - Prob. 4.25P Ch. 4 - Prob. 4.26P Ch. 4 - Prob. 4.27P Ch. 4 - Prob. 4.28P Ch. 4 - Prob. 4.29P Ch. 4 - Prob. 4.30P Ch. 4 - Prob. 4.31P Ch. 4 - Prob. 4.32P Ch. 4 - Prob. 4.33P Ch. 4 - Prob. 4.34P Ch. 4 - Prob. 4.35P Ch. 4 - Prob. 4.36P Ch. 4 - Prob. 4.37P Ch. 4 - Prob. 4.38P Ch. 4 - Prob. 4.39P Ch. 4 - Prob. 4.40P Ch. 4 - Prob. 4.41P Ch. 4 - Prob. 4.42P Ch. 4 - Prob. 4.43P Ch. 4 - Prob. 4.44P Ch. 4 - Prob. 4.45P Ch. 4 - Prob. 4.46P Ch. 4 - Prob. 4.47P Ch. 4 - Prob. 4.48P Ch. 4 - Prob. 4.49P Ch. 4 - Prob. 4.50P Ch. 4 - Prob. 4.51P Ch. 4 - Prob. 4.52P Ch. 4 - Prob. 4.53P Ch. 4 - Prob. 4.54P Ch. 4 - Prob. 4.55P Ch. 4 - Prob. 4.56P Ch. 4 - Prob. 4.57P Ch. 4 - Prob. 4.58P Ch. 4 - Prob. 4.59P Ch. 4 - Prob. 4.1PP Ch. 4 - Prob. 4.2PP

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