Explanation Concept Used: A force P having variable magnitude is placed on the beam or frame at the point and in the direction of the desired displacement using Castigliano's theorem. Write the expression for displacement at any point. Δ D = ∫ 0 L M E I ∂ M ∂ P d x = ∫ 0 4 M 1 E I ∂ M 1 ∂ P d x + ∫ 0 4 M 2 E I ∂ M 2 ∂ P d x + ∫ 0 4 M 3 E I ∂ M 3 ∂ P d x + ∫ 0 4 M 4 E I ∂ M 4 ∂ P d x ...... (I) Here, internal moment in the beam expressed as function of x caused by both force P and real loads on the beam. is M , modulus of elasticity is E and moment of inertia is I . Calculation: The free body diagram of the beam A B is shown below. Figure (1) Here, vertical component of force at point B is B y , vertical component of force at point A is A y and horizontal component of force at point A is A x . Calculate moment about point A . ∑ M A = 0 ( B y × 4 m ) − ( 60 kN × 2 m ) − ( P × 2 m ) = 0 ( B y × 4 m ) − 120 kN-m + 2 P = 0 B y = 30 kN + P 2 Calculate all forces in y-direction. ∑ F y = 0 A y + B y − 60 kN − P = 0 A y + B y = 60 kN + P ...... (II) Substitute 30 kN + P 2 for B y in Equation (II). A y + 30 kN + P 2 = 60 kN + P A y = 30 kN + P 2 Calculate all forces in x-direction. ∑ F x = 0 A x = 0 The free body diagram of the beam B C is shown below. Figure (2) Here, vertical component of force at point A is A y and horizontal component of force at point A is A x . Calculate all forces in y-direction. ∑ F y = 0 B y + C y − ( 30 kN/m × 3 m ) = 0 B y + C y = 9 0 kN ...... (III) Substitute ( 30 kN + P 2 ) for B y in Equation (III). C y + 30 kN + P 2 = 9 0 kN C y = 60 kN − P 2 Calculate moment about point. ∑ M = 0 M C = ( 30 kN/m × 3 m × 3 2 m ) − ( C y × 3 m ) = 135 kN-m − ( 60 kN − P 2 ) × 3 m = 75 kN-m + 3 P 2 Calculate all forces in x-direction. ∑ F x = 0 B x = 0 Consider the section at x 1 distance from support A such that 0 ≤ x 1 ≤ 2 . The free body diagram of the section is shown below. Figure (3) Here, shear force at x 1 distance from support A is V 1 and moment at x 1 distance from support A is M 1 Calculate the moment at section x 1 . ∑ M x 1 = 0 M 1 = A y × x 1 = ( 30 kN + P 2 ) × x 1 = 30 x 1 + P x 1 2 ...... (IV) Differentiate the Equation (IV) with respect to P . ∂ M 1 ∂ P = x 1 2 Consider the section at x 2 distance from support D such that 0 ≤ x 2 ≤ 2 . The free body diagram of the section is shown below. Figure (4) Here, shear force at x 2 distance from support D is V 2 and moment at x 2 distance from support D is M 2 . Calculate the moment at section x 2 . ∑ M x 2 = 0 M 2 = [ A y × ( 2 + x 2 ) ] − [ ( 60 kN + P ) × x 2 ] = [ ( 30 kN + P 2 ) × ( 2 + x 2 ) ] − 60 kN x 2 − ( P × x 2 ) = 60 kN + P + ( 30 kN × x 2 ) + ( P 2 × x 2 ) − 60 kN x 2 − ( P × x 2 ) M 2 = 60 + P + 30 x 2 + P 2 x 2 − 60 x 2 − P x 2 = 60 + P − 30 x 2 − P 2 x 2

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 8, Problem 8.43P

To determine

The displacement at point D.

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Chapter 8, Problem 8.42P

Chapter 8, Problem 8.44P

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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 8 Solutions

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

Ch. 8 - Determine the vertical displacement of joint A....Ch. 8 - Prob. 8.2P Ch. 8 - Prob. 8.3P Ch. 8 - Prob. 8.4P Ch. 8 - Prob. 8.5P Ch. 8 - Prob. 8.6P Ch. 8 - Prob. 8.7P Ch. 8 - Prob. 8.8P Ch. 8 - Prob. 8.9P Ch. 8 - Prob. 8.10P

Ch. 8 - Prob. 8.11P Ch. 8 - Prob. 8.12P Ch. 8 - Prob. 8.13P Ch. 8 - Prob. 8.14P Ch. 8 - Prob. 8.15P Ch. 8 - Prob. 8.16P Ch. 8 - Prob. 8.17P Ch. 8 - Prob. 8.18P Ch. 8 - Prob. 8.19P Ch. 8 - Prob. 8.20P Ch. 8 - Prob. 8.21P Ch. 8 - Prob. 8.22P Ch. 8 - Prob. 8.23P Ch. 8 - Prob. 8.24P Ch. 8 - Prob. 8.25P Ch. 8 - Prob. 8.26P Ch. 8 - Prob. 8.27P Ch. 8 - Prob. 8.28P Ch. 8 - Prob. 8.29P Ch. 8 - Prob. 8.30P Ch. 8 - Prob. 8.31P Ch. 8 - Prob. 8.32P Ch. 8 - Prob. 8.33P Ch. 8 - Prob. 8.34P Ch. 8 - Prob. 8.35P Ch. 8 - Prob. 8.36P Ch. 8 - Prob. 8.37P Ch. 8 - Prob. 8.38P Ch. 8 - Prob. 8.39P Ch. 8 - Prob. 8.40P Ch. 8 - Prob. 8.41P Ch. 8 - Prob. 8.42P Ch. 8 - Prob. 8.43P Ch. 8 - Prob. 8.44P Ch. 8 - Prob. 8.45P Ch. 8 - Prob. 8.46P Ch. 8 - Prob. 8.47P Ch. 8 - Prob. 8.48P Ch. 8 - Prob. 8.49P Ch. 8 - Prob. 8.50P Ch. 8 - Prob. 8.51P Ch. 8 - Prob. 8.52P Ch. 8 - Prob. 8.53P Ch. 8 - Prob. 8.54P Ch. 8 - Prob. 8.55P Ch. 8 - Prob. 8.56P Ch. 8 - Prob. 8.57P Ch. 8 - Prob. 8.58P Ch. 8 - Prob. 8.59P Ch. 8 - Prob. 8.60P Ch. 8 - Prob. 8.61P

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