Explanation Given information: The influence line diagram is a graph curve of response function when a unit load crosses the span of beam from left to right. Response function can be: Reaction of support SF at section BM at section Method of section is the easiest way to solving forces in the truss. The truss is a break down into the sections. The section and analysis are separated individually. For the method of section all joints in truss are pin joint and the member are consider mass less. Simply the equations of equilibrium are: ∑ F x = 0 ( Sum of all the horizontal forces is zero ) ∑ F y = 0 ( Sum of all the vertical forces is zero ) ∑ M = 0 ( Sum of all the moments is zero ) We cut the section 1-1 from the truss and the free body diagram is given below Calculation: Let the unit force is acting at point B ∑ M M = 0 ( M o m e n t a b o u t M ) R A × 300 − 1 × 275 = 0 R A = 275 300 = 11 12 ∑ F V = 0 ( s u m o f a l l v e r t i c a l f o r c e s ) 11 12 − 1 − F R N sin 45 ∘ = 0 F R N = − 2 12 = − 0.11785 Let the unit force is acting at point C ∑ M M = 0 ( M o m e n t a b o u t M ) R A × 300 − 1 × 250 = 0 R A = 250 300 = 5 6 ∑ F V = 0 ( s u m o f a l l v e r t i c a l f o r c e s ) 5 6 − 1 − F R N sin 45 ∘ = 0 F R N = − 2 6 = − 0.23570 Let the unit force is acting at point D ∑ M M = 0 ( M o m e n t a b o u t M ) R A × 300 − 1 × 225 = 0 R A = 225 300 = 3 4 ∑ F V = 0 ( s u m o f a l l v e r t i c a l f o r c e s ) 3 4 − 1 − F R N sin 45 ∘ = 0 F R N = − 2 4 = − 0.35355 Let the unit force is acting at point E ∑ M M = 0 ( M o m e n t a b o u t M ) R A × 300 − 1 × 200 = 0 R A = 200 300 = 2 3 ∑ F V = 0 ( s u m o f a l l v e r t i c a l f o r c e s ) 2 3 − 1 − F R N sin 45 ∘ = 0 F R N = − 2 3 = − 0.47140 Let the unit force is acting at point F ∑ M M = 0 ( M o m e n t a b o u t M ) R A × 300 − 1 × 175 = 0 R A = 175 300 = 7 12 ∑ F V = 0 ( s u m o f a l l v e r t i c a l f o r c e s ) 7 12 − F R N sin 45 ∘ = 0 F R N = 7 2 12 = 0.82495 Let the unit force is acting at point G ∑ M M = 0 ( M o m e n t a b o u t M ) R A × 300 − 1 × 150 = 0 R A = 150 300 = 1 2 ∑ F V = 0 ( s u m o f a l l v e r t i c a l f o r c e s ) 1 2 − F R N sin 45 ∘ = 0 F R N = 1 2 = 0

10th Edition

ISBN: 8220106821596

Author: HIBBELER

Publisher: PEARSON

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Chapter 6, Problem 6.54P

To determine

To draw: Influence line diagram for the force in the member

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Chapter 6, Problem 6.53P

Chapter 6, Problem 6.55P

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Consider, M people (aka pax) who want to travel by car from O to D. They all start working at D at Q (e.g., Q-8am). If a person departs at time t, assume the time needed to go from O to D is given by c(t)=A+Bx(t), where x(t) is the flow of people departing at time t [car/unit of time]. In addition, a is the penalty for being early at work (E(t) is how early the person arrived when departing at time t), and ẞ is the penalty for being late at work (L(t) is how late the person arrived when departing at time t). Assume 0 < a < 1 < ß. Further assume the departure time choice problem under the equilibrium conditions. Prove that the arrival time of people who depart when most of the M people start their trips is equal to Q.

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Chapter 6 Solutions

EBK STRUCTURAL ANALYSIS

Ch. 6 - Draw the influence lines for (a) the moment at C,...Ch. 6 - Prob. 6.2P Ch. 6 - Prob. 6.3P Ch. 6 - Prob. 6.4P Ch. 6 - Prob. 6.5P Ch. 6 - Prob. 6.6P Ch. 6 - Prob. 6.7P Ch. 6 - Prob. 6.8P Ch. 6 - Prob. 6.9P Ch. 6 - Prob. 6.10P

Ch. 6 - Prob. 6.11P Ch. 6 - Prob. 6.12P Ch. 6 - Prob. 6.13P Ch. 6 - Prob. 6.14P Ch. 6 - Prob. 6.15P Ch. 6 - Prob. 6.16P Ch. 6 - Prob. 6.17P Ch. 6 - Prob. 6.18P Ch. 6 - Prob. 6.19P Ch. 6 - Prob. 6.20P Ch. 6 - Prob. 6.21P Ch. 6 - Prob. 6.22P Ch. 6 - Prob. 6.23P Ch. 6 - Prob. 6.24P Ch. 6 - Prob. 6.25P Ch. 6 - Prob. 6.26P Ch. 6 - Prob. 6.27P Ch. 6 - Prob. 6.28P Ch. 6 - Prob. 6.29P Ch. 6 - Prob. 6.30P Ch. 6 - Prob. 6.31P Ch. 6 - Prob. 6.32P Ch. 6 - Prob. 6.33P Ch. 6 - Prob. 6.34P Ch. 6 - Prob. 6.35P Ch. 6 - Prob. 6.36P Ch. 6 - Prob. 6.37P Ch. 6 - Prob. 6.38P Ch. 6 - Prob. 6.39P Ch. 6 - Prob. 6.40P Ch. 6 - Prob. 6.41P Ch. 6 - Prob. 6.42P Ch. 6 - Prob. 6.43P Ch. 6 - Prob. 6.44P Ch. 6 - Prob. 6.45P Ch. 6 - Prob. 6.46P Ch. 6 - Prob. 6.47P Ch. 6 - Prob. 6.48P Ch. 6 - Prob. 6.49P Ch. 6 - Prob. 6.50P Ch. 6 - Prob. 6.51P Ch. 6 - Prob. 6.52P Ch. 6 - Prob. 6.53P Ch. 6 - Prob. 6.54P Ch. 6 - Prob. 6.55P Ch. 6 - Prob. 6.56P Ch. 6 - Prob. 6.57P Ch. 6 - Prob. 6.58P Ch. 6 - Prob. 6.59P Ch. 6 - Prob. 6.60P Ch. 6 - Prob. 6.61P Ch. 6 - Prob. 6.62P Ch. 6 - Prob. 6.63P Ch. 6 - Prob. 6.64P Ch. 6 - Prob. 6.65P Ch. 6 - Prob. 6.66P Ch. 6 - Prob. 6.67P Ch. 6 - Prob. 6.68P Ch. 6 - Prob. 6.69P Ch. 6 - Prob. 6.70P Ch. 6 - Prob. 6.71P Ch. 6 - Prob. 6.72P Ch. 6 - Prob. 6.73P Ch. 6 - Prob. 6.74P Ch. 6 - Prob. 6.75P Ch. 6 - Prob. 6.76P Ch. 6 - Prob. 6.77P Ch. 6 - Prob. 6.78P Ch. 6 - Prob. 6.79P Ch. 6 - Prob. 6.80P Ch. 6 - Prob. 6.81P Ch. 6 - Prob. 6.82P Ch. 6 - Prob. 6.83P Ch. 6 - Prob. 6.84P Ch. 6 - Prob. 6.1PP Ch. 6 - Prob. 6.2PP

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