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Explain briefly how did we get s=sP-sFD

Explain briefly how did we get s=sP-sFD

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Explain briefly how did we get s=sP-sFD
P Chapter 05.pdf File | C:/Users/Alia/Desktop/Chapter%2004.pdf | D Page view A Read aloud ▼ Draw ++ Highlight Erase O 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 4-46. If the gap between C and the rigid wall at D is initi ly 0.15 mm, determine the support reactions at A and D when the force P = 200 kN is applied. The assembly -600 mm 600 mm -0.15 mm P. is made of A36 steel. 25 mm 50 mm Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig.a, EF, = 0; 200(10°) – FD - FA = 0 (1) Compatibility Equation: Using the method of superposition, Fig. b, (+) 8 = Sp – dF, FD (600) 200(10°)(600) (0.052)(200)(10°) FD (600) 0.15 = (0.05)(200)(10°) "{(0.025°)(200)(10°) Ans. Fp = 20 365.05 N = 20.4 kN Substituting this result into Eq. (1), 100CO OF N 1001-AT
Transcribed Image Text:P Chapter 05.pdf File | C:/Users/Alia/Desktop/Chapter%2004.pdf | D Page view A Read aloud ▼ Draw ++ Highlight Erase O 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 4-46. If the gap between C and the rigid wall at D is initi ly 0.15 mm, determine the support reactions at A and D when the force P = 200 kN is applied. The assembly -600 mm 600 mm -0.15 mm P. is made of A36 steel. 25 mm 50 mm Equation of Equilibrium: Referring to the free-body diagram of the assembly shown in Fig.a, EF, = 0; 200(10°) – FD - FA = 0 (1) Compatibility Equation: Using the method of superposition, Fig. b, (+) 8 = Sp – dF, FD (600) 200(10°)(600) (0.052)(200)(10°) FD (600) 0.15 = (0.05)(200)(10°) "{(0.025°)(200)(10°) Ans. Fp = 20 365.05 N = 20.4 kN Substituting this result into Eq. (1), 100CO OF N 1001-AT
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