Explain briefly how did we get s=sP-sFD P Chapter 05.pdfFile | C:/Users/Alia/Desktop/Chapter%2004.pdf| D Page viewA Read aloud ▼ Draw++HighlightEraseO 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currentlyexist. 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 isiniti ly 0.15 mm, determine the support reactions at A andD when the force P = 200 kN is applied. The assembly-600 mm600 mm-0.15 mmP.is made of A36 steel.25 mm50 mmEquation of Equilibrium: Referring to the free-body diagram of the assemblyshown 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 kNSubstituting this result into Eq. (1),100CO OF N1001-AT

The principle of superposition states that the total internal and external forces acting over a…

Answered: Explain briefly how did we get s=sP-sFD

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P