1. The base of machine is supported by four (4) two (2) leg supports (like a shipping container crane) consisting of 2.4m long steel struts pinned at each end. The machine must remain level and aligned at all times. Due to a maintenance issue, one of the 20mm square struts is replaced with a 20mmx50mm titanium one.
1a. What is the result of this on pin “C”
2. A 500ft length of 3/4”-6×37 class bright wire rope EIPS IWRC (steel cable) is to be used to suspend some lights across a football field. It is attached to adjustable eye bolts at each support. Each eye bolt has 12 inches of adjustment available. The bolts are exactly 500 ft apart to start.
2a. How far do you need to loosen the eye bolts to initially hang just the cable? (Note: it may help to treat the distributed load as a point load at the middle with the cable acting as two thin rods initially 250 ft long)
2b. What is the final position of the bolts in order to have the cable “sag” less than or equal to 6 inches after the light) is attached?
Note Hibbeler Mechanics of Materials 10th Edition R3-4 and R3-5 may help you. Attached.
3. A steel process component is mad by welding three (3) pipes to two (2) pre drilled steel end plates. Due to a serious lack of caffeine, the center pipe is cut 0.35mm short. The worker decides to fix the problem by forcing the ends together in a large vice and then welding the center pipe (B).
3a. What is the stress in each pipe after it is removed from the vice? (Neglect any welding stress)
Note that triangle thing (all connections) on C is a welding symbol for a 6mm fillet.
3b. Discuss/explain what this did to the capacity of the finished part.
© 2017 Pearson Education, Inc., Hoboken, 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. *R3–4.1The wires each have a diameter of in., length of 2D2 ft, and are made from 304 stainless steel. If P = 6 kip, Cdetermine the angle of tilt of the rigid beam AB.2 ftP2 ft 1 ftBASolutionEquations of Equilibrium: Referring to the free-body diagram of beam AB shown in Fig. a,+S M = 0; F (3) – 6(2) = 0 F = 4 kipA BC BC+ cSM = 0; 6(1) – F (3) = 0 F = 2 kipB AD ADNormal Stress and Strain:3F 4(10 )BCs = = = 20.37 ksiBC2Ap 1BCa b4 23F 2(10 )ADs = = = 10.19 ksiAD2AAD p 1a b4 2Since s 6 s and s 6 s , Hooke’s Law can be applied.BC Y A Y3 -3s = EP ; 20.37 = 28.0(10 )P P = 0.7276(10 ) in.>in.BC BC BC BC3 -3s = EP ; 10.19 = 28.0(10 )P P = 0.3638(10 ) in.>in.AD AD AD ADThus, the elongation of cables BC and AD are given by-3d = P L = 0.7276(10 )(24) = 0.017462 in.BC BC BC-3d = P L = 0.3638(10 )(24) = 0.008731 in.AD AD ADReferring to the geometry shown in Fig. b and using small angle analysis,d – d0.017462 – 0.008731 180°BC AD-3u = = = 0.2425(10 ) rad a b = 0.0139° Ans.36 36 p radAns:u = 0.0139°177This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the World Wide Web) will destroy the integrity of the work and is not permitted.© 2017 Pearson Education, Inc., Hoboken, 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. R3–5.1The wires each have a diameter of in., length of 2D2 ft,…