Views
1 month ago

ch01-03 stress & strain & properties

02 Solutions 46060

02 Solutions 46060 5/6/10 1:45 PM Page 9 © 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. 2–15. Two bars are used to support a load P. When unloaded, AB is 5 in. long, AC is 8 in. long, and the ring at A has coordinates (0, 0). If a load is applied to the ring at A,so that it moves it to the coordinate position (0.25 in., -0.73 in.), determine the normal strain in each bar. B 60 y C 5 in. 8 in. A x P Geometry: a = 28 2 - 4.3301 2 = 6.7268 in. L A¿B = 2(2.5 + 0.25) 2 + (4.3301 + 0.73) 2 = 5.7591 in. L A¿C = 2(6.7268 - 0.25) 2 + (4.3301 + 0.73) 2 = 8.2191 in. Average Normal Strain: e AB = L A¿B - L AB L AB = 5.7591 - 5 5 = 0.152 in.>in. Ans. e AC = L A¿C - L AC L AC = 8.2191 - 8 8 = 0.0274 in.>in. Ans. 9

02 Solutions 46060 5/6/10 1:45 PM Page 10 © 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. *2–16. The square deforms into the position shown by the dashed lines. Determine the average normal strain along each diagonal, AB and CD. Side D¿B¿ remains horizontal. Geometry: D¿ y D B¿ B 3 mm AB = CD = 250 2 + 50 2 = 70.7107 mm C¿D¿ =253 2 + 58 2 - 2(53)(58) cos 91.5° = 79.5860 mm B¿D¿ =50 + 53 sin 1.5° - 3 = 48.3874 mm 53 mm A 91.5 C 50 mm C¿ x AB¿ =253 2 + 48.3874 2 - 2(53)(48.3874) cos 88.5° = 70.8243 mm 50 mm 8 mm Average Normal Strain: e AB = AB¿ -AB AB = 70.8243 - 70.7107 70.7107 = 1.61A10 - 3 B mm>mm Ans. e CD = C¿D¿ -CD CD = 79.5860 - 70.7107 70.7107 = 126A10 - 3 B mm>mm Ans. •2–17. The three cords are attached to the ring at B.When a force is applied to the ring it moves it to point B¿ , such that the normal strain in AB is P AB and the normal strain in CB is P CB . Provided these strains are small, determine the normal strain in DB. Note that AB and CB remain horizontal and vertical, respectively, due to the roller guides at A and C. A A¿ L B B¿ Coordinates of B (L cos u, L sin u) Coordinates of B¿ (L cos u + e AB L cos u, L sin u + e CB L sin u) L DB¿ = 2(L cos u + e AB L cos u) 2 + (L sin u + e CB L sin u) 2 D u C C¿ L DB¿ = L2cos 2 u(1 + 2e AB + e 2 AB) + sin 2 u(1 + 2e CB + e 2 CB) e AB e CB Since and are small, L DB¿ = L21 + (2 e AB cos 2 u + 2e CB sin 2 u) Use the binomial theorem, L DB¿ = L ( 1 + 1 2 (2 e AB cos 2 u + 2e CB sin 2 u)) = L ( 1 + e AB cos 2 u + e CB sin 2 u) Thus, e DB = L( 1 + e AB cos 2 u + e CB sin 2 u) - L L e DB = e AB cos 2 u + e CB sin 2 u Ans. 10

Stress and Strain - Fast Facts
Strains and stresses in GaN heteroepitaxy – sources and ... - Laytec
VOLU ME 03, ISSU E 04 - Camden Property Trust
Strain Analysis
PROPERTY
Sprains and Strains Prevention Guide
Academic Stress
Electronic Strain Instrumentation - Tinius Olsen
Resistant Strains of Postmodernism - Ross Feller
Lensing in an elastically strained space-time
Acute Stress Disorder and Posttraumatic Stress Disorder - Australian ...
Predicting the Creep Strain of PVA-ECC at High Stress Levels ...
From Distress to De-stress
Stress in the Wild - EUPRIM-Net
Stress Diabetes Newsletter 5