Views
7 months ago

ch01-03 stress & strain & properties

01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 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. 1–15. Determine the resultant internal loading on the cross section through point C of the pliers. There is a pin at A, and the jaws at B are smooth. 20 N 120 mm 40 mm 15 mm + c ©F y = 0; -V C + 60 = 0; V C = 60 N : + ©F x = 0; N C = 0 Ans. Ans. A C B +d©M C = 0; -M C + 60(0.015) = 0; M C = 0.9 N.m Ans. D 20 N 80 mm 30 *1–16. Determine the resultant internal loading on the 20 N cross section through point D of the pliers. 120 mm 40 mm 15 mm R+©F y = 0; V D - 20 cos 30° = 0; V D = 17.3 N Ans. C +b©F x = 0; N D - 20 sin 30° = 0; N D = 10 N Ans. +d©M D = 0; M D - 20(0.08) = 0; M D = 1.60 N.m Ans. D A B 20 N 80 mm 30 9

01 Solutions 46060 5/6/10 2:43 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. •1–17. Determine resultant internal loadings acting on section a–a and section b–b. Each section passes through the centerline at point C. Referring to the FBD of the entire beam, Fig. a, a 5 kN b B 1.5 m a+ ©M A = 0; N B sin 45°(6) - 5(4.5) = 0 N B = 5.303 kN Referring to the FBD of this segment (section a–a), Fig. b, +b©F x¿ = 0; N a - a + 5.303 cos 45° = 0 N a - a = -3.75 kN +a ©F y¿ = 0; V a - a + 5.303 sin 45° - 5 = 0 V a - a = 1.25 kN Ans. Ans. A 45 C 45 b 3 m 1.5 m a a+ ©M C = 0; 5.303 sin 45°(3) - 5(1.5) - M a - a = 0 M a - a = 3.75 kN # m Ans. Referring to the FBD (section b–b) in Fig. c, ; + ©F x = 0; N b - b - 5 cos 45° + 5.303 = 0 N b - b = -1.768 kN = -1.77 kN + c ©F y = 0; V b - b - 5 sin 45° = 0 V b - b = 3.536 kN = 3.54 kN Ans. Ans. a+©M C = 0; 5.303 sin 45° (3) - 5(1.5) - M b - b = 0 M b - b = 3.75 kN # m Ans. 10

day 68 - stress, strain, fracture
Stress-Strain Relationship in Soil
Stress and Strain - Fast Facts
Engineering Properties of Rocks: Behavior in response to stress ...
STRESS
stress
Stress
Strained Graphene
Stress • What is stress?
PROPERTY
PROPERTY
Property
PROPERTY
Properties
Predicting the Creep Strain of PVA-ECC at High Stress Levels ...
Sprains & Strains PCOC
Sources of Stress
Dealing with Stress
STRESS MANAGMENT
holiday stress
Strains and stresses in GaN heteroepitaxy – sources and ... - Laytec
Ch01 INTRO
Algae to Biofuels: From strain isolation to strain engineering - SPG
2010-03-03
FROM COMbAT STRESS TO OpERATiONAl STRESS: ThE CF's ...
When to stress patients When to stress patients with diabetes ...
Stress echocardiography 101 - AAIM
Stress Relievers - Prime Line
stress-response-report-v41
Manage and reduce stress