Views
2 months ago

# ch01-03 stress &amp; strain &amp; properties

## 03

03 Solutions 46060 5/7/10 8:45 AM 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. •3–13. A bar having a length of 5 in. and cross-sectional area of 0.7 is subjected to an axial force of 8000 lb. If the bar stretches 0.002 in., determine the modulus of elasticity of the material. The material has linear-elastic behavior. 8000 lb 5 in. 8000 lb Normal Stress and Strain: in 2 10 s = P A = 8.00 0.7 e = dL L = 0.002 5 = 11.43 ksi = 0.000400 in.>in. Modulus of Elasticity: E = s e = 11.43 0.000400 = 28.6(103 ) ksi Ans. 3–14. The rigid pipe is supported by a pin at A and an A-36 steel guy wire BD. If the wire has a diameter of 0.25 in., determine how much it stretches when a load of P = 600 lb acts on the pipe. B Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a 4 ft P a+©M A = 0; F BD A 4 5 B(3) - 600(6) = 0 F BD = 1500 lb The normal stress developed in the wire is s BD = F BD A BD = 1500 p 4 (0.252 ) = 30.56(103 ) psi = 30.56 ksi A D 3 ft 3 ft C Since s BD 6 s y = 36 ksi, Hooke’s Law can be applied to determine the strain in the wire. s BD = EP BD ; 30.56 = 29.0(10 3 )P BD P BD = 1.054(10 - 3 ) in.>in. The unstretched length of the wire is wire stretches L BD = 23 2 + 4 2 = 5ft = 60 in. Thus, the d BD =P BD L BD = 1.054(10 - 3 )(60) = 0.0632 in Ans.

03 Solutions 46060 5/7/10 8:45 AM Page 11 © 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. 3–15. The rigid pipe is supported by a pin at A and an A-36 guy wire BD. If the wire has a diameter of 0.25 in., determine the load P if the end C is displaced 0.075 in. downward. B 4 ft P A D 3 ft 3 ft C Here, we are only interested in determining the force in wire BD. Referring to the FBD in Fig. a a+©M A = 0; F BD A 4 5 B(3) - P(6) = 0 F BD = 2.50 P The unstretched length for wire BD is L BD = 23 2 + 4 2 = 5 ft = 60 in. From the geometry shown in Fig. b, the stretched length of wire BD is L BD¿ = 260 2 + 0.075 2 - 2(60)(0.075) cos 143.13° = 60.060017 Thus, the normal strain is P BD = L BD¿ - L BD 60.060017 - 60 = = 1.0003(10 - 3 ) in.>in. L BD 60 Then, the normal stress can be obtain by applying Hooke’s Law. s BD = EP BD = 29(10 3 )C1.0003(10 - 3 )D = 29.01 ksi Since s BD 6 s y = 36 ksi, the result is valid. s BD = F BD A BD ; 29.01(10 3 ) = 2.50 P p 4 (0.252 ) P = 569.57 lb = 570 lb Ans. 11

Stress-Strain Relationship in Soil
Stress and Strain - Fast Facts
Predicting the Creep Strain of PVA-ECC at High Stress Levels ...
STRESS
VOLU ME 03, ISSU E 04 - Camden Property Trust
Strain Analysis
Stress
PROPERTY
Property
Sprains and Strains Prevention Guide
Stress Management
WIPO-ESCWA/EC/BEY/03/4.B - World Intellectual Property ...
Lensing in an elastically strained space-time
Electronic Strain Instrumentation - Tinius Olsen
Resistant Strains of Postmodernism - Ross Feller
Acute Stress Disorder and Posttraumatic Stress Disorder - Australian ...
Stress and Rhythm - DLFeSCHOOL
From Distress to De-stress
Stress in the Wild - EUPRIM-Net
MPI: Stress only study
Managing Stress and Test Anxiety
STRESS MANAGEMENT IN HIGHER EDUCATION
INQÂ® Stress Platform - Integri
Making Peace with Stress
Teams Under Stress