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3 months ago

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

## 01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 39 © 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. 1–55. Rods AB and BC each have a diameter of 5 mm. If the load of P = 2 kN is applied to the ring, determine the average normal stress in each rod if u = 60° . A B u P Consider the equilibrium of joint B,Fig.a, : + ©F x = 0; 2 - F AB sin 60° = 0 F AB = 2.309 kN + c ©F y = 0; 2.309 cos 60° - F BC = 0 F BC = 1.155 kN C The cross-sectional area of wires AB and BC are A AB = A BC = p 4 (0.0052 ) = 6.25(10 - 6 )p m 2 . Thus, As avg B AB = F AB A AB = 2.309(103 ) 6.25(10 - 6 )p = 117.62(106 ) Pa = 118 MPa As avg B BC = F BC A BC = 1.155(103 ) 6.25(10 - 6 )p = 58.81(106 ) Pa = 58.8 MPa Ans. Ans. 39

01 Solutions 46060 5/6/10 2:43 PM Page 40 © 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. *1–56. Rods AB and BC each have a diameter of 5 mm. Determine the angle u of rod BC so that the average normal stress in rod AB is 1.5 times that in rod BC. What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa? A u P Consider the equilibrium of joint B,Fig.a, B + c ©F y = 0; F AB cos u - F BC = 0 (1) : + ©F x = 0; P - F AB sin u = 0 (2) The cross-sectional area of rods AB and BC are A AB = A BC = p 4 (0.0052 ) = 6.25(10 - 6 )p m 2 . Since the average normal stress in rod AB is required to be 1.5 times to that of rod BC, then C As avg B AB = 1.5 As avg B BC F AB A AB = 1.5 a F BC A BC b F AB 6.25(10 - 6 )p = 1.5 c F BC 6.25(10 - 6 )p d F AB = 1.5 F BC (3) Solving Eqs (1) and (3), u = 48.19° = 48.2° Ans. Since wire AB will achieve the average normal stress of 100 MPa first when P increases, then F AB = s allow A AB = C100(10 6 )DC6.25(10 - 6 )pD = 1963.50 N Substitute the result of F AB and u into Eq (2), P = 1.46 kN Ans. 40

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01 Hubley Int&ch01:(Q7) - Polity
The text passage provided is a technical excerpt from an engineering mechanics textbook, specifically dealing with problems related to stress and strain in rods under load. It includes detailed calculations and problem-solving steps for determining the average normal stress in rods under given conditions.
multiple_questions_with_answers

the passage contains multiple questions (problems) with their respective solutions, including the determination of average normal stress in rods and the angle and load required to achieve specific stress conditions.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Views
3 months ago

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

## 01 Solutions 46060

01 Solutions 46060 5/6/10 2:43 PM Page 39 © 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. 1–55. Rods AB and BC each have a diameter of 5 mm. If the load of P = 2 kN is applied to the ring, determine the average normal stress in each rod if u = 60° . A B u P Consider the equilibrium of joint B,Fig.a, : + ©F x = 0; 2 - F AB sin 60° = 0 F AB = 2.309 kN + c ©F y = 0; 2.309 cos 60° - F BC = 0 F BC = 1.155 kN C The cross-sectional area of wires AB and BC are A AB = A BC = p 4 (0.0052 ) = 6.25(10 - 6 )p m 2 . Thus, As avg B AB = F AB A AB = 2.309(103 ) 6.25(10 - 6 )p = 117.62(106 ) Pa = 118 MPa As avg B BC = F BC A BC = 1.155(103 ) 6.25(10 - 6 )p = 58.81(106 ) Pa = 58.8 MPa Ans. Ans. 39

01 Solutions 46060 5/6/10 2:43 PM Page 40 © 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. *1–56. Rods AB and BC each have a diameter of 5 mm. Determine the angle u of rod BC so that the average normal stress in rod AB is 1.5 times that in rod BC. What is the load P that will cause this to happen if the average normal stress in each rod is not allowed to exceed 100 MPa? A u P Consider the equilibrium of joint B,Fig.a, B + c ©F y = 0; F AB cos u - F BC = 0 (1) : + ©F x = 0; P - F AB sin u = 0 (2) The cross-sectional area of rods AB and BC are A AB = A BC = p 4 (0.0052 ) = 6.25(10 - 6 )p m 2 . Since the average normal stress in rod AB is required to be 1.5 times to that of rod BC, then C As avg B AB = 1.5 As avg B BC F AB A AB = 1.5 a F BC A BC b F AB 6.25(10 - 6 )p = 1.5 c F BC 6.25(10 - 6 )p d F AB = 1.5 F BC (3) Solving Eqs (1) and (3), u = 48.19° = 48.2° Ans. Since wire AB will achieve the average normal stress of 100 MPa first when P increases, then F AB = s allow A AB = C100(10 6 )DC6.25(10 - 6 )pD = 1963.50 N Substitute the result of F AB and u into Eq (2), P = 1.46 kN Ans. 40

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Strains and stresses in GaN heteroepitaxy â sources and ... - Laytec
WIPO-ESCWA/EC/BEY/03/4.B - World Intellectual Property ...
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Engineering Properties of Rocks: Behavior in response to stress ...
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VOLU ME 03, ISSU E 04 - Camden Property Trust
Carlson Strain Meter - MGS
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Stress incontinence
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Resistance Determination of a Respiratory Pathogenic Strain of ...
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01 Hubley Int&ch01:(Q7) - Polity
user
The text passage provided is a technical excerpt from an engineering mechanics textbook, specifically dealing with problems related to stress and strain in rods under load. It includes detailed calculations and problem-solving steps for determining the average normal stress in rods under given conditions.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

The passage contains multiple questions (problems) with their respective solutions, including the determination of average normal stress in rods and the angle and load required to achieve specific stress conditions.