Tải bản đầy đủ (.pdf) (23 trang)

Download test bank for mechanics of materials 8th edition by russell c hibbeler

Bạn đang xem bản rút gọn của tài liệu. Xem và tải ngay bản đầy đủ của tài liệu tại đây (2.18 MB, 23 trang )

Test bank For Mechanics of Materials 8th Edition by Russell C Hibbeler
Link full download:Download here

2–1. An air-filled rubber ball has a diameter of 6 in. If the air
pressure within it is increased until the ball’s diameter becomes
7 in., determine the average normal strain in the rubber.
d0 = 6 in.

d = 7 in.
pd - pd

e=

0=

= 0.167 in./in. Ans. pd0 6

2–2. A thin strip of rubber has an unstretched length of
15 in. If it
is stretched around a pipe having an outer diameter
of 5 in., determine the
average normal strain in the strip.
L0 = 15 in.
L = p(5 in.)
L - L0
e = L0

5p - 15
=

15 = 0.0472 in.>in.



Ans.

2–3. The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the
beam causes the end C to be
D

displaced 10 mm downward, determine the normal strain

E

developed in wires CE and BD.

4m
P

¢LBD
¢LCE
3
=
7

¢L BD =

A

B

C


3 (10)
7
= 4.286 mm

2m

3m

¢LCE

10
Ans. eCE = L = 4000 = 0.00250 mm>mm

¢LBD

Ans. eBD = L = 4000 = 0.00107 mm>mm

2m


1

02 Solutions 46060 5/6/10 1:45 PM Page 2

© 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.


02 Solutions 46060 5/6/10 1:45 PM


2
Page 3


© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No

2–6. Nylon strips are fused to glass plates. When moderately
heated the nylon will become soft while the glass stays
approximately rigid . Determine the average shear strain in the
nylon due to the load
P when the assembly deforms as
indicated.

y
2 mm

3 mm

P

5 mm
3 mm
5 mm
3 mm

g = tan

-1

2


x

Ans.

a10 b = 11.31° = 0.197 rad

2–7. If the unstretched length of the bowstring is 35.5 in.,
determine the average normal strain in the string when it is
stretched to the position shown.
18 in.

6 in.
18 in.

Geometry: Referring to Fig2. a, the stretched length of the string is

2

2

L = 2L¿ = 2 18 + 6 = 37.947 in.
Average Normal Strain:

=

e
avg

L - L0

L0

37.947 - 35.5
=

35.5

= 0.0689 in.>in.

Ans.

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.


02 Solutions 46060 5/6/10 1:45 PM

3


Page 4

© 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.


02 Solutions 46060 5/6/10 1:45 PM

4




02 Solutions 46060 5/6/10 1:45 PM

Page 5

© 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.


5


02 Solutions 46060 5/6/10 1:45 PM


02 Solutions 46060 5/6/10 1:45 PM Page 6

© 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.


02 Solutions 46060 5/6/10 1:45 PM Page 7

6
© 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.


02 Solutions 46060 5/6/10 1:45 PM Page 8


© 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.

7


02 Solutions 46060 5/6/10 1:45 PM Page 9

8
© 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.


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.


02 Solutions 46060 5/6/10 1:45 PM Page 11

9


1
0


02 Solutions 46060 5/6/10 1:45 PM 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.


02 Solutions 46060 5/6/10 1:45 PM Page 12

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.


02 Solutions 46060 5/6/10 1:45 PM Page 13

12


02 Solutions 46060 5/6/10 1:45 PM Page 14

© 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.


13



×