hibbeler - mechanics of materials solutions chapter 2 8th ed
02 Solutions 46060
5/6/10
1:45 PM
Page 1
© 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.
2–1. 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. e =
pd - pd0
7 - 6
=
= 0.167 in./in. pd0 6
Ans.
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.) e =
L - L0
5p - 15
=
= 0.0472 in.>in.
L0
15
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 displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
D
E
4m
P
¢LBD
¢LCE
=
3
7
A
3 (10)
= 4.286 mm
7
¢LCE
10
=
=
= 0.00250 mm>mm
L
4000
3m
¢LBD = eCE eBD =
B
Ans.
¢LBD
4.286
=
= 0.00107 mm>mm
L
4000
Ans.
1
C
2m
2m
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.
*2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.
C
œ
LAC = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm
eAC = eAB
300
œ
LAC - LAC
301.734 - 300
=
=
= 0.00578 mm>mm
LAC
300
mm
30Њ
5/6/10
1:45 PM
Page 1
© 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.
2–1. 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. e =
pd - pd0
7 - 6
=
= 0.167 in./in. pd0 6
Ans.
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.) e =
L - L0
5p - 15
=
= 0.0472 in.>in.
L0
15
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 displaced 10 mm downward, determine the normal strain developed in wires CE and BD.
D
E
4m
P
¢LBD
¢LCE
=
3
7
A
3 (10)
= 4.286 mm
7
¢LCE
10
=
=
= 0.00250 mm>mm
L
4000
3m
¢LBD = eCE eBD =
B
Ans.
¢LBD
4.286
=
= 0.00107 mm>mm
L
4000
Ans.
1
C
2m
2m
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.
*2–4. The two wires are connected together at A. If the force P causes point A to be displaced horizontally 2 mm, determine the normal strain developed in each wire.
C
œ
LAC = 23002 + 22 - 2(300)(2) cos 150° = 301.734 mm
eAC = eAB
300
œ
LAC - LAC
301.734 - 300
=
=
= 0.00578 mm>mm
LAC
300
mm
30Њ