Referring to Fig. a, Thus, Ans. Ans. = 26 - 32 32 =

advertisement
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.
2–11. The corners B and D of the square plate are given
the displacements indicated. Determine the average normal
strains along side AB and diagonal DB.
y
A
16 mm
D
B
3 mm
3 mm 16 mm
16 mm
Referring to Fig. a,
LAB = 2162 + 162 = 2512 mm
LAB¿ = 2162 + 132 = 2425 mm
LBD = 16 + 16 = 32 mm
LB¿D¿ = 13 + 13 = 26 mm
Thus,
A eavg B AB =
A eavg B BD =
LAB¿ - LAB
2425 - 2512
=
= -0.0889 mm>mm
LAB
2512
LB¿D¿ - LBD
26 - 32
=
= -0.1875 mm>mm
LBD
32
Ans.
Ans.
6
C
16 mm
x
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.
© 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.
y
2–18. The piece of plastic is originally rectangular.
5 mm
Determine the shear strain g at corners A and B if the
xy
plastic distorts as shown by the dashed lines.
*2–24. A square piece of material is deformed into the
dashed position. Determine the shear strain gxy at C.
2 mm
y
2 mm
4 mm
B
15.18 mm
C
B
C
300 mm
Geometry: For small angles,
15 mm
D
2
a = c =
= 0.00662252 rad
302
b = u =
A
2
= 0.00496278 rad
403
400 mm
89.7!
15.24
mm
2 mm
x
A
3 mm
x
D
15 mm
15.18 mm
Shear Strain:
p
89.7°
(g )xy =
- ¢
≤p
= a2 + b 180°
(gBC)xy
(gA)xy
= 0.0116 rad- 3= 11.6 A 10 - 3 B rad
= 5.24 A 10 B rad
= -(u + c)
Ans.
Ans.
= -0.0116 rad = -11.6 A 10 - 3 B rad
Ans.
2–19. The piece of plastic is originally rectangular.
•2–25. Thethe
guyshear
wire strain
AB ofgaxybuilding
frame
is originally
Determine
at corners
D and
C if the
unstretched.
Due
to an earthquake,
thelines.
two columns of the
plastic
distorts
as shown
by the dashed
frame tilt u = 2°. Determine the approximate normal strain
in the wire when the frame is in this position. Assume the
columns are rigid and rotate about their lower supports.
y
5 mm
u " 2!
2 mm
2 mm
B
C
B
u " 2!
4 mm
300 mm
Geometry: The vertical displacement is negligible
xA
3m
2°
= (1) ¢
≤ p = 0.03491 m
180°
400 mm
A
3 mm
2°
Geometry:
xB = (4) ¢ For ≤small
p = angles,
0.13963 m
180°
2
a = c =
= 0.00496278 rad
x = 4 + x403
B - xA = 4.10472 m
2
= 0.00662252 rad
b = u =
3022 + 4.104722 = 5.08416 m
A¿B¿ = 23
Shear Strain:
1m
AB = 232 + 42 = 5.00 m
(gC)xy = -(a + b)
Average=Normal
-0.0116Strain:
rad = -11.6 A 10 - 3 B rad
A¿B¿ - AB
eAB
=
+ c
(g
D)xy = u AB
=
D
Ans.
-3
= 0.0116-rad
5.08416
5 = 11.6 A 10- 3 B rad
= 16.8 A 10 B m>m
5
Ans.
Ans.
11
14
4m
2 mm
A
x
02 Solutions 46060
5/6/10
1:45 PM
Page 15
© 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.
2–26. The material distorts into the dashed position
shown. Determine (a) the average normal strains along
sides AC and CD and the shear strain gxy at F, and (b) the
average normal strain along line BE.
y
15 mm
10 mm
C
25 mm
D
B
E
75 mm
90 mm
A
Referring to Fig. a,
LBE = 2(90 - 75)2 + 802 = 26625 mm
LAC¿ = 21002 + 152 = 210225 mm
LC¿D¿ = 80 - 15 + 25 = 90 mm
f = tan-1 ¢
25
p rad
≤ = 14.04° ¢
≤ = 0.2450 rad.
100
180°
When the plate deforms, the vertical position of point B and E do not change.
LBB¿
15
=
;
LBB¿ = 13.5 mm
90
100
LEE¿
25
=
;
75
100
LEE¿ = 18.75 mm
LB¿E¿ = 2(90 - 75)2 + (80 - 13.5 + 18.75)2 = 27492.5625 mm
Thus,
A eavg B AC =
A eavg B CD =
A eavg B BE =
LAC¿ - LAC
210225 - 100
=
= 0.0112 mm>mm
LAC
100
Ans.
LC¿D¿ - LCD
90 -80
=
= 0.125 mm>mm
LCD
80
Ans.
LB¿E¿ - LBE
27492.5625 - 26625
=
= 0.0635 mm>mm
LBE
26625
Ans.
Referring to Fig. a, the angle at corner F becomes larger than 90° after the plate
deforms. Thus, the shear strain is negative.
Ans.
0.245 rad
15
80 mm
F
x
Download