Mecanica De Materiales
and Shear
1
Solution 1.2-1
PART (a)
P1 1700 dAB 1.25 tAB 0.5
dBC 2.25 tBC 0.375
AAB 1.178
sAB 1443 psi ;
sAB
P1
AAB
AAB
p[ dAB
2 (dAB 2tAB)2]
4
PART (b)
ABC 2.209 P2 ABABC P1
CHECK:
P1 +P2
ABC
1443 psi
P2 1488 lbs ;
ABC
p[ dBC
2 1dBC 2tBC22]
4
01Ch01.qxd 9/25/08 7:49 PM Page 1
2 CHAPTER 1 Tension,Compression, and Shear
Part (c)
P2 2260
dBC
2
4
p
a
P1 + P2
sAB
b
(dBC 2tBC)2
P1 + P2
sAB
2.744
P1 + P2sAB
ABC
Problem 1.2-2 A force P of 70 N is applied by
a rider to the front hand brake of a bicycle (P is
the resultant ofan evenly distributed pressure). As
the hand brake pivots at A, a tension T develops in
the 460-mm long brake cable (Ae 1.075mm2)
which elongates by 0.214 mm. Find normal
stress and strain in the brake cable.
50 mm
100 mm
P (Resultant
ofdistributed
pressure)
A
Brake cable, L = 460 mm Hand brake pivot A
37.5 mm
T
Uniform hand
brake pressure
Solution 1.2-2tBC 0.499 inches ;
tBC
dBC AdBC
2
4
p
a
P1 + P2
sAB
b
2
dBC 2tBC AdBC
2
4
p
a
P1 + P2
sAB
b
P 70 N Ae 1.075 mm2
L 460 mm 0.214 mm
Statics: sum moments about A to get T 2P
NOTE: (E for cables is approx. 140 GPa)
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