Mecanica De Materiales

Tension, Compression,
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)