Page 174 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 13] ELASTICITY 159

SOLVED PROBLEM 13.3
8
The elastic limit of aluminum is 1.3 × 10 Pa. What is the maximum mass that the wire of Prob. 13.2 can
support without exceeding its elastic limit?
2
Since the cross-sectional area of the wire is A = 7.07 × 10 −6 m and

F 8 2
= 1.3 × 10 N/m
A
max
the maximum force is
8
2
2
F max = (1.3 × 10 N/m )(7.07 × 10 −6 m ) = 919 N
which corresponds to a mass of
w F max 919 N
m = = = 2 = 94 kg
g g 9.8 m/s

SOLVED PROBLEM 13.4
2
A wire 8 ft long with a cross-sectional area of 0.01 in. stretches by 0.05 in. when a weight of 100 lb is
suspended from it. Find the stress on the wire, the resulting strain, and the value of Young’s modulus for
the wire’s material.
F 100 lb 4 2
Stress = = = 10 lb/in.
A 0.01 in. 2
L 0.05 in. −4
Strain = = = 5.2 × 10
L 0 96 in.
4
F/A 10 lb/in. 2 7 2
Y = = −4 = 1.92 × 10 lb/in.
L/L 0 5.2 × 10
SOLVED PROBLEM 13.5
A steel pipe 3.6 m long is placed vertically under a sagging floor to support it. The inside diameter of the
pipe is 80 mm, its outside diameter is 100 mm, and Y = 2 × 10 11 Pa. A sensitive strain gauge indicates
that the pipe’s length decreases by 0.1 mm. What is the magnitude of the load the pipe supports?
The cross-sectional area of the pipe (Fig. 13-3) is

2
2
2
2
A = π(R − r ) = π[(0.05 m) − (0.04 m) ] = 2.83 × 10 −3 m 2
Here L = 1 × 10 −4 m, so
L 1 × 10 −4 m
2
11
2
4
F = YA = (2 × 10 N/m )(2.83 × 10 −3 m ) = 1.57 × 10 N = 15.7kN
L 0 3.6m
2
2
A = p (R − r )
r
R

Fig. 13-3