Page 175 - Schaum's Outline of Theory and Problems of Applied Physics

160 ELASTICITY [CHAP. 13

SOLVED PROBLEM 13.6
By how much can a steel wire 3 m long and 2 mm in diameter be stretched before the elastic limit is
8
exceeded? Young’s modulus for the wire is 2 × 10 11 Pa, and its elastic limit is 2.5 × 10 Pa.
The cross-sectional area of a wire of radius r = 1mm = 10 −3 mis
2
A = πr = 3.14 × 10 −6 m 2
The maximum force that can be applied without exceeding the elastic limit is therefore

F 8 2 −6 2

F = (A) = (2.5 × 10 N/m )(3.14 × 10 m ) = 785 N
A
max
When this force is applied, the wire will stretch by
L 0 F (3m)(785 N) −3
L = = 2 = 3.75 × 10 m = 3.75 mm
Y A (2 × 10 N/m )(3.14 × 10 −6 m )
2
11

SOLVED PROBLEM 13.7
2
A steel cable whose cross-sectional area is 1 in. is used to support an elevator cab weighing 5000 lb. If
2
the stress in the cable is not to exceed 20 percent of the cable’s elastic limit of 40,000 lb/in. , find the
maximum permissible upward acceleration.
2
2
The force that corresponds to a maximum stress in the cable of (0.20)(40,000 lb/in. ) = 8000 lb/in. is
F 2 2

F = (A) = (8000 lb/in. )(1in. ) = 8000 lb
A
max
This force is to equal the weight w of the cab plus the force ma that provides it with an upward acceleration so that
F = w + ma
F − w
a =
m
Since m = w/g,
2
g(F − w) (32 ft/s )(8000 lb − 5000 lb)
a = = = 19.2 ft/s 2
w 5000 lb

SHEAR MODULUS

A shear stress changes the shape of an object, not its volume. Figure 13-4 shows a rectangular block acted on
by shear forces F. The shearing stress is equal to F/A, and the shearing strain is equal to the angle of shear φ,
expressed in radians. Because φ is always small, it is very nearly the same as the ratio s/d between the displace-
ment s of the block’s faces and the distance d between these faces. Below the elastic limit, then, there are two
equivalent expressions for the shear modulus or (modulus of rigidity):
F/A F/A
s = =
φ s/d
shear stress
Shear modulus =
shear strain