Page 176 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 13] ELASTICITY 161

s

A
d
f F

F

Fig. 13-4

SOLVED PROBLEM 13.8
8
The shearing strength of a certain steel alloy is 2.5 × 10 Pa. Two 5 mm-diameter bolts of this alloy are
used to fasten a bracket to a wall. What is the maximum load the bracket can support without shearing
off the bolts?
2
The shearing stress here is exerted perpendicular to each bolt, so for each bolt A = πr and, since r = 0.0025 m,

F 8 2 2
F = (A) = (2.5 × 10 N/m )(π)(0.0025 m) = 4.9kN
A
max
The combined load is twice this, or 9.8 kN. The corresponding mass is 1000 kg.
SOLVED PROBLEM 13.9
How much force is required to punch a hole 1 2 in. in diameter in a steel sheet 1 8 in. thick whose shearing
2
4
strength is 4 × 10 lb/in. ?
The shear stress is exerted over the cylindrical surface that is the boundary of the hole (Fig. 13-5). The area of
this surface is
A = 2πrh = (2π)(0.25 in.)(0.125 in.) = 0.196 in. 2
Since the minimum shear stress needed to rupture the steel is

F 4 2
= 4 × 10 lb/in.
A min
the required force is
F 4 lb 2

F = (A) = 4 × 10 (0.196 in. ) = 7840 lb
A in. 2
min
h

r A = 2prh

Fig. 13-5