Scribd
Upload
275 views5 pages

Shearing Stress: Solved Problems

Forces parallel to the surface cause shearing stress, which differs from tensile and compressive stresses caused by perpendicular forces. Shearing stress problems involve calculating the force required to shear materials of a given thickness and shear strength. Sample problems provide the thickness, shear strength, stress limit, and ask the reader to calculate the necessary force, hole diameter, pin diameter, or other value.

Uploaded by

GEFred
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOC, PDF, TXT or read online on Scribd
275 views5 pages

Shearing Stress: Solved Problems

Forces parallel to the surface cause shearing stress, which differs from tensile and compressive stresses caused by perpendicular forces. Shearing stress problems involve calculating the force required to shear materials of a given thickness and shear strength. Sample problems provide the thickness, shear strength, stress limit, and ask the reader to calculate the necessary force, hole diameter, pin diameter, or other value.

Uploaded by

GEFred
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOC, PDF, TXT or read online on Scribd
You are on page 1/ 5

Shearing Stress

Forces parallel to the area resisting the force cause shearing stress. It differs to tensile

and compressive stresses, which are caused by forces perpendicular to the area on

which they act. Shearing stress is also known as tangential stress.

where V is the resultant shearing force which passes which passes through the centroid

of the area A being sheared.

SOLVED PROBLEMS IN SHEARING STRESS

Problem 115

What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick?


2
The shear strength is 350 MN/m .

Solution 115
Problem 116

As in Fig. 1-11c, a hole is to be punched out of a plate having a shearing strength of 40

ksi. The compressive stress in the punch is limited to 50 ksi. (a) Compute the maximum

thickness of plate in which a hole 2.5 inches in diameter can be punched. (b) If the

plate is 0.25 inch thick, determine the diameter of the smallest hole that can be

punched.

Solution 116

Problem 117

Find the smallest diameter bolt that can be used in the clevis shown in Fig. 1-11b if P =

400 kN. The shearing strength of the bolt is 300 MPa.

Solution 117

Problem 118

A 200-mm-diameter pulley is prevented from rotating relative to 60-mm-diameter shaft

by a 70-mm-long key, as shown in Fig. P-118. If a torque T = 2.2 kN·m is applied to the

shaft, determine the width b if the

allowable shearing stress in the key is 60 MPa.


Solution 118

Problem 119

Compute the shearing stress in the pin at B for the member supported as

shown in Fig. P-119. The pin diameter is 20 mm.

Solution 119
Problem 120

The members of the structure in Fig. P-120 weigh 200 lb/ft. Determine the smallest

diameter pin that can be used at A if the shearing stress is limited to 5000 psi. Assume

single shear.

Solution 120
Problem 121

Referring to Fig. P-121, compute the maximum force P that can be applied by the

machine operator, if the shearing stress in the pin at B and the axial stress in the

control rod at C are limited to 4000 psi and 5000 psi, respectively. The diameters are

0.25 inch for the pin, and 0.5 inch for the control rod. Assume single shear for the pin

at B.

Solution 121

You might also like