STRENGTH OF MATERIALS || CHAPTER 01 || SHEAR STRESS

SHEAR STRESS

Forces parallel to the area resisting the force cause shearing stress. Usually shearing force act parallel to the surface. Shearing force is also known as tangential stress and mathematically

Here the force act on the bolt due to the applied force P, is called Shear force.

SOLVED PROBLEM AND SOLUTION

Problem: 115

What force is required to punch a 20-mm diameter hole in a plate that is 25mm thick? The shear strength is 350 MN/m^2

Solution:

The resisting area is the shaded area along the perimeter and the share force V is equal to the punching fore P

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 :

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 :

Problem: 118

A 200 mm diameter pulley is prevented from rotation relative to 60 mm diameter shaft by a 70 mm long key, as shown in Fig. P-118. If a torque T=2.2kN-m is applied to the shaft, determine the width b if the allowable shearing stress in the key is 60 MPa.

Solution :

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 :

Problem-120

The members of the structure in Fig. P-120 weigh 200 Ib/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 :

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 axial stress in the control rod at C are limited to 4000 psi and 5000 psi, respectively. The diameters ae 0.25 inch for the pin, and 0.5 inch for the control r4od. Assume single shear for the pin at B.

Solution :