Lecture on Strength of Materials

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  • Topic: Shear stress, Force, Tensile stress
  • Pages : 5 (465 words )
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  • Published : February 24, 2013
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Shear 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 through the centroid of the area A being sheared.  

Problem 115
What force is required to punch a 20-mm-diameter hole in a plate that is 25 mm thick? The shear strength is 350 MN/m2. The resisting area is the shaded area along the perimeter and the shear force  is equal to the punching force .  



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.  

(a) Maximum thickness of plate:
Based on puncher strength:

       → Equivalent shear force of the plate
Based on shear strength of plate:

(b) Diameter of smallest hole:
Based on compression of puncher:

       → Equivalent shear force for plate
Based on shearing of plate:


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
The bolt is subject to double shear.


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
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From the FBD:




       → shear force of pin at B
       → double shear


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
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For member AB:


       → Equation (1)
For member BC:


       → Equation (2)
Add equations (1) and (2)
       → Equation (1)
       → Equation (2)

From equation (1):

From the FBD of member AB



       → shear force of pin at A

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