Lecture on Strength of Materials
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 .
answer
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: →
answer (b) Diameter of smallest hole:
Based on compression of puncher:
→ Equivalent shear force for plate Based on shearing of plate: →
answer
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.
answer
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
Where:
Thus,
answer
Problem 119
Compute the
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 .
answer
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: →
answer (b) Diameter of smallest hole:
Based on compression of puncher:
→ Equivalent shear force for plate Based on shearing of plate: →
answer
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.
answer
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
Where:
Thus,
answer
Problem 119
Compute the