WORLD FOR CIVIL

Topic 3.4a: Shear Stress & Strain - Example 1

Example 1
Two metal plates, as shown in Diagram 1, are bolted together with two 3/4" inch diameter steel bolts. The plates are loaded in tension with a force of 20,000 lb., as shown. What is the shearing stress that develops in the steel bolts? If the Modulus of Rigidity for Steel in 12 x 10⁶ lb/in², what shear strain develops in the steel bolts?

As we examine the structure we see the area of the bolt where the two plates surfaces come together are in shear. That is, if we examine one bolt, the top of the bolt experiences a force to the left while the bottom of the bolt experiences and an equal force to the right. (See Diagram 2.) The surface area between the top and bottom interface is in shear. We assume the bolts carry the load equally, and so each bolt "carries" 10,000 lb. From this we can calculate the shear stress in a straight forward manner from one of our Shear Stress/Strain Relationships:

Stress = Force parallel to area/ area
Stress = F / (p * r²) = 10,000 lb/ (3.14 * .375"²) = 22, 650 lb/in².
In similar manner the Shear Strain can be found from the appropriate form of Hooke's Law:

G = (Shear Stress) / (Shear Strain), or (Shear Strain) = (Shear Stress)/ G = (22, 650 lb/in²)/(12 x 106 lb/in²) = .00189