Question

EXERCISE NUMBER 2

1. Find the average shearing stress in a bolt. The bolt holding together the plates is 1 in, in diameter and the load P is 12 kips 2. Determine the average tensile stress in a plate where the load is 20 kips. The plates are each 4 in. by 0.5 in. in cross section and the hole in each plate is 3/4 in. 2 + 3 -> 2 3. Determine its average unit strain in a reinforced concrete column that is 40 ft tall, and under load shortens 2 in.

Answer 1

1.

The bolt is subjected to single shear and the average shear stress can be given as :

$\tau$ = P/A

Where P is the load and A is the area of cross section of bolt.

i.e. A = ($\pi$/4)×(0.0254 m)² [ 1 in = 0.0254 m]

A = 0.019939 m²

load is given , P = 12 kips = 12×4448.22 N = 53378.64 N

Hence the average shearing stress is :

$\tau$_avg = P/A = (53378.64 N)/(0.019939 m²)= 2677.097 kPa