WORLD FOR CIVIL

Topic 8.4a: Principal Stresses - Example 1

In Diagram 1 we have shown a shaft attached to a wall at end A with a torque of 2000 ft-lb. and an axial force of 20,000 lb. acting at end B. We wish to determine the principal planes, principal stresses and the maximum shear stress. (The principal, maximum, stress will occur at an element on the outer edge of the shaft as that is where the shear stress is a maximum, while the normal axial stress we will assume is uniform across the area of the shaft.)

Solution:
We first review briefly the static equilibrium conditions for the shaft. The applied torque of 2000 lb/in² at end B is balance by an equal torque which develops in the wall acting on the shaft. The applied horizontal force of 20,000 lb. also applied at end B is balance by an equal 20,000 lb. force which the wall exerts on the shaft. Thus the shaft is in equilibrium. (We have ignored any weight of the shaft.)

We next determine the axial (normal) stress due to the applied 20,000 lb. force by:
= F/A = 20,000 lb. / (3.1416 * 1"²) in² = 6,370 lb/in². (We assume the stress in uniform across the circular cross section.)
We then determine the maximum transverse shear stress by:
Maximum Transverse Shear Stress = Tr/J = (2000 ft-lb * 12 in/ft) * (1") / (pi * d⁴/32) = (24,000 in-lb.)*(1") / (3.1416 * 2"⁴/32) = 15,300 lb/in².

We next show a shaft element with these values indicated in Diagram 2. Also remembering our sign conventions, shown below.
1. Tensile Stress will be considered positive, and Compressive Stresses will be considered negative.
2. The Shear Stress will be considered positive when a pair of shear stress acting on opposite sides of the element will produce a counterclockwise torque (couple). (Some text use the opposite direction for the positive shear stress. This changes a sign in several equations, so we must be somewhat careful of signs when working problems and examples.)
3. The incline plane angle will be measure from the vertical, counterclockwise to the plane. This will be the positive direction for the angle.

And finally we apply our plane stress and transformation equations.
We first apply the formula for determining the angle of the principal planes:
= 15,300 lb/in²/[(6370 lb/in² - 0)/2] = 4.8, then 2(theta) = 78.4^o, and 258.4^o, and theta = 39.2^o, and 129.2^o. These are the angles of the principal planes.

We calculate the principal stress from specific formula for the maximum/minimum stresses:
= (6370 lb/in² + 0)/2 +/- Sqrt[{(6370 lb/in² - 0)/2}² + (15,300 lb/in²)²]
= 3185 lb/in² + 15,630 lb/in² = 18,815 lb/in²
= 3185 lb/in² - 15,630 lb/in² = -12445 lb/in²
And finally we calculate the maximum shear stress from:
= Sqrt[{(6370 lb/in² - 0)/2}² + (15,300 lb/in²)²] = +/- 15,630 lb/in².