Topic 8.4: Special Topics I - Principal Stresses
We next look at the general case where we have several axial stresses and shear stresses acting. We will look at the plane stresses on an inclined plane section to determine what are called the principal stresses and the principal planes.
In Diagram 1 we have shown a structure element with both normal axial stresses and shear stresses acting on the element. We remember at this point that for static equilibrium the shear stresses Tauxy and Tauyx must be equal in magnitude.
In Diagram 2 we have shown the structure element with a plane cut through it at angle theta. Acting on this plane will be both an axial stress 
and a shear stress 
, as shown in Diagram 2a. We would like to write relationships which allow us to calculate the value of these two stress for any arbitrary plane section.
In Diagram 2b, we have shown a triangular element with axial and shear stresses shown. If we multiply these stresses by the appropriate areas, we have the forces on each surface. We may then apply static equilibrium conditions and write the equilibrium equations. Before we do so, we need to establish a sign convention as follows:
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 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.
We now write the following equilibrium equations:
Sum of Forcesx 
Sum of Forcesy 
where we have used that face that magnitude of 
= magnitude of 
.
The two equation above may be solved for two "unknowns". In this case we solve for , and ; the stresses acting on the incline plane shown in Diagram 2. The details of solving these two simultaneous equations involve a number of trigonometric identities and some extended algebraic manipulations, and will not be presented. The results of this process are as follows:
The equations may be referred to as transformation equations. In the above equations, it is clear that we will may have both maximum and minimum stress values. The maximum and minimum normal axial stresses are known as the Principal Stresses, and the planes at which they occur are known as the Principal Planes. At the principal planes, where the axial stress is a maximum or minimum, the shear stress will be zero. The value of the principal(maximum/minimum) stresses are given by:
and
for the principal plane. The planes for maximum shear stress vary by 45o from the principal planes.
The principal stresses may also be related as follows:
Example. In Diagram 3 a structural element is shown with axial and shear stress given by: normal x-stress = 4000 lb/in2, normal y-stress = 3000 lb/in2, shear stress = 1000 lb/in2. We would like to find the principal planes, the principal stresses, and the maximum shear stress.
We first apply the formula for determining the angle of the principal planes:
= 1000 lb/in2/[(4000 lb/in2 - 3000 lb/in2)/2] = 2, then
2(theta) = 63.4o, and 243.4o, and theta = 31.7o, and 121.7o. These are the angles of the principal planes. We will calculate the principal stress two ways. First from the general formula for plane stresses= (4000 lb/in2 + 3000 lb/in2)/2 + [(4000 lb/in2 - 3000 lb/in2)/2]cos(63.4o) + 1000 lb/in2*sin(63.4o)
= 3500 lb/in2 + 224 lb/in2 + 894 lb/in2 = 4618 lb/in2
and for 2nd principal plane
= (4000 lb/in2 + 3000 lb/in2)/2 + [(4000 lb/in2 - 3000 lb/in2)/2]cos(243.4o) + 1000 lb/in2*sin(243.4o)
= 3500 lb/in2 - 224 lb/in2 - 894 lb/in2 = 2382 lb/in2
The second method is from specific formula for the maximum/minimum stresses:
= (4000 lb/in2 + 3000 lb/in2)/2 +/- Sqrt[{(4000 lb/in2 - 3000 lb/in2)/2}2 + (1000 lb/in2)2]
= 3500 lb/in2 + 1118 lb/in2 = 4618 lb/in2
= 3500 lb/in2 - 1118 lb/in2 = 2382 lb/in2
Note that we arrive at the same result by both methods. And finally we calculate the maximum shear stress from:= Sqrt[{(4000 lb/in2 - 3000 lb/in2)/2}2 + (1000 lb/in2)2] = +/- 1118 lb/in2.
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