WORLD FOR CIVIL

Topic 8.1: Special Topics I - Combined Stress

Up to this point we have considered only or mainly one type of applied stress acting on a structure, member of a structure, beam, shaft, rivet, or weld. Many situations involve more than one type of stress occurring simultaneously in a structure. These problems can become relatively complicated. We will look at several examples of relatively simple combined stress problems.

One of the principles we will apply in these examples is the principle of superposition, that is, the resultant stress will be the algebraic sum of the individual stresses - at least in the case of similar stresses acting along the same line, such as the axial stress due to an axial load and a bending stress. We look at this type of problem in the example below.

Example. A four foot long cantilever beam (shown in Diagram 1) is attached to the wall at point A, and has a load of 10, 000 lb. acting (at the centroid of the beam) at angle of 20^o below the horizontal. We would like to determine the maximum axial stress acting in the beam cross section.

Solution:
We first apply static equilibrium conditions to the beam and determine the external support reactions, and the external moment acting on the beam at point A. Notice we have resolved the 10,000 lb. load into its perpendicular x and y components. The horizontal component of the load (9,400 lb.) produces a normal horizontal axial stress in the beam. The vertical component of the load (-3,420 lb.) causes a torque about point A (13,700 ft-lb) to act on the beam (balanced by the external moment). The resulting internal bending moment(s) in the beam produces an axial bending stress. The total axial stress at a point in the beam will be the sum of the normal axial stress and the axial bending stress.

The Normal Axial Stress = Force/Area = 9,400 lb. / (2" x 4") = 1175 lb/in². We note that this stress will be tensile and constant through out the length of the beam. So the maximum normal Axial Stress is 1175 lb/in², and is the same everywhere in the beam.

The maximum bending stress occurs at the outer edge of the beam, and at the point in the beam where the bending moment is a maximum. In the cantilever beam, the maximum bending moment occurs at the wall and is equal to the (negative of) external bending moment. (M = -13,700 ft-lb. = -164,400 in-lb.) We can then calculate the maximum bending moment by:
Maximum Bending Stress = M y/I = (164,400 in-lb.)(2")/(10.67 in⁴) = 30,820 lb/in².
Since the bending moment was negative, this means that the top of the beam (above the centroid) is in tension, and the bottom on the beam is in compression.

We can now combine (sum) the axial stress at the very top and bottom of the beam to determine the maximum axial stress. We see in the beam section in Diagram 2, that the stresses at the top of the beam are both tensile, and so add to a total tensile stress of 30,820 lb./in² + 1,175 lb./in² = 31,995 lb./in². At the bottom of the beam, the bending stress is compressive and the normal axial stress in tensile so the resultant bottom stress is - 30,820 lb./in² + 1,175 lb./in² = -29,645 lb./in² (compression).

We will now look at several additional examples of combined stresses.