Stress analysis and combined loading on the PE Mechanical exam

Combined-loading problems on the Machine Design & Materials exam give you a part, a load and a geometry, and ask for the stress at one point: the maximum normal stress, the maximum shear stress, or a von Mises stress and a factor of safety. Most of the work is resolving the load into internal forces at the right section and deciding which stresses add at which fiber. The Mechanical Reference Handbook covers these in Chapter 2: uniaxial loading, Mohr's circle, torsion, beam stresses, the ductile failure theories, and a solid-shaft formula for bending, axial load and torsion together.

What you actually need to own

• Move the load to the section. An offset load equals the same load at the centroid plus a moment M = P·e. A force at the end of an arm can bend and twist at once: its offset along the member sets the bending moment, and its offset from the member's axis sets the torque.
• Normal stresses superpose: σ = P/A ± Mc/I (or ± M/S). On one face the direct and bending stresses add; on the other they subtract. For a rectangle, S = bh²/6, where h is the depth in the plane of bending.
• Biaxial bending. A load offset in both directions bends about both centroidal axes. On a rectangle the two bending stresses add at the corner nearest the load, each with its own section modulus. On a round section the two moments add as vectors, M = √(M₁² + M₂²).
• Round shafts. σ = 32M/(πd³) and τ = 16T/(πd³) for a solid section; the torsion formula uses the polar moment J = πd⁴/32, twice I.
• Combine through Mohr's circle, not arithmetic. For a point with σ and τ: τ_max = √[(σ/2)² + τ²]. For a solid shaft this is τ_max = 16√(M² + T²)/(πd³).
• Into a failure theory. Maximum-shear-stress: yield when σ₁ − σ₃ = S_y, so n = S_y/(2τ_max). Distortion energy: σ′ = √(σ² + 3τ²). The failure theories page covers the choice between them and the fatigue side.
• The no-tension limit. For a material weak in tension, the offset at which tension first appears is e = I/(A·c), whatever the load. For a rectangle of depth h that's h/6, the middle third.

Where people lose points

• Direct or bending stress alone. An eccentric 20,000-lb load (A = 10 in², S = 12 in³, e = 2 in) gives 5,300 psi on the near face. Reporting 2,000 or 3,333 psi drops one term; subtracting them gives the far face.
• The wrong face. On a C-frame the inside edge, nearest the load line, carries direct tension plus bending tension. That's where the maximum is, and where C-frames crack.
• The wrong section modulus. In a 3 × 6 in post with the 6-in side in the plane of bending, the maximum compressive stress is about 6,700 psi. Using the weak-axis modulus gives about 12,700.
• Bending mistaken for torsion. In one actuator bracket, an 800-lbf force sits 10 in from the wall along the bar and 6 in off its axis on an arm. The torque is 4,800 lbf·in; the 8,000 lbf·in moment is bending.
• σ instead of σ/2 under the root. With σ = 40 MPa and τ = 15 MPa, τ_max = 25 MPa. Putting the full σ under the root gives 42.7, and adding σ/2 + τ gives 35.
• Adding a normal stress and a shear stress. A 2-in shaft with 6,000 in·lbf of bending and torque and S_y = 40 ksi has a maximum-shear-stress safety factor of 3.7. Adding 7.64 + 3.82 ksi gives 3.5; using √(σ² + τ²) for τ_max gives 2.3.
• Adding moments that act about perpendicular axes. Two bending moments on a round section combine as √(M₁² + M₂²), never as M₁ + M₂. A bending moment and a torque combine under a root too, never as M + T: √(M² + T²) for maximum shear, √(M² + ¾T²) for von Mises, as in the Handbook's solid-shaft formulas.
• A pressure vessel's third stress. At the outer surface of a thin shell the radial stress is about zero, and it's the third principal stress. The maximum shear uses the extreme pair, which is larger than the in-plane value.

How to study it

Take one bracket and load it three ways: offset axially, offset in both directions, then on an arm so it bends and twists. For each, draw the section, mark the critical point, and write the stress there before you compute anything. Then run the same shaft through both failure theories and compare the margins. Shafts are where bending and torsion meet most often, so continue with power transmission; for loads that vary, go to failure theories and fatigue.

Part of the Mechanics of Materials area of the Machine Design & Materials exam. → Practice it on CastorPrep