Combined loading occurs when a machine part carries more than one type of load at the same time, such as an axial force, a bending moment, and a torque. Real shafts, brackets, bolts, and frames rarely experience a single simple load, so engineers must calculate the stresses together to predict failure. The main goal is to find the point in the part where the combined stress state is most severe.
That point is then checked against the material yield strength using a safety factor.
Key Facts
- Axial normal stress: σa = P/A, where P is axial force and A is cross-sectional area.
- Bending normal stress: σb = Mc/I, where M is bending moment, c is distance from the neutral axis, and I is area moment of inertia.
- Torsional shear stress in a circular shaft: τ = Tc/J, where T is torque and J is polar moment of inertia.
- For a solid circular shaft: A = πd^2/4, I = πd^4/64, and J = πd^4/32.
- Normal stresses from axial load and bending superpose algebraically: σ = P/A ± Mc/I.
- Von Mises yield check for plane stress with one normal stress and one shear stress: σvm = sqrt(σ^2 + 3τ^2), and safety factor n = Sy/σvm.
Vocabulary
- Combined loading
- Combined loading is a situation where a part experiences multiple load types, such as axial force, bending, and torsion, at the same time.
- Critical point
- The critical point is the location in a part where the combined stresses create the greatest risk of yielding or failure.
- Normal stress
- Normal stress is stress that acts perpendicular to a cross section and is commonly produced by axial load or bending.
- Shear stress
- Shear stress is stress that acts parallel to a surface and is commonly produced by torsion or transverse shear.
- Von Mises stress
- Von Mises stress is an equivalent stress used to predict yielding in ductile materials under combined normal and shear stresses.
Common Mistakes to Avoid
- Adding axial, bending, and torsional stresses as if they are all the same kind of stress is wrong because normal stress and shear stress affect yielding differently.
- Checking only the center of a shaft is wrong because bending and torsional stresses are usually largest at the outer surface.
- Using the wrong sign for bending stress is wrong because one side of the member may be in tension while the opposite side is in compression.
- Comparing maximum shear stress directly to tensile yield strength is wrong unless the chosen yield criterion specifically supports that comparison.
Practice Questions
- 1 A solid circular steel shaft has diameter d = 40 mm and carries an axial tensile force P = 20 kN. Find the axial normal stress σa in MPa.
- 2 A solid circular shaft has d = 50 mm, bending moment M = 300 N·m, and torque T = 200 N·m. Compute the outer-surface bending stress σb, torsional shear stress τ, and von Mises stress σvm.
- 3 A bracket has an axial tensile load, a bending moment, and a torque. Explain how you would choose the critical point on the cross section and why that point may not be where the axial stress alone is largest.