Mixed Mechanics and Materials Practice
Key Takeaways
- Classify each item as statics, dynamics, mechanics-of-materials, or materials selection before opening the FE Reference Handbook.
- A bending-stress answer is incomplete until it is compared to yield strength through a factor of safety FS = Sy / sigma.
- Combined loading on shafts pairs torsion tau = Tr/J with bending sigma = Mc/I, then uses a maximum-shear or von Mises check.
- Material selection questions are failure-mode questions: brittle fracture, fatigue, creep, and corrosion each point to a different property.
- FE distractors reuse the correct formula with the wrong axis, radius-versus-diameter, boundary condition, or mass-versus-weight unit.
Classify the model in the first 20 seconds
A mixed mechanics set should feel nothing like a chapter homework set. One item is a pin reaction, the next a rotating disk, then a fatigue clue, a centroid lookup, and a material-selection decision. The FE Mechanical computer-based exam delivers 110 questions in a 5 hour 20 minute exam clock (about 2.9 minutes per item), so the candidates who pass are the ones who classify quickly. Before calculating, label the model out loud: statics equilibrium, particle dynamics, rigid-body rotation, axial/torsion/bending stress, beam deflection, column buckling, pressure vessel, fatigue, or materials/process selection.
Use one fixed setup routine even when the arithmetic looks trivial. Draw the free-body diagram, write the sign convention, list knowns with units, and state exactly what the answer asks for. This routine defends against the four classic FE traps: using mass where weight belongs, using diameter where radius belongs, taking a moment about the wrong point, and reading a Handbook beam case with the wrong support condition.
| Prompt cue | Governing model | Main trap |
|---|---|---|
| Pin, roller, cable, distributed load | Statics equilibrium | Resultant location of a triangular load |
| Velocity, acceleration, curved path | Particle dynamics | Assuming constant acceleration when it is not |
| Torque, power, shaft diameter | Torsion tau = Tr/J | rpm not converted to rad/s |
| Beam load + support type | Bending sigma = Mc/I or deflection | Wrong Handbook boundary case |
| Slender compression member | Euler buckling | Effective-length factor K ignored |
| Repeated load below yield | Fatigue | Static FS used alone |
| Hot or corrosive service | Materials selection | Choosing by strength only |
Worked problem: combined torsion and bending on a shaft
A solid steel shaft of diameter d = 40 mm transmits P = 15 kW at N = 1500 rpm while a gear applies a transverse force producing a bending moment M = 120 N·m at the critical section. Find the maximum bending stress and the torsional shear stress.
First get torque from power. Angular speed omega = 2piN/60 = 2pi1500/60 = 157.1 rad/s. Then T = P/omega = 15000 / 157.1 = 95.5 N·m.
Section properties for a solid circular shaft (FE Handbook): I = pid^4/64 and J = pid^4/32, with c = d/2 = 0.020 m. So I = pi*(0.040)^4/64 = 1.257e-7 m^4 and J = 2.513e-7 m^4.
Bending stress sigma = Mc/I = 1200.020 / 1.257e-7 = 19.1 MPa. Torsional shear tau = Tc/J = 95.50.020 / 2.513e-7 = 7.60 MPa. A finished design check would feed these into the maximum-shear-stress theory, tau_max = sqrt((sigma/2)^2 + tau^2) = sqrt(9.55^2 + 7.60^2) = 12.2 MPa, then compare to Sy/2.
Finish with a property or factor of safety
Many FE misses occur because the candidate stops at a stress number. A bending stress is not finished until compared with yield strength, an allowable, or endurance logic. 2% offset point is yield strength Sy (controls permanent set for ductile metals), and ultimate strength Su is not the normal yielding limit. If a part failed suddenly with little plastic deformation, suspect brittle fracture or a stress concentration; if it failed after many cycles below yield, suspect fatigue even when nominal stress is below Sy.
Close every mixed set with a five-column error log: model choice, diagram/geometry, Handbook lookup case, units, and execution. Remediate narrowly — three resultant-location drills for a triangular-load miss, three USCS dynamics problems in slugs for an lbm-as-lbf miss. The FE rewards fast recognition and clean execution under topic switching, not encyclopedic memory.
Worked problem: statics resultant and reaction
A simply supported beam spans L = 6 m and carries a triangular distributed load that grows from 0 at the left support to w0 = 12 kN/m at the right support. Find the right-support reaction. e. 4 m from the left support. Taking moments about the left support: R_right * 6 = 36 * 4, so R_right = 144/6 = 24 kN, and by vertical equilibrium R_left = 36 - 24 = 12 kN. The favorite distractor places the resultant at midspan (3 m), which would give 18 kN — wrong because the load is not uniform. Always locate a distributed-load resultant at its area centroid, not the beam center.
Worked problem: material selection as a failure mode
A component runs in a hot, mildly corrosive chemical line and the question lists four alloys by tensile strength. The trap is choosing the strongest. The governing requirements here are creep resistance at elevated temperature and corrosion resistance, so a stainless or nickel alloy with adequate (not maximum) strength outperforms a high-strength carbon steel that corrodes and softens. * Brittle fracture points to toughness and transition temperature; many cycles point to fatigue and endurance limit; sustained high temperature points to creep; aggressive chemistry points to corrosion resistance.
Strength alone answers a static-yielding question and almost nothing else.
Failure-mode trigger recap
- Sudden, low-strain fracture → brittle behavior; check toughness and transition temperature.
- Cyclic loading below yield → fatigue; compare alternating stress to the endurance limit.
- Sustained load at high temperature → creep; strength data alone will not answer it.
- Slender member in compression → buckling governs long before the material yields.
A solid shaft transmits 15 kW at 1500 rpm. Using T = P/omega, what is the transmitted torque?
A simply supported shaft carries a gear that transmits torque and also applies a transverse tooth force at midspan. Which design check best matches the loading?
A ductile bracket has a maximum normal stress of 120 MPa and a yield strength of 300 MPa. What static factor of safety against yielding is indicated?
A steel link failed after roughly 10^6 repeated cycles even though the peak nominal stress stayed below yield. Which review category identifies the missed concept?