Failure theories, springs, fasteners, gears, bearings and welds: where everything before this chapter comes together.
Choose any section below to get a free practice problem. Work it with the FE Reference Handbook open, tap your answer, then check the solution.
For ductile metals under combined stress, use the distortion-energy (von Mises) theory: σ′ = √(σ2 + 3τ2) for a shaft with bending and torsion, and failure when σ′ reaches Sy.
A ductile steel shaft (Sy = 250 MPa) sees σ = 80 MPa bending and τ = 40 MPa torsion at one point. What is the factor of safety by the maximum-shear-stress theory?
τmax = √(402 + 402) = 56.6 MPa. n = Sy/(2τmax) = 250/113.1 = 2.21.
Common trap: (D) is the von Mises result, which is always a little less conservative.
Mechanical Designvon Misesmaximum shear stressstress concentrationThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
A helical compression spring’s rate is k = Gd4/(8D3N), with wire diameter d, mean coil diameter D and N active coils.
Springs of 100 N/mm and 300 N/mm are mounted side by side (in parallel). What is the combined rate?
Parallel springs add: 400 N/mm.
Common trap: (B) is the series combination.
Springshelical springbolt preloadtorqueThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Gear speed ratios follow the tooth counts: Nout = Nin × (driver teeth/driven teeth) for each mesh.
A shaft transmits 5 kW at 200 rpm. What is its torque?
ω = 200(2π/60) = 20.94 rad/s. T = 5000/20.94 = 239 N·m.
Common trap: (B) divides by rpm without converting.
Gearsgear trainpitch diameterpower transmissionThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Rolling bearing life scales steeply with load: L10 = (C/P)a million revolutions, with a = 3 for ball bearings and 10/3 for roller bearings.
A roller bearing runs at half its rated load (C/P = 2). What is its L10 life?
(2)10/3 = 10.1 million rev.
Common trap: (C) uses the ball-bearing exponent, 3.
bearing lifeL10fitstolerancesThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
A fillet weld’s strength comes from its throat, 0.707 times the leg size h.
A bracket uses two 8 mm fillet welds, each 100 mm long, with an allowable shear stress of 140 MPa. What is the joint capacity?
Each weld: 0.707(8)(100)(140) = 79.2 kN. Two welds: 158 kN.
Common trap: (D) skips the 0.707 throat factor.
weldsfillet weldthroatdouble shearThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
A power screw’s lead L is its pitch times its number of starts.
A single-start square-thread screw has a 25 mm mean diameter, a 5 mm pitch and μ = 0.15; collar friction is negligible. What torque is needed to lower an 8 kN load?
πμdm = π(0.15)(25) = 11.78 mm and πdm = 78.54 mm. TL = (8,000 × 0.025/2)(11.78 − 5)/(78.54 + 0.75) = 100(0.0855) = 8.55 N·m.
Common trap: (A) uses the raising formula; lowering flips the signs of the L terms.
Power ScrewsPump Power EquationHydraulic CylinderThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
A worst-case tolerance stack walks the chain of dimensions: the gap’s nominal comes from adding and subtracting the nominal sizes, and its ± band is the sum of every individual ± band.
A pin is dimensioned ⌀12.00/11.90 mm with a perpendicularity tolerance of ⌀0.04 mm at MMC. The pin is produced at ⌀11.94 mm. What perpendicularity tolerance is allowed?
For a pin, MMC is the largest size, ⌀12.00 mm. Bonus = 12.00 − 11.94 = 0.06 mm, so the allowed zone is 0.04 + 0.06 = ⌀0.10 mm.
Common trap: (D) measures the bonus from LMC, ⌀11.90 mm; for a pin, MMC is the largest size.
Geometric Dimensioning and TolerancingLimits and FitsToleranceThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Reliability R is the probability that a part works for a stated time.
A conveyor drive has a motor (R = 0.98) and a gearbox (R = 0.96) in series, followed by two redundant speed sensors in parallel (R = 0.90 each). What is the system reliability?
Sensor pair: 1 − (1 − 0.90)2 = 0.99. System: 0.98 × 0.96 × 0.99 = 0.931.
Common trap: (A) multiplies all four reliabilities, treating the redundant sensors as series.
ReliabilityProcess CapabilityStatistical Quality ControlThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
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