Treat model selection as a separate skill from calculation. For each practice problem, state in one written line the load class, material class, and named criterion before touching numbers. Two worked scenarios show how a wrong model choice changes the result even when the arithmetic is perfect.
Static Failure Theories vs. Fatigue: Deciding Which Question You Are Answering
Classify the load first. Steady, single-application loads call for a static yield criterion such as von Mises or Tresca for ductile materials, or modified Mohr for brittle ones. Cyclic loads call for stress-life fatigue instead.
The two frameworks answer different questions. A static criterion asks whether one load application exceeds the material's yield strength or fracture strength. Fatigue analysis asks whether repeated stress cycles, each individually harmless, accumulate enough damage to initiate and grow a crack over thousands or millions of cycles. A part can comfortably pass a static check and still fail in fatigue, which is why the classification step cannot be skipped or guessed.
Apply the distinction by reading the problem statement for signal words. Phrases about reversed rotation, operating cycles, or service life point to stress-life fatigue: an S-N curve, an endurance strength corrected by modifiers, and a fatigue criterion on a mean-alternating diagram. Phrases about proof load, overload, or a single press-fit event point to a yield criterion. The material callout matters too: ductile behavior (significant elongation) suggests von Mises or Tresca; brittle behavior suggests modified Mohr.
Combining Bending and Torsion on a Shaft without Mixing Stress Types
On a shaft under combined bending and torsion, compute the bending stress and torsional shear stress separately, then combine them with an equivalent-stress formula. Adding the two numbers directly mixes a normal stress with a shear stress.
Scenario: a solid steel shaft of 50 mm diameter carries a steady bending moment of 400 N·m and a steady torque of 600 N·m. A plausible mistake is to check bending alone, compute 32M/(πd³) ≈ 32.6 MPa, compare it with the yield strength, and declare the shaft safe. The error is ignoring the torsional shear stress entirely, and the omission is easy to make because bending produces the more familiar formula and the more visible failure mode in textbooks.
The better decision is to compute both components and combine them with the distortion-energy relation. Bending gives σ ≈ 32.6 MPa; torsion gives τ = 16T/(πd³) ≈ 24.5 MPa. The von Mises equivalent is √(σ² + 3τ²) ≈ 53 MPa — roughly 64 percent higher than the bending-only value. This matters because the safe diameter, the design factor, and the pass/fail conclusion all shift. If the load were rotating rather than steady, the bending stress itself would alternate, and the problem would move into the fatigue framework of the next sections.
Choosing a Fatigue Criterion: Goodman, Soderberg, Gerber, or the Yield Line
Each criterion draws a different boundary on the mean-stress versus alternating-stress diagram. Match the criterion to what the problem asks: conservative yield protection, a straight-line fatigue boundary, or a curved fit to test data.
Once a fluctuating load is decomposed into a mean component and an alternating component, you must pick the boundary against which the load point is judged. Soderberg connects the endurance strength to the yield strength and is the most conservative of the three fatigue lines because it also guards against yield at the mean stress level. Modified Goodman connects endurance strength to ultimate strength and is the standard straight-line choice when the problem asks for a fatigue factor of safety without further qualification.
Gerber connects the same endpoints with a parabola that follows fatigue test data more closely and is therefore less conservative than Goodman. A fourth reference, the yield line joining the yield strength on both axes, acts as a separate check for first-cycle yielding. The practical habit is to name your choice explicitly in your solution and, when the problem leaves the choice open, state your assumption — the criteria differ enough that the numerical factor of safety changes with each one.
The table below summarizes the trade-offs.
| Criterion | Boundary on the diagram | Typical use | Watch out for |
|---|---|---|---|
| Soderberg | Straight line from endurance strength to yield strength | When the problem emphasizes guarding against yielding as well as fatigue | Most conservative; may force a larger section than needed |
| Modified Goodman | Straight line from endurance strength to ultimate strength | Standard straight-line fatigue check when none is specified | Does not by itself prevent first-cycle yielding; check the yield line too |
| Gerber | Parabola from endurance strength to ultimate strength | When the problem calls for the least conservative data-fitting curve | Easy to apply the wrong root or sign in the quadratic solution |
| Yield (Langer-type) line | Line joining yield strength on both axes | Checking that the first cycle does not yield | Not a fatigue criterion; it complements Goodman or Gerber |
Correcting Endurance Strength and Handling Fluctuating Loads
Start from the uncorrected endurance strength, apply named modifying factors to get a corrected value, then decompose the stress history into mean and alternating parts before reading a criterion off the diagram.
Endurance strengths quoted for rotating-beam specimens do not transfer directly to real parts. The standard correction chain applies named factors — surface condition, size, load type, temperature, and miscellaneous effects such as notch sensitivity — each multiplying the base value. A useful habit is to write the chain symbolically before plugging in numbers, so a factor you forgot to justify is visible rather than buried in arithmetic. For example, a rotating shaft in bending might use most factors near unity, while a transversely loaded or non-rotating member needs the load-type and size factors applied carefully.
Decomposing the load is a separate step. A stress that fluctuates between, say, 40 MPa and 120 MPa has an alternating component of (120 − 40)/2 = 40 MPa and a mean of (120 + 40)/2 = 80 MPa. For multi-level duty cycles, Miner's rule accumulates damage as the sum of nᵢ/Nᵢ terms across stress levels, and failure is assessed when the sum approaches one. Keep the two ideas separate in your notes: modifiers change the boundary on the diagram, while decomposition and damage accumulation position the point or points being judged.
Bolted Joints: Splitting the External Load between Bolt and Members
In a preloaded joint, an external load is shared according to the stiffness ratio of the bolt and clamped members. The bolt sees only the load fraction C·P, while the members lose the complementary fraction.
Scenario: a bolt preloaded to 25 kN clamps members whose combined stiffness is four times the bolt stiffness, and an external load of 15 kN is applied. A plausible mistake is to add the full external load to the bolt and report a bolt force of 40 kN. That overstates the bolt load, inflates the bolt size or the reported safety factor problem, and — worse for fatigue — overstates the alternating stress the bolt actually experiences.
The better decision uses the stiffness fraction: C = k_b/(k_b + k_m) = 1/(1 + 4) = 0.2. The bolt force becomes Fᵢ + C·P = 25 + 0.2(15) = 28 kN, and the clamped members drop to 25 − 0.8(15) = 13 kN. Separation is checked by comparing the external load with the preload: 15 kN < 25 kN, so the joint stays clamped. This matters most in fatigue work, where the alternating component on the bolt is only 3 kN, not 15 kN — a fivefold difference in the stress amplitude that drives the Goodman check.
Bearing Life Conversions and Energy Storage in Flywheels and Brakes
Bearing ratings follow a power-law life equation, so convert life between speeds and loads consistently, watching units of revolutions versus hours. Flywheel and brake problems center on energy bookkeeping, tracked through speed change and coefficient of fluctuation.
Rolling-element bearing life follows L in millions of revolutions equal to (C/P) raised to a constant exponent, with the exponent around 3 for ball bearings and 10/3 for roller bearings. The common trap is unit confusion: catalog ratings are tied to revolutions, while duty cycles are stated in hours at a given speed. Convert hours to revolutions using speed before applying the equation, and when scaling from one condition to another, use the power-law ratio directly rather than re-deriving from scratch. A small load change compounds quickly because of the cube — a 20 percent load increase cuts ball-bearing life to roughly 58 percent of its former value.
Flywheel and brake problems are energy problems. A flywheel sized for a coefficient of fluctuation Cₛ stores kinetic energy proportional to the square of angular speed, so the required inertia comes from the energy swing divided by the allowed speed-band difference of the squares. Brake problems ask how much kinetic energy at a given speed must be absorbed, then spread that energy over the braking event to estimate heat input. In both, keep energy, torque, and speed linked through work–energy relations rather than switching between them ad hoc.
A Model-Classification Drill, a Self-Check Rubric, and an Adaptable Study Sequence
Separate classification practice from calculation practice. Run a drill where you only classify problems and name the model, score yourself against a rubric, then follow a phased sequence that ends with mixed, timed sets using the electronic handbook.
Drill: collect ten mixed machine-design problems from any mechanics-of-materials or machine-design textbook. For each, spend two minutes writing three lines — load class (static, fully reversed, or fluctuating with mean), material class (ductile or brittle), and the named model you would apply with a one-sentence reason — then stop without solving. Repeat the set on three separate days. Expected observations: on the first pass you will hesitate between Goodman and Soderberg and occasionally miss that a 'steady torque plus rotating bending' shaft is a fatigue problem; by the third pass, hesitation should shrink to borderline cases.
Rubric: award one point each for correct load class, correct named model, and a stated reason consistent with the problem wording, giving 30 possible points across ten problems. A reasonable milestone is 8 of 10 on the second pass and 9 of 10 by the third — treat these as learning checkpoints only, not predictions of exam performance. Once classification is stable, follow an adaptable sequence: weeks one and two, map the failure-theory and fatigue frameworks and redo the drill; week three, fasteners, bolted joints, and welded joints; week four, shafts, bearings, lubrication; week five, gears, springs, clutches, brakes, flywheels; remaining time, mixed timed sets in which you practice searching the supplied electronic reference handbook, since that searchable PDF and the listed design standards are what you will actually use on exam day.
Readiness checks: you can state the load class of any textbook problem within about a minute; you can sketch the mean–alternating diagram from memory and place a load point on it; you can compute a bolt-load split and a separation check without notes; and you can locate each formula family in the handbook search function quickly. Administrative matters such as scheduling, current specifications, and policies are maintained by NCEES at its PE Mechanical page.
References and further reading
Use these references to explore the concepts and check the latest information from the relevant organizations.