Mechanical Design and Analysis · Study · FE Mechanical · FE → PE Prep
Mechanical Design and Analysis
10% of exam
Stress analysis of machine elements, static and fatigue failure theories, springs, pressure vessels, bearings, power screws, power transmission, and joining methods.
7 concepts
B. Failure theories and analysis
Static and Fatigue Failure Theories
When does a part actually break — static yielding by von Mises or Tresca, and fatigue life from the endurance limit, Marin factors, and the Goodman line.
Almost every Mechanical Design question on the FE comes down to one judgment: given a stress state, will the part survive? The handbook splits that judgment cleanly — static loading uses a yield/fracture theory (distortion-energy, maximum-shear-stress, or maximum-normal-stress), while fluctuating loading uses a fatigue criterion (modified Goodman or Soderberg) built on a fatigue-corrected endurance limit. Points are lost when candidates blur the two, plug a uniaxial yield strength straight against a biaxial stress, or forget that the rotating-beam endurance limit must be knocked down by the Marin factors before it means anything. Everything here lives in the FE Reference Handbook under Mechanical Engineering — 'Static Loading Failure Theories' and 'Variable Loading Failure Theories'.
From a stress state to a single number
Real parts see combined stress: an axle carries bending and torsion at once, so a point on its surface has both a normal stress σx
. A failure theory collapses that multiaxial state into one scalar — an effective stress — that you compare against the material's uniaxial strength. The first move is always to find the principal stresses. For plane stress they come from the same Mohr's-circle algebra you used in Mechanics of Materials, and the third principal stress is simply
Ductile metals yield by slip, which is driven by shear, so the maximum-shear-stress theory says yielding begins when the largest shear stress reaches the value it had when a tension specimen yielded. In a tension test τmax=Sy/2 at yield, which sets the threshold. The catch examinees miss: τmax must be the three-dimensional maximum, (σ1−σ3)/2 over the ordered principals including the zero one, not the in-plane radius of Mohr's circle. Tresca is the conservative choice and the easier hand calculation.
The distortion-energy theory says yielding starts when the distortion (shape-change) energy per unit volume equals that in a tension specimen at yield. It agrees best with test data and is the default for ductile parts. The left side of the criterion is the von Mises or effective stress σ′; for a biaxial state with nonzero principals σA and σB it reduces to the form below, and you can compute it directly from σx,σy,τxy without finding principals. Yielding is predicted when σ′≥Sy, so the factor of safety is n=Sy/σ′.
σ′=σA2−σAσB+σB2=σx2−σxσy+σy2+3τxy2
Brittle static failure: maximum-normal-stress
Brittle materials (gray cast iron, ceramics) fracture rather than yield, and they fail on the largest normal stress, not shear. The maximum-normal-stress theory predicts failure when σ1≥Sut in tension or σ3≤−Suc in compression, using separate tensile and compressive strengths because brittle materials are far stronger in compression. The Coulomb-Mohr theory refines this for the in-between quadrant. Use a brittle theory only when the problem names a brittle material — applying von Mises to cast iron is a classic trap.
σ1≥Sutorσ3≤−Suc
Stress concentration
Geometric discontinuities — holes, fillets, grooves — raise the local stress far above the nominal value. The geometric (theoretical) factor Kt multiplies the nominal stress: σmax=Ktσnom. For static loading of a ductile part, local yielding redistributes stress, so Kt is often ignored; for brittle parts and for any fatigue calculation it is essential. In fatigue you use the fatigue stress-concentration factor Kf, related to Kt through the notch sensitivity q by Kf=1+q(Kt−1), and apply it to the alternating stress.
σmax=Ktσnom,Kf=1+q(Kt−1)
Fatigue and the endurance limit
Most failures in service are fatigue: cracks nucleate and grow under stress cycles well below yield. Plotting stress amplitude against cycles to failure gives the S-N curve; for steels it flattens near 106 cycles into an endurance limit Se — a stress below which life is effectively infinite. The standard estimate from a polished rotating-beam test, Se′, must then be corrected to the real part with the Marin modifying factors for surface, size, load, temperature, and miscellaneous effects.
Se′={0.5Sut700MPaSut≤1400MPaSut>1400MPa
Marin factors
The corrected endurance limit is Se=kakbkckdkeSe′. The surface factor is ka=aSutb with table constants per finish (ground, machined/cold-drawn, hot-rolled, as-forged). The size factor for bending/torsion is kb=1.189d−0.097 for 8≤d≤250 mm and kb=1 in axial loading. The load factor is kc=1 bending, 0.923 axial (for Sut≤1520 MPa; kc=1 above that), 0.577 torsion; kd corrects for temperature (1 below 450∘C) and ke bundles everything else. Miss any factor and your Se — and your fatigue safety factor — are wrong.
A fluctuating load has a mean stress σm=(σmax+σmin)/2 and an alternating stress σa=(σmax−σmin)/2. Both matter: tensile mean stress shortens fatigue life. The modified Goodman line connects Se on the alternating axis to Sut on the mean axis; the more conservative Soderberg line uses Sy instead. A load point on the line means failure (n=1); inside is safe. Solve either for the fatigue factor of safety n.
Read the material first: ductile means von Mises or Tresca, brittle means maximum-normal-stress. For ductile static problems, von Mises (n=Sy/σ′) is the default and Tresca (n=Sy/(σ1−σ3)) is the conservative check — Tresca always gives the smaller (safer) n. If the load fluctuates, you are in fatigue: build Se from Se′=0.5Sut times every Marin factor the problem hands you, split the load into σm and σa, then apply Goodman (Sut) or Soderberg (Sy). Watch the units on ka=aSutb — the table constants differ for MPa versus kpsi. Carry the alternating-stress concentration factor Kf in fatigue even when static analysis would ignore it.
Key equations
Principal stresses (plane stress)σ1,2=2σx+σy±(2σx−σy)2+τxy2,σ3=0
Reduces a plane-stress state to principals; the third principal is zero. Inputs in MPa or psi.
Surface and size factorska=aSutb,kb=1.189deff−0.097(8≤d≤250mm)
Modified Goodman criterionSeσa+Sutσm=n1
Soderberg criterionSeσa+Syσm=n1
Worked examples
von Mises vs. Tresca on a combined-stress point
Problem. A ductile steel part (Sy=350 MPa) has a surface stress state σx=120 MPa, σy=−40 MPa, τxy=50 MPa. Find the factor of safety against yield by both the distortion-energy and maximum-shear-stress theories.
Solution. Principal stresses: center =(120−40)/2=40 MPa, radius R=((120+40)/2)2+502=802+502=94.3
σ′=1202−(120)(−40)+(−40)2+3(50)2=168MPa
Corrected endurance limit with Marin factors
Problem. Estimate the endurance limit of a machined steel shaft, Sut=700 MPa, diameter d=25 mm, loaded in rotating bending at room temperature.
Solution. Base: Sut=700≤1400
Fatigue safety factor by modified Goodman
Problem. The shaft above (Se=242 MPa, Sut=700 MPa) sees an alternating stress σa=80
Common pitfalls
•Using the in-plane Mohr radius for Tresca. τmax must be (σ1−σ3)/2 over the 3-D ordered principals, including σ3=0 in plane stress — otherwise a part with both principals positive looks safer than it is.
•Applying von Mises to a brittle material. Cast iron and ceramics fracture on normal stress; use maximum-normal-stress or Coulomb-Mohr with separate Sut and Suc.
•Forgetting the Marin factors. The polished Se′=0.5Sut is not the part's endurance limit — multiply by kakbkckdke
•Mixing the ka table units. The surface-factor constants a,b differ for MPa versus kpsi; pick the row that matches your Sut units.
•Putting mean stress on the wrong axis. Goodman uses Sut for σm and Se
•Confusing Goodman and Soderberg. Goodman uses Sut (fatigue only); Soderberg uses Sy (also guards yield) and is always more conservative.
•Dropping the stress-concentration factor in fatigue. Kf multiplies the alternating stress even when ductile static analysis would let local yielding wash Kt out.
References
NCEES FE Reference Handbook — Mechanical Engineering (Static and Variable Loading Failure Theories)
NCEES FE Reference Handbook — Mechanics of Materials (combined stress, Mohr's circle)
Shigley's Mechanical Engineering Design (Budynas & Nisbett) — fatigue and the Marin factors
D. Springs
Springs, Power Screws and Bearings
Helical spring rate and Wahl-corrected stress, surge and buckling limits, power-screw raising torque and efficiency, and rolling-bearing L10 life.
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H. Power transmission
Power Transmission: Gears, Belts, Chains, Clutches and Brakes
Spur-gear geometry and forces, gear-train ratios, belt and chain drives, clutch and brake torque/energy, all anchored by the single identity P = Tω.
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I. Joining methods
Bolted/Welded Joints and Thin-Wall Pressure Vessels
Bolt preload and the joint stiffness diagram, eccentric bolt- and weld-group shear, thread stripping, and thin-wall hoop and longitudinal stress.
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K. Quality and reliability
Quality and Reliability
Statistical process control with X-bar/R charts, process capability Cp and Cpk, the exponential reliability function and failure rate, and series/parallel system reliability with MTBF.
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L. Components (hydraulic, pneumatic, electromechanical)
Hydraulic, Pneumatic and Electromechanical Components
Sizing hydraulic cylinders, pumps, and valves through pressure-flow relations, pneumatic actuators, and electromechanical components — solenoids, DC motors, and relays — in a machine.
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M. Engineering drawing interpretations and geometric dimensioning and tolerancing (GD&T)
Manufacturability, Limits and Fits, and GD&T
Read a toleranced drawing: compute tolerance stack-up, classify ANSI clearance/interference fits, and decode GD&T feature control frames, datums, and material-condition modifiers.
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Ductile yield criterion. Use the 3-D ordered principals; n=Sy/(σ1−σ3). Sy = yield strength.
Effective stress for ductile yielding; n=Sy/σ′. Most accurate of the ductile theories.
Brittle fracture; uses separate tensile Sut and compressive Suc strengths.
Kt geometric factor; Kf fatigue factor with notch sensitivity q (0 to 1).
Mean and alternating stress for fluctuating loads; R=−1 is fully reversed.
Rotating-beam endurance limit from ultimate strength, before Marin correction.
ka surface, kb size, kc load, kd temperature, ke misc. Each is dimensionless.
a,b from finish table (MPa or kpsi). kb=1 for axial loading; kc=1 bend, 0.923 axial (Sut≤1520 MPa), 0.577 torsion.
Fatigue failure line for tensile mean stress; solve for fatigue factor of safety n.
More conservative than Goodman; guards against yield as well as fatigue.
MPa. The two in-plane principals are
40+94.3=134.3
MPa and
40−94.3=−54.3
MPa, and the out-of-plane principal is
0
.
Ordering all three principals (
134.3
,
0
,
−54.3
) from largest to smallest renames the most negative one
as expected; both are dimensionless ratios of stresses.
MPa, so
Se′=0.5(700)=350
MPa.
Surface (machined):
ka=4.51(700)−0.265=0.795
.
Size (bending,
8≤25≤250
mm):
kb=1.189(25)−0.097=0.870
.
Load (bending):
kc=1
. Temperature:
kd=1
. Misc:
ke=1
.
Se=(0.795)(0.870)(1)(1)(1)(350)=242
MPa.
Final:
Se=242
MPa. Sanity: the correction factors knock the polished
350
MPa down by about
31%
, a typical machined-part reduction; units stay in MPa.
Se=(0.795)(0.870)(1)(1)(1)(350MPa)=242MPa
MPa about a mean stress
σm=120
MPa. Find the fatigue factor of safety by the modified Goodman criterion.
Solution. Apply Goodman: n1=Seσa+Sutσm=24280+700120=0.3306+0.1714=0.5020.
n=1/0.5020=1.99.
Final: n=1.99. Sanity: n>1 so the load point sits inside the Goodman line (infinite life predicted); a Soderberg check would use Sy in place of Sut and give a smaller, more conservative n.