Mechanical Design and Analysis

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

\sigma_x

and a shear stress

\tau_{xy}

. 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

\sigma_3 = 0

.

\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}

Ductile static failure: maximum-shear-stress (Tresca)

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

\tau_{max} = S_y/2

at yield, which sets the threshold. The catch examinees miss:

\tau_{max}

must be the three-dimensional maximum,

(\sigma_1-\sigma_3)/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.

\tau_{max} = \frac{\sigma_1 - \sigma_3}{2} \ge \frac{S_y}{2} \quad\Rightarrow\quad n = \frac{S_y}{\sigma_1 - \sigma_3}

Ductile static failure: distortion-energy (von Mises)

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

\sigma'

; for a biaxial state with nonzero principals

\sigma_A

and

\sigma_B

it reduces to the form below, and you can compute it directly from

\sigma_x,\sigma_y,\tau_{xy}

without finding principals. Yielding is predicted when

\sigma' \ge S_y

, so the factor of safety is

n = S_y/\sigma'

.

\sigma' = \sqrt{\sigma_A^2 - \sigma_A\sigma_B + \sigma_B^2} = \sqrt{\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2}

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

\sigma_1 \ge S_{ut}

in tension or

\sigma_3 \le -S_{uc}

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.

\sigma_1 \ge S_{ut} \quad\text{or}\quad \sigma_3 \le -S_{uc}

Stress concentration

Geometric discontinuities — holes, fillets, grooves — raise the local stress far above the nominal value. The geometric (theoretical) factor

K_t

multiplies the nominal stress:

\sigma_{max} = K_t\,\sigma_{nom}

. For static loading of a ductile part, local yielding redistributes stress, so

K_t

is often ignored; for brittle parts and for any fatigue calculation it is essential. In fatigue you use the fatigue stress-concentration factor

K_f

, related to

K_t

through the notch sensitivity

q

by

K_f = 1 + q(K_t - 1)

, and apply it to the alternating stress.

\sigma_{max} = K_t\,\sigma_{nom}, \qquad K_f = 1 + q\,(K_t - 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

10^6

cycles into an endurance limit

S_e

— a stress below which life is effectively infinite. The standard estimate from a polished rotating-beam test,

S_e'

, must then be corrected to the real part with the Marin modifying factors for surface, size, load, temperature, and miscellaneous effects.

S_e' = \begin{cases} 0.5\,S_{ut} & S_{ut} \le 1400\ \text{MPa} \\ 700\ \text{MPa} & S_{ut} > 1400\ \text{MPa} \end{cases}

Marin factors

The corrected endurance limit is

S_e = k_a k_b k_c k_d k_e\,S_e'

. The surface factor is

k_a = a\,S_{ut}^{\,b}

with table constants per finish (ground, machined/cold-drawn, hot-rolled, as-forged). The size factor for bending/torsion is

k_b = 1.189\,d^{-0.097}

for

8 \le d \le 250

mm and

k_b = 1

in axial loading. The load factor is

k_c = 1

bending,

0.923

axial (for

S_{ut} \le 1520

MPa;

k_c = 1

above that),

0.577

torsion;

k_d

corrects for temperature (

1

below

450^\circ

C) and

k_e

bundles everything else. Miss any factor and your

S_e

— and your fatigue safety factor — are wrong.

S_e = k_a\,k_b\,k_c\,k_d\,k_e\,S_e', \qquad k_a = a\,S_{ut}^{\,b}, \qquad k_b = 1.189\,d^{-0.097}

Fluctuating stress: Goodman and Soderberg

A fluctuating load has a mean stress

\sigma_m = (\sigma_{max}+\sigma_{min})/2

and an alternating stress

\sigma_a = (\sigma_{max}-\sigma_{min})/2

. Both matter: tensile mean stress shortens fatigue life. The modified Goodman line connects

S_e

on the alternating axis to

S_{ut}

on the mean axis; the more conservative Soderberg line uses

S_y

instead. A load point on the line means failure (

n=1

); inside is safe. Solve either for the fatigue factor of safety

n

.

\underbrace{\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n}}_{\text{modified Goodman}} \qquad \underbrace{\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{n}}_{\text{Soderberg}}

Exam strategy

Read the material first: ductile means von Mises or Tresca, brittle means maximum-normal-stress. For ductile static problems, von Mises (

n = S_y/\sigma'

) is the default and Tresca (

n = S_y/(\sigma_1-\sigma_3)

) is the conservative check — Tresca always gives the smaller (safer)

n

. If the load fluctuates, you are in fatigue: build

S_e

from

S_e' = 0.5 S_{ut}

times every Marin factor the problem hands you, split the load into

\sigma_m

and

\sigma_a

, then apply Goodman (

S_{ut}

) or Soderberg (

S_y

). Watch the units on

k_a = a S_{ut}^{\,b}

— the table constants differ for MPa versus kpsi. Carry the alternating-stress concentration factor

K_f

in fatigue even when static analysis would ignore it.

Key equations

Principal stresses (plane stress)

\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2 + \tau_{xy}^2}, \quad \sigma_3 = 0

Reduces a plane-stress state to principals; the third principal is zero. Inputs in MPa or psi.

Maximum-shear-stress (Tresca)

\tau_{max} = \frac{\sigma_1-\sigma_3}{2} \ge \frac{S_y}{2}

Ductile yield criterion. Use the 3-D ordered principals; $n = S_y/(\sigma_1-\sigma_3)$ . $S_y$ = yield strength.

Distortion-energy (von Mises)

\sigma' = \sqrt{\sigma_x^2 - \sigma_x\sigma_y + \sigma_y^2 + 3\tau_{xy}^2} \ge S_y

Effective stress for ductile yielding; $n = S_y/\sigma'$ . Most accurate of the ductile theories.

Maximum-normal-stress (brittle)

\sigma_1 \ge S_{ut} \quad\text{or}\quad \sigma_3 \le -S_{uc}

Brittle fracture; uses separate tensile $S_{ut}$ and compressive $S_{uc}$ strengths.

Stress concentration

\sigma_{max} = K_t\,\sigma_{nom}, \qquad K_f = 1 + q(K_t-1)

$K_t$ geometric factor; $K_f$ fatigue factor with notch sensitivity $q$ ( $0$ to $1$ ).

Stress ratio and amplitudes

\sigma_m = \frac{\sigma_{max}+\sigma_{min}}{2}, \quad \sigma_a = \frac{\sigma_{max}-\sigma_{min}}{2}, \quad R = \frac{\sigma_{min}}{\sigma_{max}}

Mean and alternating stress for fluctuating loads; $R=-1$ is fully reversed.

Endurance limit estimate (steel)

S_e' = 0.5\,S_{ut}\ (S_{ut}\le 1400\ \text{MPa}); \quad S_e' = 700\ \text{MPa}\ \text{otherwise}

Rotating-beam endurance limit from ultimate strength, before Marin correction.

Marin corrected endurance limit

S_e = k_a k_b k_c k_d k_e\,S_e'

$k_a$ surface, $k_b$ size, $k_c$ load, $k_d$ temperature, $k_e$ misc. Each is dimensionless.

Surface and size factors

k_a = a\,S_{ut}^{\,b}, \qquad k_b = 1.189\,d_{eff}^{-0.097}\ (8\le d\le 250\ \text{mm})

$a,b$ from finish table (MPa or kpsi). $k_b=1$ for axial loading; $k_c=1$ bend, $0.923$ axial ( $S_{ut}\le1520$ MPa), $0.577$ torsion.

Modified Goodman criterion

\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_{ut}} = \frac{1}{n}

Fatigue failure line for tensile mean stress; solve for fatigue factor of safety $n$ .

Soderberg criterion

\frac{\sigma_a}{S_e} + \frac{\sigma_m}{S_y} = \frac{1}{n}

More conservative than Goodman; guards against yield as well as fatigue.

Worked examples

von Mises vs. Tresca on a combined-stress point

Problem. A ductile steel part ( $S_y = 350$ MPa) has a surface stress state $\sigma_x = 120$ MPa, $\sigma_y = -40$ MPa, $\tau_{xy} = 50$ 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=\sqrt{((120+40)/2)^2 + 50^2}=\sqrt{80^2+50^2}=94.3

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

\sigma_3

:

\sigma_1 = 134.3

,

\sigma_2 = 0

,

\sigma_3 = -54.3

MPa. von Mises:

\sigma' = \sqrt{120^2 - (120)(-40) + (-40)^2 + 3(50)^2} = \sqrt{14400+4800+1600+7500} = \sqrt{28300} = 168.2

MPa, so

n_{vM} = 350/168.2 = 2.08

. Tresca:

\tau_{max} = (134.3-(-54.3))/2 = 94.3

MPa, so

n_{T} = S_y/(\sigma_1-\sigma_3) = 350/188.7 = 1.85

. Final:

n_{vM} = 2.08

,

n_{T} = 1.85

. Sanity: Tresca is the conservative theory, and

1.85 < 2.08

as expected; both are dimensionless ratios of stresses.

\sigma' = \sqrt{120^2-(120)(-40)+(-40)^2+3(50)^2} = 168\ \text{MPa}

Corrected endurance limit with Marin factors

Problem. Estimate the endurance limit of a machined steel shaft, $S_{ut} = 700$ MPa, diameter $d = 25$ mm, loaded in rotating bending at room temperature.

Solution. Base:

S_{ut} = 700 \le 1400

MPa, so

S_e' = 0.5(700) = 350

MPa. Surface (machined):

k_a = 4.51\,(700)^{-0.265} = 0.795

. Size (bending,

8\le 25\le 250

mm):

k_b = 1.189(25)^{-0.097} = 0.870

. Load (bending):

k_c = 1

. Temperature:

k_d = 1

. Misc:

k_e = 1

.

S_e = (0.795)(0.870)(1)(1)(1)(350) = 242

MPa. Final:

S_e = 242

MPa. Sanity: the correction factors knock the polished

350

MPa down by about

31\%

, a typical machined-part reduction; units stay in MPa.

S_e = (0.795)(0.870)(1)(1)(1)(350\ \text{MPa}) = 242\ \text{MPa}

Fatigue safety factor by modified Goodman

Problem. The shaft above ( $S_e = 242$ MPa, $S_{ut} = 700$ MPa) sees an alternating stress $\sigma_a = 80$ MPa about a mean stress $\sigma_m = 120$ MPa. Find the fatigue factor of safety by the modified Goodman criterion.

Solution. Apply Goodman:

\dfrac{1}{n} = \dfrac{\sigma_a}{S_e} + \dfrac{\sigma_m}{S_{ut}} = \dfrac{80}{242} + \dfrac{120}{700} = 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

S_y

in place of

S_{ut}

and give a smaller, more conservative

n

.

\frac{1}{n} = \frac{80}{242} + \frac{120}{700} = 0.502 \;\Rightarrow\; n = 1.99

Common pitfalls

•Using the in-plane Mohr radius for Tresca. $\tau_{max}$ must be $(\sigma_1-\sigma_3)/2$ over the 3-D ordered principals, including $\sigma_3=0$ 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 $S_{ut}$ and $S_{uc}$ .
•Forgetting the Marin factors. The polished $S_e' = 0.5 S_{ut}$ is not the part's endurance limit — multiply by $k_a k_b k_c k_d k_e$ first, or you overestimate fatigue life badly.
•Mixing the $k_a$ table units. The surface-factor constants $a,b$ differ for MPa versus kpsi; pick the row that matches your $S_{ut}$ units.
•Putting mean stress on the wrong axis. Goodman uses $S_{ut}$ for $\sigma_m$ and $S_e$ for $\sigma_a$ ; swapping them inverts the criterion.
•Confusing Goodman and Soderberg. Goodman uses $S_{ut}$ (fatigue only); Soderberg uses $S_y$ (also guards yield) and is always more conservative.
•Dropping the stress-concentration factor in fatigue. $K_f$ multiplies the alternating stress even when ductile static analysis would let local yielding wash $K_t$ 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.