10.1 Static Failure Theories & Variable/Fatigue Loading

Key Takeaways

  • Maximum Shear Stress Theory (Tresca/Guest) provides a conservative hexagonal boundary for ductile materials, whereas Distortion Energy Theory (Von Mises/Hencky) offers an accurate elliptical envelope predicting shear yield strength as $S_{sy} = S_{yt}/\sqrt{3} \approx 0.577 S_{yt}$.
  • Maximum Principal Stress Theory (Rankine) is applicable solely to brittle materials failing along normal tension cleavage planes ($S_{ut}$ criteria), while failing completely for pure shear in ductile metals.
  • Theoretical stress concentration factor $K_t$ depends solely on geometry and loading, but fatigue notch factor $K_f = 1 + q(K_t - 1)$ incorporates material notch sensitivity $q$, where ductile materials exhibit lower sensitivity ($q \to 0$) than hardened brittle alloys ($q \to 1$).
  • Standard endurance limits ($S_e' \approx 0.5 S_{ut}$ for steels) are degraded on actual machine components by Marin factors ($k_a, k_b, k_c, k_d, k_e, k_f$) accounting for surface roughness, size, loading mode, temperature, reliability, and environmental effects.
  • Under fluctuating stresses, the Soderberg line yields the most conservative design envelope based on yield strength ($S_{yt}$), the Modified Goodman line is the standard design criterion for ductile fatigue based on ultimate strength ($S_{ut}$), and the Gerber parabola best fits empirical experimental fracture data.
Last updated: August 2026

Static Failure Theories & Variable/Fatigue Loading

In mechanical engineering design, machine components are rarely subjected to simple uniaxial static tension. Heavy machinery used in open-cast and underground coal mining—such as draglines, shearers, haul trucks, continuous miners, and conveyor driveheads—endure complex multi-axial states of stress combined with severe cyclic, reversing, and shock loading. To prevent catastrophic failure, a mechanical design engineer must apply appropriate static failure theories for non-cyclic multi-axial stresses and fatigue failure models for fluctuating service loads.


1. Multi-Axial Static Failure Theories

A static failure theory correlates a multi-axial stress state $(\sigma_1, \sigma_2, \sigma_3)$ in a component to mechanical property limits determined from a standard uniaxial tensile test: the yield strength $(S_{yt})$ for ductile materials and the ultimate tensile strength $(S_{ut})$ for brittle materials.

1.1 Maximum Principal Stress Theory (Rankine's Theory / MPST)

  • Statement: Failure occurs when the maximum principal tensile stress in a bi-axial or tri-axial stress system reaches the yield strength or ultimate tensile strength under simple tension. σ1SutNorσ3SucN\sigma_1 \ge \frac{S_{ut}}{N} \quad \text{or} \quad |\sigma_3| \ge \frac{S_{uc}}{N}
  • Failure Envelope: In the 2D principal stress plane $(\sigma_1, \sigma_2)$, the boundary is a square/rectangle defined by $\sigma_1 = \pm S_{ut}$ and $\sigma_2 = \pm S_{ut}$.
  • Application: Valid only for brittle materials (e.g., cast iron, high-carbon hardened steels) where failure occurs by tensile cleavage across planes of maximum normal stress.
  • Limitation: Totally unsuitable for ductile materials. Under pure shear $(\sigma_1 = \tau, \sigma_2 = -\tau)$, Rankine predicts shear yield strength $\tau_y = S_{yt}$, whereas ductile metals actually yield at $\tau_y \approx 0.50 S_{yt}$ to $0.577 S_{yt}$, creating an unsafe 73% to 100% overestimation of shear load capacity.

1.2 Maximum Shear Stress Theory (Tresca's / Guest's Theory / MSST)

  • Statement: Yielding begins when the maximum shear stress in a multi-axial stress state equals the maximum shear stress at the onset of yielding in a standard uniaxial tension test. τmax=max(σ1σ22,σ2σ32,σ3σ12)Syt2N\tau_{\max} = \max\left(\left|\frac{\sigma_1 - \sigma_2}{2}\right|, \left|\frac{\sigma_2 - \sigma_3}{2}\right|, \left|\frac{\sigma_3 - \sigma_1}{2}\right|\right) \le \frac{S_{yt}}{2N}
  • Pure Shear Prediction: For a state of pure shear $(\sigma_1 = \tau, \sigma_2 = -\tau, \sigma_3 = 0)$: τmax=τ(τ)2=τSyt2N    Ssy=0.5Syt\tau_{\max} = \frac{\tau - (-\tau)}{2} = \tau \le \frac{S_{yt}}{2N} \implies S_{sy} = 0.5 S_{yt}
  • Failure Envelope: A regular hexagon oriented at $45^{\circ}$ within the $\sigma_1-\sigma_2$ coordinate system.
  • Application & Characteristic: Standard conservative theory for ductile materials (e.g., mild steel, structural alloys). Because the hexagonal locus lies entirely within the Distortion Energy ellipse, Tresca always yields a conservative component dimension.

1.3 Distortion Energy Theory (Von Mises / Hencky Theory / DET)

  • Statement: Yielding occurs when the distortion energy (shear strain energy causing shape distortion without volume change) per unit volume under multi-axial stress reaches the distortion strain energy at yield in simple uniaxial tension. Total Strain Energy Density U=Uv(volumetric)+Ud(distortion)\text{Total Strain Energy Density } U = U_v (\text{volumetric}) + U_d (\text{distortion}) Ud=1+ν6E[(σ1σ2)2+(σ2σ3)2+(σ3σ1)2]U_d = \frac{1+\nu}{6E}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right]
  • Von Mises Equivalent Stress ($\sigma_v$ / $\sigma'$): σ=12[(σ1σ2)2+(σ2σ3)2+(σ3σ1)2]SytN\sigma' = \sqrt{\frac{1}{2}\left[(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2\right]} \le \frac{S_{yt}}{N}
  • For 2D plane stress states with $(\sigma_x, \sigma_y, \tau_{xy})$: σ=σx2σxσy+σy2+3τxy2SytN\sigma' = \sqrt{\sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3\tau_{xy}^2} \le \frac{S_{yt}}{N}
  • Pure Shear Prediction: For pure shear $(\sigma_x = 0, \sigma_y = 0, \tau_{xy} = \tau)$: σ=3τ2=3τSyt    Ssy=Syt30.577Syt\sigma' = \sqrt{3\tau^2} = \sqrt{3}\tau \le S_{yt} \implies S_{sy} = \frac{S_{yt}}{\sqrt{3}} \approx 0.577 S_{yt}
  • Failure Envelope: An ellipse inclined at $45^{\circ}$ with major axis along $\sigma_1 = \sigma_2$ and minor axis along $\sigma_1 = -\sigma_2$.
  • Application: Most accurate failure theory for ductile materials. Predicts $15.5%$ higher allowable shear strength than Tresca $(0.577 / 0.500 = 1.155)$, allowing optimum weight and material efficiency.

1.4 Maximum Principal Strain Theory (St. Venant's Theory / MPET)

  • Statement: Failure occurs when the maximum principal strain in a multi-axial stress state reaches the yield strain in uniaxial tension: $\epsilon_1 = \frac{\sigma_1}{E} - \nu\frac{\sigma_2 + \sigma_3}{E} \le \frac{S_{yt}}{E}$.
  • Failure Envelope: A rhomboid / rhombus.
  • Application: Historically used for thick-walled cylinders and gun barrels, but generally superseded by Von Mises and Tresca due to inconsistent experimental correlation.

1.5 Total Strain Energy Theory (Haigh's / Beltrami's Theory / TSET)

  • Statement: Yielding occurs when the total strain energy per unit volume absorbs an amount equal to that stored at yield in simple tension: Utotal=12E[σ12+σ22+σ322ν(σ1σ2+σ2σ3+σ3σ1)]Syt22EU_{\text{total}} = \frac{1}{2E}\left[\sigma_1^2 + \sigma_2^2 + \sigma_3^2 - 2\nu(\sigma_1\sigma_2 + \sigma_2\sigma_3 + \sigma_3\sigma_1)\right] \le \frac{S_{yt}^2}{2E}
  • Failure Envelope: An ellipse with semi-axes distinct from Von Mises.
  • Application: Suitable for ductile materials under hydrostatic compression/tension; over-predicts failure when volumetric dilation is large.
Failure TheoryAssociated NameFailure Criterion Formula (2D)2D Locus ShapeRecommended MaterialShear Yield Strength $S_{sy}$
Max. Principal StressRankine$\sigma_1 \le S_{ut}/N$Square / RectangleBrittle (Cast Iron, Ceramics)$1.00 S_{yt}$ (Unsafe)
Max. Shear StressTresca / Guest$\sigma_1 - \sigma_2 \le S_{yt}/N$HexagonDuctile (Conservative)$0.500 S_{yt}$
Distortion EnergyVon Mises / Hencky$\sqrt{\sigma_1^2 - \sigma_1\sigma_2 + \sigma_2^2} \le S_{yt}/N$EllipseDuctile (Most Accurate)$0.577 S_{yt}$
Max. Principal StrainSt. Venant$\sigma_1 - \nu \sigma_2 \le S_{yt}/N$RhomboidDuctile (Obsolete)$\frac{S_{yt}}{1+\nu} \approx 0.77 S_{yt}$
Total Strain EnergyHaigh / Beltrami$\sigma_1^2 + \sigma_2^2 - 2\nu \sigma_1\sigma_2 \le S_{yt}^2/N^2$EllipseElastic analysis$\frac{S_{yt}}{\sqrt{2(1+\nu)}} \approx 0.62 S_{yt}$

2. Stress Concentration & Notch Sensitivity

Geometric discontinuities (fillets, keyways, oil holes, shoulders, notches) disturb uniform stress trajectories, generating localized peak stresses far exceeding nominal cross-sectional values.

2.1 Theoretical Stress Concentration Factor $(K_t)$

Kt=σmaxσ0orKts=τmaxτ0K_t = \frac{\sigma_{\max}}{\sigma_0} \quad \text{or} \quad K_{ts} = \frac{\tau_{\max}}{\tau_0} where $\sigma_{\max}$ is the true localized peak stress and $\sigma_0$ is the nominal stress calculated from elementary mechanics $(P/A, M/Z, T/Z_p)$ on the net cross-section. $K_t$ depends purely on component geometry and loading mode, independent of the material.

  • Behavior under Static Loading:
    • In ductile materials, static stress concentration is generally neglected because localized plastic yielding at the notch root causes stress redistribution across the net section without macroscopic fracture.
    • In brittle materials, localized yielding cannot occur; stress concentration must be applied directly in static design: $\sigma_{\text{design}} = K_t \sigma_0$.

2.2 Fatigue Notch Factor $(K_f)$ and Notch Sensitivity $(q)$

Under cyclic fatigue loading, even ductile materials fail by progressive brittle crack growth initiated at notches. However, the actual fatigue degradation is frequently less severe than $K_t$. Kf=Endurance limit of unnotched specimen (Se)Endurance limit of notched specimen (Sef)K_f = \frac{\text{Endurance limit of unnotched specimen } (S_e)}{\text{Endurance limit of notched specimen } (S_{ef})} Notch Sensitivity Index q=Kf1Kt1    Kf=1+q(Kt1)\text{Notch Sensitivity Index } q = \frac{K_f - 1}{K_t - 1} \implies K_f = 1 + q(K_t - 1)

  • $q = 0 \implies K_f = 1$ (Material is completely insensitive to notches; typical for coarse cast iron).
  • $q = 1 \implies K_f = K_t$ (Full theoretical notch sensitivity; typical for ultra-high-strength brittle alloy steels).
  • Neuber's Equation: $q = \frac{1}{1 + \sqrt{a/r}}$, where $r$ is the notch root radius and $a$ is Neuber's material constant.

3. Variable Loading, Fatigue Mechanics & S-N Curves

Fatigue failure accounts for over $80%$ of all mechanical component breakdowns in mining haulage, gearboxes, and shafting. It occurs under dynamic repeating loads at stresses substantially below static yield strength.

3.1 Fatigue Mechanics & The Wöhler S-N Curve

Fatigue fracture evolves through three distinct physical phases: (1) Crack initiation at microscopic surface imperfections or slip bands; (2) Stage II striation propagation perpendicular to maximum tensile stress; (3) Catastrophic sudden fracture across the remaining net ligament.

  • Low-Cycle Fatigue (LCF): Occurs at $N < 10^3$ cycles, characterized by significant macroscopic plastic deformation per cycle.
  • High-Cycle Fatigue (HCF): Occurs at $N > 10^3$ to $10^6$ cycles, governed by elastic strain ranges.
  • Endurance Limit ($S_e'$): For ferrous metals (steels, cast irons) and titanium alloys, the S-N curve displays a horizontal knee between $10^6$ and $10^7$ cycles. Below this stress amplitude, the specimen can withstand infinite cycles $(N = \infty)$ without failure.
    • For standard steels: $S_e' \approx 0.50 S_{ut}$ (for $S_{ut} \le 1400\text{ MPa}$); $S_e' = 700\text{ MPa}$ (for $S_{ut} > 1400\text{ MPa}$).
    • For cast iron: $S_e' \approx 0.40 S_{ut}$.
    • Non-ferrous alloys (aluminum, brass, copper) do not exhibit a distinct endurance limit; their fatigue strength is reported at a specified life (typically $S_{f}$ at $N = 5 \times 10^8$ cycles).

3.2 Marin Factors for Modified Endurance Limit

The laboratory endurance limit $S_e'$ is measured on a pristine, polished, unnotched R.R. Moore rotating-beam specimen ($d = 7.62\text{ mm}$) at room temperature. The actual component endurance limit $S_e$ is derived using Marin modification factors: Se=kakbkckdkekfSeS_e = k_a \cdot k_b \cdot k_c \cdot k_d \cdot k_e \cdot k_f \cdot S_e'

  1. Surface Finish Factor $(k_a)$: Micro-roughness acts as crack initiation sites. ka=aSutbk_a = a \cdot S_{ut}^b
    • Ground: $k_a \approx 0.90 - 0.95$
    • Machined / Cold-drawn: $k_a \approx 0.70 - 0.85$
    • Hot-rolled: $k_a \approx 0.50 - 0.70$
    • As-forged: $k_a \approx 0.30 - 0.50$
  2. Size Factor $(k_b)$: Larger components have a higher statistical probability of flaws within the outer high-stress layer.
    • For bending/torsion: $k_b = 1.0$ for $d \le 7.62\text{ mm}$; $k_b = 1.24 d^{-0.107}$ for $7.62 < d \le 51\text{ mm}$; $k_b = 1.51 d^{-0.157}$ for $51 < d \le 250\text{ mm}$.
    • For axial loading: $k_b = 1.0$ (no gradient effect).
  3. Load Factor $(k_c)$: Accounts for stress state difference relative to standard rotating beam bending.
    • Bending: $k_c = 1.0$
    • Axial loading: $k_c = 0.85$
    • Torsion / Pure shear: $k_c = 0.59$ (consistent with Von Mises $1/\sqrt{3}$)
  4. Temperature Factor $(k_d)$: Accounts for creep and reduction in tensile strength at elevated operating temperatures ($k_d = 1.0$ for $T \le 350^{\circ}\text{C}$).
  5. Reliability Factor $(k_e)$: Standard S-N curves represent $50%$ survival probability ($z_a = 0, k_e = 1.0$). For $90%$ reliability: $k_e = 0.897$; for $99%$ reliability: $k_e = 0.814$; for $99.9%$ reliability: $k_e = 0.753$.
  6. Miscellaneous Effects Factor $(k_f)$: Includes corrosion pitting ($k_f \approx 0.3 - 0.5$), fretting fatigue, surface coatings (galvanizing/electroplating), and residual stresses (shot peening increases $k_f > 1.0$).

4. Fluctuating Stress Criteria & Fatigue Design Envelopes

When a machine member experiences cyclic loads where stress oscillates between a maximum value $\sigma_{\max}$ and a minimum value $\sigma_{\min}$:

  • Mean Stress (Static component): $\sigma_m = \frac{\sigma_{\max} + \sigma_{\min}}{2}$
  • Alternating Stress Amplitude (Dynamic component): $\sigma_a = \frac{\sigma_{\max} - \sigma_{\min}}{2}$
  • Stress Ratio: $R = \frac{\sigma_{\min}}{\sigma_{\max}}$
    • Completely reversed: $R = -1, \sigma_m = 0$
    • Repeated loading (zero-to-peak): $R = 0, \sigma_m = \sigma_a$
    • Fluctuating tensile: $0 < R < 1$
  • Amplitude Ratio: $A = \frac{\sigma_a}{\sigma_m} = \frac{1-R}{1+R}$

4.1 Classical Design Equations on the $\sigma_m - \sigma_a$ Diagram

     Alternating Stress (sigma_a)
     ^
 Se  + . . . . . . . . . . . . . . . .
     |\   Gerber Parabola             .
     | \.  (Passes through Se & Sut)   .
     |  \ '.                           .
     |   \   '.   Modified Goodman     .
     |    \    '. (Passes through Se & Sut)
     |     \     '.                    .
     |      \      '.  Soderberg Line  .
     |       \       '.(Passes through .
     |        \        '. Se & Syt)    .
     |         \         '.            .
     +----------+----------+-----------+---> Mean Stress (sigma_m)
     0         Syt        Sut
  1. Soderberg Line (Yield-Based / Most Conservative): Connects endurance limit $S_e$ on the alternating axis to yield strength $S_{yt}$ on the mean axis. It eliminates any possibility of yielding during startup cycles. σmSyt+σaSe=1N\frac{\sigma_m}{S_{yt}} + \frac{\sigma_a}{S_e} = \frac{1}{N}
  2. Modified Goodman Line (Ultimate-Based / Standard Industrial Design): Connects endurance limit $S_e$ to ultimate tensile strength $S_{ut}$. Standard design criterion for ductile metals in mechanical engineering. σmSut+σaSe=1N\frac{\sigma_m}{S_{ut}} + \frac{\sigma_a}{S_e} = \frac{1}{N}
  3. Gerber Parabola (Mean Experimental Fit): Represents parabolic regression curve passing through mean test fracture data. (NσmSut)2+NσaSe=1    σaSe=1N[1(NσmSut)2]\left(\frac{N \sigma_m}{S_{ut}}\right)^2 + \frac{N \sigma_a}{S_e} = 1 \implies \frac{\sigma_a}{S_e} = \frac{1}{N}\left[1 - \left(\frac{N \sigma_m}{S_{ut}}\right)^2\right]
  4. ASME Elliptic Line: Combines yield criteria with endurance limits in an elliptic envelope: (NσmSyt)2+(NσaSe)2=1\left(\frac{N \sigma_m}{S_{yt}}\right)^2 + \left(\frac{N \sigma_a}{S_e}\right)^2 = 1
  5. Langer Static Yield Line (Yield Check for Goodman/Gerber): Because the Goodman line extends to $S_{ut}$, the peak stress $\sigma_{\max} = \sigma_m + \sigma_a$ can exceed yield strength $S_{yt}$, producing static plastic deformation on the very first cycle. Therefore, the design must be capped by the Langer line: σm+σaSyt=1N    σmaxSytN\frac{\sigma_m + \sigma_a}{S_{yt}} = \frac{1}{N} \implies \sigma_{\max} \le \frac{S_{yt}}{N}

4.2 Comprehensive Comparison Table

Fatigue CriterionEquation (Factor of Safety $N$)Mean Stress AnchorFailure Mode PredictedRelative Conservativeness
Soderberg$\frac{\sigma_m}{S_{yt}} + \frac{\sigma_a}{S_e} = \frac{1}{N}$$S_{yt}$ (Yield)Prevents gross yielding & fatigueMost conservative (Highest Safety)
Goodman$\frac{\sigma_m}{S_{ut}} + \frac{\sigma_a}{S_e} = \frac{1}{N}$$S_{ut}$ (Ultimate)High-cycle fatigue fractureIndustry standard for ductile alloys
Gerber$\left(\frac{N\sigma_m}{S_{ut}}\right)^2 + \frac{N\sigma_a}{S_e} = 1$$S_{ut}$ (Ultimate)Empirical average fatigue failureLeast conservative (Best data fit)
ASME Elliptic$\left(\frac{N\sigma_m}{S_{yt}}\right)^2 + \left(\frac{N\sigma_a}{S_e}\right)^2 = 1$$S_{yt}$ (Yield)Combined yield and fatigueBalanced smooth transition
Langer Yield$\frac{\sigma_m + \sigma_a}{S_{yt}} = \frac{1}{N}$$S_{yt}$ (Yield)Static yielding on initial cycleUpper boundary check for Goodman/Gerber

5. Step-by-Step Worked Numerical Example

Problem: A transmission tie-rod forged from structural alloy steel has $S_{ut} = 600\text{ MPa}$, $S_{yt} = 400\text{ MPa}$, and a corrected endurance limit $S_e = 200\text{ MPa}$. It is subjected to a cyclic tensile force oscillating between $P_{\min} = 20\text{ kN}$ and $P_{\max} = 100\text{ kN}$. Determine the required rod diameter $d$ for infinite life using (a) Soderberg criterion and (b) Modified Goodman criterion with a factor of safety $N = 2.0$.

Solution:

  1. Mean and Alternating Loads: Pm=100+202=60 kN=60,000 NP_m = \frac{100 + 20}{2} = 60\text{ kN} = 60,000\text{ N} Pa=100202=40 kN=40,000 NP_a = \frac{100 - 20}{2} = 40\text{ kN} = 40,000\text{ N} σm=PmA=60000π4d2=76394d2 MPa,σa=PaA=40000π4d2=50930d2 MPa\sigma_m = \frac{P_m}{A} = \frac{60000}{\frac{\pi}{4} d^2} = \frac{76394}{d^2}\text{ MPa}, \quad \sigma_a = \frac{P_a}{A} = \frac{40000}{\frac{\pi}{4} d^2} = \frac{50930}{d^2}\text{ MPa}
  2. (a) Soderberg Design: σmSyt+σaSe=1N    76394400d2+50930200d2=12.0\frac{\sigma_m}{S_{yt}} + \frac{\sigma_a}{S_e} = \frac{1}{N} \implies \frac{76394}{400 d^2} + \frac{50930}{200 d^2} = \frac{1}{2.0} 190.99+254.65d2=0.50    445.64d2=0.50    d2=891.28    d=29.85 mm\frac{190.99 + 254.65}{d^2} = 0.50 \implies \frac{445.64}{d^2} = 0.50 \implies d^2 = 891.28 \implies d = 29.85\text{ mm}
  3. (b) Modified Goodman Design: σmSut+σaSe=1N    76394600d2+50930200d2=12.0\frac{\sigma_m}{S_{ut}} + \frac{\sigma_a}{S_e} = \frac{1}{N} \implies \frac{76394}{600 d^2} + \frac{50930}{200 d^2} = \frac{1}{2.0} 127.32+254.65d2=0.50    381.97d2=0.50    d2=763.94    d=27.64 mm\frac{127.32 + 254.65}{d^2} = 0.50 \implies \frac{381.97}{d^2} = 0.50 \implies d^2 = 763.94 \implies d = 27.64\text{ mm}
  4. Static Yield (Langer) Verification for Goodman: σmax=σm+σa=76394+50930(27.64)2=127324763.97=166.66 MPa\sigma_{\max} = \sigma_m + \sigma_a = \frac{76394 + 50930}{(27.64)^2} = \frac{127324}{763.97} = 166.66\text{ MPa} Allowable Yield Stress =SytN=4002=200 MPa>166.66 MPa(Safe against initial yield)\text{Allowable Yield Stress } = \frac{S_{yt}}{N} = \frac{400}{2} = 200\text{ MPa} > 166.66\text{ MPa} \quad (\text{Safe against initial yield})
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Fatigue Design Failure Envelopes on Mean-Alternating Stress Coordinates
Test Your Knowledge

For a machine component subjected to pure shear stress tau under static loading, what is the ratio of allowable shear stress predicted by the Distortion Energy Theory (Von Mises) to that predicted by the Maximum Shear Stress Theory (Tresca)?

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Test Your Knowledge

A notched steel shaft with a theoretical geometric stress concentration factor Kt = 2.6 is manufactured from an alloy steel having a notch sensitivity index q = 0.75. What is the effective fatigue stress concentration factor Kf to be applied in cyclic fatigue analysis?

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Test Your Knowledge

A mechanical link is subjected to a fluctuating axial stress varying from 0 to 180 MPa. The material has yield strength Syt = 360 MPa, ultimate tensile strength Sut = 540 MPa, and corrected endurance limit Se = 180 MPa. Using the Soderberg failure criterion, what is the design factor of safety N?

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