4.4 Multiaxial Yield Criteria & Combined-Load Analysis of Welded Members

5. Multiaxial Failure Theories for Structural Alloys

When structural weldments endure multiaxial stresses, yield criteria predict the onset of plastic deformation by relating the complex stress tensor to the material's uniaxial tensile yield strength $\sigma_y$.

                 σ_2 ^
                     |        von Mises Ellipse:
                     |        σ_1² - σ_1·σ_2 + σ_2² = σ_y²
                +----+----+ 
               /|    |    |\   Tresca Hexagon:
              / |    |    | \  Max(|σ_1 - σ_2|, |σ_2 - σ_3|, |σ_3 - σ_1|) = σ_y
             /  |    |    |  \
  -----------+---+----+----+---+-----------> σ_1
             \  |    |    |  /
              \ |    |    | /
               \|    |    |/
                +----+----+
                     |

1. Maximum Normal Stress Theory (Rankine)

• Criterion: Yielding or fracture occurs when the maximum principal stress reaches the uniaxial tensile strength: $\sigma_1 = \sigma_{ut}$.
• Application: Valid only for brittle materials (gray cast iron, ceramics, un-tempered martensite). It completely neglects shear stress and produces dangerously non-conservative predictions for ductile structural steels.

2. Maximum Shear Stress Theory (Tresca / Guest)

• Criterion: Yielding occurs when the maximum shear stress equals the shear stress at yield in a uniaxial tensile test ($\tau_{\max} = \sigma_y / 2$): $$\tau_{\text{abs, max}} = \frac{\sigma_{\max} - \sigma_{\min}}{2} \ge \frac{\sigma_y}{2} \implies \sigma_{\max} - \sigma_{\min} \ge \sigma_y$$
• Pure Shear Yield: In pure shear ($\sigma_1 = \tau, \sigma_2 = -\tau, \sigma_3 = 0$), $\sigma_1 - \sigma_2 = 2\tau = \sigma_y \implies \tau_y = 0.50 \sigma_y$.
• Application: Conservative yield criterion adopted in ASME Section VIII Division 1 and Division 2 (stress intensity method).

3. Distortion Energy Theory (von Mises / Huber-Hencky)

• Criterion: Yielding occurs when the distortional strain energy per unit volume ($U_d$) equals that at yield in uniaxial tension. Distortional energy is the energy required to change shape without changing volume.
• General 3D Formulation: $$\sigma_{vM} = \frac{1}{\sqrt{2}} \sqrt{ (\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2 } \ge \sigma_y$$
• Plane Stress Formulation ($(\sigma_x, \sigma_y, \tau_{xy})$): $$\sigma_{vM} = \sqrt{ \sigma_x^2 - \sigma_x \sigma_y + \sigma_y^2 + 3\tau_{xy}^2 } \ge \sigma_y$$
• Pure Shear Yield: In pure shear ($\sigma_x = 0, \sigma_y = 0, \tau_{xy} = \tau$), $\sigma_{vM} = \sqrt{3\tau^2} = \tau\sqrt{3} = \sigma_y$: $$\tau_y = \frac{\sigma_y}{\sqrt{3}} \approx 0.57735 \sigma_y$$
• Application: The standard, most accurate criterion for ductile metals (carbon steels, HSLA, stainless steels, aluminum), matching experimental yield data within $1\text{--}2\%$.

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6. Comparative Synthesis of Failure Theories

Table 4.2-1: Classical Static Failure Theories in Engineering Design

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7. Comprehensive Worked Numerical Example: Combined Bending, Torsion & Direct Shear on a Welded Cantilever Tube

Problem Statement

A cylindrical boom member in a materials handling rig is fabricated from S355 structural steel tubing ($\sigma_y = 355\text{ MPa}$) and welded to a rigid base column with a full-penetration groove weld. The tubular dimensions are:

• Outer diameter $D_o = 120.0\text{ mm}$ (outer radius $c = 60.0\text{ mm}$)
• Wall thickness $t = 6.0\text{ mm}$
• Inner diameter $D_i = D_o - 2t = 120.0 - 12.0 = 108.0\text{ mm}$
• Cantilever length $L = 1.20\text{ m} = 1,200.0\text{ mm}$

At the free end, the boom sustains combined simultaneous loads:

• Transverse downward vertical point load $P = 12.0\text{ kN} = 12,000\text{ N}$
• Twisting torque $T = 4.50\text{ kN}\cdot\text{m} = 4,500,000\text{ N}\cdot\text{mm}$

Calculate:

1. The cross-sectional area $A$, moment of inertia $I$, elastic section modulus $S$, and polar moment of inertia $J$.
2. The state of stress $(\sigma_x, \sigma_y, \tau_{xy})$ at Point A located on the uppermost outer surface of the tube at the fixed support.
3. The principal normal stresses $\sigma_1, \sigma_2$, maximum in-plane shear stress $\tau_{\max}$, and principal angle $\theta_p$ at Point A using Mohr's circle.
4. The von Mises equivalent stress $\sigma_{vM}$ and Tresca stress $\sigma_{\text{Tresca}}$ at Point A, and determine the structural factor of safety ($SF$) against initial yield.

Step-by-Step Solution

Step 1: Compute geometric section properties

$$A = \frac{\pi}{4} (D_o^2 - D_i^2) = \frac{\pi}{4} (120.0^2 - 108.0^2) = \frac{\pi}{4} (14,400.0 - 11,664.0) = \frac{\pi}{4} (2,736.0) = 2,148.85\text{ mm}^2$$ $$I = \frac{\pi}{64} (D_o^4 - D_i^4) = \frac{\pi}{64} (120.0^4 - 108.0^4) = \frac{\pi}{64} (207,360,000.0 - 136,048,896.0) = \frac{\pi}{64} (71,311,104.0) = 3,500,438.0\text{ mm}^4$$ $$S = \frac{I}{c} = \frac{3,500,438.0\text{ mm}^4}{60.0\text{ mm}} = 58,340.63\text{ mm}^3$$ $$J = 2 I = 2 \times 3,500,438.0 = 7,000,876.0\text{ mm}^4$$

Step 2: Determine resultant forces and state of stress at Point A At the fixed root ($x = 0$):

• Bending moment $M = P \times L = 12,000\text{ N} \times 1,200\text{ mm} = 14,400,000\text{ N}\cdot\text{mm}$
• Torsional torque $T = 4,500,000\text{ N}\cdot\text{mm}$
• Vertical shear force $V = 12,000\text{ N}$

Evaluate stresses at Point A (top extreme fiber, $y = +60.0\text{ mm}$):

• Normal Bending Stress ($\sigma_x$): $$\sigma_x = \frac{M}{S} = \frac{14,400,000\text{ N}\cdot\text{mm}}{58,340.63\text{ mm}^3} = 246.83\text{ MPa} \quad (\text{Tension})$$
• Normal Transverse Stress ($\sigma_y$): Unconfined outer surface: $\sigma_y = 0$.
• Transverse Shear Stress ($\tau_V$): At the extreme top fiber, $Q = 0$, so vertical shear stress is zero: $\tau_V = 0$.
• Torsional Shear Stress ($\tau_{\text{torsion}}$): $$\tau_{xy} = \frac{T \cdot c}{J} = \frac{4,500,000\text{ N}\cdot\text{mm} \times 60.0\text{ mm}}{7,000,876.0\text{ mm}^4} = 38.57\text{ MPa}$$

Thus, the stress state at Point A is: $\sigma_x = 246.83\text{ MPa}, \sigma_y = 0, \tau_{xy} = 38.57\text{ MPa}$.

Step 3: Mohr's circle and principal stresses at Point A

$$\sigma_{\text{avg}} = \frac{\sigma_x + \sigma_y}{2} = \frac{246.83 + 0}{2} = 123.415\text{ MPa}$$ $$R = \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 } = \sqrt{ (123.415)^2 + (38.57)^2 } = \sqrt{ 15,231.26 + 1,487.64 } = \sqrt{ 16,718.90 } = 129.30\text{ MPa}$$

Principal normal stresses:

$$\sigma_1 = \sigma_{\text{avg}} + R = 123.415 + 129.30 = 252.72\text{ MPa}$$ $$\sigma_2 = \sigma_{\text{avg}} - R = 123.415 - 129.30 = -5.89\text{ MPa} \quad (\text{Compression})$$

Maximum in-plane shear stress:

$$\tau_{\max} = R = 129.30\text{ MPa}$$

Principal plane angle $\theta_p$:

$$\tan(2\theta_p) = \frac{2 \tau_{xy}}{\sigma_x - \sigma_y} = \frac{2 \times 38.57}{246.83} = \frac{77.14}{246.83} = 0.3125 \implies 2\theta_p = 17.35^\circ \implies \theta_p = 8.68^\circ$$

Step 4: Failure theories and factor of safety Because $\sigma_1 > 0$ and $\sigma_2 < 0$, the out-of-plane stress $\sigma_3 = 0$ is the intermediate principal stress ($\sigma_1 > \sigma_3 > \sigma_2$).

1. Tresca (Maximum Shear Stress):

   $$\sigma_{\text{Tresca}} = \sigma_1 - \sigma_2 = 252.72 - (-5.89) = 258.61\text{ MPa}$$ $$SF_{\text{Tresca}} = \frac{\sigma_y}{\sigma_{\text{Tresca}}} = \frac{355.0}{258.61} = 1.373$$

2. von Mises (Distortion Energy):

   $$\sigma_{vM} = \sqrt{ \sigma_x^2 + 3\tau_{xy}^2 } = \sqrt{ (246.83)^2 + 3(38.57)^2 } = \sqrt{ 60,925.05 + 3(1,487.64) } = \sqrt{ 65,387.97 } = 255.71\text{ MPa}$$ $$SF_{vM} = \frac{\sigma_y}{\sigma_{vM}} = \frac{355.0}{255.71} = 1.388$$

Evaluation: The joint is structurally adequate with a factor of safety of $1.39$ per von Mises and $1.37$ per Tresca.

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8. Real-World Engineering Scenarios & Exam Pitfalls

Industrial Scenario: Torsional Dynamic Failure of an Open Crane Boom

A mobile lattice crane experienced catastrophic collapse during slewing (swinging) with a rated load. The original boom design utilized four main corner chord members braced with welded back-to-back angle latticing forming an open cross-section. Dynamic inertial slewing generated a cyclical torque of $180\text{ kN}\cdot\text{m}$. Because the open latticed section possessed an effective torsional constant $J$ less than $1/50$th of an equivalent closed box, the boom underwent severe warping and out-of-plane twist. The resulting combined flexure, warping torsion, and secondary bending stresses exceeded the yield strength at the chord-to-lattice fillet welds, triggering weld root peeling and total boom failure. The remediation required redesigning the boom with a continuously welded tubular truss (closed space frame) where closed-cell shear flow prevented warping and reduced peak torsional shear stress by $84\%$.

Common Exam Traps

Exam Trap 1: Calculating Absolute Maximum Shear Stress in 3D When finding $\tau_{\text{abs, max}}$ for a plane stress problem where both in-plane principal stresses are tensile ($\sigma_1 > 0, \sigma_2 > 0$), candidates often mistakenly calculate $\tau_{\max} = (\sigma_1 - \sigma_2)/2$. Because $\sigma_3 = 0$, the minimum principal stress is $\sigma_3$, meaning the true 3D maximum shear stress is $\tau_{\text{abs, max}} = (\sigma_1 - 0)/2 = \sigma_1 / 2$. Confusing the in-plane circle with the outer 3D circle causes severe under-prediction of maximum shear stress.

Exam Trap 2: Using Flange Width Instead of Web Thickness in Jourawski's Formula When calculating transverse shear stress $\tau = VQ/(It)$ in an I-beam, $t$ represents the width of the material at the plane of the cut. To find web shear stress, $t$ must be the web thickness $t_w$. Substituting flange width $b_f$ underestimates web shear stress by a factor of $10\text{ to }30$.

Exam Trap 3: Confusing Pure Shear Yield Factors Between Tresca and von Mises Exam questions frequently test the exact shear yield multiplier. Under Tresca, $\tau_y = 0.500 \sigma_y$ ($50\%$ of tensile yield). Under von Mises, $\tau_y = \sigma_y / \sqrt{3} \approx 0.577 \sigma_y$ ($57.7\%$ of tensile yield). Reversing these factors introduces a $15.5\%$ error.