4.2 Plane Stress vs. Plane Strain, Triaxial Restraint & Alloy Property Comparison

5. Plane Stress vs. Plane Strain Conditions in Weldments

A critical responsibility of the welding engineer is assessing whether a welded joint operates under plane stress or plane strain, as this mechanical constraint dictates whether a flaw will fail via benign ductile yielding or catastrophic brittle fracture.

      PLANE STRESS (Thin Sheet)                PLANE STRAIN (Thick Plate)
        σ_z = 0,  ε_z ≠ 0                        ε_z = 0,  σ_z = ν(σ_x + σ_y)

             ^ σ_y                                     ^ σ_y
             |                                         |           σ_z (Induced Tensile
       +-----+-----+                             +-----+-----+     /    Constraint Stress)
       |     |     |                             |     |     |    / 
<------+-----+-----+------> σ_x           <------+-----+-----+---+---> σ_x
       |     |     |                             |     |     |  /
       +-----+-----+                             +-----+-----+ /
             |                                         |
             v                                         v
     Through-thickness                         Through-thickness contraction
     contraction occurs freely                 PREVENTED by surrounding cold metal
     (Ductile 45° Shear Lips)                  --> Triaxial Tensile State (Cleavage)

Mathematical Comparison of Constraint Regimes

Why Plane Strain Promotes Brittle Cleavage in Weldments

Plastic deformation in crystalline metals requires dislocation slip, which is driven exclusively by deviatoric (shear) stresses. Hydrostatic stress (equal tension or compression in all three principal directions) causes volume change but produces zero shear stress on slip planes:

$$\tau_{\text{oct}} = \frac{1}{3} \sqrt{(\sigma_1 - \sigma_2)^2 + (\sigma_2 - \sigma_3)^2 + (\sigma_3 - \sigma_1)^2}$$

Under plane stress ($\sigma_z = 0$), the difference between the in-plane tensile stresses and $\sigma_z$ is large, generating high shear stresses that drive widespread plastic yielding and energy absorption.

In thick plate weldments, high mechanical constraint prevents through-thickness Poisson contraction ($\varepsilon_z = 0$). By Generalized Hooke's law:

$$\varepsilon_z = \frac{1}{E} \left[ \sigma_z - \nu(\sigma_x + \sigma_y) \right] = 0 \implies \sigma_z = \nu(\sigma_x + \sigma_y)$$

Because structural steels have $\nu \approx 0.30$, a tensile stress $\sigma_z$ equal to roughly $30%$ of the sum of in-plane stresses is automatically generated in the through-thickness direction! This triaxial tension elevates the mean hydrostatic stress $\sigma_m$ without increasing shear stress. As external load increases, the peak normal stress reaches the critical cleavage fracture stress ($\sigma_f^*$) before the shear stress reaches the critical resolved shear stress for yielding, resulting in sudden, catastrophic brittle fracture with zero macroscopic deformation.

ASTM E399 Validity Criterion for Plane Strain

To ensure valid plane strain fracture toughness ($K_{Ic}$) measurements, the specimen thickness $B$ must exceed:

$$B \ge 2.5 \left( \frac{K_{Ic}}{\sigma_y} \right)^2$$

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6. Comparative Mechanical Properties of Structural Alloys

Table 4.1-1: Elastic Constants & Tensile Properties of Standard Structural Metals

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7. Comprehensive Worked Numerical Example: Triaxial Restraint, Volumetric Strain & True Stress Analysis

Problem Statement

A heavy structural base column connection is fabricated from ASTM A572 Grade 50 steel with:

• Young's modulus $E = 205\text{ GPa} = 205,000\text{ MPa}$
• Poisson's ratio $\nu = 0.29$
• Yield strength $\sigma_y = 345\text{ MPa}$

During high-load service, multi-pass full-penetration groove welds develop an in-plane triaxial stress field at the weld root with:

• Longitudinal normal stress $\sigma_x = 220.0\text{ MPa}$
• Transverse normal stress $\sigma_y = 180.0\text{ MPa}$
• High rigid structural constraint preventing strain in the through-thickness direction: $\varepsilon_z = 0$ (Plane Strain)

Concurrently, a standard round tensile specimen ($d_0 = 12.8\text{ mm}$, gauge length $L_0 = 50.0\text{ mm}$) of the same plate is pulled to fracture in a qualification lab, reaching maximum tensile load $P_{\max} = 65.0\text{ kN}$ at an engineering strain $e = 0.18$, and finally fracturing at load $P_f = 48.0\text{ kN}$ with a final neck diameter $d_f = 8.2\text{ mm}$.

Calculate:

1. The induced out-of-plane constraint stress $\sigma_z$ in the welded joint.
2. The mean hydrostatic stress $\sigma_m$ and volumetric strain $e_v$ at the weld root.
3. The von Mises equivalent stress $\sigma_{vM}$ in the joint and evaluate whether plastic yielding initiates.
4. The true stress $\sigma_{\text{uts}}$ and true strain $\varepsilon_{\text{uts}}$ at maximum load for the lab tensile specimen.
5. The percentage reduction of area ($RA%$) and true fracture strain $\varepsilon_f$.

Step-by-Step Solution

Step 1: Calculate induced constraint stress $\sigma_z$ Under ideal plane strain ($\varepsilon_z = 0$):

$$\varepsilon_z = \frac{1}{E}\left[ \sigma_z - \nu(\sigma_x + \sigma_y) \right] = 0 \implies \sigma_z = \nu(\sigma_x + \sigma_y)$$ $$\sigma_z = 0.29 \times (220.0\text{ MPa} + 180.0\text{ MPa}) = 0.29 \times 400.0\text{ MPa} = 116.0\text{ MPa}$$

Step 2: Calculate mean hydrostatic stress $\sigma_m$ and volumetric strain $e_v$

$$\sigma_m = \frac{\sigma_x + \sigma_y + \sigma_z}{3} = \frac{220.0 + 180.0 + 116.0}{3} = \frac{516.0}{3} = 172.0\text{ MPa}$$

Calculate the material bulk modulus $K$:

$$K = \frac{E}{3(1 - 2\nu)} = \frac{205,000}{3(1 - 2(0.29))} = \frac{205,000}{3(0.42)} = \frac{205,000}{1.26} = 162,698.4\text{ MPa} = 162.70\text{ GPa}$$

Calculate the volumetric strain $e_v$:

$$e_v = \frac{\sigma_m}{K} = \frac{172.0\text{ MPa}}{162,698.4\text{ MPa}} = 1.0572 \times 10^{-3} = 0.1057\%$$

Step 3: Calculate von Mises equivalent stress $\sigma_{vM}$

$$\sigma_{vM} = \frac{1}{\sqrt{2}} \sqrt{(\sigma_x - \sigma_y)^2 + (\sigma_y - \sigma_z)^2 + (\sigma_z - \sigma_x)^2}$$

Evaluate individual stress differences:

• $(\sigma_x - \sigma_y)^2 = (220.0 - 180.0)^2 = (40.0)^2 = 1,600.0\text{ MPa}^2$
• $(\sigma_y - \sigma_z)^2 = (180.0 - 116.0)^2 = (64.0)^2 = 4,096.0\text{ MPa}^2$
• $(\sigma_z - \sigma_x)^2 = (116.0 - 220.0)^2 = (-104.0)^2 = 10,816.0\text{ MPa}^2$

$$\sum (\Delta\sigma)^2 = 1,600.0 + 4,096.0 + 10,816.0 = 16,512.0\text{ MPa}^2$$ $$\sigma_{vM} = \sqrt{\frac{16,512.0}{2}} = \sqrt{8,256.0} = 90.86\text{ MPa}$$

Engineering Evaluation: Although the maximum applied normal stress is $220.0\text{ MPa}$, the von Mises equivalent stress is only $90.86\text{ MPa}$, far below the yield strength $\sigma_y = 345\text{ MPa}$. No plastic yielding occurs! The severe triaxial tension suppresses shear stress, leaving the joint vulnerable to brittle cleavage if a micro-flaw exists.

Step 4: Calculate true stress and true strain at UTS for the tensile bar Original cross-sectional area $A_0$:

$$A_0 = \frac{\pi}{4} d_0^2 = \frac{\pi}{4} (12.8\text{ mm})^2 = 128.68\text{ mm}^2$$

Engineering stress at maximum load:

$$s = \frac{P_{\max}}{A_0} = \frac{65,000\text{ N}}{128.68\text{ mm}^2} = 505.13\text{ MPa}$$

True stress $\sigma_{\text{uts}}$ (valid at peak load prior to necking):

$$\sigma_{\text{uts}} = s(1 + e) = 505.13 \times (1 + 0.18) = 505.13 \times 1.18 = 596.05\text{ MPa}$$

True strain $\varepsilon_{\text{uts}}$:

$$\varepsilon_{\text{uts}} = \ln(1 + e) = \ln(1.18) = 0.1655$$

Step 5: Calculate reduction of area ($RA%$) and true fracture strain ($\varepsilon_f$) Final fractured cross-sectional area $A_f$:

$$A_f = \frac{\pi}{4} d_f^2 = \frac{\pi}{4} (8.2\text{ mm})^2 = 52.81\text{ mm}^2$$

Percentage reduction of area:

$$RA\% = \frac{A_0 - A_f}{A_0} \times 100\% = \frac{128.68 - 52.81}{128.68} \times 100\% = \frac{75.87}{128.68} \times 100\% = 58.96\%$$

True fracture strain $\varepsilon_f$:

$$\varepsilon_f = \ln\left( \frac{A_0}{A_f} \right) = \ln\left( \frac{128.68}{52.81} \right) = \ln(2.4367) = 0.8907$$

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8. Real-World Engineering Scenarios & Exam Pitfalls

Industrial Scenario: Lamellar Tearing in an Offshore Jacket Leg Node

During fabrication of a heavy tubular jacket leg node for a North Sea platform, a $65\text{ mm}$ thick ASTM A572 Grade 50 flange plate was joined to an internal diaphragm plate using full-penetration double-bevel groove welds. The weldment cooled under heavy external restraint without preheat maintenance. Ultrasonic testing (UT) revealed continuous planar separations running $4\text{ to }6\text{ mm}$ beneath the plate surface, parallel to the fusion line. Metallographic sectioning identified classic stepped lamellar tearing along flattened $\text{MnS}$ stringers. Certified mill test reports showed the plate had a sulfur content of $0.024\text{ wt}%$ and had not been specified with through-thickness testing. The engineering team replaced the joint by excavating the damaged region, buttering the joint face with two layers of low-strength, high-ductility AWS E7018 filler metal to buffer transverse shrinkage strains, and procuring replacement plates meeting ASTM A770 with guaranteed $Z35$ through-thickness reduction of area.

Common Exam Traps

Exam Trap 1: Extrapolating $\sigma = s(1+e)$ Beyond the UTS The equations $\sigma = s(1+e)$ and $\varepsilon = \ln(1+e)$ are derived strictly from the assumption of uniform elongation along the gauge length. Once maximum load (UTS) is reached, necking initiates and deformation localizes. Using these formulas at the fracture point produces massive errors; true fracture stress must be calculated directly from final neck area ($P_f / A_f$) and corrected for triaxiality via Bridgman's factor.

Exam Trap 2: Neglecting the Out-of-Plane Constraint Stress in Plane Strain Examination candidates frequently assume that in "plane strain", the out-of-plane stress $\sigma_z$ is zero. This is false! In plane strain, out-of-plane strain is zero ($\varepsilon_z = 0$), which forces an out-of-plane stress $\sigma_z = \nu(\sigma_x + \sigma_y)$ to develop. It is plane stress where $\sigma_z = 0$.

Exam Trap 3: Direct Comparison of Elongation Values Across Differing Gauge Lengths Comparing $EL%$ values between a $50\text{ mm}$ ($2\text{ in}$) gauge length bar and a $200\text{ mm}$ ($8\text{ in}$) plate specimen directly will lead to incorrect weld procedure qualification. Because necking strain is localized, shorter gauge lengths yield significantly higher total elongation percentages for the exact same steel.