--- title: "Principal stresses from a 3D stress tensor" subject: "Mechanics of Solids" chapter: 2 key: "semester_5/mechanics_of_solids:three-d-stress-invariants-numerical" source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/three-d-stress-invariants-numerical site: BCT Engineering --- # Principal stresses from a 3D stress tensor Mechanics of Solids · Chapter 2: Stress Tensor - Asked in: year not given, 8 marks - Video: [MechMat 20.1: Triaxial Stress - Finding Principal Stresses](https://www.youtube.com/watch?v=66LLpuDRCFU) — engineering-content - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/three-d-stress-invariants-numerical - Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-2 ## Question The state of stress at a point is given by $\sigma_x = 100$, $\sigma_y = -20$, $\sigma_z = 60$, $\tau_{xy} = 40$, $\tau_{yz} = 30$, $\tau_{zx} = 0$ (all in MPa). (a) Find the stress invariants and the principal stresses, and the absolute maximum shear stress. (b) Find the normal and shear stress on a plane whose unit normal is $n = (2/3,\ 1/3,\ 2/3)$. ## Short answer Stress invariants: I1=140 MPa, I2=300 (MPa)^2, I3=-306000 (MPa)^3. Principal stresses: σ1=113.7 MPa, σ2=66.7 MPa, σ3=-40.4 MPa. Absolute maximum shear stress τ_max=77.0 MPa. On plane with normal (2/3,1/3,2/3), σ_n=100.0 MPa, τ=22.4 MPa. ## Answer #### (a) Invariants and principal stresses $$ [\sigma] = \begin{bmatrix} 100 & 40 & 0 \\ 40 & -20 & 30 \\ 0 & 30 & 60 \end{bmatrix}\ \text{MPa} $$ The characteristic equation is $\sigma^3 - I_1\sigma^2 + I_2\sigma - I_3 = 0$. $$ \begin{aligned} I_1 &= \sigma_x+\sigma_y+\sigma_z = 100 - 20 + 60 = 140 \\ I_2 &= (\sigma_x\sigma_y - \tau_{xy}^2) + (\sigma_y\sigma_z - \tau_{yz}^2) + (\sigma_z\sigma_x - \tau_{zx}^2) \\ &= (-2000-1600) + (-1200-900) + (6000-0) = 300 \\ I_3 &= \det[\sigma] = 100(-1200-900) - 40(2400-0) + 0 = -306000 \end{aligned} $$ So $\sigma^3 - 140\sigma^2 + 300\sigma + 306000 = 0$. Solving (trigonometric method or trial and error): $$ \sigma_1 = 113.68,\quad \sigma_2 = 66.68,\quad \sigma_3 = -40.37\ \text{MPa} $$ Check: sum $= 140.0 = I_1$; product $= -306000 = I_3$. Absolute maximum shear stress: $$ \tau_{max} = \frac{\sigma_1 - \sigma_3}{2} = \frac{113.68 - (-40.37)}{2} = 77.03\ \text{MPa} $$ #### (b) Plane with $n = (2/3, 1/3, 2/3)$ Cauchy's formula $\mathbf{t} = [\sigma]\mathbf{n}$: $$ \begin{aligned} t_x &= 100\left(\tfrac23\right) + 40\left(\tfrac13\right) + 0 = 80.00 \\ t_y &= 40\left(\tfrac23\right) - 20\left(\tfrac13\right) + 30\left(\tfrac23\right) = 40.00 \\ t_z &= 0 + 30\left(\tfrac13\right) + 60\left(\tfrac23\right) = 50.00 \end{aligned} $$ $$ \begin{aligned} \sigma_n &= \mathbf{t}\cdot\mathbf{n} = 80\left(\tfrac23\right) + 40\left(\tfrac13\right) + 50\left(\tfrac23\right) = 100.00\ \text{MPa} \\ |\mathbf{t}| &= \sqrt{80^2+40^2+50^2} = 102.47\ \text{MPa} \\ \tau &= \sqrt{|\mathbf{t}|^2 - \sigma_n^2} = \sqrt{10500 - 10000} = 22.36\ \text{MPa} \end{aligned} $$ **Answer:** $I_1 = 140$, $I_2 = 300$, $I_3 = -306000$ (MPa units); $\sigma_1 = 113.7$, $\sigma_2 = 66.7$, $\sigma_3 = -40.4$ MPa; $\tau_{max} = 77.0$ MPa; on the given plane $\sigma_n = 100.0$ MPa and $\tau = 22.4$ MPa. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which equation gives the principal stresses? A. sigma^3 - I1 sigma^2 + I2 sigma - I3 = 0 B. sigma^2 - I1 sigma + I2 = 0 C. sigma^3 + I1 sigma = I3 D. det[sigma] = 0 only Answer: A. The principal stresses are roots of sigma^3 - I1 sigma^2 + I2 sigma - I3 = 0. (Key point: Characteristic equation) 2. What is I1 here and what must the principal stresses add up to? A. 60, since the sum equals sigma_z B. 140, and sigma_1 + sigma_2 + sigma_3 = 140 C. 300, the sum of principal stresses D. 100, the largest normal stress Answer: B. I1 = 100 - 20 + 60 = 140, and the invariant is unchanged by rotation so the principal stresses sum to 140. (Key point: First invariant) 3. How is the absolute maximum shear stress found? A. (sigma_1 + sigma_3)/2 B. (sigma_1 - sigma_2)/2 C. sigma_1 - sigma_3 D. (sigma_1 - sigma_3)/2 = 77.03 MPa Answer: D. It is half the difference between largest and smallest principal stresses: (113.68 + 40.37)/2 = 77.03 MPa. (Key point: Absolute maximum shear) 4. After finding traction t = [sigma]n, how are sigma_n and tau obtained? A. sigma_n = |t|, tau = 0 B. sigma_n = t.n, tau = sqrt(|t|^2 - sigma_n^2) C. sigma_n = t x n, tau = t.n D. sigma_n = sqrt(|t|^2 - tau^2), tau = t.n Answer: B. Normal stress is the component of t along n; shear is the remaining component: 22.36 MPa here. 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