--- title: "Uniaxial, biaxial and pure shear stress systems" subject: "Strength of Materials" chapter: 6 key: "semester_4/strength_of_materials_mechanical:uniaxial-biaxial-pure-shear-systems" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/uniaxial-biaxial-pure-shear-systems site: BCT Engineering --- # Uniaxial, biaxial and pure shear stress systems Strength of Materials · Chapter 6: Transformation of stress and strain - Asked in: year not given, 5 marks - Video: [Stresses on an Inclined Plane | Bi-axial and Shear Stress Condition](https://www.youtube.com/watch?v=YieNtiaOTX4) — MechE Design Course - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/uniaxial-biaxial-pure-shear-systems - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-6 ## Question Explain uniaxial, biaxial and pure shear stress systems. For each system write the expressions for the normal and shear stress on an inclined plane and the maximum shear stress. Show that pure shear is equivalent to equal tension and compression on planes at 45 degrees. ## Short answer Uniaxial: σθ = σ cos²θ, τθ = –(σ/2) sin 2θ, τmax = σ/2. Biaxial: σθ = (σx+σy)/2 + ((σx-σy)/2) cos 2θ, τθ = –((σx-σy)/2) sin 2θ, τmax = (σx-σy)/2. Pure shear: σθ = τ sin 2θ, τθ = τ cos 2θ; principal stresses ±τ on 45° planes, showing equivalence to equal tension and compression. ## Answer All three are special cases of the plane stress equations of transformation: $$ \sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta, \quad \tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta $$ #### Uniaxial stress ($\sigma_y = \tau_{xy} = 0$) A bar under a single axial stress $\sigma$. $$ \sigma_\theta = \frac{\sigma}{2}(1+\cos2\theta) = \sigma\cos^2\theta, \qquad \tau_\theta = -\frac{\sigma}{2}\sin2\theta $$ $\tau_{max} = \sigma/2$ at $\theta = 45^\circ$. Principal stresses: $\sigma$ and $0$. #### Biaxial stress ($\tau_{xy} = 0$) Normal stresses $\sigma_x$ and $\sigma_y$ in two perpendicular directions, e.g. a thin plate or shell. $$ \sigma_\theta = \frac{\sigma_x+\sigma_y}{2}+\frac{\sigma_x-\sigma_y}{2}\cos2\theta, \qquad \tau_\theta = -\frac{\sigma_x-\sigma_y}{2}\sin2\theta $$ $\tau_{max} = \dfrac{\sigma_x-\sigma_y}{2}$ at $45^\circ$. The x and y planes are principal planes. #### Pure shear ($\sigma_x = \sigma_y = 0$, $\tau_{xy} = \tau$) Occurs in a twisted circular shaft. $$ \sigma_\theta = \tau\sin2\theta, \qquad \tau_\theta = \tau\cos2\theta $$ Principal planes: $\tan2\theta_p = \infty$, so $\theta_p = 45^\circ$ and $135^\circ$. $$ \sigma_1 = \tau\ (\text{at }45^\circ), \qquad \sigma_2 = -\tau\ (\text{at }135^\circ) $$ So a state of pure shear $\tau$ is the same as tension $\tau$ and compression $\tau$ on planes inclined at $45^\circ$, with no shear on those planes. Maximum shear stress $= \tau$. This is why brittle materials fail along a $45^\circ$ helix in torsion (tension failure), while ductile materials fail by shear on the cross-section. | System | Principal stresses | $\tau_{max}$ | |---|---|---| | Uniaxial | $\sigma,\ 0$ | $\sigma/2$ | | Biaxial | $\sigma_x,\ \sigma_y$ | $(\sigma_x-\sigma_y)/2$ | | Pure shear | $+\tau,\ -\tau$ | $\tau$ | ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For uniaxial stress σ, what is the maximum shear stress and where does it act? A. σ on planes at 45° to the axis B. σ/2 on the cross-section plane C. σ on planes at 90° to the axis D. σ/2 on planes at 45° to the axis Answer: D. τθ = -(σ/2)sin2θ is largest in magnitude at θ = 45°, giving σ/2. (Key point: Uniaxial maximum shear) 2. For biaxial stress with τxy = 0, what is τmax? A. (σx - σy)/2 at 45° B. (σx + σy)/2 at 45° C. (σx - σy) at 45° D. (σx - σy)/2 at 0° Answer: A. Setting τxy = 0 in the shear equation gives τθ = -((σx-σy)/2)sin2θ, with maximum at 45°. (Key point: Biaxial maximum shear) 3. What are the principal stresses and planes in pure shear τ? A. Both +τ, at 0° and 90° B. +τ at 45° and -τ at 135° C. +τ at 0° and -τ at 90° D. +2τ at 45° and zero at 135° Answer: B. Pure shear is equivalent to tension τ and compression τ on planes at 45° and 135°, with no shear on them. (Key point: Pure shear principal stresses) 4. Why do brittle materials fail along a 45° helix in torsion? A. The maximum shear acts only on 45° planes in brittle materials B. Compression of magnitude 2τ acts at 45° C. The principal tensile stress equals τ on the 45° plane and brittle materials are weak in tension D. Normal stress is zero so shear alone breaks it Answer: C. In pure shear the 45° plane carries tension τ; brittle materials fail by tension, ductile ones by shear on the cross-section. (Key point: Brittle failure in torsion) ## Related in Strength of Materials - [Stresses on inclined plane and principal stresses](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ainclined-plane-stresses-principal-stresses) — Strength of Materials, chapter 6 - [Derivation of plane stress transformation equations](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Aplane-stress-transformation-principal-derivation) — Strength of Materials, chapter 6 - [Mohr's circle for a plane stress element](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Amohr-circle-plane-stress-construction) — Strength of Materials, chapter 6 - [Normal, shear, bearing stress and factor of safety](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Astress-types-factor-of-safety) — Strength of Materials, chapter 1 - [Shear stress in rectangular and circular beams](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Ashear-distribution-rectangle-circle-ratio) — Strength of Materials, chapter 5 - [Derivation of shear stress formula in beams](https://www.bctengineer.com/question.md?key=semester_4%2Fstrength_of_materials_mechanical%3Abeam-shear-stress-formula-derivation) — Strength of Materials, chapter 5 ## Related in other subjects - [Normal/shear stress on 25 degree plane, principal stresses](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ainclined-plane-30-25deg-mohr-verify) — Strength of Materials, chapter 4, asked 3× - [Plane stress: principal planes and max shear](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-15-30-60-principal) — Strength of Materials, chapter 4, asked 2× - [Plane stress: principal, max shear, normal](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aplane-stress-80-60-50-mpa) — Strength of Materials, chapter 4, asked 2× - [Principal planes, stresses on plane AC](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aprincipal-planes-max-shear-120-180) — Strength of Materials, chapter 4