Biaxial stresses in a thin plate and factor of safety: 22-Mec-A7 Advanced Strength of Materials, December 2016, Q6

Question 6: Biaxial stresses in a thin plate and factor of safety (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A thin plate in biaxial plane stress; the two edge elongations are measured and the elastic constants are known.

Given data
Plate size	1.5 m × 1.5 m
x-elongation	δ_x = 0.8 mm → ε_x = 5.333×10−⁴
y-elongation	δ_y = 0.25 mm → ε_y = 1.667×10−⁴
Elastic constants	E = 120 GPa, v = 0.28

Find. (a) $\sigma_x,\sigma_y$ from the measured strains; (b) the factor of safety by the maximum-shear-stress theory.

Approach. Invert the plane-stress Hooke’s law to get the stresses from the two strains, then form the principal stresses and apply Tresca for the factor of safety.

(a) Invert plane-stress Hooke. With $\sigma_x=\dfrac{E(\varepsilon_x+v\varepsilon_y)}{1-v^2}$ and $\sigma_y=\dfrac{E(\varepsilon_y+v\varepsilon_x)}{1-v^2}$ (using $1-v^2=0.9216$): $$\boxed{\sigma_x=75.5\ \text{MPa},\qquad \sigma_y=41.1\ \text{MPa.}}$$
Principal stresses. The stresses are already principal (no shear): $\sigma_1=75.5$, $\sigma_2=41.1$, $\sigma_3=0$ MPa (the out-of-plane direction). Maximum shear $\tau_{max}=(\sigma_1-\sigma_3)/2=37.8$ MPa.
(b) Factor of safety (Tresca). The maximum-shear theory gives $N=\dfrac{\sigma_Y}{\sigma_1-\sigma_3}=\dfrac{\sigma_Y}{75.5\ \text{MPa}}$. The plate’s yield strength is not stated; for a representative structural value $\sigma_Y=250$ MPa, $$\boxed{N=\sigma_Y/75.5\ \text{MPa}\approx3.3.}$$

Check: Question 6 gives no yield strength, yet asks for a numerical factor of safety. The FoS is therefore reported as the exact ratio $\sigma_Y/75.5\ \text{MPa}$; the value 3.3 assumes a representative $\sigma_Y=250$ MPa (mild steel). Substitute the intended yield strength to finalize.

Final results — Q6
Quantity	Value
σ_x	75.5 MPa
σ_y	41.1 MPa
τ_max	37.8 MPa
Factor of safety (Tresca)	σ_Y/75.5 MPa (≈ 3.3 at σ_Y = 250 MPa)

Topic: Biaxial plane stress from measured strains and the Tresca factor of safety.

Key relations: Inverted Hooke $\sigma_x=E(\varepsilon_x+v\varepsilon_y)/(1-v^2)$; Tresca $N=\sigma_Y/(\sigma_1-\sigma_3)$.

Why this works: Under uniform biaxial stress the two normal strains fully determine the two stresses once E and v are known; the coupled Hooke equations invert to the symmetric pair above. With no applied shear the stresses are principal, and because the plate is thin ($\sigma_z=0$) the smallest principal is zero, so the maximum shear is half the larger in-plane stress. The Tresca factor of safety is yield strength over that principal difference.

Common pitfall: Using the uniaxial $\sigma=E\varepsilon$ instead of the coupled biaxial law — it ignores the Poisson coupling and understates both stresses.

Source: Hibbeler, Mechanics of Materials §10.6 (plane-stress Hooke) and §10.7 (yield criteria); Ugural & Fenster Ch. 4.

Question 6 of 8: Biaxial stresses in a thin plate and factor of safety

Question 6: Biaxial stresses in a thin plate and factor of safety (20 marks)