Elasticity — Question 2

Question 2

A minimum-area design must satisfy both stiffness and strength

A $1\,\mathrm m$ axial tie carries $10\,\mathrm{kN}$ and may extend at most $1\,\mathrm{mm}$. Material A has $E=70\,\mathrm{GPa}$ and allowable stress $140\,\mathrm{MPa}$; material B has $E=200\,\mathrm{GPa}$ and allowable stress $250\,\mathrm{MPa}$. Ignore self-weight and use linear elasticity.

1. Find the minimum cross-sectional area for each material.

2. Identify the controlling condition for each, and explain why modulus alone cannot establish strength.

Question 2 - Solution

A minimum-area design must satisfy both stiffness and strength

(a) Derivation and calculation

Strength and stiffness impose separate lower bounds: $$A\ge A_\sigma=\frac{F}{\sigma_{\mathrm{allow}}},\qquad A\ge A_\delta=\frac{FL}{E\delta_{\max}},\quad A_{\min}=\max(A_\sigma,A_\delta).$$ For material A, $$A_\sigma=\frac{10^4}{140\times10^6}=71.4\ \mathrm{mm^2},\quad A_\delta=\frac{10^4(1)}{70\times10^9(.001)}=142.9\ \mathrm{mm^2}.$$ Thus $\boxed{A_{\min,A}=142.9\ \mathrm{mm^2}}$.

(b) Result and interpretation

For material B, $$A_\sigma=\frac{10^4}{250\times10^6}=40.0\ \mathrm{mm^2},\quad A_\delta=\frac{10^4}{200\times10^9(.001)}=50.0\ \mathrm{mm^2}.$$ Therefore $\boxed{A_{\min,B}=50.0\ \mathrm{mm^2}}$. Stiffness controls both designs. At these areas, $$\sigma_A=70\ \mathrm{MPa}<140\ \mathrm{MPa},\qquad \sigma_B=200\ \mathrm{MPa}<250\ \mathrm{MPa}.$$ Young’s modulus specifies elastic slope; allowable stress independently specifies the strength limit.

Physical check

Both selected areas give exactly 1 mm extension and stresses below their respective allowable values.