Question A long flat aluminum bar is subjected to a tensile load. The bar has a hole drilled through it. The width of the bar = 3 in The thickness of the bar = 0.4 in The hole diameter = 0.6 in The tensile load = 17 kips Gross area (before the hole is drilled? Average Stress before the hole is drilled? Net width (d) (after the hole is drilled)? r/d ? K? Net area (after the hole is drilled)? Average net stress at the hole? Maximum stress adjacent to the hole ?

Ans:Given DataWidth of the bar =3 inches\begin{aligned}\text { Thickness of bar } & =0.4 \text { inches } \\\text { Hole diameter } & =0.6 \text { inches } \\\text { Tensile load } & =17 \mathrm{kips} \\& =17000 \mathrm{lbs}\end{aligned}Gross stress:gross crosssectional Area of the bar away from hole =3 inches \times 0.4 inches=1.2 \text { inches }{ }^{2}\begin{aligned}\therefore \text { Gross stress } & =\frac{\text { Tensile load }}{\text { cross sectional Area }} \\& =\frac{17000}{1.2} \\\Rightarrow \text { Gross stress } & =14166.66 \mathrm{Psi} \\& =14.166 \mathrm{kPsi} \\\therefore \text { Gross stress } & =14.166 \mathrm{kPsi}\end{aligned}Average Net stress at hole:\begin{aligned}\text { Net area } & =2[1.2 \times 0.4]+\left[\frac{\pi \times 0.6}{2} \times 0.4\right] \\& =0.96+0.37699 \\& =1.336 \mathrm{in}^{2} \\\therefore \text { Net Area } & =1.336 \text { inches }^{2}\end{aligned}\begin{aligned}\therefore \text { Average Net stress } & =\frac{\text { Tensile load }}{\text { Net area }} \\\Rightarrow \text { Average Net stress } & =\frac{17000}{1.336} \\& =12715.11 \mathrm{Psi} \\& =12.715 \mathrm{kPsi}\end{aligned}\begin{aligned}\therefore \text { Maximum Tensile stress } & =\text { gross stress } \\& =14.166 \mathrm{kPsi}\end{aligned}\rightarrow Gross stress =14.166 \mathrm{KPsi}\rightarrow Average stress away =14.166 \mathrm{kPsi}\rightarrow Net width =1.9 \times 2+\frac{\pi \times 0.6}{2}=3.3424 inches\rightarrow \quad r / d=\frac{0.3}{0.6}=\frac{1}{2}\rightarrow Net Area =1.336 \mathrm{in}^{2}\rightarrow Average stress @ hole =12.715 \mathrm{kPsi}\rightarrow Maximum stress =14.166 KPsi ...