Question 7-2. Determine the resultant internal normal and shear force in the member at (a) section a-a and (b) section b-b. each of which passes through point A. The 500-1b load is applied along the centroidal axis of the member. b 30° 500 lb 500 lb А A b Prob. 7-2

Answer a) Normal Force=500 lbs,   shear force=0 answer b) Normal force= 433.02 lbs,   shear force= 250 lbs   \thetacase aLet the Area of crols section A B C D=A \mathrm{~m}^{2}\Rightarrow \sigma_{x}=500 / \mathrm{A} \quad \mathrm{lb} / \mathrm{m}^{2}So \sqrt{6_{n}}=\frac{6 x+6 y}{2}+\frac{6 x-6 y}{2} \cos 2 \theta+2 \sin 2 \thetaHere \theta=0 as no inclination.\begin{array}{l}\Rightarrow \sigma_{n}=\frac{6 x}{2}+\frac{6 x}{2}+0=6 x \\\Rightarrow \sigma_{n}=6_{x}=\frac{500 \mathrm{lb}}{\text { Area }} \\\Rightarrow F_{n}=6_{n} \times \text { Arean }=500 \mathrm{lb} \text { Ans }\end{array}sly\begin{aligned}\tau_{a a} & =\frac{(6 x-6 y)}{2} \sin 2 \theta-2 \cos 2 \theta \\& =0-0=0 \text { An }\end{aligned}\Rightarrow F_{\text {suear }}=0 \text { Ar }Case b\text { Here } \sigma_{x}=\frac{500 \mathrm{~b}}{\text { Area }}\begin{array}{l} \sigma_{n}=\frac{6 x+6 y}{2}+\frac{6 x-6 y}{2} \cos 2 \theta+\pi \sin 2 \theta \\=\frac{6 x}{2}+\frac{6 x}{2} \times \cos \left(2 \times 30^{\circ}\right) \\=\frac{6 x}{2}+\frac{6 x}{2} \times \frac{1}{2}=\frac{3}{4} 6 x \\\Rightarrow 6_{n}=\frac{3}{4} \times \frac{500}{\text { Area }} A B C D \\F_{n}=6_{n} \times \text { Area bb } \\\text { Area }\end{array}\begin{array}{l}\tau_{b b}=\frac{6 x-6 y}{2} \sin 2 \theta-\pi \cos 2 \theta \\=\frac{6 x}{2} \times \sin \left(60^{\circ}\right) \\=\frac{6 x}{2} \times 0.866=\frac{500}{2 \times(A r e a)_{A B C D}} \times 0.866 \\\Rightarrow F_{\text {shear }}=Z_{b b} \times(\text { Area })_{b b} \quad\left(A r e a_{b b}=\frac{\text { Area } \left._{A B C 0}\right)}{\cos \theta}\right) \\=\frac{500}{2 \times(\text { Area }) A B C D} \times \frac{0.866}{\cos \theta} \\=\frac{500}{2} \times 0.866 \times \frac{1}{0.866}=250 \mathrm{lbs} \\\end{array}An ...