Question I HAVE 10 MINS REMAINING, PLEASE GIVE ME AN ANSWER For the beam shown, calculate the magnitude of the bending stress (in psi) at the bottom of the beam on a section 1.17 feet to the right of B if P = 2044 lb, Q = 8215 lb, w = 1.16 in, L = 8.09 in, b = 3.69 in, and h = 1.1 in. Round off the final answer to two decimal places. P P W B D E L A F 2 ft 3 ft——3 ft— 2 ft | b → h

XJUW69 The Asker · Mechanical Engineering

I HAVE 10 MINS REMAINING, PLEASE GIVE ME AN ANSWER

Transcribed Image Text: For the beam shown, calculate the magnitude of the bending stress (in psi) at the bottom of the beam on a section 1.17 feet to the right of B if P = 2044 lb, Q = 8215 lb, w = 1.16 in, L = 8.09 in, b = 3.69 in, and h = 1.1 in. Round off the final answer to two decimal places. P P W B D E L A F 2 ft 3 ft——3 ft— 2 ft | b → h

More

Community Answer

Honor Code

Solved 1 Answer

\therefore \quad \begin{array}{l} \Sigma F_{y}=0 \\R_{B}+R_{C}=12303 \mathrm{lb} . \\m_{B}=0 \\\Rightarrow(2044 \times 2)+\left(R_{e} \times 6\right)= \\(8215) \times 3+(2044)(8) \\R_{C}=615.1 .5 \mathrm{lb} . \\R_{B}=6151.5 \mathrm{lb} .\end{array}\left(R_{B}\right)_{x}=0 \rightarrow \text { as } \sum F_{x}=0\left(R_{B},=R_{B}+(\text { Let })\right.\therefore Berding moment (m) at 1.17 feet to Right of B.\therefore m=717.775 \times 12 \mathrm{lb}-\stackrel{0}{0}\& m=717.775 l b-j t \text {. }\therefore we know Berding skess (\sigma)=\frac{m y}{I}\bar{y}=\frac{A_{1} y_{1}+A_{2} y_{2}}{A_{1}+A_{2}}\begin{array}{l}A_{1}=1.1 \times 3.69=4.059 \mathrm{lb}^{2} . \\A_{2}=1.16 \times 9.09=9.3844 \quad 1 b^{2} .\end{array}From boitom y_{1}=\frac{111}{2}=0.5516.\bar{y}=\frac{(4.059)(0.55)+(9.3844)(5.145)}{(4.059)+(9.3844} \quad y_{2}=\left(\frac{8.09}{2}+1.1\right)=5.145 \mathrm{lb}(4.059)+(9.3844 \begin{array}{l}\bar{y}=3.75762 \mathrm{lb} \text {. } \\ I_{1}=\frac{b h^{3}}{12}=\frac{3.69 \times 1.1^{3}}{12}=0.4093 \mathrm{lb}^{4} \\ I_{2}=\frac{\omega \times L^{3}}{12}=\frac{1.16 \times 8.09^{3}}{12}=51.1826 \mathrm{lb}^{4} . \\ I=\left(I_{1}\right)_{N A}+\left(I_{2}\right)_{N A} . \\ \left(I_{1}\right)_{\mathrm{NA}}=\left(I_{1}\right)+A_{1}(3.75762-0.55)^{2 .}=42.171645 \\ \left(I_{2}\right)_{N A}=\left(I_{2}\right)+A_{2}(3.75762-5.145)^{2}=69.2459 \mathrm{l} \\ I=111.417545 \mathrm{lb}^{4} \\ \Rightarrow \quad \sigma=\frac{(717.775 \times 12) \times(3.75762)}{111.417545} \\ \sigma=290.488 \frac{\mathrm{lb}}{i^{2}} \\ \Rightarrow \sigma=290 \cdot 49 \frac{1 b}{e_{n}^{2}} \\\end{array} ...