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Question Number 183524 by aurpeyz last updated on 26/Dec/22

Answered by mr W last updated on 26/Dec/22

$$M=P\times 2l+q\times l\times (l+\frac{l}{2})=2Pl+\frac{3{ql}^2}{2}$$$$W=\frac{a^3}{6}$$$${\sigma }_{max}=\frac{M}{W}=\frac{2Pl+\frac{3{ql}^2}{2}}{\frac{a^3}{6}}=\frac{12Pl+9{ql}^2}{a^3}$$$$\frac{12Pl+9{ql}^2}{a^3}=360MPa=360\times {10}^6\frac{N}{m^2}$$$$\frac{12P\times 20\times 0.02+9\times 400\times {(20\times 0.02)}^2}{0.{02}^3}=360\times {10}^6$$$$\Rightarrow P=480N✓$$

Commented by aurpeyz last updated on 27/Dec/22

$$thanks$$

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