Question #128728

Determine the maximum uniformly distributed load, a simply supported 40cm deep and 20cm wide timber beam can carry over a span of 4m. The maximum permissible bending stress for the timber is 2N/mm

Expert's answer

"I_x=\\frac{bh^3}{12}=\\frac{0.2\\cdot0.4^3}{12}=0.00107(m^4)"

"\\sigma_{max}=\\frac{M_{max}c}{I_x}\\to M_{max}=\\sigma_{max}I_x\/c="

"=2\\cdot10^6\\cdot0.00107\/0.2=10700(N\\cdot m)"

"M=2qx-qx^2\/2"

For "x=2(m)" "M=M_{max}"

"2q\\cdot2-q\\cdot2^2\/2=10700\\to q=5350(N\/m)". Answer

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