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Draw shear force and bending moment diagram for cantilever beam of 5 m span subjected to udl of 15 N/m up to mid span from fixity.
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1) Reaction Calculation:

$$ \begin{array}{c}{\sum \mathrm{Fy}=0} \\ {+\mathrm{R}_{\mathrm{A}}-(15 \times 2.5)=0} \\ {\mathrm{R}_{\mathrm{A}}-37.5=0} \\ {\mathrm{R}_{\mathrm{A}}=37.5 \mathrm{kN}}\end{array} $$

2) SF Calculation:

$$ \begin{aligned} \text { SF at } A &=+37.5 \mathrm{kN} \\ \mathrm{C} &=+37.5-(15 \times 2.5)=0 \mathrm{kN} \\ \mathrm{B} &=0 \mathrm{kN} \end{aligned} $$

3) BM Calculation

$\begin{aligned} \mathrm{BM} \text { at } \mathrm{B}=0 &(\mathrm{B} \text { is free end }) \\ \mathrm{C}=0 &(\mathrm{No} \text { load from } \mathrm{B} \text { to } \mathrm{C}) \\ \mathrm{A}=-(15 \times 2.5 \times 1.25)=-46.875 \mathrm{kN}-\mathrm{m} \end{aligned}$

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