Draw shear force and bending moment for simply supported beam

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written 5.6 years ago by

teamques10 ★ 69k

Reaction Calculation:

$\begin{array}{l}{\begin{aligned} \sum \mathrm{M}_{\mathrm{A}} &=0 \\ \mathrm{R}_{\mathrm{B}} \times 5.5 &=(2 \times 2) \times 1+1.5 \times 2+30 \\ \mathrm{R}_{\mathrm{B}} &=6.72 \mathrm{kN} \\ \Sigma \mathrm{Fy}=& 0 \end{aligned}} \\ {\mathrm{R}_{\mathrm{A}_{\mathrm{A}}}+\mathrm{R}_{\mathrm{B}}=(2 \times 2)+1.5} \\ {\mathrm{R}_{\mathrm{A}}=-1.22 \mathrm{kN}}\end{array}$

2) SF Calculation:

$\begin{array}{l}{\text { SF at } A=-1.22 \mathrm{kN}} \\ {\mathrm{C}_{\mathrm{L}}=-1.22-(2 \times 2)=-5.22 \mathrm{kN}} \\ {\mathrm{C}_{\mathrm{R}}=-5.22-1.5=-6.72 \mathrm{kN}} \\ {\mathrm{B}_{\mathrm{L}}=-6.72 \mathrm{kN}} \\ {\mathrm{B}=-6.72+6.72=0 \mathrm{kN} \quad(\therefore \text { ok })}\end{array}$

3) BM Calculation:

$\begin{array}{l}{\begin{aligned} \mathrm{BM} \text { at } \mathrm{A} \text { and } \mathrm{B}=0 &(\because \text { Supports } \mathrm{A} \text { and } \mathrm{B} \text { are simple }) \\ \mathrm{C}=&-1.22 \times 2-(2 \times 2) \times 1=-6.44 \mathrm{kN}-\mathrm{m} \\ \mathrm{D}_{\mathrm{L}} &=6.72 \times 2.5-30=-13.2 \mathrm{kN}-\mathrm{m} \\ \mathrm{D}_{\mathrm{R}} &=6.72 \times 2.5=+16.8 \mathrm{kN}-\mathrm{m} \end{aligned}}\end{array}$

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