simply supported beam AB having span 5m and two point load 30 kN & 20 kN acts at 3m & 2m from support A respectively then reaction at A & B is...

solution:

I. Support Reactions:

$\sum \mathrm{M}_{\mathrm{A}}=0$

$5 \times 1.5+7 \times 3.5-\mathrm{R}_{\mathrm{B}} \times 5=0$

$5 \times \mathrm{R}_{\mathrm{B}}=32$

$\mathrm{R}_{\mathrm{B}}=6.4 \mathrm{kN}$

$\sum \mathrm{F}_{\mathrm{y}}=0$

$\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{B}}-5+7=0$

$\mathrm{R}_{\mathrm{A}}+\mathrm{R}_{\mathrm{B}}=12$

$\mathrm{R}_{\mathrm{A}}=5.6 \mathrm{kN}$

II. SF calculations

$SF at A=+5.6 \mathrm{kN}$

$\mathrm{C}_{\mathrm{L}}=+5.6 \mathrm{kN}$

$\mathrm{C}_{\mathrm{R}}=5.6-5=0.6 \mathrm{kN}$

$\mathrm{D}_{\mathrm{L}}=+0.6 \mathrm{kN}$

$\mathrm{D}_{\mathrm{R}}=+0.6-7=-6.4 \mathrm{kN}$

$\mathrm{B}_{\mathrm{L}}=-6.4 \mathrm{kN}$

$\mathrm{B}=+6.4-6.4=0 \mathrm{kN}(\therefore \text { ok }$

B.M. calculation:-

B.M at A and B= 0 Since support A and B are simple.

B.M at C = 5.6 × 1.5=8.4 kN-m

B.M at D = 6.4× 1.5=9.6 kN-m