A simply supported beam of span 5 m carries two point loads of 5kN and 7 kN at 1.5 m and 3.5 m from the left hand support respectively. Draw S.F.D. and B.M.D. showing important values.

 A simply supported beam of span 5 m carries two point loads of 5kN and 7 kN at 1.5 m and 3.5 m from the left hand support respectively. Draw S.F.D. and B.M.D. showing important values.

solution:

I. Support Reactions:

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

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

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

RB=6.4kN${\mathrm{R}}_{\mathrm{B}}=6.4\mathrm{k}\mathrm{N}$

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

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

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

RA=5.6kN${\mathrm{R}}_{\mathrm{A}}=5.6\mathrm{k}\mathrm{N}$

$\mathrm{<br>}$

II. SF calculations

SFatA=+5.6kN$SFatA=+5.6\mathrm{k}\mathrm{N}$

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

CR=5.6−5=0.6kN${\mathrm{C}}_{\mathrm{R}}=5.6-5=0.6\mathrm{k}\mathrm{N}$

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

DR=+0.6−7=−6.4kN${\mathrm{D}}_{\mathrm{R}}=+0.6-7=-6.4\mathrm{k}\mathrm{N}$

BL=−6.4kN${\mathrm{B}}_{\mathrm{L}}=-6.4\mathrm{k}\mathrm{N}$

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

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