Q. Draw shear force and bending moment diagrams of simply supported beam loaded as shown in figure.

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Q. Draw shear force and bending moment diagrams of simply supported beam loaded as shown in figure.

 

SOLUTION:
Total load 15+10= 25 kN
To find the reactions
Taking moments about A
RB×6 = 15×2 + 10×4 =70
RB = 70/6=11.67KN
RA = 25 – 11.67 = 13.33

SFD (SHEAR FORCE DIAGRAM)
F{A}= 13,33kN (same in AC)
F{C}=( just left of C=13.33 kN)
F{C}=( just right of C)
= -1.67 kN (same in CD)
F{D}=( just left of D) = -1.67 kN
F{D}= just right of D)
= – 11.67KN ( SAME IN DB)
F{D}= – 11.67kN
Max +ve SF occurs in between AC = 13.33 kN Max – ve SF occurs in between DB= -11.67kN

BMD( BENDING MOMENT DIAGRAM)
M{A}=M{B}=0
M{C}=13.33 × 2 =26.66kNm
M{D}=11.67 ×2 =23.34 kNm
Max BM occurs at C (where SF changes its sign)
= Mmax = Mc = 26.66 kNm

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