Solved Examples Based On SFD And BMD
Example: A simply supported beam overhanging on one side is subjected to a U.D.L. of 1 kN/m. Sketch the shear force and bending moment diagrams and find the position of point of contraflexure.
Solution: Consider a section (X – X’) at a distance x from end C of the beam.
To draw the SFD and BMD we need RA and RB.
By taking moment of all the forces about point A.
RB × 3 – w/2 × (4)2 = 0
RB = 1 × (4)2 / 2 × 3 = 8/3 kN
For static equilibrium;
RA + RB – 1 × 4 = 0
RA = 4 – 8/3 = 4/3 kN
To draw SFD we need S.F. at all salient points:
Taking a section between C and B, SF at a distance x from end C. we have,
Fx = + w.x kN
At x = 0; FC = 0
x = 1 m; FB just right = 1 × 1 = + 1 kN
Now, taking section between B and A, at a distance x from end C, the SF is:
Fx = x – 8/3
When, x = 1 m; FB = 1 – 8/3 = –5/3 kN = –1.67 kN
At x = 4 m; FA = 4 – 8/3 = + 4/3 kN = + 1.33 kN
The S.F. becomes zero;
Fx = x – 8/3 = 0
⇒ x = 2.67 m
(The sign is taken positive taken when the resultant force is in downward directonon the RHS of the section).
To draw BMD we need B.M. at all salient points. Taking section between C and B, BM at a distance x from end C, we have
Mx = – wx2 / 2 = –1.x2/2 kN m
When x = 0, MC = 0
At x = 1 m. MB = –1 × (1)2 / 2 = –0.5 kN m
Taking section between B and A, at a distance x from C, the bending moment is:
Mx = –x2/2 + 8/3 (x – 1)
At x = 1 m, MB = – 0.5 kN m
x = 4 m; MA = –(4)2 /2 – 8/3 (4 – 1) = 0
The maximum bending moment occurs at a point where
dMx / dx = 0
⇒ d/dx [–x2 / 2 + 8/3 x – 8/3] = 0
⇒ –1/2 × 2x + 8/3 = 0
⇒ x = 8/3 m from end C.
⇒ Mmax = –1/2 (8/3)2 + 8/3 (8/3 – 1) = 0.89 kN m
The point of contraflexture occurs at a point, where
Mx = 0
⇒ –x2/2 + 8/3 (x – 1) = 0
⇒ x2 = 16/3 (x – 1)
⇒ x2 – 16/3 x + 16/3 = 0
x = 1.335 m or 4 m
The SFD and BMD can be drawn by joining the ordinates at salient point by straight line or smooth curves, as defined by governing equations.
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