Problem 54
Calculate the
values and draw
the diagrams for shearing force and
bending moment for the following
overhanging beam shown in figure
54(a) and find the position of point of contraflexure, if any.
Figure 54(a)
Solution:
Calculation of support reactions
The beam has three reaction components; A_{x}
, A_{y} and C_{y}. The freebodydiagram is shown in
figure 54(b).
Applying the equations of
static equilibrium
we get;
ΣF_{x} = 0; A_{x} =
0;
(eq. 1)
ΣF_{y} = 0; A_{y} +
C_{y} –10x4
– 20 = 0;
A_{y} + C_{y}
= 60 kN;
(eq. 2)
Considering zaxis passing through A, and
taking moment of all the forces about zaxis (taking clockwise –ve
and anticlockwise +ve);
ΣM_{z} = 0; C_{y}
x 8 – 20x10 – 10x4x2
= 0; (eq. 3)
Solving eq. 3, we get C_{y}
= 280/8 = 35 kN;
Substituting the value of C_{y}
in eq. 2, we get A_{y} = 25 kN;
Shear force calculations
In case of point load acting at a point, we
should calculate shear force on both sides (left and right) of the
point.
F_{A}left = 0;
F_{A}right = 25 kN;
F_{B} = 25 – 10x4
= 15 kN;
F_{C}left = 15 kN;
F_{C}right = 25 – 40 + 35 = 20 kN;
F_{D}left = 20 kN;
F_{D}right = 20 – 20 = 0;
The values of shear force are plotted in SFD
in figure 54(b)
Bending Moment Calculations
M_{A} = 0;
M_{B} = 25x4
– 10x4x2
= 20 kNm;
M_{C} = 25x8
–10x4x6
= – 40 kNm;
The values of bending moment are plotted in BMD
in figure 54(b) 
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