
Problem 52
Calculate the
values and draw
the diagrams for shearing force and
bending moment for the following beam shown in figure
52(a).
Figure 52(a)
Solution:
The
given beam has two unknown reaction components which are calculated
in
Prob
41
as
A_{y} = 66 kN and B_{y}
= 24 kN;Shearing
force values:
Wherever there is a point
load we have to calculate the values of shearing force on both the
sides (left and right) of that section.
F_{A}left = 0
F_{A}right = 66 kN;
F_{C} = 66 – 20
x 4 = 14 kN;
F_{D}left = 66 – 20
x 4 = 14 kN;
F_{D}right = 66 – 20
x 4 – 10 = 24 kN;
F_{B}left = 24 kN;
F_{B}right = 0;
The values of shear force are plotted in SFD in
figure 52(b)
Bending moment values:
M_{A} = 0;
M_{C} = 66
x 4 – 20
x 4
x 2 = 264160 = 104 kNm;
M_{D} = 66
x 8 – 20
x 4
x 6 = 528 – 480 = 48 kNm;
M_{B} = 0;
The values of bending moment are plotted in BMD.in
figure 52(b) 
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Throughout the span
the bending moment is sagging in nature. To determine the
maximum bending moment for the beam we use the relationship between
shearing force and bending moment which says that the
rate of change of bending moment (dM/dx) is
equal to the shearing force (F_{x}) at the section.
For maximum value of any function its first derivative should be
equal to zero. Therefore at the point of maximum bending moment the
shearing force must be equal to zero. From the shear force diagram
it is clear that the shear force is zero between A and C. The point
of zero shearing force can be easily determined by using the
property of similar triangles.
Hence, 66/x =
14/(4x) ;
Which yields, x =
3.3m;
Now calculate the
bending moment at x = 3.3m which will be the maximum bending moment
for this beam. Therefore,
M_{max} = 66
x 3.3 20
x (3.3)
x (3.3/2) = 108.9 kNm;
More examples on shear and moment diagrams
Example 51
Example 53
Example 54
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(iv)
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