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Mechanics
Module VII: Analysis of Beams
Lesson 35: Shear Force and Bending Moment - II
Figure 1:
Consider a beam with a transverse loading w(x), as shown in Fig. 1. We
consider a small element of length x as shown. From force equilibrium in
the transverse direction
V (x +x) V (x) + w(x)x = 0 dVdx
= w(x) taking x 0.
Thus, the shear force distribution is obtained by integrating this differential
-
equation as
V (x) = V (x0)
x
x0
w(x)dx.
We observe that, if w(x) = 0, dVdx
= 0, i.e., the shear force V (x) will remain
(locally) unchanged (constant/local extremum). Thus, the slope of the SFD
reflects the load distribution on the beam.
The moment equilibrium of the small element in Fig. 1 in the limit x 0leads to
dM
dx= V (x).
Thus,
M(x) = M(x0)
x
x0
V (x)dx.
Here, we observe that, when V (x) = 0, dMdx
= 0, i.e., the bending moment
remains (locally) unchanged (constant/local extremum). The slope of the
BMD is the negative of the shear force value at a cross-section.
Combining the differential equations for the shear force and bending mo-
ment, we have
d2M
dx2= w(x).
It may be noted that the differential equations for shear force and bending
moment are all linear. In certain cases, one can use the principle of superposi-
tion to solve for V (x) and M(x) for complex loadings involving concentrated
and distributed forces on the beam.
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Figure 2:
Problem 1
Determine the shear force and bending moment distributions and draw the
SFD and BMD of the simply supported beam loaded by the force distribution
as shown in Fig. 2.
Solution
The FBD of the beam with equivalent loading is shown in Fig. 3. The
equivalent concentrated force magnitude is easily obtained as P = qL/2,
which acts at the centroid of the distribution as shown. Using the equations
of equilibrium RA = qL/6 and RB = qL/3.
Now, for shear force distribution, we consider the FBD with the distributed
force as shown in Fig. 4. The shear force distribution is given by
dV
dx= w(x) = q
Lx V (x) = qx
2
2L+ D1
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Figure 3:
Figure 4:
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Now, V (0) = qL/6, implying D1 = qL/6. Hence,
V (x) =qx2
2L qL
6.
It may be noted that dVdx|x=0 = 0, and V (L/
3) = 0.
Figure 5:
The bending moment distribution is obtained as
d2M
dx2= w(x) = q
Lx
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M(x) = qx3
6L+ C1x + C2
Now, M(0) = 0 C2 = 0, and M(L) = 0; C1 = qL/6. Hence,
M(x) = qx3
6L+
qL
6x.
The SFD and BMD are presented in Fig. 5.
Figure 6:
Problem 2
The free end of a cantilever beam is hinged to a beam with a simply supported
end and loaded, as shown in Fig. 6. Determine the shear force and bending
moment distributions, and draw the SFD and BMD.
Solution
The composite beam is an indeterminate structure. So, we consider the
FBD of the individual beams, as shown in Fig. 7. From the equations of
equilibrium, R = F = 10 kN, V0 = 42 kN, and M0 = 208 kNm. In the
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Figure 7:
cantilever beam span (I), the shear force distribution is obtained as
dV
dx= 4 kN/m V (x) = 4x+ D1
Now, V (0) = 42 kN, implying D1 = 42 kN. Hence,
V (x) = 4x 42 kN
The bending moment distribution in (I) is obtained as
dM
dx= V (x) = 4x + 42 kN
M(x) = 2x2 + 42x+ C1
Now, M(0) = 208 kNm, implying
M(x) = 2x2 + 42x 208 kNm
In section (II), the shear force and bending moment distributions can be
obtained in a straightforward manner. The SFD and BMD are presented in
Fig. 8.
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Figure 8:
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Figure 9:
Problem 3
A thin and narrow 9 m long barge floating in water is loaded appropriately
to keep it horizontal, as shown in Fig. 9. Treating the barge as a beam,
determine the shear force and bending moment distributions, and draw the
SFD and BMD.
Solution
We assume that the barge is uniformly supported from below due to the
buoyancy force. The uniform support force distribution is obtained as
w =Total load
length of barge=
10
3kN/m.
The FBD of the barge is shown in Fig. 10. We divide the barge into four
sections, as shown.
Section I:
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Figure 10:
dVIdx
= 103
kN/m VI(x) = 103
x + D1
VI(0) = 0, implies
VI(x) = 103
x kN
dMIdx
= VI(x) = 103
x kN MI(x) = 53x2 + C1 kNm
MI(0) = 0, implies
MI(x) = 53x2 kNm
Section II:
dVIIdx
= (10
3 5
)=
5
3kN/m VII(x) = 5
3x +D1
VII(5/3) = VI(5/3) = 5 kN, implying
VII(x) =5
3x 15
2kN
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dMIIdx
= VII(x) = 53x +
15
2kN MII(x) = 5
6x2 +
15
2x + C1 kNm
MII(3/2) = MI(3/2) = 15/2 kNm, implying
MII(x) = 56x2 +
15
2x 15
8kNm
One can continue this and obtain the shear force and bending moment
distributions as shown in Fig. 11.
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Figure 11:
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