Shear force & Bending moment

1. Shear Force and Bending Moment
If a simply supported beam is loaded with vertical loads, it bends (i.e., deflects) due to the action
of the loads. The amount with which a beam bends, depends upon-
A) Type of the loads and its intensity,
B) Length of the beam,
C) Elasticity of the beam and
D) Type of the beam.
The behaviour and effect of transverse load of beam is studying by understanding the shear force
and bending moment diagrams of a beam.
In general, the beams are classified as under:
1. Cantilever beam,
2. Overhanging beam,
3. Simply supported beam,
4. Fixed beam and
5. Continuous beam.
Types of Loading
A beam may be subjected to either or in combination of the following types of loads:
1. Concentrated or point load,
2. Uniformly distributed load and
3. Uniformly varying load.
Shear Force and Bending Moment

2. Shear Force :The shear force (briefly written as S.F.) at the cross-section of a beam may be defined
as the unbalanced vertical force to the right or left of the section. therefore it tends to slide one
portion of the beam, upwards or downwards with respect to the other.
The shear force is said to be positive, at a section, when the left hand portion tends to slide
downwards or the right hand portion tends to slide upwards shown in Fig
+ve Shear Force -ve Shear Force
Left to right right to Left
Bending moment (B.M.): is the algebraic sum of the moments of the forces, to the right or left
of the section. At sections, sagging bending moment i.e , beam tend to bend in upward direction is
treated as +ve while Hogging B.M. i.e beam tend to bend in downward direction is considered as
-ve.
+ve Moment - ve Moment
NOTE. While calculating the shear force or bending moment at a section, the end reactions must
also be considered along with other external loads.
Shear Force and Bending Moment Diagrams
The shear force and bending moment can be calculated numerically at any particular section. But
sometimes, we are interested to know the manner, in which these values vary, along the length of
the beam. This can be done by plotting the shear force or the bending moment as ordinate and the

3. position of the cross as abscissa. These diagrams are very useful, as they give a clear picture of the
distribution of shear force and bending moment all along the length of beam.
Relation between Loading, Shear Force and Bending Moment
The following relations between loading, shear force and bending moment at a point or between
any two sections of a beam are:
1. If there is a point load at a section on the beam, then the shear force suddenly changes (i.e., the
shear force line is vertical). But the bending moment remains the same.
2. If there is no load between two points, then the shear force does not change (i.e., shear force
line is horizontal). But the bending moment changes linearly (i.e., bending moment line is an
inclined straight line).
3.If there is a uniformly distributed load between two points, then the shear force changes linearly
(i.e., shear force line is an inclined straight line). But the bending moment changes according to
the parabolic law. (i.e., bending moment line will be a parabola).
4. If there is a uniformly varying load between two points then the shear force changes according
to the parabolic law (i.e., shear force line will be a parabola). But the bending moment changes
according to the cubic law.
Shear Force at given section
= ∑ 𝑎𝑙𝑙 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑐𝑒𝑠 𝑏𝑦 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑙𝑒𝑓𝑡 𝑜𝑟 𝑟𝑖𝑔ℎ𝑡 𝑝𝑜𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
Bending Moment at given section
= ∑ 𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑎𝑙𝑙 𝑓𝑜𝑟𝑐𝑒𝑠 𝑏𝑦 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟𝑖𝑛𝑔 𝑙𝑒𝑓𝑡 𝑜𝑟 𝑟𝑖𝑔ℎ𝑡 𝑝𝑜𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑒𝑞𝑢𝑖𝑙𝑖𝑏𝑟𝑖𝑢𝑚
Cantilever with a Point Load
0

4. ∑ 𝐹𝑦 = 0, 𝑅𝑎 − 𝑊 = 0, 𝑅𝑎 = 𝑊 𝑘𝑁
(Vb)R = 0, (Vb)L = W kN
(Va)R = W kN, (Vb)L = W - W = 0
Moment = ∑ F X perpendicular distance
Mb = W X 0 = 0
Ma = −W X l = − Wl kNm
Example: Draw shear force and bending moment diagrams for a cantilever beam of span 1.5 m
carrying point loads as shown in Fig.(a).
SOLUTION.
∑ 𝐹𝑦 = 0, 𝑅𝑎 − 2.0 − 1.5 = 0, 𝑅𝑎 = 3.5 𝑘𝑁
(Vb)R = 0, (Vb)L = 1.5 kN
(Vc)R = 1.5 kN, (Vc)L = 1.5 + 2.0 = 3.5 kN
(Va)R = 3.5 kN, (Vb)L = 3.5 - 3.5 = 0
Moment = ∑ F X perpendicular distance
Mb = 1.5 X 0 = 0
Mc = −1.5 X 0.5 = − 0.75 kNm
Ma = −1.5 X 1.5 − 2 X 1 = − 4.25 kNm

5. Cantilever with UDL
∑ 𝐹𝑦 = 0, 𝑅𝑎 − 𝑊𝑙 = 0, 𝑅𝑎 = 𝑊𝑙 𝑘𝑁
(Vb)R = 0, (Vb)L = 0
(Va)R = Wl kN, (Vb)L = Wl - Ra = 0
Moment = ∑ F X perpendicular distance
Mb = W X 0 = 0
l
Ma = −W X l X
2
= −
Wl2
2
kNm
Example: A cantilever beam AB, 2 m long carries a uniformly distributed load of 1.5 kN/m over a
length of 1.6 m from the free end. Draw shear force and bending moment diagrams for the beam.
SOLUTION.
∑ 𝐹𝑦 = 0, 𝑅𝑎 − 1.5 𝑋 1.6 = 0, 𝑅𝑎 = 2.4 𝑘𝑁
(Vb)R = 0, (Vb)L = 0
(Vc)R = 1.5X1.6 = 2.4 kN, (Vb)L = 2.4kN
(Va)R = 2.4 kN, (Vb)L = 2.4 - Ra = 0
Moment = ∑ F X perpendicular distance
Mb = W X 0 = 0

6. l
Mc = −1.5 X l. 6 X
2
X1.6 = −1.92 kNm
l
Ma = −1.5 X l. 6 X (
2
X1.6 + 0.4) = −2.88 kNm
Ex:Acantilever beam of 1.5 m span is loaded as shown in Fig. 13.6 (a). Draw the shear force and
bending moment diagrams.
∑ 𝐹𝑦 = 0, 𝑅𝑎 − 1𝑋1.0 − 2 = 0, 𝑅𝑎 = 3 𝑘𝑁
(Vb)R = 0, (Vb)L = 2 kN
(Vc) = 2 kN as there is no point load
(Va) = 2 + 1X1 =3 kN
Moment = ∑ F X perpendicular distance
Mb = 2 X 0 = 0
Mc = −2 X 0.5 = −1.0 kNm
l
Ma = −2 x 1.5 − 1X 1.5 X (
2
X1.5) = −3.5kNm

7. Acantilever beam 4 m long carries a gradually varying load, zero at the free end to 3 kN/m at the
fixed end. Draw B.M. and S.F. diagrams for the beam.
∑ 𝐹𝑦 = 0, 𝑅𝑎 − 𝑋 𝑊𝑋𝑙 = 0,
1 1
2
𝑅𝑎 = − 𝑊𝑙 𝑘𝑁
2
as there is no point load
(Vb) = 0
a 2
(V ) = 1
𝑊𝑙 kN
BendingMoment
Mb = 0
l l l
Ma = (
2
XWl)X
3
Xl = −
6
Wl2

8. EXERCISE
1.A cantilever beam 2 m long carries a point load of 1.8 kN at its free end. Draw shear force and
bending moment diagrams for the cantilever. [Ans. Fmax = – 1.8 kN ; Mmax = – 3.6 kN-m]
2. A cantilever beam 1.5 m long carries point loads of 1 kN, 2 kN and 3 kN at 0.5 m, 1.0 m and
1.5 m from the fixed end respectively. Draw the shear force and bending moment diagrams for the
beam. [Ans. Fmax = – 6 kN ; Mmax = – 7 kN-m]
3. A cantilever beam of 1.4 m length carries a uniformly distributed load of 1.5 kN/m over its
entire length. Draw S.F. and B.M. diagrams for the cantilever. [Ans. Fmax = – 2.1 kN ; Mmax = –
1.47 kN-m]
4.A cantilever AB 1.8 m long carries a point load of 2.5 kN at its free end and a uniformly
distributed load of 1 kN/m from A to B. Draw the shear force the bending moment diagrams for
the beam. [Ans. Fmax = – 4.3 kN ; Mmax = – 6.12 kN-m]
5.A cantilever 1.5 m long is loaded with a uniformly distributed load of 2 kN/m and a point load
of 3 kN as shown in Fig.
Simply Supported Beam
1. Find support reactions, Ra and Rb by applying equilibrium equations
∑ 𝐹𝑦 = 0, 𝑎𝑛𝑑 ∑𝑀 = 0,
2. Find out S,V, at important points
(Va)L = 0 kN (Va)R = Ra =W/2
c L c R c L
2 2 2
(V ) = 𝑊
kN , (V ) = 𝑊
- W = (V ) = - 𝑊
kN

9. b L b R
2 2
(V ) = - 𝑊
kN , (V ) = - 𝑊
+ Rb = 0
3. Find Moment at important points
Ma = Mb = 0 as both are end supports
l W l Wl
Mc = Ra X
2
= (
2
X
2
) = −
4
4. Draw S.F. D and B.M.D.
A simply supported beam AB of span 2.5 m is carrying two point loads as shown in Fig

10. .

11. Asimply supported beam 6 m long is carrying a uniformly distributed load of 5 kN/m over a length
of 3 m from the right end. Draw the S.F. and B.M. diagrams for the beam and also calculate the
maximum B.M. on the section

12. Simply Supported Beam with a Triangle Load

13. 1.A simply supported beam of 3 m span carries two loads of 5 kN each at 1 m and 2 m from the
left hand support. Draw the shear force and bending moment diagrams for the beam. [Ans. Mmax
= 5 kN-m]
2.A simply supported beam of span 4.5 m carries a uniformly distributed load of 3.6 kN/m over a
length of 2 m from the left end A. Draw the shear force and bending moment diagrams for the
beam. [Ans. Mmax = 4.36 kN-m at 1.56 m fromA]
3.A simply supported beam ABCD is of 5 m span, such that AB = 2 m, BC = 1 m and CD = 2 m.
It is loaded with 5 kN/m over AB and 2 kN/m over CD. Draw shear force and bending moment
diagrams for the beam. [Ans. Mmax = 7.74 kN-m at 1.76 m from A]
4.Draw shear force and bending moment diagrams for a simply supported beam, loaded as shown
in Fig. 13.28. Fig. Find the position and value of the maximum bending moment that will occur in
the beam. [Ans. 3.47 kN-m at 1.3 m from C]

14. Point of Contraflexure
We have already discussed in the previous article that an overhanging beam is analysed as a
combination of simply supported beam and a cantilever. In the previous examples, we have seen
that the bending moment in a cantilever is negative, whereas that in a simply supported beam is
positive. It is thus obvious that in an overhanging beam, there will be a point, where the bending
moment will change sign from negative to positive or vice versa. Such a point, where the bending
moment changes sign, is known as a point of contraflexure.

16. Y = 0.33m

22. Example: Fig. shows the SFD of loaded beam. Find the Loading on the beam and also draw BMD.
Example: Fig. shows the SFD of loaded beam. Find the Loading on the beam and draw BMD. Also finfd
point of contraflexure if any.

24. A beam 6 m long rests on two supports 5 m apart. The right end is overhanging by 1 m. The beam carries
a uniformly distributed load of 1.5 kN/m over the entire length of the beam. Draw S.F. and B.M. diagram
and find the amount and position of maximum bending moment. [Ans. 4.32 kN-m at 2.4 m from left end]
2. Draw the shear force and bending moment diagrams, for the overhanging beam carrying loads as
shown in Fig. Also locate the point of contraflexure if any.
3. Draw the shear force and bending moment diagrams, for the overhanging beam carrying loads as
shown in Fig. Also locate the point of contraflexure if any.