This document discusses bending moment and shear force for beams. It contains 3 main sections: 1) An introduction to bending moment, shear force, and the relationship between loading, shear force and bending moment. 2) How to draw shear force diagrams and bending moment diagrams by calculating shear forces and bending moments at critical points along a beam. 3) How to calculate reactions for simply supported beams and cantilever beams by applying equations of equilibrium. Several examples of calculating reactions are provided.

1. BENDING
MOMENT AND
SHEAR FORCE
[16 MARKS]
Chapter 2
Chapter Details
2.1 Introduction
2.2 Shear Force Diagram (SFD) and
Bending Moment Diagram (BMD)
2.3 Reaction Calculation

2. 2-2 Strength of Materials
Graphical and Statistical representation of questions asked from this chapter
in previous years MSBTE Question Papers
16
14
12
0
2
4
6
8
10
12
14
16
TotalMarks
W-2009 S-2010 W-2010
Years
Graphical Representation
Statistical Analysis
MSBTE paper Total marks questions asked in this chapter
W-2009 16
S-2010 14
W-2010 12

3. 2-3Bending Moment & Shear Force
2.1 INTRODUCTION
Beam is a horizontal member in the structure. Due to
load on it, beam suffers bending, shear torsion etc.
Beam is generally designed for bending and checks for
shear. We have already discussed concept of bending
stresses and shear stresses.
Due to bending, beam suffers tension as well as
compression at extreme fibres according to the type of
beam.
Shear failure is sudden failure. It does not give any
prior warning before failure.
Shear force acts tangential to the cross-section.
Shearing action and bending action occurs due to load
at the same time.
Hence there is a relation between loading rate,
shearing force and bending moment.
Types of Load or Loading on a Beam
Practically beam suffers from various types of load:
a. concentrated load/point load
b. uniformly distributed load (udl)
c. uniformly varying load (uvl)
Among these, we are learning concentrated loads and
uniformly distributed loads.
a. Concentrated load / point load: A load acting at
a point on the beam is known as a concentrated
load or point load.
In practice, a load distributed over small area is
taken as concentrated load.
e.g., column
RA RB
W1 W2
L
C D
A B
Figure 2.1: Simply supported Beam subjected to point loads
S.S.B
Load
S.S. beam: Simply supported beam.
(Applied mechanics → equilibrium)
Compression
Tension
Loading
Cantilever
Tension
Compression
W
Shear on beam
acting
Bending action
on beam
Wall
Beam
Even this portion of wall on beam is
considered as concentrated load.

4. 2-4 Strength of Materials
Conversion of udl into point load
L
w/m
L
L/2
W L×
W1 W2 W3 W4
C
RA RB
C CLeft
part
of beam
Right
part
of beam
RA RB
w1 w2 w3 w4
SFc = ΣVLc = ΣVRc
SFc = RA – W1 – W2
SFc = –RB – W3 + W4
ΣVL = Summation of vertical
forces on left hand part of beam
ΣVR = Summation of vertical
forces on right part of beam.
b. Uniformly distributed load (udl): A load which is
spread up uniformly on the beam is known as a
uniformly distributed load or rectangular load and is
written as u.d.l.
e.g., self weight of a beam, or load from the floor or a
slab on the beam.
For convenience u.d.l. is converted into its equivalent
point load which is assumed to act at the centre of
gravity of the load.
Shear Force (SF)
Shear force at any cross-section of the beam is the
algebraic sum of all vertical forces on the beam, acting on
the right or left side of the section.
Sign convention for the shear force
L R
X
X
Positive
shear
L R
X
X
Negative
shear
Figure 2.2
i.e. If we consider only left part of a beam upward force is
to be considered positive (+ve) and downward force is to
be considered negative (–ve). While if right part of beam is
considered, downward force is to be considered positive
(+ve) while upward force is to be considered negative
(–ve).

5. 2-5Bending Moment & Shear Force
Bending Moment (BM)
Bending moment at any cross-section of the beam, is the
algebraic sum of the moments of all the forces acting on the
right or left side of the section.
Sign convention for the beam:
L R
X
X
Positive
BM
L R
X
X
Negative
BM
Figure 2.3
Clockwise moment to the left of the section and
anticlockwise moment on the right of the section, is taken as
positive. (i.e. Sagging BM is considered positive).
Sagging from both sides i.e. L or R; upward force will
produce +ve BM.
Anticlockwise moment to the left of the section and
clockwise moment on the right of the section is taken as
negative. (i.e. Hogging BM is consider negative).
From both side i.e. L or R down word force will produce
–ve BM.
BM = ΣWx
x2
x1
RA
Left part of beam
W
BM = R x – W xC A 1 1 2
C
L R
+ve
Sagging
From both sides i.e. L or R;
upward force will produce +ve
BM and from both sides
i.e. L or R downward force will
produce –ve BM.
L R
–ve
Hogging
Relation between Loading Shear force and Bending Moment
Relation between Rate of Loading and Shear Force
Shear force is a function of load. Hence rate of change of
shear force with respect to the distance, is equal to the
intensity of loading.
dF
dx
= W
BTE W.2010 – 2M
State the relation
between rate of loading,
shear force and bending
moment.

6. 2-6 Strength of Materials
Relation between Bending Moment and Shear Force
dM
dx
= F
The rate of change of bending moment at any section, is equal to the shear force at that
section.
If
dM
dx
= 0, ie. shear force = 0,
The bending moment (M) will be maximum i.e. The point at which Shear Force
(SF)changes its value from +ve to –ve, at that point bending moment value is maximum;
and called as maximum bending moment point or contrashear point.
2.2 SHEAR FORCE DIAGRAM (SFD) AND BENDING
MOMENT DIAGRAM (BMD)
A shear force diagram is that which shows the variation of shear force along the length of
the beam. A bending moment diagram shows the variation of bending moment along the
length of the beam.
There is a relation between rate of loading, shear force and bending moment. Hence, we
draw loaded beam, shear force diagram and bending moment diagram one below the
other as;
Loaded beam
then
Shear force diagram
then
Bending moment diagram
While drawing SFD and BMD, following points are to be
considered:
a. Base of SFD and BMD is equal to the span of beam.
(i.e. = L)
b. Positive values of SF and BM are plotted above the
base line and negative values of SF and BM are
plotted below the base line.
c. The SF and BM must be calculated for all the critical
points and written near the respective co-ordinates.
L
+ve SF
base of SFD
+ve BM
base of BMD
– ve SF
– ve BM

7. 2-7Bending Moment & Shear Force
Critical points are as follows:
i. A point at which point load acts.
ii. A point at which udl starts or ends.
iii. A point at which SF value changes from positive to negative.( max. BM point)
i.e. +ve called –ve contrashear point or max BM point.
d. Location of zero bending moment point i.e. point of contraflexure must be found
out.
2.3 REACTION CALCULATION
While calculating SFE at various points – first step is to calculate reactions.
Reaction Calculation for Simply Supported Beam
We consider that the beam is statistically determinant, hence it follows conditions of
equilibrium.
Σ Fx = 0
Σ Fy = 0
Σ M = 0
Let us consider, simply supported beam
W1 W2 W3
x1
x2
L
x3RA RB
A B
Figure 2.4
Σ Fy = 0
RA + RB = W1 + W2 + W3 = ΣW
∴ RA = ΣW – RB ...............................................................................................................(1)
If we consider ΣFy = 0
i.e. Σ ↑ forces =Σ ↓ forces
RB =
ΣWx
L

8. 2-8 Strength of Materials
As beam is in equilibrium ΣMA = 0
ΣMA = 0
+ve – ve
∴ 0 = W1x1 + W2x2 + W2x3 – RBL
∴ RBL = W1x1 + W2x2 + W3x3
+ve – ve
∴ RB =
ΣWx
L .....................................................................................................................(2)
We can consider as formula to calculate RB. then by using.
RA = ΣW – RB
We can calculate value of RA.
Solved
Reaction Calculation Practice Example
1. Calculate reactions of following.
10N 20N 30N
RA RB
2m 2m 2m 2m
Solution
By using,
RB =
ΣWx
L =
10 × 2 + 20 × 4 + 30 × 6
8
=
280
8 = 35 N
RA = ΣW – RB = (10 + 20 + 30) – 35
= 25 N
RA = 25 N
RB = 35 N
10N 20N 30N
2m
4m
8m
6m
RA RB

9. 2-9Bending Moment & Shear Force
2. Calculate reaction for given beam.
10N/m
20N 5N
4m 2m 1m
DC
A B
Solution
40N/m
20N
5N
4m
6m
7m
2m
DCRA RB
RB =
ΣWx
L =
40 × 2 + 20 × 4 + 5 × 6
7
= 27.14 N
RA = (40 + 20 + 5) – RB = 65 – 27.14
= 37.86 N
10 4×
4 m
10 N/m
2 m 2 m
CG
udl is converted into
equivalent point load whose
rate is 10 N/m and considered
4m = 10 N/m × 4m = 50 N.
Action at center of udl pass.
3. Calculate reactions for given beam.
15N
5N/m
5N
1m 4m 1m
A B
DC
Solution
RB =
ΣWx
L
=
15 × 1 + 30 × 3 – 5 × 5
6
= 13.33 N
RA = ΣW – RB = (15 + 30 + 5)–(13.33)
= 36.67 N
15N 30N 5N
1m
DC
3m
5m
6m
RA RB

10. 2-10 Strength of Materials
Reaction Calculation for Cantilever
Only one support will bear total load on the beam hence in cantilever:
R = ΣW
as ΣFy = 0
↑ = ↓
4. Find reaction at cantilever support.
10N
2m 2m 2m
A
RA
20N 10N
Solution
↑ = ↓
RA = 10 + 20 + 10
∴ RA = 40N
5. Find reaction at cantilever support
60N
2m 2m 1.5 m
RA
20N 5N
Solution
↑ = ↓
RA + 5 = 60 + 20
RA = 60 + 20 – 5
RA = 75 N
6. Find reaction at cantilever support
10N/m
5m
A B
10N
Solution
↑ = ↓
RA = 50 + 10
RA = 60 N
50N 10N
2.5m
RA
5m