Chapter 5 Bending Moments and Shear Force
Diagrams for Beams
In addition to axially loaded bars/rods (e.g. truss) and torsional shafts, the structural members may
experience some loads perpendicular to the axis of the beam and will case only shear and bending
in the beam. The current chapter together with Chapters 6 to 8 will focus on such an issue.
5.0 SHEAR FORCE AND BENDING MOMENTS DIAGRAMS FOR BEAMS
A Shear Force Diagram (SFD)indicates how a force applied perpendicular to the axis (i.e.
parallel to cross section) of a beam is transmitted along the length of that beam. A Bending
Moment Diagram (BMD) will show how the applied loads to a beam create a moment
variation along the length of the beam. These diagrams are used to determine the normal and
shear stresses as well as deflection and slopes in the following chapters.
5.1 BEAM SIGN CONVENTION (SI&4th:256-257; 5th:256-257)
At any point along its length, a beam can transmit a bending moment M(x) and a shear force
V(x). If a loaded beam is cut, the definitions of a positive distributed load, shear force and
positive bending moment are as Fig. 5.1 below:
Positive internal
bending moment
Positive internal shear force
Positive distributed load
Fig. 5.1 Beam shear force and bending moment sign convention
Where distributed load acts downward on the beam; internal shear force causes a clockwise
rotation of the beam segment on which it acts; and the internal moment causes compression
in the top fibers of the segment, or to bend the segment so that it holds water.
5.2 RELATIONSHIP BETWEEN BEAM LOADINGS (SI&4th:264-268; 5th:264268)
A beam (Fig. 5.2) is loaded with vertical forces Fi, bending moments Mi and distributed loads
w(x).
F1
F2
w(x)
M1
M2
x
dx
F.B.D. of element dx
w(x)
M(x)
M(x)+
dM(x)
dx
dx
V(x)
V(x)+
dV(x)
dx
dx
dx
Fig. 5.2 Transversely loaded beam and free body diagram of element dx
Lecture Notes of Mechanics of Solids, Chapter 5
1
Look at the FBD of an elemental length dx of the above loaded beam (Fig. 5.2). As it has an
infinitesimal length, the distributed load can be considered as a Uniformly Distributed Load
(UDL) with constant magnitude w(x) over the differential length dx.
It is now necessary to equate the equilibrium of the element. Starting with vertical equilibrium
+↑
∑F
y
dV (x ) 

= 0 = V ( x ) − w(x )dx − V (x ) +
dx  = 0
dx


(5.1)
Dividing by dx in the limit as dx→0,
dV (x )
= − w(x )
(5.2)
dx
Taking moments about the right hand edge of the element:
dM ( x ) 
dx 
+ ∑ M R .H .Edge = 0 = − M ( x ) − V ( x )dx + w( x )dx
dx  = 0 (5.3)
+  M (x ) +
2 
dx

Dividing by dx in the limit as dx→0,
dM ( x )
= V (x )
(5.4)
dx
Eqs. (5.2) & (5.3) are important when we have found one and want to determine the others.
5.3 BENDING MOMENT AND SHEAR FORCE EQUATIONS
Introductionary Example - Simply Supported Beam
By using the free body diagram technique, the bending moment and shear force distributions
can be calculated along the length of the beam. Let’s take a simply supported beam, Fig. (5.3),
as an example to shown the solutions:
F.B.D. (global equilibrium)
P
a
I
F.B.D. (method of section I-I)
II
A
M(x)
B
I
II
L
RAY=(1-a/L)P
RBY=Pa/L
A
o
V(x)
x
RAY=(1-a/L)P
Fig. 5.3 FBD of beam cut before force P
Step A: Cut beam just before the force P (i.e. Section I-I), and draw a free body diagram
including the unknown shear force and bending moment as in Fig. 5.3.
Take moments about the right hand end (O):
a
 a

+ ∑ M o = 0 = − P1 −  x + M ( x ) = 0 → M ( x ) = P 1 −  x
L
 L

To determine the shear force, use Eq. (5.4), giving that:
dM ( x ) 
a
V (x ) =
= 1 −  P
dx
L

To verify Eq. (5.6), equate vertical equilibrium:
(5.5)
(5.6)
Lecture Notes of Mechanics of Solids, Chapter 5
2
+↑
∑F
y
a

 a
= P  1 −  − V ( x ) = 0 → V ( x ) = P 1 − 
L
 L

which is the same equation as (5.6). These then, are the equations for the bending moment and
shear force variation in the range of 0 ≤ x ≤ a .
To find out the rest of the bending moment and shear force distributions, it is necessary to
now carry out a similar analysis, but cutting the beam just before the end (Section II-II).
Step B: Cut beam just before the right hand end (RHE)
F.B.D. (Section II-II)
P
a
A
II
o
II
M(x)
V(x)
x
RAY=( 1-a/L)P
Fig. 5.4 FBD of beam cut before the right hand end
Equate moments about the right side:
+
∑M
0
 a
= 0 = − P 1 −  x + P ( x − a ) + M ( x ) = 0
 L
giving:
a
 a
M (x ) = P1 −  x − P(x − a ) = − Px + Pa
L
 L
(5.7)
and using Eq. (5.4), the shear force equation is :
dM ( x )
a
V (x ) =
=− P
(5.8)
dx
L
These expressions for the bending moment and shear force can now be plotted against x to
produce the Shear Force and Bending Moment Diagrams as Fig. 5.5:
Lecture Notes of Mechanics of Solids, Chapter 5
3
P
a
Loading Diagram
(1-a/L)P
L
Pa/L
V(x)
(1-a/L)P
+ve
Shear Force Diagram
x
-ve
M(x)
-a/LP
(1-a/L)Pa
+ve
Bending Moment Diagram
x
Fig. 5.5 Shear Force and Bending Moment Diagrams for simply supported beam
Macauley's Notation (4th:590-599; 5th:590-599)
The two sets of equations for V(x) and M(x), Eqs. (5.5), (5.6), (5.7) and (5.8), can be condensed
to just one set of equations if we use a special type of notation called Macauley's Notation.
The above equations would look like this (to be derived in Example 5.0)
a 0
0

V ( x ) = P 1 −  x − P x − a
(5.9)
L

a 1
1

M ( x ) = P 1 −  x − P x − a
L

Where the notation has the following meaning:
x−a
n
for x < a
 0
=
n
(x − a ) for x ≥ a
(n ≥ 0)
(5.10)
(5.11)
when differentiating:
∂ x−a
∂x
n
n x − a

= x−a

0

n −1
for n ≥ 1
0
for n = 1
(5.12)
for n = 0
Remarks
To derive the bending moment equation by using Macauley's notation, you may need to do
the following:
1) Determine the ground reactions from global equilibrium;
2) Cut the beam just before the right hand end;
Lecture Notes of Mechanics of Solids, Chapter 5
4
3) Equate the cut FBD to equilibrium about the right hand end;
4) All length terms in the bending moment/shear force equations MUST be written using
Macauley's notation;
5) Always indicate the powers, even if they are 0 or 1.
Example 5.0: As in the introductory example, determine the shear force and bending moment
equations and plot them for a simply-supported beam as in the introductory example.
Step 1: Determine the ground reactions;
F.B.D. (global equilibrium)
P
a
I
A
B
I
L
RAY=(1-a/L)P
RBY=Pa/L
We have
RAY = (1-a/L)P and RBY = a/LP
Step 2: Draw FBD of beam cut just before the RHS (Section I-I).
F.B.D. (Section I-I)
P
a
A
I
o
I
M(x)
V(x)
x
RAY=( 1-a/L)P
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).
Take moments about RHS:
a 1
1

+ ∑ M O = 0 = − P1 −  x + P x − a + M ( x ) = 0
L

a 1
1

M ( x ) = P 1 −  x − P x − a
L

and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as:
dM ( x )
a 0
0

V (x ) =
= P 1 −  x − P x − a
dx
L

Step 4: Plotting the Shear Force and Bending Moment Diagrams
Lecture Notes of Mechanics of Solids, Chapter 5
5
According to M(x) and V(x) to depict the diagrams, Look at the equations segment by segment
When 0 ≤ x ≤ a
a 1
a
a 0
a




M ( x ) = P1 − ( x ) − P × 0 = P1 −  x and V ( x ) = P1 − ( x ) − P × 0 = P1 − 
L
L
L
L




To plot this segment in the diagram, firstly look at the boundary points as x = 0, M ( x) = 0
and x = a , M ( x) = Pa(1 − a / L ) . Draw two points and then connect them because the equation
gives a line. Likewise, one can plot Shear Force Diagram in this region.
When a ≤ x ≤ L
a 1
a
a


1
M ( x ) = P1 − ( x ) − P × ( x − a ) = P1 −  x − Px + Pa = Pa − P x
L
L
L


a 0
a
a


0
and V ( x ) = P1 − ( x ) − P × ( x − a ) = P1 −  − P = − P
L
L
L


Remarks: Please draw global FBD of the beam firstly and follow by its Sear Force and Bending
Moment Diagrams. The reason for doing this is that when you get sufficient experience, you may
be able to directly plot the Shear Force Diagram by observing the external forces as well as plot
Bending Moment Diagrams by observing the Shear Force Diagram. Nevertheless you MUST still
work out and indicate the locations and values (including +ve or –ve) at all turning points in the
diagrams in detail.
It is also interesting to note that concentrated forces (e.g. reaction forces and external forces)
correspond to inclined line in BMD and horizontal line in SFD.
P
a
Loading Diagram
(1-a/L)P
L
Pa/L
V(x)
(1-a/L)P
+ve
Shear Force Diagram
x
-ve
M(x)
Bending Moment Diagram
-a/LP
(1-a/L)Pa
+ve
x
Example 5.1: Determine the shear force and bending moment equations and plot them for a
simply-supported beam loaded with a UDL.
Lecture Notes of Mechanics of Solids, Chapter 5
6
Step 1: Determine the ground reactions;
From global equilibrium the ground reaction forces can be found to be both equal to wL/2 as,
w
F.B.D. (global equilibrium)
I
A
B
I
L
wL/2
wL/2
L
∑ M = 0 = − R L + (wL ) 2  = 0 →∴ R = wL / 2 (+ upwards)
+ ↑ ∑ F = 0 = R + R − wL = 0 →∴ R = wL / 2 (+ upwards)
+
B
AY
y
AY
AY
BY
BY
Step 2: Draw FBD of beam cut just before the RHS (Section I-I).
Rw
F.B.D. (Section I-I)
w
x/2
M(x)
A
o
V(x)
x
RAY=wL/2
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).
As far as V(x) and M(x) are concerned the UDL can be temporarily replaced by its resultant Rw
(=wx) applied at the centroid of the UDL distribution in the moment equilibrium equation. So if
we take moments about the RHS of the beam we get:
1
1
x
x
1
1
1
+
∑ M O = 0 = − R AY x + Rw 2 + M (x ) = −(wL / 2) x + w x 2 + M (x ) = 0
wL 1 w 2
∴ M (x ) =
x −
x
2
2
and differentiating w.r.t. 'x', as Eq. (5.4), gives :
dM ( x ) wL
0
1
V (x ) =
=
x −w x
dx
2
(
)
Step 4: Plotting the Shear Force and Bending Moment Diagrams
According to M(x) and V(x) to depict the diagrams
Lecture Notes of Mechanics of Solids, Chapter 5
7
w
A
Loading Diagram
B
L
wL/2
wL/2
V(x)
wL/2
Shear Force Diagram
+ve
x
-ve
-wL/2
M(x)
wL2/8
Parabola
+ve
Bending Moment Diagram
x
It is worth pointing out that one should not completely replace such a UDL by its
corresponding resultant concentrated force Rw (=wx) in the beginning of the solution. There is
a significant difference of the Shear Force and Bending Moment Diagrams between a
concentrated force (Example 5.0) and a UDL (Example 5.1). It is also interesting to note that
the UDL corresponds to an inclined line in the Shear Force Diagram and a quadratic curve
(parabola) in the Bending Moment Diagrams.
Lecture Notes of Mechanics of Solids, Chapter 5
8
Example 5.2: Determine the shear force and bending moment equations and plot them for a
beam loaded with a UDL between A and B and two concentrated forces at C and E.
F.B.D. (global equilibrium)
20kN
UDL=w=1kN/m
B
A
10m
RAY
I
D
C
5m
5m
10kN
E
RDY
I
10m
Step 1: Determine the ground reactions;
∑M
+↑∑F
+
A
y
= 0 = −1 × 10 × 5 − 20 × 15 + R DY × 20 − 10 × 30 = 0 →∴ R DY = 32.5kN
= 0 = R AY − 1 × 10 − 20 + R DY − 10 = 0 →∴ R AY = 7.5kN
Step 2: Draw FBD of beam cut just before the RHS (Section I-I).
Note: The only problem with Macauley's Notation is that it does not work when a UDL stops.
It however does work for a UDL which starts anywhere along a beam and continues to the
end. The problem can be corrected by applying a UDL of equal magnitude but opposite sense
where the first UDL ends.
F.B.D. (Section I-I) and application of equivalent UDL
20kN
w=1kN/m
M(x)
o
A
V(x)
32.5kN
7.5kN
x
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).
Take moments about RHS:
+
∑M
O
= 0 = −7.5 x
1
+1
1
M ( x ) = 7.5 x − x
2
1
2
x
2
2
−1
x − 10
1
+ x − 10
2
2
2
2
+ 20 x − 15 − 32.5 x − 20 + M (x )
1
1
− 20 x − 15 + 32.5 x − 20
1
1
and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as:
dM ( x )
0
1
1
0
V (x ) =
= 7.5 x − x + x − 10 − 20 x − 15 + 32.5 x − 20
dx
0
Lecture Notes of Mechanics of Solids, Chapter 5
9
Step 4: Plotting the Shear Force and Bending Moment Diagrams
UDL=w=1kN/m
A
Loading
Diagram
7.5kN
20kN
B
10m
5m
10kN
D
C
5m
E
32.5kN
V(x) kN
7.5
Shear
Force
Diagram
10m
10
+ve
+ve
x
-2.5
-ve
-22.5
M(x) kNm
quadratic
28.125
+ve
25
12.5
x
Bending
Moment
Diagram
-ve
-100
Again, the UDL segment corresponds to an inclined line in SFD and a quadratic curve in
BMD.
Example 5.3: Determine the shear force and bending moment equations and plot them for a
cantilever beam loaded with a moment MB = 40kNm and a force F= 10kN.
Global F.B.D.
10kN
MB=40kNm
MA
A
I
B
+
RAY
4m
I
C
1.5m
Step 1: Determine the ground reactions;
The cantilever beam is fully clamped in the left hand end A as shown. The ground reaction for
this point should have reaction force RAY and reaction moment MA.. So the global equilibrium
is given as
+ ↑ ∑ Fy = 0 = R AY + 10 = 0 →∴ R AY = −10kN (- downwards)
+
∑M
A
= 0 = M A + 40 + 10 × 5.5 = 0 → M A = −95kNm (- clockwise)
Lecture Notes of Mechanics of Solids, Chapter 5
10
Step 2: Draw FBD of beam cut just before the RHS (Section I-I).
F.B.D. (Section I-I)
MB=40kNm
95kNm
A
V(x)
4m
10kN
M(x)
O
B
+
x
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I).
Take moments about RHS:
1
0
0
+ ∑ M O = 0 = 10 x − 95 x + 40 x − 4 + M ( x )
M ( x ) = −10 x
1
+ 95 x
0
− 40 x − 4
0
and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as:
dM ( x )
0
0
V (x ) =
= −10 x + 0 − 0 = −10 x
dx
Step 4: Plotting the Shear Force and Bending Moment Diagrams
MB=40kNm
95kNm
B
+
Loading A
Diagram
10kN
C
10kN
V(x) kN
Shear
Force
Diagram
+ve
x
-ve
-10
M(x) kNm
95
Bending
Moment
Diagram
55
15
x
It is interesting to observe that due to concentrated bending moments MA (reaction moment)
and MB (external moment), there is respectively a sudden leap and drop in the Bending
Moment Diagram. In addition, such concentrated moments do not affect the Shear Force
Diagram (Note that the drop at point A (the left end) in SFD is due to the concentrated
reaction force RAY. In fact, both MA and MB do not appear in shear force equation V(x) at all).
Lecture Notes of Mechanics of Solids, Chapter 5
11