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Lecture Notes of Mechanics of Solids, Chapter 5 1 
 
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 forcePositive 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:264-
268) 
A beam (Fig. 5.2) is loaded with vertical forces Fi, bending moments Mi and distributed loads 
w(x). 
 
F1 F2
M1 M2
x dx
w(x)
dx
M(x)
V(x)
M(x)+ dM(x)dx dxM(x)+
dM(x)
dx dx
w(x)
F.B.D. of element dx
V(x)+ dV(x)dx dxV(x)+
dV(x)
dx dx
 
Fig. 5.2 Transversely loaded beam and free body diagram of element dx 
 
Lecture Notes of Mechanics of Solids, Chapter 5 2 
 
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 
( ) ( ) ( ) ( ) 00 =

 +−−==↑+ ∑ dxdxxdVxVdxxwxVFy (5.1) 
Dividing by dx in the limit as dx→0, ( ) ( )xw
dx
xdV −= (5.2) 
Taking moments about the right hand edge of the element: 
( ) ( ) ( ) ( ) ( ) 0
2
0 =

 +++−−==∑ dxdxxdMxMdxdxxwdxxVxMM Edge.H.R (5.3) 
Dividing by dx in the limit as dx→0, ( ) ( )xV
dx
xdM = (5.4) 
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: 
P
L
a
RAY=(1-a/L)P RBY=Pa/L
I
I
RAY=(1-a/L)P
x
F.B.D. (method of section I-I)
o
F.B.D. (global equilibrium) 
II
II M(x)
V(x)
A B A
 
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): 
( ) 010 =+

 −−==∑ xMxLaPM o → ( ) xLaPxM  −= 1 (5.5) 
To determine the shear force, use Eq. (5.4), giving that: 
( ) ( ) P
L
a
dx
xdMxV 

 −== 1 (5.6) 
To verify Eq. (5.6), equate vertical equilibrium: 
++
++
Lecture Notes of Mechanics of Solids, Chapter 5 3 
 
( ) 01 =−

 −=↑+ ∑ xVLaPFy → ( )  −= L
aPxV 1 
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 ax ≤≤0 . 
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) 
P
x
a
RAY=( 1-a/L)P
II
II
F.B.D. (Section II-II) 
M(x)
o
V(x)
A
 
Fig. 5.4 FBD of beam cut before the right hand end 
 
Equate moments about the right side: 
( ) ( ) 0100 =+−+

 −−==∑ xMaxPxLaPM 
giving: 
( ) ( ) PaPx
L
aaxPx
L
aPxM +−=−−

 −= 1 (5.7) 
and using Eq. (5.4), the shear force equation is : 
( ) ( ) P
L
a
dx
xdMxV −== (5.8) 
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 4 
 
P
L
a
(1-a/L)P Pa/L
Loading Diagram
V(x)
+ve
-ve
x
M(x)
+ve
x
(1-a/L)P
-a/LP
(1-a/L)Pa
Shear Force Diagram
Bending Moment Diagram
 
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) 
( ) 001 axPx
L
aPxV −−

 −= (5.9) 
( ) 111 axPx
L
aPxM −−

 −= (5.10) 
Where the notation has the following meaning: 
( ) ( )0
0 ≥≥
<


−=− naxfor
axfor
ax
ax n
n (5.11) 
 
when differentiating: 



=
=
≥
−
−
=∂
−∂
−
0
1
1
0
0
1
nfor
nfor
nfor
ax
axn
x
ax
n
n
 (5.12) 
 
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 5 
 
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; 
P
L
a
RAY=(1-a/L)P RBY=Pa/L
F.B.D. (global equilibrium) 
I
I
A B
 
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). 
P
x
a
RAY=( 1-a/L)P
I
I
F.B.D. (Section I-I) 
M(x)
o
V(x)
A
 
 
 
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I). 
Take moments about RHS: 
( ) 010 11O =+−+

 −−==∑ xMaxPxLaPM 
 ( ) 111 axPx
L
aPxM −−

 −= 
 
and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as: 
( ) ( ) 001 axPx
L
aP
dx
xdMxV −−

 −== 
 
Step 4: Plotting the Shear Force and Bending Moment Diagrams 
++
Lecture Notes of Mechanics of Solids, Chapter 5 6 
 
According to M(x) and V(x) to depict the diagrams, Look at the equations segment by segment 
 
 When ax ≤≤0 
( ) ( ) x
L
aPPx
L
aPxM 

 −=×−

 −= 101 1 and ( ) ( ) 

 −=×−
 −=
L
aPPx
L
aPxV 101 0 
To plot this segment in the diagram, firstly look at the boundary points as 0)(0 == xM,x 
and ( )L/aPaxM,ax −== 1)( . Draw two points and then connect them because the equation 
gives a line. Likewise, one can plot Shear Force Diagram in this region. 
 
 When Lxa ≤≤ 
( ) ( ) ( ) x
L
aPPaPaPxx
L
aPaxPx
L
aPxM −=+−

 −=−×−

 −= 11 11 
and ( ) ( ) ( )
L
aPP
L
aPaxPx
L
aPxV −=−

 −=−×−

 −= 11 00 
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
L
a
(1-a/L)P Pa/L
Loading Diagram
V(x)
+ve
-ve
x
M(x)
+ve
x
(1-a/L)P
-a/LP
(1-a/L)Pa
Shear Force Diagram
Bending Moment Diagram
 
 
 
 
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 7 
 
 
Step 1: Determine the ground reactions; 
From global equilibrium the ground reaction forces can be found to be both equal to wL/2 as, 
 
L
wL/2
F.B.D. (global equilibrium) 
I
I
wL/2
BA
w
 
 
( ) 0
2
0 =

+−==∑ LwLLRM AYB → 2/wLRAY =∴ (+ upwards) 
 00 =−+==↑+ ∑ wLRRF BYAYy → 2/wLRBY =∴ (+ upwards) 
 
 
Step 2: Draw FBD of beam cut just before the RHS (Section I-I). 
 
x
RAY=wL/2
F.B.D. (Section I-I) 
A o
M(x)
V(x)
w
Rw
x/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: 
( ) ( ) ( ) ( ) 0
2
2
2
0
1
11
1
1
O =++−=++−==∑ xMxxwx/wLxMxRxRM wAY 
 ( ) 21
22
xwxwLxM −=∴ 
and differentiating w.r.t. 'x', as Eq. (5.4), gives : 
 ( ) ( ) 10
2
xwxwL
dx
xdMxV −== 
 
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 8 
 
LwL/2 wL/2
BA
w
V(x)
+ve
-ve
x
M(x)
+ve
x
wL/2
Shear Force Diagram
Bending Moment Diagram
-wL/2
Loading Diagram
wL2/8
Parabola
 
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 9 
 
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. 
UDL=w=1kN/m 10kN
A
B C ED
20kN
RAY RDY
10m 5m 5m 10m
F.B.D. (global equilibrium) 
I
I
 
 
Step 1: Determine the ground reactions; 
0301020152051010 =×−×+×−××−==∑ DYA RM → N532 k.RDY =∴ 
 010201010 =−+−×−==↑+ ∑ DYAYy RRF → N57 k.RAY =∴ 
 
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. 
 
w=1kN/m
A
20kN
7.5kN 32.5kN
F.B.D. (Section I-I) and application of equivalent UDL
o
M(x)
V(x)
x
 
 
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I). 
 
Take moments about RHS: 
( )xMx.xxxx.M +−−−+−−+−==∑ 11
22
1
O 2053215202
10
1
2
1570 
 ( ) 11221 20532152010
2
1
2
157 −+−−−+−= x.xxxx.xM 
 
and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as: 
( ) ( ) 00110 2053215201057 −+−−−+−== x.xxxx.
dx
xdMxV 
++
++
Lecture Notes of Mechanics of Solids, Chapter 5 10 
 
Step 4: Plotting the Shear Force and Bending Moment Diagrams 
 
UDL=w=1kN/m 10kN
A
B C ED
20kN
7.5kN 32.5kN
10m 5m 5m 10m
V(x) kN
+ve
-ve
x
M(x) kNm
+ve
x
Shear
Force
Diagram
Bending
Moment 
Diagram -ve
+ve
28.125
25
12.5
-100
-22.5
10
-2.5
7.5
Loading
Diagram
quadratic
 
 
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. 
 
10kN
RAY
A B
C
I
I
4m
Global F.B.D.
MA
1.5m
+
MB=40kNm
 
 
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 
 0100 =+==↑+ ∑ AYy RF → N10kRAY −=∴ (- downwards) 
05510400 =×++==∑ .MM AA → kNmM A 95−= (- clockwise) ++
Lecture Notes of Mechanics of Solids, Chapter 5 11 
 
 
Step 2: Draw FBD of beam cut just before the RHS (Section I-I). 
10kN
A
B
4m
F.B.D. (Section I-I)
95kNm
x
+
MB=40kNm M(x)
V(x)
O
 
 
Step 3: Equilibrium for FBD of beam cut just before the RHS (Section I-I). 
Take moments about RHS: 
( )xMxxxM +−+−==∑ 001O 44095100 
( ) 001 4409510 −−+−= xxxxM 
and differentiating w.r.t. 'x', as Eq. (5.4), gives the shear force equation as: 
( ) ( ) 00 100010 xx
dx
xdMxV −=−+−== 
 
Step 4: Plotting the Shear Force and Bending Moment Diagrams 
10kN
A
B
95kNm
+
MB=40kNm 10kN
C
V(x) kN
+ve
-ve
x
M(x) kNm
x
Shear
Force
Diagram
Bending
Moment 
Diagram
-10
95
55
15
Loading
Diagram
 
 
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). 
++