Shear Force and Bending Moment. Fundamentals and Problem Solving

1

SHEAR FORCE AND BENDING MOMENT

Fundamentals and Problem Solving

Authors

Herliana Hassan
Yusnita Yusof

Politeknik Sultan Salahuddin Abdul Aziz Shah
2021

ALL RIGHTS RESERVED
No part of this publication may be reproduced, distributed or
transmitted in any form or by any means, including
photocopying, recording or other electronic or mechanical
methods, without the prior written permission of Politeknik
Sultan Salahuddin Abdul Aziz Shah.

SHEAR FORCE AND BENDING MOMENT
Fundamentals and Problem Solving

Special project by :
Herliana binti Hassan
Yusnita binti Yusof

eISBN No:

First Published in 2021 by:

UNIT PENERBITAN

Politeknik Sultan Salahuddin Abdul Aziz Shah
Persiaran Usahawan,
Seksyen U1,
40150 Shah Alam
Selangor

Telephone No. : 03 5163 4000

Fax No. : 03 5569 1903

BIBLIOGRAPHY

Herliana Hassan is a lecturer at Civil Engineering
Department, Politeknik Sultan Salahuddin Abdul Aziz
Shah, Shah Alam. Graduate Bachelor of Civil and
Structural Engineeering UKM (1999) and Master of
Technical and Vocational Education UTM
(2001). She has 19 years of teaching experience in the
field of Mechanic of Civil Engineering Structures and
Industrialised Building System.

Yusnita Yusof is a lecturer at Civil Engineering
Department, Politeknik Sultan Salahuddin Abdul Aziz
Shah, Shah Alam. Previously she was a lecturer at
Politeknik Kuching Sarawak (2004-2005) and Politeknik
Sultan Azlan Shah (2005-2008). Graduate Bachelor of
Civil Engineering KUiTHHO (2002) and Master of
Technical and Vocational Education (2003). She has 17
years of teaching experience in the field of Mechanic of
Civil Engineering Structures, Theory of Structures and
Basic Building Services.

PREFACE

The purpose of writing this eBook is to provide the students with a clear
and thorough understanding of the basic principles and applications of
shear force and bending moment in engineering structures. This eBook
presents the principles behind the methods of solving problem for
determinate beam when subjected to different types of loads.
With these responsibilities in mind, the objective for this eBook is to
develop the student's ability to recognize basic knowledge of engineering
structures. We provide an adequate number of self-test and problem
solving to enhance student knowledge and understanding.
Any suggestions, comments and feedback for further improvement are
most welcome.

Herliana binti Hassan
Yusnita binti Yusof

ACKNOWLEDGEMENTS

We would like to acknowledge the assistance and encouragement of our
families and friends who have actively contributed either directly or
indirectly to the completion of this eBook We are very grateful to the Head
of Civil Engineering Department and colleagues for the encouragement
and support and giving us the opportunity to produce this eBook.

We hope that this eBook can benefit students in increasing their
understanding of the basic principles of mechanics of civil engineering
structures

Herliana binti Hassan
Yusnita binti Yusof

TABLE OF CONTENT v
vi
Preface
Acknowledgements 1
2
1.0 EQUILIBRIUM FORCES 2
3
1.1 Principles of Equilibrium Forces 5
1.2 Statically Determinate and Indeterminate Beam 6
1.3 Beam 7
1.4 Type of Supports 8
1.5 Type of Beams 11
1.6 Main Types of External Load 14
1.7 Load Distribution 17
1.8 Free Body Diagram
SELF-TEST 1A 18
SELF-TEST 1B
SELF-TEST 1C 20
23
2.0 SHEAR FORCE AND BENDING MOMENT 24
43
2.1 What is Shear Force and Bending Moment? 53
2.2 Rules To Construct Shear Force and Bending Moment 66
68
Diagram 77
2.3 Variation of Shear Force and Bending Moments

Diagram of Shear Force and Bending
2.4 Typical Diagram of Shear Force and Bending Moment for

determinate beam
SELF-TEST 2A
SELF-TEST 2B
SELF-TEST 2C
PROBLEM SOLVING
REFERENCES
ANSWER

1.0 EQUILIBRIUM FORCE

Learning Objective

Explain the principles and concepts of equilibrium forces.

Calculate values and directions of vertical reaction,
horizontal reaction and moment.

1.1 PRINCIPLES OF EQUILIBRIUM FORCES

Engineering structures must remain in equilibrium both externally
and internally when subjected to a system of forces. The principle of
Equilibrium Forces states that if a structure is in equilibrium, all its
members are also in equilibrium.

a. Summation of all PRINCIPLE 1
forces is equal to zero Total leftwards forces is equals to the
 fx = 0 total of rightwards forces.
 fy = 0
PRINCIPLE 2
b. Summation of all Total upwards forces is equals to
moments in the the total of downwards forces.
structure is equal to
zero PRINCIPLE 3
M=0 Total clockwise moments is equals to
the total of anticlockwise moments.

1

EQUILIBRIUM FORCE

1.2 STATICALLY DETERMINATE AND INDETERMINATE 1.0
BEAM

Classification of beams according to the equilibrium condition is
categorized as:

Statically determinate Statically indeterminate
beam beam

Beams that have three Beams that have more than
unknown support reactions three unknown support
and can be analyzed by using reactions and moments and
basic equilibrium conditions. cannot be analyzed by using
basic equilibrium conditions.
Example:
Example:




Three unknowns
Four unknowns

1.3 BEAM

Beam can be defined as a horizontal structural member
spanning a distance between one or more supports and
carrying vertical loads across its longitudinal axis. It is
constructed to support the slab and withstand the loads
coming from the building.

Beam Longitudinal
axis

Supports
2

1.4 TYPES OF SUPPORT EQUILIBRIUM FORCE

1.0

Type of Total of Description
support unknown

Roller 1 unknown A support condition that resists movement
support Fy = 0 perpendicular to the point of contact. This
support develops one unknown force
Hinge or 2 component .Usually it reacts vertically.
pin unknowns
support A support condition that that can resist
Fx = 0 forces and prevent movement both in
Fy = 0 horizontally and vertically, but it does not
prevent rotations.

Fixed 3 A support condition that does not permit
Support unknowns movement or rotation of the joining
elements in any direction. This support
Fx = 0 develops three unknown reactions,
Fy = 0 including a vertical and horizontal
M = 0 component and a resisting moment.

Roller Types of Fixed
Fy support
M
Fx

Pin or Hinge Fy
Fx Where :
M = Moment
Fy Fx = Horizontal force
Fy = vertical force

3

EQUILIBRIUM FORCE

1.0
How does a support of beam look in real
life?

Here are the examples application of a support in structures

Simple pin connections using two plates

Fixed Support - Beam Fixed in Wall

Source: Terri Meyer, B., & Vincent, H. (2020)

4

EQUILIBRIUM FORCE

1.5 TYPE OF BEAMS 1.0

A beam is placed on supports and it is classified according to the
support conditions as follows.

1. Simply Supported Beam

A beam in which one end is supported by a
hinge/pin and the other end is supported by a
roller.

2. Cantilever Beam

A beam that is fixed or anchored at one end and
free at the other end.

3. Overhanging Beam

A beam that has one or both of its ends
extending beyond the supports.

4. Continuous Beam A beam that spans over more than two supports.

Simply supported beam, cantilever beam and overhanging beam are
categorized as Statically Determinate Beam

Continuous beam is categorized as Statically Indeterminate Beam

5

1.6 MAIN TYPES OF EXTERNAL LOAD EQUILIBRIUM FORCE

1.0

Structures are designed to withstand various types of loads.

DEAD LOAD
Dead loads are due to self weight of the structure. Dead loads
are the permanent loads which are always present, and it
depends upon the unit weight of the material. Example Self
weight of walls, floors, beams, columns etc.

LIVE LOAD
Live loads (also called as imposed loads) on floors and roofs
consists of all the loads which are temporarily placed on the
structure. Live loads keep on changing from time to time
and it depend upon the use of building. Example, loads of
people, furniture, machines etc.

WIND LOAD

The force exerted by the horizontal component of wind is to be
considered in the design of building. Wind loads depends
upon the velocity of wind, shape and size of the building.

Slab Load transfer
Beam
Column
Foundation

Source : Namita (2017)

General classification of loads
6

EQUILIBRIUM FORCE

1.7 LOAD DISTRIBUTION 1.0

There are three main methods how loads are distributed on a
structure. However, loads distributed in a structure are the
combination of many types of loads.

Point Load Point Load,
F
 This load acts on a very small
area and can be considered as
acting on a point.

 Its symbol is an arrow, and its
unit is N or kN

UDL, Uniformly Distributed
w Load (UDL)

 This type of loads acts along
the whole length of a beam or
some part of it and the
magnitude of the load is
equal.

 Its unit is N/m or kN/m.

Moment Moment,
M
 Moment is generated by a pair of
force.

 These forces are acting on a
point and this results in a twist at
the point.

 The direction of the twist is either
clockwise or anticlockwise.

 Its unit is Nm or kNm.

7

1.8 FREE BODY DIAGRAM EQUILIBRIUM FORCE

1.0

Any force system acting on a structure is easily analyzed if

the appropriate reaction required to maintain equilibrium are
inserted in a diagram is called Free Body Diagram.

12kNm 15kN

H RD
D
A

RA

A 2m B 3m C 3m

Digital Link

https://youtu.be/C-FEVzI8oe8

8

EQUILIBRIUM FORCE

ExaEmxpalem1.p1 le 1.1 Simply Supported Beam 1.0

Based on figure 1.1, determine reactions at the support of the simply
supported beam.

FFigiguurree11.1.1

Solution:
Step 1: Sketch Free Body Diagram.

Pin Support : Roller Support :
•2 • 1 unknown
• fy
unknowns
• fy and fx −+

anticlockwise clockwise
moment moment

Step 2: Determine reaction at support D by using principle of equilibrium forces.

Use a summation of Horizontal forces to get the horizontal reaction at A, HA

→ = ←;
= 0

Use summation of Moment at point A to get the vertical reaction at D, RD

= 0; 1 1 + 2 2 − = 0

20 1 + (15 )(3 ) − ( )(5 ) = 0 Things to keep in
mind
= 13
Moment = Force × distance
Use summation of Vertical forces to get the M = F×d
vertical reaction at A, RA
F
↑= ↓;

+ = 20 + 15 d

+ 13 = 35 Substitute = 13 **Force is perpendicular
to the distance where
= 22 moment is to be taken

9

EQUILIBRIUM FORCE

ExamExpalem1.1ple 1.2 Simply Supported Beam 1.0

Based on figure 1.2, determine reactions at the support of the simply
supported beam.

FigFuigreu1re.11.2

Solution:
Step 1: Sketch Free Body Diagram.

Pin Support : Roller Support :
•2 • 1 unknown
• fy
unknowns
• fy and fx

Step 2: Determine reaction at support D by using principle of equilibrium forces.

Use a summation of Horizontal forces to get the horizontal reaction at A, HA

→ = ←;
= 0

Use summation of Moment at point A to get the vertical reaction at D, RD

= 0; 1 1 + 2 2 − = 0

30 1.5 + (25 )(5 )(4 ) − ( )(8 ) = 0

= 68.125

Use summation of Vertical forces to get the vertical reaction at A, RA

↑= ↓;
+ = 30 + (25 )(5 )

+ 68.125 = 155 Substitute = 68.125

= 86.875

10

EQUILIBRIUM FORCE

SELF-TEST 1A Simply Supported Beam 1.0

Determine reactions at the support for each of the simply supported
beam as shown in figure below.

1.

2.

3.

4.
11

EQUILIBRIUM FORCE

ExamEpxlaem1.1ple 1.3 Cantilever Beam 1.0

Based on figure 1.3, determine reactions at the support of the cantilever
beam.

FigFuigreu1re.11.3

Solution:
Step 1: Sketch Free Body Diagram.

Fixed Support :
•3

unknowns
• fy , fx and M

Step 2: Determine reaction at support D by using principle of equilibrium forces.

Use a summation of Horizontal forces to get the horizontal reaction at A, HA

→ = ←;
= 0

Use summation of Vertical forces to get the vertical reaction at A, RA

↑= ↓;

= 20 3 + 15 1


= 75

Use summation of Moment at point D to get the Moment at A, MA

= 0;

20 3 1 1 + 2 2 − = 0
1.5 + (15 )(5 ) − = 0

= 165

12

EQUILIBRIUM FORCE

ExamEpxlaem1.1ple 1.4 Cantilever Beam 1.0

Based on figure 1.4, determine reactions at the support of the cantilever
beam.

FFiigurree11.1.4

Solution:
Step 1: Sketch Free Body Diagram.

Fixed Support :
•3

unknowns
• fy , fx and M

Step 2: Determine reaction at support D by using principle of equilibrium forces.

Use a summation of Horizontal forces to get the horizontal reaction at D, HD
→ = ←;

= 0

Use summation of Vertical forces to get the vertical reaction at D, RD

↑= ↓;

= 20 + 15 2 2


= 50

Use summation of Moment at point D to get the Moment at D, MD

= 0; − 2 2 + − 2 2 − = 0

− 20 5 + 10 − 15 2 2 + = 0


= 150

13

EQUILIBRIUM FORCE

SELF-TEST 1B Cantilever Beam 1.0

Determine reactions at the support of cantilever beam as shown in
figure below.

1.

2.
3.

4.
14

EQUILIBRIUM FORCE

ExamEpxlea1m.1ple 1.5 Cantilever Beam 1.0

Based on figure 1.5, determine reactions at the support of the overhanging

beam.

FiFgiugrue r1e.11.5

Solution:
Step 1: Sketch Free Body Diagram.

Step 2: Determine reaction at support D by using principle of equilibrium forces.

Use a summation of Horizontal forces to get the horizontal reaction at A, HA

→ = ←;
= 0

Use summation of Moment at point A to get the vertical reaction at D, RD

= 0;
1 1 + 2 2 − = 0

35 3 + (10 )(6 )(4.5 ) − ( )(6 ) = 0
= 62.5

Use summation of Vertical forces to get the vertical reaction at A, RA

↑= ↓;
+ = 35 + (10 )(6 )

+ 62.5 = 95 Substitute = 62.5

= 32.5

15

EQUILIBRIUM FORCE

ExamEpxlaem1.1ple 1.6 Cantilever Beam 1.0

Based on figure 1.6, determine reactions at the support of the
overhanging beam.

Figure 1.6

Solution:
Step 1: Sketch Free Body Diagram.

Step 2: Determine reaction at support D by using principle of equilibrium forces.

Use a summation of Horizontal forces to get the horizontal reaction at B, HB
→ = ←;
= 0

Use summation of Moment at point B to get the vertical reaction at E, RE

= 0; − 1 1 + 2 2 + 3 3 − − + 4 4 = 0
3 2.5 − 13 − ( ) 5 + 10 7 = 0
− 15 2 + 25 1 + 10
= 25.4

Use summation of Vertical forces to get the vertical reaction at B, RB

↑= ↓;

+ = 15 N + 25 + 10 3 + 10


+ 11 = 80 Substitute = 25.4

= 54.6

16

EQUILIBRIUM FORCE

SELF-TEST 1C Overhanging Beam 1.0

Determine reactions at the support for each of the overhanging beam
as shown in figure below.

1.

2.

3.

4.
17

2.0 SHEAR FORCE AND BENDING
MOMENT

Learning Objective

Calculate shear force and bending moment in beam.

Draw shear force and bending moment diagrams
for statically determinate beams.

2.1 WHAT IS SHEAR FORCE AND BENDING MOMENT?

Shear Force

Shear forces are internal forces developed in the beam
material to balance the externally applied forces in
order to secure the equilibrium of all members of the
beam.

Bending Moment

Bending moments are internal moments that are
developed in the beam material to balance the
tendency of external forces to cause rotation of any
part of the beam.

18

SHEAR FORCE AND BENDING MOMENTDigital Link

Shear force diagram and bending moment diagram is the
most important first step toward design calculations of
structural members.

Shear Force diagram indicates the shear force resisted by the 2.0

beam section along the length of that beam.
Bending Moment Diagram is the diagram to show how the
applied loads to a beam create a moment variation along the
length of the beam.

Why do we need to calculate Shear
Force and Bending Moment?

1. As analytical tools used in conjunction with structural
analysis to help perform structural design by
determining the value of shear force and bending
moment at a given point of a structural element such as
a beam.

2. These diagrams can be used to provide an appropriate
size of beam or column which can resist the applied
forces when a given set of loads, it can be
supported without structural failure.

https://youtu.be/5K27dJqGpf8

19

SHEAR FORCE AND BENDING MOMENT

2.2 RULES TO CONSTRUCT SHEAR FORCE AND
BENDING MOMENT DIAGRAM

Rule 1 Start with the far-LEFT SIDE of the beam. 2.0

P

If there is an upward Aa bB
force at a support, then RA RB
the Shear force diagram
will start at this force SFD(kN B
above the x-axis. )

If there is a downward 0
point load and no
support, then the shear AC
force diagram will start as
a negative at the value of
the point load

Rule 2 Move across the beam to right side, stop at every load
that acts on the beam

When come across the loads,

simply add (or subtract) these

loads from the value you already
have (keeping a cumulative total)

20

SHEAR FORCE AND BENDING MOMENT

Rule 3 Draw shear force and bending moment diagram by
following the sign convention as below:

SFD 2.0

POSITIVE Shear Force +ve
Diagram is ABOVE the x-x
axis while NEGATIVE Shear 0 -ve
Force Diagram is BELOW
the x-x axis BMD

0

+ve

AC B

POSITIVE Bending Moment SFD +ve
Diagram is BELOW the x-x
axis while NEGATIVE 0
Bending Moment Diagram is
ABOVE the x-x axis -ve

BMD -ve B

0 C

A

21

SHEAR FORCE AND BENDING MOMENT

Rule 4 Value of bending moment between any two point is
equal to the AREA of the shear force diagram

SFD 2.0

When come across loads, 10kN

simply add (or subtract)

these loads from the value +ve

you already have, keeping 0

a cumulative total -ve
-10kN
Consider a shear force 2m
diagram as shown in figure, 3m
therefore, bending BMD
moment at each point is +ve
calculated as follow: 0 20kNm

MA = 0 A C B
MC = (10 x 2) = 20
MB = 20 – (10 x 2) = 0

22

SHEAR FORCE AND BENDING MOMENT

2.3 VARIATION OF SHEAR FORCE AND BENDING 2.0
MOMENTS

1. Where a beam experiences point loads:
The shear force diagram comprises horizontal straight lines.
The bending moment diagram is sloping straight lines.

2. Where a beam experiences uniformly distributed loads:
The shear force diagram comprises sloping straight lines.
The bending moment diagram is curved (parabolic).

3. Where a beam experiences moment:
The shear force diagram comprises horizontal straight lines.
The bending moment diagram is vertical line (downward or
upward).

Types of load Point Load UDL Moment
(kN) (kN/m) (kNm)

Shear Force Vertical & Sloping straight Horizontal
Diagram, SFD Horizontal lines straight lines
straight lines
(kN)

Bending Sloping straight Curved line Vertical line
Moment lines (parabolic)
Diagram,

BMD
(kNm)

23

SHEAR FORCE AND BENDING MOMENT

2.4 TYPICAL DIAGRAM OF SHEAR FORCE AND BENDING
MOMENT FOR DETERMINATE BEAM

IMPORTANT concepts of SIMPLY 2.0
SUPPORTED BEAM

Types of load Simply supported beam Simply supported beam
subjected to point Load subjected to uniformly

distributed Load

Shear Force Diagram,
SFD
(kN)

Bending Moment
Diagram, BMD
(kNm)

24

SHEAR FORCE AND BENDING MOMENT

IMPORTANT concepts of
CANTILEVER BEAM

Cantilever beam Cantilever beam subjected 2.0
subjected to point Load to uniformly distributed
Types of load Load

Shear Force Diagram, + +
SFD 0 0
(kN)
−
Bending Moment − 0
Diagram, BMD 0
(kNm)

25

Example 2.1 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

CASE 1: Beam with a non-central point load

Construct shear force and bending moment diagrams for a beam as

shown in Figure 2.1. 45kN 2.0

A 2m 3m B
RA Figure 2.1 RB

Solution:

45kN

Step 1: Calculate reactions at
support A and B.

=
45(2) – =
A 2m 3m B = kN
RA =
27kN RB = 18kN ↑ = ↓

SFD (kN) + = 45
=
27

Step 2: Drop vertical lines at
each load and draw horizontal
lines on which to construct
0 shear force diagrams (SFD).

AC Step 3: Draw SFD from the left
side of the beam A and move
across the beam. In this case it
is a +27kN due to the reaction
at point A because the reaction
B arrow is pointed up 27kN.

26

Example 2.1 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

45kN

2.0

A 2m 3m B Step 4: Keep moving across
RA = RB = 18kN the beam, stopping at every
2S7FkDN(kN) load that acts on the beam.
B The value on the shear
27 diagram remains unchanged
from point A to point C
– 18
There is a downward 45kN
AC force at point C, so, minus
45kN from the existing 27kN.

Therefore;
27kN – 45kN = – 18kN
which pushes the shear
diagram down from its
present value by 45kN.

27

Example 2.1 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

45kN

Step 5: Moving across the 2.0
beam again, a positive
A 2m 3m B reaction of 18kN acting at
RA = 27kN RB = 18kN support B.

SFD (kN) Again, add this +18kN to
the shear force diagram
27 (which is currently at –
18kN) which will bring to
– 18 0 a shear force of 0.

AC B –18kN + 18kN = 0kN
which pushes the shear
diagram up to zero.

Since we are at the end of
the beam, we will go no
further, and we have our
final Shear Force Diagram
(SFD)

28

Example 2.1 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

45kN

2.0

A 2m 3m B Step 6: Calculate bending
RA = 27kN RB = 18kN moment at each point on
-ve the beam.
27 – 18
+ve +ve BMD (kNm)

0 MA = 0
SFD MC = (27 x 2) = 54
0 (kN) MB = 54 – (18 x 3) = 0

BMD To construct Bending
(kNm) Moment Diagram, start at
A where the moments at
the support A is zero.

54 B Note that the bending
moment diagram
AC comprises sloping straight
line due to the point
loaded between two
points.

29

Example 2.2 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

CASE 2: Beam carrying a uniformly distributed load over its
entire beam span

Construct shear force and bending moment diagrams for a beam as

shown in Figure 2.2. 20kN/m 2.0

A 5m B
RA Figure 2.2 RB

Solution: 20kN/m

A 5m B Step 1: Calculate reactions
RB = 50kN at support A and B.
RA = 50kN In this case, the reaction at
SFD(kN) A and B is 50kN.

50 Step 2: Drop vertical lines at
each load and draw
horizontal lines on which
to construct shear force
diagrams.

0 Step 3: Draw SFD from the
left side of the beam A and
A move across the beam. In
this case it is a +50kN due
to the reaction at point A
because the reaction
arrow is pointed up 50kN.
B

30

Example 2.2 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

20kN/m

A 5m B 2.0
RB = 50kN
RA = 50kN Step 4: Keep moving across
SFD (kN) – 50 the beam, stop at every
50 load that acts on the beam.

0 A uniform distributed load of
20kN/m along the span, so
we will minus 20kN/m from
the existing 50kN.

50kN – 20(5)kN = – 50kN
which pushes the shear
diagram down as sloping
straight line from its present
value by 20kN/m.

AC B

31

Example 2.2 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

20kN/m

Moving across the beam 2.0
again, a positive reaction of
A 5m B 50kN acting at support B.
RB = 50kN
RA = 50kN
SFD (kN) Again, add this +50kN to the
50 shear force diagram (which
is currently at -50kN) which
0 will bring us to a shear force
of 0.
AC
– 50kN + 50 kN = 0 kN
which pushes the shear
diagram up to zero.

– 50

Since we are at the end of
the beam, we will go no
further and we have our
final Shear Force Diagram
(SFD)

B

32

Example 2.2 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

20kN/m

A B To construct Bending 2.0
RA = 50kN RB = 50kN moment Diagram, start at A
5m where the moments at the
50 SFD supports A is zero.
(kN)
+ve -ve Then, move across the beam
0 and calculate the bending
– 50 moment at each point.
0 BMD
+ve (kNm) BMD (kNm)
MA = 0
MC = (21 x 2.5 x 50)

= 62.5
MB = 62.5 – (21 x 2.5 x 50)

=0

62.5 Note that the bending
moment diagram
AC comprises curved line due
B to the uniformly distributed
load between two points.

33

SHEAR FORCE AND BENDING MOMENT

Example 2.3 Simply Supported Beam 2.0

A simply supported beam loaded as shown in Figure 2.3. If reaction at
support A are RA = 22kN and RD = 13kN.
i. Calculate the value of shear force and bending moment.
ii. Illustrate the Shear Force Diagram (SFD) and Bending Moment

Diagram (BMD) that indicate all the values at important points.
iii. Determine the maximum bending moment.

Figure 2.3

Solution:

Step 1: Calculate Shear Force, kN.

Point/ Figure Shear Force
Interval
A = = +22
A +
= − 1
B RA = 22kN = 22 − 20kN = 2

F1 = − 2
20kN = 2 − 15 = −13kN

−
AB

F2
15kN
C−
BC

D = − D
D + = −13 + 13 = 0

RD = 13kN

34

SHEAR FORCE AND BENDING MOMENT

Example 2.3 Simply Supported Beam

Step 2: Illustrate the Shear Force Diagram (SFD) .

2.0

Things to keep in
mind

“The SFD should always

equal to zero at both ends”.

Step 3: Based on SFD and Figure 2.3, calculate Bending Moment, kNm.

Point/ Figure Bending Moment
Interval
= 0
A TIP: Moment at support of a simply supported beam
equal to zero.

= Area of rectangle
= (22kN) 1
A–B = 22kNm

B–C = + 2m
= 22 + 2
= 26kNm

C–D = + 2m
= 26 + −13
= 0

35

SHEAR FORCE AND BENDING MOMENT

Example 2.3 Simply Supported Beam

Step 4: Illustrate the Bending Moment Diagram (BMD).

Things to keep in 2.0
mind

“The BMD will be zero at
the support of a simply

supported beam”.

The maximum bending
moment, Mmax = 26kNm

The bending moment can also be calculated using Figure 2.3.

Point/ Figure Bending Moment
Interval

A = 0

A–B = 1
= + 22
= 22kNm

A–C = − 1
= + 22 3m − (20kN)(2m)
= 26kNm

A–D = − 1 − 2
= + 22 5m − (20kN)(4m) − (15kN)(2m)
= 0

36

SHEAR FORCE AND BENDING MOMENT

Example 2.4 Simply Supported Beam

A simply supported beam loaded as shown in Figure 2.4. If reaction at 2.0
support A are RA = 22kN and RD = 13kN.
i. Calculate the value of shear force and bending moment.
ii. Illustrate the Shear Force Diagram (SFD) and Bending Moment

Diagram (BMD) that indicate all the values at important points.
iii. Determine the maximum bending moment.

Figure 2.4

Solution:

Step 1: Calculate Shear Force, kN.

Point/ Figure Shear Force
Interval

A = = +86.875

B = − 1
= 86.875 − 30kN = 56.875

B–C = − 2 BC 25 5
= 56.875 −

= −68.125kN

D = − D
= −68.125 + 68.125 = 0

37

SHEAR FORCE AND BENDING MOMENT

Example 2.4 Simply Supported Beam

Step 2: Illustrate the Shear Force Diagram (SFD) .

Things to keep 2.0
in mind

“Any points where the SFD
cross the x-axis (intersection
point), will be a max or min

Bending Moment”.

1 = 2
ℎ1 ℎ2
h1 = 56.875kN l2 = 5 – x
1 2 = 5 − Therefore;
l1 = x 56.875 68.125 5 − = 5 − 2.275
h2 = 68.125kN
= 5(56.875) = 2.725
(56.875 + 68.125)

= 2.275

Step 3: Based on SFD in Figure 2.4, calculate Bending Moment, kNm.

Point/ Figure Bending Moment
Interval

AA = 0

FA =86.875kN =
= (86.875kN) 1.5
A–B + = 130.313kNm

0 38

A dAB = B
1.5m

Example 2.4 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

Point/ Figure Bending Moment
Interval

1 Area of triangle 2.0
2
= +

B–x = 130.313 m + 1 56.875 2.275m
x–C 2
C–D
= 195.008kNm

= + 1
2

= 195.008 m + 1 −68.125 2.725m
2

= 102.188kNm

= + 1
2

= 102.188 m + −68.125 1.5m

= 0.0005kNm

≈ 0kNm

Step 4: Illustrate the Bending Moment Diagram (BMD) .

The maximum
bending moment,
Mmax = 195.008kNm

39

Example 2.4 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

The bending moment can also be calculated using Figure 2.4.

Point/ Figure Bending Moment
Interval = 0
2.0
A

A–B =
= + 86.875 1.5
= 130.313kNm

= − 1 − 2
2

A–x = + 86.875 3.775 − (30kN)(2.275m)
A–C
A–D − 25 2.275 2.275
2

= 195.093kNm

= − 1 − 2
2

= + 86.875 6.5 − 30kN 5m

− 25 5 5
2

= 102.188kNm

= − 1 − 2 +
2

= + 86.875 8 − (30kN)(6.5m)

− 25 5 5 + 1.5
2

= 0kNm

40

SHEAR FORCE AND BENDING MOMENT

Example 2.5 Simply Supported Beam 2.0

A simply supported beam loaded as shown in Figure 2.5.
i. Calculate reactions at the support of beam
ii. Calculate the value of shear force and bending moment.
iii. Illustrate the Shear Force Diagram (SFD) and Bending Moment

Diagram (BMD) that indicate all the values at important points.
iv. Determine the maximum bending moment.

Figure 2.5

Solution:
Step 1: Sketch Free Body Diagram.

Calculate the resolution force of
20kN at an angle of 300 above the
horizontal using trigonometric.

20kN =
300 = kN

=
= 7.321kN

Step 2: Determine reaction at support A and D by using principle of equilibrium forces.

→ = ←;
= 17.321

= 0;

18 + 15 2 2 + (10 )(3 ) − (5 ) = 0
= 21.6

↑= ↓; 15 2 + 10
+ 21.6 =

= 18.4

41

Example 2.5 SHEAR FORCE AND BENDING MOMENT

Simply Supported Beam

Step 3: Calculate the Shear Force.

= = 18.4 2.0
= = 18.4

= −

= 18.4 − 15

= −11.6kN

′ = −
= −11.6 − 10kN
= −21.6kN

= ′ +
= −21.6 N + 21.6kN
=0

Step 4 : Calculate the Bending Moment.

= 2 −
18.4 11.6

= 1.227

= 0
= 18.4 1m

= 18.4kNm

′ = +
= 18.4kNm + 18kNm

= 36.4kNm

= ′ + 1
2
1
= 36.4 m + 2 18.4 1.227

= 47.688kNm

= + 1
= 2
1
47.688 m + 2 −11.6 0.773

= 43.205kNm The maximum
= + ′
= 43.205 Nm + (−21.6kN)(2m) bending moment,
= 5 × 10−3 Mmax = 47.688kNm

≈0

42

SHEAR FORCE AND BENDING MOMENT

SELF-TEST 2A Simply Supported Beam

A simply supported beam loaded as shown in figure below. 2.0
i. Calculate the reactions at the support of beam.
ii. Calculate the value of shear force and bending moment.
iii. Illustrate the Shear Force Diagram (SFD) and Bending Moment

Diagram (BMD) that indicate all the values at important points.
iv. Determine the maximum bending moment.

1.

2.

3.

4.

43