UNIVERSITY OF THE CORDILLERAS COLLEGE OF ENGINEERING AND ARCHITECTURE

MECH 3 MECHANICS OF DEFORMABLE BODIES

MODULE 4: SHEAR AND MOMENT OF BEAMS

UNIT 1: SHEAR AND MOMENT

EQUATIONS AND DIAGRAMS
Prepared by:

Brian Jhay Guzman, CE, RMP, ME-1 Melkisidick Angloan, CE, ME-1
 UNIT LEARNING OUTCOMES
In this unit, here are the following desired learning outcomes:
✘ Define shearing force and bending moment
✘ Draw shear and moment diagrams using shear and moment
equations.

2
INTRODUCTION

3
 BEAM
A beam is a member which resist transverse loads and forces by bending elastically.
A bar subject to forces or couples that lie in a plane containing the longitudinal axis of the bar is called a
beam. The forces are understood to act perpendicular to the longitudinal axis.
The beams discussed in this module will be limited to those which meet the following limitations.
1. All forces acting on the beam lie in the same plane; this (longitudinal) plane passes through the
centroids of all the cross sections of the beam.
2. The elastic limit of the beam material is not exceeded.
3. The cross-section of the beam is uniform in size and shape for the entire length.
4. The beam is thick enough to prevent localized buckling or wrinkling.
5. Vibration, shock, and impact loading do not occur.

4
 CANTILEVER BEAM
If a beam is supported at only one end and in such
a manner that the axis of the beam cannot rotate
at that point, it is called a cantilever beam. This
type of beam is illustrated in Fig. 6-1. The left end
of the bar is free to deflect but the right end is
rigidly clamped. The reaction of the supporting
wall upon the beam consists of a vertical force
together with a couple acting in the plane of the
applied loads shown.

5
 SIMPLE BEAM (SIMPLY SUPPORTED)
A beam that is freely supported at both ends is called a simple
beam. The term “freely supported” implies that the end supports
are capable of exerting only forces upon the bar and are not
capable of exerting any moments. Thus there is no restraint
offered to the angular rotation of the ends of the bar at the
supports as the bar deflects under the loads. Two simple beams
are sketched in Fig. 6-2.
It is to be observed that at least one of the supports must be
capable of undergoing horizontal movement so that no force will
exist in the direction of the axis of the beam. Thus, a roller is
shown as one of the supports. If neither end were free to move
horizontally, then some axial force would arise in the beam as it
deforms under the load. Problems of this nature are not
considered in this module. The beam of Fig. 6-2(a) is subjected to
a concentrated force; that of Fig. 6-2(b) is loaded by a uniformly
distributed load as well as a couple.

6
 OVERHANGING BEAM

A beam freely supported at two points

and having one or both ends
extending beyond these supports is
termed an overhanging beam. Two
examples are given in Fig. 6-3.

7
STATICALLY DETERMINATE BEAM

All the beams considered above, the cantilevers, simple

beams, and overhanging beams, are ones in which the
reactions of the supports may be determined by use of the
equations of static equilibrium. The values of these reactions
are independent of the deformations of the beam. Such
beams are said to be statically determinate.

8
STATICALLY INDETERMINATE BEAM

If the number of reactions exerted upon the beam exceeds the number of equations of static equilibrium,
then the statics equations must be supplemented by equations based upon the deformations of the
beam. In this case the beam is said to be statically indeterminate. Examples are shown in Fig. 6-4.

9
INTERNAL FORCES
AND MOMENTS IN
BEAMS
10
 INTERNAL FORCES AND MOMENTS IN BEAMS

When a beam is loaded by forces and couples, internal stresses arise in

the bar. In general, both normal and shearing stresses will occur. In order
to determine the magnitude of these stresses at any section of the beam,
it is necessary to know the resultant force and moment acting at that
section. These may be found by applying the equations of static
equilibrium.

11
 INTERNAL FORCES AND MOMENTS IN BEAMS
Suppose several concentrated forces act on a simple
beam as in Fig. 6-6(a). It is desired to study the internal
stresses across the section at D, located at a distance 𝑥
from the left end of the beam. To do this let us consider the
beam to be cut at D and the portion of the beam to the
right of D removed.
The portion removed must then be replaced by the effect
it exerted upon the portion to the left of D and this effect
will consist of a vertical shearing force together with a
moment as represented by 𝑉 and 𝑀, respectively, in the
free-body diagram of the left portion of the beam shown in
Fig. 6-6(b).
The force 𝑉 and the couple 𝑀 hold the left portion of the
bar in equilibrium under the action of the forces 𝑅1 , 𝑃1 , 𝑃2 .
The quantities 𝑉 and 𝑀 are taken to be positive if they
have the senses indicated to the right.

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 RESISTING MOMENT
The couple 𝑀 shown in Fig. 6-6(b) at section D is called the
resisting moment. The magnitude of 𝑀 may be found by
use of a statics equation which states that the sum of the
moments of all forces about an axis through D and
perpendicular to the plane of the page is zero. Thus

Σ𝑀𝐷 = 𝑀 − 𝑅1 𝑥 + 𝑃1 𝑥 − 𝑎 + 𝑃2 𝑥 − 𝑏 = 0
or
𝑀 = 𝑅1 𝑥 − 𝑃1 𝑥 − 𝑎 − 𝑃2 𝑥 − 𝑏

Thus the resisting moment 𝑀 is the moment at point D

created by the moments of the reaction at A and the
applied forces 𝑃1 and 𝑃2 . The resisting moment 𝑀 is due to
stresses that are distributed over the vertical section at D.
These stresses act in a horizontal direction and are tensile
in certain portions of the cross section and compressive in
others. Their nature will be discussed in detail in Stresses in
Beams. 13
 RESISTING SHEAR
The vertical force 𝑉 shown in Fig. 6-6(b) is called the
resisting shear at section D. For equilibrium of forces in the
vertical direction,

Σ𝐹𝑦 = 𝑅1 − 𝑃1 − 𝑃2 − 𝑉 = 0
or
𝑉 = 𝑅1 − 𝑃1 − 𝑃2
This force 𝑉 is actually the resultant of shearing stresses
distributed over the vertical section at D. The nature of
these stresses will be studied in Stresses in Beams.

14
 BENDING MOMENT
The algebraic sum of the moments of the external forces
to one side of the section D about an axis through D is
called the bending moment at D. This is represented by

𝑅1 𝑥 − 𝑃1 𝑥 − 𝑎 − 𝑃2 𝑥 − 𝑏

𝑀 = Σ𝑀 𝐿𝑒𝑓𝑡 = Σ𝑀 𝑅𝑖𝑔ℎ𝑡

for the loading considered to the right. Thus the bending

moment is opposite in direction to the resisting moment
but is of the same magnitude. It is usually denoted by 𝑀
also. Ordinarily the bending moment rather than the
resisting moment is used in calculations because it can be
represented directly in terms of the external loads.

15
 SHEARING FORCE
The algebraic sum of all the vertical forces to one side, say
the left side, of section D is called the shearing force at that
section. This is represented by

𝑅1 − 𝑃1 − 𝑃2

𝑉 = Σ𝐹𝑦 = Σ𝐹𝑦
𝐿𝑒𝑓𝑡 𝑅𝑖𝑔ℎ𝑡

for the above to the right. The shearing force is opposite in

direction to the resisting shear but of the same magnitude.
Usually it is denoted by 𝑉. It is ordinarily used in
calculations, rather than the resisting shear.

16
 SIGN CONVENTIONS
The customary sign conventions for shearing force and bending moment are represented in Fig. 6-7. Thus
a force that tends to bend the beam so that it is concave upward is said to produce a positive bending
moment. A force that tends to shear the left portion of the beam upward with respect to the right portion
is said to produce a positive shearing force.

An easier method for determining the algebraic sign of the bending moment at any section is to say that
upward external forces produce positive bending moments, downward forces yield negative bending
moments.

17
SHEAR AND MOMENT
EQUATIONS WITH
DIAGRAMS
18
SHEAR AND MOMENT EQUATIONS WITH DIAGRAMS

Usually it is convenient to introduce a coordinate system along the beam, with the origin at one
end of the beam. It will be desirable to know the shearing force and bending moment at all
sections along the beam and for this purpose two equations are written, one specifying the
shearing force 𝑉 as a function of the distance, say 𝑥 , from one end of the beam, the other giving the
bending moment 𝑀 as a function of 𝑥 .

The plots of the equations for 𝑉 and 𝑀 are known as shearing force and bending moment
diagrams, respectively. In these plots the abscissas (horizontals) indicate the position of the
section along the beam and the ordinates (verticals) represent the values of the shearing force
and bending moment, respectively. Thus these diagrams represent graphically the variation of
shearing force and bending moment at any section along the length of the bar. From these plots it
is quite easy to determine the maximum value of each of these quantities.

19
EXAMPLE 1.
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:

Write the shear and moment equations of each Solve for reactions
segment of the beam shown. Consider both the Σ𝑀𝐴 = 0
left and the right side. Identify the value of shear 2
15 2 0.5 + − 𝐷𝑉 (4.5) = 0; 𝐷𝑉 = 10 𝑘𝑁
and moment at each of the change of load 2
points. Σ𝑀𝐷 = 0
2
−15 2 2+ + 𝐴𝑉 (4.5) = 0; 𝐴𝑉 = 20 𝑘𝑁
2

20
 Write the shear and moment equations of each segment of the beam shown. Consider both
EXAMPLE 1. the left and the right side. Identify the value of shear and moment at each of the change of
load points.

Shear and Moment Equations Segment AB

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (0 < 𝑥 < 0.5)

VAB
𝑽𝑨𝑩 = 𝟐𝟎

MAB 𝑖𝑓 𝑥 = 0, 𝑉𝐴 = 20 𝑘𝑁; 𝑖𝑓 𝑥 = 0.5, 𝑉𝐵 = 20 𝑘𝑁

MAB
𝑴𝑨𝑩 = 𝟐𝟎𝒙
VAB
𝑖𝑓 𝑥 = 0, 𝑀𝐴 = 0 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 0.5, 𝑀𝐵 = 10 𝑘𝑁. 𝑚

𝑅𝑖𝑔ℎ𝑡 𝑆𝑖𝑑𝑒 (4 < 𝑥 < 4.5)

𝑽𝑨𝑩 = −𝟏𝟎 + 𝟏𝟓 𝟐 = 𝟐𝟎

𝑖𝑓 𝑥 = 4, 𝑉𝐵 = 20 𝑘𝑁; 𝑖𝑓 𝑥 = 4.5, 𝑉𝐴 = 20 𝑘𝑁

𝑴𝑨𝑩 = 𝟏𝟎𝒙 − 𝟏𝟓 𝟐 𝒙 − 𝟑 = 𝟏𝟎𝒙 − 𝟑𝟎 𝒙 − 𝟑

𝑖𝑓 𝑥 = 4, 𝑀𝐵 = 10 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 4.5, 𝑀𝐴 = 0 𝑘𝑁. 𝑚

21
 Write the shear and moment equations of each segment of the beam shown. Consider both
EXAMPLE 1. the left and the right side. Identify the value of shear and moment at each of the change of
load points.

Shear and Moment Equations Segment BC

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (0.5 < 𝑥 < 2.5)

𝑽𝑩𝑪 = 𝟐𝟎 − 𝟏𝟓 𝒙 − 𝟎. 𝟓
VBC
𝑖𝑓 𝑥 = 0.5, 𝑉𝐵 = 20 𝑘𝑁; 𝑖𝑓 𝑥 = 2.5, 𝑉𝐶 = −10 𝑘𝑁
MBC
MBC 𝒙 − 𝟎. 𝟓 𝟏𝟓 𝒙 − 𝟎. 𝟓 𝟐
𝑴𝑩𝑪 = 𝟐𝟎𝒙 − 𝟏𝟓 𝒙 − 𝟎. 𝟓 = 𝟐𝟎𝒙 −
𝟐 𝟐
VBC
𝑖𝑓 𝑥 = 0.5, 𝑀𝐵 = 10 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 4.5, 𝑀𝐶 = 20 𝑘𝑁. 𝑚

𝑅𝑖𝑔ℎ𝑡 𝑆𝑖𝑑𝑒 (2 < 𝑥 < 4)

𝑽𝑩𝑪 = −𝟏𝟎 + 𝟏𝟓 𝒙 − 𝟐

𝑖𝑓 𝑥 = 2, 𝑉𝐶 = −10 𝑘𝑁; 𝑖𝑓 𝑥 = 4, 𝑉𝐵 = 20 𝑘𝑁

𝟐
𝒙−𝟐 𝟏𝟓 𝒙 − 𝟐
𝑴𝑩𝑪 = 𝟏𝟎𝒙 − 𝟏𝟓 𝒙 − 𝟐 = 𝟏𝟎𝒙 −
𝟐 𝟐

𝑖𝑓 𝑥 = 2, 𝑀𝐶 = 20 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 4, 𝑀𝐵 = 10 𝑘𝑁. 𝑚
 Write the shear and moment equations of each segment of the beam shown. Consider both
EXAMPLE 1. the left and the right side. Identify the value of shear and moment at each of the change of
load points.

Shear and Moment Equations Segment CD

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (2.5 < 𝑥 < 4.5)

𝑽𝑪𝑫 = 𝟐𝟎 − 𝟏𝟓 𝟐 = −𝟏𝟎
VCD
𝑖𝑓 𝑥 = 2.5, 𝑉𝐶 = −10 𝑘𝑁; 𝑖𝑓 𝑥 = 4.5, 𝑉𝐷 = −10 𝑘𝑁
MCD
𝑴𝑪𝑫 = 𝟐𝟎𝒙 − 𝟏𝟓 𝟐 𝒙 − 𝟏. 𝟓 = 𝟐𝟎𝒙 − 𝟑𝟎 𝒙 − 𝟏. 𝟓
MCD
VCD 𝑖𝑓 𝑥 = 2.5, 𝑀𝐶 = 20 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 4.5, 𝑀𝐷 = 0 𝑘𝑁. 𝑚

𝑅𝑖𝑔ℎ𝑡 𝑆𝑖𝑑𝑒 (0 < 𝑥 < 2)

𝑽𝑪𝑫 = −𝟏𝟎

𝑖𝑓 𝑥 = 0, 𝑉𝐷 = −10 𝑘𝑁; 𝑖𝑓 𝑥 = 2, 𝑉𝐶 = −10 𝑘𝑁

𝑴𝑪𝑫 = 𝟏𝟎𝒙

𝑖𝑓 𝑥 = 0, 𝑀𝐷 = 0 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 2, 𝑀𝐶 = 20 𝑘𝑁. 𝑚

23
EXAMPLE 2.
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:

Write the shear and moment equations of each Solve for reactions
segment of the beam shown. Identify the value
of shear and moment at each of the change of Σ𝑀𝐴 = 0
load points, as these values will be used to draw 10 1 +5 2 + 15 3 − 𝐸𝑉 5 = 0; 𝑬𝑽 = 𝟏𝟑 𝒌𝑵
the shear and moment diagram later.
Σ𝑀𝐸 = 0
−10 4 −5 3 − 15 2 + 𝐴𝑉 5 = 0; 𝑨𝑽 = 𝟏𝟕 𝒌𝑵
Shear and Moment Equations Segment AB

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (0 < 𝑥 < 1)

𝑽𝑨𝑩 = 𝟏𝟕

𝑖𝑓 𝑥 = 0, 𝑉𝐴 = 17 𝑘𝑁; 𝑖𝑓 𝑥 = 1, 𝑉𝐵 = 17 𝑘𝑁

𝑴𝑨𝑩 = 𝟏𝟕𝒙

𝑖𝑓 𝑥 = 0, 𝑀𝐴 = 0 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 1, 𝑀𝐵 = 17 𝑘𝑁. 𝑚

24
EXAMPLE 2. Shear and Moment Equations Segment BC

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (1 < 𝑥 < 2)

Write the shear and moment equations of each 𝑽𝑩𝑪 = 𝟏𝟕 − 𝟏𝟎 = 𝟕

segment of the beam shown. Identify the value
of shear and moment at each of the change of 𝑖𝑓 𝑥 = 1, 𝑉𝐵 = 7 𝑘𝑁; 𝑖𝑓 𝑥 = 2, 𝑉𝐶 = 7 𝑘𝑁
load points, as these values will be used to draw
the shear and moment diagram later. 𝑴𝑩𝑪 = 𝟏𝟕𝒙 − 𝟏𝟎 𝒙 − 𝟏

𝑖𝑓 𝑥 = 1, 𝑀𝐵 = 17 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 2, 𝑀𝐶 = 24 𝑘𝑁. 𝑚

Shear and Moment Equations Segment CD

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (2 < 𝑥 < 3)

𝑽𝑪𝑫 = 𝟏𝟕 − 𝟏𝟎 − 𝟓 = 𝟐

𝑖𝑓 𝑥 = 2, 𝑉𝐶 = 2 𝑘𝑁; 𝑖𝑓 𝑥 = 3, 𝑉𝐷 = 2 𝑘𝑁

𝑴𝑪𝑫 = 𝟏𝟕𝒙 − 𝟏𝟎 𝒙 − 𝟏 − 𝟓 𝒙 − 𝟐

𝑖𝑓 𝑥 = 2, 𝑀𝐶 = 24 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 3, 𝑀𝐷 = 26 𝑘𝑁. 𝑚

25
EXAMPLE 2.
Shear and Moment Equations Segment DE
Write the shear and moment equations of each 𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (3 < 𝑥 < 5)
segment of the beam shown. Identify the value
of shear and moment at each of the change of 𝑽𝑫𝑬 = 𝟏𝟕 − 𝟏𝟎 − 𝟓 − 𝟏𝟓 = −𝟏𝟑
load points, as these values will be used to draw
the shear and moment diagram later. 𝑖𝑓 𝑥 = 3, 𝑉𝐷 = −13 𝑘𝑁; 𝑖𝑓 𝑥 = 5, 𝑉𝐸 = −13 𝑘𝑁

𝑴𝑫𝑬 = 𝟏𝟕𝒙 − 𝟏𝟎 𝒙 − 𝟏 − 𝟓 𝒙 − 𝟐 − 𝟏𝟓 𝒙 − 𝟑

𝑖𝑓 𝑥 = 3, 𝑀𝐷 = 26 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 5, 𝑀𝐸 = 0 𝑘𝑁. 𝑚

26
EXAMPLE 2.
Shear Equations

@ member AB; 𝑽𝑨 = 𝑽𝑩 = 𝟏𝟕

@ member AB; 𝑽𝑩 = 𝑽𝑪 = 𝟕

@ member AB; 𝑽𝑪 = 𝑽𝑫 = 𝟐

@ member AB; 𝑽𝑫 = 𝑽𝑬 = −𝟏𝟑

Moment Equations

𝑴𝑨𝑩 = 𝟏𝟕𝒙

𝑴𝑩𝑪 = 𝟏𝟕𝒙 − 𝟏𝟎 𝒙 − 𝟏

𝑴𝑪𝑫 = 𝟏𝟕𝒙 − 𝟏𝟎 𝒙 − 𝟏 − 𝟓 𝒙 − 𝟐

𝑴𝑫𝑬 = 𝟏𝟕𝒙 − 𝟏𝟎 𝒙 − 𝟏 − 𝟓 𝒙 − 𝟐 − 𝟏𝟓 𝒙 − 𝟑

27
EXAMPLE 3.
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:

Write the shear and moment equations of each Solve for reactions
segment of the beam shown. Identify the value
of shear and moment at each of the change of Σ𝑀𝐵 = 0
load points, as these values will be used to draw 4000 + 3500 14 −𝐷𝑉 16 = 0; 𝑫𝑽 = 𝟑𝟑𝟏𝟐. 𝟓 𝑵
the shear and moment diagram later.
Σ𝑀𝐷 = 0
4000 − 3500 2 +𝐵𝑉 16 = 0; 𝑩𝑽 = 𝟏𝟖𝟕. 𝟓 𝑵
Shear and Moment Equations Segment AB

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (0 < 𝑥 < 2)

𝑽𝑨𝑩 = 𝟎

𝑖𝑓 𝑥 = 0, 𝑉𝐴 = 0 𝑁; 𝑖𝑓 𝑥 = 2, 𝑉𝐵 = 0 𝑁

𝑴𝑨𝑩 = 𝟒𝟎𝟎𝟎

𝑖𝑓 𝑥 = 0, 𝑀𝐴 = 4000 𝑁. 𝑚; 𝑖𝑓 𝑥 = 2, 𝑀𝐵 = 4000 𝑁. 𝑚

28
EXAMPLE 3. Shear and Moment Equations Segment BC

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (2 < 𝑥 < 16)

Write the shear and moment equations of each 𝑽𝑩𝑪 = 𝟏𝟖𝟕. 𝟓

segment of the beam shown. Identify the value
of shear and moment at each of the change of 𝑖𝑓 𝑥 = 2, 𝑉𝐵 = 187.5 𝑁; 𝑖𝑓 𝑥 = 16, 𝑉𝐶 = 187.5 𝑁
load points, as these values will be used to draw
the shear and moment diagram later. 𝑴𝑩𝑪 = 𝟒𝟎𝟎𝟎 + 𝟏𝟖𝟕. 𝟓 𝒙 − 𝟐

𝑖𝑓 𝑥 = 2, 𝑀𝐵 = 4000 𝑁. 𝑚; 𝑖𝑓 𝑥 = 16, 𝑀𝐶 = 6625 𝑁. 𝑚

Shear and Moment Equations Segment CD

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (16 < 𝑥 < 18)

𝑽𝑪𝑫 = 𝟏𝟖𝟕. 𝟓 − 𝟑𝟓𝟎𝟎 = −𝟑𝟑𝟏𝟐. 𝟓

𝑖𝑓 𝑥 = 16, 𝑉𝐶 = −3312.5 𝑁; 𝑖𝑓 𝑥 = 18, 𝑉𝐷 = −𝟑𝟑𝟏𝟐. 𝟓 𝑁

𝑴𝑪𝑫 = 𝟒𝟎𝟎𝟎 + 𝟏𝟖𝟕. 𝟓 𝒙 − 𝟐 − 𝟑𝟓𝟎𝟎 𝒙 − 𝟏𝟔

𝑖𝑓 𝑥 = 16, 𝑀𝐶 = 6625 𝑁. 𝑚; 𝑖𝑓 𝑥 = 18, 𝑀𝐷 = 0 𝑁. 𝑚

29
EXAMPLE 3.
Shear Equations

@ member AB; 𝑽𝑨 = 𝑽𝑩 = 𝟎

@ member AB; 𝑽𝑩 = 𝑽𝑪 = 𝟏𝟖𝟕. 𝟓

@ member AB; 𝑽𝑪 = 𝑽𝑫 = −𝟑𝟑𝟏𝟐. 𝟓

Moment Equations

𝑴𝑨𝑩 = 𝟒𝟎𝟎𝟎

𝑴𝑩𝑪 = 𝟒𝟎𝟎𝟎 + 𝟏𝟖𝟕. 𝟓 𝒙 − 𝟐

𝑴𝑪𝑫 = 𝟒𝟎𝟎𝟎 + 𝟏𝟖𝟕. 𝟓 𝒙 − 𝟐 − 𝟑𝟓𝟎𝟎 𝒙 − 𝟏𝟔

30
EXAMPLE 4.
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
Solve for reactions
Write the shear and moment equations of each
segment of the beam shown. Identify the value Σ𝑀𝐴 = 0
of shear and moment at each of the change of 1 2
load points, as these values will be used to draw 200 20 10 + 20 −𝐶𝑉 20 = 0; 𝑪𝑽 = 𝟐𝟑𝟑𝟑. 𝟑𝟑 𝑵
2 3
the shear and moment diagram later.
Σ𝑀𝐶 = 0
1 2
200 20 20 − 10 +𝐴𝑉 20 = 0; 𝑨𝑽 = −𝟑𝟑𝟑. 𝟑𝟑 𝑵
2 3

Shear and Moment Equations Segment AB

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (0 < 𝑥 < 10)

𝑽𝑨𝑩 = −𝟑𝟑𝟑. 𝟑𝟑

𝑖𝑓 𝑥 = 0, 𝑉𝐴 = −333.33 𝑁; 𝑖𝑓 𝑥 = 10, 𝑉𝐵 = −333.33 𝑁

𝑴𝑨𝑩 = −𝟑𝟑𝟑. 𝟑𝟑𝒙

𝑖𝑓 𝑥 = 0, 𝑀𝐴 = 0 𝑁. 𝑚; 𝑖𝑓 𝑥 = 10, 𝑀𝐵 = −3333.33 𝑁. 𝑚

31
EXAMPLE 4.

Write the shear and moment equations of each Shear and Moment Equations Segment BC
segment of the beam shown. Identify the value
of shear and moment at each of the change of 𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (10 < 𝑥 < 20)
load points, as these values will be used to draw
the shear and moment diagram later. 𝟏 𝟐
𝑽𝑩𝑪 = −𝟑𝟑𝟑. 𝟑𝟑 − 𝟏𝟎 𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 = −𝟑𝟑𝟑. 𝟑𝟑 − 𝟓 𝒙 − 𝟏𝟎
𝟐

𝑖𝑓 𝑥 = 10, 𝑉𝐵 = −333.33 𝑁; 𝑖𝑓 𝑥 = 20, 𝑉𝐶 = −833.33 𝑁

𝟏 𝟏
𝑴𝑩𝑪 = −𝟑𝟑𝟑. 𝟑𝟑𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎
𝟐 𝟑

𝟓 𝟑
𝑴𝑩𝑪 = −𝟑𝟑𝟑. 𝟑𝟑𝒙 − 𝒙 − 𝟏𝟎
𝟑

𝑖𝑓 𝑥 = 10, 𝑀𝐵 = −3333.33 𝑁. 𝑚; 𝑖𝑓 𝑥 = 20, 𝑀𝐶 = −8333.33 𝑁. 𝑚

32
EXAMPLE 4.
Shear and Moment Equations Segment CD

Write the shear and moment equations of each 𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (20 < 𝑥 < 30)
segment of the beam shown. Identify the value
of shear and moment at each of the change of 𝟏
𝑽𝑪𝑫 = −𝟑𝟑𝟑. 𝟑𝟑 − 𝟏𝟎 𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 + 𝟐𝟑𝟑𝟑. 𝟑𝟑
load points, as these values will be used to draw 𝟐
the shear and moment diagram later.
𝟐
𝑽𝑪𝑫 = −𝟑𝟑𝟑. 𝟑𝟑 − 𝟓 𝒙 − 𝟏𝟎 + 𝟐𝟑𝟑𝟑. 𝟑𝟑

𝑖𝑓 𝑥 = 20, 𝑉𝐶 = 1500 𝑁; 𝑖𝑓 𝑥 = 30, 𝑉𝐷 = 0 𝑁

𝟏 𝟏
𝑴𝑪𝑫 = −𝟑𝟑𝟑. 𝟑𝟑𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 𝒙 − 𝟏𝟎 + 𝟐𝟑𝟑𝟑. 𝟑𝟑 𝒙 − 𝟐𝟎
𝟐 𝟑

𝟓 𝟑
𝑴𝑪𝑫 = −𝟑𝟑𝟑. 𝟑𝟑𝒙 − 𝒙 − 𝟏𝟎 + 𝟐𝟑𝟑𝟑. 𝟑𝟑 𝒙 − 𝟐𝟎
𝟑

𝑖𝑓 𝑥 = 20, 𝑀𝐶 = −8333.33 𝑁. 𝑚; 𝑖𝑓 𝑥 = 30, 𝑀𝐷 = 0 𝑁. 𝑚

33
 EXAMPLE 4.

Shear Equations

@ member AB; 𝑽𝑨 = 𝑽𝑩 = −𝟑𝟑𝟑. 𝟑𝟑

𝟐
𝑽𝑩𝑪 = −𝟑𝟑𝟑. 𝟑𝟑 − 𝟓 𝒙 − 𝟏𝟎

𝟐
𝑽𝑪𝑫 = −𝟑𝟑𝟑. 𝟑𝟑 − 𝟓 𝒙 − 𝟏𝟎 + 𝟐𝟑𝟑𝟑. 𝟑𝟑

Moment Equations

𝑴𝑨𝑩 = −𝟑𝟑𝟑. 𝟑𝟑𝒙

𝟓 𝟑
𝑴𝑩𝑪 = −𝟑𝟑𝟑. 𝟑𝟑𝒙 − 𝒙 − 𝟏𝟎
𝟑

𝟓 𝟑
𝑴𝑪𝑫 = −𝟑𝟑𝟑. 𝟑𝟑𝒙 − 𝒙 − 𝟏𝟎 + 𝟐𝟑𝟑𝟑. 𝟑𝟑 𝒙 − 𝟐𝟎
𝟑

34
 UNIVERSITY OF THE CORDILLERAS COLLEGE OF ENGINEERING AND ARCHITECTURE

MECH 3 MECHANICS OF DEFORMABLE BODIES

MODULE 4: SHEAR AND MOMENT OF BEAMS

UNIT 2: RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT

UNIT 3: POINT OF ZERO SHEAR AND MOMENT

Prepared by:

Brian Jhay Guzman, CE, RMP, ME-1 Melkisidick Angloan, CE, ME-1
 UNIT LEARNING OUTCOMES
In this unit, here are the following desired learning outcomes:
✘ Establish relationships between load, shear and moment.
✘ Draw shear and moment diagrams using equations and semi-
graphical methods.
✘ Solve for point of zero shear and point of inflection.

38
 RECALL
The fundamental definition of shear is expressed The fundamental definition of bending moment is
as expressed as

𝑉 = Σ𝐹𝑦 𝑀 = Σ𝑀 𝐿𝑒𝑓𝑡 = Σ𝑀 𝑅𝑖𝑔ℎ𝑡

𝐿𝑒𝑓𝑡

The shearing force 𝑉 should be computed only in The bending moment 𝑀 may be computed in
terms of the forces to the left of the section terms of the forces to either left or right of the
(change of load point) being considered for section, depending on which requires less
uniformity purposes. arithmetical work.

𝑈𝑝𝑤𝑎𝑟𝑑 𝐹𝑜𝑟𝑐𝑒𝑠 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑆ℎ𝑒𝑎𝑟 𝑈𝑝𝑤𝑎𝑟𝑑 𝐹𝑜𝑟𝑐𝑒𝑠 = 𝑃𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑀𝑜𝑚𝑒𝑛𝑡

39
 RELATIONSHIP
BETWEEN LOAD,
SHEAR AND MOMENT

40
 RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT

A simple beam with a varying load indicated by

𝑤(𝑥) is sketched in Fig. 6-8. The coordinate
system with origin at the left end 𝐴 is established
and distances to various sections in the beam are
denoted by the variable x. Let us derive
relationships between 𝑤(𝑥), 𝑉(𝑥), and 𝑀(𝑥) at any
section of the beam, shown in Fig. 6-8.
The beam is subjected to any type of transverse
load of the general form shown in Fig. 6-8(a).
Simple supports are illustrated but the following
consideration holds for all types of beams. We will
isolate from the beam the element of length 𝑑𝑥
shown and draw a free-body diagram of it.

41
 RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT

The shearing force 𝑉 acts on the left side of the

element, and in passing through the distance 𝑑𝑥 the
shearing force 𝑉 will in general change to 𝑉 + 𝑑𝑉.
The bending moment 𝑀 acts on the left side of the
element and 𝑀 + 𝑑𝑀 on the right side. Since 𝑑𝑥 is
extremely small, the applied load may be taken as
uniform over the top of the beam and equal to 𝑤.
The free-body diagram of this element thus appears
as in Fig. 6-8(b). For equilibrium of moments about 𝑂,
we have

Σ𝑀𝑂 = 𝑀 − 𝑀 + 𝑑𝑀 + 𝑉𝑑𝑥 + 𝑤𝑑𝑥 𝑑𝑥 2 = 0

Or
1 2
𝑑𝑀 = 𝑉𝑑𝑥 + 𝑤 𝑑𝑥
2 42
 RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT
Since the last term consists of the product of two
differentials, it is negligible compared with the
other terms involving only one differential. Hence

𝑑𝑀 = 𝑉𝑑𝑥
𝑂𝑟
𝑑𝑀
𝑉=
𝑑𝑥

𝑀2 𝑥2
𝑑𝑀 = 𝑉𝑑𝑥
𝑀1 𝑥1
𝑀2 − 𝑀1 = 𝑉(𝑥2 − 𝑥1 )

∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟

43
RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT
For vertical equilibrium of the element we have

Σ𝐹𝑦 = 𝑤𝑑𝑥 + 𝑉 − 𝑉 + 𝑑𝑉 = 0

𝑑𝑉 = 𝑤𝑑𝑥
𝑂𝑟
𝑑𝑉
𝑤=
𝑑𝑥

𝑉2 𝑥2
𝑑𝑉 = 𝑤𝑑𝑥
𝑉1 𝑥1
𝑉2 − 𝑉1 = 𝑤 (𝑥2 − 𝑥1 )

∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑

This relations will be of value in establishing shear and

moment diagrams.
44
RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT

𝑉2 − 𝑉1 = ∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 𝑀2 − 𝑀1 = ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟

“The change in shear ∆𝑉 between any two “The change in bending moment ∆𝑀 between any
points/sections is equal to the area of the load two points/sections is equal to the area of the
diagram between these two points/sections.” shear diagram between these two
points/sections.”

45
RELATIONSHIP BETWEEN LOAD, SHEAR AND MOMENT

𝑑𝑉 𝑑𝑀
𝑤= 𝑉=
𝑑𝑥 𝑑𝑥

Intensity of load = corresponding slope of shear Intensity of shear = corresponding slope of

diagram. moment diagram.

The relations are amplified to provide a semigraphical method of computing shear and
moment which supplements shear and moment equations and enable us to sketch the
proper shapes of the shear and moment diagrams rapidly and correctly.

46
 SIGN CONVENTION FOR SLOPES
POSITIVE SLOPES (+ SLOPE)= UP TO THE RIGHT NEGATIVE SLOPES (- SLOPE)= DOWN TO THE LEFT
“The larger the value, the steeper the slope” “The larger the value, the steeper the slope”

These interpretations of the slopes will be used to establish the direction of the concavity of
the various curves in the shear and moment diagrams.

47
POINT OF ZERO SHEAR
AND MOMENT

48
POINT OF ZERO SHEAR AND MOMENT
Shear and moment diagrams are merely the graphical visualization of the shear and
moment equations plotted on 𝑉 − 𝑥 and 𝑀 − 𝑥 axes, usually located below the loading
diagram.
The discontinuities in the shear diagram are joined by vertical lines drawn up or down to
represent the abrupt changes in shear caused respectively by upward or downward
concentrated loads.
A final point to be observed this time is that the highest and lowest points on the moment
diagram always corresponds to sections of zero shear. The value of 𝑥 making 𝑀 maximum
can be found by differentiating 𝑀 with respect to 𝑥 and equating the result to zero. This
result will be the shear equation 𝑉, thus we see that the maximum moment corresponds to
the section of zero shear.
Where the beam changes its shape from concave up to concave down, we have what is
called the point of inflection; it corresponds to the section of zero bending moment. Its
position may be calculated by letting the 𝑀 equation equal to zero.
49
 SUMMARY
The procedure for the construction of shear and moment diagrams:
1. Compute the reactions.
2. Compute values of shear at the change of load points (segments) using either 𝑉 = Σ𝐹𝑦
𝐿𝑒𝑓𝑡
or ∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 .
𝑑𝑉
3. Sketch the diagram, determining the shape from 𝑤 = that is, the intensity of the load
𝑑𝑥
ordinate equals the slope at the corresponding ordinate of the shear diagram.
4. Locate the point(s) of zero shear.
5. Compute the values of bending moment at the change of load points and at the points of
zero shear, using either 𝑀 = Σ𝑀 𝐿𝑒𝑓𝑡 = Σ𝑀 𝑅𝑖𝑔ℎ𝑡 or ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 , whichever is more
convenient.
6. Sketch the moment diagram through the ordinates of the bending moments computed in
𝑑𝑀
step 5. The shape of the diagram is determined from 𝑉 = that is, the intensity of the shear
𝑑𝑥
ordinate equals the slope at the corresponding ordinate of the moment diagram.
50
EXAMPLE 1.
USING SHEAR AND
MOMENT EQUATIONS
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛:
Consider the given loaded beam below, draw Solve for reactions
the shear and moment diagrams and
determine the locations of the point of zero Σ𝑀𝐵 = 0
shear and point of inflection. 3+9
2× 3+9 − 3 +8 9 − 𝐷𝑉 9 + 3 = 0; 𝑫𝑽 = 𝟏𝟐 𝒌𝑵
a. Using shear and moment equations. 2
Σ𝑀𝐸 = 0
b. Using the relationship of load, shear and
3+9
moment. − 2× 3+9 + 3 −8 3 + 𝐵𝑉 9 + 3 = 0; 𝑩𝑽 = 𝟐𝟎 𝒌𝑵
2

Shear and Moment Equations Segment AB

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (0 < 𝑥 < 3)

𝑽𝑨𝑩 = −𝟐𝒙

𝑖𝑓 𝑥 = 0, 𝑉𝐴 = 0 𝑘𝑁; 𝑖𝑓 𝑥 = 3, 𝑉𝐵 = −6 𝑘𝑁

𝒙
𝑴𝑨𝑩 = −𝟐𝒙 ∙ = −𝒙𝟐
𝟐

𝑖𝑓 𝑥 = 0, 𝑀𝐴 = 0 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 3, 𝑀𝐵 = −9 𝑘𝑁. 𝑚

51
EXAMPLE 1.
USING SHEAR AND
MOMENT EQUATIONS Shear and Moment Equations Segment BC

Consider the given loaded beam below, draw 𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (3 < 𝑥 < 12)
the shear and moment diagrams and
determine the locations of the point of zero 𝑽𝑩𝑪 = −𝟐𝒙 + 𝟐𝟎
shear and point of inflection.
𝑖𝑓 𝑥 = 3, 𝑉𝐵 = 14 𝑘𝑁; 𝑖𝑓 𝑥 = 12, 𝑉𝐶 = −4 𝑘𝑁
a. Using shear and moment equations.
𝒙
b. Using the relationship of load, shear and 𝑴𝑩𝑪 = −𝟐𝒙 ∙ + 𝟐𝟎 𝒙 − 𝟑 = −𝒙𝟐 + 𝟐𝟎(𝒙 − 𝟑)
moment. 𝟐

𝑖𝑓 𝑥 = 3, 𝑀𝐵 = −9 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 12, 𝑀𝐶 = 36 𝑘𝑁. 𝑚

Shear and Moment Equations Segment CD

𝐿𝑒𝑓𝑡 𝑆𝑖𝑑𝑒 (12 < 𝑥 < 15)

𝑽𝑪𝑫 = −𝟐 𝟑 + 𝟗 + 𝟐𝟎 − 𝟖 = −𝟏𝟐

𝑖𝑓 𝑥 = 12, 𝑉𝐶 = −12 𝑘𝑁; 𝑖𝑓 𝑥 = 15, 𝑉𝐷 = −12 𝑘𝑁

𝟑+𝟗
𝑴𝑪𝑫 = −𝟐 𝟑 + 𝟗 𝒙− + 𝟐𝟎 𝒙 − 𝟑 − 𝟖(𝒙 − 𝟏𝟐)
𝟐
𝑴𝑪𝑫 = −𝟐𝟒 𝒙 − 𝟔 + 𝟐𝟎 𝒙 − 𝟑 − 𝟖(𝒙 − 𝟏𝟐)

𝑖𝑓 𝑥 = 12, 𝑀𝐶 = 36 𝑘𝑁. 𝑚; 𝑖𝑓 𝑥 = 15, 𝑀𝐷 = 0 𝑘𝑁. 𝑚

EXAMPLE 1.
USING SHEAR AND
MOMENT EQUATIONS

Consider the given loaded beam below, draw

the shear and moment diagrams and Solving for points of zero shear:
determine the locations of the point of zero
shear and point of inflection. Equate the shear equations to zero

a. Using shear and moment equations. 𝑽𝑨𝑩 = −𝟐𝒙 = 𝟎

b. Using the relationship of load, shear and
moment. 𝑽𝑩𝑪 = −𝟐𝒙 + 𝟐𝟎 = 𝟎 → 𝒙 = 𝟏𝟎 𝒎 (𝒇𝒓𝒐𝒎 𝑨)

𝑽𝑪𝑫 = −𝟏𝟐 = 𝟎

Recall that point of zero shear contains point of maximum moments

EXAMPLE 1.
USING SHEAR AND
MOMENT EQUATIONS
Solving for points of zero moment (Point of Inflection):
Consider the given loaded beam below, draw
the shear and moment diagrams and Equate the moment equations to zero
determine the locations of the point of zero
shear and point of inflection. 𝑴𝑨𝑩 = −𝒙𝟐 = 𝟎 → 𝒙 = 𝟎

a. Using shear and moment equations. Since 𝒙 = 𝟎 is located at point A (start of the beam) it is not a valid point of
b. Using the relationship of load, shear and inflection. The change in curvature at this point is not really significant.
moment.
𝑴𝑩𝑪 = −𝒙𝟐 + 𝟐𝟎 𝒙 − 𝟑 = 𝟎 → 𝒙𝟏 = 𝟑. 𝟔𝟕𝟓 𝒎, 𝒙𝟐 = 𝟏𝟔. 𝟑𝟐𝟓 𝒎

Since segment BC is only from 3 m to 12 m from A only, 𝒙𝟐 = 𝟏𝟔. 𝟑𝟐𝟓 𝒎 is

not a valid point of inflection. Therefore use 𝒙𝟏 = 𝟑. 𝟔𝟕𝟓 𝒎 (𝒇𝒓𝒐𝒎 𝑨)
as the point of inflection.

𝑴𝑪𝑫 = −𝟐𝟒 𝒙 − 𝟔 + 𝟐𝟎 𝒙 − 𝟑 − 𝟖 𝒙 − 𝟏𝟐 = 𝟎 → 𝒙 = 𝟏𝟓 𝒎

Since 𝒙 = 𝟏𝟓 is located at point D (end of the beam) it is not a valid point of

inflection. The change in curvature at this point is not really significant.
 EXAMPLE 1.
USING SHEAR AND
MOMENT EQUATIONS

Shear Equations Moment Equations

𝑽𝑨𝑩 = −𝟐𝒙 𝑴𝑨𝑩 = −𝒙𝟐

𝑽𝑩𝑪 = −𝟐𝒙 + 𝟐𝟎 𝑴𝑩𝑪 = −𝒙𝟐 + 𝟐𝟎(𝒙 − 𝟑)

𝑽𝑪𝑫 = −𝟏𝟐 𝑴𝑪𝑫 = −𝟐𝟒 𝒙 − 𝟔 + 𝟐𝟎 𝒙 − 𝟑 − 𝟖(𝒙 − 𝟏𝟐)

Solving for the value of the moment at the point of zero shear:

𝑽𝑩𝑪 = −𝟐𝒙 + 𝟐𝟎 = 𝟎 → 𝒙 = 𝟏𝟎 𝒎 (𝒇𝒓𝒐𝒎 𝑨)

Recall that point of zero shear contains point of maximum moments

Therefore 𝑴𝑩𝑪 @ 𝒙 = 𝟏𝟎 𝒎 (𝒇𝒓𝒐𝒎 𝑨)

𝑴𝑩𝑪 = −𝟏𝟎𝟐 + 𝟐𝟎(𝟏𝟎 − 𝟑)

𝑴𝑩𝑪 = 𝟒𝟎 𝑘𝑁. 𝑚
55
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝑈𝑠𝑖𝑛𝑔 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑙𝑜𝑎𝑑, 𝑠ℎ𝑒𝑎𝑟 𝑎𝑛𝑑 𝑚𝑜𝑚𝑒𝑛𝑡
Consider the given loaded beam below, draw (Semi-graphical solution)
the shear and moment diagrams and
determine the locations of the point of zero Solve for reactions
shear and point of inflection.
Σ𝑀𝐵 = 0
a. Using shear and moment equations. 3+9
2× 3+9 − 3 +8 9 − 𝐷𝑉 9 + 3 = 0; 𝑫𝑽 = 𝟏𝟐 𝒌𝑵
b. Using the relationship of load, shear and 2
moment. Σ𝑀𝐷 = 0
3+9
− 2× 3+9 + 3 −8 3 + 𝐵𝑉 9 + 3 = 0; 𝑩𝑽 = 𝟐𝟎 𝒌𝑵
2

For the Shear Diagram

Recall:

𝑉2 − 𝑉1 = ∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑

“The change in shear ∆𝑉 between any two points/sections is equal to the

area of the load diagram between these two points/segments.”

“The discontinuities in the shear diagram are joined by vertical lines

drawn up or down to represent the abrupt changes in shear caused
respectively by upward or downward concentrated loads.”

56
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION
𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝑅𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑠ℎ𝑒𝑎𝑟 𝑑𝑖𝑎𝑔𝑟𝑎𝑚𝑠 𝑎𝑟𝑒 𝑑𝑟𝑎𝑤𝑛 𝑓𝑟𝑜𝑚 𝑙𝑒𝑓𝑡 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡.
@A
𝑉𝐴 = 0 → "𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑜 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑙𝑜𝑎𝑑𝑠/𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛𝑠 𝑎𝑡 𝐴

@B
𝑉𝐵 − 𝑉𝐴 = ∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 → 𝑉𝐵 = ∆𝑉 + 𝑉𝐴
→ From A to B the Area of the Load diagram is the Area of the Rectangular
Distributed Load, with an intensity of 2kN/m ↓ and a span of 3 m.
𝑘𝑁
∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 = −2 3 𝑚 = −6 𝑘𝑁
𝑚
𝑉𝐵 = ∆𝑉 + 𝑉𝐴 = −6 𝑘𝑁 + 0 → 𝑉𝐵 = −6 𝑘𝑁 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
*Still @ B, before proceeding to C:
𝑉𝐵 = −6 𝑘𝑁 + 20 𝑘𝑁 → 𝑉𝐵 = 14 𝑘𝑁 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)

“Abrupt change in shear caused by the reaction at B.” “𝑉𝐵 to be carried over to point C.”
EXAMPLE 1.
SEMI-GRAPHICAL @C
SOLUTION
𝑉𝐶 − 𝑉𝐵 = ∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 → 𝑉𝐶 = ∆𝑉 + 𝑉𝐵

𝑘𝑁
∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 = −2 9 𝑚 = −18 𝑘𝑁
𝑚
𝑉𝐶 = ∆𝑉 + 𝑉𝐵 = −18 𝑘𝑁 + 14 𝑘𝑁
𝑉𝐶 = −4 𝑘𝑁 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
*Still @ C, before proceeding to D:
𝑉𝐶 = −4 𝑘𝑁 − 8 𝑘𝑁 → 𝑉𝐶 = −12 𝑘𝑁 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
“Abrupt change in
shear caused by “𝑉𝐶 to be carried over to point D.”
concentrated load at
C.”
@D
𝑉𝐷 − 𝑉𝐶 = ∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 → 𝑉𝐷 = ∆𝑉 + 𝑉𝐶
∆𝑉 = 𝐴𝑟𝑒𝑎 𝐿𝑜𝑎𝑑 = 0 → "𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑜 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑙𝑜𝑎𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐶 𝑎𝑛𝑑 𝐷.
𝑉𝐷 = ∆𝑉 + 𝑉𝐶 = 0 𝑘𝑁 − 12 𝑘𝑁
𝑉𝐷 = −12 𝑘𝑁 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
*Still at D
𝑉𝐷 = −12 𝑘𝑁 + 12 𝑘𝑁 → 𝑉𝐷 = 0 𝑘𝑁 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
“Abrupt change in shear caused by the reaction at D.”
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION
Recall:
𝑑𝑉
𝑤=
𝑑𝑥
“Intensity of load = corresponding slope of shear diagram.”

↗ POSITIVE SLOPES (+ SLOPE)= UP TO THE RIGHT

↘ NEGATIVE SLOPES (- SLOPE)= DOWN TO THE RIGHT

Segment AB
To determine the curvature of the shear diagram from A to B. Take the
intensity of the load diagram as the slope of the curve at any given point in
Segment AB. At point A, the slope of the curve is -2 because the intensity of
the load diagram at that point is 2 kN/m (downward). Therefore the slope is
down to the right. At point B, the slope is also -2. In fact all through out the
span of Segment AB the slope is -2 because of the rectangular distributed
load of 2 kN/m (downward). And from basic calculus we know that a
polynomial curve of constant slope is a STRAIGHT LINE.
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION

Segment BC
For Segment BC the slope is also constant as -2. Therefore the curve
between B and C is also a STRAIGHT LINE.
Notice that the line connecting B and C crosses the horizontal line
representing zero (0) ordinate. This tells us that this intersection is the
location of the point of zero shear.

Segment CD
For Segment CD no distributed load present, or we can say that the intensity
of the distributed load is zero (0). And in calculus a slope of 0 is represented
by a HORIZONTAL LINE.

*Do not forget to connect with vertical lines the different values of shear for
points B and C. These vertical lines are the representation of the abrupt
changes in shear caused by concentrated loads at those points.
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION

Solving for points of zero shear:

The point of zero shear can be solved using ratio and proportion

14 14 + 4
= → 𝑥 = 7 𝑅𝑒𝑐𝑘𝑜𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 𝐵
𝑥 9

If reckoned from A,

The point of zero shear = 7 + 3 = 10 𝑚 (𝑓𝑟𝑜𝑚 𝐴)

Recall that point of zero shear contains point of maximum moments

𝒙
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION
For the Moment Diagram
Recall:

𝑀2 − 𝑀1 = ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟

“The change in bending moment ∆𝑀 between any two points/sections is

equal to the area of the shear diagram between these two points/segments.”

𝑆𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝑅𝑒𝑚𝑒𝑚𝑏𝑒𝑟 𝑡ℎ𝑎𝑡 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑖𝑎𝑔𝑟𝑎𝑚𝑠 𝑎𝑟𝑒 𝑑𝑟𝑎𝑤𝑛 𝑓𝑟𝑜𝑚 𝑙𝑒𝑓𝑡 𝑡𝑜 𝑟𝑖𝑔ℎ𝑡.
@A
𝑀𝐴 = 0 → "𝑇ℎ𝑒𝑟𝑒 𝑎𝑟𝑒 𝑛𝑜 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑚𝑜𝑚𝑒𝑛𝑡𝑠/𝑐𝑜𝑢𝑝𝑙𝑒𝑠 𝑎𝑡 𝐴
@B
𝑀𝐵 − 𝑀𝐴 = ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 → 𝑀𝐵 = ∆𝑀 + 𝑀𝐴
→ From A to B the Area of the Shear diagram is the Area of a triangle, with
an base of 3 m and height of -6 kN.
1
∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 = 3 𝑚 −6 𝑘𝑁 = −9 𝑘𝑁 ∙ 𝑚
2
𝑀𝐵 = ∆𝑀 + 𝑀𝐴 = −9 𝑘𝑁 ∙ 𝑚 + 0
𝑀𝐵 = −9 𝑘𝑁 ∙ 𝑚 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION Before we proceed to point C, consider the point of zero shear which will
give a maximum moment. Hence, point E.
@E
𝑀𝐸 − 𝑀𝐵 = ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 → 𝑀𝐸 = ∆𝑀 + 𝑀𝐵
→ From B to E the Area of the Shear diagram is the Area of a triangle, with
an base of 7 m and height of 14 kN.
1
∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 = 7 𝑚 14 𝑘𝑁 = 49 𝑘𝑁 ∙ 𝑚
2
𝑀𝐸 = ∆𝑀 + 𝑀𝐵 = 49 𝑘𝑁 ∙ 𝑚 + (−9 𝑘𝑁 ∙ 𝑚)
𝟐𝒎
𝑀𝐸 = 40 𝑘𝑁 ∙ 𝑚 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
𝟕𝒎
@C
𝑀𝐶 − 𝑀𝐸 = ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 → 𝑀𝐶 = ∆𝑀 + 𝑀𝐸
→ From E to C the Area of the Shear diagram is the Area of a triangle, with
an base of 2 m and height of -4 kN.
1
∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 = 2 𝑚 −4 𝑘𝑁 = −4 𝑘𝑁 ∙ 𝑚
2
𝑀𝐶 = ∆𝑀 + 𝑀𝐸 = 40 𝑘𝑁 ∙ 𝑚 + (−4 𝑘𝑁 ∙ 𝑚)
𝑀𝐶 = 36 𝑘𝑁 ∙ 𝑚 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION @D
𝑀𝐷 − 𝑀𝐶 = ∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 → 𝑀𝐷 = ∆𝑀 + 𝑀𝐶
→ From B to E the Area of the Shear diagram is the Area of a rectangle, with
an base of 3 m and height of −12 kN.
∆𝑀 = 𝐴𝑟𝑒𝑎 𝑆ℎ𝑒𝑎𝑟 = 3 𝑚 −12 𝑘𝑁 = −36 𝑘𝑁 ∙ 𝑚
𝑀𝐷 = ∆𝑀 + 𝑀𝐶 = 36 𝑘𝑁 ∙ 𝑚 + (−36 𝑘𝑁 ∙ 𝑚)
𝑀𝐷 = 0 𝑘𝑁 ∙ 𝑚 (𝑝𝑙𝑜𝑡 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚)

𝟐𝒎 Recall:
𝑑𝑀
𝟕𝒎 𝑉=
𝑑𝑥

“Intensity of shear = corresponding slope of moment diagram.”

↗ POSITIVE SLOPES (+ SLOPE)= UP TO THE RIGHT

↘ NEGATIVE SLOPES (- SLOPE)= DOWN TO THE RIGHT
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION
Segment AB
To determine the curvature of the moment diagram from A to B. Take the
intensity of the shear diagram as the slope of the curve at any given point in
Segment AB. At point A, the slope of the curve is 0 because the intensity of
the shear diagram at that point is 0 kN. Therefore the slope is horizontal. At
point B, the slope is -6. Therefore down to the right.
In fact all through out the span of Segment AB the slope is gradually
increasing negatively, starting from 0 slope (horizontal) to a -6 slope (down
to the right), therefore if we create tangent lines for the slope at each point
between A and B, we will observe lines starting horizontal and then
gradually going down to the right while increasing in steepness. Hence the
𝟐𝒎 shape of the curve at Segment AB.
𝟕𝒎 Segment BE
To determine the curvature of the moment diagram from B to E. Take the
intensity of the shear diagram as the slope of the curve at any given point in
Segment BE. At point B, the slope of the curve is 14 because the intensity of
the shear diagram at that point is 14 kN. Therefore the slope is up to the
right. At point E, the slope is 0 kN (point of zero shear). Therefore,
horizontal.
If we create tangent lines for the slope at each point between B and E, we
will observe lines starting up to the right and then gradually going
horizontal decreasing in steepness. Hence the shape of the curve at Segment
BE.
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION
Segment EC
To determine the curvature of the moment diagram from E to C. Take the
intensity of the shear diagram as the slope of the curve at any given point in
Segment EC. At point E, the slope of the curve is 0 because the intensity of
the shear diagram at that point is 0 kN. Therefore the slope is horizontal. At
point B, the slope is -4. Therefore down to the right.
In fact all through out the span of Segment EC the slope is gradually
increasing negatively, starting from 0 slope (horizontal) to a -4slope (down
to the right), therefore if we create tangent lines for the slope at each point
between E and C, we will observe lines starting horizontal and then
gradually going down to the right while increasing in steepness. Hence the
𝟐𝒎 shape of the curve at Segment EC.
𝟕𝒎 Segment CD
To determine the curvature of the moment diagram from C to D. Take the
intensity of the shear diagram as the slope of the curve at any given point in
Segment CD. At point C, the slope of the curve is -12 because the intensity of
the shear diagram at that point is -12 kN. Therefore the slope is down to the
right. At point D, the slope is also -12.
In fact all through out the span of Segment CD the slope is -12 because of the
rectangle on the shear diagram of height -12 kN. And from basic calculus we
know that a polynomial curve of constant slope is a STRAIGHT LINE.
EXAMPLE 1.
SEMI-GRAPHICAL
SOLUTION

Solving for the point of inflection:

The point of inflection can be solved using squared property of parabola.

𝑥1 2 𝑥2 2 𝑥1 2 72
= → = → 𝑥1 = 6.325 𝑅𝑒𝑐𝑘𝑜𝑛𝑒𝑑 𝑓𝑟𝑜𝑚 𝐸
𝑦1 𝑦2 40 40 + 9
𝟕𝒎
If reckoned from A,
𝒙𝟐
𝒙𝟏 The point of i𝑛𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 = 7 + 3 − 6.325 = 3.675 𝑚 (𝑓𝑟𝑜𝑚 𝐴)

𝒚𝟏 𝒚𝟐

Beam Deformation

Point of Inflection Lowest Point

 EXAMPLE 1.
Considering both solutions, we can observe that the degree of the
equation (n) for a particular segment dictates the shape of the
diagram.

It is also suggested for the student to use both equations and the semi-
graphical solution simultaneously to draw the diagrams and solve for
important points as fast as possible.

Shear Equations Degree of the Equations

𝑽𝑨𝑩 = −𝟐𝒙 𝒏 = 𝟏 (𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆)

𝑽𝑩𝑪 = −𝟐𝒙 + 𝟐𝟎 𝒏 = 𝟏 (𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆) 𝒏=𝟎

𝑽𝑪𝑫 = −𝟏𝟐 𝒏 = 𝟎 (𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒍𝒊𝒏𝒆)

Moment Equations Degree of the Equations

𝑴𝑨𝑩 = −𝒙𝟐 𝒏 = 𝟐 (𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒂)

𝑴𝑩𝑪 = −𝒙𝟐 + 𝟐𝟎(𝒙 − 𝟑) 𝒏 = 𝟐 (𝒑𝒂𝒓𝒂𝒃𝒐𝒍𝒂)

𝑴𝑪𝑫 = −𝟐𝟒 𝒙 − 𝟔 + 𝟐𝟎 𝒙 − 𝟑 − 𝟖(𝒙 − 𝟏𝟐) 𝒏 = 𝟏 (𝒔𝒕𝒓𝒂𝒊𝒈𝒉𝒕 𝒍𝒊𝒏𝒆)

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