### Chapter 2 - Dr. ZM Nizam

```BFC 20903 (Mechanics of Materials)
Chapter 2: Shear Force and Bending Moment
Shahrul Niza Mokhatar
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Chapter Learning Outcome
1. Draw the shear and moment diagrams.
2. Determine the largest shear and moment in a
member and specify the location.
Introduction
• Changes of shear force and bending moments in beams as
reactions to several types of loading or combinations of
• Value of shear force and bending moment as well as the
location of maximum values can be determined.
• These maximum values are important because they will be
used in the analysis of structural design.
Equilibrium system
Beam
Roller
• How to make this system to be balanced?
Types of beam
• Beams can be defined as a structural component that has
a smaller dimension of cross section than its length.
– Statically determinate beam
– Statically indeterminate beam
• These classifications of beams depend on the unknown
total reactions to the static equilibrium equation.
• There are 3 static equilibrium equations: classified into
two types based on their analyses
Types of beam
• These classifications of beams depend on the
unknown total reactions to the static equilibrium
equation. There are 3 static equilibrium equations:
• if unknown total reaction is more than 3, thus the
beam is “Statically indeterminate beam”. Below is
how the stability and capability of beams are being
determined:
Type of support
• Reaction can be determined according to its support.
– Simply support
• This beam is supported by rollers at one end and by pins
at the other end.
• This beam is a statically determinate beam.
– Cantilever beam
• This beam has a fixed support at one end and a free
support at the other end.
• The beam is a statically determinate beam
– Continuous Beam
• This beam has several supports.
• This beam is a statically indeterminate beam
Type of support
– Loading is fixed and very close to the approximation values (very little
change). Some examples are structure self weight, ceiling and fixed
equipment such as the piping system services, lighting and air
conditioning. For a sample of construction elements please refer to BS /
EUCode.
– It is a variable load on a structure for example; human weight, furniture,
and other equipment based on structure purpose. Values for the loading
can be referred to BS EN 1991.
– The design of buildings must account for wind load especially in highrise buildings. However the effects of this wind load depend on the
location, shape, building dimension, and wind speed in the particular
region.
– This load acts at one position only whether in vertical or inclined direction. The
unit for this type of loading is N, kN.
– This load acts at a certain distance or a distributed load. This loading is measured
in unit load per distance which is N/mm, kN/m and so on. The value can be
constant or it can increase and decrease along its reaction distance .
– Resulting load made by a pair of forces and this force will create moment on
the point of loading The unit for this load is the unit of load multiple with the
perpendicular distance (Nmm, kNm and so on).
Reactions on supports
• Free Body Diagram
– In order to calculate the reaction from any structure, first,
draw the actual structure again in the form of a Free Body
Diagram.
– Free body diagram is the sketch of the actual structure in
lines form that are free from supports as well as from
loads acting on it. Draw all reactions at the supports &
the directions of the reactions & their accurate
dimensions. Draw all forces or loads acting on the
structure elements.
Equilibrium equation
• This consists of 3 equations, where by using these three
equations, reactions of certain structures can be determined.
The equation is:
• To avoid a simultaneous equation, let’s use moment equation,
ΣM = 0 first on any one support point, followed by other
reactions on support point that can be determined using this
equation ΣFx = 0 ; Σ Fy = 0.
• If the solution of this simultaneous equation gives the
negative value, it means the earlier assumption of reaction
direction is incorrect and the actual direction is the opposite
direction.
Example 1
• Determine the reaction of beam stressed with a load as
shown in Figure
Example 1: Solution
Example 2
Determine the reaction on beam
Example 2: Solution
Example 3: Tutorial (Discuss in
Group)
Find all the reaction force.
4 kN/m
C
D
3m
4m
2 kN
B
E
3m
A
5m
Example 3: Solution
RAx = -2 kN
RAy = 8.8 kN
RCy = 8.8 kN, MC = 6 kNm
REy = 11.2 kN
RDy = 11.2 kN, MD = 0
Shear Force Diagram (SFD) Bending
Moment Diagram (BMD)?
Why we need SFD & BMD?
Moment (-ve)
Moment (+ve)
Shear Force and Bending Moment
– Shear force, V at a section is the total algebra force taken
only at one side of the section. Any section can be taken to
be totaled provided that both sections must have the same
values and they follow the marking direction.
– Bending moment, Mx is the total algebra of moments force
acting on the section.
Shear Force and Bending Moment
 To make sure that the beam is in a
balanced vertical direction, there should
exist a V force that balances the RA force
and one moment M with an anti clock
wise movement to prevent the beam from
rotating. Both the force and moment must
be balanced on the section that has been
cut. The total force and moment at this
section equals to zero.
 The V force is known as a shear force, the
force that causes the member to slide and
separate into two sections . Meanwhile
moment M is known as the bending
moment, which is a reaction moment at a
certain point for all forces as well as
combinations reacting on the left or right
side of the point.
Natural sign for shear force
and bending moment
• Common sign
– Horizontal direction
– Vertical direction
– Moment direction
Shape of SFD and BMD
Procedure to draw SFD and BMD
– Support reactions – determine all the reactive force & couple
moments acting on the beam.
– Shear & moment functions – specify separate coordinates x
having an origin at the beam’s left & extending to regions of the
beam between concentrated forces &/or couple moments.
Section beam at each distance x, & draw the FBD of one of the
segments.
– Shear & moment diagram – Plot the shear diagram (V versus x)
& moment diagram (M versus x). If the numerical values of the
functions describing V & M are positive, the values are plotted
above the x axis, whereas negative values are plotted below the
axis. It is convenient to show the shear & moment diagram
below the FBD of the beam.
Equation & Diagram of
Shear Force & Bending Moment
1. Simply supported beam with point load acting.
To get the V and M equations,
make sections on each part
between the point load,
distributed load and support.
The above equation shows that shear
force caused by point load is constant or
fixed and the bending moment changes
linearly with x.
Note:
– SFD positive, +ve is
drawn on top of the
beam.
– BMD positive, +ve is
drawn on top of the
beam.
– To draw SFD & BMD, it
is enough to determine
V & M value on the
important points only
such as at the support,
point under the point
load & at both ends &
in the middle of the
2. Distributed uniform load acting on simply supported
beam.
The above equation shows that shear force caused by the distributed
uniform load will change linearly with x. SFD needs to drawn first to see
the relation with BMD.
From SFD above, there is one contra point, where the SFD line cuts across the
beam axis (shear force = 0). The distance of the point from point A can be obtained
as:
3. Uniform increase load acting on simply supported
beam.
•
To determine the value of gx which is the value of uniform increase load when the
distance x is from A, the intensity of the load on the section needs to obtained. From
the figure above;
• plot every point on SFD & BMD, thus SFD & BMD will appear as
shown below.
• When shear force, V = 0 ; bending moment is at the maximum
and the point distance from A is:
4. Point load acting on cantilever beam
5. Distributed uniform load acting on cantilever beam.
6. Uniform increase load acting on cantilever beam
7. Several types of distribution load acting on beam.
The relationship between shear force and
bending moment
SFD
SFD
SFD
SFD
SFD
SFD
- END -
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