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Shear Force and Bending Moment

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STUDY UNIT 8
Shear force and Bending Moment
Synopsis
This study unit develops the fundamental theory of shear forces and bending
moments. It determines these mechanical phenomena via graphical method.
Pre-requisites
Mechanics I, Mathematics I and II
Objectives
To enable the learner to understand to draw shear force and bending moment
diagrams.
Learning
Outcomes
Estimated
study period
•
•
•
•
4 hrs
To draw free body diagrams of different beams
To determine the reactions at different supports
To draw shear force and bending moment diagrams and determine the
maximum values and their locations
To determine the points of contra-flexure on the beam
Keywords
Shear forces; Bending moment; Point of contraflexure; Free body diagram; Beams
Prescribed
TextBook(s)
Drotsky JA. Strength of Materials for Technicians, Pearson Education South Africa
(Pty) Ltd, Cape Town, 2011. ISBN: 9781775782421.
Gere JM, Goodno BJ. Mechanics of Materials, Cengage Learning, Stamford, 2009.
ISBN – 13: 9780495438076
Open Rubric
8.1. Introduction
Beams: a member or bar that is subjected to lateral forces or moments that lie in a plane that
contains the longitudinal axis. Figure 8.1 shows the four types of beams which are so classified
according to their support conditions (Drotsky under Beams; Gere and Goodno under Types of
Beams, Loads, and Reactions).
W
W
RR
RL
M
R
Simply supported beam: supported at the
ends by at least one roller and a pin. They
are both replaced by vertical reactions. The
roller ensures that there are no longitudinal
forces induced in the beam by the lateral
forces. d by a fixing moment M and a
reaction P for static equilibrium.
Cantilever: rigidly fixed at one
end. The fixed end is replaced
by a fixing moment M and a
reaction R for static
equilibrium.
WL
W
MR
ML
RL
RL
WR
RR
RR
Encastre’ or Built-in beam: both ends
are rigidly fixed. This is statically
indeterminate and will not be dealt
with in this module.
Overhanging beam: One or both of the
ends of the beam extend beyond the
supports.
Figure 8.1: Beam types
Loads: Beam loads are classified as: Point or concentrated loads, Uniformly distributed loads, and
Uniformly varying loads. In this module we mainly deal with the first two types of loads [See Drotsky
under Types of Loads for a diagram of the uniformly varying loads].
8.2. Shear Force and Bending moment
The lateral loads acting on the beams in Figure 8.1 will induce direct and shear stresses which may
result in the deflection of the beams. The deflections and their attendant stresses are not uniform
along the span of the beam and for cases of design one may want to know where their maximum
values occur. Shear force and bending moment diagrams provide a means of determining such
critical points on the beam graphically.
Drotsky under Shear force and bending moment diagram shows the following ideas with the aid of
diagrams.
•
•
The shear force at any section of a beam is numerically equal to the algebraic sum of the
lateral components of the forces to one side of the section, and
The bending moment at any section of a beam is equal to the algebraic sum of the moment
of the forces to one side of the section.
Sign conventions: Shear forces are positive if an applied force tends to cause the left hand section to
slide upwards relative to the right hand section, and the bending moment is positive if an applied
load tends to cause the beam to sag or bend upwards [For clarity, see diagrams in Drotsky under
Shear force and bending moment].
Calculating Shear Forces
For Point Loads:
You always use the method of sections when determining the shear force at a particular point on the
beam. Assume that a beam is cut at that point, then for the section to be in static equilibrium, the
forces must balance on both sides of the cutting line. Consider the beam in Figure 8.2.
5 kN
15 kN
2 kN/m
C
B
A
0.75
E
D
1
1
2
Step 1: Calculate the reactions at the supports
First Condition of Static Equilibrium:
F = 0
RA + RE − 15 − 5 − 2  2 = 0
R A = 24 − RE
Second condition of static equilibrium:
[8.1]
M
A
=0
RE  3.75 − 15  0.75 − 5  2.75 − (2  2)  2.75 = 0
3.75RE = 36
RE =
36
= 9.6 kN
3.75
Using Eqn 8.1: R A = 24 − 9.6 = 14.4 kN
The shear forces between the end of the
8.3. Point of Contra-Flexure of Inflexion
The point on the beam where the bending moment changes sign (i.e. the bending moment at that
point is zero). It can be located graphically from the properly scaled bending moment diagram.
Alternatively, it can be calculated mathematically from the bending moment equation taken at that
point.
Illustrative example 8
Go through Examples 6.1and 6.2 in Drotsky. The given steps are the appropriate steps for
solving these problems. A good solution must start with neat free body diagrams and then
solving for the reactions. These are mandatory steps if a learner has to acquire the full score.
Note that Example 2 is for a uniformly distributed load and the bending moment yields a
parabolic curve. Whenever you have a uniformly distributed load over any section of the beam,
you should know that the plot over that section becomes a curved line not linear as for point
loads.
Activity 8
Draw the free body diagrams and calculate the reactions for the following problems in Drotsky
Self-assessment: Figure 6.30; Figure 6.34 and Figure 6.35.
Practice Exercises 8
•
Do the following problems in Drotsky under Self-assessment: Question 6.1 to 6.16.
•
Do the following problems in Gere and Goodno under Shear-Force and Bending-Moment
Diagrams: 4.5-1 to 4.5-7; 4.5-9 to 4.5-17. [skip problem 4.5-8 please].
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