Download Compare the shear force and bending moment diagrams.

Lecture Notes
Engineering Sciences BTE1013
TOPIC 8
SHEAR FORCE AND BENDING MOMENT
Learning Outcomes
At the end of this chapter, student should be able to :
8.1. Describe the shear force and bending moment along a beam
8.2. Understand and apply the principle of moment to problems involving beam
8.3. Apply formula for shear force on beams
PRINCIPLE OF BENDING
Take a beam 1m long: if it starts as a straight beam then the length measured along the
top face with the aid of a tape measure is one metre. Similarly the bottom face will also be
one metre:
1m
A
C
i.e.
D
1m
AB = 1m
BC = 1m
Now try flexing the beam:
A
X
C
D
Y
When the beam was straight, AB was equal to CD, but now by visual inspection it is now obvious that
this is not the case, once the beam is flexed. The beam as shown is said to be ‘hogging’ so that the
upper face is longer than the bottom face.
How can this be? If the beam is solid then there cannot be any migration of material from the bottom to
the top reaches of the beam (because solids are generally rigid or semi-rigid). The only way the hogging
can occur is if AB is stretched further than CD by the bending action.
BTE1013
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Date : 1 Jun 2012
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Lecture Notes
Engineering Sciences BTE1013
If AB remains 1m then CD must be shorter, the bottom face is therefore compressed and the fibres beneath the
surface of the bottom face experience compressive stress.
If however CD remains 1m then the upper face AB must have been stretched and the fibres beneath the upper
surface are subject to tensile stress.
The two conditions just described represent two extremes. In most flexing situations, reality is somewhere
between the two and in many cases is close to halfway between the two, such that length XY midway between
AB / CD remains 1m. Therefore AB is stretched and CD is compressed.
The 3rd situation is that assumed in the theory of simple bending generally used for analysis of bending and
buckling.
PRINCIPLE OF MOMENTS
3000 Kg
5000 Kg
Consider a simply supported beam. Assume the beam is 2m long and rests freely on top
of two knife edge supports located right and the very ends of the beam. The beam is
lightweight so that in comparison to the loads supported the mass of the beam is
negligible. The beam carries two point Loads. The first is located 0.5m from the left
BTE1013
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Date : 1 Jun 2012
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Lecture Notes
Engineering Sciences BTE1013
hands end, where a mass of 3000Kg is supported: the other, a mass of 5000 kg is located
1.25 m from the left hand end.
The arrangement can be sketched in the following simple manner.
1.25m
F1
F2
F3
F4
0.5 m
2m
Loads F1 and F2 are known:
Force = mass x gravity
F1 = 3000 x 10 = 30,000 N
F2 = 5000 x 10 = 50,000N
Therefore the sum of the downward forces is Σ F down = 80,000N
Because the beam is in equilibrium (i.e. not moving) the vector sum of all forces applied must be zero. In
simple terms:
Σ F up = Σ F down
If the beam arrangement had been symmetrical about the mid point of the beam, (with equal weights
used) then by inspection the solution would have been easy.
F 3 = F 4 = 40,000N
The symmetrical arrangement is common in building structures, the engineer has to know the theory for the
general case. The non-symmetrical arrangement we are dealing with does not have F 3 = F 4 , but the actual values
can be found using the principle of moments.
BTE1013
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Lecture Notes
Engineering Sciences BTE1013
Remember the beam is not moving. Not only is it not moving up and down (which would be caused by an
imbalance in the up and down forces but it is not rotating (for example about the right hand end) which would be
caused in an imbalance in the leverage created by the up and down forces i.e. the moments have no imbalance
so that clockwise moment = anti-clockwise moment.
Taking moments about the right hand end will allow us to eliminate
the one of the unknowns. This is because one of the unknown fiorces is
applied at that right hand end (F 4) and so the lever arm (i.e. the
distance from the force to the pivot) is zero.
M@RHE :
i.e. this is not
happening!
moments c/w = F3 x 2m (c/w = clockwise)
moments a/cw = (F1 x 1.5m) + (F2 x 0.75m) (a/cw = anti-clockwise)
What happened to F4 ?
Now -
F4 x 0 (distance) = 0
moment c/w = moment a/cw
F3 x 2 = (F1 x 1.5) + (F2 x 0.75)
F3 x 2 = (30000 x 1.5) + (500001 x 0.75)
Therfore
2F3 = 45000 x 37500
2F3 = 82500
F3 =
So
82500
= 41,250 N
2
F4 = 80000 – 41250 = 38,750 N
SHEAR FORCE DIAGRAMS
BTE1013
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Lecture Notes
Engineering Sciences BTE1013
1.25m
30
50
41.25
38.75
0.5M
2m
Beams rarely fail due to shearing because of the distances involved. When the beam bends it usually induces
much larger tensile and compressive forces/stresses than shear forces/stresses. Therefore beams usually fail due
to bending, which creates stresses at the upper and lower surface that can produce cracks. Consideration of
shear forces can still be very useful however. Foe example a shear force diagram is much easier to produce than a
bending moment diagram and is capable of indicating the position of the maximum bending stresses.
Often a sketch is sufficient but for accuracy graph paper should be used. Graph 1 shows the shear force diagrams
for the above beam.
Shear force diagrams are easily created. They are drawn without complex calculations, but by simple inspection.
Assuming the beam is homogenous across its length, graph 1 indicates that the largest shear stress will be
experienced between the left hand end and 0.5m from the left hand end. That entire length is shown to be
subject to the greatest shear force by inspection of the shear force diagram.
Graph 1 shows that the largest shear force experienced by the beam is +41.25 KN but any section within the first
0.5m from the left is subject to that shear force. This means that the shear force is effectively transmitted from
one plane to another and IF the beam were to break due to that shearing, then it would do so at the weakest
point along that 0.5m length (where there was some slight deficiency in the otherwise homogenous beam).
Unless the location of such a deficiency is known, the exact point of failure could not be predicted, except to say
that it is somewhere in that 0.54 metre portion.
N.B. The sign is not important on the graph. If the right hand portion of the beam had experienced –45 KN
instead of –38.75 KN then the greatest shear force would have been 45 KN and failure would have occurred in the
right hand end.
The main reason for a shear force diagram is not to predict shearing stresses, but to eliminate the need to do a
bending moment diagram.
BENDING MOMENT DIAGRAM
BTE1013
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Lecture Notes
Engineering Sciences BTE1013
1.25m
50
30
x
y
41.25
38.75
0.5M
2m
At x (say 0.4m)
Bmx = (41.25 x 0.4) = 16.5 KNm
At y (say 0.8m)
Bmy = (41.25 x 0.8) – (30 x 0.3) = 24 KNm
The above calculation illustrates how to calculate a bending moment at any desired position. Two points are
selected (x + y). Fortunately where only one point loads exist, then bending moment only needs to be calculated
at the salient points where the loads are applied, and the graph can be created from this information.
Graph 2 illustrates the bending moment diagram for the above beam. The bending moments are calculated by
accounting for all forces and leverage arms to the left of whatever position is selected. A table can be plotted.
Distance KNm
0
0
0.25
10.3
0.4
16.5
0.5
20.6
0.8
24
0.875
24.8
1.25
29.1
2.0
0
(at 2m Bm = (41.25 x 2) – (30 x 1.5) – 50 x 0.75) = 0 KNm )
The four plot points spanning 0.5 metres from the left hand end show that bending moment diagrams
between point loads (with no additional load) produce straight lines. The portion between the 30 KN and
50 KN loads conform this, so no intermediate points need to be calculated for the right hand end.
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Lecture Notes
Engineering Sciences BTE1013
In fact, all BM diagrams without distributor loads are straight line graphs between plot points at the point of load
application.
Graph 2 shows that the maximum Bm experienced by the beam is 29.1 KNm. This is located 1.25m from the left
hand end, so this is where the maximum bending stress will occur, an if the beam is not large enough, where the
beam will fail due to bending. It is important to know the location of the maximum bending moment and its
value. With this information the beam can be designed to with stand the bending stresses. But the whole bending
moment diagram need not be sketched.
Compare the shear force and bending moment diagrams.
The maximum bending moment is located at one of the positions where the shear force diagram cuts the x axis.
If fact this is always the same case. So the location of the maximum bending moment can be deduced from the
shear force diagram (which is easier to plot than the BM diagram, especially with distributed loads) and the
maximum bending moment is then calculated at that location.
BTE1013
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Date : 1 Jun 2012
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Lecture Notes
Engineering Sciences BTE1013
Solved example (source= http://civilengineer.webinfolist.com/mech/prob52.htm)
Problem 5-2
Calculate the values and draw the diagrams for shearing force and bending moment for the following beam
shown in figure 5-2(a).
Figure 5-2(a)
Solution:
The given beam has two unknown reaction components which are calculated in Prob 4-1 as Ay = 66 kN
and By = 24 kN;
Shearing force values:
Wherever there is a point load we have to calculate the values of shearing force on both the side of that
section.
FA-left = 0
FA-right = 66 kN;
FC = 66 – 20 x 4 = -14 kN;
FD-left = 66 – 20 x 4 = -14 kN;
FD-right = 66 – 20 x 4 – 10 = -24 kN;
FB-left = -24 kN;
FB-right = 0;
The values of shear force are plotted in SFD in figure 5-2(b)
Bending moment values:
MA = 0;
MC = 66 x 4 – 20 x 4 x 2 = 264-160 = 104 kNm;
MD = 66 x 8 – 20 x 4 x 6 = 528 – 480 = 48 kNm;
MB = 0;
BTE1013
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Date : 1 Jun 2012
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Lecture Notes
Engineering Sciences BTE1013
The values of bending moment are plotted in BMD.in figure 5-2(b)
Throughout the span the bending moment is sagging in nature. To determine the maximum bending
moment for the beam we use the relationship between shearing force and bending moment which says
that the rate of change of bending moment (dM/dx) is equal to the shearing force (Fx) at the section. For
maximum value of any function its first derivative should be equal to zero. Therefore at the point of
maximum bending moment the shearing force must be equal to zero. From the shear force diagram it is
clear that the shear force is zero between A and C. The point of zero shearing force can be easily
determined by using the property of similar triangles.
Hence, 66/x = 14/(4-x) ;
Which yields, x = 3.3m;
Now calculate the bending moment at x = 3.3m which will be the maximum bending moment for this
beam. Therefore,
Mmax = 66 x 3.3 -20 x (3.3) x (3.3/2) = 108.9 kNm;
THANK YOU
BTE1013
Rev#:01
Date : 1 Jun 2012
Sem 1
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