PRINCIPLE OF BENDING

8/7/2019 PRINCIPLE OF BENDING

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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:

i.e. AB = 1m

BC = 1m

Now try flexing the beam:

When the beam was straight, AB

was equal to CD, but mow by visual

inspection it is now obvious that

this is not the case, once the beam isflexed. 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.

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.

1m

1m

A

C D

B

AC D

BX Y



3000 Kg 5000 Kg

2m

1.25m

0.5M

F1

F4

F3

F2

PRINCIPLE OF MOMENTS

Consider a simply supported beam. Assume the beam is 2m long and rests freely on

ton 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 isnegligible. The beam carries two point Loads. The first is located 0.5m from the left

hands end, where a mass of 3000Kg is supported: the other, a mass of 5000Kg is

located 1.25 m from the left hand end.

The arrangement can be sketched in the following simple manner.

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.

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 sothat clockwise moment = anti-clockwise moment.

Taking moments about the right hand end will

allow us to eliminate the one of the unknowns.

Thjis 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 : moments c/w = F3 x 2m

moments a/cw = (F1 x 1.5m) + (F2 x 0.75m)

What happened to F4 ? F4 x 0 (distance) = 0

Now - mom c/w = mom 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 =2

82500= 41,250 N

So F4 = 80000 – 41250 = 38,750 N

i.e. this is not

happening!



SHEAR FORCE DIAGRAMS

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 complexcalculations, 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.

2m

1.25m

0.5M

30

38.7541.25

50



BENDING MOMENT DIAGRAM

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.

(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.

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 tobending. It is important to know the location of the maximum bending moment and its

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

2m

1.25m

0.5M

30

38.7541.25

50

x y



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. In 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.

PRINCIPLE OF BENDING

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