Welcome to the first lesson of module 5 which
is on Bending of Beams part 1. In fact the last four modules which we have
looked into were on the effect of axial force on bars
and correspondingly we have looked into the aspects of stresses and strengths. Subsequently,
we have looked into the effect of the twisting
moment on bar. In this particular lesson, we are going
to study some other aspects of loading on this kind of system.
. .. So long we were talking about the forces in a
bar. Now we have brought in a specific term which
is called beam. We will look into what is meant by that and the different forces which
act on beam. Then one should be able to understand
different types of supports and types of beams. One should be able to understand the concept
of shear force and bending moment and be in a
position to evaluate reactive forces for different kinds of loading on different types of beams.
. .In this lesson
we will look into what a beam means and what the different types of supports are
on which a beam member is supported and the different types of loading that the member
will be subjected to. The scope of this lesson includes
the evaluation of reactive forces in different types
of beams, the concept of shear force and bending moment in beams and the examples for the
evaluation of reactive forces for different beams under different loading conditions.
We will evaluate the reactive forces in beam members
and consequently we will look into how to evaluate the shear force and the bending moment.
. Before we go to the bending of beams or bars
let us first examine the aspects of the answers to
the questions which were given last time. The questions which were set were related
to the torsional aspects. The first question posed
was: how is the strength to weight ratio of a bar
subjected to torsion defined. We did an example wherein we had evaluated the strength to weight
ratio. Basically by the phrase ‘the strength of a bar against twisting motion’, we mean
the strength of the bar to resist the twisting
moment. The carrying capacity of the bar of the twisting
moment is basically its strength, and the weight of a bar is its cross sectional area
multiplied by length and unit weight of the material. The
ratio of these two parameters will give us the strength
to weight ratio. .. The strength to weight ratio of a bar is defined
as the capacity of carrying twisting moment to the
weight of the bar. If twisting moment is defined as T or designated as T and weight is designated
as W then the ratio of the strength to weight is T/W, where W is the weight given by the
cross sectional area multiplied by length multiplied
by the unit weight. T divided by this ratio gives us
the strength to weight ratio. A is the cross sectional area, L is the length of the member
and Gamma is the unit weight of the material.
This is what we defined as the strength to weight ratio.
. .The second question is: what is torsion equation?
After going through the module on torsion you know that the torsion equation is defined
by this expression where this is given as T/J = Tao/rho
= G theta/L. Now T here is the twisting moment or the torsional moment which acts on the
shaft. J is the polar moment of inertia and for the solid
section the polar moment of inertia is pi d to the power of 4 by 32 where d is the diameter
of the bar and for tubular section this becomes pi
by 32 times do to the power of 4 – di to the power of
4. For solid section this is the polar moment
of inertia and for tubular section this is the polar
moment of inertia. Tao is the shear stress; rho is the radius at the point where we are
trying to define the shear stress, G is called the shear
modulus and theta is the rotation of the bar and L is
the length. So, these are the terms and this is defined as the torsion equation which is
T/J= tao/rho = G theta/L and we are aware that
all the terms have their usual meanings. The third
question is: what is the value of maximum normal stress in a solid circular shaft that
is subjected to a twisting moment?
. We have also discussed this through a problem.
Let us examine it once again. This particular bar
is fixed at this end subjected to a twisting moment T. T is a positive moment, as we have
defined that the twisting moment acts in the bar in
such a way that the thumb projects towards the
positive x direction which is the positive twisting moment.
So, when it is subjected to the positive twisting moment and if we look into a small element
this particular element is subjected to shearing
stress which is of this form. What happens when we try to plot this shearing
stress of this element in terms of Mohr circle? Sigma x is of the Mohr circle; this is the
shearing stress axis of the Mohr circle and these are the
positive directions for Tao and Sigma. Here the shearing stress acting on this particular
element .according to our convention is negative,
as this particular shear along with this complimentary shear makes a moment in the clock vise direction
which is negative. So, on this particular axis it
represents this particular point whereas the shear on the other plane is positive which
makes a clock wise moment. This is represented by
this particular point. Now if we plot the circle then this is the
Mohr circle of the stress and this particular point which
is on the Mohr circle represents the plane where we have the maximum normal stress. We
call this as Sigma1 and incidentally the value
of it is equal to Tao. This is Tao; this is also Tao which
is the radius of the circle and since normal stress is 0 on this which is a state of pear
shear the normal stress is equal to Tao. This particular
point is at an angle of 90 degrees with reference to
the reference plane. In the physical plane (since it is Mohr’s plane 2 Theta is 90
degrees, Theta is 45 degrees) if we move clockwise by 45 degrees
which is shown here. Perpendicular to this the
plane which you get is the plane along which it will fail and the direction of the normal
stress Sigma1 is this. This is the normal stress
Sigma1 and consequently the other normal stress which
is the minimum value will be Sigma2 which is equal to Tao.
These are the values of the normal stresses which get generated because of the action
of the twisting moment acting in a bar. So long we
have discussed the forces that act in a bar in the
direction of its long axis and we have called the load which acts as the axial load or axial
force. At any cross section if we try to find out
the stress corresponding to that, we have looked into
this normal stress as equal to the load divided by the area. We have also subsequently seen
that when a bar is subjected to a twisting moment,
the vectorial notation of this particular twisting
moment is in the direction of the axis which is the long axis of the bar.
We are going to discuss the loading in a bar where the vectorial directions of this loading,
either the direct load or the moment is perpendicular
to the axis of the bar and these particular kinds of
loads are transverse to the bar and we call those kinds of members which are subjected
to these transverse loads as beams. In the case of
bars, the loads were axial and their vectorial direction is
along the axis of the bar whereas for the members which we are terming as beam, the
loads are transverse to the member and the vectorial
direction of these forces, either the load directly or the
vectorial direction of the moment, are perpendicular to the axis of the bar. .. If we have a bar in which the rhos act transverse
to its direction, the vectorial direction is
perpendicular to the axis of the bar. Also we may have a moment which acts in the beam.
If we look into this particular moment in this beam,
you have a moment in this plane; so its vectorial direction is perpendicular to the axis of
the beam. The loading of x on a bar is what we designate
as beam. Members which are subjected to loads that are transverse to the longitudinal axis
are termed as the beam and the members are either
subjected to the forces or the moment having their vectors perpendicular to the axis of
the bar. All these forces that act in the beam member
which are transverse to this particular member, act
in the same plane. This is the plane of the beam and the forces which are acting on it
along with the moment are in the same plane and that
is why we term these kinds of structures as the planar
structures. In the same plane, we have the bar and the loading and consequently we will
see that when this bar or the beam is subjected to
this transverse load it will undergo deformation and this
deformation also will be in this particular plane. When the forces acting on this beam
along with the plane of the beam and its deformation
are in the same plane, we call them as a planar structure.
We already know that the moment about an axis is perpendicular to the cross section. Now
we are going to talk about the moments about
the axis which are in the plane of the cross section. If
the cross section is a rectangular one then the axis which lies in this particular plane,
the bending of the moment, will act about this axis. When
this moment acts about the axis it lies in the plane
of the beam itself. If we look into both the moments the vectorial
direction of both the moments is perpendicular to
the long axis of the beam. When we are dealing with the bending moments which are in the
plane of the beam along with the forces we call
that plane as the plane of bending and because of this .particular moment, we call the beam member
or the member which is subjected to the transverse loading undergoing a bending, as the bending
moments. Before we really go to the details of
beams let us look into some kind of supporting arrangement on which a bar is supported. We
call one of the supports as pinned support or hinged
support. . . .Here you see diagrammatically we have shown
some figures. This part is supported by a joint in
which this particular end of the beam cannot undergo displacement in the horizontal direction;
it cannot undergo displacement in the vertical
direction, but this beam allows the bar to move in
this form. That means it can undergo a rotation. It cannot move horizontally, it cannot move
vertically but it can rotate. The rotation of this particular end of the bar is allowed
and it can have some amount of rotation which you may designate
as Theta. Generally when we study problems we designate
them in such a way that in this particular form,
this particular type of support indicates that it is either a hinged support or pinned
support. We have a support of this particular form wherein
it is rests on to rollers. When this part of the
segment is loaded then this particular support can move. This can move in the horizontal
direction since it is supported on the roller. Hence horizontal movement is possible but
the movement in the vertical direction is restricted.
Also it can have rotation as we had in the previous case for a hinged support. We call
this kind of support, a roller support. Take these two kinds of supports; the hinged
support and the roller support. In the case of hinged
support the movement of this particular end both in the horizontal direction and the vertical
direction is restricted. Since they are restricted it is expected that some amount of reactive
forces will be generated and these reactive forces
will be in the vertical direction and in the horizontal
direction. In the case of a roller support since restriction of movement is only in the
vertical direction and it is free to move in the horizontal
direction, there will be a vertical reactive force
which is perpendicular to this roller plane. We call this kind of support, a fixed support.
This particular end of the beam is fixed at this point. The meaning of this fixity is
that this particular point cannot have any displacement
in the horizontal direction and cannot have any
displacement in the vertical direction. This part is not allowed to rotate freely. If we
look into this particular axis it will remain straight over
here. No rotation as we had in the previous cases is
allowed in this case. Here the horizontal moment, the vertical moment
and the rotational moment are restricted. When
these three motions are restricted, naturally corresponding to all these three motions we
will have the reactive forces. Since the vertical motion
is constrained there will be a vertical reaction and
since the horizontal moment is constrained there will be a horizontal reaction. Since
it is not allowed to rotate it has to be held back.
A hinged support has two reactive forces; one vertical and one horizontal because it
is not allowed to move in these two directions. In
the case of a roller support since it is allowed to
move in one horizontal direction there will be only a vertical support or vertical reactive
force and for a fixed support we will have three
reactive forces. They are the vertical reactive force, the
horizontal reactive force and a moment at that particular point. These are the different
kinds of supporting arrangements that we have. The
supports which have been shown over here are the
pinned support or hinged support, the roller support and the fixed support.
Based on these kinds of supports we classify the beams as well. A member or a beam member
is supported at its two ends. Let us say on one
end it is supported by a hinged connection and on
the other end it is supported by a roller connection. When they are subjected to the
transverse .loads it resists them by generating the reactive
forces in the supports. The reactive forces generated in these supports will be of this
particular form and we call this kind of beam, a simply
supported beam. So, when we talk about a simply supported beam it means that one end of it
is hinged and the other end is on the roller.
. Notice that we have two unknown reactive forces
on this particular end A. On this particular end
B we have one reactive force. There are three unknown reactive forces and as we know that
in the equations of static equilibrium we have
three equations, we can solve these three unknown
reactive forces using three equilibrium equations. These kinds of members are called statically
determinate because the three unknown reactive forces can be readily determined using the
three equations of statics. We encounter another
kind of beam called cantilever beam where the beam
member is fixed at one end and free at the other.
At this fixed support it is expected that there will be three reactive forces generated;
one is the vertical force, one is the horizontal force
and another one is the moment. These three reactive
forces will be generated at this support and again if we look into this particular member
since we have three reactive forces they can be evaluated
using equations of statics and hence this member
is also a statically determinate member. Sometimes we use a beam member which may not
be in any of these categories but could be a
combination of them. For example, we have a beam which is hinged at this place and placed
on a roller at this particular place and subjected
to some kind of loading on the transverse plane. This
part of the beam is extended beyond the support point. This part is generally called the overhang
part of the beam. This is a simply supported beam
with an overhang. .This part is like a cantilever beam but in
this cantilever beam the rotation is allowed because it is
supported at this point and has a vertical constraint. But then because of this overhang
it is expected that it can generate moment and it
can also generate rotation. But being a roller support
it cannot resist any rotational aspect over there and there is expected to be a rotation.
We generally call these kinds of beam members
as beams with overhang. So, the different kinds of
designations are simply supported beams, cantilever beams or beams with overhang.
Henceforth whenever we mean that a simply supported beam is loaded with transverse load,
it means that the beam is supported on a hinged
support at one end and the roller support at the
other or if we say that the cantilever beam is loaded with such and such loading, it means
that the one end of the beam is on a fixed support
and the other end is free and when there is a fixed
support, it is expected that there will be three unknown reactive forces which have to
be evaluated using equations of statics.
. Having looked into the types of supports and
the types of beams we designated, let us look into
the different types of loads that a beam member encounters. Let us suppose we have a simply
supported beam which is hinged at one end and supported on roller at the other. Then
we have designated one load which is known as concentrated
load. The meaning of this is that if a load which acts on an infinitesimally small area
is distributed in that particular small area then we call
that force or load as a concentrated load. .(Refer Slide Time: 25:40) Many times the hole of the beam may be subjected
to a load which is distributed over the entire length. (Refer Slide Time: 25:40) This is
the length of the beam and the entire length of the beam
is subjected to a load which is uniform and the intensity of this load could be say q
bar unit length. Now this load is called a uniformly
distributed load as it is distributed over the entire
span having uniform intensity. So, we call such a loading system a uniformly distributed
load, or in short many times it is referred to as udl.
When the simply supported beam is subjected to udl,
it indicates that a beam, which is hinged at one end and roller supported at the other,
is subjected to a uniformly distributed load over the entire
span of the beam or the length of the beam. There is another kind of loading called linearly
varying loading. We may have a beam member or
a simply supported beam which is hinged at one end and roller supported at the other
and this has a load which varies in this rectangular pattern.
Here the intensity is 0 and here say maximum intensity is q; this span of the beam or the
length of the beam in its entire length is subjected to a
load which is linearly varying. It can be of different types. It could be
in the form that we have indicated or we could have a
loading and a member that vary in a trapezoidal form. There could be different kinds of
variations. Here we have intensity q1, we have intensity q2 and between these two it
varies in this form. It is a trapezoidal variation but between
these two points q1 is constant over here and from
one point to another it varies linearly. Everywhere the intensity of the loading varies in a linear
form. Other then this distributed loading or concentrated
loading we get loading in the form of a concentrated moment. Let us say that we have
a beam and let us say this is a cantilever beam
which is fixed at this end. At a particular point in the beam it is subjected to a moment
M at this .point. This will have effects or there will
be reactive forces generated because of the application
of this moment and we call this kind of loading as concentrated moment.
In general a beam member supported on certain kinds of support (as we have seen the different
kinds of supports) can be subjected to different kinds of loading. They could be concentrated
load, the linearly varying load, uniformly distributed load or concentrated moment either
individually or a combination of these different kinds of loads. When this member is subjected
to all these transverse loading in the plain
of the member we call it a beam member. Finally, our objective is to evaluate the
stresses in this member but before we really go into the
evaluation of stresses we will have to examine how internal forces are generated in this
member because of the transverse loading. In the
previous modules we have seen that when a bar element
is subjected to the axial pull how the stresses get generated within the internal part of
the body or because of the twisting moment how the stresses
get generated in the bar. We will now look into
how the internal forces or internal stresses get generated when a beam element is subjected
to this transverse loading.
. Based on this discussion that we had we have
to evaluate the reactive forces. Whenever a beam
member is to be evaluated for the stresses, we will have to find out the internal forces.
For evaluating the internal forces the first step
is to evaluate the values of the support reactions or the
values of the reactive forces. (Refer Slide Time: 30:51) For example, this is a simply
supported beam which is supported on a hinged support
on one end and a roller supporter on the other having a length L which we call span and is
subjected to uniformly distributed load over this part
say of intensity q bar unit length. Let us say this particular length is ‘a’. .(Refer Slide Time: 30:51) Vertical concentrated load acts at a distance
of b from here. Let us say an inclined load at an
angle of Theta acts at a distance of c from here and this gap could be d. Let us draw
the free body diagram that means removing these supports
and drawing the corresponding reactive forces. We will then have one vertical reactive force
and one horizontal reactive force for the hinged
support. If we call this end as A, this end as B then let us call this as RA, this as
HA and this end will have a vertical force only, as a horizontal
moment is allowed and we will have this as RB.
So, the three reactive forces RA HA and RB have to be evaluated from the equations of
statics. We have three equations of statics where the
summation of vertical forces is 0, summation of
horizontal forces in the beam is 0 and summation of moment at any point is equal to 0. If we
employ these three equations we can evaluate three unknown quantities. This particular
member acts at an angle of Theta. So, it has components
in the vertical and horizontal direction. Note
there are no other horizontal forces. Let us take the summation of horizontal force
as 0. We have HA, p1 and p2. The vertical component is p1sin theta and the horizontal
component is p1cos theta. The equation HA minus
p1cos theta equals to 0 means the direction of HA is in the positive x direction and p1cos
theta is in the negative x direction which is why this
minus sign comes in. This gives us the value of HA.
Likewise, we can take the help of other two equations to evaluate RA and RB. Once we know
the support reactions then we can calculate other
internal forces as well. Let us compute the internal forces. When this
beam member is subjected to transverse loading it
will be subjected to stresses just like some kinds of loading in the bar that are subjected
to stresses. Our objective is to evaluate those
stresses. But before we go into the evaluation of
stresses we need to know what the acting internal forces are.
For example; let us take a beam which is subjected to a transverse loading and we have a
cantilever beam which is fixed at one end and free at the other. Let us say it is subjected
to load .P. We are interested in finding out the internal
forces at a section which is at a distance of x from
this support. Let us suppose we separate out this particular part which you call as a free
body diagram. This particular part is subjected
to the load P where this has to be in equilibrium with
the internal stresses that are generated at this cross section.
. Since at this moment we cannot evaluate the
stresses, we need to know that when a bar is
subjected to an axial pull the summation of stresses at a point over the cross section
gives us a stress resultant. When this is subjected to
the transverse load there will be stresses in the cross
section and thereby we will get stress resultants. Let us call those stress resultants as the
different kinds of forces that exist here. We have the
axial force, the vertical force, the force in the
horizontal direction, the force in the vertical direction and a moment. .(Refer Slide Time: 35:03) Depending on the type of loading of the forces
the beam member will be subjected to, we will have these quantities. Since we do not have
any horizontal force component acting on this beam,
the horizontal reactive force will be 0. We get this vertical internal force or the stress
resultant and let us call this vertical stress resultant
as the shearing force V. At any cross section we take in this beam
member, we get some stress resultant, which will hold
this particular part of the body, which is free from the whole body, in equilibrium under
the action of the load. We call those stress resultants
which act in the vertical direction to equilibrate the vertical forces as shear forces which
have internal stress resultant. We call the one that acts in
the horizontal x direction as axial force and we call a moment that resists the effect
of the transverse loading in terms of the bending
as the bending moment. The stress resultants that act at a section
are the shear force V and the bending moment M. Let us
have a proper sense of the sign and cut of this particular bar at a section which is
at distance of x1 from the left end and let us take the free
body part of this end which is fixed here. This is the
positive x direction and this is the positive y direction and along the positive x direction
we have taken normal stress. This is the positive
y direction and we have taken this as the positive z
direction. To match with this criterion our axial force
on this particular section is positive. The shearing
force is positive over here and the moment which is anti-clockwise perpendicular to the
one which projects out of the plane is positive.
When we take the other part of the beam where we
have the load and the corresponding actions which will balance this, we have the shear
force which acts downwards, the axial force which
is in the opposite direction and the moment which
was anti-clockwise will be clockwise here. Let us suppose we just take an element out
from this particular beam and if we look into that
particular element then we get the kinds of forces which we call as positive. This shear
on the .right face is upward, on the left face is
downward and the axial face on the right side points
towards the positive x direction and on this side towards the negative x direction and
this is the moment which is anti-clockwise here and clockwise
on this face. This is what we call as the positive sign
convention and its reverse is the negative sign
convention. This is a small part where on the right phase we have the positive shear
which acts in the upward direction. This is the moment which
is anti-clockwise, this is the shear which goes
down and this is the moment on the other face. So, this is the positive action and its reverse
is the negative action.
Shear forces and the bending moments like axial forces in bars and internal torques
in shafts, are the resultants of stresses which are distributed
over the cross section and these quantities are
known as stress resultants. When they are subjected to the transverse loading and when
we need to evaluate the internal stresses if we take
sections, there will be stress resultants because we
cannot at the moment evaluate the stresses directly.
We are dealing with the resulting force that acts in the cross section and they are the
forces in the vertical direction which is the shear force.
The force in the horizontal direction is axial and the
one which acts along the axis of the bar and the moment which resists this external load
in terms of the bending, is the bending moment. So,
we are interested in a beam to evaluate the value of
such shear force and the bending moments because of the transverse load that the beam is
subjected to for different kinds of supports. Once we compute this shear force and the bending
moment, then subsequently we can compute the values of the stresses in the bar or the beam. Let us take some examples where we need to
evaluate the reactive forces and consequently the
shear force and the bending moment. Determine the reaction components of the beam which
are .caused by the applied loads as shown in the
figure. The beam on one end is supported on a hinge
and on the other end it is supported on a roller. This is a simply supported beam as
we have designated. Let us look at the type of loading
that it is subjected to. Here it is subjected to a linearly varying
load from 0 at this point to 2 N/mm at this point and to 0
over here. These are all in millimeters and this particular length of 900 mms from 0 is
going to 2 N/mm and again it comes back to 0. At this
particular point it is subjected to a constant moment
which is 150 Nm which we have designated as a concentrated moment. We will have to
determine the reaction components for this beam. First let us draw the free body diagram
of this particular beam.
. If we look into the free body diagram of the
beam we have reactive components. For a hinge support we have a vertical reaction and a
horizontal reaction. Let us call this reactive force as RA
and this as HA and here we have the vertical reactive force over the roller which is RB.
Since in the horizontal direction it is allowed to
move and there is no horizontal force component over
there we will have only the vertical component. Since at both the ends the beam is allowed
to rotate there is no need for having any moment
to hold this rotation back. Hence there are no
reactive moments. The reactive forces are HA, RA and RB. These
are the three unknown quantities which have to be
evaluated from the given loading, employing the equations of statics. Summation of horizontal
forces equals to 0, summation of vertical forces equals to 0 and summation of moment
equals to 0, which are the three equations of our static
equilibrium. If we employ the summation of horizontal force which equals to 0 then here
since we do not have any horizontal loading in the
beam we get HA = 0. If we take the summation of vertical forces equals to 0 then the vertical .forces we have are the reactive forces RA
+ RB and since RA and RB are pointing upwards both
are added. Then let us add the other quantities. The loads triangularly vary as this is 0,
this is 2 over a length of 300 and so the load is 1/2
multiplied by 300 multiplied by 2 n per mm. This particular load points downward and this
reactive load RA and RB point upwards and with the minus sign. If we take this particular
part of the triangular loading then ½ multiplied
by 600 mm multiplied by 2 is the loading and since this
loading is directed downwards this is negative and so this is 0. From this we get RA + RB
= 2; 2 and 2 get canceled and we get 300 + 600 = 900n
which is equation 1. We can generate another equation from the
summation of moment which equals to 0. Let us take
the moment of all the forces with reference to this particular point say point A. If
we start from this end the moment of RB causes an anti-clockwise
moment. We have RB multiplied by 1500 mm which is anti-clockwise. Then we have a
concentrated moment which is also anti-clockwise. This is plus 150 multiplied by 10 to the power
of 3 nmm. Then we have these forces which will
have the moment with reference to A and at this moment the load acts vertically downwards
which is why this with reference to point A will cause a clockwise moment. We have considered
these two anti-clockwise moments as positive. So, contribution of this moment will be negative.
Let us take this particular load which is ½ (300) multiplied by 2 which is 300 and
300 acts at a point which is one third from here and so
the distance from here is 2/3 (300). The load is ½
multiplied by 2 multiplied by 300 and so this is -300 multiplied by 2/3 (300). Thus the
contribution of this particular loading is ½ (600) multiplied by 2. This is again 600
in a clockwise direction. The external of this
load is 1/3 from this end. Now from here to here it is
600 and this particular distance is 200 and the distance from here is 300.
So, the total distance of this load from this end is 500 and this is multiplied by 500 which
equals to 0. If we remove this particular part wherein
we have one equation, RA + RB = 900n the second equation RB = 0 gives us a value of RB = 140n.
If you compute this we will get the value of RB
from this expression as 140 n. Here RA + RB = 900 gives us RA value as 900 - RB which
is 140n and this is equal to 760 n. This is the value
of the reactive forces RA and RB and HA which equals
to 0. The results or the three reactive forces that we have are RA = 760 n, RB = 140 n and
HA = 0. .. Let us take another example where we have
to evaluate the reactive forces, the shear force V and
bending moment M at C which is at a distance of 0.5 meters from point A. So, we will have
to evaluate the shear force and bending moment
at this point and we will have to find out the
reactive forces. This particular beam is fixed at this end and free at the other and as designated,
this particular beam is known as a cantilever beam.
This beam is fixed at one end, free at the other and is subjected to uniformly distributed
loading over a distance of 2m and a concentrated loading
at a distance of 1m from support A. So, a cantilever beam is subjected to this load.
First we will have to find out the reactive forces and
secondly at this section we will have to find out the shear force, the stress resultants,
shear force V and bending moment M. .. Let us look at these values and draw the free
body diagram of this. On this free fix support we
have the reactive force which we may call as RA, the horizontal force which is HA and
the moment which is M or MA. We will have to compute
the reactive forces. Since we do not have any horizontal forces, from the summation
of horizontal forces, we get HA = 0. From the
summation of vertical forces, RA is vertically upward, other loads are vertically downward
and so - 4 - 3 kilo n (kilo Newton) per meter
is the distributed load over a length of 2 = 0.
We get the value RA = 10 kilo n. We have the support moment MA which we will
have to evaluate and we have the 4kn force. We
have assumed this moment to be anti-clockwise which is positive. So, 4kn contributes as
a clockwise moment and we have MA - 4 multiplied
by 1 knm and this distributed load is 3 kn per
meter multiplied by 2 which is 6nm acting at the center of this length, which is at
distance of 1m. The distance from here is 3 meters and this
is also in a clockwise direction. Here we have -6
multiplied by 3 and so MA gives us 22 knm. When we take the reaction forces the vertical
reaction is 10 kn, horizontal reaction is 0 and the moment value is 22 knm. These are
the three values
at this particular support. Let us compute the values of the bending moment
and the shear force at that particular end. Take
the free body of the cantilever beam from the right hand side. This is the free body
diagram where we have the distributed load over the
length of 2 meters which is 3 kn per meter. We have
a concentrated load over here which is 4 kn. This distance is 2m, this distance is 1m and
we will have to find out the bending moment and shear
force which is at a distance of 0.5m from here.
We have the positive values of shear, the bending moment and since there are no horizontal
forces we can ignore the axial force. .If we take the summation of the vertical
forces then V + 4 kn + 3 multiplied by 2 is the
distributed loading. This gives us a value of V as equal to -10 kn. The sign negative
indicates that it acts in the opposite direction as we have
assumed. Take the moment of the forces about this
particular point say MC. This is in the anti-clockwise direction and 4 kn also causes a moment in
the anti-clockwise direction and we have + 4 multiplied by 0.5. This load also causes
a moment in an anti-clockwise direction. So, MC + 4
multiplied by 0.5 + 3 multiplied by 2 multiplied by the
distance which is 2.5 equals to 0. This gives us a value of MC + 2 + 15 = 0. So, MC = -17
knm. The shear force which acts at this particular
cross section is 10 kn and the bending moment at
this cross section is 17 knm. From the values of the reactive forces that we have computed,
we can find out the values of the shear force
and the bending moment which should be identical. We
can take the free body of either part and compute the values of the shear force and
the bending moment at that particular location. We have
another problem in which the beam ABCD has overhangs at each end and carries a uniform
load of intensity q. For what ratio b/L will the
bending moment at the midpoint of the beam be 0? Try to solve this problem.
. .. We have another problem in which we have to
determine the value of shear force and the bending moment at a point which is at a distance
of 2m from here and here a force which acts at
a distance of 1m from A will produce a moment over here. For these you have to compute the
values of the reactive forces and thereby the shear force and the bending moment. In
this particular lesson we have introduced the concept
of beam, its type and the different types of
supports and loads that the beam member is subjected to. Then we have evaluated the shear
force and the bending moment in beams. We have looked
at some examples to evaluate the reactive forces, shear force and bending moment for
different types of loading. .. The following questions are set for you. What
is the difference between a bar and a beam? What are the different types of supports that
are used in beams and what is the sign convention for shear force bending moment and axial force?
You will find the answers to these questions once you go through this particular lesson. We will discuss the answers in the next lecture. .
