WEBVTT
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welcome to
the lecture 11. So, far we have done that
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limit state of collapse flexure or bending.
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Today we shall start learning shear. So, our
today’s lecture limit state of collapse
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shear and how shall we resist shear. We do
it we are doing the problem on beam. If you
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take a simply supported beam may be
it is resting over a brick wall.
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So, let us take a simply supported beam resting
over a brick wall. If we have uniformly distributed
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load let us apply the uniformly distributed
load. We have the bending moment to resist
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bending moment. What we have to provide? We
have to provide longitudinal reinforcement.
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This is for longitudinal reinforcement
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to resist flexure or bending. In addition
to that you can have that shear and diagonal
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stresses you can have shear and diagonal stresses.
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So, you can have the diagonal tension if we
take 1 element and you can have diagonal tension.
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So that, the crack it will just split and
other thing also possible it should be the
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other way the diagonal compression. So, I
think it should be the other way in this particular
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one. So, we can have the diagonal compression
and similarly we can have tension in beams.
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That if we take, these are the longitudinal
reinforcement. So, it can have tension. So,
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this is and we have 2 that we have to resist
these cracks and how shall I how shall we
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make it.
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The shear failure we can have different types
shear failure this is due to say shear tension
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this is 1 case this could be due to shear
tension.
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We can have shear bending this type of say
in your case shear bending.
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We can have that is 1 the shear bond apply
load here and that is a shear bond that also
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you can resist.
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We can have that shear compression. So, these
are the 4 different cases possible. If we
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just go back once more let me just go back
once more let me just go back once more. So,
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we can have shear tension. The crack pattern
may look like this. Then we can have shear
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bending. We can have that due to bond that
also we shall take care separately that bond.
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Today we shall consider we shall take care
that how to resist shear and shear compression.
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So, you can see this 1 shear compression. What we can do what we can do here that we can do it here itself.
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We use tie or we use stirrup. How it looks
likes? The stirrup is made this way; that
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means, you take a bar and you start bend.
So that you can make this type of thing and
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generally it comes generally it comes. If
we take the bar which can come and this 1
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the longitudinal 1 we are looking from this
side we are looking from this side. So, this
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is your longitudinal bar and from the central
line of longitudinal bar which will come in
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the range of say 8 phi 8 times the diameter
of this stirrup.
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The stirrup can have diameter say 6 millimeter
it can have diameter 8 millimeter 10 millimeter
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12 millimeter. And the other side also we
have 2 sides. So, other side also it will
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come say 8 phi depending on the link that
it can hook. So, you can have 6 phi also.
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So, we have to provide this reinforcement,
who will resist these out of these four sides;
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who will resist shear the shear will be resisted
by these 2 members.
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So, we have to provide that both sides the
stirrup that the stirrup contains 2 legs we
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will call it 2 legs. So, far the longitudinal
reinforcement concern for bending we have
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say 320 millimeter diameter bar or 420 millimeter
diameter bar or 520millimeter diameter bar
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the number of bars and the diameter.
But here what we do we provide may be we could
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say like this 8 torque that it means that
eight torque that it means that 8 dia fe 415
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with this symbol we designate. And let us
say at the rate of 150 millimeter center to
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center. So, here we are not specifying number
of bars in this state we are specifying the
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spacing of those stirrups. And here in addition
to that we provide to make it very clear we
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say 2 L. So, we provide like this 8 torque
2 legged; it is 2 legged and at the rate of
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say 150 millimeter center to center.
So, since we are having 2 legs both sides.
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So, we can have 3 legged also we can have
4 legged also. So, depending on the shear
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that whatever shear that force applied. So,
that you have to resist. So, we have to provide
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number of bars.
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And the different types of bars that stirrups
could be 1 simple 1 that what we generally
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provide this 1. If we need more then we can
have single link like this also; that means,
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we are providing the reinforcement say and this bar will come like this. Then we shall consider this 1 as 3 legged
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whereas, this 1 2 legged. So, asv the area of steel if we designate. So, far we have done ast
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the tensile reinforcement longitudinal 1 and
here we specify as say asv this is your shear
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reinforcement.
So, far 2 legged it should be twice times
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pie by 4 times say let us say pie square for
2 legged. If it is 3 legged then 3 into pie
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by 4 times pie square
so this is you are that how you will calculate
that asv. So, you have to provide asv and
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the spacing which we specify as sv, sv spacing
of stirrups. So, we have to provide that stirrup
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spacing and stirrup reinforcement the diameter.
And we have choice that whether we have to
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provide 2 legged or 3 legged or 4 legged.
Generally, most of the cases for beam we find
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that 2 legged sufficient.
So, in that case we shall increase the diameter.
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We can start with eight millimeter if it is
not sufficient then you can go to 10 millimeter
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diameter and, but most of the cases we shall
restrict ourselves to 2 legged stirrups only.
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Now, what about the shear stress? The shear
stress distribution for rectangular section
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it comes parabolic in the elastic stage say
we are getting in this pattern, ultimate stage
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that these portion becomes linear. But in
our calculation, but in our calculation we
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shall not use this formula what we shall not
use this stress distribution. What our IS
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code says let us find out.
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So, we shall take care that nominal shear
stress according to is 456 that our code says
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that we are shall not go to that parabolic
distribution. In a state we shall take this
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nominal shear stress and we will find out
in this clause the clause 40.1 IS 456 2000.
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The nominal shear stress in beams of uniform
depth shall be obtained by tow v, that is
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the shear stress developed equal to vu; vu
is the design shear force vu is the design
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shear force.
So, far we are always in our calculation in
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limits state design in limits state design
we care taking care of that your say design
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load the design moment design force. Assuming
that we have already multiplied that 1 with
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a proper partial safety factor for forces
that all ready we have taken care of that
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1.5 times or 1.2 times depending on the situation.
That we shall take care separately when we
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shall go for actual problem; practical problem
we shall take a problem separately.
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Because, right now we are doing component
wise we are taking how to take care say beam
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flexure now, how to take care say beam that
shear. Similarly, say for column for slab
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then footing staircase all the components
we shall do it separately; we are doing individually.
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Then we shall take care 1 problem say building
and then we shall analyze and then solve the
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problem and then design. There we shall do
that proper multiplication of that factors
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all those things we shall do. Right now we
are assuming that the final that vu that is
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given to us.
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So, here that vu the force, so vu the shear
force that is the design shear force divided
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by b that width of the beam and d that is
the effective depth. So, we shall get the
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shear stress tow v we shall get the shear
stress tow v that our code says that do not
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go for that parabolic distribution all those
things. So, we can take the nominal shear
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stress and we shall we can do our calculation.
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So, vu the shear force due to design loads
after proper multiplication of different factors
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if it is dead load, live load, earthquake
load or say your wind load whatever it is,
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that we do not know. After taking care of
that load we are getting that vu. So, we have
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to design that for that vu; the b breadth
of the member which for flanged section shall
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be taken as the breadth of the web that is
clearly mentioned. That if it is say flanged
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section that t beam. So, we shall not take
the bf that width of the planes we shall take
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the width of the web and d is the effective
depth and d is the effective depth.
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So, how can we resist how can we resist that
beam shear if we take care this is your longitudinal
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bar this 1 also longitudinal bar. And we have
to and this is the assumed cracks say from
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the support generally it happens because shear
force is maximum near the support. So, the
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crack all those things due to shear will get
near the support. And most of the cases we
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will get this type of the diagonal cracks
and that we have to resist by stirrups. So,
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we have different components the bent-up tension
bar let us say we can provide the beam.
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This is your say beam what we can do if we
take say simply supported beam generally say
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simply supported beam that when you are in
0 that in the support. So that means, we do
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not need that much of tension your say tensile
reinforcement. What we can do the bar which
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is coming from this side the bar which is
coming from this side. Because, that we are
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assuming the tensile force that tensile is
in the bottom in the most of the cases.
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Then we can take these bar say 45 degree and
we can go up that bent up bars. So, bent up
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tension bar, so which we have mentioned here
that bent up tension bar; that means, this
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tension bar here I have specifying this tension
bar which is going this way.
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This is 1 bar this is 1 bar. So, then this
bar is coming from this side then it is bent
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going upward and then this way. So; that means,
it will also resist certain portion or part
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of the shear. The second 1 system of vertical
stirrups the system of vertical stirrups means
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these bars which we have to provide. And the
third 1 system of inclined stirrups the stirrups
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also we can provide in the inclined manner
in this fashion.
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So, we can have 3 different components bent
up tension bar system of vertical stirrups
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and system of inclined stirrups. So, shall
we go the other 1? So, let me give you some
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time if you would like to take it. So, this
is the general format. But generally, most
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of the cases nowadays we do not go for bent
up bars we do not go for say inclined 1. Most
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of the cases we use for detailing purpose
for reinforcement detailing is also 1 important
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aspect in reinforced concrete design.
So, in that case what we do we take care the
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bars that 1 just straight we do not bend bars
because, it will take a lot of time also.
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That is why we always nowadays we use say
straight bars in very rare case we use the
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say bent up bars. So, what we have to provide?
We have to provide the say longitudinal reinforcement
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that on straight we are not bending those
bars. Inclined bars we hardly we use it most
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of the cases we use straight that say vertical
stirrups only not inclined stirrups. So, we
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have to provide this say vertical stirrups,
vertical stirrups and the longitudinal reinforcement.
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So, the shear forces are resisted by combined
action of concrete and shear steel shear steel
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is nothing, but the stirrups. So, it is resisted
by concrete and shear steel.
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So, resistance in concrete resistance in concrete
is due to the combined effect of 3 factors.
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So, we are now taking care say there is 1
concrete and you are say other 1 that is your
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stirrup. So, concrete has three different
factors. 1, which we designate by say vcz
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shear carried by uncracked section 20 to 40
percent of total shear in compression, so
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this it means that uncracked section we shall
take care that uncracked section.
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We can draw a figure. So, if we take care
of this bar the uncracked portion is at the
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top because of due to in the compression side.
So, this 1 we can say that vcz vcz is the
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uncracked portion. So, this 1 will be this
portion of the concrete will be taken care
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of by concrete only and that 1 () vcz. The
next 1 we have that va shear carried by aggregate
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interlock across the diagonal crack aggregate
interlock even it cracks there is an aggregate
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interlock and that 1 will be taken care of
almost say thirty 2 three to fifty percent.
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So, this is called va.
So, you can say va means here this is va.
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And what about the third component the third
component here vd shear carried by double
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action of longitudinal tensile steel. That
means, tensile steel also will take certain
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percentage say 15 to 25 percent of the case.
So, we shall get that fifteen to 25 percent.
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So, that is your say vd. So, vd will be taken
care of by this longitudinal reinforcement.
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So, this is vd. So, vcz va and vd. So, shear
carried by concrete the summation of all those
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3. So, you can write down here.
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So, V c equal to V cz plus V a plus V d. Vc
equal to V cz plus V a plus V d. But we do
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not know all the different components; we
do not know all the different components.
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So, we have to find out V c. Now, vu that
shear force due to design load so, we can
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say V s; that means, for which we have to
provide the stirrup. So, V s equal to V u
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minus V c. So, V c what we are doing that
whatever the force applied we are not designing
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that we are not providing stirrup for the
whole shear force that which is applied. Instead,
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we are taking care of; that means, we know
the certain portion will be taken care of
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by concrete which has three different parts.
So, V cz V a and V d in combination which
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comes say V c. If we can find out V c so,
we can subtract the V c from the vu the remaining
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balance V s if it is there at all for that
we have to provide the shear reinforcement.
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However, we provide nominal shear reinforcement
if and if we get that less than that then
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the vu if it is less then V c. The V c; that
means, say critical shear force the concrete
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can take care even it is if comes less in
even then we provide nominal shear reinforcement.
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So, what I have told total that total has
to be carried. So, v concrete plus v stirrup;
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so, vc plus vs or vu whatever. Now, what we
shall find out the tow c this is I have just
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simply given from the table 19 of IS code
456 of table there are. So, many other things
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say m fifteen also there m 30 lot of others
and also it goes beyond 1 0.25 percentage
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of the steel.
26:05.540 --> 26:12.010
So, to find out that vc we do not know how
shall we find out? Because, we do not know
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what are the different components for say
double action; we do not know what is the
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aggregate interlock how much will be the for
that compression part. Instead of that depending
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on the percentage of longitudinal steel the
tensile steel we shall get the tow c.
26:27.260 --> 26:32.980
So, the tow c is the critical shear stress
the concrete section can take which is dependent
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on the percentage of longitudinal tensile
reinforcement. And which is given in table
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19 of IS 456. So, because all ready we have
designed for flexure where you are providing
26:46.210 --> 26:51.740
the tensile reinforcement. So, on the basis
of that you know what is the how much is that
26:51.740 --> 26:57.280
percentage of tensile reinforcement and depending
on that for the grade of concrete you can
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find out what is tow c.
So; that means, we are not getting directly
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the component basis we are getting lump sum
this much we can get that which is which will
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give us that vc. So, vc is nothing, but tow
c bd. So, vc will be nothing, but tow c bd.
27:14.650 --> 27:22.800
And remaining portion we can take care by
the stirrup. So, this 1 you will get it in
27:22.800 --> 27:25.870
IS 456 at table 19.
27:25.870 --> 27:35.940
There is one more table that is your say table
20. Here, we shall get say maximum shear stress
27:35.940 --> 27:40.610
we are getting that one say shear stress that
nominal shear stress or tow c we are getting
27:40.610 --> 27:45.830
the critical shear stress. And the other side
that we are having the tow c max that, which
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is called as the maximum shear stress for
a particular grade of concrete.
27:50.650 --> 28:01.430
So, which comes as a for m 15 it is 2.5 m
22.8 m 253.1 m 33.5 like that it means that
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if you have a section say tow v we can get
the tow v.
28:08.000 --> 28:16.309
So, vu is the design force. So, tow v equal
to vu by bd that is the nominal shear stress.
28:16.309 --> 28:23.970
So, vu we are getting from our calculation
after giving the proper factor all those things.
28:23.970 --> 28:29.190
Vu we know that width d all ready we have
given that d because depth of the section
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all ready we have done it for the flexure
point of view. So, i can get tow v.
28:33.770 --> 28:42.559
So, this tow v should not be more than that
your say tow c max which is given in table
28:42.559 --> 28:53.360
twenty of IS 456. So, table 20 IS 456 which
is given here it says that m 20 that we shall
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get we shall get 2.8 maximum 2.8 Newton per
square millimeter.
28:57.420 --> 29:03.780
So, if it is more than 2.8 then we have to
redesign the section; that means, we have
29:03.780 --> 29:09.110
to change the section we have to increase
the depth. That means though we have done
29:09.110 --> 29:13.480
it for say from the flexure point of view.
Even then from the shear point of view it
29:13.480 --> 29:18.650
fails and therefore, we have to change the
section we have to increase the depth.
29:18.650 --> 29:25.260
So; that means, it is the beam is shear dominated.
So, that means, this is another aspect that
29:25.260 --> 29:29.080
we have to consider; that means, only flexure
is not sufficient. So, we have to take care
29:29.080 --> 29:35.340
that and there are also we have to increase
the depth of the section.
29:35.340 --> 29:43.850
So, now 1 more thing I have to tell because
these are the things which we should know
29:43.850 --> 29:50.750
when we are doing the design. So, according
to IS 456 what is the maximum spacing of shear
29:50.750 --> 29:58.860
reinforcement that that is a there is a limit
it should not be shall not exceed 0.75 d for
29:58.860 --> 30:04.640
vertical stirrups. When we are providing that
vertical stirrups, when we are providing vertical
30:04.640 --> 30:11.890
stirrups it should not be more than 0.75 d
d is the effective depth.
30:11.890 --> 30:24.500
Then for inclined stirrups say at 45 degree
it should not exceed d that effective depth
30:24.500 --> 30:29.260
and where d is the effective depth under consideration.
That means, where we are considering in and
30:29.260 --> 30:38.010
also 1 more clause that in no case shall the
spacing exceed 300 millimeter. So, even then
30:38.010 --> 30:44.370
say your depth is more say 0.75 d. So, you
are getting say something say 500, 500 with
30:44.370 --> 30:49.350
the effective depth. That means, that your
spacing will be more than 300. So, in that
30:49.350 --> 30:55.610
case it should be restricted by the 300 millimeter.
So, in our design calculation whenever we
30:55.610 --> 31:01.360
do the design; so, these are the different
restriction that we have to follow. And on
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the basis of that only we can design.
31:05.140 --> 31:42.510
Now, let us find out the shear capacity. So,
shear capacity of stirrups
31:42.510 --> 31:53.220
and let us say this is your longitudinal bar
and there is a crack say at 45 degree the
31:53.220 --> 32:20.770
diagonal tension and we have to resist. So,
that are diagonal tension. So, we have to
32:20.770 --> 32:52.820
provide the stirrups. So, asv we have to provide
stirrup asv total area of
32:52.820 --> 32:59.020
the legs shear links. So, we can have say
for 2 legged 2 times that area of the personal
32:59.020 --> 33:21.710
area of the bar provided sv spacing of links.
So, how many stirrups we can write down number
33:21.710 --> 33:53.750
of stirrups cut by a 45 degree crack line
this is the a crack equals say n say equal
33:53.750 --> 34:02.220
to I can right down d effective depth by sv.
So, what we can do it here in this case the
34:02.220 --> 34:10.040
number of if we have spacing is v and number
of stirrups. So, I can get that these 45 degree
34:10.040 --> 34:14.440
crack we can get because you are starting
from the effective depth here equal to d by
34:14.440 --> 34:21.769
sv. The number of stirrups which we shall
get here within this zone that because we
34:21.769 --> 34:28.059
have to resist.
Now, we are assuming that in this case we
34:28.059 --> 34:34.440
are assuming that we are giving the same sv
because, that for further calculation that
34:34.440 --> 34:40.940
we have to take care that sv same for the
throughout the section; throughout the length
34:40.940 --> 34:46.329
that we are considering. Otherwise we cannot
provide. But we restrict our self say we can
34:46.329 --> 34:52.609
go up to say 1 third distance the certain
spacing and depending on the shear force available.
34:52.609 --> 35:00.390
So, we cannot just simply, we cannot provide
we have to provide certain spacing up to certain
35:00.390 --> 35:03.960
length. And then again we can say that the
middle portion say shear force is less and
35:03.960 --> 35:08.980
that is why we provide that different spacing.
But when you are doing the calculation then
35:08.980 --> 35:12.859
we have to take care this spacing uniform.
35:12.859 --> 35:44.039
So, we can find out here the total shear resistance
of the vertical stirrup system across the
35:44.039 --> 36:36.390
section is given by vs equal to 0.87 times
fy asv times number of stirrups d by sv. This
36:36.390 --> 36:43.190
we have all ready have done. So, 0.87 fy asv
is the for the for the 1 say stirrup which
36:43.190 --> 36:48.420
is say it may be 2 legged 4 legged that which
taken care of any asv times. That number of
36:48.420 --> 36:54.470
stirrups that spacing on the basis of that
we have already done say n, so, d by sv.
36:54.470 --> 37:10.680
So, we can write down here vs equal to tow
v minus tow c bd. So, vs equal to tow v times
37:10.680 --> 37:16.460
bd because all ready have done tow v times
bd minus tow c times bd because this 1 we
37:16.460 --> 37:22.279
are getting from the depending on the longitudinal
reinforcement. So, we shall get vs equal to
37:22.279 --> 37:41.160
this 1. So, we can right down tow v minus
tow c bd equal to 0.87 fy asv times d by sv
37:41.160 --> 37:51.690
which is nothing, but this 1 equals this 1.
So, we can right down tow v minus tow c equal
37:51.690 --> 38:07.829
to 0.87 fy asv by bsv.
So, tow v minus tow c equal to 0.87 fy asv
38:07.829 --> 38:18.989
by bsv. We can write down in a different manner
we can also write down the expression here
38:18.989 --> 38:40.359
say asv by sv equal to tow v minus tow c times
b divided by 0.87 fy. So, asv by sv equal
38:40.359 --> 38:46.880
to tow v minus tow c d 0.87 fy. What we can
do; that means, this side we can find we know
38:46.880 --> 38:54.999
tow v we know tow c depending on the longitudinal
reinforcement. We know that b 0.87 fy because,
38:54.999 --> 39:01.259
which particular reinforcement you are providing
grade fe 415 or fe 250.
39:01.259 --> 39:07.650
So, we know this site asv; that means, you
are providing asv that we are choosing the
39:07.650 --> 39:12.720
bar diameter 8 millimeter 10 millimeter are
also you are choosing the 2 legged. So that
39:12.720 --> 39:21.720
means, you know asv. So, we can get sv, so
we can get sv. And we can find out and depending
39:21.720 --> 39:27.460
on the limitation, the restriction we can
provide that reinforcement that spacing. So,
39:27.460 --> 39:33.190
this is thing what we do we can solve ones
small problem I think in the short time.
39:33.190 --> 40:01.940
So, what we shall do let us take 1 example.
Let us take the rectangular beam b equal to
40:01.940 --> 40:22.109
300 millimeter D equal to 500 millimeters.
Vu 350 kilo Newton tensile reinforcement please
40:22.109 --> 40:28.559
note tensile reinforcement. I am repeating
every time tensile reinforcement not the compression
40:28.559 --> 40:34.739
because that table 19 of IS 456 it gives on
the basis of tensile reinforcement only percentage
40:34.739 --> 41:01.319
of steel. That is 520 torque concrete m 20
steel fe 4 1 5.
41:01.319 --> 41:23.549
We have to design stirrups. So, what we have
to do design of stirrups for shear? Number
41:23.549 --> 41:49.460
1 let us find out nominal shear stress. So,
vu given as 350 kilo Newton tow v equal to
41:49.460 --> 42:05.220
vu by bd 350 into 10 to the power 3 divided
by 300 times. What about d? Let us say d equal
42:05.220 --> 42:21.259
to capital d minus clear cover minus 5 by
2 equal to 500 minus clear cover is not given.
42:21.259 --> 42:27.509
But we are assuming 25 we can assume unless
otherwise specified we can assume certain
42:27.509 --> 42:33.619
dimension here. So, 25 by 2 that which comes
as 462.5millimeter.
42:33.619 --> 42:57.930
So, let us put it here 462.5 and which is
equals 2.5225 Newton per square millimeter.
42:57.930 --> 43:06.739
So, now we can calculate what about the tow
c that we can calculate.
43:06.739 --> 43:30.940
Number 2 ast the area of longitudinal steel
ast 5 times whole square equals 5 into 491
43:30.940 --> 43:49.690
which comes 2455 square millimeter. What about
then percentage of steel p equals 2455 into
43:49.690 --> 44:20.739
hundred divided by 300 into 462.5 that effective
depth equal to 1.769 percent. From table 20
44:20.739 --> 44:38.220
tow c max maximum tow c max equal to 2.8 Newton
per square millimeter. From table 19 for tow
44:38.220 --> 44:52.589
c. P 1.75 tow c for m 20 I am just giving
the part from table 19. So, for percentage
44:52.589 --> 45:06.769
of steel p 1.75 it comes 0.75. For 2 because
I am taking 1.769 for 2.79; so, this is the
45:06.769 --> 45:14.539
part of the table of table 19.
So, where it comes p 1.75 0.75 Newton per
45:14.539 --> 45:19.809
square millimeter let us write down this 1
Newton per square millimeter, this is Newton
45:19.809 --> 45:30.599
per square millimeter. So, I shall get tow
c equal to 0.79. Let us make it elaborate
45:30.599 --> 45:40.079
divided by 0.25 times 1.769 minus 1.75 plus
0.75.
45:40.079 --> 45:50.440
0.75 is the 1 that for 1.75 which comes as
very small almost we can take the same 1.72
45:50.440 --> 45:55.930
Newton per square millimeter. Just linear
interpolation that segment you take linear
45:55.930 --> 46:01.269
interpolation and you can do it. So, it is
very simple.
46:01.269 --> 46:38.269
So, we can write down asv, asv by sv equals
we can write down asv by sv equals we have
46:38.269 --> 46:55.319
already noted down we can go to that. So,
for vertical stirrups as per our code says
46:55.319 --> 47:20.440
we can write down here. Let us write down
first clause forty point four for vertical
47:20.440 --> 47:43.729
stirrups. So, which comes as it says. Vus
equal to 0.87 fy asv times d divided by sv.
47:43.729 --> 48:00.160
So, you can write down this 1 and this is
given in page seventy three of is four five
48:00.160 --> 48:16.460
six.
So, you can write down as asv by sv equal
48:16.460 --> 48:45.969
to vus by 0.87 fy times d equals. Vu minus
tow c bd divided by 0.87 fy d equals. B times
48:45.969 --> 49:04.869
tow v minus tow c by 0.87 fy tow 1 which we
have got it. Which comes as 300 2 point five
49:04.869 --> 49:13.920
2 two five minus 0.75 2 two point five 2 two
five that is tow v we have got it nominal
49:13.920 --> 49:24.150
shear stress minus 0.75 2 divided by 0.87
times four 1 five that concrete sorry steel
49:24.150 --> 49:42.729
grade and which comes as 1.471. So, asv by
sv equal to 1.471. What about asv how much
49:42.729 --> 49:46.279
is the asv?
49:46.279 --> 50:02.089
So, asv equal to 2 legged 2 into pie by 4
let us take 10 millimeter because we can do
50:02.089 --> 50:12.309
trial and error let us take i am taking say
ten millimeter. So, which comes as 157 square
50:12.309 --> 50:33.059
millimeter asv by sv. We have already computed
as 1.471 sv equals asv by 1.471 157 divided
50:33.059 --> 50:52.150
by 1 471 equals 106 square 106 millimeter.
So, what about the spacing maximum? So, let
50:52.150 --> 51:05.079
us say maximum spacing point 0.75 d equals
0.75 times how much is your d.
51:05.079 --> 51:37.150
So, d comes as 462.462.5 equals 0.75 into
462.5.346.8 millimeter and d not more than
51:37.150 --> 51:45.069
300 millimeter. So, we have got 106 millimeter
which is less then that so; that means, we
51:45.069 --> 52:00.369
have to provide 10 torque 2 legged at the
rate of 106 is not a good 1. So, we shall
52:00.369 --> 52:07.359
provide 100 millimeter center to center. So,
10 torque 2 legged at the rate of 100 millimeter
52:07.359 --> 52:14.869
center to center that we shall provide. So,
this is your that stirrup.
52:14.869 --> 52:26.549
So, this is your, that how we generally calculate
that your stirrup. So, if we summarize that
52:26.549 --> 52:41.799
what we have done
we shall get we shall get this vu from our
calculation from our analysis. After doing
52:41.799 --> 52:48.960
the proper factor all those things we shall
get the nominal shear stress using this formula
52:48.960 --> 52:56.549
and which our code permits.
We know the longitudinal area of steel because
52:56.549 --> 53:01.489
already we have provided that reinforcement
from the flexure point of view depending on
53:01.489 --> 53:10.239
that. We shall find out the tow c that concrete
contribution in shear resistance. We can find
53:10.239 --> 53:17.630
out depending on the percentage of steel and
which we shall get from table 19 of IS 456.
53:17.630 --> 53:23.859
We shall get also get the maximum shear stress
permitted for a particular grade of concrete.
53:23.859 --> 53:32.359
For this case say m 20 grade of concrete this
is m 20 grade for m 22.8 Newton per square
53:32.359 --> 53:35.299
millimeter.
If it is more than tow v if t is more than
53:35.299 --> 53:39.799
this then what will happen then we have to
redesign the section from the shear point
53:39.799 --> 53:46.999
of view. That means, we have to change that
depth of the section. After that we can get
53:46.999 --> 53:53.089
we have already derived also we can get the
formula in clause 40.4 for vertical stirrups
53:53.089 --> 54:01.609
the same formula which we have derived.
And we can find out that asv by sv 1.471 in
54:01.609 --> 54:07.660
this particular problem. Now, we shall choose
that whether we shall provide 8 millimeter
54:07.660 --> 54:12.059
or 10 millimeter or 20 millimeter stirrups
2 legged or 3 legged whatever depending on
54:12.059 --> 54:18.690
the situation. We provide this diameter we
choose the diameter and that number of legs
54:18.690 --> 54:26.420
most of the cases it is 2 legged. We can find
out asv and we can get that asv by sv on the
54:26.420 --> 54:31.390
basis of that we shall check how much is the
asv permitted or computed.
54:31.390 --> 54:36.190
Now, we have restriction in our code that
we should not be in this case we should not
54:36.190 --> 54:44.410
be more than 0.75 d 346 millimeter or 300
millimeter. So, but in anyways it is coming
54:44.410 --> 54:48.779
less than that. So, we have to provide that.
So, which is coming say we can provide safely
54:48.779 --> 54:55.190
10 torque 2 legged at the rate of 100 millimeter
center to center that we can provide. Now,
54:55.190 --> 55:02.349
now we shall continue with other problems
with the stirrups also inclined member inclined
55:02.349 --> 55:06.130
stirrup also. So, this is the one let us stop
it here.
55:06.130 --> 55:06.670
Thank you.