WEBVTT
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Today; we shall start, limit state method
and we shall start with the topic our lecture
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number 6; the limit state of collapse and
we shall consider flexure. So, limit state
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of collapse flexure that we shall consider
today. So every method has certain few assumptions
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whatever we consider. So, that particular
equation or method is based on few assumptions.
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So, it is true for any experimental setup
also. So, whenever you’re making any setup
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which we will consider that any real life
problem there also we make certain kind of
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assumption and it will automatically will
come, when we shall consider that problem.
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So, what are those assumptions?
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The assumptions here, we shall consider the
assumptions that plane sections normal to
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the axis remain plane after bending the maximum
strain in concrete at the outer most compression
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fiber is taken as 0.0035. The strain is 0.0035
that is the maximum limiting strain in concrete
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that we shall consider, that is 0.0035; the
tensile strength of the concrete is ignored.
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So, that is the almost we can say primary
assumption in our concrete design, the reinforced
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concrete design we are assuming that we shall
not consider tensile strength of concrete.
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The stress block may be assumed to be rectangle;
I shall come within a minute. What is stress
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block, the stress block may be assumed to
be rectangle trapezoid parabola or any other
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shape, which results in predication of strength
in substantial agreement with the results
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of test. So, even if you consider a beam and
that beam, if you test it whatever the failure
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load you are getting and whether with your
model whether getting that result than you
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can say that it is in a good agreement with
your experimental result and the method you
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have proposed, the stress block that you have
proposed that is valid.
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The next assumption, the stresses in the reinforcement
is derived from representative stress-strain
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curve for the type of the steel used. So,
the stress-strain curves for steel that also
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you have to consider. I shall show you a graph,
for design purposes the partial safety factor
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gamma m that is for material that partial
safety factor steel we consider 1.15. So,
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if fy is the stress yield stress. So, we have
to consider fy by 1.15; that you have to consider
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that is the for design purposes we shall consider
we never take the ultimate limit for our design
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calculation.
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The maximum strain in the tension reinforcement
in the section at failure shall not be less
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than fy by 1.15 Es plus 0.002. So, fy by 1.15
Es plus 0.002. This your that strain that
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should not be less than this value. So, for
different grade of steel we shall get different
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limiting strength. We have 3 types of steel
fe 250, fe 415 and fe 500 most of the cases
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we use fe 415 and mild steel we use it fe
250.
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According to IS 456, this is your strain and
stress stress-strain curve of plane concrete.
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What we consider here, please note the strain
there are 2 strain 1 is 0.0035, which I have
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already told which is specified in the assumption
and given in IS 456 2000. So, that you should
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remember that 0.0035 that is the limiting
strain maximum strain allowed; in concrete,
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plane concrete and for design purposes we
shall use this curve. What we consider, 1
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this is the 1 that cube strength M 20 Newton
per square millimeter. M 25 Newton per square
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millimeter. We shall take 0. 6 7; that means,
two -third of that cube strength or characteristic
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strength we shall consider for our design
purpose.
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Further we shall consider; this value that
is 0.67 fck divided by gamma M; that is partial
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safety factor material concrete, gamma M for
concrete we shall consider 1.5 gamma M for
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steel we shall take 1.15 and, which is justified
because we are considering we take steel that
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is manufactured in the workshop in the control
environment so; obviously, we can take less,
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but whereas, concrete that is made in the
field itself. So, we are taking little bit
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higher and that 1 we are considering, we are
taking 1. 5. in that curve, you will get it
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in figure 21 for your reference in IS 456.
So, 1 more thing I would like to point out,
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this part is parabolic and another part straight,
but in real situation, that if you do the
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stress-strain curve or if you do the experiment.
If you apply the load, you will get certain
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drooping portion also; that means, it will
go up this way it will go up and then, slowly
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it will come down like this; it will go certain
level, but we are not considering for our
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design purposes, for our design purposes we
are taking this curve and when we shall take
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the actual value design value we shall consider
these curve, which is based on this curve
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we are getting from the experimental result.
Experimental data, then we are taking two
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-third of that further we are dividing by
partial safety factor for materials and then
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finally, we shall get the curve for our design
purpose.
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Figure 23A of IS 456 that is for cold worked
deformed bar high strength and we shall consider
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Es modulus of elasticity for steel, we shall
take this value. So, we shall take this value
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and we can get that different for different
strain we can get the stress value. So, we
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shall consider for high yield strength deformed
bar we shall take this stress-strain curve
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for steel.
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This is figure 23B steel bar, with definite
yield point for mild steel we shall consider,
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we shall take, these curve this is the test
data and this is after dividing by partial
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safety factor.
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So, let us come to the limit state of collapse.
We are starting with the limit state of collapse,
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ultimate strength and we shall consider bending
flexure; further we are considering that singly
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reinforced section. So, let us draw a cross section of the beam and we are having area of steel d effective
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depth d and though we don’t need D, but
even then let us write down, for the completeness
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this is your D. We have width of the beam
b and we should have the neutral axis; let us draw
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it here also further. So, that we can draw
the stress strain curve, this is the neutral
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axis position from the maximum compressive
stress. We shall get the compressive stress
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at the top fiber that is the maximum. And
from there we are measuring the depth of the
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neutral axis which is defined as x.
In working stress method; we have not considered
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we have not taken the strain criteria, but
in limit state method we are specifying the
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strain that is the limiting value of strain,
that we are specifying. For concrete the limiting
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strain 0.0035 that you should always remember
0.0035; we shall go up to the steel position.
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This is 0.0035 and this value say epsilon
s. Let us say, these value epsilon cu, what
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about the stress block, the stress block comes
the same 1 which we have shown you that curve
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we are having maximum 0.0035 somewhere here
0.002 because your are starting from 0. So,
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some where we shall get 0.002 finally, we
shall get the maximum strain allowed 0.0035,
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which will never go beyond this 0.0035.
According to this curve that mentioned in
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the code, we shall consider this curve. This
our stress block also this value comes 0.45;
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we shall consider this value as 0.45. So,
we are taking 45 percent of the cube strength
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to be more specific I can say, I am taking
45 percent of the cube strength, for our design
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purpose we are taking 45 percent the cube
strength. What about these steel this, this
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will be 0.87 fy multiplied by let us write
down here, please allow me times Ast 0.87
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fy. Ast here I think it is better, if I repeat
once more here 0.87 fy that is the stress
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permitted in steel multiplied by area of steel.
This 1 that your T. So, tensile force T on
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the steel 0.87 fy Ast we shall get the compressive
force that is C.
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What is the our interest here, that beam which
under bending, under flexure how much maximum
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load it can take. What is the moment carrying
capacity of that section that is our target?
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If we can find out that, value than we can
say this section having this many number of
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bars and the cross sections that width and
depth. We can say this is the maximum moment
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carrying capacity and that we have to find
out. So, if we can find out that 1, then from
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analysis the result you have got moment different
moment that you have got in different places.
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So, you can provide your section. That way
also you can do; you can do in another way
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also; that means, you are taking a particular
section let us see, the moment carrying capacity
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and let us provide. It should be always it
will be more than the computed value the moment
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at the particular section.
This is lever arm we specify by Z so, lever
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arm Z; so, what is the difference here in
working stress method we shall not always
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compare with the working stress method, but
since we have done in the last class the working
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stress method. Let us compare for the singly
reinforced section for flexure. We shall here
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we are considering we have started with the
strain, in working stress method we have not
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considered strain we have not taken strain.
Only difference is the stress block, only
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difference is the stress blocks in that case
the stress block linear here, we are getting
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certain portion parabolic certain portion
straight.
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But otherwise; your formulation everything
will go in the same direction here also, we
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shall consider the same lever arm and it should
be in equilibrium position compressive force
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equal to tensile force and moment carrying
capacity nothing, but compression compressive
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force times lever arm or tensile force times
the lever arm. Only the difference the stress
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value that portion is different it will be
different for this case because we are starting
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from the strain diagram.
Epsilon su as per the code to repeat fy that
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is 1 of the assumptions also
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or we can write down in a familiar way; which
is nothing but, 0.87 fy or let us write down.
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So, this is our corresponding strain this
1 will be epsilon su. This 1 will be epsilon
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su that ultimate strain in steel. We are considering
singly reinforced only reinforcement in the
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tension site and we know that, 0.0035 is the
limiting strain in concrete in compression.
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We shall get, the corresponding epsilon su
for different grade of steel we shall get
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different limiting strain.
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Let us draw once more, the stress block, the
strain and also draw the strain here, the
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strain here 0.0035 this is 0.002. So, what
about this value this 1 will be equal to this
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value let us find out 0.002 by 0.0035 and
which will be 0.57. If we take depth of the
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neutral axis from the maximum compressive
stress site x this is 0.57 x and this 0.43
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x. So, 0.57 x and 0.43 x; we can write down, total compression C area of rectangle. The full 1 minus area
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of outer parabola, area of outer parabola
means, this portion equal to 0.45 fck this
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1 multiplied by the neutral axis to make it
say for doubt let us give parenthesis it is
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not multiplication 1; x is the depth of the
neutral axis multiplied by b minus one-third
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you have to find out this area 0.45 fck. This
1 also is 0.5 fck multiplied by 0.57 x multiplied
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by the area, width b.
The b is the cross sectional width, which
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equals 0.45 fck x times b minus 0.86 fck x
b equals 0.364 fck x b. We take this 0.364
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fck x b, x is the depth of the neutral axis
fck the characteristic strength b width of
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the beam 0.364, this 0.364 we can take as
1 factor that k1.We can also write down equal
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to say k1 fck x b. This is 1 factor generally
considered that k1,x b.
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We can, further we can write down as C equal
to, I have told k1 fck x b in our case 0.364
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fck xb where k1 equal to 0.364. We write down
the equation in a different form x by d multiplied
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by bd. So, 0.364 fck x by d, this is multiplied
by bd, bd is the cross sectional area please
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note taking d as the effective depth not the
overall depth. Always we shall take, bd whenever
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we consider that is the effective depth we
shall take for area of steel what percentage
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that steel we shall consider Ast by bd b width
and d effective depth please note we shall
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always take d effective depth.
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Let us write down, once more
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the stress block 0.43 x 0.57 x compressive
force. We shall add one more factor, that
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is k 2 times x, x is the, this depth what
is the factor k 2 that we shall find out.
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We shall take moment; we shall take the moment
about the neutral axis. This is our neutral
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axis, if we take moment about the neutral
axis already we have computed that c 0.364
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fck x b that is the total force. Already we
have computed that; 0.364 fck x b. This is
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1 term, that is the total force and we are
taking moment about this line x minus k 2
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x. Which is equal to 0.45 fck x b times x
by 2; the total rectangle 0.45 fck times x
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the total rectangle and cg width and x by 2.
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So, we are getting that total rectangle minus
one-twelfth; let us make it, x whole square
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b. If we take that area so, we are taking
out this parabolic portion area times that
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cg. So, we shall get this 1, which equals
1 minus k2. If we simplify it equal to 0.584.Therefore,
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k2 equals 0.416 or we can say let us say 0.42.
So, the compressive force will act from the
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top fiber having, the maximum compressive
stress from there at a distance of k 2x which
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is nothing but, 0.42 x. So, we have got 1
factor k2 and another factor k2.
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Depth of neutral axis
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of a we shall get it again from equilibrium
of forces. Total tension T fst times Ast stress
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times Ast total compression C equal to total
compression. We have derived here to show
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you total compression C equal to 0.364 fck
x b. So, let me copy it here; 0.364 fck x
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b therefore, for equilibrium case; we also
write down instead of writing 0.364 we also
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write down as 0.36 also. We can take it you
know because that way we are estimating a
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little less we are taking a little less. So,
that also we consider; so, let us take 0.36
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fck x b; let me underline that we have noted
down 0.364 and here 0.36 we are taking a little
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less value fst Ast x equal to fst Ast divided
by 0.36 fck b.
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We can also write down other way also we can
write down this 2. As x by d equal to fst
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Ast 0.36 fcx b times d. We write the equation
in this fashion that x by d, x by d we instead
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of directly computing that value we specify
in this ratio that x by d. So, if we know
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the fst we can find out that x by d ratio.
Now, if we take the section is under reinforced;
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that means, the we shall reach the yield stress
in steel cost that case fst will be equal
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to 0.87 fy; that is the permissible stress
according to IS 456.
36:49.700 --> 37:23.529
So, we can write down the equation, for under
reinforced section fst equal to 0.87 fy xu
37:23.529 --> 37:34.009
by d. We are giving a special name xu that
is the xu is the that name ultimate that you
37:34.009 --> 37:53.470
can say that, depth of neutral axis .You can
say, su by d equal to 0.87 fy Ast by 0.36
37:53.470 --> 38:37.440
fck b d. Let us check, limiting values of
x by d specified in IS 456 epsilon su equal
38:37.440 --> 38:58.119
to fy by 1.15 Es plus 0.002 where s equal
to 2 into 10 to the power of 5 Newton per
38:58.119 --> 39:16.359
square millimeter. xu by d equal to epsilon
cu divided by epsilon cu plus epsilon su.
39:16.359 --> 39:33.329
Where, from I am getting from the strain diagram.
This is epsilon cu and this 1 epsilon su;
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epsilon cu epsilon su. So, we can get let
us say this one xu and total depth d.
39:52.620 --> 40:00.060
So, we can find out from this curve from this
similar triangle in civil we can find out
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xu by d epsilon cu by epsilon cu plus epsilon
su equal to epsilon cu. We know which is nothing
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but, 0.0035 divided by 0.0035 plus epsilon
su epsilon su is different for different grade
40:21.190 --> 40:31.420
of steel and which we shall get it which is
given in our IS 456 we can get using this
40:31.420 --> 40:34.059
formula.
40:34.059 --> 40:50.480
So, limiting values of x by d type of steel
mild steel, high yield, strength high yield
40:50.480 --> 41:03.309
strength 250 fe 250 fe 415 Fe 500 using the
formula, we can get the yield strength 0.0031
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for fe 250 0.0038 for fe 415 0.0042 for fe
500 simple calculation from there we can make
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and we should remember at least we should
be remember 0.0038. So, please note for concrete
41:24.769 --> 41:33.230
0.0035; most of the cases we use fe 415. So,
we should note the limiting strength which
41:33.230 --> 41:42.210
is nothing, but 0.0038. At least you should
remember these 2 values 0.0035 for concrete
41:42.210 --> 41:54.999
and for the steel maximum used fe 415 0.0038
and we can get the limiting values xu by d
41:54.999 --> 41:59.569
0.53 0.48 and 0.46.
41:59.569 --> 42:06.869
So, from that equation we can get it 0.53,
0. 480 point 46. So, again you please not
42:06.869 --> 42:29.579
this value 0.46 xu by d limiting value 0.46
point fe 415 is used. We shall consider, we
42:29.579 --> 42:38.740
have already derived this equation C equal
to k1 fck x by d times bd. This 2 should be
42:38.740 --> 42:49.819
suffix. So, k2 times x is the distance of
the centre of compression in the concrete
42:49.819 --> 42:58.369
from the top compression fiber that, I am
repeating number of times. So, k2, x1 parameter
42:58.369 --> 43:04.019
is k1 the other parameter is k2.
43:04.019 --> 43:16.069
And the third-one; the maximum or limiting
value of compression F times fck bd. So, 1
43:16.069 --> 43:25.049
parameter K1 another parameter k2 and the
third parameter we are taking F.
43:25.049 --> 43:35.249
So, values of constants for maximum compression
block. So, we are considering that is as a
43:35.249 --> 43:42.160
block that, 1 straight portion and another
1 parabolic portion we are considering for
43:42.160 --> 43:49.089
different steel, that already we have noted
down xu by d that we have noted down 0.53
43:49.089 --> 44:00.619
0.48 0.46 k1 that is 0.364 that is constant
for steel that k1 value similarly k2 also
44:00.619 --> 44:11.859
same 0.42 for all the cases and if that is
0.192 0.175 and 0.167 for different steel
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we are getting different limiting value. We
shall get, these values and we should at least
44:24.079 --> 44:30.349
we should note down again 0.175; that at least
you should note down the 0.175.
44:30.349 --> 44:44.619
Let us take; we are deriving those things
for our limit state method, our objective
44:44.619 --> 44:51.730
here. Let us derive that how far we can go
and what are the things we require for design
44:51.730 --> 44:56.940
purpose. When we are considering say, singly
reinforced section and for limit state of
44:56.940 --> 45:37.259
collapse for flexure. Expression of resistance
moment for a balanced section in terms of
45:37.259 --> 45:51.640
fy that is steel and percentage of steel,
the steel stress and percentage of steel xu
45:51.640 --> 46:18.880
by d equal to 0.87 fy Ast divided by 0.36
fck bd that we have already got it. Mu that
46:18.880 --> 46:31.369
limiting moment, that resisting moment, moment
of resistance 0.87 fy Ast. This is the tensile
46:31.369 --> 46:47.029
force multiplied by the lever arm.
So, lever arm here, equal to d minus we can
46:47.029 --> 46:56.359
write down 0.42 or 0.416 also. So, the exact
value also we can write down. Let me repeat
46:56.359 --> 47:15.269
once more here
47:15.269 --> 47:29.319
and some where here that compressive force
C, this 1 we are talking 0.416 or 0.42 also
47:29.319 --> 47:48.109
we can consider. The lever arm, this is your
lever arm lever arm equal to d minus 0.416
47:48.109 --> 48:09.819
0.87 fy Ast divided by 0.36 fck times b; we
are getting this xu we are putting it here
48:09.819 --> 48:24.890
and we can find out the lever arm. We can
write down it as d minus 1.005 times fy Ast
48:24.890 --> 48:37.239
divided by fcx b; we can take this one instead
of going to that position 1.005 we can simply
48:37.239 --> 48:42.130
take it as just 1.
In other way, we can write down this one as
48:42.130 --> 49:05.190
d minus fy Ast fcx times b therefore, Mu,
will be equal to this is your lever arm therefore,
49:05.190 --> 49:23.799
Mu equal to we can write down 0.87 fy Ast
times d; we are multiplying the 1 minus fy
49:23.799 --> 49:39.099
Ast divided by fck times bd 0.87 fy Ast d
times 1 minus fy Ast by fck bd. Mu that is
49:39.099 --> 49:45.089
the moment of resistance the resisting moment
that is section having area of steel Ast.
49:45.089 --> 49:55.259
We can get from this formula we can get that
moment of resistance.
49:55.259 --> 50:28.509
Therefore we can write down it as Mu 0.87
fy let me copy it once more, Mu equal to 0.87
50:28.509 --> 50:43.579
fy p that percentage of steel. I can write
down it as Ast by bd into 100; this is our
50:43.579 --> 50:56.549
percentage of steel. So, we can write down
Mu equal to 0.87 fy Ast we can put it from
50:56.549 --> 51:23.069
here. So, we can write down it as p by 100
1 minus 1.005 fy by fck p by 100 bd square.
51:23.069 --> 51:38.509
We generally write down in this fashion, Mu
by bd square equal to 0.87 fy p by 100 multiplied
51:38.509 --> 51:52.140
by 1 minus here, again I can omit this 1.005
ignore that, fy by fck p by 100.
51:52.140 --> 52:03.730
So, for a particular grade of steel and percentage
of steel given, we can find out that Mu by
52:03.730 --> 52:13.499
bd square. Where we do not need the section
size, we can get 1 value; that means, 1 can
52:13.499 --> 52:19.960
plot it also for different T also we can plot
it also. So, we can get Mu by bd square we
52:19.960 --> 52:26.619
can get it. For different percentage of steel,
we can get the corresponding Mu by bd square.
52:26.619 --> 52:34.099
So, we can plot it also and from here, if
we know this value then multiple by bd square,
52:34.099 --> 52:48.799
we can get the corresponding moment of resistance
that we can find out.
52:48.799 --> 53:05.849
One thing here; I shall tell you, we can also find out for other cases also just to repeat the 1 which we have
53:05.849 --> 53:11.529
specified in your say mentioned in the working stress method. That is your say balanced sections.
53:11.529 --> 53:23.730
So, sections in which the tension steel reaches
yield strength simultaneously; as the concrete
53:23.730 --> 53:28.349
reaches the failure strain in bending are
called balanced section. Because we are considering
53:28.349 --> 53:37.519
beam and that is mainly for bending the dominating
one. So, we shall consider that section as
53:37.519 --> 53:42.960
balanced section when concrete and steel both
of them simultaneously reaching the limiting
53:42.960 --> 53:48.970
strength specified limiting strength then
we shall take it as balanced section.
53:48.970 --> 53:57.859
Similarly, we have under reinforced section
and which is preferable most of the we always
53:57.859 --> 54:05.359
prefer this 1. So, sections in which tension
steel reaches yield strain at loads lower
54:05.359 --> 54:11.259
than the load at which concrete reaches failure
strength are called under reinforced sections.
54:11.259 --> 54:19.789
So, in this case the load applied; we can
apply the load we shall stop, when we shall
54:19.789 --> 54:27.779
stop .When we shall measure the strain, we
shall find the strain in the steel that is
54:27.779 --> 54:35.839
achieved say 0.0038 for Fe 415. We can measure
the strength and we can find out 0.0035. And
54:35.839 --> 54:40.579
then we can stop there we shall not consider
that concrete may be the concrete steel we
54:40.579 --> 54:45.989
have not reached that value. And that one
we shall take it as a under reinforced section.
54:45.989 --> 54:53.239
Similarly, we have the other side also. So,
sections in which the failure strain in concrete
54:53.239 --> 54:59.849
is reached earlier than, the yield strain
of steel is reached are called over reinforced
54:59.849 --> 55:06.680
section. So, we are having balanced section
when we use the strain compared to working
55:06.680 --> 55:14.910
stress method we consider that one from the
stress point of view, but here, we are considering
55:14.910 --> 55:21.079
from the strain point of view. Because we
are checking the strain and then from the
55:21.079 --> 55:24.069
stress-stain curve we are getting the corresponding
stress.
55:24.069 --> 55:30.499
So, this is the basic difference in the working
stress method and limit stress method and
55:30.499 --> 55:36.039
we are having balanced section, when we have
reached simultaneously and; obviously, there
55:36.039 --> 55:41.940
are 2 options only either steel reached first
or concrete reached first. If steel reached
55:41.940 --> 55:47.910
first then under reinforced if concrete reached
first then, we shall consider a over reinforcement.
55:47.910 --> 55:54.559
And over reinforced section is not preferable
because if it is under reinforced section
55:54.559 --> 56:01.380
then, it will give it will deform actually
it will have some time to deform. It will
56:01.380 --> 56:08.710
yield for some time. So, occupants will get
some time to that secure that particular says
56:08.710 --> 56:13.309
your place. So, we can say that under reinforced section
56:13.309 --> 56:21.239
is preferable. We can also get that one from
the limiting xu by d. The xu by d which is
56:21.239 --> 56:29.460
given from there also we can find out; the
limiting value in no circumstances we should
56:29.460 --> 56:35.170
not, that go beyond that xu by d value. It
should be always within that limit that xu
56:35.170 --> 56:40.660
by d from there also we can check that, whether
it is over reinforced or under reinforced
56:40.660 --> 56:46.099
whether we are getting that yielding in steel
or not that also we can find out.
56:46.099 --> 56:56.529
Just to conclude this session; we shall say
that percentage of limiting steel areas for
56:56.529 --> 57:05.089
balanced design. We can find out, the derivation
other things I shall give in later stage.
57:05.089 --> 57:11.239
We can find out for different steel the xu
by d or x by d; whatever, you say and the
57:11.239 --> 57:18.259
percentage of steel instead of percentage
of steel we have retained this one pt fy by
57:18.259 --> 57:23.880
fck. So, that is equal to 21.97. So, from
there we can find out, the corresponding value
57:23.880 --> 57:32.089
of pt. So, similarly for fe 415 we can also
find out similarly for other one also we can find out.
57:32.089 --> 57:39.069
So, we shall conclude this particular
class here and we shall continue.