# Practice English Speaking&Listening with: Working Stress Method (Contd...)

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In continuation of Working Stress Method, I shall continue todays lecture also on

Working Stress Method. So, our todays topic, we shall continue the lecture number 5, that

is Working Stress Method.

Let me just briefly tell that assumptions; yesterday I told the assumptions. So, let

me just enlist it, related to that Working Stress Method. At any cross section, plane

sections, before bending remain plane after bending. So, that means, actually your strain,

the strain will be linear, that assumption, this assumption will be valid only when the

strain is linear. All tensile stresses are taken up by reinforcement,

and none by concrete, except as otherwise specified. In the special case, because concrete

can take little bit of say 10 percent, approximately you can say, the tensile case. So, in exceptional

case, we can consider that concrete also will take tensile.

So, our next assumption, that in our case, we shall not consider that concrete will take

tension, everything will be taken care of by steel only. The stress-strain relationship

of steel and concrete under working loads is a straight line. So, this is also another

assumption; and these are all assumptions I am talking from IS 456 2000. And the modular

ratio m has the value 280 by 3 sigma cbc. Sigma cbc permissible bending stress in compression;

sigma cbc that concrete bending stress in compression. So, these are the four assumptions

and you can get it in IS 456 also.

I am not going to detail of these, because I have already listed this table; already

I have given this table, in the last lecture. So, just to make it in a complete form that

I have just written, and you will get in IS 456 also. And I have given only those four

genuinely we use it; most of the cases we shall use M 20 and M 25.

Permissible stresses in concrete. And again we have given four M 15, M 20, M 25, and M

30. And bending compression and direct compression, and bond average plain bars in tension. So,

whatever if it is, compression or if it is say your high instant deformed bar, then we

have to make some kind of multiplication factor, and then, we can get the required value. That

is the standard procedure we do in our calculation. So, we start with one value, and then, we

can multiply with certain factor, and then, we can get the adequate value. That I shall

tell you in due course when you shall come across those types of problems.

And permissible stresses in steel reinforcement. This one also, this table also I have given.

And here you will find out that difference, particularly for Fe 250 - mild steel, that

up to 20 millimeter that one kind of stress pattern and up; over 20 millimeter diameter

we are having another one.

And I have already told that in the last class. And soo, this is your, that section. We are

talking, say rectangular section. We can say that t section also possible, l section - inverted

l section - is also possible; that we shall do it. Mainly we shall do it in the limit

state of method, because that is the method, which we shall consider ,and we should shall

use it. So, we are getting that one say width d; depth - overall depth - capital D; effective

depth - small d; and area of steel that ast means area of steel in the tensile reinforcement.

And this is the screen diagram, and this one, you see that one we are getting at this level;

not at the bottom; we can get in the bottom also, but we are not interested; we have interested

up to the steel bar level. And this is your, that neutral access position

from the top, that x, and which we specify x equal to kd. So, this is one factor k, which

is important, and z equal to jd, which we shall get it that one, say lever arm; this

is your lever arm. This is the one tense; compressive force, is the cgo of this triangle,

where it is acting, so, c; and tensile, force t in the steel bar; and you will get the lever

arm z, which is again specified as z equal to jd and we know that j equal to nothing

but 1 minus k by 3. M equal to 280 by 3 sigma cbc, that we have to calculate the modular

ratio, instead of calculating from es by ec, instead of calculating from es by ec, we shall

calculate it from m equal to 280 by 3 sigma cbc.

We shall calculate from the bending stress, in compression, of that concrete, from there

we shall calculate; because we do not what is the value of ec, it is difficult to get,

but it easier to get sigma cbc also, because we can get that characteristic strength, then

we can do the required factor, and then ,we can get the sigma cbc.

And c equal to, and c, that force in the concrete part, that in compression. So, we have already

given half fcb b into x, t equal to fst ast, m could be calculated on the basis of cz - compressive

force multiplied by the lever arm. We can calculate the momentum of resistance m or

we can calculate on the basis of t. And, in fact, we do it, we calculate it, that

whenever a concrete section is given, that we shall solve one problem today. Whenever

a concrete section is given, then we have to find out the moment of resistance, and

that moment of resistance, you have to find out we have to calculate moment of resistance

due to concrete failure, that if concrete fails, and we have to calculate the moment

of resistance, if steel fails. And out of these two, which one will be the minimum,

that we have to take, because it can fail before the other one reaches the failure.

So, that is the standard procedure we have to do it, and I shall show you one problem

today, for that, say how to calculate moment of resistance.

Before going to the calculations, let me tell you that what is a balanced section. Balanced

sections, it means, a reinforced concrete section is called a balanced section, when

the maximum allowable stresses, that means, sigma cbc in concrete, and sigma st in steel,

so when the maximum allowable stresses in concrete and steel are reached simultaneously.

So; obviously, we can say that is the one optimum section, if we can reach say in steel

as well as in concrete, same time, then only we can say that we have designed optimum section,

and truly speaking, it is not possible. We cannot provide the section, what we can come

closer. But generally, we have to do that closer, but we cannot, exactly we cannot make

it. If we have to make it, then your depth of the section it will be irregular. The number

will not be, say, some regular number. So, number could be something, say, fractional.

So, that is not a good choice, because if we go for that one, if you would like to optimize

in that aspect, the other aspect is that form work, shuttering. That one also it takes lot

of that cost; that it cost much. So, that one also you have to take care and you have

to use the form work repeatedly. It is not like that, for each and every beam or column,

you will use different formwork; you have to do it in such a way, so that you can use

your formwork. Then only you can optimize your cost also. So, that is another aspect.

So, our strength criteria definitely one aspect, but we have to consider other aspect also,

other practical difficulty also. So, that is why we compromise with certain kind of

strength, if we can reach within 10 percent or 15 percent, that means, it is anthis

is an optimum design, that we can say.

Next one we call it that under-reinforced section. So, that means, when the section

is not balanced, so when the section is not balanced, we mean to say that it is also possible

that related to steel I can say that, reinforcement steel, either that could be less or that could

be more, because balanced is something that which is equal to the concrete strength.

Now, when we are talking, say, steel only. So, steel area could be more or could be less,

because we have only two other options. So, if it is, say, more, that is we call it one

part; if it is less, then we call it under-reinforced section. So, the under-reinforced section,

it means, the sections in which the steel provided is less than that necessary for the

balanced section are called under-reinforced section.

So, this is the section, where we are having the reinforcement less than the balanced one.

It means that concrete will not fail first, steel will fail first, and that one, we require

also particularly forced, that means, when you are having say steel, it will fail fast,

it means that you are giving certain time for failure, because steel is ductile material,

it will give some time for deformation. So, the occupants of the particular building

will get some time to get away from that; particularly for say earthquake. If you like

to design earthquake resistance structure, so it should be better that we should use

under-reinforcement section, and also, if we can use mild steel which is more ductile,

so we shall get, you will get some time to get away. But generally, opposite it is not

possible now, it is to extensively used - the mild steel, but the section, at least it should

be under reinforced.

Because mild steel, we shall come to that next, we shall come, we shall find out, because

strength of mild steel is different than strength of, say, your high deformed that HYSD bar.

So, obviously, you need more steel, and that is why we prefer that, say we go for HYSD

bar.

So, the other one is called over-reinforced sections. Sections in which the steel provided

is more than the balanced sections, they are called over-reinforced sections. So, we shall

consider this one as a over-reinforced section and under reinforced section.

So, now, let us see the balanced sections, because we can find out the steel also. Let

me justI think, I can go back to the previous slides on that, yes. So, this is the one we

are interested. So, x and this one d minus x; this is x and this is d minus x. So, now

let us find out that or we can write down here also, at least for this page. We are

talking this one x and this part - effective depth - d minus x. So, we can write down fcb

at the top. So, we can write down kd; x equal to kd. So,

that I am writing here. And mfcb by fst. So, we can write down k equal to, let me make

it you, please check it, because we are not going to detail of that calculation. And j

equal to 1 minus k by 3. So, for the balanced section, we can find out ka, that k d, so

that means, if we know fcb, fst, and fcb, and m; m, obviously, we can find out am equal

to 280 by 3 sigma cbc. We should remember this one; and fcb, fcb means sigma cbc; and

fst sigma st. So, for a particular grade of concrete, and

for a particular grade of steel, what we can do, we can find out the value of k. And if

you know k, then we can find out j also, and if we can find out j, then we can find out

the jd, then we can find out moment of resistance, all those things we can calculate. So, here

that fcb, fst that one you are limiting permissible values, and that one we shall get it from

code; our code gives that permissible stresses for a particular grade of steel, for a particular

So, we can go; here let us see. So, we can go. Here let us write down fcb equal to sigma

cbc, and just make it here for M 15, M 20, M 25, M 30; 5.0, 7.0, 8.5, 10.0. So, these

are permissible stresses for different grades of concrete. Fst for balanced section, it

should be sigma st. Let me complete this one with giving the proper unit, and this one

also Newton per square millimeter. So, for Fe 250, Fe 415, we have 140, we have 230.

There is one more grade that is Fe 500; that is one more grade that is Fe 500. And that

also, we can use it, but I am giving only those, because you can get those things in

your code. So, in this particular class, if we cover

all the each and every numerical value, then it will be difficult to cover all those things.

So, we are only using those grade of concrete and steel, which we generally use. And other,

say Fe 400 you can get it in IS code. So, IS 456 that also we can find out, and there

you can give it in detail.

So, k equal to m sigma cbc divided by sigma st plus m sigma cbc. I repeat once more, m

equal to 280 by 3 sigma cbc, and k equal to, please note, there is one more k will come,

that is capital K. So, but this one for the time being we are only working on say small

k - that kd. So, that one comes asand j equal to 1 minus k by 3; and capital K equal

to kj by 2; and moment of resistance m, that one will be equal to capital K sigma cbc bd

square. So, m equal to capital K sigma cbc bd square.

This K, that is one factor, if you remember for homogeneous rectangular section, in the

last lecture we have found that is 0.167 times the stress times bd square. So, here also,

this capital K also, we shall get similar type of one fraction we shall get. So, let

me go little bit faster here.

So, we can find out that beam factors for Fe 250; we can find out beam factors for Fe

250, that k we can find out for different grades of concrete; k you can find out that

all of them having same, because for Fe 250, you please note, this is your that equation.

So, in this equation, you will find out it is independent of sigma cbc; it is independent

of any concrete value. We are getting only sigma st. So, that is why you will get for

any grade of concrete, you will get that k same, and you can calculate j, and you can

calculate capital K. So, if you come back to these beam factors table, the small k,

which the factor of k d, from where you can get that say 0.4 times small d - that effective

depth - so, we shall get that x. J which will give me the lever arm, and capital K - that

coefficient - you can get 0.174. So, all of them we are getting 0.174 and this one for

mild steel. So, this table for mild steel, for Fe 250.

And I have given only M 15, M 20, M 25, M 30, just to show that for all grades of concrete

we are giving the same value.

Next one, that is for Fe 415. Fe 415 we are getting that k 0.29, j 0.9, capital K 0.131.

So, this value you get it, say 0.167, the one we have told that one, and where we are

getting point 0.131; the other one, you are getting 0.174. So, that means, you can approximately,

that is not a bad choice. One can also get some kind of result even if you get say homogeneous,

but that is notobviously, that is not a correct one. So, the correct one what we

are doing now.

The next one, I shall tell you. So, I can go little further - the balanced steel percentages.

So, if we say pt or p, whatever we consider, p equal to say 100 ast by bd. This is the

percentage of steel, what we are considering, and so, ast equal to So, we can equate

forces, half sigma cbc times kd times b. This is the compressive force b times x; this the

compressive force, b times x, that we are talking.

So, this is the one we are talking, this force - c, and this is sigma cbc; this one x; width

of the beam, that is b. So, I can get capital C half sigma cbc kdb; x is nothing but kd;

that we can get it, which should be equal to p by 100 - we are talking percentage of

steel - p by 100 bd times sigma st. So, this is the area of the steel times sigma st will

give me the tensile one. The other one, this one; this is tensile force T.

So, from here what we can get, p will be equal to 50 K times sigma cbc by sigma st. If just

multiply, just rearrange it;so p percentage of steel will be equal to 50 times k sigma

cbc by sigma st. I know that k. So, I can calculate that balanced percentage of steel.

So, I can get the balanced percentage, because this one require some times that you can approximately

you can find out also considering the sigma cbc by sigma cbc and sigma st, you find out

k, and then, you can, immediately you can find out that p also; that you can find out.

So, percentage of tensile reinforcement for balanced section, and we can get the grade

of concrete, again we have taken four grades only, and sigma cbc, just to give the value,

and this is you that Fe 250, 140, 230, we have considered. So, this is your that balanced

percentage of steel. So, for M 15, that means, sigma cbc and sigma st, so, .071; we can get

for corresponding 0.31; here it is 1, 0.44, 1.21 percentage, where it is 0.53 percentage,

almost say even less than 50 percent you can say; the tor steel you need less then 50 percent.

So, you can find out that you require less than 50 percent; obviously, that we can

that is why we go for that tor steel. And we can find out that, say even it is less

than 0.60; it is 1.43. So, this is your that percentage of tensile reinforcement for balanced

section.

I shall come to the next point. We can find out, that analysis - one we call it design;

another one we call it analysis. Analysis means that when the section is given, reinforcement,

and the depth, width, everything given, then we call it analysis. Then we are analyzing

that how much load it can take; what is moment of resistance.

And design means, that your moment is given, and you have to provide the suitable section;

that is called design. So, here we are doing the analysis. So, analysis

of a given section, and we are talking bending only. So, we can do it. We can take the moment

about the neutral axis. So, this one we are taking. So, what we are doing? We are considering,

say, neutral axis somewhere here. This part is Kd; area of steel somewhere here;

area of steel somewhere here; and that distance d minus Kd. So, this area of steel. So, we

can get the area b times Kd times Kd by 2. I am taking the area with respect to their

moment of this area, with the respect to these neutral axis, and moment of these area, because

we are considering that equivalent section; that is why you have multiplied; this is the

area of steel p by - p is the percentage of steel - p by 100 times bd, because we are

considering that one - always percentage of steel with respect to the width and times

the effective depth. Please note - width times effective depth;

not the overall depth, because there is one more term - that is capital D - which is called

overall depth; there is one more term capital D, which is called overall depth, but we are

not considering overall depth. So, whenever we shall consider the percentage of steel

of the particular section, then we shall take it b times d. So, b times d multiplied by

p by 100 will give me the area of steel, multiplied by m is giving me the corresponding, say,

equivalent concrete section; that is thefor any composite, say, wood or, say, I think

you have done in first year class, wood and, say, steel bar. So, that either you have to

make the steel bar, you have to make it say the equivalent, say, wood area, or wood you

have to consider it as steel. So, wood will become thinner, but if you make, say, steel

convert it to wood, then it will be bigger. So, here we are considering the same way,

and multiplied by d times 1 minus k, that is the distance and we shall get this equation.

From there, we can find out K square, and in the last class also we have done it, and

K equal to -

last class we have told that is a reinforcement index, and here we are talking, say, percentage

of steel, and where p equal toSo, we can write down p 100 Ast by bd. And from there,

we can find out the corresponding K. That K we can calculate, the periodicity given,

then you can calculate that K.

So, we can just repeat, moment of resistance; let me write down, moment of resistance, M

equal to p by 100 bd sigma st d 1 minus k by 3. Also, you can write down in this way,

m by bd square equal to p by 100 multiplied by 1 minus k by 3 sigma st.

So, generally in sp 16, for in any design aid or you can make your own also, that we

can use this equation, which is independent of b and d. We can get a car or make a table,

knowing that giving, say p, for a particular grade of steel, for a particular grade of

steel we can find out, and for the different percentage of steel, we can find out that

corresponding m by bd square. And, then also, we can do it; that means, we can make those

things ready, and we can immediately, it will be faster in your design process, because

every time you need not either calculator or computer, you need not do it. So, what

you can do it in this way, you can make that cards or tabular form, and immediately you

Now, let us do, say, one problem. I hope we shall be able to finish it. Let us describe

the problem first. Singly reinforced, please note, I have already mentioned one term - singly

reinforced; that means, single; so, that means, there is a term, should be something else,

generally we have double. So, that we shall tell afterwards.

So, we are having one beam, width b 250 millimeter, overall depth 400 millimeter. Let us put is

as capital D; this is capital D; clear cover we shall consider, say cc 25 millimeter; reinforcement,

525; we designate 5, we specify 5, that means, we are considering mild steel; that is one

mode actually there, that I shall tell you in due course; and Fe 250 and M 20 grade of

concrete. So, this the description of the problem. This

is 250; overall depth, 400; there are 5 bars; there are 5 bars, 525. So, this is your that

problem; we have to solve it. So, sigma Fe 250; height 20 millimeter, sigma st, it is

up to 20 millimeter. So, you have to use 140 Newton per square millimeter; not 130 Newton

per square millimeter; M 20, sigma cbc. So, always you should write down first, the

values whatever you require, and m equal toSo, we know m, sigma st, sigma cbc.

So, shall I continue with the next step? I think I can change it.

bx square by 2; d equal to 400; minus 25 is the clear cover; 20 by 2, which comes as 365

millimeter. So, 250; m is 13.33; area of steel let us calculate here, 5 pi by 4 20 square

equal toSo, 5, I think. So, 1570 square millimeter. Or x square plus 1570 divided

by 125 times x minus. So, I think I have to calculate it. Divided by 125. Or x square

plus, which is coming as 167.42 x minus 13.33 into 1570 into 365 divided by 125, which is

coming as 61110, say that is equal So, x equal to

plus minus divided by 2.

So, which comes as Let us little bit elaborate and 167.4226. So, 260 or let us

say, 261. So, x

will become 261 minus 83.71. So, 177.29 millimeter. So, we can write down M 1, number 2, due to

concrete failure sigma cbc equal to 7.0 Newton per square millimeter.

M 1 equal to b x sigma cbc by 2 d minus x by 3 equal to 250 177.29 times 7 by 2 365

minus 177.29 by 3 equal to 5 minus 177.29 by 3. So, 305.9 into 7 by 2 into 250 into

177.29. So, which comes as 47454402 Newton millimeter

which comes as 47.45 kilo Newton meter. We get 47.45 kilo Newton meter; that is due concrete

failure. So, we have to get the other one.

So, we can get it the other one. So, M 2, due to steel failure. Here sigma st equal

to 140 Newton per square millimeter. M 2 equal to Ast multiplied by sigma st multiplied by

lever arm. Ast equal to 1570 millimeter. M 2 equal to 1570 multiplied by 140 divided

by 365 minus - what was the lever arm? That 177.29 by 3, which comes asSo, we are

getting 67237553 Newton millimeter equal to 67.23 kilo Newton meter.

We have got M 1, due to concrete 47.45, and here we are getting due to steel 67.23; that

means, moment of resistance M will be equal to 47.45 kilo Newton meter. It will be governed

by that, your say concrete; it is over-reinforced; that means, it is over-reinforced section.

So, if we go; that means, concrete, that is not a desired, desired choice; that means,

you can reduce your enforcement, can reduce your enforcement, and you can find out your,

No, no. We generally we avoid that; that over reinforced, because that one also not economic

also; the other way, because if the concrete fails first, that means, that is not economic

also; I do no whether I have I have I have nothing.

So, we shall conclude with the Working Stress Method, just to give an idea, and we shall

compare with the Limit State Method also, but mainly now on we shall work on Limit State

Method. So, in the class, we shall start Limit State Method.

Thank you.

The Description of Working Stress Method (Contd...)