Welcome to lesson 3 on the course Strength

of Materials. We are going to discuss about the

analysis of the stress. We have already looked into some aspects of stress analysis.

. .Now it is expected that once this particular

lesson is completed one will able to compute stress

resultant and stresses in members subjected to axial forces, evaluate stresses at a point

on a body at any arbitrary plane, evaluate principal

stresses and locate principal planes and also compute

stress invariants. . Hence the scope of this particular lesson

includes: review of normal stress, concept of shear and

bearing stress, computation of stress on any arbitrary plane, concept of principal stress

and principal plane and concept of stress invariants.

We discussed the types of stress and specifically about the normal stress.

. .And we had noticed that if we take a body

which is axially loaded by a force P and if we take a

section and draw the free body diagram, this body is under the action of external force

P so at the chord section say “aa” there will be resulting

stress component which we call as stress resultant or the force which is resisting P. At every

point there will be a stress component and the normal

stress multiplied by the area will give the force. So integral of stress multiplied by

the small area integrated over this surface will give the

stress resultant which is equal to P. While making this kind of assumption that

every where state of the stress exists if the force acts

through the centriod of the section we assume that the particle of the material at every

point contributes to the resistance of this external

force and thereby we assume the homogeneity of this

material. We assume that at every point the same state

of stress exists. When a body is subjected to

external forces which are trying to cause traction in the member or trying to pull the

member we call these kinds of forces and thereby stresses

as tensile stresses whereas when the external forces are acting on the member trying to

push the member we call this kind of external forces

and the stresses as compressive stresses. . Let us look into the aspects of shear stresses.

Let us assume that this particular body is subjected

to the action of external forces P and the resistive forces thereby will be P by 2 and

P by 2. Now if we take the free body diagram of this

particular member if we cut it over here then we

will have this body which is acted upon by external force P by 2 hence the resistive

internal force will also be P by 2. If we take the free body

of the other part of the cut the resistive internal force

is P by 2 and this force will be resisted at this inter phase and this also will be

P by 2. Though, this force is eccentric with respect to the

centre line of this body but this thickness being smaller

we neglect the other effects because of these forces. This force will try to cause stress

at the contact between these two elements and this

contact area is so near which is the plan. When we .look from the top the top view of this body

looks like this and this shaded part indicates the

contact area between these two pieces of material. Now if we consider that this particular length

as “a” and this as “b” then you can define the shear

stress which is designated as tau equals to the force which is acting at the inter phase

which is P by 2 by a(b). So a(b) is the contact area

over which the shearing stress exits. . Many a times we come across situations where

some blocks are resting over another block and

transferring the forces from external sources. For example, if we have a body like this in

which, this is a block which is resting on a bigger

block and the smaller block is subjected to external

load P. If this block is placed, that is, the centroid of the top block is placed on

the centroid of the bottom block then at the interface between

these two blocks and at this inter phase there will be

normal stresses generated and these normal stresses we designate as bearing stress. By

the term bearing we mean that the bottom block is bearing

the load of the top block. So, if the contact area

between the smaller block and the bigger block, this equals to Ac then the bearing stress

sigma bearing can be written as the external force

P by Ac which is the contact area. So this is called as

bearing stress. Here is another example which is the concept

of the bearing stress. We have a rigid bar resting

on two supports and this is subjected to external force P.

Now if we take the free body diagram of this particular bar, the external force

is P so the resistive force at the support point, assuming

this P is acting to the centroid of this body is P by 2 and P by 2. This reactive force

will be in turn transmitted to the support. This is the

support and resistive force which is getting transmitted

on and depending on the contact area we have if this is length “a” and the width of

this body is “b” then this force P by 2 has contact

area which is a(b). Hence the bearing stress that is sigma

bearing is equal to P by 2 as the reacting force which is getting transmitted on this

support divided by a(b). .. Here are the different stress components that

act on any arbitrary plane. Let us consider that A,

B, C are any arbitrary plane and o, x, y, and z is the reference axis system. As we

have noticed earlier the plane normal to which it coincides

with the axis we designate that in the name of the

particular axis. Likewise this particular plane is the x plane on which the normal stress

sigma x acts.

Likewise the plane oBC is normal to that coincides with y direction which is y-plane and the

normal stress sigma y on this plane. And the normal stress on the plane Aoc is the z-plane

is sigma z . Also, in those planes there are

shearing stresses x-plane in the y-direction will have the

stress which we call as tau xy , which we call as tau to the power XZ; on the x-plane,

in the ydirection, so tauxy. The component which is in the z-direction is designated

as tau to the power XZ. Likewise shear stress component in the

y-plane acting in the x-direction we call that as tau yx

and tau yz . Similarly this is tau zx and tau zy .

Let us assume that this arbitrary plane has a normal which is outward drawn normal is

n. This unit vector can be designated with reference

to xyz-plane. Let us assume that this is the reference

axis xyz. The unit normal is drawn here. If this vector makes an angle of α with x-axis,

beta with y-axis, and gamma with z-axis then we define

the cosine components in the x-direction as n x

which is cosine α; n y is the cosine beta; and n z as cosine gamma. Thereby the unit

vector this distance here can be represented by (n x ) whole

square plus (n z ) whole square; (n z ) whole square

plus (n y ) whole square will give this unit. So we have in effect (n x ) whole square plus

(n y ) whole square plus (n z ) whole square is equal to

1. Let us assume that on this arbitrary plane

we have the resulting stress vector as R and the

component of this resultant stress on this plane in the x, y, and z-direction be R x

, R y , and R z . Also let us assume that

the area of the arbitrary plane is dA which is the area of the plane ABC.

Now, if we take the projection of this area on x-plane which is AOB; so area of AOB is

dA into n x which is cosine of this area ABC on AOB.

Area BOC is the projection on the area ABC on y .plane which is dA into n y ; and area AOC

is the projection of ABC on the z-plane which is dA

into n z . . Now if we take the summation of all forces in the

x-direction where in the stress components involved will be sigmatau y in the x-direction,

and tauz x in the x-direction and R x then we can

write down the equilibrium equation in x-the direction. So equilibrium equation in the

xdirection will be R x into area dA; which is a force, stress resultant multiplied by

area minus sigma x dA n x the area of x-plane minus tau

yx which is acting in the y-plane times the area of yplane dA n y minus tau zx which is

the z-plane acting in x-direction multiplied by the area dA n z is

equal to 0. So the equation is as follows: R x dA minus sigma x dA n x minus tau yx dA

n y minus tau zx dA n z is equal to 0.

If we divide the whole equation by dA or in a limiting situation we get R x is equal to

sigma x n x plus tau yx n y plus tau zx n z . Similarly,

if we take the equations in the y and z-directions and write

down the equilibrium we will get R y is equal to tau xy n x plus sigma y n y plus tau zy

n z R y. And R z the resulting stress in the z-direction is

equal to tau xz (plane in the z-direction as) n x plus tau yz

(the y-plane in the z-direction) n y plus sigma z n z . That is R z is equal to tau

xz n x plus tau yz n y plus sigma z n z. These are the three resulting

stress components. So the stress components on the

arbitrary plane which are acting in the x, y and z-directions are represented in terms

of the stress components in the rectangular co-ordinate

system. This set of equations is normally designated

as Cauchy’s stress formula. .. So Cauchy’s stress formula is R x is equal

to sigma x n x plus tau yx n y plus tau zx n z ; R y is equal to

tau xy n x plus sigma y n y plus tau zy n z and Rz is equal to tau xz n x plus tau yz

n y plus sigma z n z . . Now let us look into, if we consider a plane

which has normal n and the direction cosines for the

normal are n x , n y and n z . Also, we assume that on this particular plane only

the normal stress acts in the direction of the normal to the

plane. Hence if we take the components of this in the x,

y, and z direction then as we have designated before R x as the resulting stress in the

x-direction; R y as the resulting stress in the y-direction

and R z as the resulting stress in the z-direction they .can be written in the direction cosines as:

R x as sigma n n x ; Ry as sigma n n y ; and R z as sigma n

nz. . Exactly in the same form the way we have evaluated

Cauchy’s stress formula taking the equilibrium equations in the x, y and z-direction

we can compute the resulting forces in x, y, and

z-direction in terms of sigma n . In the previous case we had forces in the

x direction as R x dA is equal to sigma x dA n x plus tau yx

dA n y plus tau zx dA n z and by dividing the whole equation by dA we had R x is equal

to sigma x n x plus tau yx n y plus tau zx n z . Exactly

in the same form in place of R x now we have sigma n n x

the resulting stress in the x-direction and this multiplied by area gives the force in

the xdirection, this equals to sigma x dA n x plus tau yx dA n y plus tau zx dA n z

. Hence from this we can write sigma n n x is equal to sigma x n x

plus tau yx n y plus tau zx n z . .. This is what is represented here the equations

of equilibrium in the three directions x, y, and z.

Now these equations can be rearranged and can be written as (sigma x minus sigma n)

n x plus tau yx n y plus tau zx n z is equal to 0;

tau xy n x plus (sigma y minus sigma n ) n y plus tau zy n z is equal

to 0; tau xz n x plus tau yz n y plus (sigma z minus sigma n) n z is equal to 0.

We have already seen that tau yx is equal to tau xy; tau zx is equal to tau xz al; tau

zy is equal to tau yz . Now these three equations can be thought of

as simultaneous equations containing n x , n y and n z

and we can evaluate the values of n x , n y and n z . Now, if we expand this particular

equation we will get a cubical equation in sigma n .

. .This is the cubical equation in sigma n : (sigma

n ) whole cube minus l 1 (sigma n ) whole square plus

l 2 sigma n minus l 3 is equal to 0. And once we solve this cubical equation we are expected

to get three roots which eventually will turn out

be real and we designate those roots as sigma 1 , sigma 2

and sigma 3 . And corresponding to each values of sigma we will get the values of n x , n

y and n z . . So here we have the equations on the arbitrary

plane on which the stress in absolutely normal and we have observed that we can get simultaneous

equations n x , n y and n z in terms of sigma x

tau xy , tau yx and tau yz; and sigma x , sigma y and sigma z .

. .This is the cubical equation in the terms

of I 1 , I 2 , and I 3 and for each of these sigma values if we

are interested to compute the values of n x , n y , and n z we can do that using these

equations along with the expression we have n x square plus

n y square plus n z square is equal to 1. So, for a trivial

solution n x , n y , n z is equal to 0 which is not really going to give us the solution

because n x square plus n y square plus n z square is

equal to 1 now for a trivial solution of the simultaneous

solution we said that the determinant of the coefficients of n x , n y , and n z is equal

to 0. And if we do that, we have (sigma x

minus sigma n ) tau xy tau xz tau xy (sigma y minus sigma n ) tau yz tau xz tau yz

(sigma z – sigma n ) is equal to 0. So if we say that the determinant of this

equals to zero for trivial solution we get the cubical

equation in sigma n cube minus I 1 sigma n square plus I 2 sigma n minus I 3 where the

values of I 1 is equal to sigmax plus sigma y plus sigma z

; I 2 is equal to determinant sigmax sigma y tau xy plus

sigma y sigma z tau yz plus sigma z sigma x tau xz ; I 3 is equal to sigma x sigma y

sigma z plus tau xy tau xz tau yz and this is symmetrical tau xy , tau

xz , tau yz these are the three coefficients I 1 , I 2 , and I 3 .

. .Now the plane on which this stress vector

is fully normal is known as principal plane. This is

what we considered in the previous situation where we had taken the arbitrary plane and

we considered the stress vector which is along

the normal and thereby we obtained the cubical equations in sigma n and from which we obtained

the three roots sigma sigma 1 , sigma 2 and sigma 3 .

In this particular plane the normal stress is acting along the normal of the plane. So

the plane on which this stress vector is wholly normal

is called as principal plane. The stress on this principal

plane which is absolutely normal is called as principal stress. And since the stress

acting is in the normal direction there are no tangential stresses,

the principal stress is the resultant normal stress

so on a principal plane there are no tangential or shearing stresses, they are 0. . .Now we have seen the cubical equation in

sigma n from which we have obtained three roots

sigma 1 , sigma 2 and sigma 3 . And we have looked into three coefficients I 1 , I 2 and

I 3 I 1 is equal to sigma x plus sigma y plus sigma z ; I 2 if

we expand determinants I 2 is equal to sigma x sigma y

(tau xy ) whole square plus sigma y sigma z (tau yz ) whole square plus sigma z sigma

x (tau zx ) whole square; and I 3 is equal to sigma x sigma

y sigma z plus 2 tau xy tau yz tau zx minus sigma x tau yz square

minus sigma y tau xz square minus sigma z taux y square. We call these as invariants.

It is to be noted that the principal stress which we have

calculated at a particular point remains the same

irrespective of the reference axis system we take.

In this particular case we have taken x, y and z as the rectangular axis system. Supposing

at that particular point if we take different axis

system which is represented as x prime, y prime and z

prime if we write down corresponding normal stresses and shearing stresses as sigma x

prime, sigma y prime and sigma z prime then we can

observe that the values of I 1 , I 2 and I 3 which are

represented in terms of the normal stresses and shearing stresses they remain the same

because the principal stresses at that particular

point remains unchanged. . .That is why these coefficients I 1 , I 2

and I 3 are known as stress invariants. Let us see how to solve

these stresses in some physical problems? In this particular example here there are

two plates which are connected together by a bolt and

it is subjected to a pull – external force P. This is the

plan of the two plates and if we take a section we cut the plate here and cut the plate here

and view from this side then the section looks

like this. Therefore the width of the plate is 200 mm

and the thickness of the plate is given as 10 mm and

the tensile pull that the plate is subjected to is 50kN. Now we will have to compute average

normal stress at a section where there are no holes for the bolts. So, if we cut the

section here and draw the free body diagram.

Here we have the plate which is put in a three dimensional form and

this is acted on by a load 50 kN, the width of the plate is given as 200

mm and the thickness of the plate is given as 10 mm.

So, at this particular section the resistive force for this external force 50kN which is

acting through the centroid of this plate is R and

hence from the equilibrium of these forces R is equal

to 50 kN. Hence the normal stress sigma at this particular

cross section where there are no holes is equals

to R divided by the cross sectional area sigma is equal to R by A is equal to 50 into 10

cube N by 200 into 10. This gives us the normal stress

as 25 N by mm square and 1000 mm is equal to 1

meter so this is equal to 25 into (10) whole power 6 N by m square. Since we designate

N by m square as P so it is 25 into (10) whole power

6 P and since (10) whole power 6 P gives the mPa,

hence we are going to write this as 25 mPa. Also, we will have to compute the average

shear stresses in the bolts.

The force which is acting here gets transmitted from this particular plate to this plate through

this connection where these two plates are connected

by the bolts. And as a result if we draw free body at this interface the force P will be

get transmitted at this interface and the two bolts which

are connecting these two plates together will be subjected to this force P. So we have one

bolt .here and another bolt here of diameter 20

mm and these two bolts will be resisting the force P by

2 and P by 2. The plate is subjected to load P and this

P is transmitted through the top plate through the

interconnecting bolt. So the force which will be resisted by these bolts is half of P, hence

the stress which is acting in the bolt which is

the force on this particular area is called as the shearing

stress on the plane of the bolt. So the average shear stress in each bolt which is tau average,

tau is equal to P by 2 by pi by 4 (d) whole square

(divided by the area of the bolt) and P is equal to 50;

25 into 10 cube N by pi by 4 into 400 is equal to 250 by pi N by m square which is so much

of mPa. So this is the average stress in the

bolt. Thirdly, we will have to compute the bearing

stress between the bolts and the plates. When the

force is getting transmitted from one plate to another it is getting transmitted through

these interconnecting bolts.

. And if we look into the transmission of the

force in little greater detail we will see that this is the

plate, here two bolts are present and this plate is being pulled by force P. Now the

transfer of force from the plate to the bolts which are

here when plate is being pulled, this part of the plate,

this is the bolt, the hub part of this particular plate comes in contact with this particular

bolt and there may be release in the contact between

the plate and the bolt surface. Basically the plate is resting on this surface

of the bolt. On an average sense we take the projection of this surface which is the diameter

of this bolt and the contact area is here which is

diameter times the thickness of the plate. So the contact surface which we get is half

the perimeter the projection of which is d and

the thickness of the plate at that particular interface.

This is the area which is in contact with the plate and the bolt.

The bearing stress is the function of the contact area so the sigma bg bearing is equal

to the force P and since we have two bolts each bolt will

get half the forces so P by 2 divided by the contact .area which is dt; sigma bg is equal to P

by 2 by dt is equal to 25 into 10 cube N by 20 into 10 is

equal to 125 MPa. . Here is another example where we are interested to

evaluate the principal stresses and the stress invariants. We have learnt how to compute

the principal stresses from the cubical equations and

the stress invariant components. Now the state of stress at a point is given by this, this

is the stress tensor where sigma x is equal to minus

5 units; sigma y is equal to 1 unit; and sigma z is

equal to 1 unit and tau xy is equal to minus 1; tau xz is equal to 0; and tau yz is equal

to 0. So, if we compute the stress invariants, as we have

seen, I 1 is equal to sigma x plus sigma y plus sigma z ;

sigma x is equal to minus 5 plus 1 plus 1, is equal to minus 3. I 2 is equal to sigmax

sigma y (tau xy ) whole square plus sigma y sigma z (tau yz

) whole square plus sigma z sigma x (tau zx ) whole square is

equal to (minus 5 minus 1) plus (1 minus 0) plus(minus 5 minus 0) is equal to minus 10. . .Likewise if you compute I 3 is equal to minus

6. Hence the cubical equation which you get is,

sigma n cube minus I 1 into (sigma n ) whole square here I 1 is equal to minus 3 plus 3

(sigma n ) whole square, I 2 is equal to minus 10 hence

minus 10sigma n plus 6 is equal to 0. Now sigma n is equal to 1 we will make this

function as 0, since sigma n in 1 of the root of the

equation, sigma n minus 1 if we take, sigma n square so this gives sigma n cube minus

sigma n square plus 4sigma n square minus 4sigma n

minus 6sigma n plus 6; so minus 6 into (sigma n minus

1) is equal to 0. So we get (sigma n minus 1) into (sigma n square plus 4sigma n minus

6) is equal to 0. From this we will get sigma n is equal

to 1, minus 2 plus or minus square root of 10. These

are the roots of this equation. These are the three principal stresses. These are the

three invariants I 1 , I 2 , and I 3 . That is how we can compute

principal stresses and stress invariants. . .Now in this we have another example problem

in which mass is hung by two wires AB and BC

and the cross sectional areas of these two wires are given as 200 mm square and 400 mm

square. And if the allowable tensile stress of these

wire material is limited to 100 MPa then you will

have to find out the mass M that can be safely supported by these wires. This particular

problem can be solved by taking the free body diagram

of this. . To summarize what we have learnt in this particular

lesson; first we recapitulated on different kinds of stresses and those stresses are the

normal stress, and the normal stress on an axially

loaded bar. What is the maximum normal stress that acts on an axially loaded bar is the

axial pull P or compressive force push by cross sectional

area and cross sectional area which is minimum is

normal to the axis of the bar. .Thereby we have seen the relationship between

the normal stress corresponding shearing stresses. We have evaluated the stresses on

any arbitrary plane which we generally designate as

Cauchy’s stress formula. And from this Cauchy’s formula we arrived at concepts of principal

plane and principal stresses and we have said that the principal plane is the one on which

the stress is fully normal and thereby tangential

stresses are at 0 and the shearing stresses in the

plane is 0. While computing the principal stresses we have seen different coefficients

which we designated as stress invariants I 1 , I 2

and I 3 . We have noticed that I 1 , I 2 and I 3 are

the functions of normal stress at a point which are sigma x ,

sigma y and sigma z and the corresponding shearing stresses at

the particular point. We have seen some examples to demonstrate how to evaluate

stress at different points. We have tried to give

you the concept of normal stress, we have computed the normal stress at a particular

cross section, we computed the shearing stresses

in the bolt, where the normal force which is acting on

the plate is transferred into the bolt cross section and then we have computed the bearing

stresses, the bearing stresses are acting at the contact area between the plate and

the bolt and finally we have computed the

principal stresses and stress invariants.

. .

Daniel Loreto MartínezPost authorOutsanding lectures! i've got a question: What is the recommended textbook for this course?. This course appears to be a little bit higher level tan my Beer & Jhonston's Mechanics of Materials

omogajuPost authoryou keep complaining about these lectures…did someone force you to watch them?

hoomanstavangerPost authorFirst of all, I would like to thank the one who upload these useful and informative clips and then I want to say a special thank you to the professor, it is great and I just started to understand what is the strength of material. To all critics: I cannot understand what is wrong with you guys!! Are u insane? the lectures are perfect!

pmig81Post authorthank you all your efforts, i really appreciate it

Paramvir JakharPost authorthanks so much for these informative lectures. I want to do m.tech from IIT …just hold for a year i am coming to iit …my dream

Ayesha NoorPost authorOutstanding lectures thank you soo much sir.!!

Ashok PareekPost authorThanks IIT

ku_7 sadPost author+nptelhrd worst audio when we gonna listen

Nadia MirPost authorhow to get the pdf of those lectures

Afreen KhanPost authorplease increase the volume of the video……it is too much slow

Schiner's ListPost authorNo need to waste lakhs of money in coaching . These lectures will always have extra edge over them .

Jeegar AvaiyaPost authorWhat we know? What we don't know?

Nandha KumarPost author@ 44:26 Shouldnt the contact area of the bolt be throughout the length of the plate ie. 20mm (thickness of plates together )