UNIT I ENERGY PRINCIPLES - Al-Ameen Engineering …€¦ · UNIT ± I ENERGY PRINCIPLES Strain energy and strain energy density- strain energy in traction, shear in flexure and torsion-

UNIT – I

ENERGY PRINCIPLES

Strain energy and strain energy density- strain energy in traction, shear in

flexure and torsion- Castigliano’s theorem – Principle of virtual work – application of

energy theorems for computing deflections in beams and trusses – Maxwell’s reciprocal theorem.

Two Marks Questions and Answers

1. Define strain energy and Proof stress.

Strain energy

Whenever a body is strained, the energy is absorbed in the body. The energy which is

absorbed in the body due to straining effect is known as strain energy. The strain energy stored in

the body is equal to the work done by the applied load in stretching the body

Proof stress The stress induced in an elastic body when it possesses maximum strain energy is termed

as its proof stress.

3. Define Resilience, Proof Resilience and Modulus of Resilience.

Resilience

The resilience is defined as the capacity of a strained body for doing work on the

removal of the straining force. The total strain energy stored in a body is commonly known as

resilience.

Proof Resilience The proof resilience is defined as the quantity of strain energy stored in a body when

strained up to elastic limit. The maximum strain energy stored in a body is known as proof

resilience.

Modulus of Resilience It is defined as the proof resilience of a material per unit volume.

Proof resilience

Modulus of resilience = -------------------

Volume of the body

4. State the two methods for analyzing the statically indeterminate structures.

a. Displacement method (equilibrium method (or) stiffness coefficient

method

b.Force method (compatibility method (or) flexibility coefficient method)

5. Define Castigliano’s first theorem second Theorem. First Theorem.

It states that the deflection caused by any external force is equal to the partial derivative of

the strain energy with respect to that force.

Second Theorem

It states that “If U is the total strain energy stored up in a frame work in equilibrium under an external force; its magnitude is always a minimum.

6. State the Principle of Virtual work.

It states that the workdone on a structure by external loads is equal to the internal energy

stored in a structure (Ue = Ui)

Work of external loads = work of internal loads

7. What is the strain energy stored in a rod of length l and axial rigidity AE to an axial force

P?

Strain energy stored

P2 L

U= --------

2AE

8. State the various methods for computing the joint deflection of a perfect frame.

1. The Unit Load method

2. Deflection by Castigliano’s First Theorem

3. Graphical method : Willot – Mohr Diagram

9. State the deflection of the joint due to linear deformation.

n

δv = Σ U x ∆

1

n

δH = Σ U’ x ∆

1

PL

∆ = --------- Ae

U= vertical deflection

U’= horizontal deflection

10. State the deflection of joint due to temperature variation.

n

δ = Σ U X A

1

= U1∆1 + U2 ∆2 + …………+ Un ∆n

If the change in length (∆) of certain member is zero, the product U.∆ for those members will be substituted as zero in the above equation.

11. State the deflection of a joint due to lack of fit.

n

δ = Σ U ∆ 1

= U1∆1 + U2 ∆2 + …………+ Un ∆n

If there is only one member having lack of fit ∆1, the deflection of a particular joint will

be equal to U1∆1.

12. What is the effect of change in temperature in a particular member of a redundant

frame?

When any member of the redundant frame is subjected to a change in temperature, it will

cause a change in length of that particular member, which in turn will cause lack of fit stresses in

all other members of the redundant frame.

13. State the difference between unit load and strain energy method in the determination of

structures.

In strain energy method, an imaginary load P is applied at the point where the deflection is

desired to be determined. P is equated to zero in the final step and the deflection is obtained.

In the Unit Load method, a unit load (instead of P) is applied at the point where the

deflection is desired.

14. State the assumptions made in the Unit Load method.

1. The external and internal forces are in equilibrium

2. Supports are rigid and no movement is possible

3. The material is strained well within the elastic limit.

15. State the comparison of Castigliano’s first theorem and unit load method.

The deflection by the unit load method is given by

n PUL

δ = Σ ------- 1 AE

n PL

δ = Σ ------- x U

1 AE

n

= Σ ∆ x U ----- (i) 1

The deflection by castigliano’s theorem is given by

n

W

P

AE

PL

1

--------- (ii)

By comparing (i) & (ii)

UW

P

16. State Maxwell’s Reciprocal Theorem.

The Maxwell’s Reciprocal theorem states as “ The work done by the first system of loads due to displacements caused by a second system of loads equals the work done by the

second system of loads due to displacements caused by the first system of loads.

17. Define degree of redundancy.

A frame is said to be statically indeterminate when the no of unknown reactions or stress

components exceed the total number of condition equations of equilibrium.

20. Define Perfect Frame.

If the number of unknowns is equal to the number of conditions equations available, the

frame is said to be a perfect frame.

21. State the two types of strain energies.

a. strain energy of distortion (shear strain energy)

b.strain energy of uniform compression (or) tension (volumetric strain energy)

22. State in which cases, Castigliano’s theorem can be used. 1. To determine the displacements of complicated structures.

2. To find the deflection of beams due to shearing (or) bending forces (or)

bending moments are unknown.

3. To find the deflections of curved beams springs etc.

23. Define Proof stress. The stress induced in an elastic body when it possesses maximum strain energy is termed as

its proof stress.

16 Marks Questions And Answers

1. Derive the expression for strain energy in Linear Elastic Systems for the following cases.

(i) Axial loading (ii) Flexural Loading (moment (or) couple)

(i)Axial Loading

Let us consider a straight bar of Length L, having uniform cross- sectional area A. If an

axial load P is applied gradually, and if the bar undergoes a deformation ∆, the work done, stored as strain energy (U) in the body, will be equal to average force (1/2 P)

multiplied by the deformation ∆.

Thus U = ½ P. ∆ But ∆ = PL / AE

U = ½ P. PL/AE = P2 L / 2AE ---------- (i)

If, however the bar has variable area of cross section, consider a small of length dx and

area of cross section Ax. The strain energy dU stored in this small element of length dx

will be, from equation (i)

P2 dx

dU = ---------

2Ax E

The total strain energy U can be obtained by integrating the above expression over the

length of the bar.

U = EA

dxP

x

L

2

2

0

(ii) Flexural Loading (Moment or couple )

Let us now consider a member of length L subjected to uniform bending moment M.

Consider an element of length dx and let di be the change in the slope of the element due to

applied moment M. If M is applied gradually, the strain energy stored in the small element

will be

dU = ½ Mdi

But

di d

------ = ----- (dy/dx) = d2y/d

2x = M/EI

dx dx

M

di = ------- dx

EI

Hence dU = ½ M (M/EI) dx

= (M2/2EI) dx

Integrating

U = L

EI

dxM

0

2

2

2. State and prove the expression for castigliano’s first theorem.

Castigliano’s first theorem: It states that the deflection caused by any external force is equal to the partial

derivative of the strain energy with respect to that force. A generalized statement of the

theorem is as follows:

“ If there is any elastic system in equilibrium under the action of a set of a forces W1 , W2, W3 ………….Wn and corresponding displacements δ1 , δ2, δ3…………. δn and a

set of moments M1 , M2, M3………Mn and corresponding rotations Φ1 , Φ2, Φ3,…….. Φn , then the partial derivative of the total strain energy U with respect to any one of the

forces or moments taken individually would yield its corresponding displacements in its

direction of actions.”

Expressed mathematically,

1

1

W

U ------------- (i)

1

1

M

U ------------- (ii)

Proof:

Consider an elastic body as show in fig subjected to loads W1, W2, W3

………etc. each applied independently. Let the body be supported at A, B etc. The

reactions RA ,RB etc do not work while the body deforms because the hinge reaction is

fixed and cannot move (and therefore the work done is zero) and the roller reaction is

perpendicular to the displacements of the roller. Assuming that the material follows the

Hooke’s law, the displacements of the points of loading will be linear functions of the loads and the principles of superposition will hold.

Let δ1, δ2, δ3……… etc be the deflections of points 1, 2, 3, etc in the direction of the

loads at these points. The total strain energy U is then given by

U = ½ (W1δ1 + W2 δ2 + ……….) --------- (iii)

Let the load W1 be increased by an amount dW1, after the loads have been applied.

Due to this, there will be small changes in the deformation of the body, and the strain

energy will be increased slightly by an amount dU. expressing this small increase as the

rate of change of U with respect to W1 times dW1, the new strain energy will be

U + 1

1

xdWW

U

--------- (iv)

On the assumption that the principle of superposition applies, the final strain energy

does not depend upon the order in which the forces are applied. Hence assuming that dW1

is acting on the body, prior to the application of W1, W2, W3 ………etc, the deflections will be infinitely small and the corresponding strain energy of the second order can be

neglected. Now when W1, W2, W3 ………etc, are applied (with dW1 still acting initially),

the points 1, 2, 3 etc will move through δ1, δ2, δ3……… etc. in the direction of these forces and the strain energy will be given as above. Due to the application of W1, rides

through a distance δ1 and produces the external work increment dU = dW1 . δ1. Hence the

strain energy, when the loads are applied is

U+dW1.δ1 ----------- (v)

Since the final strain energy is by equating (iv) & (v).

U+dW1.δ1= U + 1

1

xdWW

U

δ1=1W

U

Which proves the proportion. Similarly it can be proved that Φ1=1M

U

.

Deflection of beams by castigliano’s first theorem:

If a member carries an axial force the energies stored is given by

U = EA

dxP

x

L

2

2

0

In the above expression, P is the axial force in the member and is the function of external

load W1, W2,W3 etc. To compute the deflection δ1 in the direction of W1

δ1=1W

U

= dxW

p

AE

PL

10

If the strain energy is due to bending and not due to axial load

U = EI

dxML

2

2

0

δ1=1W

U

=EI

dx

W

MM

L

10

If no load is acting at the point where deflection is desired, fictitious load W is applied at

the point in the direction where the deflection is required. Then after differentiating but

before integrating the fictitious load is set to zero. This method is sometimes known as

the fictitious load method. If the rotation Φ1 is required in the direction of M1.

Φ1=1M

U

=EI

dx

M

MM

L

10

3. Calculate the central deflection and the slope at ends of a simply supported beam

carrying a UDL w/ unit length over the whole span.

Solution:

a) Central deflection:

Since no point load is acting at the center where the deflection is required, apply the

fictitious load W, then the reaction at A and B will (WL/2 + W/2)↑ each.

δc=W

U

=EI

dx

W

ML

0

Consider a section at a distance x from A.

Bending moment at x,

M=222

2wx

xWwL

2

x

x

M

dxxwx

xWwL

EI

l

c2222

2 22

0

Putting W=0,

dxxwx

xwL

EI

l

c222

2 22

0

= 2

0

43

1612

2

l

wxwLx

EI

EI

wlc

4

384

5

b) Slope at ends

To obtain the slope at the end A, say apply a frictions moment A as shown in fig. The

reactions at A and B will be

l

mwl

2 and

l

mwl

2

Measuring x from b, we get

A =

l

MxEIm

u

0

1 Dx

M

Mx.

-------------------------------- 2

Where Mx is the moment at a point distant x from the origin (ie, B) is a function of M.

Mx =

l

mwl

2 x -

2

2Wx

inl

x

m

Mx

2

A = l

EI0

1

l

mwl

2 x -

2

2Wx X/2 Dx

Putting M=0

dxl

xWXx

wl

Eia

l

2

2

2

1

0

L

AL

wxwx

EI0

43

86

1

EI

wLA

24

3

4. State and prove the Castigliano’s second Theorem.

Castigliano’s second theorem:

It states that the strain energy of a linearly elastic system that is initially

unstrained will have less strain energy stored in it when subjected to a total load system

than it would have if it were self-strained.

t

u

= 0

For example, if is small strain (or) displacement, within the elastic limit in the direction

of the redundant force T,

t

u

=

=0 when the redundant supports do not yield (or) when there is no initial lack of fit in the

redundant members.

Proof:

Consider a redundant frame as shown in fig.in which Fc is a redundant member of

geometrical length L.Let the actual length of the member Fc be (L- ), being the initial

lack of fit.F2 C represents thus the actual length (L- ) of the member. When it is fitted to

the truss, the member will have to be pulled such that F2 and F coincide.

According to Hooke’s law

F2 F1 = Deformation = )()(

approxAE

TL

AE

lT

Where T is the force (tensile) induced in the member.

Hence FF1=FF2-F1 F2

=AE

TL ------------------------------------ ( i )

Let the member Fc be removed and consider a tensile force T applied at the corners F and C

as shown in fig.

FF1 = relative deflection of F and C

= T

u

1

------------------------------------------ ( ii )

According to castigliano’s first theorem where U1 is the strain energy of the whole frame

except that of the member Fc.

Equating (i) and (ii) we get

T

u

1

= --AE

TL

(or)

T

u

1

+ AE

TL= ----------------------- ( iii )

To strain energy stored in the member Fc due to a force T is

UFC = ½ T. AE

TL =

AE

LT

2

2

T

U FC

AE

TL

Substitute the value of AE

TL in (iii) we get

T

U

T

u FC' (or)

T

U

When U= U1

+ U Fc.If there is no initial lack of fit, =0 and hence 0

T

U

Note:

i) Castigliano’s theorem of minimum strain energy is used for the for analysis of

statically indeterminate beam ands portal tranes,if the degree of redundancy is not more than

two.

ii) If the degree of redundancy is more than two, the slope deflection method or the

moment distribution method is more convenient.

5) A beam AB of span 3mis fixed at both the ends and carries a point load of 9 KN at C distant

1m from A. The M.O.I. of the portion AC of the beam is 2I and that of portion CB is I.

calculate the fixed end moments and reactions.

Solution:

There are four unknowns Ma, Ra, Mb and Rb.Only two equations of static are

available (ie) 0v and 0M

This problem is of second degree indeterminacy.

First choose MA and MB as redundant.

δA=

dx

R

M

EI

Mx

R

UA

x

A

AB 0 -----------(1)

θA= dxM

M

EI

M

M

U

A

xx

B

AA

AB

0 -------------(2)

1) For portion AC:

Taking A as the origin

Mx = -MA + RA x

1;

A

x

A

x

M

Mx

R

M

IIOM 2.. Limits of x: 0 to 1m

Hence

dxEI

dxR

M

EI

M

A

x

C

A

x

1

0

AA

2

x xR M-

232

1

3

1

2

1

2

132

AA

AA

MR

EI

RM

EI

And

dx

EIdx

R

M

EI

M

A

x

C

A

x

1

0

AA

2

1 xR M-

22

1

2

11

2

12

A

A

A

A

RM

EI

RM

EI

For portion CB, Taking A as the origin we have

xM = )1(9 XXRM AA

1;

A

x

A

x

M

Mx

R

M

M.O.I = I Limits of x : 1 to 3 m

Hence

dxEI

dxR

M

EI

M

A

x

B

C

x

3

1

AA x1)-9(x- xR M-

=

42

3

264

1AA RM

EI

And

dx

EIdx

M

M

EI

M

A

x

B

C

x

3

1

AA 1-1)-9(x- xR M-

= 18421

AA RMEI

Subs these values in (1) & (2) we get

0

A

AB

R

U

042

3

264

1

23

1

AA

AA RMEI

MR

EI

2.08 – MA = 9.88 __________ (3)

0

A

AB

M

U

01842

1

212

1

AA

AA RMEI

RM

EI

MA – 1.7RA = -7.2 -------------- (4)

Solving (3) & (4)

MA = 4.8 KN – M (assumed direction is correct)

RA = 7.05 KN

To find MB, take moments at B, and apply the condition 0M there. Taking

clockwise moment as positive and anticlockwise moment as negative. Taking MB clockwise,

we have

MB – MA =RA (3) – 9x2 = 0

MB – 4.8 + (7.05x 3) -18 = 0

MB = 1.65 KN – m (assumed direction is correct)

To find RB Apply 0V for the whole frame.

RB = 9 – RA = 9-7.05 = 1.95 KN

6.Using Castigliano’s First Theorem, determine the deflection and rotation of the

overhanging end A of the beam loaded as shown in Fig.

Sol:

Rotation of A:

RB x L = -M

RB = -M/L

RB = M/L ( )

& RC = M/L ( )

B

C

x

x

x

x

B

A

A dxM

MM

EIdx

M

MM

EIM

U..

1.

1 ____________ (1)

For any point distant x from A, between A and B (i.e.) x = 0 to x = L/3

Mx = M ; and 1

M

M x ________ (2)

For any point distant x from C, between C and B (i.e.) x = 0 to x = L

Mx = (M/L) x ; and L

x

M

M x

________ (3)

Subs (2) & (3) in (1)

LL

A dxL

xx

L

M

EIdxM

EIM

U

0

3/

0

1).1(

1

EI

ML

EI

ML

33

)(

3

2clockwise

EI

ML

b) Deflection of A:

To find the deflection at A, apply a fictitious load W at A, in upward direction

as shown in fig.

)

3

4( WLMxLRB

L

WLMRB

1)

3

4(

L

WLMRB

1)

3

4(

L

WLMRC

1)

3

1(

dxW

MM

EIW

MM

EIW

U x

x

B

C

x

B

A

xA .11

For the portion AB, x = 0 at A and x = L/3 at B

Mx = M + Wx

xW

M x

For the portion CB, x = 0 at C and x = L at B

xL

WLMM x .1

8

1

3

x

W

M x

dxx

L

xWLM

EIxWxM

EI

LL

A

0

3/

03

.3

111

Putting W = 0

dxL

Mx

EIdxMx

EI

LL

A

0

23/

03

11

LL

A

x

EI

Mx

EI

M0

33/

0

2

)3

(3

)2

(

EI

ML

EI

MLA

918

22

EI

MLA

6

2

7. Determine the vertical and horizontal displacements of the point C of the pin-jointed

frame shown in fig. The cross sectional area of AB is 100 sqmm and of AC and BC 150

mm2 each. E= 2 x 10

5 N/mm

2. (By unit load method)

Sol:

The vertical and horizontal deflections of the joint C are given by

AE

LPu

AE

PuL

H

V

'

A) Stresses due to External Loading:

AC = m543 22

Reaction:

RA = -3/4

RB = 3/4

Sin θ = 3/5 = 0.6; Cos θ = 4/5 = 0.8

Resolving vertically at the joint C, we get

6 = PAC cos θ + PBC sin θ

Resolving horizontally at the joint C, we get

PAC cos θ = PBC sin θ; PAC = PBC

PAC sin θ + PBC sin θ = 6

2 PAC sin θ = 6

PAC = 6/sin θ = 6/2 x 0.6 = 5 KN (tension)

PAC = PBC = 5 KN (tension)


PAB = PAC cos θ

PAB = 5 cos θ ; PAB = 5 x 0.8

PAB = 4 KN (comp)

B) Stresses due to unit vertical load at C:

Apply unit vertical load at C. The Stresses in each member will be 1/6 than of those

obtained due to external load.

3/26/4

6/5

AB

BCAC

u

uu

C) Stresses due to unit horizontal load at C:

Assume the horizontal load towards left as shown in fig.

Resolving vertically at the joint C, we get

sin'sin' CBCA uu

'' CBCA uu


)(8/5'

8/5'

)(8/58.02

1

cos2

1'

1cos'2

1cos'cos'

1cos'cos'

compKNu

KNu

tensionKNx

u

u

uu

uu

CA

CA

CB

CB

CBCB

CACB

Resolving horizontally at the joint B, we get

)(5.0'

5.08.08/5'

cos''

compKNu

KNxu

uu

AB

AB

BCAB

Member Length(L)

mm

Area

(mm)2

P(KN) U (kN) PUL/A U’(KN) PU’L/A

AB 8000 100 -4 -2/3 640/3 -1/2 160

BC 5000 150 5 5/6 2500/18 5/8 2500/24

CA 5000 150 5 5/6 2500/18 -5/8 2500/24

E = 2 X 105 n/mm

2= 200 KN/m

2

v mm

AE

Pul45.2

200

491

mmAE

lpuh 8.0

200

160'

8) The frame shown in fig. Consists of four panels each 25m wide, and the cross

sectional areas of the member are such that, when the frame carries equal loads at the

panel points of the lower chord, the stress in all the tension members is f n/mm2 and the

stress in all the comparison members of 0.8 f N/mm2.Determine the values of f if the

ratio of the maximum deflection to span is 1/900 Take E= 2.0 x 105 N/mm

2.

Sol:

The top chord members will be in compression and the bottom chord members,

verticals, and diagonals will be in tension. Due to symmetrical loading, the maximum

deflection occurs at C. Apply unit load at C to find u in all the members. All the members

have been numbered 1, 2, 3….. etc., by the rule u8 = u10 = u12 = 0.

Reaction RA = RB = 1/2

θ = 45º ; cos θ = sin θ = 2

1

)(2

2

cos

)(2

1

2

1.

2

2cos

)(2

2

sin

49

473

7

tensionu

u

tensionuuu

compR

u A

Also, 197 coscos uuu

)(0.1

2

1

2

2

2

1

2

21 compxxu

Member Length (L) mm P (N/mm2) U PUL

1 2500 -0.8 F -1.0 +2000F

3 2500 +F +1/2 +1250F

4 2500 +F +1/2 +1250F

7 2500 (2)0.5

-0.8F -(2)0.5

/2 +2000F

8 2500 +F 0 0

9 2500(2)0.5

+F +(2)0.5

/2 +2500F

Sum: +9000F

δC = 09.0102

290005

1

xE

PULn

F mm

9

10010000

900

1

900

1 xxspanC mm

Hence 0.09 F = 100/9 (or) F = 100/(9 x 0.09) = 123.5 N/mm2.

9. Determine the vertical deflection of the joint C of the frame shown in fig. due to

temperature rise of 60º F in the upper chords only. The coefficient of expansion = 6.0 x

10-6

per 1º F and E = 2 x 10 6 kg /cm

2.

Sol: Increase in length of each member of the upper chord = L α t = 400 x 6x 10

-6 x 60

= 0.144 cm

The vertical deflection of C is given by

u

To find u, apply unit vertical load at C. Since the change in length (∆) occurs only in the three top chord members, stresses in these members only need be found out.

Reaction at A = 4/12 = 1/3

Reaction at B = 8/12 = 2/3

Passing a section cutting members 1 and 4, and taking moments at D, we get

U1 = (1/3 x 4) 1/3 = 4/9 (comp)

Similarly, passing a section cutting members 3 and 9 and taking moments at C, we

get

Also

332211

12

3

)(9

4

)(9

8

3

14

3

2

uuu

compuu

compxu

C

cm

x

C

C

256.0

)144.0(9

8

9

4

9

4

10) Using the principle of least work, analyze the portal frame shown in Fig. Also plot

the B.M.D.

Sol:

The support is hinged. Since there are two equations at each supports. They are HA, VA,

HD, and VD. The available equilibrium equation is three. (i.e.) 0,0,0 VHM .

The structure is statically indeterminate to first degree. Let us treat the horizontal H ( )

at A as redundant. The horizontal reaction at D will evidently be = (3-H) ( ). By taking

moments at D, we get

(VA x 3) + H (3-2) + (3 x 1) (2 – 1.5) – (6 x 2) = 0

VA = 3.5 – H/3

VD = 6 – VA = 2.5 + H/3

By the theorem of minimum strain energy,

0

0

H

U

H

U

H

U

H

U

H

U

DCCEBEAB

(1)For member AB: Taking A as the origin.

dxH

MM

EIH

U

xH

M

xHx

M

AB

3

0

2

1

.2

.1

12.1091

83

1

2

1

3

0

43

23

0

HEI

xHx

EI

dxxHxx

EI

(2) For the member BE:

Taking B as the origin.

dxH

MM

EIH

U

x

H

M

HxxHM

xH

xxHM

BE

1

0

1

33

35.35.43

35.35.1133

dxxHx

xHEI

3

33

5.35.431

1

0

dxHx

xxHxHxxHEI

9

67.15.15.105.1391 2

2

1

0

dxHx

xHxxHEI

9

67.12125.1391 2

2

1

0

1

0

3322

27389.065.139

1

HxxHxxxHx

EI

27389.065.139

1 2 HHH

EI

9.791

HEI

(3) For the member CE:

Taking C as the origin

2

0

3

1

35.226

)3

5.2(2)3(

H

MM

EIH

U

HxxHM

xH

xHM

CE

=

2

03

23

5.2261 xHx

xHEI

dxHx

xHxxHxxHEI

9

833.067.6267.654121 2

2

2

0

dxHx

xxHxxHEI

9

833.0234.1334121 2

2

2

0

= EI

1(10.96H - 15.78)

(4) For the member DC:

Taking D as the origin

xx

M

HxxxHM

33

dxH

MM

EIH

U DC

2

0

1

dxxHxxEI

31

2

0

dxHxxEI

22

2

0

31

dxHxx

EI

2

0

33

33

31

dx

Hxx

EI

2

0

33

3

1

= EI

1(2.67H -8)

Subs the values

0H

U

1/EI (9-10.2) + (8.04H-7.9) + (10.96H-15.78) + (-8+2.67H) = 0

30.67H = 41.80

H = 1.36 KN

Hence

VA = 3.5 - H/3 = 3.5 - 1.36/3 = 3.05 KN

VD = 2.5 + H/3 = 2.5 + 1.36/3 = 2.95 KN

MA= MD =0

MB = (-1 x 32)/2 + (1.36 x 3) = -0.42 KN –m

MC = - (3-H) 2 = - (3-1.36)2 =-3.28KNm

Bending moment Diagram:

11) A simply supported beam of span 6m is subjected to a concentrated load of 45 KN

at 2m from the left support. Calculate the deflection under the load point. Take E = 200

x 106 KN/m

2 and I = 14 x 10

-6 m

4.

Solution:

Taking moments about B.

VA x 6 – 45 x 4=0

VA x 6 -180 = 0

VA = 30 KN

VB = Total Load – VA = 15 KN

Virtual work equation:

EI

mMdxL

c 0

V

Apply unit vertical load at c instead of 45 KN

RA x 6-1 x 4 =0

RA = 2/3 KN

RB = Total load –RA = 1/3 KN

Virtual Moment:

Consider section between AC

M1 = 2/3 X1 [limit 0 to 2]

Section between CB

M2 = 2/3 X2-1 (X2-2 ) [limit 2 to 6 ]

Real Moment:

The internal moment due to given loading

M1= 30 x X1

M2 = 30 x X2 -45 (X2 -2)

6

2

222111

2

0

VEI

dxMm

EI

dxMmc

2

0

6

2

22222

2

1

2

0

6

2

2

222

1

11

90453023

220

1

2453023

230

3

2

dxxxxxxEI

dxEI

xxxx

dxEI

xx

2

0

222

6

2

2

1 901523

201

dxxx

xEI

2

0

222

2

2

6

2

2

1 18030305201

dxxxxxEI

6

2

2

3

2

3

2

3

0

1 1802

60

3

5

3

201

x

xxx

EI

=

2161802630263

51

3

820 2233

EIEI

mmormxxxEI

EI

1.57)(0571.0101410200

160160

72096067.34633.531

66

The deflection under the load = 57.1 mm

12) Define and prove the Maxwell’s reciprocal theorem.

The Maxwell’s reciprocal theorem stated as “ The work done by the first system loads due to displacements caused by a second system of loads equals the work done by the second

system of loads due to displacements caused by the first system of loads”.

Maxwell’s theorem of reciprocal deflections has the following three versions:

1. The deflection at A due to unit force at B is equal to deflection at B due to unit

force at A.

δAB = δBA

2. The slope at A due to unit couple at B is equal to the slope at B due to unit couple

A

ΦAB = ΦBA

3. The slope at A due to unit load at B is equal to deflection at B due to unit couple.

'

' ABAB

Proof:

By unit load method,

EI

Mmdx

Where,

M= bending moment at any point x due to external load.

m= bending moment at any point x due to unit load applied at the point where

deflection is required.

Let mXA=bending moment at any point x due to unit load at A

Let mXB = bending moment at any point x due to unit load at B.

When unit load (external load) is applied at A,

M=mXA

To find deflection at B due to unit load at A, apply unit load at B.Then m= mXB

Hence,

dxEI

mm

EI

Mmdx XBXA

BA

. ____________ (i)

Similarly,

When unit load (external load) is applied at B, M=mXB

To find the deflection at A due to unit load at B, apply unit load at A.then m= mXA

dxEI

mmB

EI

Mmdx XA

AB

. ____________ (ii)

Comparing (i) & (ii) we get

δAB = δBA

13. Using Castigliano’s theorem, determine the deflection of the free end of the

cantilever beam shown in the fig. Take EI = 4.9 MN/m2. (NOV / DEC – 2003)

Solution:

Apply dummy load W at B. Since we have to determine the deflection of the free end.

Consider a section xx at a distance x from B. Then

2165.1*1*20130 xxxWxM x

dxW

M

EI

M

xxxxxxWxdxxx

xxxxWxxdxWxEI

2

1

3

2

1

0

(16*)5.1(1*20)1(30*)2

1)(1(20)1(30**

1

3

2

23

23233

2

1

2342331

0

3

31675.0

320

2230

3

23

2

410

2330

3

3

3

1

xx

xxxxWx

xxxxxWxxW

EI

Putting W =0

5

3

191675.3

3

1920

2

5

3

1930

2

3

3

14

4

1510

2

3

3

730

1

EI

3

41658.2*20

6

2330

2

710

6

530

1

EI

mmorm

x

x

64.44)(446.0

33.216.5111583.525109.4

1016

3

14. Fig shows a cantilever, 8m long, carrying a point loads 5 KN at the center and an

udl of 2 KN/m for a length 4m from the end B. If EI is the flexural rigidity of the

cantilever find the reaction at the prop. (NOV/DEC – 2004)

Solution:

To find Reaction at the prop, R (in KN)

Portion AC: ( origin at A )

EI

R

EI

R

EI

xR

EI

dxRxU

3

32

6

64

62

224

0

3224

0

1

Portion CB: ( origin at C )

Bending moment Mx = R (x+4) – 5x – 2x2/2

= R (x+4) – 5x –x2

EI

dxMU x

2

24

0

2

Total strain energy = U1 +U2

At the propped end 0

R

U

dxdR

dMx

EI

M

EI

R

R

U xx

4

03

64

= dxxxxxREIEI

R)4(54

1

3

64 22

4

0

dxxxxxxREIEI

R)4(454

1

3

64 224

0

dxxxxxxxREIEI

R)4()4(5168

1

3

64 2322

4

0

0

4

0

342

32

3

)3

4

4()2

3(5164

3

1

3

64

xxx

xxx

xR

EIEI

R

)

3

256

4

256()32

3

64(56464

3

64

3

64R

R

= 21.33 R + (149.33R – 266.67 – 149.33)

= 21.33 R + (149.33 R – 416)

21.33 R +149.33 R – 416 =0

R = 2.347 KN

15. A simply supported beam of span L is carrying a concentrated load W at the centre and a

uniformly distributed load of intensity of w per unit length. Show that Maxwell’s reciprocal theorem holds good at the centre of the beam.

Solution:

Let the load W is applied first and then the uniformly distributed load w.

Deflection due to load W at the centre of the beam is given by

EI

WlW

384

5 4

Hence work done by W due to w is given by:

EI

wlWxU BA

384

5 4

,

Deflection at a distance x from the left end due to W is given by

22 4348

xxlEI

WxW

Work done by w per unit length due to W,

dxxxlEI

WwxU

l

AB )43(48

2 22

2/

0

,

422

,222

3

24

lll

EI

WwU AB

168

3

24

44

,

ll

EI

WwU AB

EI

WwlU BA

4

,384

5

Hence proved.

CE1252- STRENGTH OF MATERIALS /UNIT-II/INDETERMINATE BEAMS

Page 1 of 34

DEPARTMENT OF CIVIL ENGINEERING/SEC/TIRUCHENGODE

CE1252- STRENGTH OF MATERIALS (FOR IV – SEMESTER)

UNIT - II

PREPARED BY

Mrs.N.SIVARANJANI.M.E. (Struct).,

LECTURER

DEPARTMENT OF CIVIL ENGINEERING

SENGUNTHAR ENGINEERING COLLEGE, TIRUCHENGODE – 637 205.


Page 2 of 34


UNIT – II

INDETERMINATE BEAMS

Propped Cantilever and fixed end moments and reactions for concentrated

load (central, non central), uniformly distributed load, triangular load (maximum at

centre and maximum at end) – Theorem of three moments – analysis of continuous

beams – shear force and bending moment diagrams for continuous beams

(qualitative study only)

S.NO 2 MARKS PAGE NO

1 Define statically indeterminate beams. 5

2 State the degree of indeterminacy in propped cantilever. 5

3 State the degree of indeterminacy in a fixed beam. 5

4 State the degree of indeterminacy in the given beam. 5

5 State the degree of indeterminacy in the given beam. 6

6 State the methods available for analyzing statically indeterminate

structures. 6

7 Write the expression fixed end moments and deflection for a fixed

beam carrying point load at centre. 6

8 Write the expression fixed end moments and deflection for a fixed

beam carrying eccentric point load. 6

9 Write the expression fixed end moments for a fixed due to sinking

of support. 7

10 State the Theorem of three moments. 7

11 Draw the shape of the BMD for a fixed beam having end moments

–M in one support and +M in the other. (NOV/DEC 2003) 7

12

What are the fixed end moments for a fixed beam of length ‘L’

subjected to a concentrated load ‘w’ at a distance ‘a’ from left end?

(Nov/Dec – 2004)

8

13 Explain the effect of settlement of supports in a continuous beam.

(Nov/Dec 2003) 8

14 What are the advantages of Continuous beams over Simply

Supported beams? 8

15

A fixed beam of length 5m carries a uniformly distributed load of 9

kN/m run over the entire span. If I = 4.5x10-4

m4 and E = 1x10

7

kN/m2, find the fixing moments at the ends and deflection at centre.

8


Page 3 of 34


16

A fixed beam AB, 6m long is carrying a point load of 40 kN at its

center. The M.O.I of the beam is 78 x 106 mm

4 and value of E for

beam material is 2.1x105 N/mm

2. Determine (i) Fixed end moments

at A and B.

9

17

A fixed beam AB of length 3m is having M.O.I I = 3 x 106 mm

4

and value of E for beam material is 2x105 N/mm

2. The support B

sinks down by 3mm. Determine (i) fixed end moments at A and B.

9

18

A fixed beam AB, 3m long is carrying a point load of 45 kN at a

distance of 2m from A. If the flexural rigidity (i.e) EI of the beam

is 1x104kNm

2. Determine (i) Deflection under the Load.

9

19

A fixed beam of 5m span carries a gradually varying load from

zero at end A to 10 kN/m at end B. Find the fixing moment and

reaction at the fixed ends.

10

20

A cantilever beam AB of span 6m is fixed at A and propped at B.

The beam carries a udl of 2kN/m over its whole length. Find the

reaction at propped end.

11


Page 4 of 34


S.NO 16 MARKS PAGENO

1

A fixed beam AB of length 6m carries point load of 160 kN and

120 kN at a distance of 2m and 4m from the left end A. Find

the fixed end moments and the reactions at the supports. Draw

B.M and S.F diagrams.

12

2

A fixed beam AB of length 6m carries two point loads of 30 kN

each at a distance of 2m from the both ends. Determine the

fixed end moments and draw the B.M diagram.

14

3

Find the fixing moments and support reactions of a fixed beam

AB of length 6m, carrying a uniformly distributed load of

4kN/m over the left half of the span.

15

4

A continuous beam ABC covers two consecutive span AB and

BC of lengths 4m and 6m, carrying uniformly distributed loads

of 6kN/m and 10kN/m respectively. If the ends A and C are

simply supported, find the support moments at A,B and C. draw

also B.M.D and S.F.D.

16

5

A continuous beam ABCD of length 15m rests on four supports

covering 3 equal spans and carries a uniformly distributed load

of 1.5 kN/m length .Calculate the moments and reactions at the

supports. Draw The S.F.D and B.M.D.

19

6

A continuous beam ABCD, simply supported at A, B, Cand D

is loaded as shown in fig. Find the moments over the beam and

draw B.M.D and S.F.D. (Nov / Dec 2003)

22

7

Using the theorem of three moments draw the shear force and

bending moment diagrams for the following continuous

beam.(April / May 2003)

24

8

A beam AB of 4m span is simply supported at the ends and is

loaded as shown in fig. Determine (i) Deflection at C (ii)

Maximum deflection (iii) Slope at the end A.

E= 200 x 106 kN/m

2 and I = 20 x 10

-6 m

4

27

9 A continuous beam is shown in fig. Draw the BMD indicating

salient points. (Nov/Dec 2004) 29

10 For the fixed beam shown in fig. draw BMD and SFD.

(Nov/ Dec 2004) 32


Page 5 of 34


Two Marks Questions and Answers

1. Define statically indeterminate beams.

If the numbers of reaction components are more than the conditions equations, the

structure is defined as statically indeterminate beams.

E = R – r

E = Degree of external redundancy

R = Total number of reaction components

r = Total number of condition equations available.

A continuous beam is a typical example of externally indeterminate structure.

2. State the degree of indeterminacy in propped cantilever.

For a general loading, the total reaction components (R) are equal to (3+2) =5,

While the total number of condition equations (r) are equal to 3. The beam is statically

indeterminate, externally to second degree. For vertical loading, the beam is statically

determinate to single degree.

E = R – r

= 5 – 3 = 2

3. State the degree of indeterminacy in a fixed beam.

For a general system of loading, a fixed beam is statically indeterminate to third

degree. For vertical loading, a fixed beam is statically indeterminate to second degree.

E = R – r

For general system of loading:

R = 3 + 3 and r = 3

E = 6-3 = 3

For vertical loading:

R = 2+2 and r = 2

E = 4 – 2 = 2

4. State the degree of indeterminacy in the given beam.

The beam is statically indeterminate to third degree of general system of loading.


Page 6 of 34


R = 3+1+1+1 = 6

E = R-r

= 6-3 = 3

5. State the degree of indeterminacy in the given beam.

The beam is statically determinate. The total numbers of condition equations are equal

to 3+2 = 5. Since, there is a link at B. The two additional condition equations are at link.

E = R-r

= 2+1+2-5

= 5-5

E = 0

6. State the methods available for analyzing statically indeterminate structures. i. Compatibility method

ii. Equilibrium method

7. Write the expression fixed end moments and deflection for a fixed beam carrying point

load at centre.

EI

WLy

WLMM BA

192

83

max

8. Write the expression fixed end moments and deflection for a fixed beam carrying

eccentric point load.

)(3 3

33

max

2

2

2

2

loadtheunderEIL

bWay

L

bWaM

L

WabM

B

A


Page 7 of 34


9. Write the expression fixed end moments for a fixed due to sinking of support.

2

6

L

EIMM BA

10. State the Theorem of three moments.

Theorem of three moments:

It states that “If BC and CD are only two consecutive span of a continuous beam subjected to an external loading, then the moments MB, MC and MD at the supports B,

C and D are given by

2

_

22

1

1

_

12211

66.)(2

L

xa

L

xaLMLLMLM DCB

Where

MB = Bending Moment at B due to external loading

MC = Bending Moment at C due to external loading

MD = Bending Moment at D due to external loading

L1 = length of span AB

L2 = length of span BC

a1 = area of B.M.D due to vertical loads on span BC

a2 = area of B.M.D due to vertical loads on span CD

1

_

x = Distance of C.G of the B.M.D due to vertical loads on BC from B

2

_

x = Distance of C.G of the B.M.D due to vertical loads on CD from D.

11. Draw the shape of the BMD for a fixed beam having end moments –M in one support

and +M in the other. (NOV/DEC 2003)


Page 8 of 34


12. What are the fixed end moments for a fixed beam of length ‘L’ subjected to a concentrated load ‘w’ at a distance ‘a’ from left end? (Nov/Dec – 2004)

Fixed End Moment:

2

2

2

2

L

WabM

L

WabM

B

A

13. Explain the effect of settlement of supports in a continuous beam. (Nov/Dec 2003)

Due to the settlement of supports in a continuous beam, the bending stresses will alters

appreciably. The maximum bending moment in case of continuous beam is less when

compare to the simply supported beam.

14. What are the advantages of Continuous beams over Simply Supported beams? (i)The maximum bending moment in case of a continuous beam is much less than in case

of a simply supported beam of same span carrying same loads.

(ii) In case of a continuous beam, the average B.M is lesser and hence lighter materials of

construction can be used it resist the bending moment.

15. A fixed beam of length 5m carries a uniformly distributed load of 9 kN/m run over the

entire span. If I = 4.5x10-4

m4 and E = 1x10

7 kN/m

2, find the fixing moments at the ends

and deflection at the centre.

Solution:

Given:

L = 5m

W = 9 kN/m2 , I = 4.5x10

-4 m

4 and E = 1x10

7 kN/m

2

(i) The fixed end moment for the beam carrying udl:

MA = MB = 12

2WL

= KNmx

75.1812

)5(9 2

(ii) The deflection at the centre due to udl:

mmxxxx

xy

EI

WLy

c

c

254.3105.4101384

)5(9

384

47

4

4

Deflection is in downward direction.


Page 9 of 34


16. A fixed beam AB, 6m long is carrying a point load of 40 kN at its center. The M.O.I of

the beam is 78 x 106 mm

4 and value of E for beam material is 2.1x10

5 N/mm

2.

Determine (i) Fixed end moments at A and B.

Solution:

Fixed end moments:

8

WLMM BA

kNmx

MM BA 5.378

650

17. A fixed beam AB of length 3m is having M.O.I I = 3 x 106 mm

4 and value of E for beam

material is 2x105 N/mm

2. The support B sinks down by 3mm. Determine (i) fixed end

moments at A and B.

Solution:

Given:

L = 3m = 3000mm

I = 3 x 106 mm

4

E = 2x105 N/mm

2

= 3mm

2

6

L

EIMM BA

=2

65

)3000(

31031026 xxxxx

=12x105 N mm = 12 kN m.

18. A fixed beam AB, 3m long is carrying a point load of 45 kN at a distance of 2m from A.

If the flexural rigidity (i.e) EI of the beam is 1x104kNm

2. Determine (i) Deflection under

the Load.

Solution:

Given:

L = 3m

W = 45 kN

EI = 1x104 kNm

2

Deflection under the load:

In fixed beam, deflection under the load due to eccentric load


Page 10 of 34


3

33

3EIL

bWayC

mmy

my

xxx

xxy

C

C

C

444.0

000444.0

)3(1013

)1()2(4524

33

The deflection is in downward direction.

19. A fixed beam of 5m span carries a gradually varying load from zero at end A to 10

kN/m at end B. Find the fixing moment and reaction at the fixed ends.

Solution:

Given:

L = 5m

W = 10 kN/m

(i) Fixing Moment:

2030

22WL

MandWL

M BA

MA = kNm33.830

250

30

)5(10 2

kNmM B 5.1220

250

20

)5(10 2

(ii) Reaction at support:

20

7

20

3 WLRand

WLR BA

kNR

kNR

B

A

5.1720

350

20

5*10*7

5.720

150

20

5*10*3


Page 11 of 34


20. A cantilever beam AB of span 6m is fixed at A and propped at B. The beam carries a

udl of 2kN/m over its whole length. Find the reaction at propped end.

Solution:

Given:

L=6m, w =2 kN/m

Downward deflection at B due to the udl neglecting prop reaction P,

EI

wlyB

8

4

Upward deflection at B due to the prop reaction P at B neglecting the udl,

EI

PlyB

3

3

Upward deflection = Downward deflection

EI

Pl

3

3

EI

wl

8

4

P = 3WL/8 = 3*2*6/8 =4.5 kN


Page 12 of 34


16 Marks Questions And Answers

1. A fixed beam AB of length 6m carries point load of 160 kN and 120 kN at a distance of

2m and 4m from the left end A. Find the fixed end moments and the reactions at the

supports. Draw B.M and S.F diagrams.

Solution:

Given: L = 6m

Load at C, WC = 160 kN

Load at D, WC = 120 kN

Distance AC = 2m

Distance AD =4m

First calculate the fixed end moments due to loads at C and D separately

and then add up the moments.

Fixed End Moments:

For the load at C, a=2m and b=4m

kNmxx

M

L

abWM

A

C

A

22.142)6(

)4(21602

2

1

2

2

1

kNmxx

M

L

baWM

B

C

B

11.71)6(

)4(21602

2

1

2

2

1

For the load at D, a = 4m and b = 2m

kNmxx

M

L

baWM

A

D

A

33.53)6(

)4(21202

2

2

2

2

2

kNmxx

M

L

baWM

B

D

B

66.106)6(

)4(21602

2

2

2

2

2

Total fixing moment at A,

MA = MA1 + MA2

= 142.22 + 53.33


Page 13 of 34


MA = 195.55 kNm

Total fixing moment at B,

MB =MB1 + MB2

= 71.11 + 106.66

= 177.77 kN m

B.M diagram due to vertical loads:

Consider the beam AB as simply supported. Let RA* and RB

* are the

reactions at A and B due to simply supported beam. Taking moments about A, we get

kNR

xxxR

B

B

33.1336

800

412021606

*

*

RA*

= Total load - RB*=(160 +120) – 133.33 = 146.67 kN

B.M at A = 0

B.M at C = RA* x 2 = 146.67 x 2 = 293.34 kN m

B.M at D = 133.33 x 2 = 266.66 kN m

B.M at B= 0

S.F Diagram:

Let RA = Resultant reaction at A due to fixed end moments and vertical

loads

RB = Resultant reaction at B

Equating the clockwise moments and anti-clockwise moments about A,

RB x 6 + MA = 160 x 2 + 120 x 4 + MB

RB= 130.37 kN

RA = total load – RB = 149.63 kN

S.F at A = RA = 149.63 kN

S.F at C = 149.63- 160 = -10.37 kN

S.F at D = -10.37 – 120 = -130.37 kN

S.F at B= 130.37 KN


Page 14 of 34


2. A fixed beam AB of length 6m carries two point loads of 30 kN each at a distance of

2m from the both ends. Determine the fixed end moments and draw the B.M diagram.

Sloution:

Given:

Length L = 6m

Point load at C = W1 = 30 kN

Point load at D = W2= 30 kN

Fixed end moments: MA = Fixing moment due to load at C + Fixing moment due to load at D

mkNxxxx

L

baW

L

baW

406

2430

6

42302

2

2

2

2

2

222

2

2

111

Since the beam is symmetrical, MA = MB = 40 kNm

B.M Diagram:

To draw the B.M diagram due to vertical loads, consider the beam AB as simply

supported. The reactions at A and B is equal to 30kN.

B.M at A and B = 0

B.M at C =30 x 2 = 60 kNm

B.M at D = 30 x 2 = 60 kNm


Page 15 of 34


3. Find the fixing moments and support reactions of a fixed beam AB of length 6m, carrying a

uniformly distributed load of 4kN/m over the left half of the span.

Solution:

Macaulay’s method can be used and directly the fixing moments and end reactions can be calculated. This method is used where the areas of B.M diagrams cannot be determined

conveniently.

For this method it is necessary that UDL should be extended up to B and then compensated for

upward UDL for length BC as shown in fig.

The bending at any section at a distance x from A is given by,

EI22

2x

wxMxRdx

ydAA +w*(x-3)

2

)3( x

=RAx – MA- (2

24x) +4(

2

)3 2x)

= RAx – MA- 2x2 +2(x-3)

2

Integrating, we get

EIdx

dy=RA

2

2x

-MAx - 23

3x

+C1 +3

)3(2 3x -------(1)

When x=0, dx

dy=0.

Substituting this value in the above equation up to dotted line,

C1 = 0

Therefore equation (1) becomes

EIdx

dy=RA

2

2x

-MAx - 23

3x

+3

)3(2 3x

Integrating we get

12

)3(2

12

2

26

4

2

423

xC

xxMxRyEI A

A

When x = 0 , y = 0

By substituting these boundary conditions upto the dotted line,

C2 = 0

6

)3(1

626

4423

xxxMxRyEI AA ________(ii)


Page 16 of 34


By subs x =6 & y = 0 in equation (ii)

6

)36(1

6

6

2

6

6

60

4423 AA MR

5.132161836 AA MR

18RA – 9 MA = 101.25 ------------- (iii)

At x =6, 0dx

dy in equation (i)

332

363

26

3

26

2

60 xxMxR AA

126618

018144618

AA

AA

MR

xMR

By solving (iii) & (iv)

MA = 8.25 kNm

By substituting MA in (iv)

126 = 18 RA – 6 (8.25)

RA = 9.75 kN

RB = Total load – RA

RB = 2.25 kN

By equating the clockwise moments and anticlockwise moments about B

MB + RA x 6 = MA + 4x3 (4.5)

MB = 3.75 kNm

Result:

MA = 8.25 kNm

MB = 3.75 kNm

RA = 9.75 kN

RB = 2.25 KN

4. A continuous beam ABC covers two consecutive span AB and BC of lengths 4m

and 6m, carrying uniformly distributed loads of 6kN/m and 10kN/m respectively. If

the ends A and C are simply supported, find the support moments at A,B and C.

draw also B.M.D and S.F.D.

Solution:

Given Data:

Length AB, L1=4m.

Length BC, L2=6m

UDL on AB, w1=6kN/m

UDL on BC, w2=10kN/m


Page 17 of 34


(i) Support Moments:

Since the ends A and C are simply supported, the support moments at A and C

will be zero.

By using cleyperon’s equation of three moments, to find the support moments at B (ie) MB.

MAL1 + 2MB(L1+L2) + MCL2 = 6

6

4

6 2211 xaxa

0 + 2MB(4+6) + 0 = 6

6

4

6 2211 xaxa

20MB = 2211

2

3xa

xa

The B.M.D on a simply supported beam is carrying UDL is a parabola having an

attitude of .8

2wL

Area of B.M.D = 3

2*L*h

= 3

2* Span *

8

2wL

The distance of C.G of this area from one end, = 2

span

. a1=Area of B.M.D due to UDL on AB,

= 3

2*4*

8

)4(6 2

=32

x1=2

1L

= 4/2

= 2 m.

a2= Area of B.M.D due to UDL on BC,

= 3

2*6*

8

)6(10 2

= 180m.

x2=L2 / 2

= 6 / 2

=3m

Substitute these values in equation(i).

We get,

20MB = )3*180(2

2*32*3

= 96+540

MB =31.8 kNm.


Page 18 of 34


(ii) B.M.D

The B.M.D due to vertical loads (UDL) on span AB and span BC.

Span AB:

=8

2

11Lw

=8

4*6 2

=12kNm

Span BC: =8

2

22 Lw

=8

6*10 2

=45kNm

(iii) S.F.D:

To calculate Reactions,

For span AB, taking moments about B, we get

(RA*4)-(6*4*2) – MB=0

4RA – 48 = 31.8 (MB=31.8, -ve sign is due to hogging moment.

RA=4.05kN

Similarly,

For span BC, taking moment about B,

(Rc*6)-(6*10*3) – MB=0

6RC – 180=-31.8

RC=24.7kN.

RB=Total load on ABC –(RA+RB)

=(6*4*(10*6))-(4.05+24.7)

=55.25kN.

RESULT:

MA=MC=0

MB=31.8kNm

RA=4.05kN

RB=55.25kN

RC=24.7kN


Page 19 of 34


5. A continuous beam ABCD of length 15m rests on four supports covering 3 equal spans

and carries a uniformly distributed load of 1.5 kN/m length .Calculate the moments and

reactions at the supports. Draw The S.F.D and B.M.D.

Solution:

Given:

Length AB = L1 = 5m

Length BC = L2 = 5m

Length CD = L3 = 5m

u.d.l w1 = w2 = w3 = 1.5 kN/m


Page 20 of 34


Since the ends A and D are simply supported, the support moments at A and D will be Zero.

MA=0 and MD=0

For symmetry MB=0

(i)To calculate support moments:

To find the support moments at B and C, by using claperon’s equations of three moments for ABC and BCD.

For ABC,

MAL1+[2MB(L1+L2)]+MCL2=2

22

1

11 66

L

xa

L

xa

0+[2MB(5+5)]+[MC(5)]= 5

6

5

6 2211 xaxa

20MB+5MC= )(5

62211 xaxa --------------------------------------(i)

a1=Area of BMD due to UDL on AB when AB is considered as simply supported

beam.

= **3

2AB Altitude of parabola (Altitude of parabola=

8

11Lw)

= 8

)5(*5.1*5*

3

2 2

=15.625

x1=L1/2

=5/2=2.5m

Due to symmetry

.a2=a1=15.625

x2=x1=2.5

subs these values in eqn(i)

20MB+5MC = )]5.2*625.15()5.2*625.15[(5

6

=93.75

Due to symmetry MB=MC

20MB+5MB=93.75

MB=3.75kNm.

MB=MC=3.75kNm.

(ii) To calculate BM due to vertical loads:

The BMD due to vertical loads(here UDL) on span AB, BC and CD (considering

each span as simply supported ) are shown by parabolas of altitude

kNmLw

6875.48

5.1*5.1

8

22

11 each.

(iii)To calculate support Reactions:

Let RA,RB,RC and RD are the support reactions at A,B,C and D.

Due to symmetry

RA=RD

RB=RC


Page 21 of 34


For span AB, Taking moments about B,

We get

MB=(RA*5)-(1.5*5*2.5)

-3.75=(RA*5)-18.75

RA=3.0kN.

Due to symmetry

RA=RD=3.0kN

RB=RC

RA+RB+RC+RD=Total load on ABCD

3+RB+RB+3=1.5*15

RB=8.25kN

RC=8.25kN.

Result:

MA = MD = 0

MB=MC=3.75kNm.

RA=RD=3.0kN

RB=8.25kN

RC=8.25kN.


Page 22 of 34


6. a continuous beam ABCD, simply supported at A,B, C and D is loaded as shown

in fig. Find the moments over the beam and draw B.M.D and S.F.D. (Nov/ Dec 2003)

Solution:

Given:

Length AB = L1 = 6m

Length BC = L2 = 5m

Length CD = L3 = 4m

Point load W1 = 9kN

Point load W2 = 8kN

u.d.l on CD, w = 3 kN/m

(i) B.M.D due to vertical loads taking each span as simply supported:

Consider beam AB, B.M at point load at E = kNmL

abW12

6

4*2*9

1

1

Similarly B.M at F = kNmL

abW6.9

6

3*2*82

2

B.M at the centre of a simply supported beam CD, carrying U.D.L

kNmwL

68

4*3

8

22

3

(ii) B.M.D due to support moments:

Since the beam is simply supported MA =MD = 0

By using Clapeyron’s Equation of Three Moments:

a) For spans AB and BC

MAL1 + 2MB(L1+L2) + MCL2 = 6

6

4

6 2211 xaxa

5

6

6

6)5()56(20 2211 xaxa

MM cB

22115

6522 xaxaMM CB ------------ (i)

a1x1 = ½*6*12*L+a/3 = ½*6*12*(6+2)/3 = 96

a2x2 = ½*5*9.6*L+b/3 = ½*5*9.6*(6+4)/3 = 64

Substitute the values in equation (i)

22MB + 5MC = 96+6/5*64

22MB + 5MC = 172.8 ------------ (ii)

b) For spans BC and CD

MBL2 + 2MC(L2+L3) + MDL3 = 3

33

2

22 66

L

xa

L

xa


Page 23 of 34


MB*5 + 2MC(5+4) +0 = 4

6

5

6 3322 xaxa

4

6

5

6185 332 xaax

MM CB ----------- (iii)

a2x2 = ½ * 5 * 9.6 *(L+a)/3 =1/2 * 5 * 9.6 *(5+2)/3 = 56

a3x3 = 2/3 * 4*6*4/2 =32

Substitute these values in equation (iii)

4

32*6

5

56*6185 CB MM

2.115185 CB MM

By solving equations (ii) &(iv)

MB = 6.84 kNm and MC = 4.48 kNm

(iii) Support Reactions:

For the span AB, Taking moment about B,

MB = RA * 6 – 9*4

= 366 AR

RA = KN86.46

84.636

For the span CD, taking moments about C

)48.4(2

4434 CDC MRM

RD = 4.88KN

For ABC taking moment about C

Mc = 3*85*45956* BA RR

11*86.424815 BR

RB = 9.41 kN

RC = Total load on ABCD – (RA +RB+RD)

RC = (9+8+4*3) – (4.86+9.41+4.88)

RC = 9.85 kN

Result:

MA = MD = 0

MB = 6.84 kNm and MC = 4.48 kNm

RA = 4.86kN

RB = 9.41kN

RC = 9.85 kN

RD = 4.88KN


Page 24 of 34


7. Using the theorem of three moments draw the shear force and bending moment

diagrams for the following continuous beam. (April / May 2003)

Solution:

Given:

Length AB, L1=4m.

Length BC, L2=3m.

Length CD, L3=4m.

UDL on AB, w=4 kN/m

Point load in BC, W1=4kN/m

Point load in CD, W1=6kN


Page 25 of 34


(i) Bending Moment to Vertical Loads:

Consider beam AB, B.M=8

4*4

8

22

wL

=8kNm.

Similarly for beam BC,

B.M=3

1*2*6

2

1 L

abW

=4kNm

Similarly for beam CD,

B.M=4

3*1*8

3

2 L

abW

=6kNm

(ii) Bending Moment to support moments:

Let MA,MB,MC And MD be the support moments at A,B,C and D. Since the

ends is simply supported, MA =MD=0.

By using Clayperon’s equation of three moments for span AB and BC,

MAL1+[2MB(L1+L2) ]+ MCL2 =2

22

1

11 66

L

xa

L

xa

0+[2MB(4+3)] MC(3) =3

6

4

6 2211 xaxa

14MB+ 3MC = 1.5a1x1 + 2a2x2 ----------------------------(i)

a1x1= Moment of area BMD due to UDL

= )*(*2

*3

2AltitudeBase

Base

= )8*4(*2

4*

3

2

=42.33

a2x2= Moment of area BMD due to point load about point B

= )4*2(*3

2*2*

2

1

=5.33

Using these values in eqn (i),

14MB + 3MC =1.5(42.33) +(2*5.33)

14MB + 3MC =63.495+10.66 -------------------------(ii)

For span BC and CD,

MBL1+[2MC(L2+L3) ]+ MDL3 =3

33

2

22 66

L

xa

L

xa


Page 26 of 34


MB(3)+[2MC(3+3) ]+ MDL3 =3

6

3

6 3322 xaxa

3MB+12MC = 2a2x2 + 2a3x3 ------------------------(iii)

a2x2= Moment of area BMD due to point load about point C

=(1/2)*2*4*3

1*2

=2.66

a3x3= Moment of area BMD due to point load about point D

= 3

3*2*6*1*

2

1

=6

Using these values in Eqn(iii),

3MB+ 12MC =2(2.66) + (2*6)

3MB + 12MC = 17.32 -------------------(iv)

Using eqn (ii) and (iii),

MB = 5.269 kN m

MC = 0.129 kN m

(iii) Support Reaction:

For span AB, taking moment about B

2*4*44* AB RM

-5.269 = RA *4 – 32

RA *4=26.731

RA = 6.68 kN

For span CD, taking moment about C

1*84* DC RM

-0.129 = RD *4-8

RD = 1.967 kN

Now taking moment about C for ABC

1*63*5*4*4)7( BAC RRM

63)20(47 BAC RRM

6380)68.6(7129.0 BR

RB = 13.037 kN

RC = Total load – (RA +RB + RC)

= 037.13967.168.6864*4

RC = 8.316 kN


Page 27 of 34


Result:

MA = MD = 0

MB = 5.269 kN m

MC = 0.129 kN m

RA = 6.68 kN

RB = 13.037 kN

RC = 8.316 kN

RD = 1.967 kN

8. A beam AB of 4m span is simply supported at the ends and is loaded as shown in

fig. Determine (i) Deflection at C (ii) Maximum deflection (iii) Slope at the end A.

E= 200 x 106 kN/m

2 and I = 20 x 10

-6 m

4

Solution:

Given:

L = 4m

E= 200 x 106 kN/m

2 and I = 20 x 10

-6 m

4


Page 28 of 34


To calculate Reaction:

Taking moment about A

)112

2(2*101*204* BR

RB *4 = 20 + 20(3)

RB = 80/4 = 20 kN

RA = Total load - RB

= (10*2+20) -20

RA = 20 kN

By using Macaulay’s method:

2

)2(10)1(2020

2

2

2

xxx

xd

ydEIM X

Integrating we get

3

)2(5)1(1010

32

1

2

xxCx

dx

dyEI

Integrating we get

12

)2(5

3

)1(10

3

10 43

21

3

xxCxC

xEIy ---------- (ii)

When x = 0, y = 0 in equation (ii) we get C2 = 0

When x = 4m, y = 0 in equation (ii)

43

1

3 2412

5)14(

3

104)4(

3

100 C

= 213.33 +4C1 – 90 -6.67

C1 = -29.16

Hence the slope and deflection equations are

Slope Equation:

3

)2(5)1(1016.2910

322

x

xxdx

dyEI

Deflection Equation:

12

)2(5

3

)1(1016.29

3

10 433

xxx

xEIy

(i) Deflection at C, yC :

Putting x = 2m in the deflection equation, we get

3

)12(10)2(16.29

3

)2(10 33 EIy

= 26.67 -58.32 -3.33

= -34.98

yc = 8.74 (downward)


Page 29 of 34


(ii) Maximum Deflection , ymax :

The maximum deflection will be very near to mid-point C.

Let us assume that it occurs in the sections between D and C. For maximum

deflection equating the slope at the section to zero, we get

22 )1(1016.2910 xxdx

dyEI

10x2 -29.16 -10(x-1)

2 = 0

10x2 -29.16 -10 (x

2 -2x+1) = 0

x = 39.16/20 =1.958 m

3

)1958.1(10)958.1(16.29

3

)958.1(10 33 EIy

ymax = -35/EI

ymax = 8.75 mm (downward)

(iii) Slope at the end A, θA:

Putting x = 0 in the slope equation,

16.29dx

dyEI

θA = dy/dx = -29.16/EI

θA = -0.00729 radians

θA = -0.417º

Result:

(i) Deflection at C = 8.74 mm

(ii) Maximum deflection = 8.75 mm

(iii) Slope at the end A, θA = -0.417º

9. A continuous beam is shown in fig. Draw the BMD indicating salient points.

(Nov/Dec 2004)

Solution:

Given:

Length L1 = 4m

Length L2 = 8m

Length L3 = 6m

Udl on BC w = 10 kN/m

Point load W1 = 40 kN

Point load W2 = 40 kN


Page 30 of 34


(i) B.M due to vertical loads:

Consider beam AB, B.M = kNmL

abW30

4

1*3*40

1

1

For beam BC,

B.M = kNmwL

808

)8(10

8

22

For beam CD,

B.M = kNmLW

604

6*40

4

32

(ii) B.M due to support moments:

Let MA, MB, MC, MD be the support moments at A, B, C, D.

Since the end A and D are simply supported MA = MD = 0

By using Clapeyron’s Equation of Three moments.

For Span AB and BC:

2

22

1

112211

66)(2

L

xa

L

xaLMLLMLM CBA

8

6

4

6)8()84(20 2211 xaxa

MM CB

2MB (12) +8 MC = -1.5a1x1 – 0.75 a2 x2

24 MB +8 MC = -1.5a1x1 – 0.75 a2 x2 ----------- (i)

a1x1 = Moment of area of B.M.D due to point load

= ½*4*30*2/3*3 = 120

a2x2 = Moment of area of B.M.D due to udl

= 2/3 (Base x Altitude) x Base/2

= 2/3 (8*80)*8/2 = 1706.67

Using these values in equation (i)

24 MB +8 MC = -1.5(120) – 0.75 (1706.67)

24 MB +8 MC = -1460.0025 ---------------- (ii)

For Span BC and CD:

3

33

2

223322

66)(2

L

xa

L

xaLMLLMLM DCB

6

6

8

60)68(2)8( 3322 xaxa

MM CB

8 MB + 28 MC = - 0.75 a2x2 - a3x3 -------------- (iii)

a2x2 = Moment of area of B.M.D due to udl

= 2/3 (Base x Altitude) x Base/2

= 2/3 (8*80)*8/2 = 1706.67

a3 x3 = Moment of area of B.M.D due to point load

= ½ * b*h*L/3

= ½ * 6*60*6/3

= 360


Page 31 of 34


Using these values in equation (iii)

8 MB + 28 MC = - 0.75 (1706.67) – 360

8 MB + 28 MC = - 1640.0025 ------------------ (iv)

From (ii) & (iv)

MC = 45.526 kNm

MB = 45.657 kNm

Result:

MA = MD = 0

MC = 45.526 kNm

MB = 45.657 kNm


Page 32 of 34


10. For the fixed beam shown in fig. draw BMD and SFD. (Nov / Dec 2004)

Solution:

(i) B.M.D due to vertical loads taking each span as simply supported:

Consider beam AB as simply supported. The B.M at the centre of AB

kNmwL

25.28

)3(*2

8

22

1

(ii) B.M.D due to support moments:

As beam is fixed at A and B, therefore introduce an imaginary

zero span AA1 and BB1 to the left of A and to the right of B. The support moments at A1 and B1

are zero.

Let M0 = Support moment at A1 and B1 and it is zero.

MA = Fixing moment at A

MB = Fixing moment at B

MC = Support moment at C

To find MA, MB and MC, Theorem of three moments is used.

(a) For the span A1A and AC,

1

11

0

00110

66)0(20*

L

xa

L

xaLMLMM CA

1

116)3()3(2

L

xaMM CA

6 MA + 3MC = - 2a1x1 ------------- (i)

a1x1 = moment of area of B.M.D due to udl on AB when it is considered as simply

supported beam about B

= 2/3 * Base * Altitude * L1/2

= 2/3 * 3 * 2.25 * 3/2

a1x1 = 6.75

subs this values in equation (i) we get

6 MA + 3 MC = -13.50 ------------ (ii)

(b) For the span AC and CB:

2

22

1

112211

66)(2

L

xa

L

xaLMLLMLM BCA


Page 33 of 34


3

6

3

6)3()33(2)3( 2211 xaxa

MMM BCA

3 MA + 12 MC + 3 MB = 2a1x1 + 2a2x2

a1x1 = moment of area of B.M.D due to udl on AB when it is considered as simply

supported beam about B

= 2/3 * Base * Altitude * L1/2

= 2/3 * 3 * 2.25 * 3/2

a1x1 = 6.75

a2x2 = 0

3 MA + 12 MC + 3 MB = 13.5 ----------- (ii)

( c ) For the span CB and BB1

0

00

2

220022

660*)(2

L

xa

L

xaMLLMLM BC

3

6)3(23 22xa

MM BC

3MC + 6MB = 2a2x2

a2x2 = 0

3MC + 6MB = 0

By solving (iii), (iv), (ii)

MC = 1.125 kNm

MA = 0.5625 kNm

MB = -0.5625 kNm

(iii) Support Reactions:

Let RA, RB , and RC are the support reactions at A, B and C.

For the span AC, taking moment about C, we get

RA x 3 – 2 x 3 x 1.5 + MA = MC

RA x 3 – 9 + 0.5625 = 1.125

RA = 3.1875 kN

For the span CB, taking moment about C, we get

RB x 3 + MC = MB

RB x 3 + 1.125 = 0.5625

RB = 0.1875 kN

RC = Total load – (RA + RB )

= 2*3*1.5 – (3.1875 + 0.1875)

RC = 5.625 kN


Page 34 of 34


Result:

MC = 1.125 kNm

MA = 0.5625 kNm

MB = -0.5625 kNm

RA = 3.1875 kN

RB = 0.1875 kN

RC = 5.625 kN

CE1252 - STRENGTH OF MATERIALS/ UNIT III/ COLUMNS

DEPARTMENT OF CIVIL ENGINEERING/ SEC/ TIRUCHENGODE

1


UNIT - III

PREPARED BY

Mrs.N.SIVARANJANI.M.E. (Struct),

LECTURER





2

UNIT – III

COLUMNS

Eccentrically loaded short columns - middle third role – core section –

Columns of unsymmetrical sections-(angle channel sections) - Euler’s theory of long columns – critical loads for prismatic columns with different end conditions;

Rankine –Gordon formula for eccentrically loaded columns – thick cylinder –

compound cylinder.


1 Define columns. 4

2 Define struts. 4

3 Mention the stresses which are responsible for column failure. 4

4 State the assumptions made in the Euler’s column theory. 4

5 What are the important end conditions of columns? 4

6 Write the expression for crippling load when the both ends of the

column are hinged. 4

7 Write the expression for buckling load (or) Crippling load when

both ends of the column are fixed? 5

8 Write the expression for crippling load when column with one

end fixed and other end hinged. 5

9 Write the expression for buckling load for the column with one

fixed and other end free. 5

10 Explain equivalent length (or) Effective length. 5

11 Write the Equivalent length (L) of the column in which both

ends hinged and write the crippling load. 5

12 Write the relation between Equivalent length and actual length

for all end conditions of column. 6

13 Define core (or) Kernel of a section.(April/May 2003) 6

14 Derive the expression for core of a rectangular section.(Nov/Dec

2003) 6

15 Derive the expression for core of a solid circular section of

diameter D. 6

16

A steel column is of length 8m and diameter 600 mm with both

ends hinged. Determine the crippling load by Euler’s formula.

Take 5101.2 E N/mm

2.

6

17 Define Slenderness ratio. 7

18 State the Limitations of Euler’s formula.(April /May 2005) 7

19 Write the Rankine’s formula for columns. 7

20 What is the middle third rule? (Nov/Dec 2003) 8



3


1 Derive the expression for crippling load for four end

conditions of columns.(April/May 2003) 9

2

A mild steel tube 4m long, 3cm internal diameter 4mm thick

with both ends hinged. Find the collapsing load, what will

be the crippling load if Both ends are built in? One end is

built –in and one end is free?

15

3

A column having a T section with a flange 120 mm x 16 mm

and web 150 mm x 16 mm is 3m long. Assuming the

column to be hinged at both ends, find the crippling load by

using Euler’s formula. E = 2 x 106 Kg/cm

2.

17

4

A steel bar of solid circular cross-section is 50 mm in

diameter. The bar is pinned at both ends and subjected to

axial compression. If the limit of proportionality of the

material is 210 MPa and E = 200 GPa, determine the

minimum length to which Euler’s formula is valid. Also determine the value of Euler’s buckling load if the column has this minimum length.

19

5

A rolled steel joist ISMB 300 is to be used a column of 3

metres length with both ends fixed. Find the safe axial load

on the column. Take factor of safety 3, fc = 320 N/mm2

and7500

1 . Properties of the column section. Area =

5626 mm2, IXX = 8.603 x 10

7 mm

4Iyy =4.539 x 10

7 mm

4

24

6

A column of circular section has 150 mm dia and 3m length.

Both ends of the column are fixed. The column carries a

load of 100 KN at an eccentricity of 15 mm from the

geometrical axis of the column. Find the maximum

compressive stress in the column section. Find also the

maximum permissible eccentricity to avoid tension in the

column section. E = 1 x 105 N/mm

2.

28

7 Derivation of Rankine’s formula for long and short columns. 30

8 Derivation of Rankine’s formula for Eccentric column. 32

9 Derivation of stresses in thick and compound cylinders. 35

10

A column with alone end hinged and the other end fixed has

a length of 5m and a hollow circular cross section of

outer diameter 100 mm and wall thickness 10 mm. If E =

1.60 x 105 N/mm

2 and crushing strength

2

0 /350 mmN , Find the load that the column may

carry with a factor of safety of 2.5 according to Euler

theory and Rankine – Gordon theory. If the column is

hinged on both ends, find the safe load according to the two

theories. (April/May 2003)

38



4

TWO MARKS QUESTIONS AND ANSWERS

1. Define columns

If the member of the structure is vertical and both of its ends are fixed rigidly

while subjected to axial compressive load, the member is known as column.

Example: A vertical pillar between the roof and floor.

2. Define struts.

If the member of the structure is not vertical and one (or) both of its ends is

Linged (or) pin jointed, the bar is known as strut.

Example: Connecting rods, piston rods etc,

3. Mention the stresses which are responsible for column failure.

i. Direct compressive stresses

ii. Buckling stresses

iii. Combined of direct compressive and buckling stresses.

4. State the assumptions made in the Euler’s column theory.

1. The column is initially perfectly straight and the load is applied axially.

2. The cross-section of the column is uniform throughout its length.

3. The column material is perfectly elastic, homogeneous and isotropic and

obeys Hooke’s law.

4. The self weight of column is negligible.

5. What are the important end conditions of columns?

1. Both the ends of the column are linged (or pinned)

2. One end is fixed and the other end is free.

3. Both the ends of the column are fixed.

4. One end is fixed and the other is pinned.

6. Write the expression for crippling load when the both ends of the column

are hinged.

2

2

l

EIP

P = Crippling load

E = Young’s Modulus

I = Moment of inertia

l = Length of column



5

7. Write the expression for buckling load (or) Crippling load when both ends

of the column are fixed?

2

24

L

EIP

P = Crippling load




8. Write the expression for crippling load when column with one end fixed

and other end linged.

2

22

l

EIP

P = Crippling load




9. Write the expression for buckling load for the column with one

fixed and other end free.

2

2

4l

EIP

P = Crippling load




10. Explain equivalent length (or) Effective length.

If l is actual length of a column, then its equivalent length (or) effective length

L may be obtained by multiplying it with some constant factor C, which depends on

the end fixation of the column (ie) L = C x l.

11. Write the Equivalent length (L) of the column in which both ends hinged

and write the crippling load.

Crippling Load 2

2

L

EIP

Equivalent length (L) = Actual length (l)

P = Crippling load



L= Length of column



6

12. Write the relation between Equivalent length and actual length for all end

conditions of column.

Both ends linged L = l Constant = 1

Both ends fixed 2

lL Constant =

2

1

One end fixed and other

end hinged 2

lL Constant =

2

1

One end fixed and other

end free lL 2 Constant = 2

13. Define core (or) Kernel of a section. (April/May 2003)

When a load acts in such a way on a region around the CG of the section So

that in that region stress everywhere is compressive and no tension is developed

anywhere, then that area is called the core (or) Kernal of a section. The kernel of the

section is the area within which the line of action of the eccentric load P must cut the

cross-section if the stress is not to become tensile.

14. Derive the expression for core of a rectangular section.(Nov/Dec 2003)

The limit of eccentricity of a rectangular section b x d on either side of XX axis

(or) YY axis is d/6 to avoid tension at the base core of the rectangular section.

Core of the rectangular section = Area of the shaded portion

632

12

db

18

bd

15. Derive the expression for core of a solid circular section of diameter D.

The limit of eccentricity on either side of both XX (or) YY axis = D/8 to avoid

tension of the base.

Core of the circular section = Area of the shaded portion

28/D

64

2D

16. A steel column is of length 8m and diameter 600 mm with both ends

hinged. Determine the crippling load by Euler’s formula. Take 5101.2 E N/mm

2.



7

49441036.6600

6464mmdI

Since the column is hinged at the both ends,

Equivalent length L = l

2

2

L

EIPcr

2952

8000

1036.6101.2

N81006.2

17. Define Slenderness ratio.

It is defined as the ratio of the effective length of the column (L) to the least

radius of gyration of its cross –section (K) (i.e) the ratio of K

L is known as slenderness

ratio.

Slenderness ratio = K

L

18. State the Limitations of Euler’s formula.(April /May 2005)

a. Euler’s formula is applicable when the slenderness ratio is greater than or

equal to 80

b. Euler’s formula is applicable only for long column

c. Euler’s formula is thus unsuitable when the slenderness ratio is less than a

certain value.

19. Write the Rankine’s formula for columns.

2

1

K

L

AfP c

K = Least radius of gyration A

I

P = Crippling load

A = Area of the column

fc = Constant value depends upon the material.

= Rankine’s constant E

fc

2



8

20. Write the Rankine’s formula for eccentric column.

2

211

k

L

k

ey

AfP

c

c

K = Least radius of gyration A

I

P = Crippling load

A = Area of the column

fc = Constant value depends upon the material.

= Rankine’s constant E

fc

2

21. Define thick cylinder.

If the ratio of thickness of the internal diameter of a cylindrical or spherical

shell exceeds 1/20, it is termed as a thick shell.

The hoop stress developed in a thick shell varies from a maximum value at the

inner circumference to a minimum value at the outer circumference.

Thickness > 1/20

22. State the assumptions involved in Lame’s Theory

i. The material of the shell is Homogeneous and isotropic.

ii. Plane section normal to the longitudinal axis of the cylinder remains

plane after the application of internal pressure.

iii. All the fibers of the material expand (or) contact independently without

being constrained by there adjacent fibers.

23. What is the middle third rule? (Nov/Dec 2003)

In rectangular sections, the eccentricity ‘e’ must be less than or equal to b/6. Hence the greatest eccentricity of the load is b/6 form the axis Y-Y and with respect to

axis X –X1 the eccentricity does not exceed d/6. Hence the load may be applied with in

the middle third of the base (or) Middle d/3.



9

16 MARKS QUESTIONS AND ANSWERS

1. Explain the failure of long column.

Solution:

A long column of uniform cross-sectional area A and of length l, subjected to

an axial compressive load P, as shown in fig. A column is known as long column if

the length of the column in comparison to its lateral dimensions is very large. Such

columns do not fail y crushing alone, but also by bending (also known buckling)

The load, at which the column just buckles, is known as buckling load and it is

less than the crushing load is less than the crushing load for a long column.

Buckling load is also known as critical just (or) crippling load. The value of

buckling load for long columns are long columns is low whereas for short columns the

value of buckling load is high.

Let

l = length of the long column

p = Load (compressive) at which the column has jus

buckled.

A = Cross-sectional area of he column

e = Maximum bending of the column at the centre.

0 = Stress due to direct load A

P

b = Stress due to bending at the centre of the column

= Z

eP

Where

Z = Section modulus about the axis of bending.

The extreme stresses on the mid-section are given by

Maximum stress = 0 + b

Minimum stress = 0 - b

The column will fail when maximum stress (i.e) 0 + b is more the crushing

stress fc. In case of long column, the direct compressive stresses are negligible as

compared to buckling stresses. Hence very long columns are subjected to buckling

stresses.



10

2. State the assumptions made in the Euler’s column Theory. And explain

the sign conventions considered in columns. (April/May2003)

The following are the assumptions made in the Euler’s column theory:

1. The column is initially perfectly straight and the load is applied

axially

2. The cross-section of the column is uniform throughout its

length.

3. The column material is perfectly elastic, homogeneous and

isotropic and obeys Hooke’s law. 4. The length of the column is very large as compared to its lateral

dimensions

5. The direct stress is very small as compared to the bending stress

6. The column will fail by buckling alone.

7. The self-weight of column is negligible.

The following are the sign conventions considered in columns:

1. A moment which will tend to bend the column with its convexity

towards its initial centre line is taken as positive.

2. A moment which will tend to bend the column with its concavity

towards its initial center line is taken as negative.

3. Derive the expression for crippling load when the both ends of the column

are hinged.

Solution:

Consider a column AB of length L hinged at both its ends A and B carries an

axial crippling load at A.

Consider any section X-X at a distance of x from B.

Let the deflection at X-X is y.



11

The bending moment at X-X due to the load P, M = yP.

ykEI

Py

dx

yd 2

2

2

Where EI

pk 2

` 02

2

2

ykdx

yd

Solution of this differential equation is

kxBkxAy sincos

EI

pxB

EI

pxAy sincos

By using Boundary conditions,

At B, x = 0, y = 0 A = 0

At A, x = l, y = 0

EI

plB sin0

0EI

pSinl

......3,2,,0 EI

pl

Now taking the lest significant value (i.e)

EI

pl ;

22

EI

pl

2

2

l

EIp

`The Euler’s crippling load for long column with both ends hinged.

2

2

l

EIp



12

4. Derive the expression for buckling load (or) crippling load when both ends of

the column are fixed.

Solution:

Consider a column AB of length l fixed at both the ends A and B and caries an

axial crippling load P at A due to which buckling occurs. Under the action of the load

P the column will deflect as shown in fig.

Consider any section X-X at a distance x from B.Let the deflection at X-X is y.

Due to fixity at the ends, let the moment at A or B is M.

Total moment at XX = M – P.y

Differential equation of the elastic curve is

PyMdx

ydEI

2

2

IE

M

EI

py

dx

yd

2

2

p

p

IE

M

EI

py

dx

yd

2

2

P

M

EI

P

EI

py

dx

yd

2

2

The general solution of the above differential equation is

P

MEIPxBEIPxAy /sin/cos (i)

Where A and B are the integration constant

At, N. x = 0 and y = 0

From (i)

p

MBA 010

p

MA

Differentiating the equation (i) with respect to x,

0./.

EI

PxCos

EI

PBEIPxSin

EI

PA

dx

dy



13

At the fixed end B, x = 0 and 0dx

dy

0EI

PB

Either B = 0 (or) 0EI

P

Since 0EI

P as p 0

B = 0

Subs p

MA and B = 0 in equation (i)

P

M

EI

Px

P

My

.cos

EI

Px

P

My ..cos1

Again at the fixed end A, x = l, y = 0

EIPlCosP

M/.10

........6,4,2,0/. EIPl

Now take the least significant value 2

2. EI

Pl

22 4.

EI

Pl

2

24

l

EIP

The crippling load for long column when both the ends of the column are fixed

2

24

L

EIP



14

5. Derive the expression for crippling load when column with one end fixed

and other end hinged. (April/May 2003)

Solution:

Consider a column AB of length l fixed at B and hinged at A. It carries an

axial crippling load P at A for which the column just buckles.

As here the column AB is fixed at B, there will be some fixed end moment at

B. Let it be M. To balance this fixing moment M, a horizontal push H will be exerted

at A.

Consider any section X-X at a distance x from the fixed end B. Let the

deflection at xx is y.

Bending moment at xx = H (l-x) - Py

Differential equation of the elastic curve is,

PyxlHdx

ydEI

2

2

EI

xly

EI

P

dx

yd

142

2

P

p

EI

xlHy

EI

P

dx

yd

2

2

EI

p

EI

xlHy

EI

P

dx

yd

2

2

The general solution of the above different equation is

P

xlH

EI

pxB

EI

pxAy

.sin.cos

Where A and B are the constants of integration. (i)

At B, x = 0, y = 0

From (i) P

HlA

P

H

EI

PB

p

EI

P

HB



15

Again at the end A, x = l, y=0. substitute these values of x, y, A and B in

equation (i)

EIPlSinP

EI

P

HEIPlCos

P

Hl/./.0

EIPlCosP

HlEIPlSin

p

EI

P

H/./..

lEIPlEIPl ././.tan

The value of lEIP ./tan in radians has to be such that its tangent is equal to

itself. The only angle whose tangent is equal to itself, is about 4.49 radians.

49.4./ lEIP

22 49.4lEI

P

22 2l

EI

P(approx)

2

22

l

EIP

The crippling load (or) buckling load for the column with one end fixed and one end

hinged.

6. Derive the expression for buckling load for the column with one end fixed

and other end free. (April/May 2003)

Solution:

Consider a column AB of length l, fixed at B and free at A, carrying an axial

rippling load P at D de to which it just buckles. The deflected form of the column AB

is shown in fig. Let the new position of A is A1.

Let a be the deflection at the free end. Consider any section X-X at a distance

x from B.

Let the deflection at xx is y.

2

22

l

EIP



16

Bending moment due to critical load P at xx,

yaPdx

ydEIM

2

2

pyPadx

ydEI

2

2

EI

pq

EI

py

dx

yd

2

2

The solution of the above differential equation is,

aEI

PxB

EI

PxAy

.sin.cos Where A and B are constants of

integration.

At B, x = 0, y = 0

From (i), A = 0

Differentiating the equation (I w.r. to x

EI

PxCos

EI

PB

EI

PxSin

EI

PA

dx

dy..

At the fixed end B, x = 0 and 0dx

dy

EI

PB0

0EI

PAs 0 p

Substitute A = -a and B = 0 in equation (i) we get,

aEI

Pxay

.cos

EI

Pxay ..cos1 (ii)

At the free end A, x = l, y = a, substitute these values in equation (ii)



17

EI

Paa ..1cos1

0..1cos

EI

P

2

5,

2

3,

21

EI

P

Now taking the least significant value,

2

1

EI

P

4

12

2

EI

P

2

2

4l

EIP

The crippling load for the columns with one end fixed and other end free.

7. A steel column is of length 8 m and diameter 600 mm with both ends hinged.

Determine the crippling load by Euler’s formula. Take E =2.1 x 105 N/mm

2

Solution:

Given,

Actual length of the column, l = 8m = 8000 mm

Diameter of the column d= 600 mm

E = 2.1 x 105 N/mm

2

464

dI

460064

491036.6 mmI

Since the column is hinged at the both ends,

2

2

4l

EIP



18

Equivalent length L =l

Euler’s crippling load,

2

2

L

EIPcr

2952

8000

1036.6101.22

= 2.06 x 108 N

8. A mild steel tube 4m long, 3cm internal diameter and 4mm thick is used as

a strut with both ends hinged. Find the collapsing load, what will be the

crippling load if

i. Both ends are built in?

ii. One end is built –in and one end is free?

Solution:

Given:

Actual length of the mild steel tube, l = 4m = 400 cm

Internal diameter of the tube, d = 3 cm

Thickness of the tube, t = 4mm = 0.4cm.

External diameter of the tube, D = d + 2t

= 3+2(0.4)

= 3.8 cm.

Assuming E for steel = 2 x 106 Kg/cm

2

M.O.I of the column section,

44

64dDI

2438.3

64

I = 6.26 cm 4

i. Since the both ends of the tube are hinged, the effective length of the column

when both ends are hinged.

L = l = 400 cm

Euler’s crippling load 2

2

L

EIPcr



19

2

62

400

26.6102

.30.772 KgPcr

The required collapsed load = 772.30 Kg.

ii. When both ends of the column are built –in ,

then effective length of the column,

cml

L 2002

400

2


2

2

L

EIPcr

262

200

26.6102

Pcr = 3089.19 Kg.

iii. When one end of the column is built in and the other end is free,

effective length of the column, L = 2l

= 2 x 400

= 800 cm


2

2

L

EIPcr

262

800

26.6102

Pcr = 193.07 Kg.

9. A column having a T section with a flange 120 mm x 16 mm and web 150

mm x 16 mm is 3m long. Assuming the column to be hinged at both ends,

find the crippling load by using Euler’s formula. E = 2 x 106 Kg/cm

2.

Solution:

Given:

Flange width = 120 mm = 12 cm

Flange thickness = 16 mm = 1.6 cm

Length of the web = 150 mm = 15cm



20

Width of the web = 16mm = 1.6cm

E = 2 106 Kg/cm

2

Length of the column, l = 3m = 300 cm.

Since the column is hinged at both ends, effective length of the column.

L = l = 300 cm.

From the fig. Y-Y is the axis of symmetry. The C.G of the whole section

lies on Y-Y axis.

Let the distance of the C.G from the 16 mm topmost fiber of the section = Y

6.1156.112

2

156.16.115

2

6.16.112

Y

cmY 41.5

Distance of C.G from bottom fibre = (15+1.6) - 5.41

= 11.19cm

Now M.O.I of the whole section about X-X axis.

2323

2

1519.11156.1

12

156.1

2

6.141.56.112

12

6.112XXI

492.1188 cmIXX

M.I of the whole section about Y-Y axis

433

52.23512

10615

12

126.1cmI yy

4

min 52.235 cmI

Euler’s Crippling load,

2

2

L

EIPcr

262

300

52.235102

; .32.51655 KgPcr



21

10. A steel bar of solid circular cross-section is 50 mm in diameter. The bar is

pinned at both ends and subjected to axial compression. If the limit of

proportionality of the material is 210 MPa and E = 200 GPa, determine the m

minimum length to which Euler’s formula is valid. Also determine the value of

Euler’s buckling load if the column has this minimum length.

Solution:

Given,

Dia of solid circular cross-section, d = 50 mm

Stress at proportional limit, f = 210 Mpa

= 210 N/mm2

Young’s Modulus, E = 200 GPa = 200 x 10 3 N/mm

2

Area of cross –section, 2249.196350

4mmA

Least moment of inertia of the column section,

4341079.6.350

64mmI

Least radius of gyration,

243

2 25.1565049.1963

1079.306mm

A

Ik

The bar is pinned at both ends,

Effective length, L = Actual length, l

Euler’s buckling load,

2

2

L

EIPcr

22

/ KL

E

A

Pcr

For Euler’s formula to be valid, value of its minimum effective length L may be found out by equating the buckling stress to f

210

2

2

K

L

E

210

222 kE

L

210

25.156102 522

L

L = 1211.89 mm = 1212 mm = 1.212 m



22

The required minimum actual length l =L = 1.212 m

For this value of minimum length,

Euler’s buckling load 2

2

L

EI

2352

1212

1075.306102

= 412254 N = 412.254 KN

Result:

Minimum actual length l = L = 1.212 m

Euler’s buckling Load =412.254 KN

11. Explain Rankine’s Formula and Derive the Rankine’s formula for both short and long column.

Solution:

Rankine’s Formula:

Euler’s formula gives correct results only for long columns, which fail mainly

due to buckling. Whereas Rankine’s devised an empirical formula base don practical

experiments for determining the crippling or critical load which is applicable to all

columns irrespective of whether they a short or long.

If P is the crippling load by Rankine’s formula.

Pc is the crushing load of the column material

PE is the crippling load by Euler’s formula. Then the Empirical formula devised by Rankine known as Rankine’s formula stand as:

Ee PPP

111

For a short column, if the effective length is small, the value of PE will be very

high and the value of EP

1 will be very small as compared to

CP

1and is negligible.

For the short column, (i.e) P = PC

cPP

11



23

Thus for the short column, value of crippling load by Rankine is more or less

equal to the value of crushing load:

For long column having higher effective length, the value of PE is small and

EP

1will be large enough in comparison to

CP

1. So

CP

1 is ignored.

For the long column, CP

1

EP

1 (i.e) p PE

Thus for the long column the value of crippling load by Rankine is more or less

equal to the value of crippling load by Euler.

Ec PPP

111

Ec

cE

PP

PP

P

1

cE

Ec

PP

PPp

;

E

c

c

P

P

Pp

1

Substitute the value of Pc = fc A and 2

2

L

EIPE

in the above equation,

22 /1

LEI

Af

Afp

c

c

Where,

fc = Ultimate crushing stress of the column material.

A = Cross-sectional are of the column

L = Effective length of the column

I = Ak2

Where k = Least radius of gyration.

22

2

22 1/

1EAk

LAf

Af

LEI

Af

Afp

c

c

c

c

2

1

K

L

Afp c



24

where = Rankine’s constant E

f c

2

P = 2

/1 kL

LoadCrushing

When Rankine’s constant is not given then find

E

f c

2

The following table shows the value of fc and for different materials.

Material fc N/mm2

E

f c

2

Wrought iron 250 9000

1

Cast iron 550 1600

1

Mild steel 320 7500

1

Timber 50 750

1

12. A rolled steel joist ISMB 300 is to be used a column of 3 meters length with

both ends fixed. Find the safe axial load on the column. Take factor of

safety 3, fc = 320 N/mm2 and

7500

1 . Properties of the column section.

Area = 5626 mm2, IXX = 8.603 x 10

7 mm

4

Iyy =4.539 x 10

7 mm

4

Solution:

Given: Length of the column, l = 3m = 3000 mm

Factor of safety = 3

fc = 320 N/mm2,

7500

1

Area, A = 5626 mm2

IXX = 8.603 x 107 mm

4

Iyy =4.539 x 10

7 mm

4

The column is fixed at both the ends,



25

Effective length, mml

L 15002

3000

2

Since Iyy is less then Ixx, The column section,

47

min 10539.4 mmIII yy

Least radius of gyration of the column section,

mmA

IK 82.89

5626

10539.4 7

Crippling load as given by Rakine’s formula,

22

82.89

1500

7500

11

5626320

1

K

L

Afp c

cr

Pcr = 1343522.38 N

Allowing factor of safety 3,

Safe load = safetyofFactor

LoadCrippling

N79.4478403

38.1343522

Result:

i. Crippling Load (Pcr) = 1343522.38 N

ii. Safe load =447840.79N

13. A built up column consisting of rolled steel beam ISWB 300 with two

plates 200 mm x 10 mm connected at the top and bottom flanges.

Calculate the safe load the column carry, if the length is 3m and both ends

are fixed. Take factor of safety 3 fc = 320 N/mm2 and

7500

1

Take properties of joist: A = 6133 mm2

IXX = 9821.6 x 104 mm

4 ; Iyy = 990.1 x 10

4 mm

4

Solution:



26

Given:

Length of the built up column, l = 3m = 3000 mm


fc =320 N/mm2

7500

1

Sectional area of the built up column,

2101331020026133 mmA

Moment of inertia of the built up column section abut xx axis,

2

34 15510200

12

102002106.9821XXI

= 1.94 x 108 mm

4

Moment of inertia of the built up column section abut YY axis,

12

200102101.990

34

YYI

= 0.23 x 108 mm

4

Since Iyy is less than Ixx , The column will tend to buckle about Y-Y axis.

Least moment of inertia of the column section,

48

min 1023.0 mmIII YY

The column is fixed at both ends.

Effective length,

mml

L 15002

3000

2

Least radius of gyration o the column section,

mmA

JK 64.47

10133

1023.0 8

Crippling load as given by Rankine’s formula,

22

64.47

1500

7500

11

10133320

1

K

L

Afp c

cr



27

= 2864023.3 N

Safe load = N43.9546743

3.2864023

Result:

i. Crippling load = 2864023.3 N

ii. Safe load = 954674.43 N

14. Derive Rankine’s and Euler formula for long columns under long columns under Eccentric Loading?

i. Rankine’s formula:

Consider a short column subjected to an eccentric load P with an eccentricity e

form the axis.

Maximum stress = Direct Stress + Bending stress

Z

M

A

Pf c

y

IZ

2

..

Ak

yep

A

P c 2

AkI

A

Ik

where

A = Sectional are of the column

Z = Sectional modulus of the column

yc = Distance of extreme fibre from N.A

k = Least radius of gyration.

21

k

ey

A

Pf c

c

Crippling load

Factor of safety

Eccentric load,

21

k

ey

AfP

c

c



28

Where

2

1k

eyc

is the reduction factor for eccentricity of loading.

For long column, loaded with axial loading, the crippling load,

2

1

K

L

AfP c

Where

2

1K

L is the reduction factor for buckling of long column.

Hence for a long column loaded with eccentric loading, the safe load,

ii. Euler’s formula

Maximum stress n the column = Direct stress + Bending stress

Z

lEIPeP

A

P 2/sec

Hence, the maximum stress induced in the column having both ends hinged

and an eccentricity of e is

2/sec

lEIP

Z

Pe

A

P

The maximum stress induced in the column with other end conditions are

determined by changing the length in terms of effective length.

15. A column of circular section has 150 mm dia and 3m length. Both ends of

the column are fixed. The column carries a load of 100 KN at an

eccentricity of 15 mm from the geometrical axis of the column. Find the

maximum compressive stress in the column section. Find also the

maximum permissible eccentricity to avoid tension in the column section.

E = 1 x 105 N/mm

2

Solution:

Given,

Diameter of the column, D = 150 mm

2

211

K

L

K

ey

AfP

c

c



29

Actual length of the column, l = 3m = 3000 mm

Load on the column, P = 100 KN = 1000 x 103 N

E = 1 x 105 N/mm

2

Eccentricity, e = 15 mm

Area of the column section 4

2D

A

2150

4

= 17671 mm2

Moment of inertia of the column section N.A.,

44 1506464

DI

= 24.85 x 106 mm

4

Section modulus,

2/D

I

y

IZ

= 3

6

331339

2

150

1085.24mm

Both the ends of the column 2 are fixed.

Effective length of the column, mml

L 15002

3000

2

Now, the angle

2

1500

1085.24101

10100

2/

65

3

LEIP

= 0.1504 rad = 8.61 o

Maximum compressive stress,

2

/secL

EIPZ

eP

A

P

331339

61.8sec1510100

17671

10100 33 o

= 10.22 N/mm2

To avoid tension we know,



30

Z

M

A

P

Z

ep

A

Po61.8.sec

331339

61.8.sec10100

17671

10100 33 oe

e = 18.50 mm

Result:

i. Maximum compressive stress = 10.22 N/mm2

ii. Maximum eccentricity = 18.50 mm

16. State the assumptions and derive Lame’s Theory?

1. The assumptions involved in Lame’s Theory.

i. The material of the shell is homogenous and isotropic

ii. Plane sections normal to the longitudinal axis of the cylinder remain

plane after the application of internal pressure.

iii. All the fibres of the material expand (or) contract independently

without being constrained by their adjacent fibres.

2 Derivation of Lame’s Theory

Consider a thick cylinder

Let

rc = Inner radius of the cylinder

r0 = Outer radius of the cylinder

Pi = Internal radial pressure

Po = External radial pressure

L = Length of the cylinder

f2 = Longitudinal stress.

Lame’s Equation:

axx pf 2

axx

bP

2

aax

bf x 2

2



31

ax

bf x

2

where

fx = hoop stress induced in the ring.

px = Internal radial pressure in the fig.

Px + dPx = External radial pressure in the ring.

The values of the two constants a and to b are found out using the following

boundary conditions:

i. Since the internal radial pressure is Pi,

At x = ri, Px = Pi

ii. Since the external radial pressure is P0

,

At x = r0, Px = P0

17. A thick steel cylinder having an internal diameter of 100 mm an external

diameter of 200 mm is subjected to an internal pressure of 55 M pa and an

external pressure of 7 Mpa. Find the maximum hoop stress.

Solution:

Given,

Inner radius of the cylinder, mmri 502

100

Outer radius of the cylinder, mmro 1002

200

Internal pressure, Pi = 55 Mpa

External pressure, P0 = 7 Mpa

In the hoop stress and radial stress in the cylinder at a distance of x from the

centre is fx and px respectively, using Lame’s equations,

ax

bf x

2 (i)

ax

bPx

2 (ii)

where a and b are constants,

Now by equation, at x = 50 mm, Px = 55 MPa (Boundary condition)

Using these boundary condition in equation (ii)

ax

bPx

2



32

a

b

250

55 (iii)

Then x = 100 mm, px = 7 Mpa

Using these boundary condition is equation (ii)

ab

2100

7 (iv)

Solving (iii) & (iv)

7100/2 ab

5550/2 ab

(- ) (+)

10000

3b = - 48

Substitute a & b in equation (i)

9160000

2

xf x

The value of fx is maximum when x is minimum

Thus fx is maximum for x = ri = 50 mm

Maximum hoop stress

950

1600002

= 73 Mpa (tensile)

Result:

Maximum hoop stress = 73 MPa (tensile)

18. A cast iron pipe has 200 mm internal diameter and 50 mm metal

thickness. It carries water under a pressure of 5 N/mm2. Find the maximum and

minimum intensities of circumferential stress. Also sketch the distribution of

circumferential stress and radial stress across the section.

Solution:

Given:

Internal diameter, di = 200 mm

b = 160000

a = 9



33

Wall thickness, t = 50 mm

Internal pressure, Pi = 5 N/mm2

External pressure, P0 = 0.

Internal radius mmdi

ri 1002

200

2

External radius mmtrr i 150501000

Let fx and Px be the circumferential stress and radial stress at a distance of x from the

centre of the pipe respectively.

Using Lame’s equations,

ax

bf x

2 (i)

ax

bpx

2 (ii)

where, a & b are arbitrary constants.

Now at x = 100 mm, Px = 5 N/mm2

At x = 150 mm, Px = 0

Using boundary condition is (ii)

ab

2

1005 (ii)

ab

2

1500 (iv)

By solving (iii) & (iv) a = 4 ; b = 90000

,490000

2

xf x ,4

900002

x

Px

Putting x = 100 mm, maxi circumferential stress.

tensilemmNf x

2

2/134

100

90000

Putting x = 150 mm, mini circumferential stress.

tensilemmNf x

2

2/84

150

90000



34

19. Explain the stresses in compound thick cylinders.

Solution: Consider a compound thick cylinder as shown in fig.

Let,

r1 = Inner radius of the compound cylinder

r2 = Radius at the junction of the two cylinders

r3 = Outer radius of the compound cylinder

When one cylinder is shrunk over the other, thinner cylinder is under

compression and the outer cylinder is under tension. Due to fluid pressure inside the

cylinder, hoop stress will develop. The resultant hoop stress in the compound stress is

that algebraic sum of the hoop stress due to initial shrinkage and that due to fluid

pressure.

a. Stresses due to initial shrinkage:

Applying Lame’s Equations for the outer cylinder,

12

1 ax

bPx

12

1 ax

bf x

At x = r3, Px = 0 and at x = r2, px = p

Applying Lame’s Equations for the inner cylinder

22

2 ax

bPx

22

2 ax

bf x

At x = r2, Px = p and at x = r3, px = 0

b. Stresses due to Internal fluid pressure.

To find the stress in the compound cylinder due to internal fluid pressure alone,

the inner and outer cylinders will be considered together as one thick shell. Now

applying Lame’s Equation,

Ax

BPx

2



35

Ax

Bf x

2

At x = r1, Px = pf ( Pf being the internal fluid pressure)

At x = r3, px = 0

The resultant hoop stress is the algebraic sum of the hoop stress due to

shrinking and due internal fluid pressure.

20. A compound cylinder is composed of a tube of 250 mm internal diameter

at 25 mm wall thickness. It is shrunk on to a tube of 200 mm internal

diameter. The radial pressure at the junction is 8 N/mm2.

Find the

variation of hoop stress across the wall of the compound cylinder, if it is

under an internal fluid pressure of 60 N/mm2

Solution:

Given: Internal diameter of the outer tube, d1 = 250 mm

Wall thickness of the outer tuber , t = 25 mm

Internal diameter of the inner tube , d2 = 200 mm

Radial pressure at the junction P = 8 N/mm2

Internal fluid pressure within the cylinder Pf = 60 N/mm2

External radius of the compound cylinder,

2

212

tdr

mm1502522502

1

Internal radius of the compound cylinder,

mmd

r 1002

200

2

21

Radius at the junction, mmd

r 1252

250

2

11

Let the radial stress and hoop stress at a distance of x from the centre of the

cylinder be px and fx respectively.

i. Hoop stresses due to shrinking of the outer and inner cylinders before fluid

pressure is admitted.

a. Four outer cylinder:

Applying Lame’s Equation

12

1 ax

bPx (i)



36

12

1 ax

bf x (ii)

Where a1 and b1 are arbitrary constants for the outer cylinder.

Now at x = 150 mm, Px = 0

X = 125 mm, Px = 8 N/mm2

12

1

150a

bo (iii)

12

1

1258 a

b (iv)

Solving equation (iii) & (iv) a1 = 18 ; b1 = 409091

18409091

2

xf x (v)

Putting x = 150 mm in the above equation stress at the outer surface,

2

2/3618

150

409091mmNf x (tensile)

Again putting x = 125 mm in equation (v), stress at junction,

2

2/4418

125

409091mmNf x (tensile)

b). For inner cylinder: Applying Lame’s Equation with usual Notations.

22

2 ax

bPx (iv)

22

2 ax

bf x (v)

Now at x = 125 mm, Px = 8 N/mm2

x =100 mm, Px = 0

22

2

1258 a

b (vi)

22

2

100a

bo (vii)

By solving (vi) & (vii) a2 = -22

b2 = -222222



37

2

2/2.4422

100

222222mmNf x

(comp)

2

2/2.3622

125

222222mmNf x

(comp)

iii. Hoop stresses due to internal fluid pressure alone for the compound cylinder:

In this case, the two tubes will be taken as a single thick cylinder. Applying

Lame’s equations with usual notations.

Ax

BPx

2 (viii)

Ax

Bf x

2 (ix)

At x = 150 mm, Px = 0

x = 100 mm, Px = pf = 60 N/mm2

From Equation (viii)

A

BO

2150

(x)

A

B

2100

60 (xi)

By solving (x) & (xi)

A = 133, B = 3 x 106

1331032

6

x

f x

Putting x = 150 mm, hoop stress at the outer surface

2

2

6

/266133150

103mmNf x

(Tensile)

Again putting x = 125 mm, hoop stress at the junction

TensilemmNf x

2

2

6

/325133125

103

Putting x = 100 mm, hoop stress at the inner surface



38

TensilemmNf x

2

2

6

/433133100

103

iii. Resultant hoop stress (shrinkage +Fluid pressure):

a. Outer cylinder

Resultant hoop stress at the outer surface = 36 + 266

= 302 N/ mm2 (Tensile)

Resultant hoop stress at the junction = 44 + 325 = 369 N/mm2 (tensile)

b. Inner cylinder;

Resultant hoop stress at the inner face = - 44.2 + 433

= 388.8 N/mm2 (Tensile)

Resultant hoop stress at the junction = - 36.2 + 325

= 288.8 N/mm2

(Tensile)

21. A column with alone end hinged and the other end fixed has a length of

5m and a hollow circular cross section of outer diameter 100 mm and wall

thickness 10 mm. If E = 1.60 x 105 N/mm

2 and crushing strength

2

0 /350 mmN , Find the load that the column may carry with a factor of

safety of 2.5 according to Euler theory and Rankine – Gordon theory. If the

column is hinged on both ends, find the safe load according to the two theories.

(April/May 2003)

Solution:

Given: L = 5 m = 5000 mm

Outer diameter D = 100 mm

Inner diameter d = D-2t = 100 – 2 (10) = 80 mm

Thickness = 10 mm

I = 1.60 x 105 N/mm

2

2

0 /350 mmN

f = 2.5

i. Calculation of load by Euler’s Theory:

Column with one end fixed and other end hinged.

2

22

L

EIP

mm

lL 60.3536

2

5000

2



39

2

52

06.3536

1060.114.32 IP

44

64dDI

44 8010064

4096000010000000064

I = 28.96 x 105 mm

4

14.12503716

1096.281060.114.32 552 P

p = 73.074 x 103 N

ii. Calculation of load by Rankine-Gordon Theory:

Rankine’s Constant 7500

1a (assume the column material is mild steel.)

2

1

K

La

Afp c

K = lest radius of Gyration

01.322826

1096.28 5

A

I

22 801004

A

6400100004

fc = c

= 2826 mm2

2

01.32

06.3536

7500

11

26.28350

P

036.122031033.1

9891004

P

NP41094.60



40

iii. Both ends are hinged

Euler’s theory

2

2

L

EIP

L = l

2

552

5000

1096.281060.114.3

P = 18.274 x 104 N ; Safe Load =

5.2

10274.18 4

= 73096 N

Rankine’s Theory

2

1

K

La

Afp c

2

01.32

5000

7500

11

2826350

81.243981033.1

9891004

Safe load

5.2

10480.30 4 = 121920 N

P = 30.480 x 104

Result: i. Euler’s Theory

One end fixed & one end hinged P = 73.074 x 103 N

Both ends hinged P = 18.274 x 104 N

ii. Rankine’s Theory

One end fixed & one end hinged P = 60.94 x 104 N

Both ends hinged P = 30.480 x 104 N

iii. Safe Load

Euler’s Theory = 73096 N

Rankine’s theory = 121920 N

22. A column is made up of two channel ISJC 200 mm and two 25 cm x 1 cm

flange plate as shown in fig. Determine by Rankine’s formula the safe load, the column of 6m length, with both ends fixed, can carry with a

factor of safety 4. The properties of one channel are A = 17.77 cm2, Ixx =

1,161.2 cm4 and Iyy = 84.2 cm

4. Distance of centroid from back of web =

1.97 cm. Take fc = 0.32 KN/mm2 and Rankine’s Constant

7500

1 (April

/May 2003)



41

Solution: Given:

Length of the column l = 6 m = 600 mm


Yield stress, fc = 0.32 KN/mm2

Rankine’s constant, 7500

1a

Area of column,

A = 2 (17.77+25 x 1)

A = 85.54 cm2

A = 8554 mm2

Moment of inertia of the column about X-X axis

2

3

5.1012512

1252.161,12XXI = 7839.0 cm

4

23

97.1577.1742.812

2512YYI = 4,499.0 cm

4

Iyy < IXX The column will tend to buckle in yy-direction

I = Iyy =4499.0 cm4

Column is fixed at both the ends

mml

L 30002

6000

2

mmA

IK 5.72

855

1044994

4

2

1

.

L

Ka

AfP

c

2

5.72

3000

75000

11

.855432.0

A = 2228 KN

Safe load of column SOF

P

..

4

2228 =557 KN

Result:

Safe load = 557 KN

CE1252 - STRENGTH OF MATERIALS/ UNITIV/ STATE OF STRESS

1



UNIT - IV

PREPARED BY


LECTURER




2


UNIT – IV

STATE OF STRESS IN THREE DIMENSIONS

Spherical and deviatory components of stress tensor- determination of

principal of principal stresses and principal planes – volumetric strain- dilation

and distortion – Theories of failure – principal stress dilatation. Principal strain

– shear stress - strain energy and distortion energy theories - application in

analysis of stress. Load carrying capacity and design of members – interaction

problems and interaction curves – residual stresses.


1 Define stress 4

2 Define principal planes. 4

3 Define spherical tensor. 4

4 Define Deviator stress tensor. 4

5 Define volumetric strain 4

6 State the principal theories of failure. 4

7 State the Limitations of Maximum principal stress theory. 5

8 Explain maximum principal stress theory. 5

9 Define maximum shear stress theory. 5

10 State the limitations of maximum shear stress theory. 5

11 Explain shear strain Energy theory. 5

12 State the limitations of Distortion energy theory. 5

13 Explain Maximum principal strain theory. 6

14 State the Limitations in maximum principal strain theory. 6

15 State the stress tensor in Cartesian components. 6

16 Explain the three stress invariants. 6

17 State the two types of strain energy 6

18 Explain Mohr’s Theory. 6

19 State the total strain energy theory. 7

20 State the shear strain energy per unit volume 7

7


3



1 Calculate the principal stress and principal planes for the

given normal stress and shear stress. 8


given normal stress and shear stress. 11


given normal stress and shear stress. (April / May 2003) 13

4 Explain the Energy of Distortion ( shear strain energy ) and

Dilatation. 16

5 Explain the Maximum Principal stress Theory:

( Rankine’s Theory) 17

6 Explain the Maximum shear stress (or) Stress Difference

theory (April / May 2003) 18

7 Explain the Shear strain Energy Theory (April / May

2003) 19

8 Explain the Maximum principal strain Theory? 20

9

In a material the principal stresses are 60 MN/m2, 48 MN/m

2

and - 36 MN/m2. Calculate (i)Total strain energy

(ii)Volumetric strain energy (iii)Shear strain energy

(iv)Factor of safety on the total strain energy criteria if the

material yields at 120 MN/m2. Take E = 200 GN/m

2 and

1/m = 0.3 (April / May 2005)

22

10 Explain Mohr’s Theory? 23

11

In a material the principal stresses are 70 MN/m2, 58MN/m

2

and -30 MN/m2. Calculate (i)Total strain energy




2 and

1/m = 0.3

24

12

In a material the principal stresses are 50 MN/m2, 40 MN/m

2

and - 30 MN/m2. Calculate (i)Total strain energy




2 and

1/m = 0.3

27


4


UNIT – IV


1. Define stress When a certain system of external forces act on a body then the body offers

resistance to these forces. This internal resistance offered by the body per unit area is

called the stress induced in the body.

2. Define principal planes. The plane in which the shear stress is zero is called principal planes. The plane

which is independent of shear stress is known as principal plane.

3. Define spherical tensor.

0

0

m

iiij

0

0

m

m0

0

It is also known as hydrostatic stress tensor

zyxm

3

1

m is the mean stress.

4. Define Deviator stress tensor

xz

xy

mx

ij

1

yz

my

xy

l

l

mz

yz

xz

5. Define volumetric strain It is defined as the ratio between change in volume and original volume of the

body and is denoted by e v

6. State the principal theories of failure.

1. Maximum principal stress theory

2. Maximum shear stress (or) stress difference theory

3. Strain energy theory

Change in volume v

e v =

Original volume v


5


4. Shear strain energy theory

5. Maximum principal strain theory

6. Mohr’s Theory

7. State the Limitations of Maximum principal stress theory

1. On a mild steel specimen when spiel tension test is carried out sliding occurs

approximately 45o to the axis of the specimen; this shows that the failure in

this case is due to maximum shear stress rather than the direct tensile stress.

2. It has been found that a material which is even though weak in simple

compression yet can sustain hydrostatic pressure for in excess of the elastic

limit in simple compression.

8. Explain maximum principal stress theory. According to this theory failure will occur when the maximum principle tensile

stress (1) in the complex system reaches the value of the maximum stress at the

elastic limit (et) in the simple tension.

9. Define maximum shear stress theory This theory implies that failure will occur when the maximum shear stress

maximum in the complex system reaches the value of the maximum shear stress in

simple tension at elastic limit (i.e)

22

31

max

etl

(or) et 31

10. State the limitations of maximum shear stress theory.

i. The theory does not give accurate results for the state of stress of pure shear

in which the maximum amount of shear is developed (i.e) Torsion test.

ii. The theory does not give us close results as found by experiments on

ductile materials. However, it gives safe results.

11. Explain shear strain Energy theory.

This theory is also called “ Distortion energy Theory” or “Von Mises - Henky

Theory.”

According to this theory the elastic failure occurs where the shear strain energy

per unit volume in the stressed material reaches a value equal to the shear strain energy

per unit volume at the elastic limit point in the simple tension test.

12. State the limitations of Distortion energy theory.

1. The theory does to agree the experiment results for the material for which

at is quite different etc.

2. This theory is regarded as one to which conform most of the ductile

material under the action of various types of loading.


6


13. Explain Maximum principal strain theory

The theory states that the failure of a material occurs when the principal tensile

strain in the material reaches the strain at the elastic limit in simple tension (or) when

the min minimum principal strain (ie ) maximum principal compressive strain reaches

the elastic limit in simple compression.

14. State the Limitations in maximum principal strain theory

i. The theory overestimates the behaviour of ductile materials.

ii. The theory does no fit well with the experimental results except for

brittle materials for biaxial tension.

15. State the stress tensor in Cartesian components

xz

xy

x

ij

.

'

yz

y

xy

z

yz

xz

16. Explain the three stress invariants.

The principal stresses are the roots of the cubic equation,

032

2

1

3 III

where

zyxI 1

xzyxyI zzxzyy222

2

xzyzxyxyzxzyxyxZyxI 22223

17. State the two types of strain energy

i. Strain energy of distortion (shear strain energy)

ii. Strain energy of dilatation.

18. Explain Mohr’s Theory

Let f

The enveloping curve f must represent in this abscissa and

ordinates e, the normal and shearing stresses in the plane of slip.

2

312

2

31

22


7


Let 312

1 P

312

1 m

222lmp

19. State the total strain energy theory.

The total strain energy of deformation is given by

133221

2

3

2

2

2

1 22

1 vE

U

and strain energy in simple tension is

E

U2

20

20. State the shear strain energy per unit volume

213

232

221

12

1 C

s

where

m

EC

112

21. Explain the concept of stress? When certain system of external forces act on a body then the body offers

resistance to these forces. This internal resistance offered by the body per unit area is

called the stress induced in the body.

The stress may be resolved into two components. The first one is the normal

stress n, which is the perpendicular to the section under examination and the second

one is the shear stress , which is operating in the plane of the section.


8


22. State the Theories of failure.

The principal theories are:

1. Maximum principal stress theory

2. Maximum shear stress (or) stress difference theory

3. Strain energy theory

4. Shear strain energy theory

5. Maximum principal strain theory

6. Mohr’s Theory

SIXTEEN MARKS QUESTIONS AND ANSWERS:

1. The stress components at a point are given by the following array.

6

5

10

10

8

5

6

10

6

Mpa

Calculate the principal stress and principal planes.

Solution:

The principal stresses are the roots of the cubic equation

0322

13 III (1)

where,

zyxI 1

222

2 zxzyyxxzzyyxI

xzyzxyxyzxzyyzxzyxI 22223

are three stress invariants

The stress tensor

zx

yx

x

ij

.

zy

y

xy

z

yz

xz

By comparing stress tensor and the given away,

zyxI 1

= 10 + 8 +6 =24


9


xzyzxyxzzyyxI222

2

= (10 x 8) + (8 x 6) + (6 x 10) - (5)2 – (10)

2 – (6)

2

=80 + 48 + 60 - 25 – 100 -36

=27


= 10 x 8 x 6 -10 (10)2 -8 (6 )

2 - 6 (5)

2 + 2(5) (10) (6)

=480 -1000-288-150+600

=-358

Substitute these values in (1) equation

03582724 23 (2)

We know that

From this

CosCosCos 3334

04

33

4

13 CosCosCos (3)

put,

3

1IrCos

3

24 rCos

8 rCos

Equation (2) becomes

82642419224512 222233 rCosCosrrCosCosrCosr

27 (r cos + 8) + 358 =0

r3 Cos3 + 512 - 24 r

2 Cos

2+

+ 192 r Cos - 24 r2 Cos

2 - 1536 -

384 r Cos + 27 r Cos + 216 + 358 =0

r3

Cos3 - 165 r Cos - 450 = 0

Divided by r3

0450165

32

3 r

Cosr

Cos (4)

Comparing equation (3) and (4) ,w e get,

CosCosCos 3343


10


4

31652

r

r = 14.8324

and

4

34503

Cos

r

38324.14

44503

Cos

Cos 3 = 0.551618

1 = 18.84o

2 = 1 + 120

2 = 138.84o

3 = 2 +120

3 = 258.84o

1 = r Cos 1 + 8

= 14.8324 Cos (18.84o) + 8

1 = 22.04 MPa

= 14.8324 Cos 138. 84o + 8

= - 3.17 MPa

3 = r cos 3 + 8

= 14.8324 Cos 258. 84o + 8

= 5.13 MPa

Result:

1 = 18.84 o 1 = 22.04 MPa

2 = 138.84 o 2 = -3.17 MPa

3 = 258.84 o 3 = 5.13 MPa


11


2. Obtain the principal stresses and the related direction cosines for the

following state of stress.(April / May 2003)

6

4

.3

5

2

4

MPa

1

5

6

Solution:

The principal stresses are the roots of the cubic equation.

0322

13 III (1)

zyxI 1

= 3 + 2 + 1 = 6

xzyzxzyyx yxI222

2

= (3 x 2 ) + (2 x 1) + (1 x 3) - (4)2 - (5)

2 - (6)

2

= 11 – 16 - 25 - 36

I2 = -66


=(3 x 2 x 1) - 3(5)2 - 2(6)

2 - 1 (4)

3 + 2 (4 x 6 x 5)

= 6 - 75 - 72 - 16 + 240

I3 = 83

Substitute these values in equation (1)

083666 23 (2)

We know that

CosCosCos 3334

CosCosCos4

33

4

13

CosCosCos4

33

4

13 (3)

Put 3

1IrCos

2 rCos


CosCosCos 3343


12


083666 23

083266262333 rCosrCosrCos

2232233 643238 CosrrCosCosrCosr

083132cos6624cos24 rr

0179662733 rCosrCosCosr

01793933 rCosCosr

Divided by r3

017939

32

3 r

Cosr

Cos (4)

By comparing (3) and (4)

4

1392

r

r2 = 156

r = 12.48

and 4

31793

Cos

r

716 = Cos 3 x (12.48 )3

765.1943

7163 Cos

Cos 3 = 0.3683573

3 = 68.38565

1 = 22.79o

2 = 1 + 120

2 = 142.79

3 = 2 +120

3 = 262.79

2cos 11 r

= 12.48 Cos (22.790) + 2

MPa506.131


13


222 rCos

= 12.48 Cos (142.79) + 2

MPa939.72

23 3 rCos

= 12.48 Cos (262.79) + 2

= 0.433680 MPa

Result:

1 = 22. 79o 1 = 13.506 MPa

2 = 142. 79o 2 = -7.939 MPa

3 = 262. 79o 3 = 0.433680 MPa

3. The state of stress at a point is given by

10

6

.20

8

10

6

MPa

7

8

10

Determine the principal stresses and principal direction.

Solution:

The cubic equation

032

2

1

3 III (1)

zyxI 1

= 20 + 10 + 7 = 37

222

2 zxyzxyxzzyyxI

=(20 x 10) + (10 x 7) + (7) x 20 + (36) + (64) + (100)

=200 + 70 + 140 + 26 + 64 + 100

I2=610

zxyzxyxyzxzyyzxzyxI 2222

3

=(20 x 10 x 7) - 20 (64) - 10 (100) - 7 (36) + 2 (6) (8) (10)

=1400 - 1280 - 1000 – 252 + 960

=1308

Substitute these values in equation (1)


14


0130861037 23 (2)

We know that

CosCosCos 334 3

CosCosCos4

33

4

13

CosCosCos4

33

4

13 (3)

Put 3

1IrCos

33.12 rCos


0130861037 23

0130833.1261033.123733.1223 rCosrCosrCos

0289.15266.2437087.45699.36516.1874 222233 rCosCosrrCosCosrCosr

160 r Cos + 1972.80 - 1308 = 0

0693.562512.937087.45699.36516.1874 222222233CosrCosrrCosCosrCosr

160 r Cos + 1972.80 - 1308 = 0

02693.496029533 Cosr 3

r

02693.496029532

3 r

Cosr

Cos (4)

By comparing (3) & (4)

2

295

4

1

r

CosCosCos 343 3


15


r2 = 1180

r = 34.35

and

3

2693.4960

4

3

r

Cos

331.40534

2693.4960

4

3

Cos

3 = 60.6930

1 = 20.231o

2 = 1 + 120

2 = 140 .23 o

3 = 26.231 o

33.1211 rCos

= 34.35 Cos (140.23o) + 12.33

MPa530.441

33.1222 rCos

= 34.35 Cos (140.231o) + 12.33

MPa217.142

33.1233 rCos

= 34.35 Cos (260.231o) + 12.33

5016.63

Result:

1 = 20.231o 3 = 260.231

o 2 = - 14.217 MPa

2 = 140.23o 1 = 44.530 MPa

3 = 6.5016 MPa


16


4. Explain the Energy of Distortion ( shear strain energy ) and Dilatation

The strain energy can be split up on the following two strain energies.

i. Strain energy of distortion (shear strain energy)

ii. Strain energy of Dilatation (Strain energy of uniform compression (or))

tension (or) volumetric strain energy )

Let e1 e2 an d e3 be the principal strain in the directions of principal stresses 1,

2 and 3.

Then

3211

1 E

e

1322

1 E

e

2133

1 E

e

Adding the above equation we get,

321321321 21 E

eee

21321

E

But e1 + e2 + e3 = e v (Volumetric strain)

321

21

Eev

If 0,0321 ve . This means that if sum of the three principal

stress is zero there is no volumetric change, but only the distortion occurs.

From the above discussion,

1. When the sum of three principal stresses is zero, there is no volumetric

change but only the distortion occurs.

2. When the three principal stresses are equal to one another there is no

distortion but only volumetric change occurs.

Note:


17


In the above six theories,

et , ec = Tensile stress at the elastic limit in simple tension and

compression;

1, 2, 3 = Principal stresses in any complex system

(such that e1 > e2 > e3 )

It may be assumed that the loading is gradual (or) static (and there is no cyclic

(or) impact load.)

5. Explain the Maximum Principal stress Theory: ( Rankine’s Theory) This is the simplest and the oldest theory of failure

According to this theory failure will occur when the maximum principle

tensile stress (1) in the complex system reaches the value of the

maximum stress at the elastic limit (et) in the simple tension (or) the

minimum principal stress (that is, the maximum principal compressive

stress), reaches the elastic limit stress () in simple compression.

(ie.) 1 = et (in simple tension)

ac 3 (In simple compression)

3 Means numerical value of 3

If the maximum principal stress is the design criterion, the maximum

principal stress must not exceed the working for the material. Hence,

1

This theory disregards the effect of other principal stresses and of the

shearing stresses on other plane through the element. For brittle materials

which do not fail by yielding but fail by brittle fracture, the maximum

principal stress theory is considered to be reasonably satisfactory.

This theory appears to be approximately correct for ordinary cast – irons and

brittle metals.

The maximum principal stress theory is contradicted in the following cases:

1. On a mild steel specimen when simple tension test is carried out sliding

occurs approximately 45o to the axis of the specimen; this shows that

the failure in the case is due to maximum shear stress rather than the

direct tensile stress.

2. It has been found that a material which is even though weak in simple

compression yet can sustain hydrostatic pressure for in excess of the

elastic limit in simple compression.


18


6. Explain the Maximum shear stress (or) Stress Difference theory (April /

May 2003)

This theory is also called Guesti’s (or) Tresca’s theory. This theory implies that failure will occur when the maximum shear

stress maximum in the complex system reaches the value of the

maximum shear stress in simple tension at the elastic limit i.e.

22

31

max

et

in simple tension.

(or) et 31

In actual design et in the above equation is replaced by the safe stress.

This theory gives good correlation with results of experiments on ductile

materials. In the case of two dimensional tensile stress and then the

maximum stress difference calculated to equate it to et.

Limitations of this theory:

i. The theory does not give accurate results for the state of stress of pure

shear in which the maximum amount of shear is developed (ie) Torsion

test.

ii. The theory is not applicable in the case where the state of stress consists

of triaxial tensile stresses of nearly equal magnitude reducing, the

shearing stress to a small magnitude, so that failure would be by brittle

facture rather than by yielding.

iii. The theory does not give as close results as found by experiments on

ductile materials. However, it gives safe results.

7. Explain the Shear strain Energy Theory (April / May 2003)

This theory is also called “Distortion Energy Theory”: (or) “Von Mises –

Henky Theory”

According to this theory the elastic failure occurs where the shear strain

energy per unit volume in the stressed material reaches a value equal to

the shear strain energy per unit volume at the elastic limit point in the

simple tension test.

Shear strain energy due to the principal stresses 1, 2, and 3 per unit

volume of the stress material.

2

13

2

32

2

2112

1 C

U S


19


But for the simple tension test at the elastic limit point where there is only one

principal stress (ie) et we have the shear strain energy per unit volume which is

given by

2221 000012

1tates

CU

Equating the two energies, we get

0

0

3

2

1

et

22

13

2

32

2

21 2 et

The above theory has been found to give best results for ductile material for which

ecet approximately.

Limitations of Distortion energy theory:

1. Te theory does to agree with the experimental results for the material for

which et is quite different from ec.

2. The theory gives 0et for hydrostatic pressure (or) tension, which

means that the material will never fail under any hydrostatic pressure (or)

tension. When three equal tensions are applied in three principal directions,

brittle facture occurs and as such maximum principal stress will give

reliable results in this case.

3. This theory is regarded as one to which conform most of the ductile

material under the action of various types of loading.

8. Explain the Maximum principal strain Theory?

This theory associated with St Venent

The theory states that the failure of a material occurs when the principal

tensile strain in the material reaches the strain at the elastic limit in

simple tension (or) when the minimum principal strain (ie) maximum

principal compressive strain reaches the elastic limit in simple

compression.

Principal strain in the direction of principal stress 1,

3211

11 mE

e

Principal strain in the direction of the principal stress 3,

2133

11 mE

e


20


The conditions to cause failure according to eh maximum principal strain

theory are:

E

eet

1 (e1 must be +Ve)

and

E

eec

3 (e3 must be -Ve)

EmE

et

321

11

EmE

et

213

11

etm

311

1

ecm

313

1

To prevent failure:

etm

321

1

cem

213

1

At the point of elastic failure:

etm

321

1

and cem

213

1

For design purposes,

tm

213

1

cm

213

1

(where, t and c are the safe stresses)

Limitations:

i. The theory overestimates the behavior of ductile materials.

ii. Te theory does not fit well with the experimental results except for brittle

materials for biaxial tension.


21


9. Explain the Strain energy theory?

The total stain energy of deformation is given by

133221

2

3

2

2

2

1 22

1 vE

U

and the strain energy under simple tension is

E

Ue

2

2

Hence for the material to yield,

133221

2

3

2

2

2

1 2 v

The total elastic energy stored in a material before it reaches the plastic state

can have no significance as a limiting condition, since under high hydrostatic pressure,

large amount of strain energy ma be stored without causing either fracture (or)

permanent deformation.

10. Explain Mohr’s Theory?

A material may fail either through plastic slip (or) by fracture when either the

shearing stress in the planes of slip has increased.

Let f

The enveloping curve f must represent in their abscissa and

ordinates , the normal and shearing stresses in the plane of slip. Now

2

312

2

31

22

Let 312

1 P

312

1 m

then

222mp

This equation represents the family of major principal stress circles in

parameter form. The equation of this envelope is obtained by partially differentiating

with respect to P

222mP

2222 2 mPp

dp

dp m

m

.

dp

d m

m

2

1.

This is to equation of Mohr’s envelope of the

major principal stress in parameter form.


22


11. In a steel member, at a point the major principal stress is 180 MN/m2 and

the minor principal stresses is compressive. If the tensile yield point of the

steel is 225 MN/m2, find the value of the minor principal stress at which

yielding will commence, according to each of the following criteria of

failure.

i. Maximum shearing stress

ii. Maximum total strain energy

iii. Maximum shear strain energy

Take Poisson’s ratio = 0.26

Solution:

Major principal stress, 2

1 /180 mMN

Yield point stress 2

2 /225 mMN

26.01

m

To calculate minor principal stress (2)

(i) Maximum shearing stress criterion

e 12

= 180 - 225

2 = - 45 MN/m2

2 = 45 MN/m2

(comp)

ii. Maximum total strain energy criterion:

2

133221

2

3

2

2

2

1

2e

m

3 = 0

(180)2 + 2

2 - 2 x 0.26 x 180 2 = (225)

2

32400 + 2

2 -93.6 2 = 50625

22 - 93.6 2 - 18225 = 0

e 21

2

21

2

2

2

1

2e

m


23


2

1822546.9336.92

2

2/08.962

76.28536.9mMN

(Only –Ve sign is taken as 2 is compressive)

2 = 96.08 MN/ m2 (compressive)

iii. Maximum shear strain energy criterion:

putting 3 = 0

221

22

221 2 e

22

121

2

2

2

1 22 e

(180)2 + (2)2+ - 180 2 = (225)

2

(2)2 - 180 2 - 18225 = 0

2

1822541801802

2

2

2 /25.722

5.324180mMN

2 = 72.25 MN/m2 (Compressive)

12. In a material the principal stresses are 60 MN/m2, 48 MN/m

2 and - 36

MN/m2. Calculate

i. Total strain energy

ii. Volumetric strain energy

iii. Shear strain energy

iv. Factor of safety on the total strain energy criteria if the

material yields at 120 MN/m2.

Take E = 200 GN/m2+ and 1/m = 0.3

2213

232

221 2 e


24


Solution: Given Data:

Principal stresses:

1 = + 60 MN/m2

2 = + 48 MN/m2

3 = - 36 MN/m2

Yield stress, e = 120 MN /m2

E = 200 GN/m2, 1/m = 0.3

i. Total strain energy per unit volume:

133221

2

3

2

2

2

1

2

2

1 mE

U

60364848603.02364860102002

10 222

9

12

U

2160172828806.01296230436005.2U

U = 19.51 KNm/m3

ii. Volumetric strain energy per unit volume:

3

9

12210

102002

3.02110364860

3

1

ve

e v = 1.728 KN/m3

iii. shear strain energy per unit volume

E

mev

2

/21

3

1 2

321

213

232

221

12

1 c

es


25


Where, 2/923.76

3.012

200

112

mGN

m

EC

222

9

12

60363648486010923.7612

101

se

31092167056144083.1 se

3/78.17 mKNmes

iv. Factor of safety (F.O.S)

Strain energy per unit volume under uniaxial loading is

33

9

262

/3610102002

10120

2mKNm

E

e

F.O.S 845.151.19

36

13. In a material the principal stresses are 50 N/mm2, 40 N/mm

2 and - 30

N/mm2, calculate:

i. Total strain energy

ii. Volumetric strain energy

iii. Shear strain energy and

iv. Factor of safety on the total strain energy criterion if the

material yield at 100 N/mm2.

Take E = 200 x 103 N/mm

2 and poission ratio = 0 .28

Solution:

Given,

Principal stresses:

2

1 /50 mmN

2

2 /40 mmN

2

3 /30 mmN

Yield stress, 2/100 mmNe


26



133221

2

3

2

2

2

1

2

2

1 mE

U

5030304040503.02304050102002

1 222

3

1500120020006.09001600250010400

13

7006.0500010400

13

542010400

13

U = 13.55 KNm/m3

ii)Volumetric strain energy per unit volume:

E

mev

2

/21

3

1 2221

3

2

102002

3.021304050

3

1

ve

3

2

10400

4.060

3

1

310

001.03600

3

1ve

ev = 1.2 K N m / m3


2

13

2

32

2

2112

1 C

es

where 23

3

/10923.763.012

10200

/112mmN

m

EC


27


222

3503030404050

10923.7612

1

se

6400490010010076.923

13

se

3/35.12 mKNnes



3

3

22

/254102002

100

2mKNm

E

e

845.155.13

25.. SOF

14. In a material the principal stresses are 50 N/mm

2, 40 N/mm

2 and - 30

N/mm2, calculate:

v. Total strain energy

vi. Volumetric strain energy

vii. Shear strain energy and

viii. Factor of safety on the total strain energy criterion if the

material yield at 100 N/mm2.

Take E = 200 x 103 N/mm

2 and poission ratio = 0 .28

Solution:

Given,

Principal stresses:

2

1 /50 mmN

2

2 /40 mmN

2

3 /30 mmN

Yield stress, 2/100 mmNe



28


133221

2

3

2

2

2

1

2

2

1 mE

U

5030304040503.02304050102002

1 222

3

1500120020006.09001600250010400

13

7006.0500010400

13

542010400

13

U = 13.55 KNm/m3

ii)Volumetric strain energy per unit volume:

E

mev

2

/21

3

1 2221

3

2

102002

3.021304050

3

1

ve

3

2

10400

4.060

3

1

310

001.03600

3

1ve

ev = 1.2 K N m / m3


2

13

2

32

2

2112

1 C

es

where 23

3

/10923.763.012

10200

/112mmN

m

EC

222

3503030404050

10923.7612

1

se


29


6400490010010076.923

13

se

3/35.12 mKNnes



3

3

22

/254102002

100

2mKNm

E

e

845.155.13

25.. SOF

CE1252 - STRENGTH OF MATERIALS/ UNITV/ ADVANCED TOPICS

1



UNIT - V

PREPARED BY


LECTURER




2


UNIT – V

ADVANCED TOPICS IN BENDING OF BEAMS

Unsymmetrical bending of beams of symmetrical and unsymmetrical

sections- curved beams- Winkler Bach formula- stress concentration- fatigue and

fracture.


1 Define Unsymmetrical bending. 4

2 State the two reasons for unsymmetrical bending. 4

3 Define shear centre. 4

4 Write the shear centre equation for channel section. 4

5 A channel Section has flanges 12 cm x 2 cm and web 16 cm x 1

cm. Determine the shear centre of the channel. 4

6 Write the shear centre equation for unsymmetrical I section. 4

7 State the assumptions made in Winkler’s Bach Theory. 5

8 State the parallel Axes and Principal Moment of inertia. 5

9 Define stress concentration. 5

10 Define stress – concentration factor. 5

11 Define fatigue stress concentration factor. 5

12 Define shear flow. 5

13 Explain the position of shear centre in various sections. 6

14 State the principles involved in locating the shear centre. 6

15 Determine the position of shear centre of the section of the beam

shown in fig. 6

16 State the stresses due to unsymmetrical bending. 6

17 Define the term Fatigue. 6

18 State the types of fatigue stress. 6

19 State the reasons for stress- concentration. 7

20 Define creep. 7


3



1 Explain the stresses induced due to unsymmetrical bending. 7

2 Derive the equation of Shear centre for channel section.

April/May 2005. 8

3 Derive the equation of Shear center for unequal I-section 10

4 Derive the stresses in curved bars using Winkler – Bach

Theory. 11

5

The curved member shown in fig. has a solid circular cross –section 0.01 m in diameter. If the maximum tensile and

compressive stresses in the member are not to exceed 150

MPa and 200 MPa. Determine the value of load P that can

safely be carried by the member.

14

6

Fig. shows a frame subjected to a load of 2.4 kN. Find (i)

The resultant stresses at a point 1 and 2;(ii) Position of

neutral axis. (April/May 2003)

16

7 Fig. shows a ring carrying a load of 30 kN. Calculate the

stresses at 1 and 2. 17

8

.A curved bar is formed of a tube of 120 mm outside

diameter and 7.5 mm thickness. The centre line of this is a

circular arc of radius 225 mm. The bending moment of 3

kNm tending to increase curvature of the bar is applied.

Calculate the maximum tensile and compressive stresses set

up in the bar.

18

9 A curved beam has a T-section (shown in fig.). The inner

radius is 300 mm. what is the eccentricity of the section? 19

10 Fig. shows a C- frame subjected to a load of 120 kN.

Determine the stresses at A and B. 20

11 Derive the formula for the deflection of beams due to

unsymmetrical bending. 21

12

A 80 mm x 80 mm x 10 mm angle section shown in fig. is

used as a simply supported beam over a span of 2.4 m. It

carries a load of 400 kN along the line YG, where G is the

centroid of the section. Calculate (i) Stresses at the points A,

B and C of the mid – section of the beam (ii) Deflection of

the beam at the mid-section and its direction with the load

line (iii) Position of the neutral axis. Take E = 200 GN/m2

23


4



1.Define Unsymmetrical bending

The plane of loading (or) that of bending does not lie in (or) a plane that

contains the principle centroidal axis of the cross- section; the bending is called

Unsymmetrical bending.

2. State the two reasons for unsymmetrical bending.

(i) The section is symmetrical (viz. Rectangular, circular, I section) but

the load line is inclined to both the principal axes.

(ii) The section is unsymmetrical (viz. Angle section (or) channel

section vertical web) and the load line is along any centroidal axes.

3. Define shear centre.

The shear centre (for any transverse section of the beam) is the point of

intersection of the bending axis and the plane of the transverse section. Shear

centre is also known as “centre of twist”

4. Write the shear centre equation for channel section.

f

w

A

A

be

6

3

e = Distance of the shear centre (SC ) from the web along the

symmetric axis XX

Aw = Area of the web

Af = Area of the flange

5. A channel Section has flanges 12 cm x 2 cm and web 16 cm x 1 cm.

Determine the shear centre of the channel.

Solution:

b= 12-0.5 = 11.5 cm

t1 = 2cm, t2 = 1cm, h= 18 cm

Af = bt1 = 11.5 x 2 = 23 cm2

Aw = ht2 = 18 x 1= 18 cm2

f

w

A

A

be

6

3

cme 086.5

23

186

)5.11(3

6. Write the shear centre equation for unsymmetrical I section.

xxI

bbhte

4

)( 212

21

e = Distance of the shear centre (SC) from the web along the symmetric

axis XX

t1 = thickness of the flange

h = height of the web


5


b1 = width of the flange in right portion.

b2 = width of the flange in left portion.

Ixx = M.O.I of the section about XX axis.

7. State the assumptions made in Winkler’s Bach Theory. (1) Plane sections (transverse) remain plane during bending.

(2) The material obeys Hooke’s law (limit state of proportionality is not exceeded)

(3) Radial strain is negligible.

(4) The fibres are free to expand (or) contract without any constraining

effect from the adjacent fibres.

8. State the parallel Axes and Principal Moment of inertia.

If the two axes about which the product of inertia is found, are such ,

that the product of inertia becomes zero, the two axes are then called the

principle axes. The moment of inertia about a principal axes is called the

principal moment of inertia.

9. Define stress concentration.

The term stress gradient is used to indicate the rate of increase of stress

as a stress raiser is approached. These localized stresses are called stress

concentration.

10. Define stress – concentration factor.

It is defined as the ratio of the maximum stress to the nominal stress.

nom

tK max

max = maximum stress

nom = nominal stress

11. Define fatigue stress concentration factor.

The fatigue stress – concentration factor (Kf ) is defined as the ratio of

flange limit of unnotched specimen to the fatigue limit of notched specimen

under axial (or) bending loads. )1(1 tf KqK

Value of q ranges from zero to one.

12. Define shear flow.

Shear flow is defined as the ratio of horizontal shear force H over

length of the beam x. Shear flow is acting along the longitudinal surface

located at discharge y1.Shear flow is defined by q.

z

zy

I

QV

x

Hq

H = horizontal shear force


6


13. Explain the position of shear centre in various sections.

(i) In case of a beam having two axes of symmetry, the shear centre

coincides with the centroid.

(ii) In case of sections having one axis of symmetry, the shear centre

does not coincide with the centroid but lies on the axis of symmetry.

14. State the principles involved in locating the shear centre.

The principle involved in locating the shear centre for a cross – section

of a beam is that the loads acting on the beam must lie in a plane which

contains the resultant shear force on each cross-section of the beam as

computed from the shearing stresses.

15. Determine the position of shear centre of the section of the beam shown

in fig.

Solution:

t1 = 4 cm, b1 = 6 cm, b2 = 8 cm

h1 = 30 – 4 = 26 cm

xxI

bbhte

4

)( 212

21

Ixx = 43

33

2085212

222)13(414

12

4142 cm

xx

x

cmx

e 9077.020852(4

)68(264 22

16. State the stresses due to unsymmetrical bending.

VVUU

bI

u

I

vM

sincos

σb = bending stress in the curved bar

M = moment due to the load applied

IUU = Principal moment of inertia in the principal axes UU

IVV = Principal moment of inertia in the principal axes VV

17. Define the term Fatigue.

Fatigue is defined as the failure of a material under varying loads, well

below the ultimate static load, after a finite number of cycles of loading and

unloading.

18. State the types of fatigue stress.

(i) Direct stress

(ii) Plane bending

(iii) Rotating bending

(iv) Torsion

(v) Combined stresses

(a) Fluctuating or alternating stress

(b) Reversed stress.


7


19. State the reasons for stress- concentration.

When a large stress gradient occurs in a small, localized area of a

structure, the high stress is referred to as a stress concentration. The reasons for

stress concentration are (i) discontinuities in continuum (ii) contact forces.

20. Define creep.

Creep can be defined as the slow and progressive deformation of a

material with time under a constant stress.

SIXTEEN MARKS QUESTIONS AND ANSWERS

1. Explain the stresses induced due to unsymmetrical bending.

Fig. shows the cross-section of a beam under the action of a bending

moment M acting in plane YY.

Also G = centroid of the section,

XX, YY = Co-ordinate axes passing through G,

UU, VV = Principal axes inclined at an angle θ to XX and YY axes respectively

The moment M in the plane YY can be resolved into its components in

the planes UU and VV as follows:

Moment in the plane UU, M’ = M sinθ

Moment in the plane VV, M’ = M cosθ

The components M’ and M” have their axes along VV and UU respectively. The resultant bending stress at the point (u,v) is given by,

UUVVUUVV

bI

M

I

M

I

vM

I

uM cossin"'

vvUU

bI

uSin

I

VCosM


8


At any point the nature of σb will depend upon the quadrant in which it lies.

The equation of the neutral axis (N.A) can be found by finding the locus of the

points on which the resultant stress is zero. Thus the points lying on neutral axis

satisfy the condition that σb = 0

0

vvUU I

uSin

I

VCosM

0vvUU I

uSin

I

VCos

uCos

Sin

I

Iv

vv

UU

(or) uI

Iv

vv

UU

tan

This is an equation of a straight line passing through the centroid G of the

section and inclined at an angle with UU where

tantan

vv

UU

I

I

Following points are worth noting:

i. The maximum stress will occur at a point which is at the greatest

distance form the neutral

ii. All the points of the section on one side of neutral axis will carry

stresses of the same nature and on the other side of its axis, of opposite

nature.

iii. In the case where there is direct stress in addition to the bending stress,

the neutral axis will still be a straight line but will not pass through G

(centroid of section.)

2. Derive the equation of Shear centre for channel section. April/May 2005

Fig shows a channel section (flanges: b x t1 ; Web h x t2) with XX as the

horizontal symmetric axis.


9


Let S = Applied shear force. (Vertical downward X)

(Then S is the shear force in the web in the upward direction)

S1 = Shear force in the top flange (there will be equal and opposite

shear force in the bottom flange as shown.)

Now, shear stress () in the flange at a distance of x from the right hand edge

(of the top flange)

tI

ySA

xa

2

.1

hxtyA (where t = t1 , thickness of flange)

xx

xh

xx I

Sh

tI

xSt

22.

.

.

1

1

Shear force is elementary area

dztddxtd AA 11 ..

Total shear force in top flange

dxt

b

..

0

1 (where b = breadth of the flange)

b

xx

b

xx

xdxI

shtdxt

I

hSS

0

11

0

12

.;2

(or) 4

.2

11

b

I

ShtS

xx

Let e = Distance of the shear centre (sc) from taking moments of shear forces

about the centre O of the web,We get

hSeS .. 1

xxxx I

bhtSh

b

I

Sht

4

..

4.

221

21

xxI

thbe

4

122

(1)

Now, 122.

122

32

2

1

31 hth

tbtb

Ixx

122

.

6

32

21

31 hthtbbt

122

32

21 hthbt

(neglecting the term 3

3

1bt, being negligible in comparison

to other terms)(or) 12

2

12bbtht

hI xx

Substitute the value of Ixx in equation (1) we get,


10


12

12

122

122

6

3

6

12

4 htht

tb

bthth

thbe

Let bt1 = Af (area of the flange)

ht2 = A (area of the web)

Then

f

wfw

f

A

A

b

AA

bAe

6

3

6

3

i.e

3. Derive the equation of Shear center for unequal I-section

Solution:

Fig. shows an unequal I – section which is symmetrical about XX axis.

Shear stress in any layer,

It

ySA

where I = IXX =

12122

3

121

31

21

hxtbb

tbb

Shear force S1 :

f

w

A

A

be

6

3


11


2

... 11

hxtyAdxtdA

S1 = 1

0

11

1

2

..b

XX

dxxth

tI

txSdA

= 1

0

12

..b

XX

dxth

I

xS =

XX

b

XX I

bShtx

I

Sht

422

211

0

21

1

Similarly the shear force (S2) in the other part of the flange,

S2 =XXI

bSht

4

221

Taking moments of the shear forces about the centre of the web O, we get

S2. h = S1. h + S .e (S3 = S for equilibrium)

(where, e = distance of shear centre from the centre of the web)

or, (S2 – S1) h = S.e

eSI

bbtSh

XX

.4

)( 21

221

2

4. Derive the stresses in curved bars using Winkler – Bach Theory.

The simple bending formula, however, is not applicable for deeply curved

beams where the neutral and centroidal axes do not coincide. To deal with such cases

Winkler – Bach Theory is used.

Fig shows a bar ABCD initially; in its unstrained state. Let AB’CD’ be the strained position of the bar.

xxI

bbhte

4

21

22

21


12


Let R = Radius of curvature of the centroidal axis HG.

Y = Distance of the fiber EF from the centroidal layer HG.

R’ = Radius of curvature of HG’ M = Uniform bending moment applied to the beam (assumed

positive when tending to increase the curvature)

= Original angle subtended by the centroidal axis HG at its

centre of curvature O and

’ = Angle subtended by HG’ (after bending) a t the center of

curvature ’ For finding the strain and stress normal to the section, consider the fibre EF at a

distance y from the centroidal axis.

Let σ be the stress in the strained layer EF’ under the bending moment M and e is strain in the same layer.

Strain,

)(

)(')''('

yR

yRyR

EF

EFEFe

or 1'

.''

yR

yRe

e0 = strain in the centroidal layer i.e. when y = 0

1'

.'

R

R or

'

.''

1yR

yRe

--------- (1)

and 1+e = '

.'

R

R --------- (2)

Dividing equation (1) and (2) , we get

01

1

e

e

'.

''

R

R

yR

yR

or

R

y

R

ye

R

y

R

ye

e

1

'

'

'

'. 00

According to assumption (3) , radial strain is zero i.e. y = y’

Strain,

R

y

R

ye

R

y

R

ye

e

1

''. 00

Adding and subtracting the term e0. y/R, we get

R

y

R

ye

R

ye

R

ye

R

y

R

ye

e

1

.''

. 0000

R

y

yRR

e

ee

1

)1

'

1)(1( 0

0 ------------- (3)

From the fig. the layers above the centroidal layer is in tension and the layers below

the centroidal layer is in compression.

Stress , σ = Ee = )

1

)1

'

1)(1(

(0

0

R

y

yRR

e

eE

___________ (4)

Total force on the section, F = dA.

Considering a small strip of elementary area dA, at a distance of y from the centroidal

layer HG, we have


13


dA

R

y

yRR

e

EdAeEF

1

)1

'

1)(1(

.0

0

dA

R

y

y

RReEdAeEF

1

)1

,

1(1. 00

dA

R

y

y

RReEAeEF

1

)1

,

1(1. 00 ____________ (5)

where A = cross section of the bar

The total resisting moment is given given by

dA

R

y

yRR

e

EydAeEdAyM

1

)1

'

1)(1(

...

20

0

dA

R

y

y

RReEeEM

1

)1

,

1(10.

2

00 (since )0ydA

M = E (1+e0)

dA

R

y

y

RR1

1

'

12

Let

22

1

AhdA

R

y

y

Where h2

= a constant for the cross section of the bar

M = E (1+e0)21

'

1Ah

RR

----------- (6)

Now,

dA

yR

yydA

yR

RydA

R

y

y2

..

1

= dA

yR

yydA .

2

dA

R

y

y

1

dA

R

y

y

R.

1

10

2

= 21Ah

R ---------- (7)

Hence equation (5) becomes

F = Ee0 .A – E (1+e0 )R

Ah

RR

21

'

1

Since transverse plane sections remain plane during bending

F = 0

0 = Ee0 .A – E (1+e0 )R

Ah

RR

21

'

1

E e0 .A = E (1+e0 )R

Ah

RR

21

'

1

e0 = (1+e0 )R

Ah

RR

21

'

1

(or)

2

0

h

Re(1+e0 )

RR

1

'

1

Substituting the value of 2

0

h

Re(1+e0 )

RR

1

'

1 in the equation (6)


14


M = E 2

2

0Ah

h

Re = e0 EAR

Or EAR

Me 0 substituting the value of e0 in equation (4)

2

0*

1

*h

Re

R

y

yE

AR

M

(or)

EAR

M

h

R

R

y

yE

AR

M**

1

*2

2

1*

1

*h

R

y

Ry

AR

M

AR

M

yR

y

h

R

AR

M2

2

1 (Tensile)

yR

y

h

R

AR

M2

2

1 (Compressive)

5. The curved member shown in fig. has a solid circular cross –section 0.01 m

in diameter. If the maximum tensile and compressive stresses in the

member are not to exceed 150 MPa and 200 MPa. Determine the value of

load P that can safely be carried by the member.

Solution:

Given,

d = 0.10 m; R = 0.10 m; G = 150 MPa = 150 MN / m2 (tensile )

2 = 200 MPa = 200 MN / m2 (Compressive)

Load P:

Refer to the fig . Area of cross section,

2322

10854.710.044

md

A


15


Bending moment, m = P (0.15 + 0.10) =0.25 P

2

422

10.0

10.0.

128

1

16

dh = 7.031 x 10

-4 m

2

Direct stress, compA

pd

Bending stress at point 1 due to M:

yR

y

h

R

AR

Mb 2

2

1 1 (tensile)

Total stress at point 1,

11 bd

yR

y

h

R

AR

M

A

P2

2

1150 (tensile)

05.010.0

05.0

10031.7

10.01

10.010854.7

25.0

10854.7150

4

2

33

PP

= -127.32 P + 318.31 P x 5. 74

= 1699.78 P

KNP 25.8878.1699

10150 3

(i)

Bending stress at point 2 due to M:

1

2

2

2yR

y

h

R

AR

Mb (comp)

Total stress at point 2,

22 bd

1200

2

2

yR

y

h

R

AR

M

A

P

1

05.010.0

05.0

10031.7

10.0

10.010854.7

25.0

10854.7 4

2

33

PP

=127.32 P + 318. 31 P x 13.22

= 4335.38 P

MNP38.4335

200

KNP 13.4638.4335

10200 3

(ii)

By comparing (i) & (ii) the safe load P will be lesser of two values

Safe load = 46.13 KN.


16


6. Fig. shows a frame subjected to a load of 2.4 kN. Find (i) The resultant

stresses at a point 1 and 2;(ii) Position of neutral axis. (April/May 2003)

Solution:

Area of section 1-2,

A = 48 * 18*10-6

= 8.64 * 10-4

m2

Bending moment,

M = -2.4*103*(120+48) * = -403.2 Nm

M is taken as –ve because it tends to decrease the curvature.

(i) Direct stress:

Direct stress σd = 26

4

3

/77.210*10*64.8

10*4.2mMN

A

P

23

2

2

2log R

DR

DR

D

Rh e

Here R = 48 mm = 0.048 m, D = 48 mm = 0.048 m

23

2 )048.0(048.0)048.0(2

048.0)048.0(2log

048.0

048.0

eh

= 0.0482 (loge3 – 1) = 2.27 * 10

-4 m

2

(ii) Bending stress due to M at point 2:

yR

y

h

R

AR

Mb 2

2

2 1 ;

26

4

2

4/10*

024.0048.0

024.0

10*27.2

048.01

048.0*10*64.8

2.403mMN

= -9.722 (1-10.149) = 88.95 MN/m2 (tensile)

(iii) Bending stress due to M at point 1:

yR

y

h

R

AR

Mb 2

2

1 1

26

4

2

4/10*

024.0_048.0

024.0

10*27.2

048.01

048.0*10*64.8

2.403mMN

= -42.61 MN/m2 = 42.61 MN/m

2 (comp)


17


(iv) Resultant stress:

Resultant stress at point 2,

σ2 = σd + σb2 = 2.77 + 88.95 = 91.72 MN/m2 (tensile)

Resultant stress at point 1,

σ1 = σd + σb1 = 2.77 -42.61 = 39.84 MN/m2 (comp)

(v) Position of the neutral axis:

42

4

22

2

10*27.2048.0

10*27.2*048.0y

hR

Rhy

= -0.00435 m = - 4.35 mm

Hence, neutral axis is at a radius of 4.35 mm

7. Fig. shows a ring carrying a load of 30 kN. Calculate the stresses at 1 and

2.

Solution:

Area of cross-section = 2222 01131.01.113124

mcmcmx

Bending moment M = 30*103 * (13.5*10-2

)Nm = 4050 Nm

h2

= ......*128

1

16 2

42

R

dd

Here d = 12 cm, R = 7.5 +6 = 13.5 cm

h2

=2

42

5.13

12*

128

1

16

12 = 9.89 cm

2 = 9.89*10

-4 m

2

Direct Stress σd = 263

/65.210*01131.0

10*30mMN

A

P

Bending stress at point 1 due to M,

yR

y

h

R

AR

Mb 2

2

1 1


18


6

4

2

1 10*06.0135.0

06.0

10*89.9

135.01

135.0*01131.0

4050

b

2.65*6.67 = 17.675 MN/m2 (tensile)

Bending stress at point 2 due to M,

yR

y

h

R

AR

Mb 2

2

2 1

6

4

2

1 10*06.0135.0

06.0

10*89.9

135.01

135.0*01131.0

4050

b

2.65*13.74 = 36.41 MN/m2 (comp)

Hence σ1 = σd + σb1 = -2.65 + 17.675

= 15.05 MN /m2 (tensile)

and σ2 = σd + σb2 = -2.65 – 36.41

= 39.06 MN/m2 (comp)

8. A curved bar is formed of a tube of 120 mm outside diameter and 7.5 mm

thickness. The centre line of this is a circular arc of radius 225 mm. The bending

moment of 3 kNm tending to increase curvature of the bar is applied. Calculate

the maximum tensile and compressive stresses set up in the bar.

Solution:

Outside diameter of the tube, d2 = 120 mm = 0.12 m

Thickness of the tube = 7.5 mm

Inside diameter of the tube, d1 = 120-2*7.5 = 105 mm = 0.105m

Area of cross-section,

222 00265.015.012.04

mA

Bending moment M = 3 kNm

Area of inner circle,

221 00866.0105.0

4mA

Area of outer circle,

222 01131.012.0

4mA

For circular section,

h2

= ......*128

1

16 2

42

R

dd

For inner circle,

h2

= ......*128

1

16 2

41

21

R

dd

h2

= 4

2

42

10*08.7225.0

105.0*

128

1

16

105.0

For outer circle,

h2

= ......*128

1

16 2

42

22

R

dd; h

2= 4

2

42

10*32.9225.0

12.0*

128

1

16

12.0


19


211

222

2hAhAAh

0.00265 h2 = 0.01131*9.32*10

-4 – 0.00866*7.078*10

-4

h2 = 0.00166 m

2, and R

2/h

2 = 0.225

2/0.00166 = 30.49

Maximum stress at A,

yR

y

h

R

AR

MA 2

2

1 (where, y = 60 mm = 0.06 m)

263

/10*06.0225.0

06.049.301

225.0*00265.0

10*3mMNA

σA = 37.32 MN/m2 (tensile)

Maximum stress at B,

yR

y

h

R

AR

MB 2

2

1

263

/10*06.0225.0

06.049.301

225.0*00265.0

10*3mMNB

σB = 50.75 MN/m2

(comp)

9. A curved beam has a T-section (shown in fig.). The inner radius is 300 mm.

what is the eccentricity of the section?

Solution:


20


Area of T-section, = b1t1 + b2t2

= 60*20 + 80*20 = 2800 mm2

To find c.g of T- section, taking moments about the edge LL, we get

21

2211

AA

xAxAx

)20*80()20*60(

)10*20*80)(20*80()202

60)(20*60(

x =27.14 mm

Now R1 = 300 mm; R2 = 320 mm; R= 327.14 mm; R3 = 380 mm

Using the Relation:

2

2

31

1

22

32 log.log. R

R

Rt

R

Rb

A

Rh ee

23

2 )14.327()320

380(log*20)

300

320(log*80

2800

)14.327(

eeh

= 12503.8(5.16+3.44) – 107020.6 = 512.08

y = )(56.108.512)14.327(

08.512*14.327222

2

mm

hR

Rh

where y = e (eccentricity) = distance of the neutral axis from the centroidal axis.

Negative sign indicates that neutral axis is locates below the centroidal axis.

10. Fig. shows a C- frame subjected to a load of 120 kN. Determine the stresses at

A and B.

Solution:

Load (P) = 120 kN

Area of cross – section = b1t1 +b2t2+ b3t3

= 120*30 + 150*30 +180*30 = 0.0135 mm2

To find c.g of the section about the edge LL,

21

2211

AA

xAxAx


21


)30*180()30*150()30*120(

)120*30*180()15*30*150()225*20*120(1

y =113 mm=0.113 m

y2 = 240 – 113 = 127 mm = 0.127 m

R1 = 225 mm = 0.225 m

R2 = 225 + 30 = 255 mm = 0.255 m

R = 225 + 113 = 338 mm = 0.338 m

R3 = 225 +210 = 435 mm = 0.435 m

R4= 225 + 240 = 465 mm = 0.465 m

2

3

41

2

33

1

22

32 logloglog R

R

Rb

R

Rt

R

Rb

A

Rh eee

23

2 338.0435.0

465.0log12.0

255.0

435.0log03.0

225.0

255.0log15.0

0135.0

)338.0(

eeeh

= 2.86 (0.01877 +0.016 +0.008) – 0.1142 = 0.008122 m2

Direct stress, σd = )(/89.810*0135.0

10*120 263

compmMNA

P

Bending moment, M = P*R

Bending stress at A due to the bending moment,

2

2

2

2

1)(yR

y

h

R

AR

MAb

127.0338.0

127.0

008122.0

338.01

*)(

2

2

AR

RPAb

= 8.89 (1+3.842) = 43.04 MN/m2 (tensile)

Bending stress at B due to the bending moment:

1

1

2

2

1)(yR

y

h

R

AR

MAb

113.0338.0

113.0

008122.0

338.01

*)(

2

AR

RPAb

= 8.89 ( 1- 7.064)

= -53.9 MN /m2 = 53.9 MN/m

2 (comp)

Stress at A, σA = σd + (σb)A

= -8.89 + 43.04 = 34.15 MN/m2 (tensile)

Stress at B, σB = σd + (σb)B

= -8.89 – 53.9 = 62.79 MN/m2 (comp)

11. Derive the formula for the deflection of beams due to unsymmetrical bending.

Solution:

Fig. shows the transverse section of the beam with centroid G. XX and

YY are two rectangular co-ordinate axes and UU and VV are the principal axes

inclined at an angle θ to the XY set of co-ordinates axes. W is the load acting along the

line YY on the section of the beam. The load W can be resolved into the following two

components:


22


(i) W sin θ …… along UG

(ii) W cos θ …… along VG

Let, δu = Deflection caused by the component W sin θ along the line GU for its bending about VV axis, and

Δv = Deflection caused by the component W cos θ along the line GV due to bending abodt UU axis.

Then depending upon the end conditions of the beam, the values of δu and δv are

given by

VV

uEI

lWK3sin

UU

vEI

lWK3cos

where, K = A constant depending on the end conditions

of the beam and position of the load along the beam, and

l = length of the beam

The total or resultant deflection δ can then be found as follows: 22

vu

223 cossin

UUVV I

W

I

W

E

Kl

UUVV IIE

Kl2

2

2

23 cossin

The inclination β of the deflection δ, with the line GV is given by:

tan

tanVV

UU

I

I

v

u


23


12. A 80 mm x 80 mm x 10 mm angle section shown in fig. is used as a simply

supported beam over a span of 2.4 m. It carries a load of 400 kN along the line

YG, where G is the centroid of the section. Calculate (i) Stresses at the points A, B

and C of the mid – section of the beam (ii) Deflection of the beam at the mid-

section and its direction with the load line (iii) Position of the neutral axis. Take

E = 200 GN/m2

Solution: Let (X,Y) be the co-ordinate of centroid G, with respect to the

rectangular axes BX1 and BY1.

Now X = Y = mm66.23700800

350032000

10*7010*80

5*10*7040*10*80

Moment of inertia about XX axis:

2

32

3

)66.2345(*10*7012

70*10)566.23(*10*80

12

10*80XXI

= (6666.66 + 278556) + (285833.33 + 318777) = 889833 mm4

= 8.898 * 105 mm

4 = IYY (since it is an equal angle section)

Co-ordinates of G1 = + (40-23.66), - (23.66-5) = (16.34,- 18.66)

Co-ordinates of G2 = -(23.66-5). + (45 – 23.66) = (-18.66, + 21.34)

(Product of inertia about the centroid axes is zero because portions 1 and 2 are

rectangular strips)

If θ is the inclination of principal axes with GX, passing through G then,

90tan2

2tan

XXXY

XY

II

I (since Ixx =Iyy)

2θ = 90º i.e. θ1 = 45º and θ2 = 90º + 45º = 135º are the inclinations of the principal axes GU

and GV respectively.

Principal moment of inertia:

IUU = 22 )()2

()(2

1XY

XXYYYYXX I

IIII

= 25255

55 )10*226.5()2

10*898.810*895.8()10*898.810*895.8(

2

1


24


= (8.898 + 5.2266) *105 = 14.1245*10

5 mm

4

IUU + IVV = IXX + IYY

IVV = IXX IYY – IUU

= 2*8.898 x 105 – 14.1246 x 10

5 = 3.67 x 10

5 mm

4

(i) Stresses at the points A, B and C:

Bending moment at the mid-section,

Nmm

WlM

53

10*4.24

10*4.2*400

4

The components of the bending moments are;

M’ = M sin θ = 2.4 x 105 sin 45º = 1.697 x 10

5 Nmm

M’’ = M cos θ = 2.4 x 105 cos 45º = 1.697 x 10

5 Nmm

u,v co-ordinates:

Point A: x = -23.66, y = 80-23.66 = 56.34 mm

u = x cos θ + y sin θ

= -23.66 x cos 45º + 56.34 x sin 45º = 23.1 mm

v = y cosθ + x sin θ

= 56.34 cos 45º - (-23.66 x sin 45º) = 56.56 mm

Point B:

x = -23.66, y = -23.66


= -23.66 x cos 45º + (-23.66 x sin 45º ) = - 33.45 mm


= -23.66 cos 45º - (-23.66 x sin 45º) = 0

Point C ; x = 80 – 23.66 = 56.34, y = -23.66


= 56.34 cos 45º -23.66 x sin 45º = 23.1 mm


= -23.66 cos 45º - 56.34 sin 45º) =- 56.56 mm

UUVV

AI

vM

I

uM "'

2

5

5

5

5

/47.17101246.14

)56.56(10*697.1

1067.3

)1.23(10*697.1mmN

xxA

2

55

5

/47.15101246.14

0

1067.3

)45.33(10*697.1mmN

xxB

2

55

5

/788.3101246.14

56.56

1067.3

)1.23(10*697.1mmN

xxB

(ii) Deflection of the beam, δ: The deflection δ is given by:

UUVV IIE

KWl2

2

2

23 cossin


25


where K = 1/48 for a beam with simply supported ends and carrying a

point load at the centre.

Load , W = 400 N

Length l = 2.4 m

E = 200 x 103 N/mm

2

IUU = 14.1246 x 105 mm

4

IVV = 3.67 x 105 mm

4

Substituting the values, we get

25

2

25

233

)101246.14(

45cos

)1067.3(

45sin)104.2(400

48

1

xxE

xx

δ = 1.1466 mm

The deflection δ will be inclined at an angle β clockwise with the kine GV, given by

848.345tan1067.3

101246.14tantan

5

5

x

x

I

I

VV

UU

β = 75.43º - 45º = 30.43º clockwise with the load line GY’. (iii) Position of the neutral axis:

The neutral axis will be at 90º - 30.43º = 59.57º anti-clockwise with the

load line, because the neutral axis is perpendicular to the line of deflection.

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