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Strength of Material Syllabus:

1. Introduction

2. Stress-Strain [1 question]

3. a. Principle Stress [1/2 question]

3. b. Moment of Inertia (M. I.) [1/2 question]

4. B.M., S.F. diagram [12 marks, 1 full question]

5. Theory of flexure [1 question]

6. Torsion [1/2 question]

7. Thin walled vessel [1/2 question]

8. Column (succeing of column) [1/2 question]

Direct Stress [1/2 question]

9. Theory of failure [1/2 question]

M.I. α Strength

so,

� Section design ubf{ M.I. a9L cfpg] ul/ ul/G5

� Depth a9fof] eg] M.I. a9\5 .

k

E

y

f

I

M == Theory of flexure



Chapter 1

Introduction 1.1 Types of Load

(i) Static Load: Load acting on a body in equilibrium is called static load.

(ii) Dynamic Load: The load acting on a body in motion is called dynamic load. Its

effect depends on time.

(iii) Dead Load: It includes the wf of all permanent components of the structure such

as beams, columns, floors, slabs, etc and any other immovable loads that are

constant in magnitude and permanently attached to the structure.

(iv) Live Load: It includes any external loads imposed on the structure during it’s

service such as the wf. Of the stored material, furniture and people.

Dynamic Load: Wind load, earthquake, hydrostatic load, uplift pressure, snow and rain

load.

1.2 Supports (i) Roller support:

Fx = 0

Fy = acted

m = 0

Example : 3/sf] blnt

(ii) Hinge support:

Fx = acted

Fy = acted

m = 0

(iii) Fixed support:

Fx , Fy , and m are acted.

(iv) Ball and socket support:

- Fx , Fy and Fz are acted

- m = 0

MFx

Fy

Ry

Rz

Rx



- This is example of hinge support in 3-D.

(v) Fixed support:

Determinant and Indeterminant Structure:

Determinate Structure:

If number of unknown = number of equilibrium equation then the structure is called

determinate structure.

Indeterminate Structure:

If number of unknown is not equal to the number of equilibrium equation then the

structure is called indeterminate.

Number of Indeterminacy: Let,

R = no. of unknown reaction

r = no. of equilibrium equation

Degree of indeterminacy, E = R – r

Here,

no. of unknown reaction, (R) = 7

no. of equilibrium equation, (r) = 3

so,

Degree of indeterminacy, E = R – r = 7 – 3 = 4

Also,

no. of unknown reaction, R = 5

no. of equilibrium equation, r = 3

then,

Degree of indeterminacy, E = R – r = 5 – 3 = 2

Rx, Ry & Rz # j 6} 5 .mx,my & mz # j 6} 5 .

Hinge Support Fixed Support

Fig - 1 Fig - 2



Here, Here, no. of unknown = 2 = R no. of unknown, R = 4 no. of equilibrium equation = 3 =1 no. of equilibrium equation , r = 3 Now, Now, E = R – r =2 – 3 = –1 E = R – r = 4 – 3 = 1 Hence, Hence, E > 0 E > 0 So, it is determinatre. So, it is indeterminate and stable.

Conditions: When, E = 0 = R – r = 0 (structure is determinate)

E > 0 = indeterminate and stable

E < 0 = Indeterminate and un-stable

Direct stress and strain: Stress: When a body is subjected to any external load then there is a deformation of a body.

During a deformation the percicles of a body exerts resisting force and the deformation stops

when the resisting force becomes equal to the applied external load. This resisting force per

unit area is called stresses.

A

F==Area

Forcey ResistivitStress i.e.

Strain: Change in length per unit length when applying a force on a obdy.

L

L∆=Strain i.e.

Normal stress: When the force acts normal to he surface of a body is called normal stress.

P PCompressive Stress

P PTangential Stress



Tangential Stress (Shear Stress): When the force acts (transverse) i.e. tangentially the surface of the body then the

resulting stress is called shear stress (rf}8f kl6 act x'g] force)

Longitudinally (t/jf kl6) act x'g] force nfO{ axial force elgG5 / beam tail x'bf longitudinal df w/} sd

x'G5 / transverse df Hofbf tail x'G5 .

Hook’s Law: It states that stress is directly proportional to strain within elastic limit.

So, Stress (σ) α Strain (e)

σ = e.E

where, E = permittivity constant and is called Yong’s modulus of elasticity.

Since,

Ee.=σ

EL

L

A

P ⋅∆=

AE

PLL =∆∴

==Strain

Stress

e E ,elasticity of modulusor modulus sYong'

σ

Modulus of rigidity (c): It is ratio shear stress and shear strain.

So, StrainShear

Stress Stressc =

φq

c =∴

Bulk Modulus (K): It is the ratio of volumetric stress to the volumetric strain.

So, V

V

eK

σ=

Stress given by all six (6) directions normally.

Stress and strain diagram of mild steel:

Shear Strain

A B

C

U

L

Rasabolic(Strain hardening)region

Strain

Max Limit



B Ct

L1L2 L3

Mild steel – 250 M Pa (N/mm2)

Tor Steel – 495 M Pa

TMT (Thermo Mechanical Treated) – 500 M Pa

Dutility for earth quake = 16% - 25% desisable.

AB = sd load (stress) df klg Strain a9]sf]

A = elastic limit

B = Yield point (upper) = YUS

C = Lower end point

= YL (stress dropped on const strain)

(load glbPklg of] action x'G5)

UL = pure plastic state (stress const and strain increase)

U = ultimate stress

U = Stress ga9]klg strain a9\5 / meening x'b} x'G5 .

Working load = OA region = elastic limit

Max capacity = ultimate load

Ultimate Stress: Maximum load per unit original cross-sectional area is called ultimate stress.

Working Stress (or safe load): It is a safe load within elastic limit.

safely offactor

load Ultimate stress working =

OR

safety offactor

stress Yield stress working =

Factor of Safety: F:

stress working

stress ultimateor

stress working

stress Yield F=

Factor of safety for,

R.C.C. work = 3

Timer = 4 to 6

Steel = 1.85

Principle of Superposition: It states that when the no. of loads are acting

on a body then the resulting strain will be the

algebraic sum of strains caused by the individual

loads.



Prob Soln:

Cross-sectional area of the all body i.e. A, B & C are same.

Now,

Now, for a body A:

AEAEAE

PLLA

20210 =×==∆

For body, B:

AEAEAE

PLLB

2437 =×==∆

For body, C:

AEAEAE

PLLC

3649 =×==∆

Now,

∴total elongating AEAE

LLLL CBA

77362420 =++=∆+∆+∆=∆

Bar of Varying Cross-Section:

Let, a force P is acting on a body of varying cross-section as shown in figure.

AB C10KN 3KN

2KN9KN

2m 3m 4m

A10KN

(3-2+9)=10KN

10-3=7KNB C

9-2=7 10+2-3=9KN 9KN

PA2 A3

L1 L2 L3



Extension of bar (I) EA

PL

1

11 =∆

Extension of bar (II) EA

PL

2

22 =∆

Extension of bavr (III) EA

PL

3

33 =∆

Net extension EA

PL

EA

PL

EA

PL

3

3

2

2

1

1321 ++=∆+∆+∆=∆

++=

3

3

2

2

1

1

A

L

A

L

A

L

E

P

If no. bars are these then,

++++==∆

n

n

A

L

A

L

A

L

A

L

E

P...................

3

3

2

2

1

1

Tapered Section:

Suppose a uniformly tapered section from diameter d1 to d2 of length L. Consider a strip

of length dx at a distance ‘x’ from face AA|

xL

ddddx ⋅

−+= 12

1 (from similar ∆ s relation)

Extension of δx length of a stress

ExL

ddd

xP

AE

PL

⋅

−+

×==∆ 2

1214

πδδ

( ) tkxd

xP2

1

4

+⋅=

πδ

( )letkL

ddwhere =

− 12,

Hence, total elongalation,

∫ ∆=∆L

O

δ

A

A1

L

d2d1d

B

B1x

dx

p



160xN

180 xN

( )∫ ⋅+=

L

O Ekxd

xP2

1

4

πδ

( )∫ ⋅+= −L

O

xkxdE

P δπ

221

4

( )

( )

L

OK

kxd

E

P

+−+

=+−

12

4122

1

π

L

OkxdEk

P

+−=

1

14

π

−

+−=

11

114

dkLdEk

P

π

( )

−

−+−−=

112112

114

ddddddE

PL

π

( )

−−

−=21

21

12

4

dd

dd

ddE

PL

π

( )( )

21

12

12

4

dd

dd

ddE

PL −⋅

−−=

π

21

4

ddE

PL

⋅=∆∴

π

Check if it is rectangular section then, (solid cylindrical x'bf)

ddd == 21

AE

PL

dE

PL

dE

PL ==⋅

=∆

4

422 ππ

Q. A specimen of steel 25 mm diameter with gauge length of 200 m is tested to destruction. It

has an extension of 0.16 mm under a load of 80 KN and the load at elastic limit is 160 KN.

The maximum load is 180 KN.

The total extension at fracture is 56 mm and diameter at neck is 180 mm. Find

(i) The stress at elastic limit

(ii) Percentage elongation

(iii) Yong’s modulus of elasticity

(iv) Percentage reduction in area

(v) Ultimate tensile stress.

AE

PL=∆∴



53 grade cement = characteristics strength of cement i.e. 57% eGbf sd sample n] dfq Tof] strength gb]vfpbg Test fulfill x'G5 100 j6f sample with c/s ratio 1:13. 53v M Pa load up to 95 sample n] lbg' k5{ .

So|n:

( ) 2

22

874.4904

25

4mm

dA =×== ππ

(i) The stress at elastic limit area sectional-cross original

limit elasticat load=

22

3

/949.325874.490

10460mmN

mm

N =×=

(ii) Percentage elongation 100length original

extension Final ×=

100200

56 ×=

%28=

(iii) Yong’s modulus of elasticity (within elastic limit only) Strain

Stress=

Since, AE

PL=∆

( ) 16.0874.490

1020080

16.04

25

20080 3

2 ×××=

×××=

∆⋅= N

A

PLE

π

25 /1003.2 mmN×=

(iv) Percentage reduction in area %100area Initial

area Final - area Initial ×=

%100

4

254

18

4

25

2

22

××

×−×=

π

ππ

%16.48=

(v) Ultimate stress area sectional-cross Original

load ultimateor load Maximum=

23

/693.366874.490

10180mmN=×=

PaM693.366=

[here, E = 2 × 105 N/mm2 = 200 KN/mm2 always]

In 28 days, 7 days 66% strength 28 days

M15 = 15 M Pa

M20 = 20 M Pa

53 grade = quick section



dx

x

L

ωx

43 grade = less quick than 53 grade

Q. A steel bar of 25 mm diameter is acted upon by forceless shown in figure. Determine the

focal elongation of the bar if, (i.e. take)

E = 200 KN/m2

So|n:

For body (i):

52

111 1029.490

100010005.1600

2004

54.160

×××××=

×

×==∆dAE

LP

π

mm917.01 =∆∴

For body (ii):

5

222 1029.490

10001100070

×××××==∆

AE

LP

mm713.02 =∆∴

For body (iii):

5

333

3 10209.490

100021050

×××××==∆

AE

LP

mm019.13 =∆∴

∴ Total elongation ( ) ( )mm019.1713.0917.0321 ++=∆+∆+∆=∆

mm649.2=∆∴

Elongation due to self fω :

(Bar of uniform section):

Let a bar of length ‘L’ hanging freely. Suppose a strip of

length dx, its extension ∆δ , is given by,

AE

dxWx⋅=∆δ

Where, Wx = wf. of portion belone the strip

= Ax

Now,

E

dxx

AE

dxAx ⋅⋅=⋅⋅=∆γγδ

E

dxx ⋅⋅=∆γδ .

Total elongation of the bar,

1 2 360 xN 20 xN 20 xN 50 xN

1.5 m 1.0 m 2.0 m

60 xN 50 + 20 - 10 = 60 xN1

1.5 m

60 + 10 = 70 20 + 50 = 70 xN2

I m



dx x y

L

d

x

∫ ∫ ∫ ⋅=

=⋅=⋅⋅==∆ ∆

L

O

L

O

L

O

L

O

L

E

x

Edxx

EE

dxx

22

22 γγγγδ

E

gL

E

L

22

22 ργ ==∆∴

If w be the total wf of the bar,

LAw ⋅⋅= γ

AL

w=γ A

w==L

Pstress

EAL

wL

2

2

=∆∴ E

PL

aE

wL

22==∆∴

AE

wL

2=∆∴ (in terms of area)

wf = wf. of the bar [in Newton (N) or Kilo Newton (KN)]

Elongation of bar of tapered shape due to self fω :

Let a tapering bar as shown in figure. Consider an

elementary area of length dx. Let, Ax be the area of cross

section at face x.y.

Total elongation of the strip,

EAx

axwx

⋅⋅=∆δ

Where, wx = wf of the position below the strip

Wx = (x × Ax) × 1/3

Note:

Vol(V) + Vol(VV) + Vol(Vvv) = whole Vol of rectangle

volumerectangle of 3

1 Volumebar Tapered =∴

Now,

EAx

dxxAx

×

××××=∆

γδ 3

1

E

dxx

3

×= γ

∴ Total elongation,

∫ ∫ ⋅=⋅⋅=∆L

O

L

O

dxxEE

dxx

33

γγ

L

O

x

E

=

23

2γ



L

23

2L

E⋅= γ

E

gL

E

L

66

22 ργ ==

Let, w = total wf. of the bar

γπ ⋅= Ld

w43

1 2

Ld

w2

12

πγ =

Here,

Ed

wL

Ed

wL

E

L

Ld

w

42

2

6

1222

2

2 πππ==⋅=∆

AE

wL

2=∆∴

Temperature Stress: When the temperature of the body is raised oe lowered and the body not allowed to

expand or contract freely, the stress are setup in the body. This stress is known as temperature

stress.

Single Bar: Suppose a bar of length L is placed between two rigid

suppose and temperature is raised through tºC then the

extension of the bar,

tL ⋅⋅=∆ α

Where, α = coefficient linear expansion

Also,

E

L

AE

P ⋅==∆ σ2

=A

PSince σ,

Where, δ = Temp. Stress

Now,

tLE

L ⋅⋅= ασ

Let, the case when the end fields by an amount a, then the stress will be due to

extension of a−∆ ,

( )

L

ta−∆=σ (since Stress = e × E)

Et ⋅⋅=∴ ασ

a



( )L

EatL ⋅−⋅⋅=∴ ασ

But strain, L

ae

−∆==length Actual

expansion Actual

( ) ( )

L

EaE

L

e

L

ae

−∆=×=−∆=∴ σ&

Composite Bar:

Consider two bar of length ‘L’ of different materials suppose steel and copper as

shown in figure, are composite. Let the composite bar subjected to a temperature. As a result

due to different capacity of expansion of each bar, there is setup opposite kinds of stresses i.e.

(Tensile and compressive in the bar). Then,

Ps = Pc = P

i.e. Tensile force in steel = Compressive force in steel = P

Also, ∆=∆=∆ CS

From figure,

PS

tSS ∆+∆=∆

PC

tCC ∆−∆=∆

Where, ∆ S = ∆ C = ∆ = final extension

tS∆ = free extension of steel due to temperature

tC∆ = free extension of copper due to temperature

PS∆ = Expansion of steel due to temperature stress

PC∆ = Compression of copper due to temperature stress.

Since,

CS ∆=∆

PC

tC

PS

tS ∆−∆=∆+∆

tS

tC

PC

PSor ∆−∆=∆+∆,

tLtL SC αα −=

Steel, αsAs

Copper, αc, Ac

Ds Ds

L x

Dc

D pDc

t p



1 2A1 A2

P

L

( )SCLt αα −=

Where, αc = coefiicient of linear expansion of copper

αs = coefficientr of linear expansion of steel

Composite section:

Now, ( )SCtS

tC

PC

PS Lt αα −=∆−∆=∆+∆ (From above proff)

( )SCCCSS

LtEA

PL

EA

PL αα −=+

( )SCCCSS

LtEAEA

PPor αα −=

+ 11

,

( ) )(11

itEAEA

P SCCCSS

−−−−−−=

+ αα

( )

)(11

ii

EAEA

tP

CCSS

SC −−−−−+

−=∴

αα

Also we know,

C

CS

S A

P

A

P == σσ &

From equation (i), we get,

( ) )(iiitEE SC

C

C

S

S −−−−−−=+ αασσ

( ) )(ivtee SCCS −−−−−−=+ αα [From formula]

Equations (ii), (iii), and (iv) are required expression for composite section for

temperature stress.

2.5 compound Bars subjected to axial tension compression: Consider two bars of different material having

equal lengths are rigidly fixed at one unit and let P is

applied as shown in figure, then the load ‘P’ which equal

to sum of the loads carried by each material.

P = P1 + P2

Also, the extension of each bar is same i.e. B1 = B2

So,

22

2

11

1

EA

LP

EA

LP=

)(22

1121 i

EA

EAPP −−−−−=



d

BLL

D

So, from (i),

+=+=

22

1122

22

112 1

EA

EAPP

EA

EAPP

22

112

1EA

EAP

P+

=∴

Similarly,

P1 = P2

11

222

1EA

EAP

P+

=∴

Poisson’s Ratio: It is the ratio of lateral strain to the longitudinal strain.

i.e. al strainlongitudin

rainlateral st=µ

Suppose a solid circular bar of length ‘L’ and

diameter D. Due to stress, the length increases by L∆

and diameter reduces by ( )d−∆

∆−∆= d

Strain Lateral

L

L∆=Strain alLongitudin

If the strain in the direction of load is σ/E then in other two direction is,

EE

σµσµ −− &

Q. A rod as shown in figure is subjected to poll of 500 KN on the ends. Take E = 2.05×105

N/mm2. Find extension of rod.

So|n:

)(3

3

2

2

1

1 iA

L

A

L

A

L

E

P −−−−−

++=∆

120 80 100 80 100 80 120

200 100 250 150 250 150 350 20 0



A D

CB C1B1

z

z

z

F

Now, 113104

120

4

22

1 =×== ππdA

50274

802

2 =×= πA

78544

1002

3 =×= πA

Then, equation (i) written as,

+++++++×

=∆7854

200250250

5027

150150100

11310

200200

1005.2

8005

mm796.0=∆∴

Relation between modulus of elasticity(E), modulus of rigidity(C) and bulk modulus(K):

Relation between E and C: Consider a square element ABCD deformed to ADC|B|

due to stress.

Due to pure shear, there is tensile stress along the

diagonal BD and compressive stress at right angle to the

diagonal.

−−=E

q

E

q µ BD diagonal ofStrain

( ) )(1 iE

q −−−−−+= µ

2

BD diagonalin Strain ||

AB

FB

BD

FB ==

Since, the deformation is very small so we can take,

º45| ≈∠ FBB then,

FB| = B|BCos45º -------- (a)

From equation (a), putting the value of FB| in equation (i),

2

º45

2 BD diagonalin Strain

||

AB

CosBB

AB

FB ==

AB

BB

AB

BB

222

1|

|

==

AB

BB|

2

1 ⋅=

Ø2

1= (since Ø is very small)



Since, c

qØ = rigidity) of modulus(

(Ø)Strain Shear

(q) SressShear C=

So, )(2

BD diagonalin Strain iiC

q −−−−−=

Since, equation (i) and (ii) are same,

So, ( )µ+= 12 E

q

C

q

( ) )(12 iiiCE −−−−−+=∴ µ

Which is required relation between E and C.

Relation between E and K: Consider a cube subjected to three mutually

perpendicular tensile stress along X,Y,Z directions.

EEE

eX

σµσµσ −−=

( )µσ21−=

E

EEE

eY

σµσµσ −−=

( )µσ21−=

E

EEE

eZ

σµσµσ −−=

( )µσ21−=

E

Now,

( ) ( ) ( )µσµσµσ212121 −+−+−=++=

EEEeeee ZYXV

( )µσ21

3 −=E

Since,

( )µγ

γ

213Strain Volumetric

Stress Volumetric

−==

E

K

( )µ213 −= E

K

( ) )(213 ivKE −−−−−−= µ

Which is required relation between E and K.



Putting the value of µ from equation (iii) to equation (iv) then,

−−= 132C

E21KE

213

+−=C

E

K

E

C

E

K

E −= 33

C

EKKE

39 −=

KC

EKE 9

3 =+

KC

KEor 9

31, =

+

+=

C

KC

KE

39

KC

KCE

3

(

+=∴

Which is required relation between E, C, and K.

Q. A copper rod 25 mm in diameter is inclosed in steel 30 mm internal diameter and 35 mm

external diameter. The ends are rigidly attached. The composite bar is 500 mm long and is

subjected to an axial pull of 300 KN. Find the stresses induced in the rod and the tube. Take

E for steel 2×105 N/mm2 and E for copper as 1×105 N/mm2.

So|n:

AC = Area of copper in cross-section = (25)2×π/4

∴AC = 490.9 mm2

AS = cross section area of steel = {(35)2×π/4} – {(30)2×π/4}

∴AS = 255.25 mm2

Now, PC + PS = P

)(1030 3 iAA SSCC −−−−−×=+ σσ

Elongation in steel = Elongation in copper

25 30 3530 KN 30 KN



C

C

S

S

E

t

E

tor

σσ=,

C

C

S

S

EEor

σσ=,

5

5

101

102,

×××=×= C

C

SCS E

Eor σσσ

.26cS =∴ σ

From equations (i),

3103025.255269.490 ×=×+× cCσ

23

/95.294.1001

1030, mmNor C =×=σ

2/95.29 mmNC =∴ σ

And, 91.5995.29226 =×== cSσ

2/91.59 mmNS =∴ σ

Principal stresses: The planes having no shear stresses are known as principal planes. The normal stresses

acting on a principal plane are known as principal stress.

CASE-I: Stress acting on a plane inclined to the direction of a applied forces:

Consider a rectangular member of unit thickness and of uniform cross-sectional area.

Let, P = Axial force acting on the member

A = Area of cross-section, which is perpendicular to the line of a action of the force P

Here, the area of section EF which is perpendicular to the line action of force, P is,

A = EF × 1

Now, let us consider an oblique plane FG inclined at an angle θ with the section EF.

The area of section FG = FG × 1

θθθ Cos

A

Cos

EF

Cos

EF =×=×= 11

θSecA⋅

Stresses on the section FG,

θ

θ

E G

P

F

P



A

P=

FG ofSection

P i.e.

θSecA

P

⋅=

θσ Cos⋅=

Since, this stress is not normal to the FG, it is parallel to the axis of the member. So, it

has normal and tangential components on the sections FG.

PX = P Cosθ

Pt = P Sinθ

Hence, normal stress (σn) on the plane FG,

θθσ

θθ

Sec

Cos

SecA

CosP ×=⋅⋅==

FG of area

force normal

= σA

PSince,

θσ 2Cos⋅=

θσσ 2Cosn ⋅=∴

Tangential Stress (σt) on the plane FG,

θσθθσθθσθθσ 2

22

2FG of area

Force TangentialSinCosSinCosSin

SecA

SinPt =××=××=

⋅⋅==

θσσ 22

Sint =∴

The normal stress which be maximum when Cosθ = 1 i.e. θ = 0º

Maximum normal stress = σ

Also, the tangential stress which be maximum when Sin2θ = 1

or, Sin2θ = Sin 90º or Sin270º

or, θ = 45º or 155º

Hence, maximum tangential stress = 1×σ/2

= σ/2

For principal plane, σt = 0

022

=θσSin

º0=θ

LEFT A LOT OF PAGES

Case:III:

(Left A Lot Of Lines)

112

×−=

FC

CosQSinQ θθ

FC

CosBCSinFB θθ ××−××= 22



θθ CosFC

BCqSin

FC

FB −×= 2

θθθθ CosCosSinSin ⋅−⋅−= 22

( )θθ 222 SinCos −−=

θσ 2qCost −=

Case-IV:

A member subjected to direct stresses in two mutually perpendicular directions

accompanied by a simple shear stress.

Consider a rectangular plate ABCD of unit thickness which is subjected to tensile

stresses σ1 and σ2 and shear stresses 1 at their faces as shown in figure.

Suppose oblique plane FC inclined at an angle θ.

Let, P1 = tensile force on face BC due to stresses σ1

= σ1×BC×1

P2 = tensile force on face BF due to stresses σ2

= σ2×BF×1

Q1 = shear force on face BC due to shear stress q

= q×BC×1

Q2 = shear force on face BF due to shear stress q

= q×BF×1

Hence, resolving all forces total normal force on oblique plane FC;

θθθθ CosQSinQSinPCosP 2121 +++=

θθθσθσ CosBFqSinBCqSinBFCosBD ××+××+××+××= 21

And, total tangential force on oblique plane FC;

θθθθ CosQSinQCosPSinP 1221 −+−=

θθθσθσ CosBCqSinBFqCosBFSinBC ××−××+××−××= 21

Hence, total normal stress (σn) on oblique plane FC;

1

2

1

2

F

q

q

θ

D

A F B

C

Q1 Cos θ

P1 Sin θ

P2 Cos θ

Q2 Sin θ

Q2 = Q x BC x Z

P2 = 2 x FB x 1

P1 Cosθ +P2 Sin θ

q x BC Sin θ

Q1 Cos θ

P1 = 1 x BC 1

B1 = q x BC x 1



1

21

×××+××+××+××

=FC

CosBFqSinBCqSinFCosBCn

θθθσθσσ

θθθσθσ CosFC

BFqSin

FC

BCqSin

FC

BFCos

FC

BC ×+×+×+×= 21

θθθθθθσθθσ CosqSinCosqSinSinSinCosCos ×+×+×+×= 21

( ) ( ) θθθσθσCosqSinCosCos ×+−++= 221

221

221

θθσσσσσ 22222

2121 SinCosn +−++=∴

And, Tangential stress across FC;

1

21

×××−××+××−××=

FC

CosBCqSinBFqCosFBSinBCt

θθθσθσσ

θθθσθσ CosFC

BCqSin

FC

BFqCos

FC

FBSin

FC

BC ×−×+×−×= 21

θθθθθθσθθσ CosqCosSinqSinCosSinSinCos ×−×+××−×= 21

θθσσσ 2222

21 CosSint −−=∴

Major and minor principle stress:

For principle plane;

0=tσ

02222

21 =−− θθσσCosSin

θθσσ22

221 qCosSin =−

21

2

2

2

σσθθ

−= q

Cos

Sin

b

pq =−

=21

22tan

σσθ

Hence, diagonal of right angle triangle,

( ) ( )2221 2q+−±= σσ

either ( ) ( )2221 2q+−σσ or ( ) ( )22

21 2q+−− σσ

Let,

( ) 2221 4qdiagonal +−= σσ

( ) 22

21 4

22

q

qSin

+−=

σσθ

For major principle stress:



θσσσσσ22

222121 qSinCos +

−+

+=

( ) ( ) 22

2122

21

212121

4

2

422 q

qq

q +−⋅+

+−

−⋅

−+

+=

σσσσ

σσσσσσ

( )

( ) 2221

222121

42

4

2 q

q

+−

+−+

+=

σσ

σσσσ

( ) 2221

21 42

1

2q+−++= σσσσ

2

2

2121

22q+

−+

+=

σσσσ

Minor principle stress:

Let, ( ) 2221 4qdiagonal +−−= σσ

( ) 22

21

21

42

qCos

+−

−−=

σσ

σσθ

( ) 22

21 4

22

q

qSin

+−−=

σσθ

θθσσσσ22

22 stress principleMinor 2121 qSinCos +−++=∴

( )

( ) ( ) 2221

2221

212121

4

2

422 q

qq

q +−−⋅+

+−−

−⋅

−+

+=

σσσσ

σσσσσσ

( )

( ) 2221

222121

42

4

2 q

q

+−

+−−

+=

σσ

σσσσ

( ) 2221

21 42

1

2q+−−+= σσσσ

2

2

2121

22q+

−−

+=

σσσσ

For maximum and minimum shear stress:

The shear stress will be maximum or minimum when,

( )

0=θσ

d

d t

02222

21 =

−− θθσσ

θCosSin

d

d

( ) 022222

, 21 =−−×− θθσσSinqCosor



( ) 0222, 21 =+− θθσσ qSinCosor

( ) θσσθ 2222, 1 CosqSinor −−=

( )

also22

2, 1221

qqTTaor

σσσσθ −+=

−−=

( ) 22

12

12

42

qSin

+−

−±=

σσ

σσθ

( ) 22

12 4

22

q

qCos

+−±=

σσθ

Hence, maximum or minimum shear stress is,

θθσσ22

221 qCosSin −

−

( ) ( ) 22

1222

12

1221

4

2

42 q

qq

q +−±

+−

−⋅

−±=

σσσσ

σσσσ

( )

( ) ( ) 2212

2

2212

212

4

2

42 q

q

q +−±

+−

−±=

σσσσ

σσ

( )

( ) 2212

2212

42

4

q

q

+−

+−±=

σσ

σσ

( ) 2212 4

2

1q+−±= σσ

2

2

12

2q+

−±=

σσ

Hence, major shear stress 2

2

12

2q+

−=

σσ

And, minor shear stress 2

2

12

2q+

−−=

σσ

Left a page

Mohr Circle method: It is a graphical method of determination of normal tangential and resultant stress. It can

be used for the following cases:

(A) A body subjected to two mutually perpendicular principle tensile stress of unequal

intensities.

(B) A body subjected to two mutually perpendicular principal stress which are unequal

and unlike (i.e. one is tensile and other is compressive).



N x

y

oAC B2 1

y1

1 = - ve2 = + ve

A = origin

(C) A body is subjected to two mutually perpendicular tensile stresses accompanied by

simple shear stress.

Case (A):

Let, σ1 = major tensile stress

σ2 = minor tensile stress

θ = angle made by oblique plane with minor stress

Proof:

2

21 σσ −=== OEOBCO

θσσσ 22

Now, 212 OECosODCOACAD +

−+=++=

θσσσσσ 222

21212 Cos

−+

−+=

θσσσσσ2

22

2 21212 Cos−

+−+

=

nCos σθσσσσ=

−+

+= 2

222121 i.e. normal stress

tSinSinOEED σθσσθ =−

=×= 22

2 21

i.e. tangential stress

Now, AD = normal stress on oblique plane

DE = tangential stress on oblique plane

AE = resultant stress

Case (B):

σ1 = major tensile stress

σ2 = minor compressive stress

θ = angle made by oblique plane with minor B

stress

AD = Normal stress on oblique plane

DE = Tangential stress on oblique plane

AE = Resultant stress on oblique plane

Case (C):

σ1 = major tensile stress

σ2 = minor tensile stress

q = shear stress

θ = angle made by oblique plane with minor stress

now, AD = normal stress on oblique plane

1

2

1

2

y

A

2 2

θo DC

E



DE = tangential stress on oblique plane

AE = resultant stress on oblique plane

C.G. � whole wf

Centroid � whole area

Moment of inertia (M.I.):

Centre of gravity (C.G.) = The point where the whole wf of the body is concentrated.

Centroid = The point where the whole area of the body is concentrated

Centroid of plane lamina:

nnxaxaxaxaXA ++++= ...............332211

n

nn

aaaa

xaxaxaxaX

++++++++

=..........

...............

321

332211

∫

∫

∑

∑ ⋅==

=

=

da

dax

a

xa

n

ii

n

iii

1

1

Integration = summation of area

Derivative = slope or any inter vol of time

Similarly,

x1

x2

x3

x4x5

x6x7

x8

o

y



hL M

A

dAyG G

∫

∫

∑

∑ ⋅==

=

=

da

day

a

yaY

n

ii

n

iii

1

1

For line element;

∫

∫

∑

∑ ⋅==

=

=

dL

dLx

L

xLX

n

ii

n

iii

1

1

And,

∫

∫

∑

∑ ⋅==

=

=

dL

dLy

L

yLY

n

ii

n

iii

1

1

Moment of Inertia (M.I.):

The moment of this plane lamina about Y-axis = Ax

The moment of this plane lamina about X-axis = Ay

The moment of moment of this plane lamina about Y-axis = Ax × x = Ax2

and, The moment of moment of this plane lamina about X-axis = Ay × y = Ay2

so, moment of moment of an area is called moment of inertia. It is represented by I. It is

also called 2nd moment of an area.

[xfdLn] o; subject study ug{] ;a} M.I. for area xf] . of] mass df klg lgsfNf ;lsG5 t/ xfdL oxf+ area sf]

af/]df dfq k9\b5f} .]

Parallel axis Theorem: 2AkII GGLM +=

Statement and Proof: It states that “the moment of inertia of a plane

lamina about any axis in the plane of lamina is equal

to the sum of the moment of inertia of that lamina

about centroidal axis and the product of the area and

x Ay

y

o

IGG

Ah

LM



A

ry

x da

y

x

z

square of perpendicular distance between the two axis.”

Proof:

Let a lamina of area ‘A’. Let the centroidal axis GG about which moment of inertia is

known i.e. IGG. Let axis “LM” parallel to axis “GG” about which M.I. is to be found out.

The distance between these two axis be “h”. Suppose an elementary area “dA” which is at a

distance ‘y’ from centroidal axis “GG”.

M.I. of elementary area about axis LM,

( )2yxdAdI LM +=

( )22 2or, yxyxdAdI LM ++=

Now, M.I. of whole area about axis LM;

∑= LMLM dII

( )∑ ++= 22 2 yxyxdA

∑∑∑ +⋅+= dAydAyxdAx 22 2

AyxAx 22 02 +×+=

0i.e.GGGG about dA of M.I.. snce, ==dAzy

GGLM IAxAyAxI +=+=∴ 222

Perpendicular axis theorem:

Statement:

It states that “the M.I. of plane lamina about centroidal axis perpendicular to the plane

of lamina is equal to the sum of its M.I. about two mutually perpendicular centroidal axis in

the plane of the lamina.

YYXXZZ IIIei +=..

Proof:

Let, A plane laminma of Area A, X and Y axis are

its centroidal axis lying ni its plane. Z-axis is its centroidal

axis lying in the plane perpendicular to it.

Let, elementary area dA at a distance y, x and r from X,

Y, and Z axis respectively.

The M.I. of this elementary area dA about Z-axis,

2rdAdI ZZ ⋅=

( ) 2222 dAydAxyxdA +=+=

M.I. of whole area about Z-axis is

∑= ZZZZ dII

( )∑ += 22 dAydAx

∑∑ += dAydAx 22



22 AyAx +=

YYXX II +=

LEFT 3 PAGES

(ii) M.I. of triangular section:

Now, h

b

yh

x =−

(by the relation of similar triangle)

( )yhh

bx −=∴

( )dyyhh

bxdy −== (dA) Strip of Area

Moment of inertia of this strip about base;

2ydAdIb ⋅=

( ) dyyyhh

b ⋅⋅−= 2

M.I. of whole triangular section about its base,

( )∫∫ ⋅⋅−==hh

bb dyyyhh

bdII

0

2

0

( )∫ −=h

dyyhyh

b0

32

h

yyh

h

b

0

43

43

−=

−=

43

44 hh

h

b

=

12

4h

h

b

12

3bhI b =∴

M.I. about centroidal axis We know,

2AxII GXb +=

2, AxIIor bGX −= ( ) ( )22 3tan.. hAcedisAei ×=×

YYXXZZ III +=∴

h

x

s

C.G.

h/3



23

32

1

12

×××−= hhb

bh

2812

33 bhbh −=

36

3bhI GX =∴

M.I. of circular section:

Let, a circular section of radius r and diameter D. Suppose an elementary ring of

thickness dr and radius ‘r’.

Now, area of elementary strip ring = 2πr×dr

M.I. of this elementary ring (dIzz) = 2πr×dr×r2

or, dIzz = 2πr3×dr

M.I. of whole circular area about 2-axis:

∫=R

ZZZZ dII0

∫ ⋅=R

drr0

32π

24

24

0

4 RrR

ππ =

=

Since, 2DR =

( )

322

2 44 DDI ZZ

ππ ==∴

32

4DI ZZ

π=∴

We know,

YYXXZZ III +=

Since, IXX and IYY are symmetrical,

So, IXX = IYY

r

R

drx x

y

y



XXXXZZ III +=∴

322

1

2

1,

4DIIor ZZXX

π⋅==

64

4DII YYXX

π==∴

Q-1. The flanges and web of a 15cm × 7.5cm is classed section are 9mm and 6mm

respectively. Find the position of C.G. of section and its IXX and IYY.

So|n:

fig (1): A1 = 75×9 = 675mm2

x1 = 75/2 = 37.5

fig (2): A2 = (150-2×9)×6

x2 = 6/2 = 3mm

fig(3): A3 = 675mm2 (since A1 and A3 are symmetrical)

x3 = 37.5mm

675675792

5.3767537925.37675

321

332211

+×+×+×=

++++

=∴AAA

xAxAxAx

mmx 7.24=∴

Now, for fig(1);

23

12Ax

bdI GX +=

( ) 23

2

9

2

150975

12

975

−××+×= ( )5.475.. −ei

5.4758725.506 +=

75.48093=

For fig(2);

( ) 2

3

12

181506AxI GX +−×= [since, h = 0, Ax2= 0]

1149984=

For fig(3);

6mm

9mm

1.5 cm

7.5 cm

9 mm1

2

3



( ) ( )[ ]

hAI GX

23 5.475975

12

975 −−×+×=

5.4758725.506 +=

75.48093=

Again for fig(v);

( ) ( )

2

23 7.245.37759

12

759

hAI GY

−×+×=

=∴ GYI

For fig(v);

( ) ( ) ( )

2

23 37.24618150

12

618150

hAI GY

−××−+×−=

=∴ GYI

For fig(3);

23

12Ax

dbI GY +=

( )[ ]23

7.245.3775912

759 −−×+×=

=∴ GYI

Product of inertia:

Strength of Material

Find the centroidal M.I. of shaded area as shown in figure:

So|n: Since, the figure is symmetrical in the Y-axis,

So, mma

ayy 8.15

8.996

15748===∑∑

2AxII GGLM +=

Component Area Distance from L-L ay

1) ½ ×15×30 225 10 2250

2) 30×30 900 15 1550

1

2

3 cm

3 4

y

30mm30mm

yy



3) π ×(15)2/2 353.40 6.37 (-)2251.2

4) ½ ×15×30 225 10 2250

Σa = 996.8 Σay =

15748

mma

ayy 8.95

8.996

15748===∴∑∑

321 XXXXXXXX IIII ++=

( ) ( ) ( ) ( ) ( ) ( )

−+−−++−×+= 2

42

32

3

37.68.154.353428

30158.15900

12

3030108.15225

6

3015 π

468691mm=

Also, ( ) ( ) ( ) ( )2343

23

51522530

1530

428

30

12

3030515225

6.3

1030 ++×+

−×+++×=YYI

46.233244 mm=

# Find the centroidal M.I. of the given figure:

So|n:

Fig(1): 21 4503030

2

1mma =××=

mmx 853

304020151 =+++=

mmy 10303

11 =×=

Fig(2): 22 18003060 mma =×=

( ) mmx 4540202

1152 =++=

4

15mm 20mm 40mm 30mm

y

xI xII

yI

30mm32

1

C.G

15



mmy 152

302 ==

Fig(3): ( ) 2

2

3 71.1764

15mma == π

mmx 3520153 =+=

mmy 153 =

Fig(4): ( ) 2

22

4 43.3532

15

2

1mmra ==×= ππ

mmR

x 63.83

15445

3

4154 =×−=−=

ππ

mmy 154 =

Now, mmaaaa

xaxaxaxax 84.47

4321

44332211 =+−+

+−+=

mmaaaa

yayayayay 07.14

4321

44332211 =+−+

+−+=

so, coordinate of C.G. = (47.84 , 14.07)

=+ 2AxI GG

Now, for fig(2)

432

|1

||| XXXXXXXXXXIIIII +++=

Now, ( )2

11

32

3

36

3030

361yyaAx

bhI XX −×+×=+=

( )23

1007.4445036

3030 −×+×=

=

For fig(2);

( )2

2

22

3

306012

3060

122yyAx

bdI XX −××+×=+=

( )21507.14180012

90060 −×+×=

=

For fig(6);

24

643Ax

DI XX += π

( )23

4

1507.1464

15 −×+×= aπ

( )24

07.141571.17664

15 −+×= π



=

For fig(4);

( )07.141543.353128

4

4−+= D

I XX

π

So, =+−+=4321 XXXXXXXXXX IIIII

Again, 4321

|| yyyyyyyyyyIIIII +−+=

( ) ( ) =−+×=−+×= 85.478545036

3030450

36

3030 3

1

3

1xxI yy

( ) ( ) =−+×=−+×= 23

23

4585.47180012

60301800

12

60302

xxI yy

( ) ( )2

4

3585.4771.17664

153

−+= πyyI

( ) 24

3

48515.4743.353

128

304

++=π

π RI yy [due to half circle]

LEFT A LOT OF PAGES

Q-1. Find the product of inertia (POI) for the plane hatched area about the axes XX and YY

as shown in figure.

So|n: For figure (1);

( ) ( ) 1111 bAaII XgYgxy +=

( )( )203060400 ×+=

4410144 mm×=

Since, AabII XgYgxy +=

69.2676321 =+=∑ aaaa

33.29=X and 75.27=y

For fig (2);

( ) ( ) 22222 baAII XgYgxy +=

22

22

72bAa

hb +−=

( ) ( ) ( )( )50206030

2

1

72

6030 22

××+−=

44105.85 mm×=

For fig (3);

( ) ( ) 33333 baAII XgYgxy +=

20

1

2

3

10mm

10mm

60mm

x

Y



( )

+××+=π

π3

41030

2

200

2 R [since, R = 20mm]

44085.34 mm×=

Now, ( ) ( ) ( )321 xyxyxyxy IIII ++=

444 1085.34105.8510144 ×−×+×=

mmI xy41065.194 ×=∴

Since,

yxAII XYYX+=||

yxa ××+×= ∑41065.194

75.2733.2969.26711065.194 4 ××××=

44101.42 mm×=

Q-2. Find principle moments of inertia and directions of principal axes for angle section

shown in figure.

So|n: Since,

XXYY

XY

II

I

−=

22tan θ and

( ) ( )22

22 XYXXYYYYXX

UU IIIII

I +−

++

=

Now, 21 24212 cma =×=

cmx 62

121 ==

cmy 12

21 −=−=

And, ( ) 22 362220 cma =×−=

cmx 12 =

cmy 112

1822 −=

−−=

Now, cmaa

xaxax 3

21

2211 =++

=

cmaa

yayay 7

21

2211 −=++

=

Now, ( ) ( )21 XXXXXX III +=

2

23

1

23

7212

++

+= Ax

bdAx

bd

( ) ( ) ( ) ( )

−×+×+

−+×= 2

32

3

71118212

1821724

12

212

30mm40mm

C.G

12 cm

1

230 cm

2 cm

M

2 cm

M



42420cm=

And, ( ) ( )21 YYYYYY III +=

( ) ( ) ( ) ( )

−×+×+

−×+×= 2

32

3

1318212

21836122

12

122

4660cm=

Now, ( ) ( ) ( )21 XYXYXY III +=

[ ] [ ]2211 00 bAabAa +++=

( )( )[ ] ( )( )[ ]241823617122 ×+−−×=

4720cmI XY =∴

now, 810.02420660

720222tan −=

−×=

−=

XXYY

XY

II

Iθ

810.2tan, −=θor

( )810.0tan2, 1 −= −θor

( )2

810.0tan,

1 −=−

θor

º7.19−=∴θ or º3.70

Again, ( )22

22 XYXXYYYYXX

UU IIIII

I +

−+

+=

42677cm=

And, ( )22

22 XYXXYYYYXX

VV IIIII

I +

−−

+=

4403cm=

Also, θθ 2222

|| SinICosIIII

I XYXXYYYYXX

XX−

−−

+=

Since, θ = -19.7º

42677|| cmIXX

=

B.M.D., S.F.D. and A.F.D: Bending Moment (B.M.):

Turning effect of a force is called moment (B.M.). It is represented by M.

It is also defined as product of force and perpendicular distance.

Shear force (S.F.):

Algebraic sum of vertical force is called S.F.

Axial force = Normal force

Shear force = Tangential force



In cruss, generally,

Bottom member = Tensile

Top member = compressive x'G5.

aLrsf] member calculation ug'{ k5{ / dfq yfxf x'G5

Axial Force (A.F.):

The force acting longitudinal axis of the beam is called axial force (A.F.)

In Beam for S.F. calculation:

From left (+)ve and (-)ve (to right)

From right (-)ve and (+)ve (to left)

Draw B.M. and S.F. for given figure.

So|n: ∑ = 0BM

021054107228 =×−+×−××−×AR

020540288, =−+−=ARor

8

588,

−=ARor

37.108

83 ==∴ AR

Now, ∑ = 0dy (i.e. vertical load)

010104 =−−−+ BA RR

0248

833 =−+R

63.138

109

8

83192 ==−=∴ BR

37.10=∴ AR

0.5

10XN

10KN 10KN

0.5M

2M 2M 2M 2M

10×0.5=5KN-M sf] moment

E

2M 2M 2M 2M

5KN-M

CA

CA

2 KN/M



63.13=∴ BR

B.M.D. (+)ve

Span AC:

2

2x

xxRM AX ⋅=×=

2

237.10

2xx −×=

237.10, xxMor X −=

Since, this equation is quadratic, so, its shape is parabolic.

At, x = 0, M0 = 0

At, x=1, M1=9.37

At, x=2, M2=16.74

Span CD:

( )122 −×−×= xxRM AX

4437.10 +−= xx

437.6 += x

At, x=2, M2=16.74 [Same as above. Hence checked.]

At, x=4(left), M4=10.37×4-4(4-1)=29.48

Span DE:

( ) ( ) 541012 +−−−−= xxhxCM AX

( ) ( ) 54101437.10 +−−−−= xxx

Now, at x=4(right) M4=29.48+5=34.48

At, x=6, M6=10.37×6-4(6-1)-10(6-4)+5=27.22

Span EB:

( ) ( ) ( )61054101437.10 −−+−−−−×= xxxxM X

At, x=6, M6=27.22

x=8, M8=82.96-28-40-20+5=87.96-88=-0.04≈ 0

CA

x

KA

2 M

CA

x

2 M 2 M

D

CA

x

2 M 2 M

D E B

2 M 2 M 2 M

CA

x

2 M 2 M

D E B

2 M 2 M

CA

D E B

o

9.3716.14 29.48

34.48

27.22

10.37

6.37 6.37

-3.37 3.36

13.63

2M

x

RD



For S.F.

Span AC:

xxRF AX 237.102 −=×−=

At, x=0(right), F0(right)=10.37

X=2, F210.37-2×2=6.37

Span-CD:

37.6437.102 =−=×−= AX RF

At, x=2, F2=6.37

x=4(left), F4(left)=6.37

Span DE:

63.31022 −=−×−= AX RF

At, x=4(right), F4(right)=-3.63

X=6(left), F6(left)=-3.63

Span EB:

63.132437.10101022 −=−=−−×−= AX RF

At, x=6(right), F6(right)=-13.63

x=8(left), F8(left)=-13.63

Now, at x=8(right)

BA RRrightF +−−×−= 101022)(8

063.132437.10 =+−=

LEFT 3 PAGES

Draw B.M.D and S.F.D for given figure:

CA D

2M 2M

x

CA

2M 2M 2M

x

CA

x

D E

Io Io

2M 2M 2M

B

RB

2M

Ra

AD

10 KNInternalHInge

2 XN/m

2 XN/m

B E

C

2 m 2 m 2 m 2 m 2 m

3.67 3.67

4.67

0.67++

-

-

parabilic

A B E CD

1.33

6.33 6.33

S.F.D

0.5320.4625



So|n:

67.3=∴ AR

11=∴ BR

=∴ CR

For bending moment diagram (B.M.D.):

Span AD:

xxRM AX 67.3==

0,0, 0 == Mxat

34.7,2, 2 == Mxat Span DB: ( )210 −−×= xxRM AX

34.7,2, 2 == Mxat

32.5,4, 4 −== Mxat Span BE:

( ) ( ) ( ) ( )2

4424210

−×−×−−×+−×−×= xxxRxxRM BAX

( ) ( ) ( )2441121067.3, −−−+−−= xxxxMor X

32.5,4, 4 −== Mxat

65.1,5, 5 −== Mxat

0,6, 6 == Mxat

Span EC:

( ) ( ) ( ) ( ) ( )63

166

2

1

2

222224210 −××−×−×−

++−×−−+−−= xxxxxRxxRM BAX

( ) ( ) ( ) ( )366

15441121067.3 −−−−−+−−= xxxxx

0,6, 6 == Mxat

523.0,7, 7 == Mxat

4625.0,5.7, 5.7 == Mxat

A

x

RA

10 KN

D

A

x

RA

D

11 KN

B

2

A

x

RA

D

10 KN

2M

B

RB

2M

A

x

D

10 KN

2M

B

RA RB

2M

E

2M

2 KN/m

y 2

x-6

2



1498.0,9.7, 9.7 == Mxat

0026.0,8, 8 ≈== Mxat

For S.F.D:

Sign Convention:

From left (+)ve and from right (-)ve

From left (-)ve and from right (+)ve

Now, S.F. at lrft of A = 0

Span AB:

67.3== AX RF [(+)ve due to upward]

67.3)(),(0, == rightFrightxat

67.3)(),(2, 2 == leftFleftxat

Span DB:

33.61067.310 −=−=−= AX RF

33.6,2, 2 −== rightFrightxat

33.6,4, 4 −== leftFleftxat

Span BE:

( )4210 −×−+−= xBAX RRF

( ) ( )4267.442111067.3 −−=−−+−= xxFX

67.4,4, 4 == rightFrightxat

67.0,6, 6 == Fxat

Span EC:

( ) ( )662

12210 −×−×−×−+−= xxRRF BAX

( )

2

64111067.3

2−−−+−= x

( )

2

667.0

2−−=∴ xFX

67.0,6, 6 == Fxat

545.0,5.6, 5.6 == Fxat

388.0,75.6, 75.6 == Fxat

33.1,8, 8 −== leftFleftxat

033.133.133.1,8, 8 =+−=+−== CRrightFrightxat

FRAME:

A

x

10 KN

2

x-4

2

A

x

2 2

D E

RA RB

2 x-6

B D

D

10KN

A E2M 2M

2 KN

4 m

RA V RE V



So|n: ΣME = 0

02

4422104 =××−×−×AVR

94

1620, =+=AVRor

9=∴ AVR

Now, ∑ = ;0YF then,

010 =+− EVAV RR

1910, =−=EVRor

∑ = ;0XF then,

042 =×−AHR

8=∴ AHR

Now,

Span AB:

xxRM AHX 8−=×−= Since, xM X 8−=

0,0, 0 == Mxat

32,4, 4 −== Mxat

Span BC:

4894 ×−=×−×= xRxRM AHAVX

329 −=∴ xM X

32,0, 0 −== Mxat

14,2, 2 −== Mxat

Span CD:

( )2104 −−×−×= xRxRM AHAVX

( )210329 −−−=∴ xxM X 14,2, 2 −== Mxat

16,4, 4 −== Mxat

Span ED: (+)ve

2

22 x

xxM X −=××−=

0,0, 0 == Mxat

RAVRAH

RAV=9

RAH

x

4m

B

A

x

4m

2

(x-2)

B D

C

10

2 KN

x

A E

RAV = 9 REV 21

4M

RAH



4,2, 2 −== Mxat

16,4, 4 −== Mxat

�Q� Draw B.M., S.F. and A.F. diagram

[Hint or clue:

When hinge is not given i.e. as internal hinge then,

no. of unknown = 4

no. of equilibrium = 3

∴Degree of indetermacy = 4 – 3 = 1

If internal hinge is given then,

no. of unknown = 4

no. of determacy = 3+1 =4

∴Degree of determacy = 4 – 4 = 0

Hence structure is determinate.]

� Internal hinge cfpbf hinge af6 left / right port equilibrium u/]/ So|n ug]{ .

Now, ΣMF = 0

Sign Convention:

And force is (+)ve

And force is (-)ve

So, ΣMF = 0

023

122

2

122

3

222

2

12104 =

××××−

×××××−×+×AVR

033.433.20204, =−−+AVRor

-32

-32

-14-16

-16

-

--

-

parabolic

B.M.D 2KN/m2KN/m

InternalHInge

ECD

10KN

A F

2M

2M

2M 2MRAV=3

RFV = 7

RFH = 2.84



[ ]

34

12

4

830.20 −==−+−=∴ AVR

3−=∴ AVR i.e. downward direction.

Now, ΣFY = 0;

0222

122

2

13 =+××−××−− FVR

34, +=FVRor

7=∴ FVR i.e. upward direction.

For horizontal force calculation:

Let, left of internal hinge is in equilibrium and moment about D;

ΣMD = 0

023

222

2

12

2

1210423 =

×××−×−×−×+×− AHR

03

8206, =−−+− AHRor

16.734

832, =

×+×=AHRor

16.7=∴ AHR

Now, ΣFX = 0;

010 =+− FHAH RR

01076.7 =+− FHR

84.2=∴ FHR [(+)ve , direction]

B.M. calculation:

Sign convention:

& � (+)ve

& � (-)ve

Span AB:

xxRM AHX 16.7=×=

0,0, 0 == Mxat

32.14,2, 2 == Mxat

For Span BC:

( ) ( )21016.7210 −−=−−×= xxxRM AHX

32.14,2, 2 == Mxat

64.8,4, 4 == Mxat

RAVRAH

x

B

x

2

2

A

C

RAHRAV



2KN/m2KN/m

ECD

10KN

A FB

RAH RAV

RFH

RFV

For Span CD:

×××−××−×−×+×−=32

1

222104

xxx

xxRxRM AHAVX

−−−×+−=

620416.73,

32 x

xxMor X

64.802064.280,0, 0 =−−+== Mxat

002.097.293033.164.28306

842064.286,2, 2 ≈−=+−=++−=+−−+−== Mxat

For Span DE:

( ) ( )

−×−×−×−

×−××−×−×+×−=3

222

2

12

3

122

2

12104

xxxxRxRM AHAVX

( )

6

2

3

222064.283,

3−−

−−−+−= xxxMor X

x

C D

BA

x

A3

3

x 2-x

2

2

g

B

C

D

2

x 2-x

F

ARAH= 7.16

RFH= 2.84

RFN

D

anti clockwisedirection

x

(x-2)

(x-2)3



+

-

-+

+

8.64

8.69

14.32

C

B

A F

ED

-11.36

11.36

0,2, 2 == Mxat

36.1133.167.664.812,4, 4 −=−−+== Mxat

Span FE:

xxRM FHX 84.2−=×−=

0,0, 0 == Mxat

36.11454.2,4, 4 −=×−== Mxat

Dflysf calculation at Pt. E / ofxf+sf] calucaltion at PT. E

value same ePsf]n] calculation correct 5 .

After dia plot:

(+)ve aflx/ / dfly /

(-)ve nfO{ eLq / tn plot ug]{ .

For S.F.

S.F. at bottom

Span AB:

S.F. at bottom of A = 0 from left (+)ve

16.7== AHX RF

16.7,0, 0 == topFtopxat

16.7,2, 2 == bottomFbottomxat

Span BC:

84.21016.710 −=−=−= AHX RF

84.2,2, 2 −== topFtopxat

84.2,4, 4 −== Fxat

Span CD:

××−×−−= xxxRF AVX 2

12

−=

223

2xx

3,0, 0 −== Fxat

5234,2, 2 −=+−== Fxat

Span DE:

( ) ( ) ( )2

22322

2

122

2

1 2−−−−=−×−×−×−−= xxxRF AVX

5,2, 2 −== Fxat

725,4, 4 −=−−== Fxat



[Since, EFFE ≠ ]

Span FE:

At bottom of F, 0=bottomFF

84.2== FHX RF

F RFH

RFH

A3710

3

7.15

2.84

+

-

-

2.824 2.824

3 7

3 7

10

2 2



LEFT MANY PAGES

Q. A beam of rectangular section 20cm wide and 40cm deep is simply supported over a span

of 5m. It is loaded with a point load od 10KN at the centre. Calculate the maximum shear

stress.

So|n: Given,

b = 20cm, d = 40cm

KNF 52

10 == (due to symmetrical)

( ) 4

33

67.10666612

4020

12cm

bdI =×==

Now, ( ) ( ) =

××××=

⋅=

2067.106666

22020205)stress(shear

bI

yFAq

Since, For maximum shear in section, take neutral axis unless specified.

Now,

Q. A timber beam 100mm wide and 250mm deep is simply supported over a span of 4m.

Find the uniformly distributed load that can be applied over the whole span so that the

deflected of the centre may not exceed 6mm. Take E = 1.12×104N/mm2.

So|n: Since, EI

wly

384

5 4

=

Now, given,

y = 6mm

l = 4m = 4×1000 = 4000mm

E = 1.12×104N/mm2

I = ?

Now, ( ) =×==

12

200100

12

33bdI

So, we know,

EI

wly

384

5 4

=

( )

4

4

1012.1384

40006,

×××= w

mmor

=wor,

Q. A timber contilever beam 200mm wide and 300mm deep is 3m long. It is loaded with udl

of 3KN/m over the entire length. A point load of 2.7KN is placed at the free end of

contilever. Find the maximum bending stress (σmax) produced.

So|n: given,

b = 200

40 cm

20 cm

w/unit legk

4d= 300mm

b = 200mm

3 KN/m

3 m

2.7 KN



l = 1200

w

d = 300

w = 3KN/m

D = 2.7KN

σmax = ? (bending Stress)

now, ( ) 48

33

510.412

300200

12mm

bdI =×==

48105.4 mmI ×=∴

2

33337.2max ××+×=M [Since, B.M. maxm & always at fixed end for

contilever]

NKNM 4max 1026.26.21 ×==∴

Now, yI

M maxmax σ=

I

yM ×= max

maxσ

8

4

105.42

3001016.2

×

××= [since, y = d/2]

mmN /102.7 3max

−×=∴σ

Q. A beam of size 25mm × 25mm is carrying the maxm contrd load of 20KN as a simply

supported span of 600mm. The beam of same material but of size 25mm wide and

100mmdecs.

So|n: Given,

l = 600mm

For simply supported beam,

yI

M maxmax σ=

But, 4max

wLM =

4222.B.M. Since, max

wLLwLFS =×=×=

?max =σ

Now, I

dWL

I

yM 24maxmax

×=

×=σ

?max =∴σ

30 KN

l=600

b = 25 mm

d = 25 mm



For contilever,

same=maxσ [Since, σmax is same for same material for both simply and contilever

beam]

1200max ×=×= wLwM

( )12

10025 3×=I

2dy =

Now, yI

M σ=

II

wor

σ=×1200,

y

Iw

×=× σ1200

Derive pure bending, elastic bending and plastic bending.

Elastic Bending = Elastic Limit leqsf] load g} x'g] bending nfO{ g} Elastic Bending elgG5 . Plastic Bending = Plastic Stage df x'g] Bending x?n] laser df x'g] bending stress same x'G5 .

Q. Three equal loads of 52KN have been supported by a simply supported beam of 12m

dividing the beam in four equal parts. Does the beam support the load is the allowable

bending stress is 50N/mm2 and section is 100mm(b) and

50mm(d).

So|n:

( )

12

50100 3×=I

12..

3bdei

=∴ I

And, 152

30

2=== d

y

Elastic B. Stress diagramPlastic Stress B. Stress diagram

5 KN5 KN5 KN

7.5

b = 100

d = 50



3525.7 ×−×=CM (moment at centre i.e. Mmax)

=∴ CM

Now, yI

M σ=

I

yM ×=σ

=σ

If σ < 50 then support

If σ > 50 then not support

Q. A cast iron 540mm dia & 80mm wall thickness is running full of wqater and supported

over a length of 8m. Determine the maxm stress intensity in the metal if the density of C.I. is

72KN/m3 and that of water is 10KN/m3

So|n: σ = ?

==64

4dI

π

== 2Dy

==8

2

max

wlM

yI

M σ=

d = 540 = 0.54 (dia)

D = 0.57

( )

4

54.0 2πWW SM = (mass of water)

( ) ( )

−= 22 54.0

457.0

4

ππρWPm

88.1=

17.4=+= PW mMw

Q. A rectangular beam 100mm wide and 200mm deep and 4m long is simply supported at

ends. It carries a UDL of 5KN/m run over the entire span. If this load is removed and two

loads wKN each are placed at 1 meter from each end, calculate the greatest value which may

be assigned to the load so that the maxm B.S. (bending stress) remain same as beam.

So|n: 8

.2

max

wlMB =

==12

3bdI

w KN/m

Part- I 100

200

Part- II

w N

W Rent W Rent

2M1M 1M



2

200

2== d

y

yI

M σ=∴

=maxσ

Part-II

B.M. maxm at centre ==×−×= mKNwww /12

==12

3bdI

σ = (known from above) =

yI

M σ=∴

yI

wor

σ=,

=×=∴y

Iw

σ

Determine the equation of elastic curve of (i.e. deflection curve line) contilever beam

supporting a UDL of intensity w over its port of length as shown below.

Solution:

At 0, ==dx

dyLx

[ ]EIMMdx

ydEI xx =∴==

2

2

xM sf] 1st integrant → slope

Slope sf] integrant → deflection

TORSION: Torque/ Twisting moment/ Turning moment/ Torsion/ Process Torsion:

Assumptions: (i) plane normal section of shaft remain plane twisting i.e. no wrapping or distortion

of parallel plane normal to the axis of the member takes place.

(ii) Torsion is uniform through the length i.e. all the normal cross-sections which are

at the axil distance suffer equal relative rotation.

(iii) Radia remain straight after torsion.

(iv) A stress is proportional to strain i.e. all the stresses are within the elastic limit.

(v) The material is homogeneous and isotropic.

3 m

x

w/units

L

(L-a)

A



Relation between twisting moment, twist and shear stress:

Let a torque T applied at the free end of shaft. So, balancing torque of equal magnitude

but opposite in direction in induced at the other end.

Due to the torque the radial ive OA shifts to OA1 and CA shift to CA1.

Let, θ=∠ 1AOA

φ=∠ 1ACA

11 φ=∠BDB

Let, qs = intensity of shear at the surface of the shaft.

Then, cqS=φ

Where, c = modulus of rigidity

And, L

R

CA

R

CA

OA

CA

AA θθθφ ==⋅

== 11

Hence, )(iL

R

c

qS −−−−−= θ

φSq

c ==strainshear

stressshear Since,

Similarly,

Let, ‘DB’ shift to ‘DB1’

Let, ‘DB’ is at radius ‘r’ from the axis of shaft.

Let, ‘q’ be the shear stress at layer of radius ‘r’, then,

c

q=φ

L

r

DB

BB θφ == 11&

So, L

r

c

q θ=

)(iiL

c

r

q −−−−−= θ

From equation (i) and (ii),

B

Bθ o

O

O

T

A1

L

A

T CD

A1B1

θ

A

B

Torque

Note: D & B line are joined so that DB//CA. Also DB=CA



L

c

r

q

R

qS θ==

Maximum torque transmitted by a solid circular shaft: Suppose a solid circular shaft of radius ‘R’. Let an elementary ring of thickness ‘dr’ at

radius ‘r’.

Let, qs = shear stress at outer most surface i.e. at layer R

q = shear stress at radius r.

Area of elementary ring = drr ×π2

Since, r

q

R

qS =

R

rqq S ⋅

=∴

Turning force this elementary ring = shear stress (q) × Area of ring (A)

drrR

rqS ××⋅

= π2

drrR

qS ××= 22π

Turning moment of this elementary ring about centre of the shaft = Turning force × r

rdrrR

qS ×××= 22π

drrR

qS ××= 32π

Hence, Turning moment (or Torque) on the whole circular shaft;

∫ ××=R

S drrR

qT

0

32π

∫=R

S drrR

q

0

32π

R

S r

R

q

0

4

4

2

=

π

4

2 4

×=

R

RqSπ

2

3RqS=

2

3RqT Sπ

=∴ for radius of shaft

16

3DqT Sπ

=∴ for diameter of shaft

rR

dr



r

dr

Ri

Ro

Torque transmitted by a hollow circular shaft: Consider a hollow circular shaft of inner radius ‘Ri’ and outer

radius ‘Ro’ subjected to a Torque ‘T’. Let, an elementary strip of

thickness ‘dr’ at a radius ‘r’.

Let, qs = maxm shear stress at the outer most surface

q = shear stress on a surface of radius ‘r’

now, Area of elementary strip ring ‘dr’ drr ×= π2

Turning force on this ring rR

qdrrqdrr

O

S ×××=××= ππ 22

Hence, Turning moment on the whole hollow circular shaft,

∫=Ro

Ri O

S drrR

qT 32π

∫=Ro

RiO

S drrR

q 32π

Ro

RiO

S r

R

q

=

4

2 4π

( ) ( )

−

O

iOS

R

RRq 44

4

2π

( ) ( )

−=∴

O

iOS

R

RRqT

44

2

π

Since, 2

&2

ii

OO

DR

DR ==

( ) ( )

−=∴

O

iOS

D

DDqT

44

16

π

Torque in terms of polar moment of inertia: Polar M.I. of a plane area is M.I. of the area about an axis perpendicular to the plane of

the figure and passing through the ‘C.G.’ of the area and it is denoted by j.

Turning moment on the ring;

drrR

qS 32π=

∴Total torque (or twisting moment);

∫=R

S drrR

qT

0

32π



( )∫×=

RS

dA

drrr

R

q

0

2

also

2π

∫=R

S dArR

q

0

2

where, ∫R

dAr0

2 = Polar moment of inertia (J)

for circular shaft,

32

4DJ

π= [in Z-axis]

Now, JR

qT S=

R

q

J

Tor S=,

Since, L

c

r

q

R

qS θ==

Hence, L

c

r

q

R

q

J

T S θ===

Polar modulus: It is defined as ratio of polar M.I. to the radius of the shaft. It is also called for sional

section modulus and is denoted by ‘zp’.

(a) For solid circular shaft:

32

4DJ

π=

2DR =

34

162

32D

D

D

R

JZP

ππ ===∴

(b) For a hallow circular shaft:

[ ]44

32 iO DDJ −= π

2ODR =

[ ]

[ ]44

44

162

32

iOOO

iOP

DDDD

DD

R

JZ

−=

−==∴ ππ

[ ]44

16 iOO

P DDD

Z −=∴ π

Perpendicular axis theorem = Polar axis theorem



Torsional rigidity:

Since, l

c

J

T θ=

θTl

Jcor =×, where, c × J = Torsional rigidity

L = L

θTL

Jc =×∴ rigidity, Torsional

Expression of strain energy stored in a body due to torsion: Let a solid circular shaft of radius ‘R’ in which torque (T) is applied. Let an elementrary

ring of radius ‘r’ and thickness dr,

Let, D = Dia. of shaft

l = length of shaft

c = modulus of rigidity

U = total strain energy stored in shaft.

Now, r

q

R

qS =

R

rqq S ×

=∴

Area of ring, drrdA ⋅= π2

Vol. of ring, ldrrdV ×⋅= π2

( )

volume2

stressShear energystrain Shear

2

×=c

Hence, shear strain energy in the ring,

ldrrc

rR

q

du

S

×⋅×

= π22

2

drrcR

lqS ⋅×= π22 2

2

∫=∴R

duU0

energy,strain Total

∫⋅×××

=R

S

cR

drrrlq

02

22

2

2π

∫ ⋅=R

S dArcR

lq

0

22

2

2

JcR

lqS2

2

2=

⋅= ∫

R

dArJ0

2where,

r R

dr



For solid circular shaft,

32

4DT

π=

322

4

2

2 D

cR

lqU S π×=∴

( )32

2

2,

4

2

2 R

cR

lqUor S π×=

32

16

2

2

2

2 R

cR

lqS ××= π

lRc

qS 22

4π×=

Vc

qS ×=4

2

Where, V = volume of shaft

vc

qU S ×=∴

4

2

for solid shaft

For hollow circular shaft;

[ ]222 iOO

S DDvcD

qU +×=

Q. The shearing stress in a solid shaft is not to exceed 40N/mm2 when the torque transmitted

is 2000N-m. Determine the diameter of the shaft.

So|n: Given,

qs = 40N/mm2

Torque (T) = 2000N-m

Now, 3

16DqT S

π

34016

2000, Dor ××= π

mmmD 2.13640162000 3

1

=

××=∴ π

Q. A solid circular shaft and hollow circular shaft whose inside dia is ¾ th of the outside dia,

are of same material of equal lengths and are required to transmit a given torque. Compare

the weights of these two shafts if the maximum shear stress developed in the two shafts are

equal.

So|n: Given,

Din = ¾ 30

Di

Hollow Solid

D



Dout = 0.75DO

Torque transmitted by solid circular shaft,

3

16DqT S

π=

Torque transmitted by hollow circular shaft,

O

iOS D

DDqT

44

16

−= π

( )

O

OOS D

DDq

44 75.0

16

−××= π

36836.016 OS Dq ×××π

Now, Torque transmitted by solid circular shaft = Torque transmitted by hollow circular shaft

6836.01616

.. 33 ×= DqDqei SS

ππ

ODD 8809.0=∴

Now, wf. of solid shaft gvρ= since, mg =ρ & vmwf ×=

)(4

2

iLD

g −−−−−××= πρ

wf. of hollow circular shaft ( ) LDDg iO ×−×= 22

4

πρ

)(4375.04

2 iiLDg O −−−−−××= πρ

Dividing equation (i) by (ii):

7737.14375.0

4

42

2

4

3 =××××

××=

LDg

LD

g

w

w

O

πρ

πρ

7737.14

3 =∴w

w

Q. Calculate the maxm intensity of shear stress induced and the angle of twist in degrees for a

length of 10m for a solid shaft of 100mm dia. transmitting 112.5KW at 150R.P.M.. Toxe c =

8.2×104N/mm2 for material of shaft.

Hint: since, N

PTTorque

π2

60)( =

Where, N is in R.P.M.

P = Power in watt

N = R.P.M.

T = Torque



So|n: Given, P = 112.5KW = 112.5×103

N = 150 R.P.M.

c = 8.2×104N/mm2

now, mKNN

PTTorque −=

×××== 162.7

1502

105.11260

2

60)(

3

ππ

again, 3

16)( DqTTorque S

π= (for solid circular shaft)

( )

23

6

/5.361001416.3

10162.716, mmNTor =

×××=

Now, l

c

R

qS θ=

=2

,given then is Since,D

RD

41082

100010

50

5.36

×××=×=

c

l

R

qSθ

radian089.0=∴θ

Q. Select a suitable dia. of solid shaft of circular section is transmit 112.5KW of power at

200R.P.M.. If the allowable shear stress is 75 N/mm2 and allowable twist is 1º in a length of

3m. Take c = 0.082×106N/mm2.

So|n: Given,

P = 112.5KW

N = 200R.P.M.

c = 0.082×106N/mm2

θ = 1º (maxm)

l = 3.0m

qs = 75N/mm2

now, we know,

2002

5.11260

2

60)(

××==

ππN

PTTorque

mmNmKNT −×=−=∴ 10637.537.2

Again, for strength;

3

16DqS

πτ = (for solid circular shaft)

36

161037.5, Dqor S ××=× π

37516

D××= π

mmD 4.7175

161037.5 3

16

=

×××=∴

π

Again, we know,



l

c

J

T θ= 32

4D

J

T π=

Where, T, c, θ, l are known by this relation, D is known.

Now, cπ=º180

180

º1π=∴

Since, L

c

J

T θ=

18010082.0

1000310372.5

6

6

πθ ××

×××==c

TLJ

..........32

,4

=Dor

π

mmD 5.103=∴

Hence, this diameter is greater than diameter from strength criteria. So,

Diameter, mmD 5.103=

Q. A hallow circular shaftof external diameter 150mm and transmits 200KW power at

200R.P.M.. Determine maxm internal diameter if the maxm stress in this shaft is not to exceed

60N/mm2

So|n: Given:

N = 100R.P.M.

P = 200KW

qs = 60N/mm2

DE or DO = 150mm

Now, we know,

mKNN

DT −=

××=×= 95.190

1002

20060

2

60

ππ

mmNT −×=∴ 61098.190

Now, O

inOS D

DDqT

44

16

−= π

[for hollow circular shaft]

( )

−××=×

150

15060

161098.190,

446 inD

orπ

=inDor,

473.209=

Q. A hollow shaft 3mm outer dia rans at 400R.P.M. against a power of 50KW. Find the inner

diameter of the shaft so that shear strain does not exceed 1/1000 . Take c = 8×106N/cm2.

So|n: Given,



1000

1=φ

226 /108/108 mmNcmNc ×=×=

Now, N

DT

π2

60=

4002

5060

××=

πT

Again, strainshear ×= cqS φ×= cqei S..

1000

1108, 6 ××=qor

23 /108 mmNq ×=∴

Now, O

iOS D

DDqT

44

16

−= π

=∴ inD

Q. A shaft of dia 5m and length 40m by power 200KW at 20R.P.M.. Does the shaft transmits

the power safely if the permissible stress is 50N/mm2.

So|n: 202

60

2

60

××==

ππN

DT

Now, 3

16DqT S

π=

Is

=>

=N

DTDqT S π

π2

60

163 then the shaft is safe otherwise non-safe.

i.e. Applied T should be less than design T.

i.e. Desgin, 3

16DqT S ××= π

and, applied, N

DT

π2

60=

Q. A solid shaft of 150mm diameter is to be replaced by hollow shaft of the same

material.The internal diameter equal to 60% external diameter. Find the saving in material if

the maxm allowable shear stress is the same for both the shaft.

So|n: Given,

Now, 3

16DqT S

π= (for solid shaft)

=STor,

And, O

iOSH D

DDqT

44

16

−= π

(for hollow shaft)



=Dor,

=×== lDvmsolid2

4

πρρ

And, =

−×=

44

22iO

hollow

ddm

ππρ

Now, solving material, m = msolid – mkollow

=

* Thickness < (1/15 to 1/20) of diameter (internal) � The vessel which has its thickness less than 1/15 to 1/20 of internal diameter is called thin

walled vessel. It is used to keep the fluid under pressure. Due to the pressure of the fluid the

stresses in the wall of the cylinder on the cross-section along axis and cross-section

perpendicular to the axis are set up. These stresses are tensile and called as:

- circumferencial stress or hoop stress

- longitudional stress

The stress acting along circumference of the cylinder is called circumferential stress where as

the stress acting along the length of the cylinder is calleds longitudinal stress.

Expression for circumferential stress:

Let us consider a thin cylindrical shell under fluid pr..Let the circumferential stress developed

in the cylinder as in figure.

Let, p = internal fluid pr.

d = internal dia. of the cylinder

t = thickness of the cylinder

σ1 = circumferential stress

L



Force due to fluid pressure = pressure × area on which pressure is acting

)(iLdp −−−−−××=

Now, Force due to circumferential stress

( )tLtL ×+××= 1σ

tL21 ×= σ

tLtL 11 22 σσ ==

t

pd

21 =∴σ

Expression for longitudinal stress:

Let us consider a thin cylindrical vessel under fluid pressure.

Let, longitudinal stress developed in vessel as shown in figure.

Now, Force due to fluid pressure,

= p × area on which pressure is acting

)(4

2

id

p −−−−−×= π

Again, Force due to longitudinal stress,

)(2 iide ee −−−−−×= πσ

Since, equation (i) and (ii) are same,

So, dtd

p πσπ ×=× 2

2

4

t

pd

42 =∴σ s t

pd

t

pdofei

422

1

2

1.. 12 =×== σσ

Effect of pressure on the dimension of a thin cylindrical shell:

Let us consider a thin cylindrical shell, having internal fluid pressure ‘P’

Let, σ1 = circumferential stress = pd/2t

σ2 = longitudinal stress = pd/4t

L = length of cylinder

t = thickness of cylinder

L

d L

t



E = modulus of elasticity

µ = Poisson’s ratio

δd = change in diameter

δL = change in length

δv = change in volume

now, e1 = strain along circumferential

e2 = strain along longitudional

then, EE

e 211

σµσ−=

)(2

11

2

42i

tE

pd

E

tpd

E

tpd −−−−−

−=−= µµ

Now, EE

e 122

σµσ −=

tE

pd

tE

pd

24µ−=

)(2

1

22 iitE

pdP −−−−−

−=∴ µ

But, circumferential strain,

ncecircumfere original

ncecircumferein Change1 =e

( )

d

ddd

ππδπ −+=

)(iiid

d −−−−−= δ

Longitudinal strain;

l

le

δ==length original

lengthin Change2

)(2 ivl

le −−−−−=∴ δ

Now, equating (i) & (i) and (iii) & (iv), then,

−= µδ2

11

2tE

pd

d

d

−=∴ µδ2

11

2

2

tE

pdd

Similarly from equation (ii) and (iv),

−= µδ2

1

2tE

pd

L

L

−=∴ µδ2

1

2tE

pdLL



Now, volumetric strain:

Volumetric strain V

vδ=

We know, Ld

v ×=4

2π

Now, Final volume, ( ) ( )LLdd δδπ +×+= 2

4

( )[ ] ( )LLdddd δδδπ +×⋅++= 24

22

( ) ( )[ ]LddLdLdLdddLLd δδδδδδδπ ⋅⋅+×+×+×⋅++= 224

2222

Neglecting smaller quatities,

Final volume [ ]LdddLLd δδπ ⋅+⋅+= 22 24

Now, change in volume, ( ) [ ] Ld

LdddLLdv ×−⋅+×+=4

24

222 πδδπδ

[ ]LdddLvor δδπδ ⋅+⋅= 224

,

Now, volumetric strain, [ ]

Ld

LdddLeV

×

⋅+⋅=

4

24

2

2

π

δδπ

Ld

Ld

Ld

ddL2

2

2

2 δδ ⋅+⋅=

L

L

L

d δδ += 2

2122

eeL

L

L

deV +=+=∴ δδ

� 212 eeeV +=

Also,

−+

−×= µµ2

1

22

11

2

22

tE

pd

tE

pdeV

−= µµ2

1_2

2tE

pd

−=∴ µ22

5

2tE

pdeV

Q. A cylindrical shell, 90cm long 20cm internal dia. having thickness of metal as 8mm in

failed with luid at atmospheric pressure. If 20cm3 of fluid is pumped in to the cylindewr find;

(i) the pr. exterted by fluid on the cylinder



(ii) The hoop stress induced.

Take, E = 2×105N/mm2 and µ = 0.3

So|n: Given,

L = 90cm

d = 20cm

t = 8mm

µ = 0.3

E = 2×105N/mm2

Now, volume of cylinder 32

33.282744

cmLd == π

Change in volume = 20cm3

µ = 0.3

E = 2×105N/mm2 = 2×103N/cm2

Now, (i)

−= µδ2

2

5

2tE

pd

V

v

== PP ?

(ii) t

pd

21 =σ

� A boiler is subjected to an internal pr. of 20Kg/cm2. The thickness plate is 2cm.

So|n: p = 20Kg/cm2

t = 2cm

σt = 1200Kg/cm2

nl = 90%=0.9

σnc = 40% = 0.4

since, ltn

pd

21 =σ

p

tnd ln 2σ

=

Since, 31 /1200 cmKg=σ

So, cmd 21620

9.0221200 =×××=

Similarly,

ctn

pd

42 =σ

20

9.024120042 ×××=××

=p

nd cσ

cmd 192=∴



Q. A cylindrical shell is 3M long, 1.5m internal dia and 20mm metal thickness. Calculate the

intensity of maxm shear stress incuced and also the change in dimension of the shell if it is

subjected to an internal pressure of 2N/mm2. Take E = 0.2×106N/mm2 and µ = 0.3.

{hint: maxm shear stress, t

pdq

8max = ]

So|n: Given,

L = 3m = 3000m

D = 2N/mm2

d = 1.5m = 1500mm

E = 0.2×106N/mm2

t = 20mm

µ = 0.3

now, 2max /75.18

208

15002

8mmN

t

pdq =

××==

now, for change in dimension;

for dia change,

−= µδ2

11

2

2

tE

pdd

( )

×−×××

×= 3.02

11

102.0202

150026

2

mm478.0=

mmd 478.0=∴δ

Change in length:

−= µδ2

1

2tE

pdLl

( )

−×××

××= 3.02

1

102.0202

3000150026

mml 225.0=∴δ

Also, volumetric strain,

410125.722

5

2−×=

−== µδtE

pd

V

veV

Since, 392

103.24

mmld

V ×=×= π

394 3776250103.510125.7 mmVev V =×××=×=∴ −δ

Column: According to analysis, the type of column are,

(i) short column

(ii) long column



(i) short column:

If slenderness ratio is less than 12 then the column is short column

(ii) long column:

If slenderness ratio is greater than 12 then the column is long column.

Slenderness ratio: It is ratio of affection length of the column to the least lateral dimension. The side

which is less among a and b is called least lateral dimension.

22

yxPePe

A

p ±±=σ

Type of column according to support condition:

(1) Both ends, effectively held in position and rest against in ratio.

Left = 2l

(2) Both ends hinged

l(left) = l

(3) One ends fixed and other end free

Left =

(4) One ends fixed and other ends hinged:

Left =

� Fixed ends: no deflection, no slope

� Hinge ends: no vertical and horizontal deflection but slope occurs

� Roller: no vertical deflection but there occurs horizontal deflection and slope

� Free end: horizontal deflection, vertical deflection and sloe occurs.

a

b

(i) (ii) (iii) (iv)



P

Introduction to Buckling

Column: A structure vertical member whose two ends are fixed which subjected to a axial

compressive load is called column.

Struts: A structure member which is not vertical and one or both of its ends are hinged or pis-

joined is called struts.

Failure of column: Column fails due to any one of the following stresses:

(i) Direct-compressive stresses

(ii) Buckling stresses

(iii) Combined of direct compressive and buckling stresses

Failure of short column: Let, a short column subjected to a compressive load ‘p’, then

compressive stresses is given by,

APσ

When the load is increased, the column reaches at a point of

failure by crushing. The load at this stage is called crushing at this

stage is called crushing load and the stress is called crushing stress.

So, A

Pcc =σ

Where, Pc = crushing load

σc = crushing stress

Failure of long column: Let a long column subjected to a compressive load P. This

column fails by combine action of crushing and bending stress.

Let, σo = Stress due to direct load.

σb = stress due to bending at the centre of the column

Z

Pe=

Where, e = maxm bending of the column at the centre

Z = section modulus about axis of bending

z

Pe

yI

M

I

My

yI

M

===⇒

=

σ

σ



At mid span,

maxm stress = σo + σb

minimum stress = σo - σb

The failure of the column occurs when maxm stress (σo + σb) will be more than crushing

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

compared to the buckling stresses so the very long column is only subjected to buckling

stress.

Assumptions of Euler’s column theory: (1) The column is initially perfectly straight and load is applied axially

(2) The cross-section is uniform throughout its length

(3) The material of the column is perfectly elastic, homogeneous, isotropic and obeys

Hook’s law

(4) The length of the column is very large as compared to the lateral dimension.

(5) The column will fail by buckling alone

(6) The self wf. of column is negligible.

Sign Convention:

- The moment which bends the column convexity towards the original position is taken

as (+)ve

- The moment which bends the column concavity towards the original position is taken as

(-)ve.

Expression for cripping load when both ends of column are hinged:

D

e

p

tensile zone

Conpression zone

tensile = - Vecompressive = + Ve

σo-σbσo+σb

concaveconvex



Let, ‘P’ be the cripping load. Let ‘y’ be the deflection at a section x from A. The

moment due to cripping load at this section;

Pym −= (Since, (-)ve sign is due to concavity)

Pydx

yor −=

2

2dEI,

0d

EI,2

2

=+ Pydx

yor

0EI

d,

2

2

=+ Py

dx

yor

The so|n of above differential equation.

+

=

EIEI 21

PxSinC

PxCosCy

At, x = 0, y = 0

� C1 = 0

At, x = l, y = 0

So,

+=

EI00 2

PlSinC

0EI

, 2 =

PlSinCor

So, either C2 = 0 or 0EI

=

PlSin

If C2 = 0 and C1 is already zero, then column will not but but at all which is not true.

Hence, 0EI

=

PlSin

=

EI

PlSin Sin0, Sinπ, Sin2π, ……….

Taking least practical value (i.e. Sinπ)

πSinP

lSin =

EI

π=EI

Pl

2

2

l

EIP

π=∴ which is Euler’s required equation.

Expression for crippling load when one end of the column is fixed and other end is free:

yl

P

B

A

x



Let, a crippling load is acting on the column. The point B shifts to B| due to crippling

load. Let ‘y’ be the deflection at a section ‘x’ from A and ‘a’ be the deflection at the free end

B.

The moment at this section due to crippling load;

( )yaPm −+= (Since, (+)ve sign due to convexity buckling)

( ) PyPayaPdx

yor −=−=

2

2dEI,

PxPydx

ydEI =+

2

2

aEI

Py

Ey

P

dx

yd =+2

2

The so|n of this differential equation is,

)(21 iaEI

PxSinC

EI

PxCosCy −−−−−+

+

=

( ) 01 21 =×

+

×

−=

EI

P

EI

PxCosC

EI

P

EI

PxSinC

dx

dy

)(21 iiEI

PxCos

EI

PC

EI

PxSin

EI

PC −−−−−

+

−=

At, x = 0, y = 0

So, from equation (i);

aCC +×+= 00 21

aC −=∴ 1

At, x = 0, 0=dx

dy

So, from equation (ii);

º0º00 21 CosEI

PCSin

EI

PC +−=

02 =∴C

So from (i);

)(iiiaEI

PxaCosy −−−−−+

−=

Since, at, x = l, y = a,

So from equation (iii)

aEI

PlaCosa +

−=

y

B

A

B1

a

l

x

p



0=

EI

PlaCos

Since, 0≠a so,

0=

EI

PlCos

..........,25,23,2 πππ CosCosCosEI

PlCos =

Taking least practical value,

2

πCos

EI

PlCos =

2

π=EI

Pl

( )2

2

2

2

24 l

EI

l

EIP

ππ ==∴

lL 2=∴

Expression for crippling load when both ends of the column are fixed: Let a crippling load ‘P’ as the column. Let y be the deflection at a section ‘x’ from A.

Let mo be the fixed end moment then,

Moment at the section, = mo – Py

Pymdx

ydEI o −=

2

2

omPydx

ydEI =+

2

2

EI

m

EI

Py

dx

yd o=+2

2

P

m

EI

P

P

P

EI

my

EI

P

ax

yd oo ×=×=+2

2

The solution of above differential equation is,

)(21 iP

m

EI

PxSinC

EI

PxCosCy o −−−−−+

⋅+

⋅=

Now, ( )EI

P

EI

PxCosC

EI

P

EI

PxSinC

dx

dy

+×

−= 21 1

EI

P

EI

PxCosC

EI

P

EI

PxSinC

dx

dy

+×

−=∴ 21

At, x = 0, y = 0

D

y

MG

x

l



So, from 1st equation i.e. (i)

P

mCC o+×+×= 010 21

P

mC o−=∴ 1

At, x = 0, 0=dx

dy also, then from above expression

EI

PCC ×+×−= 21 00

02 =∴C

Hence, from (i),

P

m

EI

PxCos

P

my oo +

−=

Now, at, x = l, y = 0, then above expression also written as

P

m

EI

PlCos

P

m oo +

−=0

P

mP

m

EI

PlCos

o

o

=

1=

EI

PlCos

..........,2,,º0 ππ CosCosCosEI

PlCos =

Now, taking least practical value i.e. Cos2π, then

π2CosEI

PlCos =

π2=EI

Pl

2

24

l

EIP

π=∴

Expression for crippling load when one end of the column is fixed and the other end is hinged: Proof:



Let a crippling load ‘P’ on the column. Let ‘y’ be the deflection at the section ‘x’ from

‘A’. Let mo be the fixed end moment at ‘A’ and ‘H’ be the horizontal refn at ‘B’ to balance

mo.

So, the moment at a section ‘x’,

( )xlHPym −+−=

( )xlHPydx

ydEI −+−=

2

2

( )xlHPydx

ydEI −=+

2

2

( )xlEI

Hy

EI

P

dx

yd −=+2

2

( )P

Pxl

EI

H ×−=

( )P

xlH

EI

Py

EI

P

dx

yd −⋅=+2

2

The soln of the above differential equation,

( ) )(21 ixlP

H

EI

PxSinC

EI

PxCosCy −−−−−−+

+

=

Now, differentiating the equation (i) then we get,

( )P

H

EI

P

EI

PxCosC

EI

P

EI

PxSinC

dx

dy −

+

−= 21 1

)(21 iiP

H

EI

P

EI

PxCosC

EI

P

EI

PxSinC

dx

dy −−−−−−

+

−=

At, x = 0, y = 0, then from equation (i),

( )00º00 21 −+×+= lP

HCCosC

lP

HC −=∴ 1

Again, at x = 0, 0=dx

dy, then from equation (ii),

P

H

EI

PCosCSinC −+= º0º00 21 (Since, Cos0º = 1)

EI

P

PHC =×12

P

EI

P

HC =∴ 2

D

y

B

x

l

H

p

A Mo



Hence, from equation (i),

( ) )(iiixlP

H

EI

PSinCx

P

EI

P

H

EI

PCosCxl

P

Hy −−−−−−+×+××−=

now, at, x= l, y = 0, hence from (ii),

( )llP

H

EI

PlSin

P

EI

P

H

EI

PlCosl

P

H −+

+

×−=0

=

EI

PllCos

P

H

EI

PlSin

P

EI

P

H

EI

Pl

EI

Pl =

tan

now, The total so|n of above differential equation is,

radianEI

Pl 5.4=

5.202 =EI

Pl

22 2π=EI

Pl

2

22

l

EID

π=∴ {Since, 2π2 ≈ 20.25]

Effective length:

End condition of the column Effective length(L)

(1) Both ends hinged L = l

(2) One end fixed and other end free L = 2l

(3) Both ends fixed L = l/2

(4) One end is fixed and other end is hinged 2lL =

Q-1. A column oftimer section is 15cm × 20cm is 6m long, both ends being fixed. If the

young modulus of elasticity for timber = 17.5KN/mm2 determine:

(i) crippling load

(ii) safe load for the column if factor of safety (i.e. F.S. = ?)

So|n: Given,

b = 15cm

d = 20cm

l = 6m

E = 17.5KN/mm2

now, ( ) 444

33

1010000100012

2015

12mmcm

bdI XX ×==×==

x

y

y

x20

15



and, ( ) 444

33

105625652512

1520

12mmcm

dbI YY ×==×==

Since, IYY is less than IXX so column buckles towards Y axis

Now, effective length (L) mmme

300032

6

2====

Now, ( )2

42

)(2

2

3000

1056255.17 ×××== ππLeftL

EIP

KNP 48.1079=∴

Hence, for (ii), safe load, KNF

P8.359

3

48.1079 ===

KNLoadSafe 8.359=∴

48.1079, =∴ PLoadCripling

Q-2. Determine the crippling load of a T-section of dimension 16cm × 10cm ×2cm of length

5m. When it is used as strut with both of it’s endhinged. Take yong modulus, E = 2

×105N/mm2.

So|n:

Since the column is symmetrical in Y-section. So, we only found out

the y .

Now, Let C.G. gets at a distance y from top fibre then,

( ) 682121082210 ××+××=×+× y

cmy 23.3=

Now, ( ) ( ) ( ) ( ) 42

32

3

21.31423.368212

82123.3210

12

210cmI XX =

−×+×+

−×+×=

And, ( ) ( ) 4

23

17212

28

12

102cmI YY =

×+×=

Since, IYY is less than IXX hence, the column buckles towards y-axis.

Now, L(eft) = l = 5m = 5000mm [Since, both ends hinges then Left = l]

So, ( ) ( )2

452

2 5000

10172102 ××××== ππleft

EIP

NP 7.135805=∴

2cm

10 cm

y = 3.23

6.77

10 cm

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