ce6302 notes rejinpaul_2.pdf

1

UNIT – I

STRESS STRAIN AND DEFORMATION OF SOLIDS, STATES OF

STRESS

Rigid bodies and deformable solids – stability, strength, stiffness – tension,

compression and shear stresses – strain, elasticity, Hooke’s law, limit of

proportionately, modulus of elasticity, stressstrain curve, lateral strain –

temperature stressres – deformation of simple and compound bars – shear

modulus, bulk modulus, relationship between elastic constants – biaxial state of

stress – stress at a point – stress on inclined plane – principal stresses and

principal planes – Mohr’s circle of stresses.

S.NO 2 MARKS PAGE NO

1 Define stress and its types 6

2 Define strain. 6

3 Define tensile stress and tensile strain. 6

4 Define the three Elastic moduli. (April/May 2005) 7

5 Define shear strain and Volumetric strain 7

6

A square steel rod 20 mm x 20 mm in section is to

carry an axial load (compressive) of 100 KN. Calculate

the shortening in a length of 50 mm. E = 28 /1014..2 MKN

7

7 Define Poisson’s ratio. (Nov / Dec 04) 8

8 Find the force “P”acting on the bar given in fig.(Apr/May04) 8

9 Find the force P”acting on the bar given in fig.(Apr/ May 04) 9

10 Define principal plane and principal stress. 9

11 Write down the relation between modulus of elasticity and

modulus of rigidity and that between modulus of elasticity and

bulk modulus. (Nov/Dec 02)

9

12

When a rod of diameter 20mm is subjected to a tensile force of

40 kN, thje extension is measured as 250 divisions in 200mm

extension meter. Find the modulus of elasticity if each division is

equal to 0.001mm. (Nov/Dec 02

9

13 What do you understand by the assumption, plane section remain

plane even after the application of load? 9

14

A metal bar 50 mm x 50 mm section, is subjected to

anaxial compressive load of 500 KN.The contraction of

a 200 mm gauge length is found to be 0.5 mm and the

increase in thickness 0.04 mm. find E and μ.

9

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15

Brass bar, having C.S.A of 10 cm2 is subjected to axial

forces as shown in fig. Find the total elongation of

the bar Take 23 /108.0 cmtE

10

16

Steel bar is 900 mm long its two ends are 40 mm and

30 mm in diameter and the length of each rod is 200

mm.The middle portion of the bar is 15 mm in diameter

and 500 mm long. If the bar is subjected to an axial

tensile load of 15 KN, find the total extension. 2/200 mGNE 9101 GandgigaG

11

17

Bar of 2 m length, 2 cm breadth and 1.5 cm thickness is

subjected to a tensile load of 3000 kg. Find the final

volume of the bar, if 26 /100.24

1cmkgEand

12

18 A bronze specimen has a 26 /102.1 cmkgE and

26 /1047.0 cmkgC . Determine 13

19

Two parallel walls 6 m apart, are stayed together by a

steel rod 20 mm passing through metal plates and

nuts at each end. The nuts are tightened home, when

the rod is at a temp of 1000 C. Determine the stress in

the rod, when the temperature falls down 200C if.

13

20

A bar is subjected to a tensile stress of 1000 kg / cm2.

Determine the normal and Tangential stresses on a

plane making an angle of 600 with the axis of tensile

stress.

14

S.NO 16 MARKS PAGENO

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1

A hollow cast – iron cylinder 4m long, 300 mm

outer diameter and thickness of metal 50 mm is

subjected to a central load on the top when standing

straight. The stress produced is 75000 KN / m2.

Assume Young’s Modulus for cast iron as 28 /105.1 mKN and find (i)Magnitude of the

load(ii)Longitudinal strain produced and (iii)Total

decrease in length.

18

2

The following observations were made during a

tensile test on a mild steel specimen 40 mm in

diameter and 200 mm long.Elongation with 40 KN

load (within limit of proportionality) 0304.0l

mm, yield load = 161 KN. Maximum load = 242

KN, Length of specimen at fracture = 249 mm.

Determine:

i. Young’s Modulus of Elasticity

ii. Yield point stress

iii. Ultimate stress

iv. Percentage elongation.

19

3

A steel 2m long and 3 mm in diameter is extended

by 0.75 mm when a weight W is suspended from

the wire. If the same weight is suspended from a

brass wire, 2.5 m long and 2 mm in diameter, it is

elongated by 4.64 mm. Determine the modulus of

elasticity of brass if that of steel be 25 /100.2 mkN

20

4

A member formed by connecting a steel bar to an

aluminium bar as shown in fig. Assuming that the

bars are prevented form buckling sidewise; calculate

the magnitude of force p1 that will cause the total

length of the member to decrease 0.25 mm. The

values of elastic modulus of steel and aluminum are

2101 KN / mm2 and 70 KN / mm

2 (April/May 04)

22

5

A steel tie rod 50 mm in mand 5.2 long is

subjected to a pull of 100 KN. To what length the

rod should be bored centrally so that the total

extension will increase by 15 % under the same

pull, the bore being 25 mm ?

23

6

A steel flat plate AB of 1 cm thickness tapers

uniformly from 10 cm to 5 cm width in a length of

40 cm. From first principles, determine the

elongation of the plate, if an axial tensile force of

5000 kg acts on it. Take 26 /100.2 cmkgE

24

7 A steel cube block of 50 mm side is subjected to a 26

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force of 6 KN (Tension), 8 KN (compressive) and 4

KN (tension) along x, y and z directions. Determine

the change in the volume of the block.

2/200 mmKNe and 3

10m

8

A bar of 30 mm is subjected to a pull of 60 KN.

The measured extension on gauge length of 200 mm

is 0.09 mm and the change in diameter is 0.0039

mm. Calculate and the values of the three

module. (Nov/Dec 04)

27

9

At a point within a body subjected to two mutually

perpendicular directions, the stresses are 80 N/mm2

tensile and 40 N/mm2 tensile. Each of the above

stresses is accompanied by a shear stress of 60

N/mm2. Determine the normal stress. Shear stress

and resultant stress on an oblique plane inclined at

an angle of 450 with the axis of minor tensile stress.

28

10

A point in strained material is subjected to the

stresses as shown in figure. Locate the principal

planes and evaluate the principal tresses. (Apr/May

02)

29

UNIT – I

TWO MARKS:

1. Define stress and its types

When a body is acted upon by some load (or) external force, it undergoes

deformation (i.e. change in shape or dimensions)

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Stress is defined as the internal resistance offered by the material to the

extremely applied force, expressed per unit area.

A

P

tionofAreaA

loadappliedP

stress

sec

Types of stresses:

1. Axial stress

2. Bearing stress

3. Bending stress

4. Shear stress

Types of axial stress:

1. Tensile stress

2. Compressive stress.

2. Define strain.

Strain is defined as the ratio of change in length to the original length of the

member

Change in length (dl)

Strain =

Original length (l)

3. Define tensile stress and tensile strain.

When the resistance offered by a section of a member is against an increase in

length, the section is said to offer tensile stress.

Tensile stress A

p

ASC

ceresisInternalt

..

tan

Tensile strain:

The strain corresponding to tensile stress is tensile strain.

lengthOriginal

lengthIncrease

l

lestrainTensile

4. Define the three Elastic moduli. (April/May 2005)

Young’s Modulus:

It is the ratio between tensile stress and tensile strain (or) compressive stress

and compressive strain.

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c

ct

eor

eteE

Modulus of Rigidity:

It is defined as the ratio of shear stress ( ) to shear strain and is denoted by C,

N or G It is also called shear stress modulus of elasticity.

Bulk (or) Volume Modulus of Rigidity

It is defined as the ratio of normal stress (on each face of a solid cube) to

volumetric strain and is denoted by the letter K.

V

nK

5. Define shear strain and Volumetric strain

Shear strain is defined as the ratio of transverse displacement to the distance

from the lower force.

Transverse displacement

Shear strain =

Distance from lower face

Volumetric strain:

It is defined as the ratio between change in volume and original volume of the

body

VvolumeOriginal

volumeinchange v

v

6. A square steel rod 20 mm x 20 mm in section is to carry an axial

load (compressive) of 100 KN. Calculate the shortening in a length of

50 mm. E = 28 /1014..2 MKN

Solution:

Area A = 20004.002.002.0 m

Length mml 50 (or) 0.05 m

KNP 100

81014.2 E KN / m2

Shortening of the rod :l

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2

1 /2500000004.0

100mKN

A

PStress

Strain

StressE

EE

StressStrain

81014.2

250000

81014.2

250000

l

l

05.01014.2

2500008

l

= 0.0000584 m (or) 0.0584 mm

Hence the shortening of the rod = 0.0584 mm.

7. Define Poisson’s ratio. (Nov / Dec 04)

The ratio of lateral strain to the longitudinal strain is a constant for a given

material, when the material is stressed within the elastic limit. This ratio is

called Poisson’s ratio and it is generally denoted by 1/m (or) .

strainalLongitudin

strainLateral

8. Find the force “P” acting on the bar given in fig. (April / May 04)

Sum of all left direction force = Sum of all right direction force

50 + P = 45 + 15

P = 60 – 50 = 10 kN

9. What type of stress will be induced in a bar when the ends are restrained and

subjected to i) rise in temperature and ii) a fall in temperature? (April / May 04)

Thermal stresses: These are the stresses induced in a body due to

change in temperature. Thermal stresses are set up in a body, when the

temperature of the body is not allowed to expand or contract freely. i) Rise in

temperature – compressive stress ii) a fall in temperature -tensile stress.

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10. Define principal plane and principal stress.

The plane, which have no shear stress, are known as principal planes. Hence

principal planes are the planes of zero shear stress. These planes carry only

normal stresses.

The normal stresses, acting on a principal plane, are known as principal

stresses.

11. Write down the relation between modulus of elasticity and modulus of rigidity

and that between modulus of elasticity and bulk modulus. (Nov/Dec 02)

The relation between modulus of elasticity and bulk modulus is given by

E = 3K (1-2/m)

The relation between modulus of elasticity and modulus of rigidity

E = 2G (1+(1/m))

12. When a rod of diameter 20mm is subjected to a tensile force of 40 kN, thje

extension is measured as 250 divisions in 200mm extension meter. Find the

modulus of elasticity if each division is equal to 0.001mm. (Nov/Dec 02)

Solution:

Diameter = 20mm ; P = 40 kN = 40 x 103 N ; Extension = 250 divisions

1 division = 0.001 mm; L = 200 mm

δl = 250 x 0.001 = 0.25 mm

E = σ/e = 200/25.0

16.314/1040

/

/ 3x

ll

AP

Modulus of Elasticity = 101.85 x 103 N/mm

2

13. What do you understand by the assumption, plane section remain plane even

after the application of load?

When some external load acts on the beam, the shear force and bending

moments are set up at aii sections of the beam. Due to the shear force and bending

moment, the beam undergoes certain deformation. After removal of the load the beam

will come to its original position based on that assumption.

14. A metal bar 50 mm x 50 mm section, is subjected to an axial

compressive load of 500 KN. The contraction of a 200 mm gauge

length is found to be 0.5 mm and the increase in thickness 0.04 mm.

find E and μ.

Solution:

b = 50 mm, t = 50 mm

Area = 225005050 mm

P = 500 KN

Length, l = 200 mm, .5.0 mml

Increase in thickness, mmt 04.0

Young’s Modulus:

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AE

pll

2

3

/802500

200105005.0 mmKNE

E

Poission’s Ratio:

strainLinear

strainLateral

Linear strain = 0.0025

thicknessstrainLateralt

500025.01

04.0 m

15. brass bar, having C.S.A of 10 cm2 is subjected to axial forces as

shown in fig.

Find the total elongation of the bar Take 23 /108.0 cmtE

Given,

Area A = 10 cm2

23 /108.0 cmtE

l Total elongation of the bar.

For the sake of simplification, the force of 8 tonnes acting at B, may be split

up into three forces of 5 tones, 2 tonnes and 1 tone. Now it will be seen that the part

AB of the bar is subjected to a tension of 5 tonnes part BC is subjected to a

compression of 2 tonnes and the part BD is subjected to a compression of 1 tonne as

shown in fig.

32.01

m

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Using the relation,

332211

1lplplp

AEl

22011002605108.010

13

mmcm 15.0015.0800

12

16. Steel bar is 900 mm long its two ends are 40 mm and 30 mm in

diameter and the length of each rod is 200 mm. the middle portion of

the bar is 15 mm in diameter and 500 mm long. If the bar is

subjected to an axial tensile load of 15 KN, find the total extension. 2/200 mGNE 9101 GandgigaG

P = 15 kN

A1 = 1256 mm2

A2 = 176.625 mm2

A3 = 706.50 mm2

l1 = 200 mm

l 2 = 500 mm

l 3 = 200 mm

Total extension of the bar:

321 , landll

EA

pll

1

1

1 EA

pll

2

2

2 EA

pll

3

3

3

321 llll

EA

pl

EA

pl

EA

pl

3

3

2

2

1

1

3

3

2

2

1

1

A

l

A

l

A

l

E

P

= 0.2454 mm.

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17. bar of 2 m length, 2 cm breadth and 1.5 cm thickness is subjected

to a tensile load of 3000 kg. Find the final volume of the bar, if

26 /100.24

1cmkgEand

L = 2 m = 200 cm, b = 2 cm, t = 1.5 cm.

Vol 36005.12200 cmV

P = 3000 kg.

44

1

3

1 m 26 /102 cmkgE

me

v

v 21

;

E

stresse

2/10003000

cmkgtbA

P

2000

1

102

10006

4

21

2000

1

v

v

4000

1

315.06004000

1cmv

18. A bronze specimen has a 26 /102.1 cmkgE and 26 /1047.0 cmkgC . Determine

Solution:

12

m

mEC

12

102.11047.0

66

m

m

mm 2.1194.0

mm 2.194.094.0

mm 94..02.194.0

0.94 m = 0.26 m

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277.01

94.0

26.01

mm

19. Two parallel walls 6 m apart, are stayed together by a steel rod 20

mm passing through metal plates and nuts at each end. The nuts

are tightened home, when the rod is at a temp of 1000 C. Determine

the stress in the rod, when the temperature falls down 200C if.

a. The ends do not yield and

b. The ends yield by 1mm. 26 /102 cmkgE and

c/1012 6

Solution:

Length of rod = l = 6m = 600 cm

of rod = d = 20 mm = 2 cm.

Temperature t = 1000 - 20

0 = 80

0 c

;/102 26 cmkgE C06 /1012

when the ends do not yield.

rodtheinstress1 tE

266

1 /1920102801012 cmkg

when the ends yield by 1mm

rodtheinstress2

266

2 /158102600

1.081012 cmkg

20. A bar is subjected to a tensile stress of 1000 kg / cm2.

Determine the normal and Tangential stresses on a plane making an

angle of 600 with the axis of tensile stress.

030 22 /750cos cmkgn

2/433cossin cmkgt

2/866 cmkg

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21. A point in a stained material is subjected to two mutually

perpendicular stresses of 2000 kg/cm2 and 1000 kg/cm

2.

Determine the intensities of normal and resultant stresses on a

plane inclined at 300 to the axis of the minor stress.

Major stress = 2

1 /2000 cmkg

Minor stress = 2

2 /1000 cmkg

Angle of plane, which it makes with the axis of minor, principle

stress 030

Normal stress

2cos22

2121

n

= 1750 kg/ cm2

Tangential stress

221 /4332sin

2cmkgt

Resultant stress 222 /8.1802 cmkgtnr

22. At a point in a strained material the principal stresses are 100

N/mm2

(tensile) and 60 N/mm2 (comp). Determine normal stress,

shear stress, resultant stress on a plane inclined at an 500 to the axis of

major principal stress. Also determine the maxi shear stress at the

point.

,/100 2

1 mmN ,/60 2

2 mmN

405090 00

,/9.33 2mmNn 2/8.78 mmNt

2/8.85 mmNR

Maximum shear stress;

221 /802

max mmNt

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23. A point in a strained material is subjected to a compressive

stress of 800 kg/cm2 and a shear stress of 560 kg/cm

2. Determine

graphically or otherwise, the maximum and minimum intensity of

stress.

,/800 2cmkg 2/560 cmkg

Maximum intensity of stress:

22tan

800

5602

= 1.4

'28542 0

'1427 0

2

2

122

n

= 288.2 kg/cm2 (Tensile)

2n Minimum intensity of direct stress.

22

2

2 /2.108822

cmkgn

24. A steel bolt 2.5 cm diameter is subjected to a direct tension of

1500 kg and a shearing forces of 1000 kg. Determine the intensities of

normal and shear stress across a plane inclined at an angle of 600 to

the longitudinal axis of the bolt. Also determine the resultant stress. Solution:

.5.2 cm Area = 4.91 cm2

Direct tension = 1500 kg.

Direct stress on the bolt = 2/5.30591.4

1500cmkg

Shearing force = 1000 kg.

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Shear stress = 2/7.20391.4

1000cmkg

Angle which the plane makes with the longitudinal axis of the bolt = 60 0

Angle, which the plane makes with the normal to the longitudinal axis of the

bolt

000 306090

Normal stress:

2sin2cos12

n

= 405.5 kg/ cm2

Shear stress: 2cos2sin2

1et

= 30.5 kg/cme+

Resultant stress: 22

tnr

225.305.405

= 406.6 kg/cm.2

25. A point is subjected to a tensile stress of 60 N/mm2 and a

compressive stress of 40 N/mm2, acting on two mutually

perpendicular planes and a shear stress of 10 N/mm2 on these planes.

Determine the principal stresses as well as maxi shear stress. Also

find out the value of maxi shear stress.

2

1 /60 mmN and 2

2 /40 mmN

2/10 mmN

Principal stresses:

22

2

21211 /61

22mmNn

22

2

21212 /41

22mmNn

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Maximum shear stress

221 /51

2

4161

2mmNnn

t

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SIXTEEN MARKS QUESTION AND ANSWERS:

1.A hollow cast – iron cylinder 4m long, 300 mm outer diameter and

thickness of metal 50 mm is subjected to a central load on the top

when standing straight. The stress produced is 75000 KN / m2.

Assume Young’s Modulus for cast iron as 28 /105.1 mKN and find

i. Magnitude of the load

ii. Longitudinal strain produced and

iii. Total decrease in length.

Solution: Outer diameter, D = 300 mm = 0.3 m

Thickness , t = 50 mm = 0.05 m

Length, l = 4 m

Stress produced 000,75 KN / m2

8105.1 E KN / m

2

Inner diameter of the cylinder tDd 2

05.023.0

= 0.2 m

i. Magnitude of the load P:

Using the relation,

A

P

(or) 22

475000 dDAP

222.03.0

475000

P = 2945.2 KN.

ii. Longitudinal strain produced e:

Using the relation,

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0005.0105.1

000,75,

8

E

Stressestrain

iii. Total decrease in length , :l

using the relation,

lengthOriginal

lengthinchangestrain

4

0005.0l

40005.0 l

0.002 m = 2 mm

Hence decrease in length = 2 mm.

2. The following observations were made during a tensile test on a

mild steel specimen 40 mm in diameter and 200 mm long. Elongation

with 40 KN load (within limit of proportionality)

0304.0l mm, yield load = 161 KN

Maximum load = 242 KN

Length of specimen at fracture = 249 mm.

Determine:

v. Young’s Modulus of Elasticity

vi. Yield point stress

vii. Ultimate stress

viii. Percentage elongation.

Solution:

i. Young’s Modulus of Elasticity (E) :

Stress,

24

2

/1018.3

04.04

40mKN

A

P

Strain, 000152.0200

0304.0

l

le

000152.0

1018.3 4

Strain

StressE

81009.2 KN / m2

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(ii) Yield point stress:

Yield point stress area

loadpoYield int

24

2

/108.12

04.04

161mKN

(iii) Ultimate stress:

Maximum load

Ultimate stress =

Area

24

2

/102.19

04.04

242mKN

(iv). Percentage elongation:

Length of specimen at fracture original length

Percentage elongation =

Original length

245.0200

200249

= 24.5 %

3. A steel 2m long and 3 mm in diameter are extended by 0.75 mm

when a weight W is suspended from the wire. If the same weight is

suspended from a brass wire, 2.5 m long and 2 mm in diameter, it is

elongated by 4.64 mm. Determine the modulus of elasticity of brass if

that of steel be nN /100.2 5

Solution:

Given:

,2ml s ,3mmd s mml s 75.0

,/100.2 25 mmNEs ,5.2 mlb mmd b 2

mlb 64.4

Modulus of Elasticity of brass, Eb :

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From Hooke’s law,

AE

Wll

where l Extension, l = length, A = Cross sectional area

E = Modulus of Elasticity.

Case 1:

For steel wire

ss EA

Wll 3

3

1000275.0 W

52103

4

2000

1102

4

375.0 5

2

W (1)

Case 2:

For brass wire

bb

b

bEA

lWl

bE

W

22

4

10005.264.4

2500

1

4

264.4

2

bEW

(2)

Equating equation (i) and (ii), we get

2500

1

4

264.4

2000

1102

4

375.0

2

5

2

bE

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25 /10909.0 mmNEb

4. A member formed by connecting a steel bar to an aluminium bar as

shown in fig. Assuming that the bars are prevented form buckling

sidewise; calculate the magnitude of force p1 that will cause the total

length of the member to decrease 0.25 mm. The values of elastic

modulus of steel and aluminum are 2101 KN / mm2 and 70 KN / mm

2

Solution: Given:

225005050 mmAs

2000,10100100 mmAA

Length of steel bar = l3 = 300 mm.

= l 3 = 380 mm

mml 25.0

:/10210 23 nnNEs 23 /1070 mmNEA

P = Magnitude of the required force.

22

2

11

1

EA

l

EA

lPl

P = 224.36 KN

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Area at the reduced section

222 001472.0025.005.04

m

Stress in the reduced section, 001472.0

10001001

26 /1093.67 mN

Elongation of the rod

E

x

E

x .5.2

310731.0

3

9

6

9

6

10731.010200

109.67

10200

5.21092.50

xx

3966 10731.010200109.675.21092.50 xx

61092.50

87.233.15.2 xx

x = 1.12 m.

5. A steel tie rod 50 mm in mand 5.2 long is subjected to a pull of 100

KN. To what length the rod should be bored centrally so that the

total extension will increase by 15 % under the same pull, the bore

being 25 mm ?

Solution:

Dia of steel tie rod = 50 mm = 0.05 m

Length of steel tie rod l = 2.5 m

P = 100 KN.

Dia of bore = 25 mm = 0.025 m

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29 /10200 mNe

Length of the bore x:

Stress in the solid rod,

26

2

/1092.50

05.04

1000100mN

A

P

Elongation of the solid rod 9

6

10200

5.21092.50

E

ll

= 0.000636 m (or) 0.636 mm.

Elongation after the rod is bored 636.015.1

= 0.731 mm.

6. A steel flat plate AB of 1 cm thickness tapers uniformly from 10 cm

to 5 cm width in a length of 40 cm. From first principles, determine

the elongation of the plate, if an axial tensile force of 5000 kg acts on

it. Take 26 /100.2 cmkgE

Given:

Width of the plate at a distance x from A

8

1040

51010xx

C.S.A of the bar

2

810

8101 cm

xxAx

Stress, 2/

810

5000cmkg

xA

P

x

x

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Strain,

810400

1

1028

10

5000

6 xxE

xx

Elongation of the elementary length

cmx

dxdxex

810400

.

The total extension of the bar may be found out by integrating the above

equation between the limits 0 and 40.

(ie.)

40

0

810

400

1

x

dxl

40

08

10log

8

1400

1

x

e

10log5log50

1ee

5log10log50

1ee

5

10log

50

1e

cme 2log50

1

cm2log3.250

110 10log3.2log e

50

3010.03.2

= 0.014 cm = 0.14 mm.

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7. A steel cube block of 50 mm side is subjected to a force of 6 KN

(Tension), 8 KN (compressive) and 4 KN (tension) along x, y and z

directions. Determine the change in the volume of the block.

2/200 mmKNe and 3

10m

Side of steel cube = 50 mm

225005050 mmA

and 3125000505050 mmv

;6KNPX ;8KNPY KNPz 4

ZYX eeeV

v

mEmEE

e ZYx

X

2

3

/4.22500

106mmN

A

PX

2

3

/2.32500

108mmN

A

PY

2/6.1

2500

4000mmN

A

PZZ

EEEE

eX

88.2

10

36.1

10

32.34.2

EEEE

eY

4.4

10

36.1

10

34.22.3

EEEE

eZ

84.1

10

32.3

10

34.26.1

ZYX eeev

v

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26

EEE

v 84.14.488.2

10125 3

3

3

3 2.010200

32.010125 mmv

8. A bar of 30 mm is subjected to a pull of 60 KN. The measured

extension on gauge length of 200 mm is 0.09 mm and the change in

diameter is 0.0039 mm. Calculate and the values of the three

module.

Solution:

i. Young’s Modulus:

e

E

2

2

3

/9.84

304

1060mmN

A

P

00045.0200

09.0

l

le

2/67.188 mmKNE

ii. Poisson’s ratio:

e

dd

strainLinear

strainLateral /

00013.03

00039.0/ dd

13

45

45

13

00045.0

00013.0 m

iii. Modulus of rigidity:

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27

2

3

/19.73

113

452

1067.18813

45

12mmKN

m

mEC

iv. Bulk Modulus:

2

3

/95.148

213

453

1067.18813

45

23mmKN

m

mEK

9. At a point within a body subjected to two mutually

perpendicular directions, the stresses are 80 N/mm2 tensile and 40

N/mm2

tensile. Each of the above stresses is accompanied by a shear

stress of 60 N/mm2. Determine the normal stress. Shear stress and

resultant stress on an oblique plane inclined at an angle of 450 with

the axis of minor tensile stress.

Solution: Given,

Major tensile stress, 2

1 /80 mmN

Minor tensile stress, 2

2 /40 mmN

Shear stress, q = 60 N/mm2

Angle of oblique plane, with the axis of minor tensile stress.

i. Normal stress ( n )

2sin2cos22

2121

n

452sin602sin452cos2

4080

2

4080 0

2/120 mmNn

ii. Shear (or tangential) stress t

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28

2cos2sin2

21

t

452cos60452sin2

4080

2/20 mmNt

iii.. Resultant stress R

22

tnR

= 121.665 N / mm2

10. A point in strained material is subjected to the stresses as shown

in figure. Locate the principal planes and evaluate the principal

tresses.

Solution: Given,

Stress on the face BC and AD = 600 kg / cm2

Inclination of the stress = 600

Stress normal on the face BC or BD

0

1 30cos600

030sin600

4002

Location:

'213902

2 0

21

R

lan

Principal stress

2

2

2121

122

n

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29

= 766 kg/cm2

2

2

21212

22

n

= 154 kg/cm2

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1

UNIT – II

ANALYSIS OF PLANE TRUSS, THIN CYLINDERS / SHELLS

Stability and equilibrium of plane frames – types of trusses – analysis of forces in

truss members’ method of joints, method of sections, method of tension

coefficients – thin cylinders and shells – under internal pressure – deformation of

thin cylinders and shells.

S.NO 2 MARKS PAGE NO

1

A steel penstock of 1m dia and 1 cm thick is subjected

to 100 m head of water. Calculate the hoop stress and

longitudinal stress at the bottom of the penstock.

4

2

A gas cylinder of internal diameter 1.5 m is 3 cm thick.

Find the allowable pressure of the gas inside the

cylinder, if the tensile stress in the material is not to

exceed 1000 kg/cm2

4

3

A cylinder of thickness 1.5cm , has to withstand

maximum internal pressure of 1.5 N/mm2. If the

ultimate tensile stress in the material of the cylinder is

300 N/mm2, f.o.s - 3.0 and joint efficiency – 80%.

Determine the of the cylinder.

5

4 What are the assumptions made in finding the forces in a truss?

(Apr/May 05) 5

5 Explain determinate and stable frame. (Apr/May 05) 5

6 When is a cylinder called as thin cylinder? What is t6he effect of

this on stress distribution? (Apr/May 05) 6

7 Write the expression for the determination of circumferential

stress or hoop stress in thin cylinder. 6

8

Boiler steel is to be made of 15mm thick plate having a

limiting tensile stress of 120 N/mm2. If the efficiencies

of the longitudinal and circumferential joint are 70%

and 30%. Find the maximum per of the shell for an

6

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2

internal pressure of 2 N/mm2

9

Boiler steel is to be made of 15mm thick plate having a

limiting tensile stress of 120 N/mm2. If the efficiencies

of the longitudinal and circumferential joint are 70%

and 30%. The Permissible intensity of internal pressure

when the shell - 1.5m

7

10

Write the expression for the determination of Longitudinal

stress in thin cylinder.

7

S.NO 16 MARKS PAGENO

1 Analyze the given truss using method of joints. 8

2

A boiler is subjected to an internal steam pressure of

2 N/mm2. The thickness of boiler plate is 2.0 cm

and permissible tensile stress is 120 N/mm2. Find

out the maximum diameter , when efficiency of

longitudinal joint is 90% and circumference joint is

40%.

11

3

A cylinder shell 3m long, 1m in diameter and 10

mm thick is subjected to an internal pressure of 2

Mpa. Calculate the change in dimensions of the

shell and the maximum shear stress

induced MpaE 5102 . 3.0 (Nov/ Dec 04)

12

4

Fig shows a warren grider consisting of seven members each

of 6m lengths freely supported at its end points. The grider is

loaded at B & C as shown in fig. Find the forces in all the

members of the grider.

13

5 Analyze the given truss by method of joints. (Apr/ May 04) 15

6 A truss of span 9m is loaded as shown in fig. Find the

reaction of forces in the members 1,2 and 3. 17

7 Analyze given the trusses using Tension coefficient method. 18

8 Analyse given the trusses using Tension coefficient method: 20

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UNIT – II

TWO MARKS:

1. A steel penstock of 1m dia and 1 cm thick is subjected to 100 m

head of water. Calculate the hoop stress and longitudinal stress

at the bottom of the penstock. Solution:

Given,

d = 1 m = 100 cm

t = 1 cm

Head of water, h = 100 m = 10,000 cm

Water pressure = P = wh = 1 x 10,000 gm/cm2

= 10 kg/cm2 w = 1 gm/cm

2

Hoop stress:

2

1 /50012

10010

2cmkg

t

pdf

Longitudinal stress:

2

2 /25014

10010

4cmkg

t

pdf

2. A gas cylinder of internal diameter 1.5 m is 3 cm thick. Find

the allowable pressure of the gas inside the cylinder, if the

tensile stress in the material is not to exceed 1000 kg/cm2

d = 1.5 m = 150 cm, t = 3 cm, f1 = 1000 kg/cm2

2

1 /402

cmkgpt

pdf

since the circumferential stress (ie) f1 is double the longitudinal stress (ie) f2.

There maximum stress is taken as the circumferential stress.

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4

2

1 /402

cmkgpt

pdf

2

2 /804

cmkgpt

pdf

Now the allowable pressure, in the gas cylinder is to be taken the lesser of the

two . Therefore allowable pressure P = 40 Kg/cm2

3. A cylinder of thickness 1.5cm , has to withstand maximum

internal pressure of 1.5 N/mm2. If the ultimate tensile stress in

the material of the cylinder is 300 N/mm2, f.o.s - 3.0 and joint

efficiency – 80%. Determine the of the cylinder.

t = 1.5cm , p = 1.5 N/mm

2, Ultimate tensile stress = 300 N/ mm

2

f.o.s – 3.0, 80.01

t

pdf

1

12

working stress sof

stresstensileulf

..1

5.180.02

15100

d

3

300

= 100 N/mm2

d = 160 cm = 1.6 cm

4. What are the assumptions made in finding the forces in a truss?

(Apr/May 05)

(a) All the members are pin – jointed

(b) The frame is loaded only at the joints

(c) The frame is a perfect frame

(d) The self – weight of the members is neglected

5. Explain determinate and stable frame. (Apr/May 05)

The structure, for which the reactions at the supports and the

internal forces in the members can be found out by the condition of

static equilibrium, is called a determinate frame.

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5

To find out the reaction components at the supports and internal

forces in the frame, so far we have used three static equilibrium

equations. If the static equations are satisfied then it is called stable

frame.

6. When is a cylinder called as thin cylinder? What is t6he effect of this

on stress distribution? (Apr/May 05)

If the thickness of the wall of the cylindrical vessel is less than

1/15 to 1/20 of its internal diameter the cylindrical vessel is known as

thin cylinder. In case of thin cylinders, the stress distribution is assumed

uniform over the thickness of the wall.

7. Write the expression for the determination of circumferential stress or

hoop stress in thin cylinder.

t

Pdf

21

f1 = Circumferential stress

P = internal pressure

d = diameter of thin cylinder

t = thickness of the cylinder

8. Boiler steel is to be made of 15mm thick plate having a

limiting tensile stress of 120 N/mm2. If the efficiencies of the

longitudinal and circumferential joint are 70% and 30%.

(i) The maximum per of the shell for an internal pressure

of 2 N/mm2

t = 15 mm, Limiting tensile stress 120 N/ mm

2

,70.0l 30.0c

t

Pdf

21

t

Pdf

22

(i) f1 = 120 N/mm2 f2 = 120 N / mm

2

d = 1260 mm d = 1080 mm

Maximum = 1080 mm

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6

9. Boiler steel is to be made of 15mm thick plate having a

limiting tensile stress of 120 N/mm2. If the efficiencies of the

longitudinal and circumferential joint are 70% and 30%.

(i) Permissible intensity of internal pressure when the

shell - 1.5m

t

Pdf

21

t

Pdf

21

d = 1.5 m = 1500 mm f2 = 120 N/mm2

p = 1.68 N / mm2 p = 1.44 N / mm

2

Maximum P = 1.44 N / mm2

10. Write the expression for the determination of Longitudinal stress in

thin cylinder.

t

Pdf

42

f1 = Longitudinal stress

P = internal pressure

d = diameter of thin cylinder

t = thickness of the cylinder

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7

SIXTEEN MARKS QUESTIONS AND ANSWERS:

1. Analyze the given truss using method of joints.

The truss is supported on rollers at B and hence RB Will be vertical.

The truss is hinged at A, hence the support reactions at A will consists of a

horizontal reaction HA and a vertical reaction RA. Taking moment about A, we get

86435.1812 BR = 72

KNRB 6

and RA = Total vertical loads - RB

BR 63 22cdACAC

= KN3 225.12 AC

2.5.

and HA = sum of all horizontal loads

KN8 8.05.2

2cos

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8

6.05.2

5.1sin

Joint A:

Resolving the forces vertically,

KNSinFCA 3

)(5sin

3compKNFCA

Horizontally

ACAFA HFF cos

TensileKN12

Joint C:

Resolving vertically

sinsin CFCA FF

CFCA FF

TensileKNFCF 5

Horizontally,

coscos CFCACD FFF

compKNFCD 8

Joint F:

Resolving vertically,

3sinsin5 DFF

0DFF

Resolving Horizontally,

coscos512 DFGF FF

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9

TensileKNFGF 16

Joint D


sinsin DfDG ff

0DGf

Resolving Horizontically,

KNff CDDE 8

compKNfDE 8

Joint G


6sinsin DGGE ff

TensileKNfGE 10

Resolving Horizontally,

Cosff GEGB 16

8.01016

= 8 KN (Tensile)

Joint E


sinsin BEGE ff

10 GEBE Ff

)(10 compKNf BE

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10

2. A boiler is subjected to an internal steam pressure of 2

N/mm2. The thickness of boiler plate is 2.0 cm and

permissible tensile stress is 120 N/mm2. Find out the

maximum diameter , when efficiency of longitudinal joint is

90% and circumference joint is 40%.

P = 2 N/mm2 t = 2 cm f1 = 120 N/mm

2

t

dpf

l21 90.0l

cmd 0.2162

29.02120

(i)

cmdt

dpf

c

1922

0.24.04120

42

Maximum diameter of the boiler is equal to the minimum value of diameter

given by the equation (i) and (ii)

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11

d = 192 cm

If d is taken as equal to 216.0 cm , the longitudinal stress (f2) will be more than the

given permissible value

2

2 /1354

mmNt

dpf

C

3. A cylinder shell 3m long, 1m in diameter and 10 mm thick is

subjected to an internal pressure of 2 Mpa. Calculate the

change in dimensions of the shell and the maximum shear

stress induced MpaE 5102 . 3.0

Solution:

d = 1 m, L = 3 m, t = 10 mm, p = 2 Mpa = 2 N/ mm2

;/102102 255 mmNMpaE 3.01

m

i. Change in diameter

mtE

pdd

2

11

2

2

m2

11

102102

100025

2

= 0.425 mm

ii. Chang e in length

mtE

pdLl

1

2

1

2

3.0

2

1

102102

3000100025

= 2.05.1

= 0.3 mm

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12

iii. Change in volume

L

L

d

d

V

v 2

3000100043000

3.0

1000

425.02

2

= 2238384. 766 mm3

Maximum shear stress induced:

t

pd

8

= 25 N/mm2

4. Fig shows a warren grider consisting of seven members each of 6m

lengths freely supported at its end points. The grider is loaded at B &

C as shown in fig. Find the forces in all the members of the grider.

Taking moment about A and equating the same,

(VD*6)-(2*1.5)-(4*4.5) = 0

VD=21/6

=3.5t.

VA+VD=8.

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13

And

VA=(2+4) – 3.5

=2.5t.

Methods of joints:

Consider the joint A. Let the directions of FAB and FAE be assumed as

shown in fig.

Resolving the forces vertically and equating the same,

FAB sin 60° =2.5

FAB= 2.5 / sin 60°

=2.9t (Comp)

Resolving the forces horizontally and equating the same.

FAE – FAB cos 60° =0

FAE = FAB cos 60°

=2.9*0.5

=1.45t

Now consider joint D. Let the directions of FCD and FDE

be assumed as shown.

Resolving the forces vertically and equating the same.

FCD sin 60° =3.5

FCD=3.5/sin60°

=4.04t (Comp)

Now resolving the forces horizontally and equating the same.

FDE = FCD cos 60°

=4.04*0.5

=2.02t (Tension)

Now consider joint B. Let the directions of FBC and FBE be assumed as shown.


FBE sin60°= FAB sin60° -2

=2.9*0.866 – 2

=0.5t.

FBE= 0.5 / sin60°

=0.58t (Tension)

Resolving the forces horizontally and equating the same.

FBC=2.9 cos60° +(0.58cos 60°

FBC=1.74t ( Comp)

Now resolving joint C. Let the direction of FCE is assumed as shown in fig.

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14


FCE sin60°+ 4= FCD sin60°

=-4+(4.04*0.866)

=-0.5

FCE=-0.5 / sin 60°

=-0.58 (Comp)

5. Analyze the given truss by method of joints. (Apr/ May 04)

The truss is hinged at A and hence the support reactions at A will consists of a

horizontal reactions HA and a vertical reaction RA.

AC=4 cosθ = 4*0.866 = 3.464 m.

AB=2*AC = 2*3.464 = 6.928m.

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15

Taking moments about A,

RB * 12 = (2*AC) +(1*AD) +(1*AE)

RB=1.77kN

Total vertical components of inclined loads,

=(1+2+1) sin 60° + 1

=4.464 kN

RA= Vertical component of inclined loads

= 4.464-1.77

=2.69 kN

HA= Sum of all horizontal components

=2kN


FAC sin30° + (1*sin 60°) = 2.974

FAC=4.216 kN(Comp)

Resolving horizontally,

FAE = 2+(FAC cos 30° ) – (1*cos 60°)

= 5.15 kN (Tensile)

Joint C:

FCE=2 kN (Comp)

FCD=FAC=4.216 kN (Comp)

Joint E:


1+(2*sin60) = FED sin60

FED=3.155 (Tensile)

Resolving horizontally,

5.15-(2*cos60) – FED cos 60 - FEF=0

FEF=8 kN (Tensile)

At the joint G, two forces FBG and FDG are in the same straight line and hence the third

force FGF should zero.

Joint F:

In vertical,

FDF sin 60=0

FDF=0

In horizontal,

FFB=FEF=2.58 kN (Tensile)

Joint B:

In vertical,

FBG sin 30 = 1.49

FBG=2.98 kN(Comp)

Joint G:

FGD=FBG=2.98 kN

FGD=2.98 kN(Comp)

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16

6. A truss of span 9m is loaded as shown in fig. Find the reaction of forces

in the members 1,2 and 3.

Taking moment about A,

RB*9 = (9*3) +(12*6)

RB=11 kN

RA=(9+12)-11

=10kN

Now draw a section line (1,-1) cutting the members 1,2 and 3 in which

forces are to be determined.

Taking moments of all the forces acting on the left part about point D.

(10*3) = F3*4

F3=7.5kN(Tensile)

Taking moments of all the forces acting on the left part about point G.

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17

(10*3) + (F1*4) =0

F1=-7kN (Comp)

Now taking moments about point C,

(F2*3) – (9*3) +(F3*4) = 0

Substituting the value of F3,

F2=-1.0 kN (Comp)

F1=FCD=7kN(Comp)

F2=FDG=1 kN (Comp)

F3=FGH=7.5 kN(Comp)

7. Analyze given the trusses using Tension coefficient method.

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18

(RC*8) + (50*3) = (100*4)

RC=31.25 kN

RA + RC=100

RA=68.7 kN

HC=50kN

Coefficients:

A=(0,0) B=(4,3) C=(8,0) D=(4,0)

Joint A:

tAB(XB-XA) + tAD(XD-XA)=0

tAB(YB-YA) + tAD(YD-YA)=0

tAB(4) + tAD(4) =0

tAB(3) + tAD(0) =0

Equating this equation,

tAB=0 and tAD=0;

FAB=tAB*LAB=0

FAD=tAD*LAD=0

Joint B:

tBA(XA-XB) + tBD(XD-XB)+tBC(XC-XB)-50=0

tBA(YA-YB) + tBD(YD-YB)+tBC(YC-YB)=0

tBA(-4) + tBD(4-4)+tBC(8-4)-50=0

tBA(-3) + tBD(-3)+tBC(-3)=0

tBC = 12.5 kN

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19

8. Analyse given the trusses using Tension coefficient method:

=sin-(4/5)

=53.13

Taking moment A,

(RC*6) = (8*3) +(6*4)

=8kN

RA=Total Load - RC

=0.

HA=6kN

Coefficients:

A=(0,0) B=(3,0) C=(6,0) D=(6,4) E=(3,4)

Joint A:

tAB(XB-XA) + tAE(XE-XA)=0

tAB(YB-YA) + tAE(YE-YA)=0

tAB(3) + tAE(3)=0

tAB(0) + tAE(4)=0


tAE=0 and tAB=0.

FAB=tAB*LAB=0

FAE=tAD*LAE=0

Joint B:

tBA(XA-XB) + tBC(XC-XB)+tBE(XE-XB)=0

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20

tBA(YA-YB) + tBC(YC-YB)+tBE(YE-YB)-8=0

-3tBA+ (3tBC) = 0

tBA and tBC are zero.

4tBE=8

tBE=2

FBA=tBA*LBA=0

FBC=tBC*LBC=0

FBE=tBE*LBE=2*4

=8 kN (Tension)

Joint C:

tCB(XB-XC) + tCE(XE-XC)+tCD(XD-XC)-6=0

tCB(YB-YC) + tCE(YE-YC)+tCD(YD-YC)+8=0

-3tCB – 3tCE-6 = 0

4tCE+4tCD+8=0


tBC=0;

tCE=-2;

tCD=0;

FCD=0; FCB=0; FCE=(tCE x LCE) = -2 x 5 = -10 kN (Comp)

Joint D:

tDE (XE –XD) + tDC (XC – XD) +6 =0

tDE (YE –YD) + tDC (YC – YD) + 0 =0

-3tDE + 6 = 0 . This implies tDE = 2.0

-4tDC = 0 Therefore tDC = 0

FDC = 0: FDC = tDE x LDE = 2*3 = 6.0 kN (Tension)

Result:

FAB = FAE = 0

FBE = 8 kN (T)

FBC = 0

FCE = 10 kN (c)

FCD = 0

FCE = 6 kN (T)

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ce6302 notes rejinpaul_2.pdf

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