1401802399_Strength of Materials-II 2-2 Set-4 (a)

S.50 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014

B.Tech. II-Year II-Sem. ( JNTU-Anantapur )

Code No : 9A01401/R09

B.Tech II Year II Semester Regular and Supplementary Examinations

April/May - 2013

STRENGTH MATERIALS-II( Civil Engineering )

Time: 3 Hours Max. Marks: 70

Answer any FIVE Questions

All Questions carry Equal Marks

- - -

1. A steel cylinder is 1 m inside diameter and is to be designed for an internal pressure of 8 MN/m2. Calculate thethickness if the maximum shearing stress is not to exceed 35 MN/m2. Calculate the increase in volume, due to workingpressure, if the cylinder is 6 m long with closed ends. E = 200 GN/m2, Poissons ratio = 1/3. (Unit-I, Topic No. 1.1)

2. A compound cylinder formed by shirking one tube to another in subjected to an internal pressure of 90 N/mm2.Before the fluid in admitted, the internal and external diameters of the compound cylinder are 180 mm and 300 mmrespectively and the diameter at the junction is 240 mm. If after shrinking on, the radial pressure at the commonsurface is 12 N/mm2. Determine the final stresses developed in the compound cylinder. (Unit-II, Topic No. 2.2)

3. Solid shaft is subjected to a torque of 100 Nm. Find the necessary shaft diameter if the allowable shear stress is 100MPa and the allowable twist is 3 degree per 10 m length of the shaft. Take C = 1 105 N/mm2. (Unit-III, Topic No. 3.3)

4. A closed coiled helical spring made of round steel wire is required to carry a load of 800 N for a maximum stress notto exceed 200 N/mm2. Determine the wire diameter if the stiffness of the spring is 10 N/mm and the diameter of the helixis 80 mm. Calculate also the number of turns required in the spring given G

steel = 80 kN/mm2. (Unit-IV, Topic No. 4.2)

5. A stanchion is built-up of two 325 mm 165 mm R.S. joists placed 200 mm centre to centre with two 400 mm 12 mmplates riveted to each flange. If it is 6 meters long, both ends fixed, calculate the safe axial load using Rankinesformula and a factor of safety 3. For each joist, area of section = 54.9 cm2, I

XX = 9874.6 cm4, I

YY = 510.8 cm4. Take f

c =

315 N/mm2. (Unit-V, Topic No. 5.3)

6. (a) What is the limit of eccentricity? Explain briefly. (Unit-VI, Topic No. 6.1)

(b) Explain core of section. Find out the core of a circular section? (Unit-VI, Topic No. 6.1)

7. A timber beam 250 mm wide by 300 mm deep is used as simply supported beam on a spam of 5 m. It is subjected to aconcentrated load of 30 N at the mid-section of the span. If the plane of the load makers an angle of 45 with thevertical plane of symmetry find the direction of neutral axis and the maximum stress in the beam.(Unit-VII, Topic No. 7.3)

8. A curved beam, semi circular in plan of radius 5 m, supported on three equally spaced supports. The beam carries auniformly distributed load of 30 kN/m of the circular length. Analyze the beam and sketch the bending moment andtwisting moment diagrams giving the salient values. (Unit-VIII, Topic No. 8.2)

Set-4Solutions

S.51Strength of Materials-II (April/May-2013, Set-4) JNTU-Anantapur


Q1. A steel cylinder is 1 m inside diameter andis to be designed for an internal pressure of8 MN/m2. Calculate the thickness if themaximum shearing stress is not to exceed35 MN/m2. Calculate the increase in volume,due to working pressure, if the cylinder is 6m long with closed ends. E = 200 GN/m2,Poissons ratio = 1/3.

Answer : April/May-13, Set-4, Q1

Given that,

Diameter, d = 1 m

Pressure applied, P = 8 MN/m2

Maximum shearing stress, f = 35 MN/m2

Length of cylinder, l = 6 m

= 6000 mm

Thickness, t = ?

Poissons ratio, 1/m = 1/3

Youngs modulus, E = 200 GN/m2

Longitudinal Stress,

L

= t

Pd

4

35 106 = 8.04

1108 6

t

t = 8.0435

81

t = 0.0714 m

or

= 71.4 mm

Circumferential Stress

c=

t

Pd

2

= 0714.02

1108 6

c= 56.022 106 N/m2

c= 56.02 MN/m2

Longitudinal Stress

l=

t

Pd

4

= 0714.04

1108 6

= 28.011 MN/m2

Longitudinal Strain

el

= E

l

m

21

= 9

6

10200

10011.28

3

121

el

= 4.668 105 (or) 0.4668 104

Circumferential Strain

ec=

Ec

m2

11

= 9

6

10200

1002.56

321

1

ec= 2.334 104

Increase in Volume

We know that,

eV

= VV

V

= eV V

Volumetric strain,

eV

= el + 2e

c

= (0.4668 + 2 2.334) 104

eV

= 5.1348 104

V

= 5.1348 104 4

d2l

= 5.1348 104 4

12 6 m3

V

= 2.419 103 m3

V

= 2.419 103 109 mm3

33 mm102419=V

SOLUTIONS TO APRIL/MAY-2013, SET-4, QP



Q2. A compound cylinder formed by shirking onetube to another in subjected to an internalpressure of 90 N/mm2. Before the fluid inadmitted, the internal and externaldiameters of the compound cylinder are 180mm and 300 mm respectively and thediameter at the junction is 240 mm. If aftershrinking on, the radial pressure at thecommon surface is 12 N/mm2. Determine thefinal stresses developed in the compoundcylinder.


Given that,

Internal pressure, pr = 90 N/mm2

Internal diameter, d1 = 180 mm

Internal radius, r1 = 90 mm

External diameter, d3 = 300 mm

External radius, r3 = 150 mm

Diameter at junction, d2 = 240 mm

Radius at junction, r2 = 120 mm

Inner Cylinder

Let Lames equation be,

px

= 21

x

b a

1and f

x = 2

1

x

b+ a

1

At x= r1 = 90 mm [Q d1 = 180 mm, r1 = 90 mm]

px

= 0

0 = 2

1

90

b a

1... (1)

At x = r2 = 120 mm ; p

x = 12 N/mm2

12 = 2

1

120

b a

1... (2)

From solving equation (1) and (2),

We get,

a1

= 27.428

b1

= 222171.42

Now,

fx

= 242.222171

x + ( 27.428)

f90

= 290

42.222171 27.428

= 54.856 N/mm2

f120

= 2120

42.222171 27.428

= 42.856 N/mm2

Outer Cylinder

Let Lames equation,

px

= 22

x

b a

2 ; f

x = 2

2

x

b + a

2

At x = r3 = 150 mm ; p

x = 0

x = r2 = 120 mm ; p

x = 12 N/mm2

0 = 22

150

b a

2... (3)

12 = 22

120

b a

2... (4)

Solving equation (3) and (4).

a2

= 22

150

b

a2

= 21.33 ; b2 = 480000

Now,

fx

= 22

x

b + a

2

f150

= 2150

480000+ 21.33

f150

= 42.663 N/mm2

f120

= 2120

480000 + 21.33

f120

= 54.66 N/mm2

When cylinder is subjected to internal pressure,

px

= 23

x

b a

3 and f

x =

23

x

b+ a

3

At x = r1 = 90 mm ; p

x = 90 N/mm2

90 = 23

90

b a

3... (5)



At x = r3 = 150 mm ; p

x = 0

0 = 2

3

150

b a

3...(6)

Solving equation (5) and (6).

We get,

a3

= 50.625

b3

= 1139062.5

Now,

fx

= 23

x

b + a

3

Hoop stresses are,

f90

= 290

5.1139062 + 50.625

f90

= 191.25 N/mm2

f150

= 2150

5.1139062 + 50.625

f150

= 101.25 N/mm2

f120

= 2120

5.1139062 + 50.625

f120

= 129.726 N/mm2

Pressure at common surface,

f120

= 2120

5.1139062 50.625

f120

= 28.476 N/mm2

Final Stresses

f90

= 54.856 + 191.25

f90

= 136.39 N/mm2

f120,inner

= 42.856 + 129.726

= 86.87 N/mm2

f120,outer

= 54.66 + 129.726

= 184.386 N/mm2

f150

= 42.663 + 101.25

= 143.913 N/mm2

Q3. Solid shaft is subjected to a torque of 100Nm. Find the necessary shaft diameter if theallowable shear stress is 100 MPa and theallowable twist is 3 degree per 10 m lengthof the shaft. Take C = 1 105 N/mm2.


Given that,

Torque, T = 100 Nm

Allowable shear stress, Ps = 100 N/mm2

Allowable twist, = 3

Length, l = 10 m

Modulus of rigidity, c = 1 105 N/mm2

Let the diameter be d.

We know that,

Torque = Shear stress Polar modulus

Polar modulus = 2/

inertia ofmoment Polar

d

=

2

32

4

d

d

= 16

3d

Torque = Ps Polar modulus

100 103 = 100 16

3d

d3 = 16103

d3 = 5092.95

d = 17.205 mm

Given the allowable twist is 3 per 10 m length ofshaft.

= c

l

PI

T

3 = 5101

100010

32

101004

3

d



180

3 = 510

100010 4

3 3210100

d

d4 = (101859.16 180)/3

d4 = 1945366.72

d = 37.34 mm

The diameter of shaft, d = 37.34 mm.

Q4. A closed coiled helical spring made of roundsteel wire is required to carry a load of 800 Nfor a maximum stress not to exceed 200 N/mm2. Determine the wire diameter if the stiff-ness of the spring is 10 N/mm and the diam-eter of the helix is 80 mm. Calculate also thenumber of turns required in the spring givenGsteel = 80 kN/mm

2.

Answer : April/May-13, Set-4, Q4Given that,

Load, P = 800 NMaximum stress, P

s = 200 N/mm2

Stiffness of the spring = 10 N/mmDiameter of helix, D = 80 mmG

steel = 80 kN/mm2

Mean radius, R = 2

D

= 2

80

= 40 mmStiffness of spring,

S = Deflection

load Axial

S =

P

= S

P

= 10

800

= 80 mm

Section modulus, Z = 16

3d

Maximum shear Stress = PZ

T

Ps = PZ

RP

Ps

=

16

408003d

200 = 31640800

d

d3 = 200

1640800

d = 9.342 mmWire diameter d = 9.34 mm

(ii) Number of Turns

n = 3

4

8PD

Gd

= 3

43

)80(8008

8034.91080

= 14.86 ~ 15Number of turns = 15 Wire diameter = 9.34 mm.

Q5. A stanchion is built-up of two 325 mm 165mm R.S. joists placed 200 mm centre to cen-tre with two 400 mm 12 mm plates rivetedto each flange. If it is 6 meters long, bothends fixed, calculate the safe axial load us-ing Rankines formula and a factor of safety3. For each joist, area of section = 54.9 cm2,IXX = 9874.6 cm4, IYY = 510.8 cm

4. Take fc = 315N/mm2.


200 mm 165 mm

12 mm

400 mm

225 mm

200 mm 165 mm

12 mm

400 mm

225 mm

Figure



Length of span, l = 6 m

= 6000 mm

For single section width of plate, bp = 400 mm

dp = 12 mm

Area of section, A = 54.9 cm2

IXX

= 9874.6 cm4

IYY

= 510.8 cm4

fc

= 315 N/mm2

Area of composite section, A = 2(A + bp d

p)

A = 2(54.9 102 + 400 12)

= 20580 mm2

Moment of inertia about X X

IXX

= 2 Moment of inertia of single section at X X

+ 2

++

23

2distance centre toCentre

12p

pppp ddb

db

= 2 9874.6 104 + 2

++ 2

3

)6200(1240012

12400

= 19749.2 104 + 2[57600 + 203692800]

= 19749.2 104 + 407500800

= 19749.2 104 + 40750.08 104

IXX

= 60499.28 104 mm4

Moment of inertia about Y Y,

IYY

= [2 Moment of inertia of Y Y] +

122

3pp db

= 2 510.8 104 + 2 12

12400 3

IYY

= 1021.6 104 + 11.52 104

IYY

= 1033.12 104 mm4

Here IYY

< IXX

So buckling in columns takes place in Y Y direction.

Given condition,

Both ends are fixed.



Effective length, le

= 2

l

= 2

6

= 3 m

Radius of gyration,

k2 = A

IYY

= 20580

1012.1033 4

k2 = 502 mm2

By using Rankines formulae,

PRankine

=

2

2

.1k

la

Af

e

c

+

=

7500

1Let a

=

5023

.7500

11

205803152

+

= 6482684.5 N

PRankine

= 6482.68 kN

Safe Load= safety ofFactor

RankineP

= 3

68.6482

= 2160.89 kN

Q6. (a) What is the limit of eccentricity? Ex-plain briefly.

Answer : April/May-13, Set-4, Q6(a)

Limit of Eccentricity

A load whose line of action is not coinciding withaxis of column (or) struts is called eccentric loading.

1. This eccentric load acts on a column and producesdirect as well as bending stresses.

2. There is a maximum and minimum stresses on eithersides of neutral axis.

3. If the bending stress is less than direct stress thenthe obtained resultant stress is compressive and theother side the stress will be zero in case of bendingstress equal to direct stress.

4. The tensile stresses are produced when bendingstress exceeds the direct stress.

fb

fd

Z

eP.

A

P

e A

Z

fd

Direct stress AP

e

P

fb

Bending stress

AP.e

P

fd

Direct stress AP

e

P

fb

Bending stress

AP.e

fd

Direct stress AP

e

P

fb

Bending stress

AP.e

P

fb

= Bending stress

fd

= Direct stress.

Limit of eccentricity for some section is as follows,

1. Rectangular section

2. Hollow rectangular section

3. Circular section

4. Hollow circular section.

For remaining answer refer Unit-VI, Q2, Topics: (i),(ii), (iii), (iv).

(b) Explain core of section. Find out thecore of a circular section?

Answer : April/May-13, Set-4, Q6(b)

For answer refer Unit-VI, Q2(iii).

Q7. A timber beam 250 mm wide by 300 mm deepis used as simply supported beam on a spamof 5 m. It is subjected to a concentrated loadof 30 N at the mid-section of the span. If theplane of the load makers an angle of 45 withthe vertical plane of symmetry find the di-rection of neutral axis and the maximumstress in the beam.

April/May-13, Set-4, Q7



Answer :

A D

30 kN

45

X

Y

300 m

m

N

y

x

250 mm

5 m

N

30 kN

B C

A D

30 kN

45

X

Y

300 m

m

N

y

x

250 mm

5 m

N

A D

30 kN

45

X

Y

300 m

m

N

y

x

250 mm

5 m

N

30 kN

B C

Given that,

Width, b = 250 mm

Depth, d = 300 mm

Clear span, l = 5 m

Concentric load, P = 30 N

= 45Moment of Inertia about X-axis

IXX

= 12

3bd

= 12

300250 3

= 562.5 106 mm4

Moment of Inertia about Y-axis

IYY

= 12

3bd

= 12

250300 3

= 390.62 106 mm4

Maximum bending moment is,

= 4

Pl=

4

530

= 37.5 kN

By resolving the bending moment in X and Y direc-tions, we get,

MXX

= 37.5 cos45

= 26.516 kNm

MYY

= 37.5 sin45

= 26.516 kNm.

Due to this bending moment the NA (Neutral Axis)will be inclined with Y-axis.

Here compression occurs at top of NN and tensionoccurs at the bottom of NN.

= yI

M

XX

XX . + xI

M

YY

YY .

Both X and Y should be positive for maximum bend-ing moment.

max

= 6

6

105.562

10516.26

150 + 6

6

1062.390

10516.26

125

max

= 15.556 mm2

For = 0 we get the neutral axis NN.

yI

M

XX

XX . + xI

M

YY

YY . = 0

6

6

105.562

10516.26

.y + 6

6

1062.390

10516.26

.x = 0

0.047y + 0.067x = 0

0.047y = 0.067x

x

y=

047.0

067.0

tan = 1.42

= 54.50

= 5450'



30 kNYN

XX

Y N

5457'

30 kNYN

XX

Y N

5457'

Figure

Q8. A curved beam, semi circular in plan of ra-dius 5 m, supported on three equally spacedsupports. The beam carries a uniformly dis-tributed load of 30 kN/m of the circular length.Analyze the beam and sketch the bendingmoment and twisting moment diagrams giv-ing the salient values.


/2

/2

OK

P

N

M

Q

RS

L

r = 5 m

/2

/2

OK

P

N

M

Q

RS

L

r = 5 m

Given that,

Radius of semi-circle, r = 5 m

Uniformly distributed load, P = 30 kN/m.

Reactions

Reaction at K,

RK

= 2

.rP( 2)

= 2

530( 2)

= 85.619 kN

RL

= 2 P.r

= 2 30 5

RL

= 300 kN

Shear Force

Shear force at K,

SFK

= RK

= 85.619 kN

Shear force at M,

SFM

= P.r

= 30 5

= 150 kN

Shear force at any other position,

SF = Pr

1

2

= 30 5

1801

2

= 150[0.5707 0.0174] ... (1)

at = 90

SF90

= 150[0.5707 0.0174 90]

= 149.29 kN

Now equating equation (1) with zero.

SF = 0

150[0.5707 0.0174]= 0

= 0174.0

5707.0

= 32.79 (or) 3247'

Bending Moment

Bending moment at M,

BMM

= 0.429 Pr2

[Q i.e., hogging moment]

= 0.429 30 52

= 321.75 kN-m (Hogging)

Maximum bending moment,

BMmax

= 0.1514 Pr2

= 0.1514 30 52

= 113.55 kN-m (Sagging)



Bending moment at any other position,

BM = Pr2[0.5708 sin 2 sin2/2]

= 30 52[0.5708 sin 2 sin2/2]

BM = 750[0.5708 sin 2 sin2/2]

= 30

BM = 750[0.5708 sin30 2 sin230/2]

BM30

= 113.569 kN-m

BM60

= 750[0.5708 sin60 2 sin260/2]

= 4.254 kN-m

BM45

= 750[0.5708 sin45 2 sin245/2]

= 83.04 kN-m

BM90

= 750[0.5708 sin90 2 sin290/2]

= 321.9 kN-m

BM120

= 750[0.5708 sin120 2 sin260]

= 754.24 kN-m

BM150

= 750[0.5708 sin150 2 sin275]

= 1185.46 kN-m

BM180

= 750[0.5708 sin180 2 sin290]

= 1500 kN-m

0 30 45 60 90 120 150 180

(+)

113.56

83.04

4.275

321.9

754.24

1185.46 1500 kN-m

BM diagram

()

0 30 45 60 90 120 150 180

(+)

113.56

83.04

4.275

321.9

754.24

1185.46 1500 kN-m

BM diagram

()

Figure

Twisting Moment

Maximum torsional moment,

tmM = 0.1045 Pr2

= 0.1045 30 52

= 78.375 kN-m

M = 0

750[0.5708 sin 2 sin2/2] = 00.5708 sin 2 sin2/2 = 0

We know that,

sin = 2 sin/2 . cos/2 0.5708 2 sin/2 . cos/2 = 2 sin2/2

0.5708 . cos/2 = sin/2 tan/2= 0.5708

/2 = tan1(0.5708)/2 = 29.717 = 59.43

The torsional moment distribution is given as,

M= Pr2

+ sincos

2

2

2

2

= 30 52

+

180sincos

2

2

2

2

The values of torsional moments are tabulated be-low,

Centre of beam090

78.1660

Zero bending moment point

78.1959.43

39.6530

End of beams00

PositionTorsional moment at

any other pointAngle ()

Centre of beam090

78.1660

Zero bending moment point

78.1959.43

39.6530

End of beams00

PositionTorsional moment at

any other pointAngle ())(M t

0 30 60 90 120 150 180

Twistingmomentdiagram

78.31 kN-m

78.31 kN-m

0.15

0 30 60 90 120 150 180

Twistingmomentdiagram

78.31 kN-m

78.31 kN-m

0.15

Figure

Documents

1401802399_Strength of Materials-II 2-2 Set-4 (a)