1401802399_Strength of Materials-II 2-2 Set-1 (a)

S.1Strength of Materials-II (April/May-2013, Set-1) JNTU-Anantapur

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 closed cylindrical vessel made of steel plates 5 mm thick with plane ends, carries fluid under pressure of 6 N/mm2.The diameter of the cylinder is 35 cm and length is 85 cm. Calculate the longitudinal and hoop stresses in the cylinderwall and determine the change in diameter, length and volume of the cylinder. Take E = 2.1 105 N/mm2 and 1/m =0.286. (Unit-I, Topic No. 1.1)

2. (a) What are the different methods of reducing hoop stresses? Explain the terms: Wire winding of thin cylindersand shrinkage one cylinder over another cylinder. (Unit-II, Topic No. 2.2)

(b) Derive Lames equations. (Unit-II, Topic No. 2.1)

3. A hollow steel shaft of external diameter equal to twice the internal diameter has to transmit 2250 kW power at 400r.p.m. If the angle of twist has not to exceed 1 in a length equivalent to 10 times the external diameter and themaximum turning moment is 1/4 times the mean, calculate the maximum stress and diameter of the shaft. Assume themodulus of rigidity to be 0.8 105 N/mm2. (Unit-III, Topic No. 3.2)

4. A helical spring, in which the mean diameter of the coils is 8 times the wire diameter, is to be designed to absorb 200N-m of energy with an extension of 10 cm. The maximum shear stress is not to exceed 125 MPa. Determine the meandiameter of the helix, diameter of the wire and the number of turns. Also find the load with which an extension of 4 cmcould be produced in the spring. G = 84 GPa. (Unit-IV, Topic No. 4.2)

5. (a) Derive the Eulers buckling load for a column with both ends hinged. (Unit-V, Topic No. 5.2)

(b) Find the ratio of buckling strength of a solid column to that of a hollow column of the same material and havingthe same cross-sectional area. The internal diameter of the hollow column is half of its external diameter. Boththe columns are hinged and the same length. (Unit-V, Topic No. 5.2)

6. A hollow rectangular masonry pier is 120 cm 80 cm wide and 15 cm thick. A vertical load of 200 kN is transmitted inthe vertical plane bisecting 120 cm side and at an eccentricity of 10 cm from the geometric axis of the section.Calculate the maximum and minimum stress intensities in the section. (Unit-VI, Topic No. 6.2)

7. A cantilever of rectangular section 40 mm (width) 60 mm (depth) is subjected to an inclined load P at the free end.The inclination of the load is 25 to the vertical. If the length of the cantilever is 2 meters and maximum stress due tobending is not to exceed 200 MN/m2, determine the value of P. (Unit-VII, Topic No. 7.3)

8. Find the bending moment at mid span of the semicircular beam of diameter 6 m loaded at the mid span with aconcentrated load of 80 kN. The beam is fixed at both supports. Find the maximum bending moment and maximumtorque in the beam. (Unit-VIII, Topic No. 8.2)

Set-1Solutions

S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014


SOLUTIONS TO APRIL/MAY-2013, SET-1, QPQ1. A closed cylindrical vessel made of steel

plates 5 mm thick with plane ends, carriesfluid under pressure of 6 N/mm2. Thediameter of the cylinder is 35 cm and lengthis 85 cm. Calculate the longitudinal and hoopstresses in the cylinder wall and determinethe change in diameter, length and volumeof the cylinder. Take E = 2.1 105 N/mm2 and1/m = 0.286.

Answer : April/May-13, Set-1, Q1

Given that,

Thickness, t = 5 mm = 0.5 cm

Fluid pressure, P = 6 N/mm2

Length of cylinder, L = 85 cm

Diameter of cylinder, D = 35 cm

Youngs modulus, E = 2.1 105 N/mm2

Poissons ratio, = m

1 = 0.286

Let,

1 - Hoop stress

2 - Longitudinal stress.

Hoop stress is given by,

1

= t

PD

2

= 5.02

356

1

= 210 N/mm2

Longitudinal stress is given by,

2

= t

PD

4

= 5.04

356

2

= 105 N/mm2

(i) Circumferential strain is given by,

e1

= EE

21

= E

1[

1

2]

e1

= 5101.2

1

[210 0.286 105]

= 8.57 104

Circumferential is also given by,

e1

= D

D

D

D= 8.57 104

D = 8.57 104 D

D = 8.57 104 35

D = 0.029 cm

Increase in diameter, D = 0.029 cm

(ii) Longitudinal strain is given by,

e2

= EE

12

e2

= E

1[

2

1]

e2 = E1

[2

1]

= 5101.2

1

[105 0.286 210]

= 2.14 104

Longitudinal strain is also given by,

e2

= L

L

L

L= 2.14 104

L = 2.14 104 L

L = 2.14 104 85

L = 0.018 cm

Increase in Length, L = 0.018 cm



(iii) Volumetric strain is given by,

V

V= 2

D

D +

L

L

= 2e1 + e

2

= 2 8.57 104 + 2.14 104

VV

= 1.928 103

V = 1.928 103 V

= 1.928 103 4

D2 L

= 1.928 103 4

352 85

V = 157.67 cm3

Change in volume, V = 157.67 cm3

Q2. (a) What are the different methods of reducing hoop stresses? Explain the terms: Wire windingof thin cylinders and shrinkage one cylinder over another cylinder.

Answer : April/May-13, Set-1, Q2(a)

The different methods of reducing hoop stresses are,

(i) Wire winding of thin cylinders

(ii) Shrinkage one cylinder over another cylinder.

(i) Wire Winding of Thin Cylinders

d

t

Wire Wound Round Apipe

Unit Length

w w

Wire

Cylinder

cc

d

t

Wire Wound Round Apipe

Unit Length

w w

Wire

Cylinder

cc

w

- Winding stress in wire

c

- Compression stress exerted by wire on cylinder.

1. The strong steel wire is winded on the walls of the cylinder under tension to reduce the hoop stresses.

2. By this wire winding, the walls of the cylinder are subjected to an initial compressive stress.



3. Now, if the cylinder is subjected to the internal fluid pressure, the hoop tensile stress is caused in the walls.

4. The resultant stress will be less with the net effect of initial compressive stress due to the wire winding and due tointernal fluid pressure.

5. The internal fluid pressure minus the initial compressive stress will be resultant stress in the material.

Resultant stress = Internal fluid pressure Initial compressive stress.

6. The sum of tensile stress due to internal pressure in the cylinder and initial tensile winding stress will be the stressin the wire.

Stress in the wire = Tensile stress in cylinder + Initial tensile winding stress

= c +

w.

By considering a length L of cylinder, we can obtain the relation between w and

c.

Initial tensile force in the wire for length L of cylinder = n

2

42 wd

w

Where,

n Number of turns of wire

2wd Diameter of wire.

n = wd

L

Initial tensile force = wd

L 2

4

2wd w

= L 2

d

w

w

Compressive force exerted by the wire on cylinder for length L = 2 L t c.

For equilibrium,

Initial tensile force in wire = Compressive force on the cylinder

L 2

d

w

w= 2 L t

c

c = td ww

4

Where,

'c - Circumferential stress developed in the cylinder due to fluid pressure only.

'w - Stress developed in the wire due to fluid pressure only.

Resultant stress in the cylinder = ( 'c c)

Resultant stress in the wire = (w

+ 'w )



d

L

c

dt

d

L

c

dt

Initial compressive stress in cylinder = c

Initial compressive force in cylinder for length L = 2L t c

d

L

w

dw

2nd Turn of Wire

w

1st Turn of Wire

F

d

L

w

dw

2nd Turn of Wire

w

1st Turn of Wire

F

Initial tensile stress in wire = w

Initial tensile force in wire for length L = n

2

42 wd

w

n - Number of turns in length L.

(ii) Shrinkage of One Cylinder over another Cylinder

1. Due to shrinkage, the inner and outer cylinders are subjected to the initial compressive and initial tension respec-tively.

2. The inner and outer cylinders are subjected to the hoop tensile stress when the compound cylinder is subjected tothe internal fluid pressure.

3. The resultant stresses will be more or less uniform with the net effect of initial stresses due to shrinkage and internalfluid pressure.



r2

r1

r

r2

r1

r

The figure shows the compound thick cylinder con-sisting of two cylinders.

Let,

r - Radius at junction of two cylinders

r1

- Inner radius of compound cylinder

r2

- Outer radius of compound cylinder

p - Radial pressure at junction of two cylinders.

Apply Lames equation to the outer and innercylinders.

(i) For Outer Cylinder

Lames equation at a radius x for outer cylinder aregiven by,

px

= 21

x

b a

1... (1)

x= 2

1

x

b + a

1... (2)

Where,

a1, b1 - Constant for outer cylinder.

At x = r2, p

x = 0 and at x = r, p

x = p

Substitute these in equation (1),

= 22

1

r

b a

1... (3)

px=

21

r

b a

1... (4)

From equations (3) and (4), the constants can bedetermined.

By substituting the values of a1 and b

1 in equation

(2), the hoop stresses in the outer cylinders due to shrinkagecan be obtained.

(ii) For Inner Cylinder

Lames equation at a radius x for inner cylinders aregiven by,

px= 2

2

x

b a

2... (5)

x= 2

2

x

b + a

2... (6)

Where,

a2, b

2 - Constants of inner cylinder.

At x = r1, p

x = 0 and at x = r, p

x = p

Substitute these in equation (5),

= 21

2

r

b a

2... (7)

px=

22

r

b a

2... (8)

From equations (7) and (8), the constants a2 and b

2

can be determined.

By substituting the values of a2 and b

2 in equation

(6), hoop stresses in the inner cylinders can be obtained.

(b) Derive Lames equations.

Answer : April/May-13, Set-1, Q2(b)

For answer refer Unit-II, Q2.

Q3. A hollow steel shaft of external diameterequal to twice the internal diameter has totransmit 2250 kW power at 400 r.p.m. If theangle of twist has not to exceed 1 in a lengthequivalent to 10 times the external diameterand the maximum turning moment is 1/4times the mean, calculate the maximumstress and diameter of the shaft. Assume themodulus of rigidity to be 0.8 105 N/mm2.


Given that,

External diameter = 2 (Internal diameter)

d2 = 2d

1

Power transmitted, P = 2250 kW

= 2250 103 W

Speed of rotation, N = 400 r.p.m

Length of shaft, L = 10 External diameter

L = 10 d2



Maximum turning moment,

Tmax

= 4

1 Mean turning

Tmax

= 4

1 T

mean

Angle of twist,

= 1

= 1 180

radians

= 0.017 rad

Modulus of rigidity, C = 0.8 105 N/mm2

Power transmitted is given by,

P = 60

2 NT

2250 103 = 60

4002 meanT

2250 103 60 = 800 Tmean

Tmean =

800

60102250 3

Tmean = 53714.79 N-m= 53714793.29 N-mm

Tmax

= 4

1 T

mean

= 4

1 53714793.29

= 13428698.32 N-mm

(i) Diameter of the shaft is given by,

PI

T=

L

C

Where,

T Torque

IP

Polar moment of inertia

C Modulus of rigidity

L Length of the shaft.

4

12 )(32

32.13428698

dd = 2

5

10

017.0108.0

d



412 )(

3232.13428698

dd

= 2

5

10

017.0108.0

d

0.8 105 0.017 (d2 d1)4 = 13428698.32 32 10d2 4272.56(d2 d1)4 = 429718346.2 10d2 (d2 d1)4 = 105763.164 d2

4

22 2

d

d = 1005763.164 d2

4

2

2

d

= 1005763.164 d2

16

42d = 1005763.164 d

2 16

42d = 16092210.62 d2

2

42

d

d= 16092210.62

32d = 16092210.62

d2 = 252.46 mmQ d2 = 2d1

d1 =

22d

d1 =

2

46.252

d1 = 126.23 mm

Internal diameter, d1 = 126.23 mm

External diameter, d2 = 252.46 mm

Diameter of the shaft, D = d

1 + d

2

= 126.23 + 252.46= 378.69 mm

(ii) Maximum stress is given by,

T = 16

D3

13428698.32= 16

(378.69)3

= 3)69.378(1632.13428698

= 1.259 N/mm2

Maximum stress, = 1.259 N/mm2.



Q4. A helical spring, in which the mean diameter of the coils is 8 times the wire diameter, is to bedesigned to absorb 200 N-m of energy with an extension of 10 cm. The maximum shear stress isnot to exceed 125 MPa. Determine the mean diameter of the helix, diameter of the wire and thenumber of turns. Also find the load with which an extension of 4 cm could be produced in thespring. G = 84 GPa.


Given that,

Mean diameter of coil = 8 times wire diameter

D = 8 d

Torque, T = 200 N-m = 200 103 N-mm

Extension, 1 = 10 cm = 100 mm

Maximum shear stress, = 125 MPa = 125 N/mm2

Extension, 2 = 4 cm = 40 mm

Modulus of rigidity, C or G = 84 GPa

= 84 103 N/mm2

Q Mean diameter of coil = 8 Wire diameter

Mean radius of coil = 4 Wire diameter

R = 4 d

(i) Diameter of the Wire

We known that twisting moment,

T = 16

d3

200 103 = 16

125 d3

d3 = 125

1610200 3

d3 = 8148.73

d = 20 mm

Diameter of the wire, d = 20 mm.

(ii) Mean Diameter of the Helix

Mean diameter of helix or coil,

D = 8 d

= 8 20

D = 160 mm

Mean diameter of helix, D = 160 mm



(iii) Number of Turns

We know that twisting moment,

T = W.R.

R = 4 d = 4 20 = 80 mm

200 103 = W 80

W = 80

10200 3

W = 2500 N

Deflection of the spring,

1

= 4

364

Gd

nWR

100= 433

)20(1084

)80(250064

n

n = 343

80250064

201084100

n = 16

Number of turns, n = 16

(iv) Load when the Extension is 4 cm

2 = 4 cm = 40 mm

Deflection of the spring,

2

= 4

364

Gd

nWR

40 = 433

)20(1084

16)80(64

W

W = 16)80(64

201084403

43

W = 1025.39 N

Load when the extension is 4 cm, W = 1025.39 N.

Q5. (a) Derive the Eulers buckling load for a column with both ends hinged.

Answer : April/May-13, Set-1, Q5(a)

For answer refer Unit-V, Q8.



(b) Find the ratio of buckling strength of a solid column to that of a hollow column of the samematerial and having the same cross-sectional area. The internal diameter of the hollowcolumn is half of its external diameter. Both the columns are hinged and the same length.

Answer : April/May-13, Set-1, Q5(b)

Given that,

Internal diameter of hollow column = 2

1 External diameter of hollow column

di =

2

1 d

e

Let,

D Diameter of solid column

di

Internal diameter of hollow column

de

External diameter of hollow column.

Both the columns are made of same material and have same length, cross-sectional area and the end conditions.

Area of solid column = Area of hollow column.

2

D2 =

2

[ ]22 ie dd

2

D2 =

2

22

2 ee

dd

2

D2 =

2

4

2

2 ee

dd

D2 =

2

2

4

2

2 ee

dd

D2 = 4

4 22 ee dd

D2 = 4

3 2ed

D = 2

3 ed ... (1)

Buckling load of a column is given by Eulers formula as,

P = 2

2

L

EI



Q All the properties of both the columns are same.i.e., A, E and L.

P I ... (2)

Where,

P = Buckling load

I = Moment of inertia.

Let,

P1

Buckling load of hollow column

P2 Buckling load of solid column

I1

Least moment of inertia of hollow column

I2 Least moment of inertia of solid column.

I1 = 64 [ ]44 ie dd

= 64

44

2 ee

dd

= 64

16

4

4 ee

dd

= 64

16

16 44 ee dd

I1=

64

16

15 4ed

And I2=

64

D4

From equation (2),

1

1

I

P=

2

2

I

P

2

1

P

P=

2

1

I

I

2

1

P

P=

4

4

64

16

15

64

D

de

2

1

P

P= 4

4

16

15

D

de

But from equation (1), D = 2

3 ed

2

1

P

P= 4

4

2

316

15

e

e

d

d

=

169

16

154

4

e

e

d

d

= 4

4

916

1516

e

e

d

d

2

1

P

P=

3

5

2

1

P

P= 1.667

column solid of load Buckling

column hollow of load Buckling = 1.667

Q6. A hollow rectangular masonry pier is 120 cm 80 cm wide and 15 cm thick. A vertical loadof 200 kN is transmitted in the vertical planebisecting 120 cm side and at an eccentricityof 10 cm from the geometric axis of the sec-tion. Calculate the maximum and minimumstress intensities in the section.


Given that,

Width, B = 120 cm = 1200 mm

Depth, D = 80 cm = 800 mm

Thickness, t = 15 cm = 150 mm

Load, P = 200 kN = 200 103 N

Eccentricity, e = 10 cm = 100 mm

(i) Maximum Stress Intensity in the Section

Area of pier,

A = (BD bd)

b = B t



= 1200 2 150

b = 900 mm

d = D t

= 800 2 150

d = 500 mm

A = (1200 800 900 500)

A= 510000 mm2

Section modulus,

Z = 6

1[BD2 bd2]

= 6

1[1200 8002 900 500]

Z = 90.5 106 mm3

Moment due to eccentricity of load,

M = P e

= 200 103 100

M = 200 105 N-mm

Maximum stress intensity in section,

max

= A

P +

Z

M

= 510000

10200 3 + 6

5

105.90

10200

max

= 0.613 N/mm2

(ii) Minimum Stress Intensity in the Section

min

= A

P

Z

M

= 510000

10200 3 6

5

105.90

10200

min

= 0.171 N/mm2

(b) Second Method

Area of pier, A = (BD bd)

= (1.2 0.8 0.9 0.5)

A = 0.51 m2

1.2 mX

0.9 m

Xe

Y

Y

0.8 m

0.5 m

min

ma x

1.2 mX

0.9 m

Xe

Y

Y

1.2 mX

0.9 m

Xe

Y

Y

0.8 m

0.5 m

min

ma x

Figure

Direct stress,

d=

A

W

= 51.0

200

= 392.15 kN/m2

Bending stress,

b=

Z

M

= Z

eW

Z = y

I



I = 12

3BD

12

3bd

= 12

8.02.1 3

12

5.09.0 3

I = 0.0418 m4

y = 2

D =

2

8.0 = 0.4 m

Z = y

I =

4.0

0418.0 = 0.1045 m3

b

= Z

M

= 1045.0

1.0200

b

= 191.38 kN/m2

Maximum stress,

max

= d

b0

= 392.15 + 191.38

= 583.53 kN/m2

Minimum stress,

min

= d

b

= 392.15 191.38

min

= 200.77 kN/m2

Q7. A cantilever of rectangular section 40 mm(width) 60 mm (depth) is subjected to aninclined load P at the free end. Theinclination of the load is 25 to the vertical. Ifthe length of the cantilever is 2 meters andmaximum stress due to bending is not toexceed 200 MN/m2, determine the value ofP.


Given that,

Width of section, b = 40 mm

Depth of section, d = 60 mm

Inclination angle, = 25

Length of cantilever, L = 2 m = 2000 mm

Maximum bending stress,

b= 200 MN/m2

= 200 N/mm2

X X

D C

Y

Y

25

60 mm

P

A B

40 mmA

N

2 m

P

X X

D C

Y

Y

25

60 mm

P

A B

40 mmA

N

2 m

P

For the given section,

IX

= 12

3bd

= 12

6040 3

IY

= 72 104 mm4

IY

= 12

3bd

= 12

4060 3

IY

= 32 104 mm4

The X and Y axes are the principal axes of sectionthrough centroid.

If is the inclination of Neutral Axis (NA),

tan =

Y

X

I

Itan

tan =

4

4

1032

1072tan25

tan = 1.049 = 46



Maximum bending moment = P L

MX = P cos L = P cos25 2000 = 1812.61P N-mm

MY = P sin L = P sin25 2000 = 845.23P N-mm

MX causes compression at C and D and tension at A and B

MY causes compression at B and C and tension at A and D.

B=

X

X

I

dPM

2

+ Y

Y

I

bPM

2

B

= 41072

3061.1812

P

+ 41032

2023.845

P

Equating this bending stress to given bending stress.

41072

3061.1812

P

+ 41032

2023.845

P

= 200

0.075 P + 0.050 P = 200

0.125 P = 200

P = 1600 N Inclined load, P = 1600 NStresses at different points are,

A= 41072

3061.1812

P

+ 41032

2023.845

P

= 41072

30160061.1812

+ 41032

20160023.845

A= 205.36 MPa (Tensile)

B= 41072

3061.1812

P

41032

2023.845

P

= 41072

30160061.1812

41032

20160023.845

= 36.31 MPa (Compressive)

C= 205.36 MPa (Compressive)

D= 36.31 MPa (Tensile)

Q8. Find the bending moment at mid span of the semicircular beam of diameter 6 m loaded at themid span with a concentrated load of 80 kN. The beam is fixed at both supports. Find the maxi-mum bending moment and maximum torque in the beam.


Given that,

Diameter of semi-circular beam, d = 6 m = 6000 mm

Concentrated load, P = 60 kN = 60 103 N



Let,

Rigidity of modulus, G = 80 GPa

G = 80 106 kN/m2

Modulus of elasticity, E = 200 GPa = 200 106 kN/m2

Radius, R = 2

d = 3 m

N

M

LK

K L

P

O

90 45

80 kN

N

M

LK

K L

P

O

90 45

80 kN

The maximum bending moment acts at the mid-span of the beam.

MM

= +

+2

22

sin)1()1(2

sin)sincos22( (PR)

Here,

= 90

= GJ

EI

Moment of inertia,

I = 64

d4

= 64

(6)4

= 63.61 m4

Polar moment of inertia,

J = 32

d4

= 32

(6)4

J = 127.23 m4

= 23.1271080

61.63102006

6

= 1.24



K1

= (2 2 cos sin2)= 2 2 cos90 sin90

K1

= 1

K2

= sin2= sin290

K2

= 1

K3

= 2( + 1) ( 1)sin2

= 2 90 180

(1.24 + 1) (1.24 1)sin290

K3

= 6.79

MM

= +

+90sin)1()1(902

90sin)90sin90cos22(24.12

22

[80 3]

= )380(90sin)124.1()124.1(902

90sin)90sin90cos22(24.12

22

+

+

MM

= 1.33 kNm

(MM @ = 45)=

++

2

22

sin)1()1(2

sin)sincos22( (PR)

= +

+45sin)124.1()124.1(452

45sin)45sin45cos22(24.12

22

(80 3)

= 0.722 kNm

0.722 kNm

1.33 kNm1.33 kNm

Bending moment diagram

80 kN

M

PQN

OKRK

LRL

0.722 kNm

1.33 kNm1.33 kNm

Bending moment diagram

80 kN

M

PQN

OKRK

LRL

Reactions

RK + R

L = 80 kN

Taking forces about tangent at any point M.

RK 3 80(OP OM) = 0

3RK = 80(OP OM)



OP = 3 m

OM = 3 cos45

OM = 2.12

3RK = 80(3 2.12)

3RK = 70.4 RK = 23.467 kN R

K + R

L = 80

23.467 + RL = 80 RL = 80 23.467 RL = 56.53 kN

Point of Zero Bending Moment

M = RK Rsin 80 R sin

2

For zero bending moment, M = 0

RK R sin 80 R sin

2 = 0

RK R sin = 80 R sin

2

23.467 sin = 80 sin

2

2sin

sin =

467.23

80

2sin.cos

2cos.sin

sin

= 3.40

cos

sin= 3.40

tan = 3.40 = tan1(3.40) = 73.61 = 7336'

Maximum Torsional Momenti.e., Torque in beam at = + 7336'

'maxM = RK (R R cos( 7336')) 80(R R cos( 7336' 45))

= 23.467[3 3 cos( 7336')] 80[3 3 cos( 11836')]

'maxM = 304.36 kNm

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1401802399_Strength of Materials-II 2-2 Set-1 (a)