1401802399_Strength of Materials-II 2-2 Set-3 (a)

S.32 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 cast iron cylinder of 200 mm inner diameter and 12.5 mm thick is closely wound with a layer of 4 mm diameter steelwire under a tensile stress of 55 MN/m2. Determine the stresses set up in the cylinder and steel wire if water under apressure of 3 MN/m2 is admitted in the cylinder. Take E

cast iron = 100 GN/m2, E

steel = 200 GN/m2 and Poisson’s ratio = 0.25.

(Unit-I, Topic No. 1.1)

2. (a) Stating assumptions derive Lame’s equations to find out the stresses in a thick cylindrical shell.(Unit-II, Topic No. 2.1)

(b) A hollow cylinder has an external diameter of 250 mm and thickness of the wall is 50 mm. The cylinder issubjected to an internal fluid pressure = 35 MPa and external pressure = 3.5 MPa. Calculate the maximum andminimum circumferential stresses and plot the variation of the same across the wall thickness.(Unit-II, Topic No. 2.1)

3. A hollow steel shaft of external diameter equal to twice the internal diameter, 5 m long is to transmit 160 kW of powerat 120 r.p.m. The total angle of twist is not to exceed 2° in this length and the allowable shear stress is 50 N/mm2.Calculate diameter of the shaft. (Unit-III, Topic No. 3.2)

4. A close-coiled helical spring is required to have an axial stiffness of 5 kN/m and an angular stiffness of 0.1 Nm perdegree angle of twist. If the spring is made of steel wire 6 mm diameter, find the mean diameter of the coil and thenumber of turns required. Assuming E = 200 GPa and G = 80 GPa. (Unit-IV, Topic No. 4.2)

5. (a) Derive an expression for crippling load when one end of the column is fixed and the other end is free.(Unit-V, Topic No. 5.2)

(b) Calculate the Euler’s critical load for a strut of T-section. The flange width being 10 cm, overall depth 8 cm andboth flange and stem 1 cm thick, the strut is 3 m long and is built in at both ends. Take E = 2 × 10 N/mm3.(Unit-V, Topic No. 5.2)

6. A masonry retaining wall, 7 meters high, is trapezoidal in section, 1 meter wide at the top and 3 meters at the base, withone side vertical. If the lateral pressure exerted by the retained material on the vertical face varies from zero at the topto 25 kN/m2 at the base, calculate the maximum and minimum stresses induced in the base, the weight of masonrybeing 21 kN/m3. (Unit-VI, Topic No. 6.2)

7. Find the centroidal principal moments of inertia of a equal angle section 30 × 30 × 8 mm. (Unit-VII, Topic No. 7.3)

8. Find the bending moment at mid span of the semicircular beam of diameter 9 m loaded at the mid span with aconcentrated load of 60 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-3Solutions

S.33Strength of Materials-II (April/May-2013, Set-3) JNTU-Anantapur


Q1. A cast iron cylinder of 200 mm inner diameter and 12.5 mm thick is closely wound with a layer of4 mm diameter steel wire under a tensile stress of 55 MN/m2. Determine the stresses set up in thecylinder and steel wire if water under a pressure of 3 MN/m2 is admitted in the cylinder. TakeEcast iron = 100 GN/m2, Esteel = 200 GN/m2 and Poisson’s ratio = 0.25.

Answer : April/May-13, Set-3, Q1

Given that,

Cast iron cylinder

Inner diameter, D = 200 mm

= 0.2 m

Thickness of wall, t = 12.5 mm

= 0.0125 m

Pressure admitted, P = 3 MN/m2

Young’s modulus, Ecast iron

= 100 GN/m2

Steel wire

Diameter of steel wire, Dw = 4 mm

= 0.004 m

Tension in the steel wire, σw = 55 MN/m2

Young’s modulus, Esteel

= 200 GN/m2

Poissons ratio m

1 = 0.25

Stresses Set up in the Cylinder and Steel Wire

Before the admitted water into the cylinder,

Per unit length tensile force exerted by wire = Compressive forced producing in the cylinder.

2 × 4

π × 2

wD × σw × n = 2 × t × 1 × σ

c

Circumferential compression in the cylinder is,

σc=

t4

πD

w × σ

w

= 0125.04

004.0

××π

× 55

= 13.823 MN/m2

After admitting water into the cylinder,

Longitudinal stress developed in the cylinder,

SOLUTIONS TO APRIL/MAY-2013, SET-3, QP



1l

σ = t

PD

4

= 0125.04

2.03

××

= 12 MN/m2

Bursting force = Unit length of total resisting force

P × D × 1 = 1cσ × 2t × 1 +

1wσ × 2 × 4

π × 2

wD × n

PD = 1cσ × 2t +

1wσ × 2

π 2wD ×

wD

1

=

wDn

1Q

PD = 1cσ × 2t +

1wσ × 2

πD

w

3 × 0.2 = 1cσ × 2 × 0.0125 +

1wσ × 2

π × 0.004 ... (1)

Here,

Circumferential strain in cylinder = Circumferential strain in wire

c

c

E1

σ –

c

l

mE1

σ=

w

w

E1

σ

Here,

25.010100

102–

10100

109

6

9

61 ×

××1

×

×σc = 9

6

10200

101

×

×σw

10–5 × [21cσ – 2 × 12 × 0.25] =

1wσ × 10–5

21cσ – 6 =

1wσ ... (2)

Substitute equation (2) in equation (1),

0.6 = 1cσ × 2 × 0.0125 + (2

1cσ – 6)2

π × 0.004

0.6 = 0.0251cσ + 0.0125

1cσ – 0.0376

0.6 = 0.03751cσ – 0.0376

0.6376 = 0.03751cσ

1cσ = 17 MN/m2

1wσ = 2 × 17 – 6

= 28 MN/m2



In cylinder,

Resultant stress,

= 1cσ – σ

c

= 17 – 13.82

= 3.18 MN/m2 (T)

In steel wire,

Resultant stress,

=σw +

1wσ

= 55 + 28

= 83 MN/m2 (T)

Q2. (a) Stating assumptions derive Lame’s equations to find out the stresses in a thick cylindricalshell.

Answer : April/May-13, Set-3, Q2(a)

For answer refer Unit-II, Q2.

(b) A hollow cylinder has an external diameter of 250 mm and thickness of the wall is 50 mm.The cylinder is subjected to an internal fluid pressure = 35 MPa and external pressure = 3.5MPa. Calculate the maximum and minimum circumferential stresses and plot the variationof the same across the wall thickness.

Answer : April/May-13, Set-3, Q2(b)

Given that,

External diameter, d0 = 250 m

Radius, r0 =

2

250

r0

= 125 mm

Internal fluid pressure,

P1= 35 MPa

External pressure,

P2= 3.5 MPa

External diameter = Internal diameter + Thickness

125= r1 + 50

r1= 125 – 50

r1= 75 mm

Now from Lame’s equation,

Px

= ax

b–

2

At x = 75 mm, Px = 35 N/mm2

35 = ab

–752 ... (1)



At x = 125 mm, Px = 3.5 N/mm2

3.5 = ab

–1252 ... (2)

From equation (1),

a = 35–752

b

Substitute ‘a’ in equation (2),

3.5 = 3575

–125 22

+bb

3.5 – 35 = b×

22 75

1–

125

1

– 31.5 = – b.140625

16

b = 276855.46

Substitute b in equation (2),

3.5 = a–125

46.2768552

a = 17.718 – 3.5

218.14=a

46.276855=b

Variation of radial pressure Px will be,

Px

= 2

46.276855

x – 14.218

At x = 75 mm, Px = 35 N/mm2

At x = 100 mm, Px = 2100

46.276855 – 14.218

= 13.467 N/mm2

At x = 125 mm, Px = 2125

46.276855 – 14.218

= 3.5 N/mm2

Variation of hoop stress,

At x = 75 mm

fx= 275

46.276855 + 14.218

= 63.436 N/mm2



At x = 100 mm

fx

= 2100

46.276855 + 14.218

= 41.903 N/mm2

At x = 125 m

fx

= 2125

46.276855 + 14.218

= 31.936 N/mm2

75

35 N/mm2

63.436

41.903

13.467

Hoop stress distribution

31 N/mm2

0

125

Radial pressure distribution

75

35 N/mm2

63.436

41.903

13.467

Hoop stress distribution

31 N/mm2

0

125

Radial pressure distribution

Figure

Q3. A hollow steel shaft of external diameter equal to twice the internal diameter, 5 m long is totransmit 160 kW of power at 120 r.p.m. The total angle of twist is not to exceed 2° in this length andthe allowable shear stress is 50 N/mm2. Calculate diameter of the shaft.


Let,

External diameter be ‘D’

Internal diameter be ‘d’

Given that,

External diameter = 2 × Internal diameter

D = 2d ⇒ d = 2

D

Power transmitted = 160 kW

Speed, N = 120 r.p.m

Angle of twist, φ = 2°

Allowable shear stress = 50 N/mm2

Let us assume the rigidity modulus, ‘C’ = 8 × 104 N/mm2

We know that,

Power = 4106

2

×πNT

160= 4106

1202

×××π× T



T = 1202

106160 4

×π×××

= 12732.39 N-m

mm-N1039.12732 3×=T

Polar moment of inertia,

IP

= 32

π[D4 – d4]

=

π 4

4

2–

32

DD

512

15 4DIP

π=

Polar modulus=

2DIP

=

2/512

15 4

D

Dπ

= D

D 2

512

15 4

×π

= 256

15 4Dπ

Given,

Allowable shear stress (PS) = 50 N/mm2

We know that,

Torque = PS × Polar modulus

Polar modulus = SP

T

= 50

1039.12732 3×

256

15 3Dπ= 254647.8

D3 = π××

15

2568.254647

D3 = 1383371.30

D = 111.42 mm

Angle of twist should not exceed 2° for length 5 m.

i.e.,Polar modulus = 256

15 3Dπ


IP=

256

15 3D×

Angle of twist, φ = c

l×

PI

T

2° = 4

3

4 15

121039.12732

108

5000

D×π××××

×

180

2π× = 4

63.8646070

D

D4 = π××

2

18063.8646070

π=° radians

1801Q

D4 = 247691678.5

mm45.125=D

Q4. A closed-coiled helical spring is required tohave an axial stiffness of 5 kN/m and an an-gular stiffness of 0.1 Nm per degree angle oftwist. If the spring is made of steel wire 6mm diameter, find the mean diameter of thecoil and the number of turns required. As-suming E = 200 GPa and G = 80 GPa.


Given that,

Angular stiffness, Kt = 0.1 Nm/degree

= 100 Nmm

Axial stiffness, Ka = 5 kN/m

Diameter of steel wire, d = 6 mm



Mean diameter of coil, D = ?

Number of turns, n = ?

Axial stiffness, Ka = δ

W

5 = 3

4

64nR

Gd

nR3 = 564

680 4

××

nR3= 324 ... (1)

Angular stiffness, Kt= φ

T

100 × 1° = nR

Ed

128

4

[Q 100 Nmm per degree angle of twist]

nR = π×××180100

6200 4

nR = 45.238

n = R

238.45... (2)

Put equation (2) in (1),

nR3= 324

3238.45R

R× = 324

45.238 R2 = 324

R2 = 7.162

R = 2.676 mm

Mean diameter of the coil,

D = 2 R

= 2 × 2.67

D = 5.35 mm

Substitute R in equation (2),

n = R

238.45

= 676.2

238.45

n = 16.9 –~ 17

Q5. (a) Derive an expression for crippling loadwhen one end of the column is fixedand the other end is free.

Answer : April/May-13, Set-3, Q5(a)

For answer refer Unit-V, Q10.

(b) Calculate the Euler’s critical load for astrut of T-section. The flange width be-ing 10 cm, overall depth 8 cm and bothflange and stem 1 cm thick, the strut is3 m long and is built in at both ends.Take E = 2 x 10 N/mm3.

Answer : April/May-13, Set-3, Q5(b)

Given that,

Length of strut,‘L’ = 3 m = 3000 mm

Elastic modulus, E = 200 × 103 N/mm2

Centroidal Axis from the Top

y = 21

2211

AA

yAyA

++

10 mm1

2

70 mm

y2 = 45 mm

y1 = 5 mm

100 mm

10 mm

10 mm1

2

70 mm

y2 = 45 mm

y1 = 5 mm

100 mm

10 mm

Figure

From figure,

A1= b

1 × d

1

= 100 × 10

= 1000 mm2

y1= 5 mm

A2= b

2 × d

2

= 70 × 10

= 700 mm2



y2

= 2

70 + 10

= 45 mmCentroidal axis,

Y = 7001000

4570051000

+×+×

Y = 21.470 mm

Y = yt = 21.470 mm

Centroidal axis from bottom,

yB = (70 + 10) – 21.470 y

B= 58.53 mm

Moment of Inertia about X-X axis

IXX

=

+

21

1

311

2–

12

dyA

dbt +

+

22

2

322

2–

12

dyA

dbB

=

+× 2

3

)5–470.21(100012

10100 +

+× 2

3

)35–53.58(70012

7010

IXX

= 952990.19 mm4

Moment of Inertia about Y-Y Axis

IYY

= 12

311bd

+ 12

322bd

= 12

10010 3× +

12

1070 3×

= 839166.66 mm4

When both ends are fixed,

Leff

= 2

L

= 2

3000

= 1500 mmEulers critical load,

Pcritical

= 2

2

effl

EIπ

= 2

632

1500

10839.010200 ××××π

= 736053.16 N

= 736.05 kN



Q6. A masonry retaining wall, 7 meters high, is trapezoidal in section, 1 meter wide at the top and 3meters at the base, with one side vertical. If the lateral pressure exerted by the retained materialon the vertical face varies from zero at the top to 25 kN/m2 at the base, calculate the maximumand minimum stresses induced in the base, the weight of masonry being 21 kN/m3.


Given that,

Top width of trapezoidal section = 1 m

Bottom width of trapezoidal section = 3 m

Height of trapezoidal section = 7 m

Unit weight of masonry = 21 kN/m2

O

1 m LK

M N

7 m

3 mO

1 m LK

M N

7 m

3 m

Weight of masonry when 1 m length of the wall is considered,

P = 2172

13 ×

+

= 294 kN

x =

××+×

+

××+××

7221

)71(

32

17221

)5.071(

= 1.083 m

Pressure at every 1 m interval,

= 25 × 1

= 25 kN/m

Total lateral force = 7252

1 ××

= 87.5 kN



Let the resultant of load P and thrust ‘W’ pass through ‘Q’ and its magnitude to be ‘R’.

R

P

W

O N

LK

Me

2

b

2

b

x

R

P

W

O N

LK

Me

2

b

2

b

x

Taking moments of forces about the point ‘R’.

)–(294–3

75.87 xy

× = 0

204.16 = 294 (y – 1.083) y – 1.083 = 0.694

y = 1.77 m

Eccentricity, e= y – 2

b

= 1.77 – 2

3

= 1.77 – 1.5 e = 0.277 m

Direct stress on the base,

σd = A

P

= 13

294

×= 98 kN/m2

Moment of the base, M = P × e

= 294 × 0.277= 81.438 kN-m

Modulus of section,

Z = db2

6

1 ×

= 136

1 2 ××

Z = 1.5 m3



Bending stress

σb=

Z

M

= 5.1

438.81

= 54.292 kN/m2

σmax

= σd + σ

b

= 98 + 54.292

= 152.29 kN/m2 (Compression at N)

σmin

= σd – σ

b

= 98 – 54.292

= 43.708 kN/m2 (Compression at K)

Q7. Find the centroidal principal moments of inertia of a equal angle section 30 × 30 × 8 mm.


X X

Y

Y

Y'

X'

30 mm22 mm

2

1 8 mm

ML30 mm

G

U

K8 mmV

V

U

45°

135°

X X

Y

Y

Y'

X'

30 mm22 mm

2

1 8 mm

ML30 mm

G

U

K8 mmV

V

U

45°

135°

Divide the L-section into 2-rectangles - (1) and (2),

Area, A1= 30 × 8

= 240 mm

A2= 22 × 8

= 176 mm



The coordinates of centroid ‘G’ be ),( yx with respect to rectangular axis BX' and BY',

Rectangle-1

From BY'

The distance of centre of gravity,

x1=

2

30

= 15 mm

From BX'

y1=

2

8

= 4 mm

From BY', the distance of centre of gravity,

x2=

2

8

= 4 mm

From BX'

y2= 8 +

2

22

= 19 mm

Centroid about X-axis,

x = 21

2211

AA

xAxA

++

= 176240

417615240

+×+×

x = 10.346 mm

Centroid about Y-axis,

y = 21

2211

AA

yAyA

++

= 176240

191764240

+×+×

y = 10.346 mm

Moment about X-axis

IXX

=

+

2

1

311

2–

12

dyA

db +

+

22

2

322 –

212y

dA

db

=

+× 2

3

)4–364.10(24012

830 +

+× 2

3

)346.10–11(17612

228



= 10945.211 + 7173.944= 18119.155 mm4

IXX

= 18119.155 mm4

Hence, the given section is equal angle section so the section has equal dimensions.Principal Axis

The inclination of principal axis,

tan2θ = XXYY

xy

II

I

–

2

We know that, I

XY= A

1 V

1 K

1 + A

2 V

2 K

2

Where, K

1= Horizontal distance of centre of gravity of rectangle (1) from Y-Y axis.

K1= x–

2

30

= 15 –10.346= 4.654 mm

V1= Vertical distance of centre of gravity of rectangle-(1) from X-X axis

V1= y –

2

8

= 10.346 – 4 V

1= 6.346 mm

V1= – 6.346 mm [QV

1 below X-X axis]

K2= Horizontal distance of centre of gravity of rectangle-(2) from Y-Y axis.

K2= x –

2

8

= 10.346 – 4 K

2= 6.346 mm

K2

= – 6.346 mm [Q K2 is towards the left of XY-axis]

V2= Vertical distance of centre of gravity of rectangle-(2) from X-X axis

= y–2

228

+

= 19 – 10.346 V

2= 8.654 mm

IXY

= 240 × 4.654 × (– 6.346) + 176 × 8.654 × (– 6.346)= – 16753.84 mm

tan2φ = 0

2)84.16753(– ×

2φ = 90° φ

1= 45° and

φ2

= 90 + 45°= 135°

Here φ1, φ2 are the position of principal axis through centroid.



Principal Moment of Inertia

Moment of inertia about principal axis U-U,

IUU

= 2

1[I

xx + I

yy] +

2

1[I

xx – I

yy] cos2θ – I

xy sin2θ

= 2

1[18119.155 + 18119.155] + 0 + 16753.84 × sin90°

IUU

= 34872.99 mm4

4mm34873–~UUI

Moment of inertia about principal axis V-V,

IVV

= IXX

+ IYY

– IUU

= 18119.155 + 18119.155 – 34873

4mm31.1365=VVI

Q8. Find the bending moment at mid span of the semicircular beam of diameter 9 m loaded at themid span with a concentrated load of 60 kN. The beam is fixed at both supports. Find the maxi-mum bending moment and maximum torque in the beam.


Given that,

Diameter of semicircular beam,

d = 9 m

= 9000 mm

Concentrated load,

p = 60 kN

∈ = 200 GPa

Let, G = 80 GPa

N

M

LK

K L

P

O

90° 45°

60 kN

N

M

LK

K L

P

O

90° 45°

60 kN



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

MM

= θλ+λθ

θ+θθλ2

22

sin)1–(–)1(2

sin)sin–cos2–2( (Pr)

Here,

θ = 90°

λ = GJ

EI

Moment of inertia,

I = 4

64d

π


J = 4

32d

π

λ = 43

43

)109(8064

32)109(200

××π×××××π×

λ = 4

5

K1= (2 – 2cosθ – sin2θ)

= 2 – 2cos90 – sin290°

K1= 1

K2= sin2θ

= sin290°

K2= 1

K3= 2θ (λ + 1) – (λ – 1) sin2θ

= 2 × 90 × 180

π ×

+1

4

5 –

1–4

5sin290°

= π

4

9 –

4

1

K3= 6.818

MM

= 3105.460818.6

1145

×××+

×



MM

= 0.33 × 60 × 4.5 × 103

= 89.1 × 103 kN-m= 89.1 kN-mm

°=θ 45NM = θλ+λθ

θ+θθλ2

22

sin)1–(–)1(2

sin)sin–cos2–2( (Pr)

= °

+π××

°+°°

45sin1–45

–145

180452

45sin)45sin–45cos2–2(45

2

22

× 60 × 4.5 × 103

= 125.0–534.3

5.01072.0 + × 60 × 4.5 × 103

= 48.091 × 103 kN-m

°=θ 45NM = 48.091 kN-mm

B.M Diagram

89.1 N-m

48.091 kN-mm48.091

B.M Diagram

89.1 N-m

48.091 kN-mm48.091

PQN

K LRLRK

O

M60 kN

PQN

K LRLRK

O

M60 kN

ReactionsRK + RL = 60 kNTaking forces about tangent at any point ‘M’, R

K × 4.5 – 60 (OP – OM) = 0

4.5 RK

= 60 (OP – OM)OP = 4.5 m

OM = 4.5 × cos45° OM = 3.181



RK × 4.5= 60 (4.5 – 3.181)

RK

= 5.4

14.79

RK

= 17.586 kN

RL

= 42.426 kN

Point of zero Bending Moment

Mφ = RK × R sinφ – 60R sin (φ – π/2)

For zero bending moment,

Mφ = 0

RK × R sinφ = R × 60 sin (φ – π/2)

RK sinφ = 60 sin (φ – π/2)

πφ

φ

2–sin

sin=

586.17

60

2sin.cos–

2cossin

sinπφπφ

φ= 3.411

φ

φcos–

sin= 3.411

– tanφ = 3.411

φ = – tan–1 (3.411)

φ = – 73°39'

Maximum torsional moment,

i.e., Torque in beam at φ = – 73°39'

maxφM = R

K × (R – R cos(– 73°39')) – 60(R – R cos(– 73º39' – 45°))

= 17.586 × (4.5 – 4.5 cos (– 73°39')) – 60((4.5 – 4.5 cos(– 118°39'))

= 17.586 (4.5 – 1.266) – 60(4.5 + 2.157)

= 56.85 – 399.42

tMmaxφ = – 342.57 kN-m

Documents

1401802399_Strength of Materials-II 2-2 Set-3 (a)