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

S.19Strength of Materials-II (April/May-2013, Set-2) JNTU-Anantapur

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

Code No : 9A01401/R09B.Tech II Year II Semester Regular and Supplementary Examinations

April/May - 2013STRENGTH MATERIALS-II

( Civil Engineering )Time: 3 Hours Max. Marks: 70

Answer any FIVE QuestionsAll Questions carry Equal Marks

- - -

1. A thin spherical shell, 1200 mm in diameter, is subjected to an internal pressure of 1.8 N/mm2. Find the thickness ofplate required if the permissible tensile stress is 120 N/mm2. The joint efficiency may be taken as 75%.(Unit-I, Topic No. 1.2)

2. A thick walled cylindrical pressure vessel has inner radius of 150 mm and outer radius of 185 mm. Draw a sketchshowing the radial pressure and hoop stress distribution in the section of the cylinder wall, when an internalpressure of 10 MN/m2 is applied. (Unit-II, Topic No. 2.1)

3. A solid shaft of 250 mm diameter has the same cross-sectional area as the hollow shaft of the same material withinside diameter of 200 mm.

(a) Find the ratio of power transmitted by the two shafts for the same angular velocity, and(b) Compare the angles of twist in equal lengths of these shafts, when stressed to the same intensity.

(Unit-III, Topic No. 3.3)

4. A laminated steel spring simply supported at ends with span of 0.75 m is centrally loaded with a load of 10 kN. Thecentral deflection under the above load is not to exceed 50 mm and the maximum stress is to be 400 MPa, determine,

(i) Width of plate(ii) Thickness of plate(iii) Number of plates(iv) The radius to which plates should be bent so that the spring become straight under the given 7.5 kN load.

Assume width = 12 thickness and E = 200 GPa. (Unit-IV, Topic No. 4.2)

5. Derive a formula for the maximum compressive stress induced in an initially straight, slender, uniform strut whenloaded along an axis having an eccentricity e at both ends which are pin-jointed. (Unit-V, Topic No. 5.3)

6. A 10 m high brick masonry wall of rectangular section 4 m 1.5 m is subjected to horizontal wind pressure of 15 N/mm2 on the 4 m side. Find the maximum and minimum stress intensities induced in the base. Weight density ofmasonry = 22 kN/mm2. (Unit-VI, Topic No. 6.2)

7. A 45 mm 45 mm 5 mm angle is used as a simply supported beam over a span of 2.4 meters. It carries a load of 300N along the vertical axis passing through the centroid of the section. Determine the resulting bending stresses on theouter corners of the section, along the middle section of the beam. (Unit-VII, Topic No. 7.3)

8. A horizontal circular bow girder of radius 5 m is continuous over five equally spaced supports. It carries a verticalu.d.l of 50 kN/m. Obtain the B.M torsional moment and S.F diagrams for one span indicating the critical values.(Unit-VIII, Topic No. 8.1)

Set-2Solutions

S.20 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014


Q1. A thin spherical shell, 1200 mm in diameter,is subjected to an internal pressure of 1.8 N/mm2. Find the thickness of plate required ifthe permissible tensile stress is 120 N/mm2.The joint efficiency may be taken as 75%.

Answer : April/May-13, Set-2, Q1Given,

Diameter of shell, d = 1200 mmInternal pressure, P = 1.8 N/mm2

Permissible tensile stress, = 120 N/mm2

Efficiency, = 75%Let, t-thickness of plate.Stress induced is given by,

= tPd4

120 = 75.0412008.1

t

t = 12075.04

12008.1

t = 6 mm Thickness of plate, t = 6 mm.Q2. A thick walled cylindrical pressure vessel has

inner radius of 150 mm and outer radius of185 mm. Draw a sketch showing the radialpressure and hoop stress distribution in thesection of the cylinder wall, when an internalpressure of 10 MN/m2 is applied.


Inner radius, r1 = 150 mmOuter radius, r2 = 185 mm

Internal pressure, P = 10 MN/m2

= 10 N/mm2

Radial pressure is given by,

Px= a

x

b2 ... (1)

Now, apply the boundary conditions to the equation (1).

(i) At x = r1 = 150 mm, Px = 10 N/mm2(ii) At x = r2 = 185 mm, Px = 0

Substitute these values in equation (1).

10 = ab

)150( 2

ab

22500 = 10 ... (2)

0 = ab

)185( 2

ab

34225 = 0 ... (3)

Subtract equations (2) and (3).

10 0 = ab

ab

+34225

22500

10 = 34225

22500bb

10 = 61605001808.273 bb

10 = 61605008.93 b

93.8 b = 6160500 10

b = 656769.72 mm

Substitute b in equation (3).

34225b

a = 0

3422572.656769

a = 0

19.18 a = 0

a = 19.18

a = 19.18 mm

SOLUTIONS TO APRIL/MAY-2013, SET-2, QP



The values of a and b are substituted in hoop stress.

Now, hoop stress at any radius x is given by,

x= a

x

b+2

= 272.656769

x + 19.18

At x = 150 mm, 150= 2)150(72.656769

+ 19.18

= 29.18 + 19.18

= 48.36 N/mm2

At x = 185 mm, 185= 2)185(72.656769

+ 19.18

= 19.18 + 19.18

= 38.36 N/mm2.

The radial pressure distribution and hoop stress distribution across the section is as shown in the figure below. ABis taken as horizontal line. AC = 10 N/mm2. The variation between B and C is parabolic. The curve BC shows the variation ofradial pressure across AB.

D

B

48.36 N/mm2

10 N/mm2150 m

m

185 m

m

E

C RadialPressure (Px)

Hoop S

tress (x

)

A

38.36 N/mm2

D

B

48.36 N/mm2

10 N/mm2150 m

m

185 m

m

E

C RadialPressure (Px)

Hoop S

tress (x

)

A

38.36 N/mm2

Figure

The curve DE which is also parabolic shows the variation of hoop stress across AB. The values BD = 38.36 N/mm2and AE = 48.36 N/mm2.

The radial pressure is compressive whereas the hoop stress is tensile.



Q3. A solid shaft of 250 mm diameter has thesame cross-sectional area as the hollow shaftof the same material with inside diameter of200 mm.(a) Find the ratio of power transmitted by

the two shafts for the same angularvelocity, and

(b) Compare the angles of twist in equallengths of these shafts, when stressedto the same intensity.


Diameter, d = 250 mmInside diameter, di = 200 mm

(a) Ratio of Power Transmitted by Two ShaftsCross-sectional area of solid shaft,

A1 = 4

d2

A1 = 4

2502

A1 = 49087.38 mm2 ... (1)Cross-sectional area of hollow shaft,

A2 = )(422idD

=

4

(D2 2002)

A2 = 4

(D2 40000) ... (2)

Q The cross-sectional areas of both the shafts are same,equate (1) and (2).

4

(D2 40000) = 49087.38

D2 40000 =

438.49087

D2 40000 = 62499.99

D2 = 62499.99 + 40000

D2 = 102499.99

D = 320 mm

We also know that torque transmitted by solid shaft,

T1 = 16

d3

= 16

(250)3

T1 = 3067961.57 N-mm ... (3)Hollow shaft,

T2 =

DdD i

44

16

=

320)200()320(

16

44

T2 = 5452234.05 N-mm ... (4)

shaftsolidbydtransmittePowershafthollowbydtransmittePower

=

1

2

TT

1

2

TT

=

05.545223457.3067961

1

2

TT

= 0.56

(b) Ratio of Angle of Twist in Both ShaftsAngle of twist for a shaft,

R

= lC

= RCl

Angle of twist for solid shaft,

1 = RCl

R = 2d

=

2250

= 125

1 = Cl

125

Angle of twist for hollow shaft,

2 = RCl

R = 2D

=

2320

= 160

2 = Cl

160



shaftsolidfortwistofAngleshafthollowfortwistofAngle

=

ClC

l

125

160

2

1

= lC

Cl

125160

2

1

= 160125

2

1

= 0.78

Q4. A laminated steel spring simply supportedat ends with span of 0.75 m is centrallyloaded with a load of 10 kN. The centraldeflection under the above load is not toexceed 50 mm and the maximum stress is tobe 400 MPa, determine,(i) Width of plate(ii) Thickness of plate(iii) Number of plates(iv) The radius to which plates should be

bent so that the spring become straightunder the given 7.5 kN load. Assumewidth = 12 thickness and E = 200 GPa.


Span, l = 0.75 m = 750 mmLoad, P = 10 kN = 10 103 NExtension, = 50 mmMaximum stress, b = 400 MPa = 400 N/mm2

Width = 12 Thickness b = 12 t

Modulus of elasticity, E = 200 GPa= 200 103 N/mm2

(i) Thickness of PlateWe know that central deflection,

= Etlb

4

2

50 = t

3

2

102004750400

t = 50102004

7504003

2

t = 5.62 ~ 6 mm

Thickness of plate, t = 6 mm(ii) Width of Plate

Width = 12 Thickness b = 12 t

b = 12 6

b = 72 mm

Width of plate, b = 72 mm(iii) Number of Plates

We know that bending stress,

b = 223nbtWl

400 = 23

672275010103

n

n = 2

3

672400275010103

n = 10.85 ~ 11

Number of plates, n = 11(iv) Radius of Spring Under the Load of 7.5 kN

We know that,

= 223nbtWl

= 2

3

672112750105.73

= 295.92 N/mm2

Radius of spring,

R = 2

Et

= 92.2952610200 3

R = 2027.57 mm Radius of spring, R = 2027.57 mm



Q5. Derive a formula for the maximumcompressive stress induced in an initiallystraight, slender, uniform strut when loadedalong an axis having an eccentricity e at bothends which are pin-jointed.

Answer : April/May-13, Set-2, Q5Formula for the Maximum Compressive Stress When Boththe Ends are Pin-jointed

The column when the load is applied just bends orbuckles is called crippling load.

Consider a column AB of length l and area of cross-section a which is pinned at both the ends.

Let W be the crippling load at which the column willjust buckle.

Consider the section at a distance x from the end ALet y - deflection at the section.

The moment at the section due to crippling load = W y

But, Moment = 2

2

dxydEI

x

yC

W

A

B

l

W

x

yC

W

A

B

l

W

Figure

Equating both the moments,

2

2

dxydEI = W y

2

2

dxydEI

+ W y = 0

EIW

dxyd

+2

2 y = 0

The solution of the differential equation is,

y = C1 cos

EIW

x + C2

EIW

xsin... (1)

Where, C1 and C2 are constants of integration.(i) At A, x = 0 and y = 0

Substitute these values in equation (1),

0 = C1 cos

EIW0

+ C2 sin

EIW0

C1 cos 0 + C2 sin 0 = 0

C1(1) + C2(0) = 0 C1 = 0 ... (2)

(ii) At B, x = l, y = 0Substitute these in equation (1),

0 = C2 cos

EIWl

+ C2 sin

EIWl

0 = 0 cos

EIWl

+ C2 sin

EIWl

(Q C1 = 0)

0 + C2 sin

EIWl

= 0

C2 sin

EIWl

= 0 ... (3)

From equation (3), C2 = 0

sin

EIWl = 0

As C1 = 0 and C2 = 0 then y = 0 in equation (1).i.e., the bending moment of the column will be zero or

the column will not bend at all. This is not true.

sin

EIWl

= 0



sin

EIWl

= sin 0 or sin or sin2 or ----.

EIWl

= 0 (or) (or) 2 (or) 3 (or) ----.

Taking the least practical value,

i.e.,EIWl

=

Now, squaring on both sides.

2

EIWl = 2

l2 EIW

= 2

W = 22

lEI

Maximum compressive stress, W = 22

lEI

Q6. A 10 m high brick masonry wall ofrectangular section 4 m 1.5 m is subjectedto horizontal wind pressure of 15 N/mm2 onthe 4 m side. Find the maximum andminimum stress intensities induced in thebase. Weight density of masonry = 22 kN/mm2.

Answer : April/May-13, Set-2, Q6Note: In the given question Horizontal wind pressure shouldbe 150 kN/m2 and Weight density of masonry should be22 kN/m3.

Given,

Height of wall, H = 10 m

Length of wall, L = 4 m

Breadth of wall, B = 1.5 m

Horizontal wind pressure, P = 150 kN/m2

Weight density of masonry, W = 22 kN/m3

P

a b1.5 m

c

P

PWd

4 m

10 m

2H

P

a b1.5 m

c

P

PWd

4 m

10 m

2H

Figure

Area of the wall = L B A = 4 1.5

= 6 m2 Weight of the wall = A H W

W = 6 10 22= 1320 kN

Direct stress, d = AW

= 61320

d = 220 kN/m2

Moment of inertia, I = 12

3BH

= 12105.1 3

=

121500

I = 125 m4

y = 2B

=

25.1

= 0.75 mm



Section modulus, Z = yI

=

75.0125

Z = 166.67 m3

Horizontal wind pressure, PW = k.P.APWhere, k= Coefficient of wind resistance

= 1 (usually taken) P - Horizontal intensity of wind pressure AP - Projected area

= B H

AP = B H= 1.5 10= 15 m2

PW = k.P.AP= 1 150 15= 2250 kN

Bending moment at the base,

M = PW 2H

= 2250 2

10

M = 11250 kNm

Bending stress, b = ZM

= 67.16611250

b= 67.49 kN/m2

Maximum stress intensity,

max= d + b

= 220 + 67.49

max= 287.49 kN/m2

Minimum stress intensity,

min = d b

= 220 67.49

min = 152.51 kN/m2



Q7. A 45 mm 45 mm 5 mm angle is used as a simply supported beam over a span of 2.4 meters. Itcarries a load of 300 N along the vertical axis passing through the centroid of the section. Determinethe resulting bending stresses on the outer corners of the section, along the middle section of thebeam.


Width of beam, b = 45 mmDepth of beam, d = 45 mmThickness of beam, t = 5 mmLoad, p = 300 NSpan, l = 2.4 m = 2.4 103 mm

X X

45 mm

45 mm

Y

Y

5 mmCB

A5mm

V

X1U

V

1

2

45

U

Y1

X X

45 mm

45 mm

Y

Y

5 mmCB

A5mm

V

X1U

V

11

22

45

U

Y1

Figure

Let x , y be the co-ordinates of centroid G, with respect to rectangular axes BX1 and BY1.

Now, x = y = 5405455.25405.22545

+

+

x = y = 13.08 mm

Moment of inertia about XX axis,

IXX =

++

21

22

322

21

11

311

2122

12x

xdbdbyxdbdb



IXX =

+

23

25

08.1354512

545 +

+

2308.13

250405

12405

IXX = 80738.38 mm4

IXX = 8.07 104 mm4 = IYYCo-ordinate of G1= + (22.5 13.08), (13.08 2.5)

= (9.42, 10.58)Co-ordinate of G2= (13.08 2.5), + (25 13.08)

= ( 10.58, + 11.92)Product of inertia,

IXY = 45 5 (9.42) ( 10.58) + 40 5 ( 10.58) (11.92)= 22424.31 25222.72

IXY = 47647.03 mm4

= 4.76 104 mm4

As the portions 1 and 2 are rectangular strips, the product of inertia about centroidal axes is zero.

If is the inclination of the principal axes with GX, passing through G.

tan 2 = XXYY

XY

III

2

tan 2 = (Q IXX = IYY) tan 2 = tan 90 2 = 90

1 = 45 1 + 2= 180

45 + 2= 180

2 = 180 45

2 = 135

Principal Moment of Inertia

IUU = 21

(IXX + IYY) + 21

(IXX IYY) cos 90 IXY sin 90

=

21

(80738.38 + 80738.38) + 21

(80738.38 80738.38) cos 90 ( 47647.03) sin 90

=

21

(161476.76) + 21

0 cos 90 + 47647.03 1

= 80738.38 + 47647.03

IUU = 128385.41 mm4



Also, IUU + IVV= IXX + IYY IVV = IXX + IYY IUU

= 2 80738.38 128385.41= 161476.76 128385.41

IVV = 33091.35 mm4Stresses on the Outer Corners of the Section

Bending moment at mid-section,

M = 4

Wl

=

4104.2300 3

M = 1.8 105 N-mmThe components of bending moment are, M1 = M sin = 1.8 105 sin 45

= 1.272 105 N-mm M2 = M cos = 1.8 105 cos 45

= 1.272 105 N-mmCo-ordinates of u, v

Point A: x = 13.08, y = 45 13.08 = 31.92 mm u = x cos + y sin

= 13.08 cos 45 + 31.92 sin 45 u = 13.32 mm v = y cos x sin

= 31.92 cos45 ( 13.08 sin45) v = 31.81 mm

Point B: x = 13.08, y = 13.08 u = 13.08 cos 45 + ( 13.08 sin 45)

= 18.49 mm v = 13.08 cos45 ( 13.08 sin45)

= 0Point C: x = 45 13.08 = 31.92, y = 13.08

u = 31.92 cos 45 13.08 sin 45= 13.32 mm

v = 13.08 cos45= 9.24 mm

Stress at point C,

C = UUVV I

vMI

uM 21 +

C = 1.272 105

+

41.12838524.9

35.3309132.13

C = 60.355 N/mm2

Stress at point B,

B = 1.272 105

+

41.1283850

35.3309149.18

B = 71.07 N/mm2

Stress at point A,

A = 1.272 105

+

41.12838581.31

35.3309132.13

A = 82.71 N/mm2

Q8. A horizontal circular bow girder of radius 5m is continuous over five equally spacedsupports. It carries a vertical u.d.l of 50 kN/m. Obtain the B.M torsional moment and S.Fdiagrams for one span indicating the criticalvalues.


Radius of circular girder, R = 5 mUniformly distributed load, W = 50 kN/mNumber of supports, n = 5

As it is a circular bow girder, it covers an angle of 360

2 = 5360

= 72

= 36

= 0.6283 radFrom the standard table of coefficient for bending

moment and twisting moment in circular beams, C1 = 0.108, C2 = 0.054, C3 = 0.014

m

= 15 4

1 = 15.25 = 0.267 rad

WR2.2 = 50 (5)2 (2 0.6283)= 1570.75

Mo

= C1 WR2 (2)= 0.108 50 52 2 0.6283

Mo

= 169.641 kN-m

MC = C2 WR2 (2)= 0.054 1570.75

MC = 84.82 kN-m



tmM = C3 WR2 (2)

= 0.014 1570.75

tmM = 21.99 kN-m

Fo

= W.R.

= 50 5 0.6283

Fo

= 157.075 kN

(i) Shear Force DiagramThe distribution of shear force is given by,

F = WR ( )

= 50 5 (36 ) 180

F = 4.36 (36 ) ... (1)The values of F at varying intervals are given below.

F (kN) Location0 156.96 Ends18 78.48

36 0.00 Middle

(ii) Bending Moment DiagramThe bending moment at any point is given by,

M = WR2 [ sin + cos 1]

= 50 52 [0.6283 sin + 0.6283 cos 1] M = 1250 [0.6283 sin + 0.6283 cos 1] ... (2)

The values of M at various intervals are given below,

M Location0 464.625 End of beam

15.25 1607.214 Point of maximum torsion

18 1321.210

36 2129.417 Centre of beam

(iii) Twisting Moment DiagramtM = WR2 [ cos cot . sin ( )]

= 50 52 [0.6283 cos 0.6283 cot36.sin (36 ) /180]tM = 1250 [0.6283 cos 0.864 sin 0.01745 (36 )] ... (3)



The values of tM at various intervals are given below,

tM Location0 0.125 End of beam

15.25 276.70

18 320.93 Point of maximum torsion

36 571.9 Middle of beam

18 36 54 720

156.96 kN

B

56.96 kN

Shear Force

Bending Moment

464.625 464.625

2129.417

202

202

Twisting Moment

A

18 36 54 720

156.96 kN

B

56.96 kN

Shear Force

Bending Moment

464.625 464.625

2129.417

202

202

Twisting Moment

A

Figure

Documents

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