IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (1)

1. (c)

2. (a)

3. (b)

4. (b)

5. (c)

6. (a)

7. (c)

8. (a)

9. (a)

10. (b)

11. (a)

12. (b)

13. (d)

14. (c)

15. (d)

16. (d)

17. (c)

18. (b)

19. (c)

20. (a)

21. (b)

22. (d)

23. (d)

24. (d)

25. (b)

26. (d)

27. (c)

28. (b)

29. (c)

30. (b)

31. (d)

32. (d)

33. (c)

34. (d)

35. (d)

36. (d)

37. (b)

38. (d)

39. (c)

40. (c)

41. (b)

42. (b)

43. (c)

44. (b)

45. (a)

46. (a)

47. (c)

48. (b)

49. (b)

50. (c)

51. (d)

52. (c)

53. (a)

54. (b)

55. (d)

56. (a)

57. (b)

58. (d)

59. (a)

60. (d)

61. (a)

62. (d)

63. (b)

64. (c)

65. (c)

66. (c)

67. (d)

68. (c)

69. (a)

70. (c)

71. (a)

72. (b)

73. (a)

74. (c)

75. (d)

76. (b)

77. (b)

78. (a)

79. (a)

80. (b)

81. (c)

82. (a)

83. (b)

84. (a)

85. (a)

86. (b)

87. (c)

88. (b)

89. (d)

90. (c)

91. (d)

92. (b)

93. (a)

94. (b)

95. (b)

96. (c)

97. (b)

98. (c)

99. (c)

100. (c)

101. (a)

102. (b)

103. (c)

104. (c)

105. (b)

106. (d)

107. (b)

108. (b)

109. (a)

110. (a)

111. (d)

112. (b)

113. (a)

114. (d)

115. (a)

116. (d)

117. (a)

118. (a)

119. (c)

120. (a)

Conventional Question Practice ProgrameDate: 19th March, 2016

ANSWERS

IES M

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(2) CE (Test-6), Objective Solutions, 19th March 2016

1. (c)2. (a)3. (b)4. (b)5. (c)6. (a)

For code use, method must be as simpleas possible, using deterministic rather thanprobabilistic data. Such a method calledLevel-I reliability method or first-order secondmoment reliability method.

7. (c)Unbuttoning joint failure (progressive jointfailure) more likely to happen in longconnections because end bolts resist moreforces.

8. (a)

V = 150 25 kN

6

T = 200 33.33 kN

6

2 2e

sd nd

TV 1V T

2 225 33.33 0.305 0.25, 0.56 145.3 66.6

Safe9. (a)

Throat thickness

te = 5 5t 12 7.5 mm8 8

Lw = 150 mm

Strength = w e y

mw

L t f 250 1150 7.51.25 1000

= 225 kN [Note: 0.9fu > fy]10. (b)

C

B

A 125 mm leg

75 mm leg

55

79

50 50 50dhole = 22

Path AC

Net area

= 1164 – 2 × 22 × 6 = 900 mm2

Path ABC

=

2 250 501164 3 22 6 64 55 4 79

= 883.65 mm2

11. (a)

Uniform distribution of shear is assumed inboth type of joint.

12. (b)

Design strength =

UAf3 1.5

=

2(30) 410 111.55 kN4 3 1.5 1000

13. (d)

Residual stress reduces the plastic rangeof member.

14. (c)

600×10–6 200×10–61000×10–6

xy2

xy2

= 800 × 10–6

xy = 1600 × 10–6

15. (d)16. (d)

Given,

1 = – 100 MPa

2 = – 10 MPaApplying maximum shear stress theory(MSST).

Abs max < Sys

1

2

< ytS2

Syt > –100 MPa

ytS 100 MPa

17. (c)Given, a solid circular shaft of diameter d issubjected to a combined bending moment Mand torque T.

IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (3)

T

d

T

M M MBy torsion equation,

=T .RJ =

3

T d2d

32

= 316T

dBy bending moment equation,

b =M .yI =

4

M d2d

64

b = 332M

dMaximum shear stress,

max =2

2b2

2 2

max 316 M Td

Condition for safe design,Maximum stress < Permissible stress

induced

maxSysFOS

Hence, torsional yielding strength (Ssy or Sys)will be used in design.

18. (b)As the given shafts are arranged in series, sothe torque will be the same on each part MN,NO and OP. Deflection on each part is givenby

M N O P

TT = 10 N-m

MN =MN

Tq =

1020 = 0.5 rad

NO =NO

Tq =

1030 =

13 rad

OP =OP

Tq =

1060 =

16 rad

Total deflection,

MP =1 1 12 3 6 =

66

MP 1rad

19. (c)20. (a)

For hollow shaft,

= h4

3 i0

0

16T

dd 1d

Th =43

0d 1116 2

Th =30d 15

16 16

........(1)

for solid shaft,

= S30

16Td

TS =30d

16

.......(2)

from equation (1) and (2)

h

S

T 15T 16

21. (b)Let, TA Reaction at the end AA

TB Reaction at the end B

3L/4

dT0

TA

L/4

TB

A B

TA = 0

3L4 TL

TA = 03T4

TB = 0

L4 TL

TB = 0T4

Since, TA > TB therefore maximum shear stresswill be

max = AT dJ 2

=0

4

3T d4 2

d32

0max 3

12Td

IES M

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(4) CE (Test-6), Objective Solutions, 19th March 201622. (d)

Since, the shaft is subjected to pure torsionhaving shear stress ( ) therefore, Mohr'scircle for pure shear stress state is

–

–

( , 0)(– , 0)

(0, – )

(0, )

Maximum principal stress,

max = radius of mohr circle= =

It can be written as

max 2 sin45 .cos45

23. (d)The bulk modulus of liquid

=dP P= ×d

=500.5×0.5

1= 250.25 MPa

24. (d)The variation of shear stress in various typesof fluids.

Bingha

m plas

tic

Shea

r stre

ss Non-Newtonian

Newton

ian

Ideal fluid

Velocity gradientDrilling mud and sewage sludge are Binghamplastic.

25. (b)

yV

Viscosity of oil = 9.81 poise= 0.981 N-s/m2

The shear stress from Newton's law of viscosity,

=dudy

=

220.981

10= 1.962 × 102

= 196.2 N/m2

26. (d)Assuming same height of oil and water inmanometer,

Poil + oilgh = Pw + w gh(Poil – Pw) = (w – oil)gh

If value of oil is given, then pressure differencecan be obtained. Generally most of commonliquid oil has density lower than water but somehas density more than water. So nothing isclear about oil density and we can say dataare insufficient.

Horizontal Plane

Pipes

Oil Water

Mercury

27. (c)

50cm

56cm45cm

atm

Benzene

sp. g.=0.88

Mercurysp. g = 13.6

Water

Taking reference level of pressure of left limbof mercury,

P + 0.56g = 0.5 × 0.88g + 0.45 × 13.6 g

Pg = (0.5×0.88 + 0.45 × 13.6 – 0.56)

= 6 m of water

28. (b)

AirOil

0.25

m

Gauge ‘A’

(Relative density of mercury 13.6)

Relative density of oil 0.8

4m

IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (5)Air pressure on oil.

Pair + 0.25 × 13.6 × 9.81 = Patm

Since the requirement is gauge pressure. SoPatm = 0

Pair = –0.25 × 13.6 × 9.81 kPa= –33.354 kPa

Gauge pressure at A,

PA = Pair + gh

= –33.354 + 4 × 0.8 × 9.81= – 33.354 + 31.392

= – 1.962 kPa

29. (c)Piezometer is a simple tube attached to tubecarrying water as shown in figure.

Flow

So in case of gases we do not use Piezometerbecause gas will leak out.

30. (b)Horizontal force-

Fy

Fx

Hemisphere

O

Fx = gAh

=

2R 4Rg2 3

Fx =

32Rg3 ...(i)

Vertical component= weight in hemispherical portion

F y = 31 4 R g4 3

=

3R g3

...(ii)

y

x

FF =

2

31. (d)

Z = max2

2 u rR

20 =

22 2 2 0.1

0.2

= 1 Ns/m2

f =e

64 64 64VdR 1000 2 0.4

= 0.0832. (d)

Assuming unit depth of structure.The over turning stability of dam is analysed.

F

h

A

W

b

SG= 2.56

2b3

h3

Volume of dam = 1b h 12

Weight = b h 2.56 g

2

W =2.56 ghb2

Hydrostatic force

F = hh 1 g2

= 2h g2

In limiting condition, moment about A = 0

h 2bF = W3 3

2h g h 2.56 2b= ghb×2 3 2 3

2 2h 2.56 2 b=6 6

h = 2.263 bb = 0.442h

33. (c)

2 m O

B

A

D

The vertical force acting on the cylindrical gates,= weight of water displaced by section ADB

= 21gV g A L g 2 12 4

= =

= g 1000 4.81 15409.5N.

2 2 = =

IES M

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(6) CE (Test-6), Objective Solutions, 19th March 2016Alternatively:

Vertical force = Weight of liquid supported by surface DB

= Weight in surface area

=

21 11 1 1+ 1 1 1 1 14 4

=

1 1 1+ g

4 4

= ×1000×9.842

= 15409.5 N

34. (d)

M

GB

For stable equilibrium the moment caused byweight W and Buoyancy force FB shouldbalance each other. For this condition M shouldbe above G.

If G is above M, the body will be in unstableequilibrium because moment caused by weightW and buoyancy force add up.

If M and G coincide then neutral equilibriumoccurs.

If FB > W, the body floats.

35. (d)The large metacentric height in a vessel isensured by lower center of gravity 'G' i.e. morestability.The period of oscillation,

T = 2Agh

I

T 1h

Hence larger metacentric height makes thetime period smaller.

36. (d)

3 mm

w

Weight of hydrometer, W = 0.03 NLet depth of hydrometer in oil is h0

W = B

0.03 = 3 20×(3 ×10 ) h × 750g

4

h0 = 60.03×4

×9×10 ×750×9.81= 0.5768 m

Let depth of hydrometer in alchohol is ha,

ha = 60.03× 4

×9×10 ×800×9.81= 0.5408 m

h0 – ha = 0.5768 – 0.5408

= 0.036 m = 36 mm

37. (b)

1m

0.75

= 30°

3m

hp =

2

GGI sinhAh

= 4

2

1364 41.75

3 1.754

= 1.75 + 0.08= 1.83 m

38. (d)

The period of rolling, is inversely proportionalto metacentric height. So by increasing it,period of oscillation reduces. At the same timethe frequency increases with increase inmetacentric height. The period of oscillation-

T =2k2

g×GM

By adding load below centre of gravity, thecentroid comes down and centre of buoyancygoes up. Due to this centre of buoyancyupward movement, metacentre in pitchingcomes down and height of metacentre inpitching reduces and frequency of oscillationin pitching reduces.

IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (7)Every oscillation has its own metacenter.

(GM BM BG)IGM BG

V

39. (c)

07 = FBV1

V2GW

B

m 013.6 =

Total volume of metal pieceV = V1 + V2

= Outside volume + volumeinside mercury

Buoyancy force = weight of piece.

m 2V g = gV

V2 =m

7V V 0.515 V13.6

= =

40. (c)

MB = 40 kNm

KB = KBC + KBA = EI 4E (2I) 2.5 EI2 4

B

B

M2.5 EI

B = 40 16

2.5 EI EI41. (b)

MOB =

4EIM 22 M4EI 4EI 32 4

MB = OBM M2 3

42. (b)

BA

L = 10 m

EI = 10000 kN-m2

Here B = 0.001 radiansA = 0

Using slope deflection method,

MAB = A B2EI0 2L

=

2 100000 0.001

10= 2 kN-m

43. (c)44. (b)45. (a)46. (a)

Examples of Force Method : Castigliano’s theorem (method of least

work).

Virtual work method / unit load method.

Column analogy method (used in rigidframes with fixed supports).

Flexibility matrix method.

Eamples of Displacement Methods :

Slope deflection method.

Moment distribution method.

Stiffness matrix method.

Kani’s method.

Moment distribution method is the mostsuitable manual method for analysis ofcontinous beams and plane frames. Themethod was presented by prof. Hardy crossof USA in 1930 and also sometimes referredto as Hardy Cross method.

Kani’s Method : This method was firstdeveloped by Prof. Gasper Kani of Germanyin 1947. The method is named after him.This is an indirect extension of slopedeflection method. This is an efficientmethod due to simplicity of momentdistribution. The method offers an iterativescheme for applying slope deflectionmethod of structural analysis. Whereas,the moment distribution method reducesthe number of linear simultaneous equationsand such equations needed are equal tothe number of translation displacements,the number of equations needed is zero inthe case of Kani’s method.

In this method, sum of the restrainedmoment of a joint and all rotationcontributions of the far ends of membersmeeting at that joint is multiplied byrespective rotation factors to get therequired near end rotation contribution.

Slope Deflection Method : This methodwas devised by Heinrich Manderla and OttoMohr to study the secondary stresses intrusses and was further developed by G.A.Maney and extended its application toanalyse indeterminate of this method is tomoments. It’s assumed that the effects of

IES M

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(8) CE (Test-6), Objective Solutions, 19th March 2016shear force or axial force deformations arenegligible in indeterminate beams or frames.

Analysis of statically indeterminatestructures are generally done either by forceor by displacement methods. In the forcemethod, we initially determine unknownforces and then joint displacements. In thedisplacement method, we initially determinejoint displacements and then the forcessuch as reaction, axial forces, shear forcesbending moments, etc.

47. (c)

CB EIEI

4m 3m

Relative stiffness when far end is fixed = Il

Relative stiffness when far end is hinged

=3 I4 l

Distribution factor for

BA = BA

BA BC

MM +M

= I / 43 II / 4 +4 3

= 0.5

Distribution factor for

BC = BC

BA BC

MM +M

=

3I4 3 0.5I 3 I

4 4 3

48. (b)

2

6EIL

2

6EIL

L

RB

RA

L/2 L/2

C

AM 0

2 26EI 6EI

L L = RB × L

RB = 3

12EIL

= RA

MC = B2

6EI L R2L

= 2 3

6EI L 12EI2L L

= 0

49. (b)

50. (c)

1 kN/m

Axis ofsymmetry

BC'

C

A D

CB

A

1 kN/m

The frame is symmetrical here, only 1/2 partof the frame will be analysed.

MFBC = –21 2 –12kNm

12

StiffnessJoint Member DF

factor3EI 3BA 0.612 5

B2EI 2BC 0.412 5

A C

0.6

0

+7.2

+7.2

B 0.4

– 12

+ 4.8

– 7.2

FEMbalancing

Total moment

51. (d)

52. (c)

53. (a)

3 2q m s m

depthy

ECSp. energy

A

B

C

For same E, transition will occur from A to B,if specific energy first decreases to C and thenincreases to B as in local hump.

TEL

EA EB

EC

IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (9)54. (b)

Underestimation of the future live load.

55. (d)

56. (a)

Unfinished bolts are also called ordinary,common, rough or black bolts.

57. (b)

x

=x

5Re

x

20.82.4

=x

x 3.6

x = 0.45 m

58. (d)

59. (a)

Velocity at Nose = 8 m/s

Velocity at 1m ahead = 5 m/s

Let pressure at Nose and 1m ahead are p1and p2 respectively. From Bernouli’s equation(in terms of relative velocity)

21 1

1p Vzg 2g

=

22 2

2p Vzg 2g

2 1p pg =

2 2(3) (0)2g

(Relative velocity of point 1 with respect to2 is 8 – 5 = 3 m/s)p2 – p1 =4500 N/m2

60. (d)

61. (c)

22v 1 QE y y2g 2g By

21 301.5 2.3 m2 10 5 1.5

62. (d)

63. (b)

The ratio of maximum elastic stress to averagestress is known as stress concentration factor.

The design strength of the member under theaxial tensile load is the lowest of the designstrength due to yielding of gross-section,rupture of critical (net) section and block shear.

64. (c)

40 50

50

50

100

100

A

D

F G

B

C

EGF

The net width along A – B – C – E – F

n = 3

Bn1 = (B – nd) = 300 – 3 × 25 = 225 mm

The net width along A – B – D – G

n = 2, 1n 1, p 50 mm, g, 100 mm

2 2

n2n p 1 50B B nd 300 2 254g 4 100

= 256.25 mm

The net width along A – B – D – E

n = 3, n 2 p 50 mm, g, 100 mm

Bn3 = 22 50300 3 25 237.5 mm

4 100

The least width is 225 mm.

65. (c)

The connection should be designed so as toreduce the effect of bending to a minimum andthe sections which are symmetrical about oneor both axes are, therefore, and ideal choice.

66. (c)

Celerity C = 21 2

1

y1g y y2 y

Since 0.1 m is negligible compared to 5m; 1 2y y y

Hence C = gy 9.8 5 7m s

67. (d)

A tidal bore is a phenomenon in which apositive surge travels upstream in a tidal riverwith the incoming tide.

Sluice gates used in field applications have abevel on the up stream face.

IES M

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(10) CE (Test-6), Objective Solutions, 19th March 2016

68. (c)

yyc

y0S2

Surface profile

69. (a)

Let velocity of wave = C

Let velocity of river water be v

4.2 = C + v

& 1.4 = C – v

2C = 5.6

C = 2.8 m/s

v = C – 1.4

v = 1.4 m/s

y =22.8

g = 0.8 m

70. (c)

Froude number is given by

Fr = VgD

where V = Mean velocity

Hydraulic depth, D = AT

A = Area of the section

T = Top width of the section

71. (a)

Lj = 2 16.9 y y = 6.6 (4.5 – 0.5) = 27.6 m

EL = 3

2 1

2 1

y y4y y =

34.5 0.54 4.5 0.5

= 7.11 m

72. (b)

Type of jumpFroude

Number F of incoming

flow

1 Relative Energy loss

Undular Jump 1.0<f <1.71 0

Weak Jump 11.7 F 2.5 5 – 18%

Oscillating Jump 12.5 F 4.5 18 – 45%

Steady Jump 14.5 F 9.0 45 – 70%

F > 9.01 70%

73. (a)

Hydraulic jumps are rapidly varied flows andare associated with energy losses. Watersurface connects sequent depths which areassociated with constant specific force.

74. (c)

yCH3

Zone 3

Zone 2

yn

y

Supercritical flow on horizontal bed

yn

yC

Zone 1

Zone 2

Zone 3M3

y

Supercritical flow on a wild slope

yn = yC

Zone 3

Zone 1

C3y

Supercritical flow on a critical slope

yCZone 1

Zone 2

Zone 3

yn

SZy

Supercritical flow on a steep slope

IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (11)75. (d)

CDL

NDL

S1

S3

76. (b)

For a triangular channel of small slope, specificenergy at critical depth is given by

EC = 2C

C Cv

y 1.25 y2g

2C

Cv

0.25 y2g

Hence statement (1) is incorrect.

Discharge per unit width is maximum for agiven specific force at critical section. Anyattempt to further increase discharge leads toan increase in specific energy.

At critical flow both specific force and specificenergy are minimum for a given discharge.

77. (b)

At critical flow, the specific energy is leastpossible.

3 2q m s m

depthy Critical

depth

ECE2

Sp. energy

E3

For the same flow q m3/s/m2 EC < E2 < E1

Critical depth of flow can be obtained by raisinga hump or by constricting width or both donetogether.

When width of flow is constricted the value ofq(m3/s/m2) increases and its curve on depth-sp. energy graph moves outward and upwardssp. energy E1 remains constant. This happensuntil q-curve touches sp. energy line at justone point, thereby creating a critical flow atthroat section.

E1

Sp. Energy

Depthy

q1q2q3

CyC

q3>q >q2 1

q1 = Flow rate per unit widthat up stream section

During this process the depth of flow of asubcritical flow reduces to yC while that ofsupercritical flow increases to yC. Thus bothsubcritical and super critical flows change intocritical flows. Depth of flow upstream remainssame, because upstream section isrepresented by q1 and E1.On further decreasing width q-curve movesfurther outward, water surface levels adjust;both at upstream section and throat section,so that specific energy increases to form atangent at new q-curve.

E1

E2

q1

q4

Depth of flow upstream afterreducing width

Initial depth of flow upstream

45°

Depth of water upstream either increases ordecreases based on whether the flow issubcritical or super critical. Flow at throatsection remains critical.

CDL

NOL

Steeper Slope S3CDLNOL

Steeper Slope

78. (a)The boundary layer thickness is defined asdistance from boundary where flow velocity is99% of free stream velocity V . So Boundarylayer thickness

IES M

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(12) CE (Test-6), Objective Solutions, 19th March 2016

y = at u = 0.99U

Displacement thickness

* = 0

u1 dy.U

Momentum thickness

=

0u u1 dy

U UEnergy thickness

* * =2

20

u u1 dy.U U

79. (a)80. (b)

The figure clearly explains that at separationpoint and beyond, the pressure gradient isadverse i.e. positive and velocity profile haspoint of infection i.e. velocity profile changesits sign of slope.

dP 0d

=x

dP 0d

xdP 0d

x

Solidsurface

P

u

U

uU

81. (c)

U

*

u yU

=

y

Displacement BL

Momentum BL

Displacement thickness of flow

* = 0

u1 dyU

= 0

y1 dy

=

=2 2

Momentum thickness of flow

=

0u u1 dyU U

= 0

y y1 dyδ

=

2 3 6

* / 2= =3

/ 6

82. (a)The thickness of boundary layer over flat plateat distance x-

x = 16cm

U U

u

= 5 Blassius SolutionRex

=5 16 5 16

16025600 =

= 0.5 cm = 5 mm

83. (b)

Co-axial cylinder Viscometer: It is basedupon Newton law of viscosity. The gapbetween two cylinders is very small.

Capillary tube viscometer: HagenPoiseulle viscosity.

Say-bolt viscometer: In this viscometerviscous fluid falls through capillary. Thisfalling of fluid is called efflux type viscometer.

Falling sphere viscometer: Here a smallsphere is gently droped in liquid at rest anddistance is measured in time i.e. velocity.

84. (a)

Since pressure drop in flow is directlyproportional to friction loss

P hf and hf = 24fL V×

D 2gFor laminar flow, the friction factor

f =16 16=Re VD

h f =2

264 V×

2gVD

h f = 264 × V

2 gD

h f V

IES M

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CE (Test-6), Objective Solutions, 19th March 2016 (13)For laminar flow P V.

Hence flow is laminar.

85. (a)

rVmax

R

U

u

Average velocity = 5 m/sec

Maximum velocity Vmax= 2×5=10 m/sec

Velocity distribution in pipe, or velocity atdistance 'r' from centre

u =

2

max 2rV 1R

=2510 1

10

=

110 14

=304

= 7.5 m/sec

86. (b)Assuming the flow in pipe is laminar.

RU

r

Shear stress at radius r,

r

= x0P

R

= = constant

The pressure drop across length of the pipe

0 =

R P2 x

=

3 350 10 50 102 10

=250

2 N/m2

= 0.125 kPa

87. (c)The friction head loss in pipe,

h f =24fL V×

D 2g

In turbulent flow, f is constant so hf V2.

88. (b)

Because of boundary layer formation roughessis neglected friction factor in laminar flow

f =64Re

In fully developed turbulent flow in roughpipe the friction factor 'f' is independent ofReynold number.

89. (d)

dr

Vmax

17

maxyu UR

=y

r

The mean velocity at centre is nothing butmaximum velocity at centre.

m

uU =

1/7yR

=

1/7R rR

=

1/7r1R

Average velocity V,

1/7R2

m0

rR V = 2 r 1 U drR

1/7Rm2

0

2U rV = r 1 drRR

Assuming r << R

Rm2

0

2U rV r 1 dr7RR

22m 12

2U RRV2 21R

m

1 12U2 21

m

2U 121

m19 U 0.9V21

90. (c)From Darcy-Weisbach equation in terms offlow discharge, the head loss,

IES M

ASTER

(14) CE (Test-6), Objective Solutions, 19th March 2016

h f =

2

2 564L fQ2 g D

fQ2 =

2 5

f2 g D h

64 L = constant

Differentiating,

df 2dQ+ = 0f Q

dQ+2.5% + 2 = 0Q

dQ = 1.25%Q

91. (d)

The phenomenon of development of lift forrotating sphere may be used in the gamessuch as cricket, table tennis, golf and tennisin which by imparting a spin or rotation tothe ball in the backward or forward directionthe lift may be exerted on it in the upwardor downward direction and hence thetrajectory of the ball may be modified by theplayer to his best advantage. As there lift isproduced by circulation around a circularcross-section placed in a uniform stream offluid, it is commonly, known as magnuseffect.An interesting feature of flow past two-di-mensional bodies is the formation and alter-nate release of vortices behind the cylinderfor Re > 30. This vortex shedding leads tolateral vibration of two-dimensional bodies.These trails of vortices are commonly knownas karman vortex trails or Karman vortexstreet.

92. (b)

93. (a)

GM =1 BG

I = Moment inertia of the plan of the pontoon

= 3 41 455 3 m12 4

= Vol. of the body sub merged in water= 3 × 0.8 × 5 = 12 m3

BG = 0.6 – 0.4 = 0.2

GM =

45 1 0.2 0.7375 m4 12

94. (b)

CD =e

24R for Re < 1

=240.2

= 120

95. (b)The maximum shear stress theory (sys = 0.5syt) is the most conservative theory out ofvarious other theories of failure. It gives theover safe and uneconomical results. While,maximum distortion energy theory (sys = 0.577syt) gives the most accurate results (resultsare very close to experimental results). It givesthe safe and economical results.

96. (c)P. Maximum - Normal stress criterion.

y

2

1x– x

– y (M)Q. Maximum - distortion energy crierion

2

1x

– y

– x

y

(N)R. Maximum shear stress criterion.

2

y

1x

– y

– x

(L)97. (b)

Principal stresses are,

1/2 = 2

x y x y 2xy2 2

= 2 250 30 40

1/2 = 50 50

1 = 100; 2 = 0

max = 100 50 MPa

2

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CE (Test-6), Objective Solutions, 19th March 2016 (15)Shear strength of material = 100 MPa

F.O.S. = 100 250

98. (c)At A, resultant will be horizontal only.

i.e. Horizontal force, Fx = gAh

= 31000 9.81 3 42

= 176580 N

= 176.58 kN99. (c)100. (c)

1 = 360 MPa

2 = 140 MPa

3 = 0According to distortion energy theory :

2 2 21 2 2 3 3 1 = 2

s22 21 2 1 2 = 2

s

2s = (360)2 + (140)2 – 360 × 140

s 314.3 MPa

101. (a)

Concave fillet weld Convex fillet weld102. (b)

Wind has lower frequency and hence highertime period.

103. (c)The material cost of HSFG is about 50%greater than that of ordinary bolts and specialworkmanship is required in installing andtightening of these bolts.

104. (c)Bracing will be under compression or tensiondue to lateral loads (wind load).

105. (b)The small insect can sit on liquid surfacedespite having higher density than liquid. Thereason being surface tension as shown infigure.

liquid

Insect

w

TT

The liquid has viscosity but no role inphenomenon.

106. (d)The general expression for non-Newtonian fluid.

= 0dudy

x

So ' ' is not linear function of velocity gradientor distance.This is due to large size of molecules inpolymers etc. and do not behave like moleculesof Newtonian fluid.

107. (b)The distance between centre of pressure andcentroid of surface

x = h hh

G

Px

h = ahAh

x =

ah hAh

= GIAh

At great depth h is very high and the otherterms are constant. So the whole term becomenegligible.

108. (b)This is the case of floating body. Ballast loadingof bottom means addition of heavy weight inlower portion of ship. The ballast increasesweight and brings the center of gravity lower.The stability is related to metacentric height,

MG = BG 0V

I

As metacentric height Gm increases, thestability increases.

G

B

M

Ballast

109. (a)110. (a)

FB V1

V2BG

W

P

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(16) CE (Test-6), Objective Solutions, 19th March 2016Buoyant force,

FB = 2V g W =V2 = Volume of solid body submerged in liquidor volume of liquid displaced by solid body.V1 = Volume of solid body outside the liquid.So FB is weight of liquid displaced and alwaysvertically up against the weight of body.

111. (d)The stability is determined by relative positionsof meta center and center of gravity. Forstability, the relative positions are shown infigure below.The metacentric height MG should be high forhigher stability.

M

G

B

MG = BGV

I

112. (b)If ideal conditions are considered, the thicknessof boundary layer will be ever increasingwhether the boundary layer is laminar orturbulent. The boundary layer breakes whenthere is flow separation.But in practice 99% depth of boundary layerdevelops within short distance from leadingedge. This short distance depends upon flowand surface conditions.

113. (a)

U

u

U

At outer layer of boundary layer, the velocitygradient is zero

du = 0dy

i.e. at outer edge, u = USo shear stress from Newton's law of viscosity.

=duu = 0dy

114. (d)

Velocity of Surge =

Vw =

C V for downstream movementC V for upstream movement

where C = celerityV = river velocityIf C < V, surge would fail to move in upstreamdirection

115. (a)

max 3 2 1q q q q q

y2y3

y1

E1E2

Supercritical

Subcritical

q1q2q3

qmax

45°

yC

1y

Cy

q

When width of flow is reduced, discharge perunit width of flow increases from q1 to q2. Noenergy losses are involved since bed elevationsare same. If the flow is subcritical then, depthof flow decreases from y1 to y2.On further decreasing width, q2 increases toq3 at throat section and depth further reducesto y3. This goes on until a minimum throatwidth Bmin is achieved when flow per unit widthat upstream is q1, and flow is critical at throatwith depth yC.If the width at throat section is further reduced,flow becomes impossible at specific energyE1. Flow, by its natural mechanism, raises itsspecific energy levels to E2 such that flow atthroat section remains critical but with yc >yC. Depth at upstream section increases toy1.

116. (d)Introduction of hump will cause a decrease orincrease in flow depth depending upon whetherthe flow is subcritical or super critical.

E1E2 Sp. energy (E)

Z

Depth y

q

Subcritical flow

Supercritical flow

C

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CE (Test-6), Objective Solutions, 19th March 2016 (17)E2 = 1E Z

WhereE1 = sp. energy at upstream sectionE2 = sp. energy at Hump Section

Z = Height of Hump.

Discharge per unit width (q) curve remainsuncharged because width is unchanged.

117. (a)

PL/2 L/2

PL4

PL4

4EIL

2EIL

L

P

L/2 L/2

PL 2EI4 L

PL 4EI4 L

118. (a)119. (c)

It depends on discharge whether a givenchannel is mild, step or critical. Hencestatement (I) is correct.

Critical depth is a function of discharge perunit width but not normal depth for a rectangularchannel.

yc =1/32q

g

Normal depth is calculated from Manning’s orChezy’s equation.

Hence, statement II is wrong.

120. (a)

The flow is unidirectional and uniform in arectilinear flow. The flow has no vorticity andthus the velocity field is irrotational. Unlike areal fluid, the net drag on the body is zero.