F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 · 11. Thick and Thin cylinder/sphere ......
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Transcript of F-126, (Lower Basement), Katwaria Sarai, New Delhi-110016 · 11. Thick and Thin cylinder/sphere ......
First Edition : 2017
Typeset at : IES Master Publication, New Delhi-110016
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PREFACE
STRENGTH OF MATERIALS 1 – 112
STRUCTURE ANALYSIS 113 – 172
PERT CPM 173 – 216
FLUID MECHANICS 217 – 357
ENVIRONMENTAL ENGINEERING 358 – 417
HYDROLOGY 418 – 426
SOIL MECHANICS 427 – 532
IRRIGATION 533 – 537
HIGHWAY ENGINEERING 538 – 577
SURVEYING 578 – 631
RAILWAY ENGINEERING 632 – 655
AIRPORT ENGINEERING 656 – 665
CONTENT
1. Strength of Materials ................................................................................. 01 – 18
2. Shear force and bending moment ............................................................. 19 – 43
3. Deflection of Beam ................................................................................... 44 – 60
4. Transformation of stress and strain .......................................................... 61 – 67
5. Combined stress....................................................................................... 68 – 72
6. Bending stress in beam ........................................................................... 73 – 82
7. Shear stress in beams ............................................................................. 83 – 86
8. Torsion of circular shaft ............................................................................ 87 – 91
9. Columns ................................................................................................... 92 – 99
10. Springs ................................................................................................. 100 – 103
11. Thick and Thin cylinder/sphere ............................................................. 104 – 108
12. Moment of Inertia ................................................................................... 109 – 112
SYLLABUS
Elastic constants, stress, plane stress, Mohr’s circle of stress, strains, plane strain, Mohr’s circ le of strain,combined stress, Elastic theories of failure; Simple bending, shear; Torsion of circular and rectangularsections and simple members.
UNIT 1STRENGTH OF
MATERIALS
CONTENTS
IAS – 19941. The ratio of Young’s modulus to modulus of rigidity
for a material having Poisson’s ratio 0.2 is
(a)125
(b)5
12
(c)5
14(d)
145
2. A straight wire 15 m long is subjected to a tensilestress of 2000 kgf/cm2. Elastic modulus is 1.5 ×106 kgf/cm2. Coefficient of linear expansion forthe material is 16.66 × 10–6/0f. The temperaturechange (in °F) to produce the same elongation asdue to the 2000 kgf/cm2 tensile stress in thematerial is
(a) 40 (b) 80
(c) 120 (d) 160
3. Which one of the following favours brittle fracturein a ductile material?
(a) Elevated temperature
(b) Slow rate of straining
(c) Presence of notch
(d) Circular cross-section
4. A composite section made of two materials hasmoduli of elasticity in the ratio 1 : 2 and lengthsin the ratio 2 : 1. The ratio of correspondingstresses under a direct load is
(a) 2 : 1 (b) 1 : 2
(c) 4 : 1 (d) 1 : 4
5. Assertion (A) : Cast iron torsion specimen has ahelicoidal fracture at failure.
Reason (R) : Cast iron is weak in tension andfailure of the specimen due to tension takes placeat 45° to the axis of the specimen.
IAS – 19956. Match List-I with List-II and select the correct
answer using the codes given below the lists:
List-I List-II(Material type) (Stress-strain curve)
A Soft and Weak 1.
B. Hard brittle 2.
C. Hard strong 3.
D. Soft tough 4.
E. Hard tough 5.
Codes:A B C D E
(a) 5 4 3 2 1(b) 1 2 3 4 5(c) 5 3 2 4 1(d) 1 4 3 2 5
1 STRENGTH OF
MATERIALS
CSE Prelims Previous Years Solved Papers 3
IES MASTER Publications
7. The material of a rubber balloon has a poisson’sratio of 0.5. If uniform pressure is applied to blowthe balloon, the volumetric strain of the materialwill be
(a) 0.50 (b) 0.25
(c) 0.20 (d) zero
8. The ratio between the stress produced in a barby a sudden application of load (impact loading)as compared to the stress produced by thegradual application of same load is
(a) 1.5 (b) 2.0
(c) 2.5 (d) 3.0
9. A three-dimensional state of stress can be splitinto the two components, the hydrostatic stateand the deviatoric component. The magnitude ofthe hydrostatic state pressure is given by
(a) x y z
3 ( )
(b) x y z
4 ( )
(c) x y z
6 ( )
(d) x y z( )
IAS – 199610. A gradually applied load W is suspended by wire
ropes AB and CD as shown in the figure. Thewires AB and CD, made of the same material andof the same cross-section are connected to arigid block from which the load W is suspendedin such a way that both the ropes stretch by thesame amount. If the stress in AB and CD are p1
and p2 respectively then the ratio 1
2
pp will be
6 m4 m
A
B
C
D
p2p1
x
(a)32
(b)23
(c)49
(d)94
11. A bolt is threaded through a tubular sleeve andthe nut is turned up just tight. Then the nut isturned further, the bolt being put in tension andthe sleeve in compression.
The distance by which the nut is turned is equalto
(a) Deformation in the bolt
(b) Deformation in the sleeve
(c) Difference in deformation in the bolt and thesleeve
(d) Summation of the deformation in the bolt andthe sleeve
12. When the strain in a material increases with timeunder sustained constant stress, the phenomenonis known as
(a) Strain hardening (b) Hysteresis
(c) Creep (d) Visco-elasticity
13. Assertion (A): Assertion (A): Strain is afundamental behaviour of the material, while thestress is a derived concept.
Reason (R): Strain does not have a unit while thestress has a unit.
IAS – 199714. Assertion (A): The ultimate load of a structure
made of ductile material, subjected to reversiblerepeating loads and plastic deformation, is loweredwith each reversal of load.
Reason (R): When subjected to repeated reversalof loads and plastic deformation, the structuremade of a ductile material accumulates residualstrains.
15. In a plane strain problem, the tensile stressesalong two mutually perpendicular rectangularcoordinate axes x and y are x and y respectivelywith x > y and there are no shearing stresses.The Poisson’s ratio is . The stress along thethird rectangular co-ordinate axis z will be
(a) x y( ) (b) x y( )
(c) x y( ) (d) x y( )
16. A bar of elastic material is subjected to a directcompressive stress 1 in the longitudinal direction.Suitable lateral compressive stress 2 is appliedalong each of the other two lateral directions tolimit the net strain in each of the lateral directionsto half the magnitude of what it could be under 1acting alone. If is the poisson’s ratio of thematerial, then the magnitude of 2 is
(a) 12(1 )
(b) 1
1 (1 )2
(c) 112 (1 )
(d) 1212 (1 )
4 Strength of Materials
17. In a particular material, if the modulus of rigidityis equal to the bulk modulus,then the poisson’sratio will be
(a) 1/8 (b) 1/4
(c) 1/2 (d) 1
18. In the given figure showing the XY quarter plane,
x and y linear strains the directions and is
the linear strain the direction at an inclination of from X and Y. The shear strain is defined asshown. The critical value of is given by = where tan 2 is equal to
x
y
O
(a)x y
(b)x y
2
(c)x y
(d)
x y
2
IAS – 199819. A sperical ball of volume 20 m3 is placed under
a certain liquid wherein the ball is subjected to auniform hydrostatic pressure of 200 MPa. If thematerial of the ball has a bulk modulus of elasticityof 2.5 ×105 MPa and a poisson’s ratio of 0.30,then due to the hydrostatic pressure, the volumeof the ball will change by
(a) 0.0008 m3 (b) 0.0144 m3
(c) 0.0160 m3 (d) 0.048 m3
20. A Steel rod of circular section tapers from 2 cmdiameter to 1 cm diameter over a length of 50cm. If the modulus of elasticity of the material is2 × 106 kg/cm2 then the increase in length undera pull of 3000 kg will be
(a)0.3 cm2
(b)30 cm
(c)300 cm
(d) 750 cm
IAS – 199921. Match List-I with List-II and select the correct
answer:
List-I List-II
A. Ratio of lateral strain to 1. Strainlinear strain
B. Ratio of stress to strain 2. Poisson’s ratioC. Ratio of extension to 3. Tensile stress
original lengthD. Ratio of axial pull to 4. Young’s
area of sectionmodulus
Codes:A B C D
(a) 4 2 3 1(b) 4 2 1 3(c) 2 4 3 1(d) 2 4 1 3
22. Assume that Young’s modulus of steel is twicethat of brass. Two bars of brass and a bar of steelof equal cross-section form a single tensionmember with the help of rigid pins as shown inthe figure:
Brass
Steel
Brass
P P
Pin
The shear in the pin will be(a) 0.25 P (b) 0.5 P(c) 0.33 P (d) 0.4 P
23. The volumetric strain of a cylindrical shell withan intenral pressure is equal to(a) 2 longitudinal strain + net circumferential strain(b) 2 longitudinal strain + ½ net circumferential
strain(c) ½ longitudinal strain + 2 net circumferential
strain(d) longitudinal strain + 2 net circumferential
strain.
IAS – 200024. Assertion (A): When a material is subjected to
repeated tensile stress within elastic range, it isfound that the material deteriorates and fracturesafter many but finite number of repeated applicationof stress.
Reason (R) : The critical stress below whichfluctuating stresses cannot cause a fatigue failureis termed as ‘endurance limit’
CSE Prelims Previous Years Solved Papers 9
IES MASTER Publications
1. (a)
2. (b)
3. (c)
4. (d)
5. (a)
6. (a)
7. (d)
8. (b)
9. (a)
10. (a)
11. (d)
12. (c)
13. (b)
14. (a)
15. (b)
16. (c)
17. (a)
18. (a)
19. (c)
20. (a)
21. (d)
22. (a)
23. (d)
24. (b)
25. (b)
26. (a)
27. (d)
28. (b)
29. (a)
30. (c)
31. (b)
32. (c)
33. (a)
34. (b)
35. (c)
36. (c)
37. (b)
38. (c)
39. (b)
40. (a)
41. (b)
42. (c)
43. (a)
44. (c)
45. (a)
ANSWER KEY
46. (c)
47. (d)
48. (a)
49. (d)
50. (c)
51. (d)
52. (a)
53. (b)
54. (a)
EXPLANATIONS
Sol–1: (a)Here Poisson’s ratio, = 0.2 E = 2G(1 + )
E 122(1 0.2) 2 1.2 2.4G 5
Sol–2: (b)Elastic modulus, E = 1.5 × 106 Kgf/cm2
Coefficient of linear expansion, = 16.66× 10–6/°F
l = 15 m
s = 2000 Kgf/cm2
We have
E = E E
l
= 2
6 22000 Kgf/cm 15 m
1.5 10 Kgf/cm
= 0.02 m
By question,
= tl t = 60.02
16.66 10 15 = 80.03°F
Sol–3: (c)Fracture : separation of a body into piecesdue to stress at temperatures below meltingpoint.
Depending on the ability of material to undergoplastic deformation before the fracture twofracture modes can be defined–ductile orbrittle.
Brittle fracture – ceramics, ice, cold metal- Relatively little plastic deformation- Crack is unstable propagates rapidly
without increase in applied stress. Ductile fracture –most metals (not too
cold)- Extensive plastic deformation ahead
of crack- Crack is stable resists further
extension unless applied stress isincreased.
Ductile materials – extensive plasticdeformation and energy absorption(toughness) before fracture.
Brittle materials– little plastic deformationand low energy absorption beforefracture.
Brittle
Stress
B
Ductile
A C
StrainC1
B1
Ductile to brittle transitionAs temperature decreases, a ductile materialcan become brittle–ductile to brittle transition.
10 Strength of Materials
Alloying usually increases the ductile to brittletransition temperature. The ductile to brittletransition can be measured by impact testing.
Two standard tests, the charpy and Izod,measure the impact energy (the energyrequired to fracture a test piece under animpact load), also called the notch toughness.
Fracture strength of a brittle solid is relatedto the cohesive forces between atoms. Theapplied stress is amplified at the tips of micro-cracks, voids, notches, surface scratches,corners, etc. that are called stress raisers.The magnitude of this amplification dependson micro-crack orientatious, geometry anddimensious.
Sol–4: (d)
PA
= EL
= EL
for the same deflection
1
2
=
1 2
2 1
E LE L
= 1 1 1 1: 42 2 4 = =
Sol–5: (a)
Failure in Torsion
A round bar of a low carbon steel can betwisted through several revolutious before itfails. When it does fail, the failure is in aplane of maximum shear stress at right anglesto the axis of the bar and there is littlereduction in area and little, if any, elongation.On the other hand, when a round bar of brittlematerial such as gray cast iron is twisted tofracture, the strains involved are much lessand the failure is of a helical nature. Thefracture plane is at 45° to the axis of the bar,perpendicular to the direction of principaltensile stress. It has been noted that whenthe hardness of a tool steel is above 720vickers hardness (HV), it behaves more likecast iron and fractures in a helical manner.
Sol–6: (a)
Toughness : The ability of a metal to deformplastically and to absorb energy in the processbefore fracture is termed toughness. Theemphasis of this definition should be placedon the ability to absorb energy before fracture.
The ductility is a measure of how muchsomething deforms plastically before fracturebut just because a material is ductile doesnot make it tough. The key to toughness isa good combination of strength and ductility.A material with high strength and high ductilitywill have more toughness than a material withlow strength and high ductility. Therefore oneway to measure toughness is by calculatingthe area under stress-strain curve from atensile test. This value is simply calledmaterial toughness and it has units of energyper volume.
Soft and weak
Soft and tough
Hard and strong
Hard and tough
Hard and brittle
Stress-strain curves for different materials
Hardness : It is the property of a materialthat enables it to resist plastic deformation,penetration, indentation and scratching.Hardness tests serve an important need inindustry even though they do not measure aunique quality that can be termed hardness.
Sol–7: (d)Assuming the dia of the balloon = D
We have volumetric strain of the balloon,V = 3 D
where D = Diametric strain = DD
V = 3p (1 2 )E
CSE Prelims Previous Years Solved Papers 11
IES MASTER Publications
Dx
z
y
here = 0.5 and E is elastic modulus
V = 3pE
(1 – 2 × 0.5) = 0
Sol–8: (b)When a load P is applied gradually to amaterial, the material deforms and stress, is introduced in the material where
= PA
but due to sudden impact or shockload,the maximum stress will be
max = 2PA
Sol–9: (a)Hydrostatic and deviatoric stresses are twosubsets of any given stress tensor whichwhen added together give the original stresstensor back. The hydrostatic stress is relatedto volume change while the deviatoric stressis related to shape change.
Hydrostatic stressHydrostatic stress is simply the average ofthe three normal stress components of anystress tensor.
Hyd = 11 22 33
3
It is a scalar quantity, althrough it is regularlyused in tensor form as
Hyd =Hyd
Hyd
Hyd
0 00 00 0
Hydrostatic stress and pressurePressure is simply the negative of hydrostaticstress. The negative aspect is often confusing.It is why we talk about atmospheric pressureas 30 inches of Hg, a positive number eventhough atmospheric pressure is in fact anegative stress because it is compressive.So using p for pressure
P = 11 22 33
Hyd 3
=
The shears tensor containing pressure, P is
Hyd =P 0 00 P 00 0 P
of course, it is rare to talk about pressureunless the hydrostatic stress is compressive,which corresponds to a positive pressure.
Deviatoric stressDev iatoric stress is what’s lef t af tersubstracting out the hydrostatic stress. Thedeviatoric stress will be represented by 1.for example
1 = Hyd
In tensor notation, it is written as
1ij = ij ij kk
13
And in terms of pressure, it is written as
1ij = ij ij
Example
Given the following stress tensor
=50 30 2030 20 1020 10 10
The hydrostatic stress is
Hyd = 50 1020 13.3
3 =
which can be written as
Hyd =13.3 0 0
0 13.3 00 0 13.3
substracting the hydrostatic stress tensorfrom the total stress gives
1 = 50 30 2030 20 1020 10 10
13.3 0 00 13.3 00 0 13.3
= 36.7 30.0 20.030.0 33.3 10.020.0 10.0 3.3
Sol–10: (a) Elongation in wire AB = AB
Elongation in wire CD = CD
12 Strength of Materials
AB = CD
=E l
and 1 2 21
p pE E
ll
1 2
2 1
p 6 3p 4 2
ll
Sol–11: (d)As the nut is tightened, bolt being in tensionwill elongate and sleeve being in compressionwill shorten. Thus, the distance by which thenut is turned equals to summation of thedeformations in the bolt and the sleeve.
Sol–12: (c)Strain hardening : Strain hardening is whena metal is strained beyond the yield point. Anincreasing stress is required to produceadditional plastic deformation and the metalapparently becomes stronger and more difficultto deform. Strain harderning reduces ductility,which increases the chances of brittle failure.
Hysteresis : All piezoelectric materials exhibita mechanical hystersis as the strain doesnot follow the same track upon charging anddischarging. The hysteresis is expressed asthe maximum strain divided by the maximumdifference between the two tracks.
Creep : Creep describes the slow, progressivedeformation of a material under a sustainedloading.
Visco-Elasticity : It is the property of materialsthat exhibit both v iscous and elasticcharacteristics when undergoing deformation.Viscoelastic materials exhibit time-dependentstrain. Some phenomena in visoc-elasticmaterials are :
if the stress is held constant, the strainincreases with time (creep).
if the strain is held constant, the stressdecreases with time (relaxation)
if cyclic loading is applied, hysteresis (aphase lag) occurs, leading to a dissipationof mechanical energy.
Sol–13: (b)Consider the fact that first we have to applya force to an object and it is only then theobject is observed to have been deformed orstrained. Accordingly, one may say that forcesproduce strain and therefore it seems that
stress has to be more fundamental but thestress can not be measured directly. It is thestrain which can be measured directly andconsequently strain can be obtained. Thusstrain is a fundamental quantity.
Sol–14: (a)A material can fail after repeated stressloadings even if the stress level exceeds theultimate strength, a condition known as fatiquefailure. The yield strength is an irrelevant factorin cyclic loading. Fatique failures are fracturefailures, not yielding failures. They start withmicroscopic cracks at the material surface.These cracks grow minutely with eachloading. Since cracks start at the location ofsurface defects, the endurance limit isincreased by proper treatment of the surface.The endurance limit is not a true property ofthe material since the other significantinfluences particularly surface finish, are nevereliminated.
Sol–15: (b)
y
y
x
x
z
Strain along z- axis = yxz – –
E E
or, z x y x y– ( )
i.e. stress along z-axis,
z = x y( )
Sol–16: (c)
y2
x
1
2z
1
CSE Prelims Previous Years Solved Papers 13
IES MASTER Publications
If 1 is acting alone in z-direction
1z E
If stresses are acting in all direction, thenstrain in z-direction.
z
2 2 1
E E E
By question,
1 2 2 112 E E E E
or, 12 1( 1)
2
or, 11 2( 1)
2
or, 12(1 )
2
1
2 2(1 )
Sol–17: (a)
We have 3K 2G6K 2G
Where = Poisson’s ratio
K = Bulk modulus
G = Modulus of rigidity
Here by equation G = K
3K 2G K 16K 2G 8K 8
Sol–18: (a)
x y
tan2
Where is maximum or minimum straini.e. principal strains.
Sol–19: (c)
Volumetric strain of the spherical ball
v = 3 = p3 (1 2 )E
E = 3K(1 – 2)
= 3 × 2.5 × 105(1 – 2 × 0.3)
= 3 × 105 MPa
5v 3 200 (1 2 0.3)
V 3 10
= –80 × 10–5
v = –80 × 10–5 × 20m3 = 0.016 m3
Sol–20: (a)
d2 Pd1P
Elongation = 61 2
4PL 4 3000 50d d E 2 1 2 10
= 0.15 0.30 cm
2
here d1 = 2 cm
d2 = 1 cmL = 50 cmP = 3000 kgE = 2 × 106 kg/cm2
Sol–21: (d) Normal strain is defined as deformation
permit length.
Normal strain, = L
For homogeneous and isotropic material,elongation/(contraction) produced by axialany force in the direction of force isaccompanied by contraction/ (elongation)in all transverse direction and all suchcontractions/(elongations) are same.
Poisson’s ratio () = –Lateral strainAxial strain
young’s modulus = StressStrain within elastic
limitSol–22: (a)
Brass 0.25 P
Steel
Brass
0.5 P
0.25 P
Sol–23: (d)Let volume of the cylinder = V
V = 2d4 l
z
y
x d
l
