Test-1-obj -14-3-15 Sol- CE
-
Upload
akshay-kumar -
Category
Documents
-
view
229 -
download
4
description
Transcript of Test-1-obj -14-3-15 Sol- CE
1. (d)
2. (c)
3. (d)
4. (b)
5. (d)
6. (c)
7. (b)
8. (c)
9. (b)
10. (b)
11. (a)
12. (c)
13. (b)
14. (d)
15. (d)
16. (a)
17. (c)
18. (d)
19. (c)
20. (c)
21. (a)
22. (a)
23. (a)
24. (b)
25. (d)
26. (b)
27. (a)
28. (a)
29. (b)
30. (c)
31. (b)
32. (b)
33. (d)
34. (d)
35. (d)
36. (b)
37. (a)
38. (c)
39. (a)
40. (c)
41. (d)
42. (d)
43. (a)
44. (b)
45. (d)
46. (d)
47. (d)
48. (a)
49. (d)
50. (c)
51. (d)
52. (c)
53. (b)
54. (a)
55. (a)
56. (d)
57. (a)
58. (b)
59. (d)
60. (d)
61. (b)
62. (d)
63. (c)
64. (a)
65. (a)
66. (c)
67. (a)
68. (b)
69. (d)
70. (b)
71. (b)
72. (b)
73. (a)
74. (a)
75. (d)
760. (a)
77. (b)
78. (b)
79. (b)
80. (d)
81. (c)
82. (a)
83. (c)
84. (a)
85. (b)
86. (b)
87. (b)
88. (d)
89. (c)
90. (d)
91. (b)
92. (d)
93. (d)
94. (c)
95. (c)
96. (a)
97. (b)
98. (a)
99. (c)
100. (a)
101. (b)
102. (a)
103. (d)
104. (d)
105. (d)
106. (d)
107. (c)
108. (d)
109. (a)
110. (a)
111. (d)
112. (b)
113. (d)
114. (a)
115. (c)
116. (d)
117. (b)
118. (a)
119. (a)
120. (d)
ANSWERS
TEST-1 OBEJECTIVE ANSWER AND SOLUTION
IES M
ASTER
(2)
1. (d)
tmax = 1 22
where 1 and 2 are principal stressesMaximum shear stress for set 1 =
0
2 2Maximum shear stress for set 2 =
02
Maximum shear stress for set 3 =
2
Maximum shear stress for set 4 =
22 4
2. (c) Since, shaft is arranged in parallel, thereforeangular deflection in both the shafts will bethe same.
T
L
2dd
L
A B C
TA TC
AB = BC
A
AB
T LGJ =
C
BC
T .LG.J
A
4
T
.d32
=
C4
T
2d32
TA = CT16
C AT 16T
3. (d) Given,
1 = – 100 MPa
2 = – 10 MPa
Applying maximum shear stress theory(MSST).
Abs max < Sys
12
< ytS2
Syt > –100 MPa
ytS 100MPa
4. (b)
At the point within the element, value ofhydrostatic pressure is same in all the di-
rections so x y and hydrostatic pres-
sure acts normal to the surface therefore
xy 0 .
Radius of the Mohr’s circle,
R =
2x y 2
xy2 = 0
5. (d) Shear stress distribution in a rectangularcross-section beam is shown below.
maxNeutral
axis
The figure clearly indicates that the trans-verse shear stress is variable with maxi-mum on the neutral axis.
6. (c) Given, a solid circular shaft of diameter d issubjected to a combined bending momentM and torque T.
T
d
T
M M M
By torsion equation,
=T .RJ
= 3
T d2d
32
= 316T
d
TEST-1 SOLUTION
IES M
ASTER
(3)
By bending moment equation,
b =M.yI
= 4
M d2d
64
b = 332M
dMaximum shear stress,
max =
22b
2
2 2max 3
16 M Td
Condition for safe design,
Maximum stress < Permissible stressinduced
ysmax
SFOS
Hence, torsional yielding strength (Ssy orSys) will be used in design.
7. (b)y
(– , 0) (0, 0) (+ , 0)
Mohr’s circle for the given conditions
8. (c) Given,
Emild steel= 206 GPaEcast Iron = 100 GPa
Elongation in bar due to force P,
=PLAE
Since P, L and A are constant.
1E
mild steel
cast Iron
= cast Iron
mild steel
EE
=100 1206
mild steel cast Iron
9. (b)0.2 m × 0.2 m
50 kN
A
15 m
Cable
T TT
T
T
The cable cannot carry any shear force ormoment it can carry only axial force
10. (b) Drop of weight W causes an impact on theflange and transfers the potential energyWh to the vertical rod PQ of length L
Wh =12
× stress × strain × volume
=12
.E
.A × L
=21 AL
2 E
From above equation it is clear that, todecrease the stress in the rod PQ, lengthof the rod should be increased or area ofcross section should be increased ormodulus of elasticity should be decreased.
However, for increasing the stress, converseof above holds true.
11. (a) Outer rods are of same length and samecross section,
IES M
ASTER
(4)
Let force resisted by each outer rod is P1
Let force resisted by central rod is P2
2P1 + P2 = 50 ...(i)
50 kN
2AA,L
2A
3
1
22L
All rods are connected by rigid bar andthere is symmetry in loading so deflectionin each rod will be same
1P (2L)2A E
= 2P LA E
P1 = P2
3P1 = 50
P1 = 50 16.67 kN3
P2 = 50 16.67 kN3
12. (c) Length of bar = 200 mm
Increase in temperature,t = 20°C
Young’s modulus,E = 2 × 105 MPa
Coefficient of thermal expansion, = 12 × 10–6 per °C
Stress in the bar
=EL
× (Deflection prevented)
=EL
× Lt
= 2 × 105 × 12 × 10–6 × 20
= 48 MPa
13. (b)
2A 3A A
= I II III
= PL PL PL
2AE 3AE AE
=
PL(3 2 6) 11PL
6AE 6AE
14. (d)
P
P
2.5 AE
P
P
2AE 1
2
P L / 22.5 AE =
PL5AE
15. (d) Major principal stress
1 = 3 MPaMinor principal stress
3 = – 6 MPaMaximum shear stress
=
3 ( 6) 4.5 MPa
2
16. (a) Principal plane is the plane on which onlynormal stress acts and shear stress is zero.Isotropic state of stress is indistinguishablewith respect to the frame of reference
17. (c)stress on horizontal axis
(0, )
(– ,0) (0,0) ( ,0)
(0,– )
270°
18. (d) State of stress
50 00 50
=
xx xy
yx yy
IES M
ASTER
(5)
xx = 50yy = 50 xy = 0
Co-ordinate of centre of Mohr circle,
(x, y) =
50 50 ,02 = (50, 0)
radius r =2
xx yy 2xy2
=
2250 50 0
2
= 0
19. (c)
20. (c)
Vol. change Shrinkage limitPotential %low 12Moderate 10 12high 10
Dual symbol as per IS code is used persoil having fines between 5 to 12%.
21. (a)
7.5
7.5
7.5
–7.5
7.5 N/mm2
7.5 N/mm2
Diagram (II)Diagram (I)
22. (a)MI
=fy =
ER
f =EyR
= 3 2200 10 N/mm 1mm
1000 mm= 200 N/mm2
23. (a) In a simply supported beam two types ofstress acts
P
Q
NA
L
L1
L1
W
W
W
L
Shear force diagram
L1 BMD
1. Flexural stress MY
I &
2. Shear stress b
VAYI
All the symbols have their usual meaningPoint P
Point P lies on NA
y = 0
Hence, 0
Point P also lies at mid span, So shear ithas force, VHence, it has shear stressPoint QAt point Q flexural stress is acting andnature of which is tensile due to downwardloading.As the point Q lies on bottom of beam
Q
Hence, 0
24. (b)
C.G
Maximum shear stress occurs at thecentroid of the section.
25. (d) In lime stabilisation, the liquid limit of thesoil generally decreases but the plastic limitincreases. Lime reacts chemically withavailable silica and alumina in soils. A naturalcement composed of calcium aluminosilicate complexes is formed, which causesa cementing action.
IES M
ASTER
(6)
Lime addition leads to addition of Ca2+ ionswhich replace Na+ ion and makes soil moreflocculated.
Note that degree of flocculation is governedby the valency of cation and concentrationand cation increase in valency leads toflocculation
26. (b) A BC D
T TL LL
Structure and loading is symmetrical aboutmid point so torsional moment generatedat C and D will be equal to T.
C
TA
TUsing Torsion formula,
TJ
=GL
1 =1P
TLGI
27. (a) Activity of soil depends on plasticity indexwhich in turn depends on type and amountof clay mineral. It also depends on % claysized particles.
28. (a)
Theory of Failures Also known as
(a) Maximum principal stress theory
Rankine theory, Lame’stheory, max stress theory
(b) Maximum principal strain theory
(c) Maximum shear stress theory
(d) Maximum strain energy theory
(e) Maximum shear strain energy theory
St. Venant theory
Tresca, Guest, columbtheory
Beltrami Haigh theory
Distortion energy theoryVon-Mises theory
29. (b) According to maximum principal straintheory,
maximum strain caused by stresses100 kN/mm2 and 60 N/mm2 = Maximumstrain caused by equivalent stress in simpletension.
100E
– 60
E=
E
= 100 – 0.25 × 60
= 85 N/mm2
30. (c)Cast iron is brittle material so most appropriatetheory is Rankine theory (maximum principalstress theory).
31. (b) Strain variation across the cross sectionremains linear but due to change in materialin the filtched beam, stress changesabruptly although strain variation is linearthroughout.
32. (b) In the derivation of formula, M f EI y R we
consider
(i) Linear variation of strain(ii) Plane bending
Under pure bending
y = 2 xz zx 0
33. (d)
Bending stress, = MyI
At the extreme fibre y is maximum sobending stress is also maximum.
At neutral axis, y = 0 = 0
hence bending stress is zero at NA. Neutral axis passes through the centroid of
cross-section (If Hook’s law is valid).
34. (d)Max. bending stress max = max maxM yI
For same cross-sectional shape dimensionsand loading, Mmax, ymax and I will remainsame. Therefore bending stress will remainunaltered.
35. (d) In simply supported beam loading isdownward therefore sagging bendingmoment will be generated in beam. Topfibre will be in compression and bottom fibrewill be in tension.
IES M
ASTER
(7)
AN
c
t(Stress diagram)
40 mm
100 mm60 mm
In stress diagram from similar trianglet60
=c40
t
c
=
3 1.52
36. (b) Given,
WL = 38%
WP = 27%
WS = 24.5%
Wn = 30%
Volu
me
LiquidPlasticSemisolidSolid
24.5% 27% 30% 38%Water content
Hence, the clay is in plastic stage.
37. (a)
(a) Loose (b) Dense
Structure in loosest and densest states assuming spherical particle size.
e = 0.91max e = 0.35min
38. (c)
Given,
e = 1.5
Porosity
n =e
1 e =
1.5
1 1.5 =
1.52.5
n 60%
39. (a) Standard penetration Test (SPT) is basicallymeant to determine the relative density ofthe sandy grounds; but has been empiricallyextended to be used for determining theallowable bearing pressure for a givensettlement and also to approximate theshear parameters of the ground.
When the test is performed below thewater-table, then care must be taken toavoid entry of water through the bottomof the bore hole, as this would tend toloosen the sand due to upward seepagepressure. Water, on the other hand,should be added in the bore hole asnecessary, to maintain the water-tablelevel in the bore hole so as to balanceany excess pore water pressure.
Terzaghi has also recommended that theSPT-N values should be measuredbetween the foundation level and a depthapprox im ately equal to width offoundation, B in different bore holes, boredat different points of the building area.The average value for each bore holeshould be worked out and then theminimum of these average values shouldbe taken as the observed N-value whichis further corrected for over-burdencorrection and dilatancy correction, as toobtain the corrected N-value for furtherfinding out the soil characteristics towhich these N-v alues hav e beenempirically connected.
40. (c) If =P
f
II
41. (d)Relative density
Dr = max
max min
e ee e
= 0.6 0.50.6 0.2
= 0.25 = 25%
IES M
ASTER
(8)
42. (d) Properties of coarse grained soil(cohesionless) to a greater extent dependon grain size distribution. Properties of finegrained soil depends little on grain sizedistribution. They rather depend on structure,shape, geological origin etc. Particle sizedistribution curves helps in determininggradation and uniformity of coarse grainedsoils. This knowledge helps in constructionof earth dams, embankement filters etc.
43. (a) A uniformly graded soil is a soil that hasmost of its particles at about the samesize.
A well graded soil is a soil that containsparticles of a wide range of sizes and goodrepresentation of all sizes of particles.
A poorly graded soil is a soil that doesnot have a good representation of all sizesof particles.
A gap graded soil is a soil that has anexcess or deficiency of certain particle sizesor soil that has at least one particle sizemissing.
44. (b) The Group index value of soils vary in therange of 0 to 20.
45. (d)
Undistrubred sample is the one thatpreserves the particle size distribution as wellas the soil structure of the in-situ valve. Suchundisturbed soil samples are required forshear strength and consolidation tests. Sucha sample can be lifted by stopping the boringprocess at a certain level and then insertingthe appropriate sampler below the bottom ofthe bore.
Representative or disturbed sample is thatwhich contains the same particle sizedistribution as in the insitu stratum fromwhich it is collected, though the soil structuremay be seriously disturbed. The watercontent may also have changed. Suchdisturbed samples can be used foridentification of soil types of different stratafor determining Atterberg limits, specificgravity of solids,organic and carbonatecontent and for compaction test, etc.
California Bearing ratio method (CBRMethod) : This method is applicable to thedesign of flexible pavements only and is
considered to give quite reliable resultsprovided the tests are carried out underspecified conditons. The soil sample to beused in this test is drawn either in anundisturbed condition from the subgradgeground or can be remouldedin the laboratoryfrom the subgrade soil. An undisturbedsample is to be tested when the sub-gradeis to be used in natural condition withoutartificial compaction; and a remoulded sampleis to be used when the subgrade is to becompacted.
46. (d) Dry density,
d = a w(1 n ) G1 Gw
For zero air void density
na = 0 and S = 1
(d)theoretical = wG1 Gw
…(i)
We know that,
eS = Gw
e = Gw
From (i)
(d)theoretical =
wG1 e
Thus, dry density at 100% saturation beinga non-linear function of void ratio and watercontent. Hence the zero-air voids curve isnon-linear.
47. (d) Zero air void density
d =
wG
1 Gwand e = wG
Hence, the correct option is (d).
48. (a) vv 2
C tTH
For the same soil and same degree ofconsolidation Tv and Cv remains same
2t H
21 1t H
When additional drainage layer is presentat the middle of clay layer, length of drainagepath will be reduced to H1/2,
IES M
ASTER
(9)
21
2Ht2
211
22 1
Htt H
2
= 4
Given t1 = 8 years
t2 = 1t4
= 2 years
49. (d) mv = 0 v
0
e1 e a
1 e
=(1.0 0.9) 100
2.0
=0.1
2.0 100
= 5 × 10–4 m2/kN
50. (c) Terzaghi’s one-dimensional consolidationtheory assumes that coef f icient ofcompressibility (av) is constant throughoutthe soil.
Co-efficient of compressibility is the slopeof e Vs P plot. Hence the av to be constante Vs P relationship is linear.
e1
e2
P1 P2
P
e
av = 1 2
2 1
e eP P
av = eP
51. (d) Power = Torque × Angular velocity
A
B
PP
= A A
B B
T 2 NT 2 N
= 1 3 1.52
52. (c) Cast iron is brittle material so it will fail dueto tensile stress. In case of torsion, tensilestress is maximum at an angle of 45° to
the axis of the specimen, so failure surfacewill be in helicoidal shape at 45° to theaxis of specimen.
Tension
Compression45°
Failure surface
53. (b)
A
B
C+ 30 kN
ba
RA = b 3LP 30 18 kN Tension
a b 2L 3L
RC = a 2LP 30 12 kN (Comp)
a b 2L 3L
54. (a) Hydrostatic and deviatoric stresses are twosubsets of any given stress tensor whichwhen added together give the original stresstensor back. The hydrostatic stress isrelated to volume change while thedeviatoric stress is related to shape change.
Hydrostatic stressHydrostatic stress is simply the averageof the three normal stress componentsof any stress 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 negat iv e aspect is of tenconfusing. It is why we talk about atmosphericpressure as 30 inches of Hg, a positivenumber even though atmospheric pressureis in fact a negative stress because it iscompressive. So using p for pressure
P = 11 22 33
Hyd 3
=
IES M
ASTER
(10)
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 iator ic st ress is what ’s lef t af tersubstracting out the hydrostatic stress. The
deviatoric 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
55. (a)
Material Modulus ofElasticity (N/mm2)
Steel –––––––––––––––– 2 × 105
Cast iron –––––––––––––––– 1 × 105
Aluminium –––––––––––––––– 0.6 × 105
Timber –––––––––––––––– 0.1 × 105
Copper –––––––––––––––– 1.2 × 105
Diamond –––––––––––––––– 12 × 105
Rubber –––––––––––––––– 20
Concrete –––––––––––––––– 0.165 × 105
56. (d) Poisson’s ratio is defined as the negativeof the ratio of the lateral strain to the axialstrain for a uniaxial stress state.
= – lateral
axial
Poisson’s ratio is sometimes also defined asthe ratio of the absolute values of lateral andaxial strain. For stresses within the elasticrange, this ratio is aproximately constant. Fora perfectly isotropic elastic material, poisson’sratio is 0.25, but for most materials the valuelies in the range of 0.28 to 0.33. Generally forsteels, poisson’s ratio will have a value ofapproximately 0.3.
57. (a) Physical properties of carbon steel
Material Density 3 310 kgm
Thermal Conductivity
1 1 1Jm K 5
Thermal Expansion
6 110 K
Young’s Modulus
2GNm
Tensile strength
2MNm
% Explanation
0.2% C Steel 7.86 50 11.7 210 350 30
0.4% C Steel 7.85 48 11.3 210 600 20
0.8% C Steel 7.84 46 10.8 210 800 8
Corrosion resistance reduces with moreaddition of carbon.
Carbon content reduces the fracture tough-ness
With more addition of carbon leads fromductile to brittleness
Both yield strength and ultimate strength canbe increased with the carbon content.
58. (b)
12
A
B C
12
Considering equilibrium of the element,
(AC)n = (AB) cos
n = cos ABAC
n
A
B Cor, n = cos2
59. (d) The Mohr Circle will be a point. At all pointson any plane, normal stress will be 100 N/mm2.
Alternatively,
n =
100 100 100 100 cos60
2 2
= 100 N/mm2
IES M
ASTER
(11)
60. (d) The strength of a section depends uponthe section modulus. Higher the sectionmodulus, higher is the strength.
(A) 200 mm
600 mm AN
Section modulus
AZ = 2bd
6
= 2200 6006
= 6 312 10 mm
800mm
60mm
(B)
Section modulus, BZ = 2bd
6 =
= 2800 606
= 4 348 10 mm
A
B
ZZ =
6
412 1048 10
= 25
61. (b) Volume at liquid limit > Volume at Plasticlimit > Volume at Shrinkage limit.
62. (d) Terzaghi stated that effective stress controlscertain aspects of the soil behavior, notablycompression and strength. This means thatthe compression dependes on the effectivestress only Namely
e = f ()
Where e is the void ratio and f () standsfor a function describing the consolidation.By using the effective stress principle,Terzahi derived the consolidation theory forsaturated soils.
63. (c) Casagrande’s logarithm of time fittingmethod is used to determine coefficient ofconsolidation Cv.
64. (a) e – log p curve is convex upward for overconsolidated clay and it is straight line fornormally consolidated clay.
65. (a)
66. (c) Area ratio,
2 22 1
r 21
D DA 100D
D4
D3
D1
D2
D1 Inside diameter of cutting edge
D2 Outside diameter of cutting edge
D3 Inside diameter of sampling tube
D4 Outside diameter of sampling tube
It is desirable < 10%
but / 20%
Inside clearance
i3 1
1
D DC 100D
= 1 to 3%
Outside clearance
2 40
4
D DC 100D
= 0 to 2%
67. (a)Auger Boring Exploration for shallow
foundation
Wash Boring Below water table inall soil types excepthard soils and rocks
Percussion Bouldery andDrilling gravelly strata
Rotary large diameter boreDrilling holes over 150 mm in
size
IES M
ASTER
(12)
68. (b) All types of soils carried and deposited bywater are known as alluvial deposits.Deposits made in lakes are called lacustrinedeposits. Marine deposits are formed whenthe flowing water carries soils to ocean orsea. Soils deposited by wind are known asAeolian deposits.
Correct option is (b).
69. (d) Aeolian soil is depositied by wind. Itconsists of uniformly graded particles. Theyare in loose state so void ratio andpermeability of soil is high. These soils havehigh compressibility and low density.
70. (b) Activity (A) of soil is the ratio of the plasticityindex and the percentage of clay fraction(minus 2 size)
A = PIF
... (i)
IP = Plasticity indexF = Clay fraction (percentage finer than
2 size )IP = WL – WP
WL = Liquid limit = 65%WP = Plastic limit = 29%
IP = 65 – 29 = 36%F = 48
Using (i)
A = 36 0.7548
71 . (b)
1
1/6
2/3
vs
Air
Water
Solid
Void ratio e =Volume of voidsVolume of solids
=
a w
a w
1 2V V 6 3
1 2V (V V ) 16 3
= 5
72. (b) Liquid limt %
20–30 Low plasticity
35–50 Intermediate plasticity
> 50 High plasticity
Above A line clay
Below A line Silt or organic soil
73. (a) At shrinkage limit soil will be fully saturated(S = 1)
Se = wGs
w =
s
Se 1 0.27G 2.7 = 0.1 = 10%
74. (a) Volume of solids will remain same in filland barrow pit
Vs =V
1 e
fill
V1 e
=borrow pit
V1 e
1200001 0.8
=V
1 1.4V = 160000 cum
75. (d) Consistency is a term which is used todescribe the degree of firmness of a soil ina qualitative manner by using descriptionssuch as soft, medium, firm, stiff or hard. Itrefers to the resistance offered by it againstforces that tend to deform or rupture thesoil aggregate. Consistency is directlyrelated to strength and depends on moisturecontent of soil.
76. (a) Bulk density, t = wG Se1 e
Dry density, d =
wG1 e
{For dry condition S = 0}
Saturated density,
sat = w
G 1 e1 e
{For saturated soil S = 1}
Submerged density,sub = sat – w
= w w
G e1 e
= w
G 11 e
IES M
ASTER
(13)
77. (b)
Expansive soil can imbibe more water soits liquid limit is more.
It shows change in volume at lesser watercontent so shrinkage limit is less.
Shrinkage and swelling is more in expansivesoil.
Plastic limit is less in expansive soil.
Activity is more in expansive soil.
Plasticity index is more in expansive soil.
78. (b) Relative compaction= Degree of compactionachieved as a percentage of the laboratorycompaction
R(%) =
d
d
in the field 100Maximum from the proctor test
Relative density =
max natural
max min
e e 100e e
Note : Relative density can be zero but relativecompaction can never be zero.
79. (b) Coefficient of compressibility
a v =e
=
(0.6 0.8)(19 17)
= 0.1 m2/t
80. (d) Settement,
H =
0
e H1 e
=
0 f
0
e e 0.8 0.44H 41 e 1 0.8
= 0.8 m = 80 cm
81. (c) We know, coefficient of permeability,
k = Cvmvw
k Cv
or Cv k
82. (a)
For 100% consolidation
Time factor, Tv
hence time take for 100% consolidation t
According to Terzaghi’s one dimensionaltheory of consolidation there is a uniquerelationship between void ratio and effective
stress. But in secondary compressionrelationship between void ratio and effectivestress is not unique, as secondarycompression occurs at constant effectivestress.
So secondary consolidation does not obeyTerzaghi’s 1 D theory of consolidation.
Init ial excess pore water pressuredistribution leads to change in the timefactor. For example time factor for the initialpore water pressure as shown in Fig. (a) isless than that for Fig. (b)
Fig (a) Fig (b)
83. (c) According to Terzaghi’s one dimensionalconsolidation theory
Time factor,
Tv = v2
C tH
where Cv is coefficient of consolidation andis given as
Cv =
0
v w
K(1 e )a
where K is permeability and av is coefficientof compressibility.
For a particular degree of consolidation anddrainage condition, Tv and H are constant,hence time required for consolidation isinversely proportional to Cv
v
1tC
(i) As compressibility (av) increases, Cvdecreases hence consolidation timeincreases.
(ii) As permeabil i ty (K) increases, Cvincreases, consolidation time decreases.
(iii) Cv also depends on magnitude of effectivestress, so consolidation time will alsodepend on magnitude of stress increase.
84. (a) The standard practice to take significantdepth of exploration is upto the depth at
IES M
ASTER
(14)
which the excess vertical stress caused bya fully loaded foundation is of the order of20% or less of the net imposed verticalstress at the foundation base level. Thedepth as per this practice works out toabout 1.5 times the least width of thefoundation from the base level of thefoundation.
But in the case of square loaded areas, theisobar of 10% intensity of loading atfoundation level extents to a depth of abouttwice the width of foundation below the baseof foundation. It is therefore useful toinvestigate the subsoil to a depth of atleasttwice the width of foundation below the baseof foundation.
Hence, the correct option is (d).
85. (b) Recovery ratios
= Recovered length of sample
Penetration length of samplerLr = 1 good recoveryLr < 1 Soil in sample is compressedLr > 1 Soil has swelled
86. (b) Clay soil rich in montmorillonite exhibitsmore swelling characteristics.
Free swelling of the soil is defined as theincrease in volume of soil without anyexternal constraints on submergence inwater (IS : 2720, 1997) and in general freeswelling ceases at plastic limit.
87. (b)
< 20
20 – 3535 – 50
> 50
Differentialfree soil (%)
Degree ofexpansiveness
LowModerateHighVery high
88. (d) As per IS : 2720 (Part III – 1980)
Differential free Degree ofswell expansion
< 20% Low
20 – 35% Moderate
35 – 50% High
> 50% Very high
The swelling pressure of a soil does nothave a unique value because it depends
upon a number of factors such as initialmoisture content, initial dry density, methodof compaction, height of specimen,surcharge pressure etc.
If the swelling pressure of a soil is lessthan 20 kN/m2, i t indicates theexpansiveness is low and the conventionaldesign of shallow foundation can beadopted.
The swelling pressure test is conducted inan oedometer.
89. (c) S =w
v
vv
During consolidation process, the soil samplealways remains saturated.
90. (d) The temperature lower than 110° 5°C maynot cause complete evaporation of waterand a temperature higher than thistemperature may cause the breaking downof the crystalline structure of the soilparticles and loss of chemically bound,structural water. This temperature is notsuitable for soils containing significantamount of organic matter. For all such soils,a temperature of 60° to 80°C isrecommended. At higher temperature,gypsum loses its water of crystaliline andthe organic soils tend to decompose andget oxidised.
91. (b) Soil A Soil B
e = 0.5 e = 0.7
v = 1.5 m3 v = 1.7 m3
vv = 0.5 vv = 0.7
vS = 1 vS = 1
e of soil sample ‘c’ = 0.5 0.7 1.2 0.6
2 2
92. (d)93. (d)
Clay clay 50%
sand
(%) 2
0%
30% Silt (%)
IES M
ASTER
(15)
94. (c)s.e. = w.G
e = 30 2.7 0.81100
95. (c)
R
d
pene
trat
ion
resi
stan
ce
Water Content(%)Calibration curve for penetration resistancefor proctor needle.
96. (a)97. (b)
Sand
Clay
isochronesmaxu
t=0t =
t1
t2t3
Sand
sand
98. (a) 0
=over consolidation Ratio (OCR)
where = present over burden pressure
0 = pressure under which the deposite hasbeen fully consolidated in the past.
99. (c)
HigherenergyLesserenergy
OMC2 OMC1water content %
Dry unit weight
Optimum moisture content decreases asthe compactive effort (energy) increases,compaction curve shifts to the top and left,however maximum dry density is more fortest with higher energy.
100. (a) Soil structure is dispersed on DS andflocculated on WS.
101. (b) Ductile materials are weak in shear so wecan apply max shear distortion theory
102. (a) For bending without twisting, plane of loadingmust contain one of the principal centralaxis of the section. In case the section ishav ing a plane of symmetry, thesymmetrical plane contain the principalcentral axis.Thus assertion is correct the bending axiswill be perpendicular to the plane of loadingis the case. Hence reason is true.
103. (d) Strength of beam depends upon the sectionmodulus and section modulus depend uponsection area so assertion is wrong.Reasoning is correct because dispositionof area affects the moment of inertia of thesection.have option is (d)
104. (d) From torsion formula
r
=T GJ L
=T rJ
d2ds
d1
They have same weight Area of x-section will be same for the two
shaft
2Sd
4
=2 21 2(d d )4
2Sd = 2 2
1 2d d ... (i)
From (i) we can say that d1 > dS
Assuming same max. shear stress in both case
h
s
TT =
hmax
1
Smax
S
J(d / 2)
J(d / 2)
IES M
ASTER
(16)
h
s
TT =
sh
s 1
dJJ d
h
s
TT =
4 4s1 2
s 1
dd d .d 4 d
h
s
TT =
3 41 2 1
3s
d d / dd
...
(ii)From (i) & (ii)We can say that
Th > Ts
also power transmitted by shaftP = TP T
So Ph > Ps
105. (d)
Two beams, one placed directly over theother (when they are not rigidly connected)have total moment of resistance equal tothe sum of moments of resistance ofindividual sections.
In bending the transverse sections whichare plane before bending remain plane afterbending as well.
106. (d)
z xz yz xz yz
x y xy x y z xy
xz yz z xz yz
x y z xy x y xy
Stress 0, 0, 0 0, 0
, and , , and
may be non-zero may be non-zeroStrain 0, 0 0, 0, 0
, , and , ,
may be non-zero may be non-zero
Plane stress Plane strain
yx zx E E E
y z xy E E E
yz xz E E E
If z = 0, that doesnot mean z = 0 [except,in case when = 0 i.e. ideal material orwhen x = –y]Similarly,
If z = 0, that doesnot mean z = 0 [exceptin case when = 0 i.e. ideal material
or when x = –y] Thus note that plain stress and plain strain
components are not same. In term of stress, z = 0 in plane stress
but z may not be zero in plane strain.
107. (c) Let 1 2and be two stress
n = 1 2 1 2 cos22 2
for > 45°; cos 2 will be (–)ve
So n < 1 22
R = 1 22
P ( , )PP
21O2
for > 45°, horizontal co-ordinate (n) will be lessthan the co-ordinate of centre of Mohr circle
1 22
108. (d) V
V
= x y z
E
(1 – 2)
then,
V 0
V for = 0.25
Material is incompressibleModulus of elasticity E = 3K(1 – 2)
109. (a)
110. (a)
=VQIt
Where Q represents 1st moment of the areaabove the line of interest about NA
Nh
b
A max
Q = Ay
IES M
ASTER
(17)
Consistency index =
L L
L P
w w 0w w
112. (b)
100
80
60
40
20
0 0.001 0.01 010 1.0 10.0
Perc
enta
ge fi
ner (
w)
Particle size (mm)
The semi-log plot for the particle size distribution,as shown in figure, has the following advantagesover natural plots.
(1) The soils of equal uniformity exhibit thesame shape, irrespective of the actualparticle size.
(2) As the range of the particle sizes is verylarge, for better representation, a log scaleis required.
113. (d) Permeability on wet side of optimum is lessthan the dry side of optimum.
N ot e :
Project Compactionwater content
Reason
Homogenousearth dam
Sub-gradeof pavement
Dry ofoptimum
Wet ofoptimum
To reduce permeability and prevent cracking in core.To have a stronger soil& to prevent build up ofhigh pore water pressure.
To limit volume change.
Core of anearth dam.
Wet ofoptimum
114. (a) Ultimate settlement does not depend ondrainage condition. Drainage conditionaffects the rate of consolidation and timerequired for consolidation.
115. (c) Sensitivity measures the effect of remouldingof soil on its strength without any changein its moisture content. It is defined as theratio of the unconfined compressive strengthof an undisturbed soil sample to the
t represents the width of material which isequal to width b. Notice that for typical cross-sections, the maximum shear stress occursat the Neutral axis.
Q = = =
21 11Ay bh bhh2 84
I = =31 bh and t b12
rmax = = = =
2
3 rect
1V bhVQ 3V 3V81It 2bh 2Abbh
12
To calculate this for a circular cross-section
N
d
A max
d = 2r
Q = = = 32 4r 21Ay rr
3 32
I = 41 r4
and t = 2r
c max =
=
3
4
2V rVQ 31It r 2r4
= = 2
circle
4V 4V3A3 r
For equal area and shearforce of bothrectangle and circle
c
r
max
max= = rect
circle
2A4V 8 13 A 3V 9
i.e. c rmax max
111. (d) Consistency index =
L n
L P
w ww w
When soil is at liquid limit, wn = wL
IES M
ASTER
(18)
unconfined compressive strength of thespecimen of the same soil after remouldingit at an unaltered moisture content.
From the point of view of foundationengineering, marine clays constitute someof the most troublesome soil conditions.Not only are they extremely soft and verycompressible but their high sensitivitymakes them dif f icult to handle inexcavations. The buildings on such clayeysoils exhibit too much settlement.
Ratio of undisturbed strength to disturbedstrength of marine clay is high
116. (d) Granular soil found on a particular projectsite may be appropriate for use as backfill.Such soil is typically a mixture of granularparticles with silt and clay. The mostcommonly specified sizes range from amaximum of 3 inches to the standard No.200 sieve particles size. Using the unfiedsoil classification system, well graded soilsclassified as GW, SW, GM, SM, GC or SCthat are primary non-plastic are generallyacceptable for backfilling flexible structures.However, ML and CL materials, which areprimarily silt and clay should be avoided.
117. (b)118. (a)
119. (a) The water held by electrochemical forcesexisting on the soil surface is known asadsorbed water or hygroscopic water. Thequantity of adsorbed water depend uponthe colloidal fraction in the soil, the chemicalcomposition of the clay mineral and theenvironment surrounding the particle. Theadsorbed water is important only for clayeysoils. For coarse grained soils, its amountis negligible or zero.
120. (d) If large compactive effort is applied on wetside of optimum, density will not increasesignificantly. Therefore higher compactiveeffort produces highest increase in drydensity an dry side of optimum.
d
OMC2 OMC1
2
1
80% sat line
Zero air void line
a
bB
A
Dry side of optimum
Wet side of optimum
Curve-1 Lower compactive effort,Curve-2 Large compactive effort
d AB d ab