CIVIL ENGINEERING
ESE CONVENTIONAL SOLUTION PAPER–I FROM (1995-2017)
UPSC Engineering Services Examination, GATE, State Engineering Service Examination & Public Sector Examination. (BHEL, NTPC, NHPC, DRDO, SAIL, HAL, BSNL, BPCL, NPCL, etc.)
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ISBN :
First Edition : 2017
Typeset at : IES Master Publication, New Delhi-110016
Preface It is an immense pleasure to present topic wise previous years solved paper of Engineering Services Exam. This booklet has come out after long observation and detailed interaction with the students preparing for Engineering Services Exam and includes detailed explanation to all questions. The approach has been to provide explanation in such a way that just by going through the solutions, students will be able to understand the basic concepts and will apply these concepts in solving other questions that might be asked in future exams. Engineering Services Exam is a gateway to an immensly satisfying and high exposure job in engineering sector. The exposure to challenges and opportunities of leading the diverse field of engineering has been the main reason for students opting for this service as compared to others. To facilitate selection into these services, availability of arithmetic solution to previous year paper is the need of the day. Towards this end this book becomes indispensable.
Mr. Kanchan Kumar Thakur Director–IES Master
CONTENTS 1.
Strength of Material -------------------------------------------------------------------------------- 1 – 197
2.
Structure Analysis ------------------------------------------------------------------------------- 198 – 409
3.
Steel Structure ----------------------------------------------------------------------------------- 410 – 527
4.
RCC and Prestressed Concrete ------------------------------------------------------------ 528 – 659
5.
PERT CPM --------------------------------------------------------------------------------------- 600 – 731
6.
Building Material --------------------------------------------------------------------------------- 732 – 808
C HAPTER
1 Strength of Material Syllabus Elastic constants, stress, plane stress. Mohr’s circle of stress, strains, plane strain, Mohr’s circle of stain, combined stress, Elastic theories of failure; Simple bending, shear; Torsion of circular and rectangular sections and simple members.
IES –1995 1.
The principal stresses at a point in an elastic material are 1.5 (tensile), (tensile) and 0.5 (compressive). The elastic limit in tension is 210 MPa and = 0.3. What would be the value of at failure when computed by the different theories of failure? [15 Marks]
Sol.
2 = +
Given data : 1 1.5 ;
2 ;
1 = + (1.5)
3 = – 0.5 * 3 = (–0.5)
Elastic limit in tension (f y) = 210 MPa. = 0.3
(Macroscopic View of a Point)
Determine : at failure when computed by different theories of failure. (1)
Maximum Principle Stress Theory : As per this theory for no failure maximum principal Stress should be less than yield stress under uniaxial loading. So,
(2)
1 1.5
fy
1.5 fy
fy 1.5
210 140.00 1.5
140 MPa
Maximum principal strain theory : As per this theory, for no failure maximum principal strain should be less than yield strain under uniaxial loading i.e.,
max
y E
2
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IES CONVENTIONAL SOLUTION PAPER-I 1995-2017
Among ( x , y , z ), x will be maximum because 1 is maximum 1.5 0.5 1.5 – 0.3 0.15 1.35 – E E E E E 1.35 210 210 1.35 155.55 MPa. E E
x =
(3)
Maximum shear stress theory : For no failure, maximum shear stress should be less than or equal to maximum shear stress under uniaxial loading. Since we have 3–D case, 1 – 3 1 – 2 2 – 3 , , Maximum shear stress = maximum 2 2 2 fy Maximum shear stress under uniaxial loading : y = 2 From this theory – – – 3 fy Maximum 1 3 , 1 2 , 2 2 2 2 2 1.5 – –0.5 210 1 – 3 105 2 2 2 2 210
(4)
Maximum strain energy theorem : For no failure, maximum strain energy absorbed at a point should be less than or equal to total strain energy per unit volume under uniaxial loading, when material is subjected to stress upto elastic limit. Total strain energy =
12 22 32 – 2 12 23 3 1 2E
Total strain energy per unit volume under uniaxial loading =
According to this theory,
12
2
fy
2
2E
2
2 3 – 2 1 2 2 3 3 fy 2
1.5 2 2 –0.5 2 – 2 0.3 1.5 – 0.5 – 1.5 0.5 2102
2102 114.73 MPa 3.35 Maximum distortion energy theory :- For no failure, maximum shear strain energy in a body should be less than maximum shear strain energy due to uniaxial loading. 1 1 – 2 2 2 – 3 2 3 – 1 2 fy 2 2 1 1.5 – 2 0.5 2 –0.5 –1.5 2 2102 2 2
(5)
105 MPa.
0.252 2.252 42 2 × 2102
2
2 2102 6.5
116.487 MPa
2 2102 6.5
STRENGTH
OF
|
MATERIAL
3
The strain measurements from a rectangular strain rosette were e0 = 600 × 10–6, e45 = 500 × 10—6 and e90 = 200 × 10–6. Find the magnitude and direction of principal strains. If E = 2 × 105 N/mm2 and = 0.3 find the principal stresses.
2.
[10 Marks ]
Sol. y x –6
90 = 200 × 10
–6
45 = 500 × 10 90° 45° x –6
0 = 600 × 10
We know that
and
x = max
min
=
x y 2
x y 2
x y 2
cos 2 2
xy 2
x y xy 2 2
sin 2
... (i)
2
... (ii)
Thus to determine the principal strain, we need normal strain in two mutually perpendicular direction and shear strain (xy) associated with these directions. From (i) 45 = 500 × 10–6 =
xy 0 90 0 90 cos (2 45) sin (2 45) 2 2 2
cos 90 0 sin 90 1
600 200 600 200 10 6 106 cos90 + xy 2 2 2
xy 200 10 6
From (ii)
2
[ xy 0 90 ]
max
min
=
0 90 90 xy 0 2 2 2
2
2
600 200 600 200 200 12 12 106 = 10 10 2 2 2 2 2 6 = [400 (200) (100) ] 10
= [400 223.607] 10 6
max = 623.607 × 10–6 = major principal strain min = 176.393 × 10–6 = minor principal strain
4
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IES CONVENTIONAL SOLUTION PAPER-I 1995-2017
Also, we know that, tan 2P =
xy / 2 x y / 2
200 200 1 600 200 400 2
P = 13.282° or 103.282°
One of these angles will be associated with major principal strain and other with minor principal strain. To determine which of the angle is associated with major principal strain, let us put the value of P in strain transformation eq. xy 0 90 0 90 cos 2P sin 2P x = 2 2 2 200 600 200 600 200 cos 2 (13.282) sin 2 (13.482) 10 6 = 2 2 2
x = 623.607 × 10–6
P = 13.282° is associated with major principal strain
i.e., direction of major principal strain is at 13.282° in anticlockwise direction from 0 strain direction and hence direction of minor principal strain is at 103.282° in anticlockwise direction from 0 strain direction. Calculation of principal stresses:
max (min ) = max E E min max = min E E max – 0.3 min = 623.607 ×10–6 × 2 × 105 N/mm2
max– 0.3 min = 124.72 N/mm2 min – 0.3 max = 176.393
×10–6
... (i) × 2×
105
N/mm2
... (ii)
0.3 min – 0.09 max = 35.279 × 0.3
... (iii)
From (i) + (iii) (10.09)max = 124.7 + 35.279 × 0.3
max 148.685 N/mm2
min 79.885 N/mm2
Alternative approach (Mohr circle approach): If we use Mohr transformation, we will not have to check which of the two angles 13.28° and 103.28° corresponds to major principal strain By analytical approach we have found that xy is (+)ve. This implies that it is associated with (+)ve shear stress as shown below. y x
STRENGTH
OF
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MATERIAL
5
Hence strains are shown as y xy
2
xy x 1
direction of y 6 6 200 10 200 10 , 2
xy y , i.e. 2
xy 2
–3
(623.607 × 10 , 0)
(176.393 × 10–6 , 0)
direction of minor principal strain
–6
(400 × 10 , 0)
2P
direction of major principal strain
xy 200 106 x , i.e, 600 106 , 2 2
direction of x
max = (400 × 10–6) + R min = (400 × 10–6) –R
[R = radius of circle]
200 0 106 R = (600 400)2 = 223.607 × 10–6 2 max = 623.607 × 10–6 2
min = 176.393 × 10–6 100 1 tan2P = 600 400 2 P = 13.28° Major principal strain is at 13.28° in anticlockwise direction from the direction of x and minor principal strain is 13.28 3.
180 = 103.28° in anticlockwise direction from the direction of x. 2
Draw bending moment & shear force diagrams for the beam loaded as shown in fig. 10 kN
3kN/m
1m P 2m
1m
1m
2m
2m
[ 15 Marks ]