ies master publication, · 2017. 11. 24. · alternative approach (mohr circle approach): if we use...
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UPSC Engineering Services Examination, GATE,
State Engineering Service Examination & Public Sector Examination.
(BHEL, NTPC, NHPC, DRDO, SAIL, HAL, BSNL, BPCL, NPCL, etc.)
C I V I L E N G I N E E R I N GESE CONVENTIONAL SOLUTION
PAPER–I
FROM (1995-2017)
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Typeset at : IES Master Publication, New Delhi-110016
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ISBN :
IES MASTER PublicationF-126, (Lower Basement), Katwaria Sarai, New Delhi-110016Phone : 011-26522064, Mobile : 8010009955, 9711853908E-mail : [email protected], [email protected] : iesmasterpublications.com, iesmaster.org
First Edition : 2017
Preface
It is an immense pleasure to present topic wise previous years solved paper ofEngineering Services Exam. This booklet has come out after long observation anddetailed interaction with the students preparing for Engineering Services Exam andincludes detailed explanation to all questions. The approach has been to provideexplanation in such a way that just by going through the solutions, students will beable to understand the basic concepts and will apply these concepts in solving otherquestions that might be asked in future exams.
Engineering Services Exam is a gateway to an immensly satisfying and high exposurejob in engineering sector. The exposure to challenges and opportunities of leading thediverse field of engineering has been the main reason for students opting for thisservice as compared to others. To facilitate selection into these services, availabilityof arithmetic solution to previous year paper is the need of the day. Towards this endthis book becomes indispensable.
Mr. Kanchan Kumar ThakurDirector–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
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 wouldbe the value of at failure when computed by the different theories of failure?
[15 Marks]
Sol. Given data :
1 1.5 ; 2 ; 3 = – 0.5 *
2 = +
1 = + (1.5 )
3 = (–0.5 )
(Macroscopic View of a Point)
Elastic limit in tension (fy) = 210 MPa.
= 0.3
Determine : at failure when computed by different theories of failure.
(1) Maximum Principle Stress Theory : As per this theory for no failure maximum principal Stressshould be less than yield stress under uniaxial loading.
So, 1 1.5 yf
1.5 yf yf1.5
210 140.001.5
140 MPa(2) 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
CHAPTER
Strength of Material
1Syllabus
Elastic constants, stress, plane stress. Mohr’s circle of stress, strains, plane strain, Mohr’scircle of stain, combined stress, Elastic theories of failure; Simple bending, shear; Torsion ofcircular and rectangular sections and simple members.
2 | IES CONVENTIONAL SOLUTION PAPER-I 1995-2017
Among x y z x( , , ), will be maximum because 1 is maximum
x = 1.5 0.5 1.5 – 0.3 0.15 1.35–
E E E E E
1.35
E
210E
2101.35 155.55 MPa.
(3) Maximum shear stress theory : For no failure, maximum shear stress should be less thanor equal to maximum shear stress under uniaxial loading.
Since we have 3–D case,
Maximum shear stress = 1 3 2 31 2– ––maximum , ,
2 2 2
Maximum shear stress under uniaxial loading : y = yf2
From this theory
1 3 2 31 2– ––Maximum , ,
2 2 2
yf2
1 3–2
2102
1.5 – –0.5
2
105
2 210 105 MPa.
(4) Maximum strain energy theorem : For no failure, maximum strain energy absorbed at a pointshould be less than or equal to total strain energy per unit volume under uniaxial loading, whenmaterial is subjected to stress upto elastic limit.
Total strain energy = 2 2 21 2 3 1 2 2 3 3 1– 2
2E
Total strain energy per unit volume under uniaxial loading = 2
yf2E
According to this theory, 2 2 2 21 2 3 1 2 2 3 3 y– 2 f
2 2 21.5 –0.5 – 2 0.3 1.5 – 0.5 –1.5 0.5 2102
2 2210
3.35 114.73 MPa
(5) Maximum distortion energy theory :- For no failure, maximum shear strain energy in a bodyshould be less than maximum shear strain energy due to uniaxial loading.
2 2 21 2 2 3 3 1
1 – – –2 2
yf
2 2 21 1.5 – 0.5 –0.5 –1.52 2102
2 2 20.25 2.25 4 2 × 2102
2 22 210
6.5
22 210
6.5
116.487 MPa
STRENGTH OF MATERIAL | 3
2. 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.
[10 Marks ]
Sol.
90°45°
90 = 200 × 10–6
45 = 500 × 10–6
0 = 600 × 10–6
y
x
x
We know that x = x y x y xycos 2 sin 22 2 2
... (i)
and max min =2 2
x y x y xy
2 2 2
... (ii)
Thus to determine the principal strain, we need normal strain in two mutually perpendicular directionand shear strain (xy) associated with these directions.
From (i)
45 = xy0 90 0 90 cos (2 45 ) sin (2 45 )2 2 2
xy 0 90[ ]
500 × 10–6 = 6 6600 200 600 20010 10 cos902 2
+ xy
2 cos 90 0
sin 90 1
6xy 200 10
From (ii)
max min =22
xy0 90 0 902 2 2
=2
6 12 12600 200 600 200 20010 10 102 2 2
=2 2 6[400 (200) (100) ] 10
= 6[400 223.607] 10
max = 623.607 × 10–6 = major principal strain
min = 176.393 × 10–6 = minor principal strain
4 | IES CONVENTIONAL SOLUTION PAPER-I 1995-2017
Also, we know that,
tan 2P = xy
x y
/ 2 200 200 1/ 2 600 200 400 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 straintransformation eq.
x =xy0 90 0 90
P Pcos 2 sin 22 2 2
= 6600 200 600 200 200cos 2 (13.282 ) sin 2 (13.482) 102 2 2
x = 623.607 × 10–6
P = 13.282° is associated with major principal straini.e., direction of major principal strain is at 13.282° in anticlockwise direction from 0 strain direction andhence direction of minor principal strain is at 103.282° in anticlockwise direction from 0 strain direction.
Calculation of principal stresses:
max min( )E E
= max
maxminE E
= min
max – 0.3 min = 623.607 ×10–6 × 2 × 105 N/mm2
max– 0.3 min = 124.72 N/mm2 ... (i)min – 0.3 max = 176.393 ×10–6 × 2× 105 N/mm2 ... (ii)
0.3 min – 0.09 max = 35.279 × 0.3 ... (iii)From (i) + (iii)
max(10.09) = 124.7 + 35.279 × 0.3
2max 148.685 N/mm
2min 79.885 N/mm
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 strainBy analytical approach we have found that xy is (+)ve. This implies that it is associated with (+)ve shearstress as shown below.
y
x
STRENGTH OF MATERIAL | 5
Hence strains are shown as
1
2
x
xy
xy
y
6xy 6
y200 10, i.e. 200 10 ,
2 2
xy
2
(176.393 × 10 , 0)–6
(400 × 10 , 0)–6 2P
(623.607 × 10 , 0)–3 direction of major principal
straindirection of
minor principal strain
xyx ,
2
66 200 10i.e, 600 10 ,
2
direction of x
direction of y
max = (400 × 10–6) + Rmin = (400 × 10–6) –R [R = radius of circle]
R =2
62 200 0 10(600 400)2
= 223.607 × 10–6
max = 623.607 × 10–6
min = 176.393 × 10–6
tan2P =100 1
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 18013.28
2 = 103.28° in anticlockwise direction from the direction of x.
3. Draw bending moment & shear force diagrams for the beam loaded as shown in fig.
10 kN 3kN/m1m
P
2 m 2 m2 m 1 m 1 m
[ 15 Marks ]
