MPSC Civil Engg. Main Exam

MPSC Civil Engg. Main Exam – Previous Year Question Papers 2020 Paper - I Page | 1

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MPSC CIVIL ENGG. [MES] MAIN EXAM Previous Year Questions, Answer key and Detail Explanation.

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Civil Engineering Paper I

Civil Engineering Paper II

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MPSC CIVIL ENGG. [MES] MAIN EXAM PAPER - 1

INDEX

1. BULDING CONSTRUCTION AND MATERIALS

1.1 Question Paper………………………………………………………………. 11

1.2 Answer Key…………………………………………………………………… 18

1.3 Detailed Solutions…………………………………………………………….. 19

2. STRENGTH OF MATERIALS

2.1 Question Paper……………………………………………………………….. 30

2.2 Answer Key…………………………………………………………………… 39


3. THEORY OF STRUCTURES AND STRUCTURAL ANALYSIS

3.1 Question Paper……………………………………………………………….. 58

3.2 Answer Key……………………………………………………………………. 72


4. STEEL STRUCTURES

4.1 Question Paper……………………………………………………………….. 93

4.2 Answer Key……………………………………………………………………. 100


5. DESIGN OF REINFORCED CONCRETE STRUCTURES

5.1 Question Paper……………………………………………………………….. 112

5.2 Answer Key……………………………………………………………………. 119


6. PRESTRESSED CONCRETE STRUCTURES

6.1 Question Paper……………………………………………………………….. 130

6.2 Answer Key……………………………………………………………………. 137


7. CONSTRUCTION PLANNING & MANAGEMENT

7.1 Question Paper……………………………………………………………….. 148

7.2 Answer Key……………………………………………………………………. 153


8. NUMERICAL METHOD

8.1 Question Paper……………………………………………………………….. 161

8.2 Answer Key……………………………………………………………………. 165





MPSC CIVIL ENGG. [MES] MAIN EXAM PAPER - 2 INDEX

1. SURVEYING

1.1 Question Paper 175

1.2 Answer Key 182

1.3 Detailed Solutions 183

2. ESTIMATING, COSTING AND VALUATION


2.2 Answer Key with Solutions 196


3. GEO-TECHNICAL ENGINEERING




4. FLUID MECHANICS AND MACHINES




5. ENGINEERING HYDROLOGY




6. IRRIGATION ENGINEERING




7. HIGHWAY ENGINEERING




8. BRIDGE ENGINEERING




9. TUNNELING




10. ENVIRONMENTAL ENGINEERING







1. BUILDING CONSTRUCTION AND MATERIALS - QUESTIONS

MES Civil Gr. A Preliminary Exam 2011

1. What is the minimum 7 days strength of ____ low

heat cement?

A. 24 MPa B. 16 Mpa C. 30 Mpa D. 28 Mpa

2. M-20 grade concrete means, concrete having

characteristic compressive strength at 28 days

A. Less than 20 N/mm2

B. Less than 20 Kg/mm2

C. More than 20 N/mm2

D. More than 200 N/mm2

3. Match List -I (defect in painting) with List-II (cause

of defect) and select the correct answer using the

codes given below the lists:

List -I (defect) List -II(cause)

a. Blistering 1.Too smooth surface to be

painted

b. Crawling 2. Insufficient opacity of final coat

c. Grinning 3. Application of too thick paint

d. Running 4. Formation of bubbles under the

film of paint

a b c d

A 4 3 2 1

B 3 4 2 1

C 4 3 1 2

D 4 2 3 1

4. Which liquid thins the consistency of the paint?

A. Primer B. Solvent

C. Filler D. Drier

5. King post roof truss is used for spans.

A. 3 to 4 m B. 5 to 8 m

C. 10 to 15m D. 16 to 20 m

6. The estimated cost of a building is Rs.20, 00,000

(exclusive of water supply and sanitary works, electri-

fication works, contingencies and work charged es-

tablishment). If 8% for water supply and sanitary

work, 8% for electrification work, 3% for contingen-

cies and 2% for work charged establishments are

considered the total estimated cost of the building is

A. Rs.24, 36,392 B. Rs. 24, 20,000

C. Rs.24, 37,392 D. Rs. 22, 05,000

7. In which year was National Building Code pub-

lished?

A. 1950 B. 1978 C. 1952 D. 1970

8. Original cost of property minus depreciation is

A. Book value B. Salvage value

C. Market value D. Obsolescence value


9. The ratio of tensile strength to compressive

strength of concrete is about:

A. 1/5 B. 1/10 C. 1/2 D. 1/20

10. At what height non-combustible material shall be

used in construction of a building?

A. 20 m above B. 15 m above

C. 30 m above D. 45 m above

11. The minimum aggregate area of openings, ex-

cluding doors in residential buildings in wet hot cli-

mate shall not be less than:

A. 1/10𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟 𝑎𝑟𝑒𝑎

B. 1/6𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟 𝑎𝑟𝑒𝑎

C. 1/8𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟 𝑎𝑟𝑒𝑎

D. 1/12𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑜𝑟 𝑎𝑟𝑒𝑎

12. What is the main constituent in fire proof paints?

A. Aluminum powder B. Red lead

C. Copper powder D. Asbestos fibers

MPSC CIVIL MAINS EXAM PREVIOUS YEAR QUESTION PAPER

BUILDING CONSTRUCTION AND MATERIALSPREVIOUS YEAR QUESTIONS1.1

PAPER - I




13. A brick moulded with a rounded angle is called

as

A. Bull nose B. Horse nose

C. Cow nose D. Donkey nose

14. The guidelines for Concrete Mix Design are cov-

ered in:

A. IS: 10262 – 1982 B. IS: 14272 - 1985

C. IS: 10272 - 1983 D. IS 14273 – 1985

15. A rough estimate of the quantity of dynamite re-

quired in grams for blasting rocks is given by:

𝐴. 𝐿2/0.008 𝐵. 𝐿2/3/340 C. 𝐿2/500 D. 𝐿/0.008

16. Lean to Roof is suitable for the span

A. Upto 1.5 m B. Upto 2.5 m

C. Upto 3.5 m D. Upto 4.5 m

MES Civil Gr. B Preliminary Exam 2012

17. In Residential building, kitchen should have ____

aspect.

A. Eastern B. Southern

C. South – eastern D. Northern

18. Workability of concrete can be measured by

A. Slump test B. Compaction factor test

C. Kelly ball test D. All the above

19. For a rectangular room, better proportion is to

adopt length as____ times of breadth

A. 1 to 1.2 B. 1.2 to 1.7

C. 1.2 to 1.5 D. 1.5 to 1.7

20. The laboratory slump test result of the fresh con-

crete is between 25 - 30 mm. The degree of worka-

bility of such concrete is

A. Very low B. Low C. Medium D. High

21. Black cotton soil is a product of decomposition of

A. Granite B. Marble C. Basalt D. Sandstone

22. The strength achieved by a brick depends on

A. Composition of brick earth

B. Nature of moulding adopted

C. Burning and cooling process

D. All the above

23. Capacity of concrete to bear imposed stresses

safely is called as

A. Compressive strength B. Shear strength

C. Durability D. Resistance

24. State whether the following statements are true or

false:

a) Consistency test is used to determine the percent-

age of water required for preparing cement paste.

b) Vicat Apparatus is used for determining the con-

sistency of cement.

A. a true, b true B. a false, b false

C. a true, b false D. a false, b true

25. Durability of construction material is

A. Resistance to crushing B. Resistance to weathering

C. Shear strength D. Compressive strength

26. Seasoning of timber means

A. Removing the moisture content

B. Reducing weight of timber

C. Both (A) and (B)

D. None of the above

27. Artificial method of seasoning timber is

A. Boiling B. Chemical seasoning

C. Water seasoning D. All of the above

28. Laterite is used in

A. Carving and ornamental works

B. Fire resistance works

C. Electrical switchboards

D. Heavy engineering works

29. Light weight concrete is also known as

A. Low concrete B. Lean concrete

C. Transparent concrete D. Cellular concrete

MES Civil Gr. B Mains Exam 2013

30. For what span is the Queen Post roof truss suita-

ble?

A. 5 to 9 m B. 9 to 14 m

C. 14 to 18 m D. None of the above

31. What is a Header as seen in elevation of wall?

A. Longer face of brick

B. Horizontal distance between vertical joints of suc-

cessive brick courses

C. Lower surface of brick when laid flat

D. Shorter face of brick

32. What is the temperature range in the low temper-

ature tempering process?

A. 150°C to 200°C B. 200°C to 250°C

C. 100°C to 150°C D. 250°C to 300°C

33. A distemper is composed of a base with:




1. BUILDING CONSTRUCTION AND MATERIALS ANSWER KEY

QUE ANS QUE ANS QUE ANS QUE ANS QUE ANS

01 B 26 A 51 C 76 B 101 B

02 C 27 D 52 B 77 C 102 D

03 A 28 A 53 C 78 D 103 A

04 B 29 D 54 C 79 C 104 C

05 B 30 B 55 C 80 B 105 C

06 B 31 D 56 B 81 D 106 C

07 D 32 A 57 B 82 A 107 B

08 A 33 A 58 A 83 A 108 B

09 B 34 A 59 C 84 D 109 B

10 B 35 A 60 C 85 A

11 B 36 B 61 D 86 C

12 D 37 C 62 D 87 B

13 A 38 C 63 C 88 C

14 A 39 D 64 C 89 C

15 A 40 D 65 A 90 D

16 B 41 A 66 A 91 D

17 A 42 B 67 C 92 A

18 D 43 B 68 A 93 C

19 C 44 D 69 D 94 C

20 B 45 C 70 C 95 B

21 C 46 B 71 B 96 C

22 D 47 B 72 B 97 B

23 A 48 D 73 D 98 A

24 A 49 C 74 C 99 B 25 B 50 C 75 D 100 C


BUILDING CONSTRUCTION AND MATERIALSANSWER KEY1.2

PAPER - I





1. BUILDING CONSTRUCTION AND MATERIALS DETAILED SOLUTIONS

1. ANSWER: B

As per IS 4031 − 1968 compressive strength shall be as follows

Days/Duration Ordinary Ce-

ment 𝑁/𝑚𝑚2

Rapid Hardening

Cement 𝑁/𝑚𝑚2

Low Heat

Cement

A)1𝑑𝑎𝑦/24 ± 30 𝑚𝑖𝑛 𝑛𝑜𝑡 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 - 16 -

B)3𝑑𝑎𝑦/72 ± 1𝐻 𝑛𝑜𝑡 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 16 27.5 10

C)7𝑑𝑎𝑦/168 ± 2𝐻 𝑛𝑜𝑡 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 22 - 16

D)28𝑑𝑎𝑦/672 ± 4𝐻 𝑛𝑜𝑡 𝑙𝑒𝑠𝑠 𝑡ℎ𝑎𝑛 - - 35

2. ANSWER: C

As per IS code 456 − 2000

The seven grade of concrete designated as

M10,M15,M20,M25, and M30,M35,M40.In the desig-

nation at concrete mix later M factors as the mix and

number to be specified characteristic compressive

strength (Fck) at 15cm cube at 28days expressed in

N/mm2

The characteristic strength is define as the strength of

material below which not more than 5% of the test

results are expected to fail.

M20 Grade concrete means concrete having charac-

teristic strength at 28days more than20N/mm2.

3. ANSWER: A

Defects in Painting:-

i) Blistering:-It is the defect caused due to the for-

mation of bubbles under film of water paint. The

bubbles are formed by water vapors trapped behind

the painted surface.

ii) Bloom:-In this defect dull patches are formed on

finished polished surface. This may be either due to

defect in paints or due to bad ventilation.

iii) Crawling or sagging:-This defect occurs due to the

application not too thick paint.

iv) Fading:-This is the gradual loss of color of paints

due to effect of sunlight on pigments of paints.

v) Flaking:-Flaking is the dislocation or loosening of

some portion of the painted surface resulting from a

poor adhesion.

vi) Flashing:-It is the formation of glossy patches on

the painted surface resulting from bad workmanship

cheap paint or weather action.

vii) Ginning:-This defect is caused when the surface fi-

nal coat does not have sufficient capacity so that

background is clearly seen.

viii) Running:-This defect occur when the surface to be

painted is too smooth due to this the paint runs back

& leaves small area of the surface uncovered.

ix) Saponification:-This is the formation of soap

patches on the painted surface due to chemical ac-

tion of alkalis.

4. ANSWER: B

i) Primer:-Primer is a paint product that is designed to

adhere to surface and to form a binding layer that is

better prepared to receive the paint. It is applied be-

fore the paint.

ii) Solvent:-Are added to paint to make it thin so that

it can be easily applied on surface.

iii) Drier:-Drier are used to accelerate the process of

drying & hardening by extracting oxygen from the at-

mosphere and transferring it to vehicle.

iv) Filler:-Filler are the special type of pigment that

serve to thicken the film supports structure and in-

crease the volume of paint fillers are usually cheap &

1.3MPSC CIVIL MAINS EXAM PREVIOUS YEAR QUESTION PAPER

BUILDING CONSTRUCTION AND MATERIALSDETAIL EXPLANATION




inert materials such as diatomaceous earth talc lime

etc.

5. ANSWER: B

i) King post(5 − 9 𝑚)

ii) Queen post(9 𝑡𝑜 14 𝑚)

iii) Sow tooth truss(5 𝑡𝑜 8𝑚)

iv) Compound fink(10 − 20𝑚)or French truss

v) Single fan(12𝑚)

vi) North light roof truss(8 − 10𝑚)

6. ANSWER: B

Total cost of building=

20,00,000 (𝑒𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑎𝑙𝑙 𝑜𝑡ℎ𝑒𝑟)

Cost required for 𝑊𝑆, 𝑆𝑊 = 8%

Electrification work= 8%, 3%

Contingencies & 2% for work charge establishment.

Total cost= 𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 + 𝑜𝑡ℎ𝑒𝑟

= 20,00,000 +21

100× 20,00,000

= 20,00,000 + 4,20,000

= 24,20,000

7. ANSWER: D

National building code published in1970 current ver-

tion(2005)

8. ANSWER: A

Book value→the book value of a property at a partic-

ular year is the original cost minus amount of depre-

ciation upto previous year.

Salvage value→it is value at end of useful life without

being dismantled

Obsolescence value→is the value which is generally

when other option are there for this particular prod-

uct is known as obsolescence value.

Market value→market value of an assist is determined

by fluctuation in supply & demand.

9. ANSWER: B

Tensile strength is almost 1

10𝑡ℎ of compressive

strength.

10. ANSWER: B

The height of non-combustible material shall be used

in construction of building= 15𝑚 above.

11. ANSWER: B

The minimum aggregate area of opening excluding

doors in residential building in wet hot climate shall

not be less than 1 6⁄𝑡ℎ of the floor area and for dry

climate < 1 10⁄𝑡ℎ

of floor area.

12. ANSWER: D

Asbestos fibers-

A. Noncombustible.

B. Better fire resistance

C. Low coefficient of expansion

Aluminum powder→Use as air-entraining admixture.

Red lead→ Inorganic compound with the formula

𝑃𝑏304 uses in rust proof primer paints

Copper powder→ pure copper powder is used in the

electrical & the electronics industries.

13. ANSWER: A

It is typical/special molded brick with one edge found

called as single bull nose or with two edges rounded

(double bull nose). They are used in coping or in such

position where rounded corner are preferred

Double Bull Nose Single Bull Nose Cow Nose

14. ANSWER: A

IS 10262 − 1982 → 𝑀𝑖𝑥 𝑑𝑒𝑠𝑖𝑔𝑛

IS 14272 − 1985 → 𝐴𝑢𝑡𝑜𝑚𝑜𝑡𝑖𝑣𝑒 𝑣𝑒ℎ𝑖𝑐𝑙𝑒

IS 10272 − 1983 → 𝑁𝐴

IS 14273 − 1985 → 𝑁𝐴

15. ANSWER: A

Rough estimate of the quantity of dynamite required

in grames for blasting rocks is given by𝐿2

0.008⁄

16. ANSWER: B

Lean to roof is suitable for span upto 2.5𝑚.

17. ANSWER: A

Location of room

i) Kitchen -East or SE

II) Puja room -NE (North East)

iii) Living -SE/South

iv) Window -Northern side of a room

v) Bedroom ⟶ West

18. ANSWER: D

Measurement of workability:-




i) Slump Test-is most common method in the field

suitable for concrete of high and medium worka-

bility

ii) Compaction Factor Test-suitable for concrete

having medium to low workability

iii) Flow Table Test- Concrete having very low

workability

iv) Vee-Bee Consistometer- Concrete having very

low workability cannot measure concrete having

workability more than 50𝑚𝑚

19. ANSWER: C

Principle of planning (Roominess)

Refers to the effect product by deriving the maximum

benefit from the minimum dimension of room.

length ∶ width Shall be 1.2 ∶ 1 𝑡𝑜 1.5 ∶ 1

Height of room 3𝑀

20. ANSWER: B

Slump Test:-

Slump Degree of Workability

i)25 − 50 Low

ii)50 − 100 Medium

iii)100 − 150 High

21. ANSWER: C

i) Granite: It is an igneous rock is mainly composed

of quartz, feldspar and mica specific gravity= 2.64

ii) Marble: It is a metamorphic rock of calcareous ver-

ity it specific gravity is 2.7and is available in many

colors

iii) Basalt: Parent rock of black cotton soil

iv) Sandstone: It is a sedimentary rock of siliceous va-

riety it is mainly composed of quartz lime and silica

22. ANSWER: D

Strength achieved by a brick depends on

1. Composition of brick earth.

2. Nature of moulding adopted.

3. Burning and cooling process.

23. ANSWER: A

1. Strength-The hardened concrete should have com-

pressive strength so that it can resist imposed load

2. Durability-Concrete must be durable to resist effect

of rain frost action etc.

3. Resistance- Resistance may be to weathering then it

called durability resistance to imposed load we

may called compressive strength

4. Shear strength - ability of concrete/body to resist

tangential force without failure called shear

strength

24. ANSWER: A

Consistency Test:-

This test is performed to find the moisture content

which is required to produce the cement paste to

standard consistency.

Define as the cement paste that permit the vicat’s

plunger to dia 10𝑚𝑚 height 50𝑚𝑚 to penetrate into

mould upto the depth of 33 𝑡𝑜 35𝑚𝑚 from top

Instrument used−Vicat Apparatus.

25. ANSWER: B

Durability−Resistance to Weathering.

Resistance to crushing and compressive strength are

same.

Compressive strength− Resistance or oppose to im-

posed load without failure

Shear strength −maximum shear load body can with-

stand before failure or resistance to sliding offered by

the material beam.

26. ANSWER: A

Seasoning of timber is done in order to

i) To allow timber to burn readily if used as a fuel

ii) To reduce its weight so as to reduce its trans-

portation cost

iii) To make it more workable

iv) To reduce its tendency against cracking, shrink-

age & wrapping

Seasoning of timber is process of drying of the timber

Seasoning of timber is done either naturally or artifi-

cially

27. ANSWER: D

The methods of artificial seasoning

i) Boiling: it is one of the quickest method available to

carry out seasoning

ii) Chemical Seasoning: This method is also referred

as salt seasoning in which timber section is im-

mersed in the solution of soluble salts that in-

crease rate of evaporation

iii) Electrical Seasoning: it is most rapid method for

the seasoning of the timber section in which it is

subjected to high frequency alternating current It is

rarely used as it is uneconomical.

iv) Kiln Seasoning: In This method of seasoning the

timber is carried out in an air tight chamber or

oven.

v) Water Seasoning: In this method timber s/c is cut

into pieces of suitable sizes & is immersed in the

stream of following water in such way that larger

portion faces upstream & smaller portion faces




2. STRENGTH OF MATERIALS QUESTIONS


1. If bulk modulus and modulus of rigidity for a ma-

terial are K and G respectively, then what will be

Poisson's ratio

A. 3𝐾 +4𝐺

6K − 4G B.

3𝐾 − 4𝐺

6K+ 4G C.

3𝐾 − 2𝐺

6K+ 2G D.

3𝐾 + 2𝐺

6K − 2G

2. A bar of (𝐸 = 2.1 × 106 𝑘𝑔/𝑐𝑚2) steel 70 cm long

varies in its cross section with 2.5 cm dia for the first

20 cm, 2 cm dia for next 30 cm and 1.5 cm dia for

the rest of the length. Find the elongation if the bar is

subjected to a tensile load of 15 tonnes.

A. 0.118 cms B. 0.25 cms

C. 0.178 cms D. 0.354 cms

3. A solid circular shaft of diameter 100 mm is sub-

jected to a torque of 25 kNm. The angle of twist over

a length of 3m is observed to be 0.09 rad. The mod-

ulus of rigidity of

A. 7.64 × 10−5 𝑁/𝑚𝑚2 B. 8.49 × 104 𝑁/𝑚𝑚2

C. 2.1 × 105 𝑁/𝑚𝑚2 D. 6.74 × 104𝑁/𝑚𝑚2

4. Find the Euler's crippling load for a hollow cylinder

column (E-205 GPa) of 38 mm external diameter and

2.5 mm thick, with a length of 2.3 m and fixed at

both of its ends.

A. 17.16Kn B. 14.29kN C. 16.88kN D. 21.49 kN

5. A bar of 400 mm length and of uniform cross-sec-

tion 20 mm diameter is subjected to tensile load of

50 kN, assuming E = 200 GPa the elongation of bar

is

A. 2.489 mm B. 3.203 mm

C. 0.318 mm D. 0.187 mm

6. Reaction at a prop in a propped cantilever when it

is subjected to UDL of 'w/m' over an entire span 'L' is

equal to

A. 3𝑤𝐿

8 B.

5𝑤𝐿

8 C.

𝑤𝐿2

8 D.

𝑤𝐿4

8𝐸𝐼

7. The slope of cantilever beam having span `L', at

the free end due to concentrated load `P' applied at

free end is

A. 𝑃𝐿2

𝐸𝐼 B.

𝑃𝐿2

2𝐸𝐼 C.

𝑃𝐿3

3𝐸𝐼 D.

𝑃𝐿3

2𝐸𝐼

8. If the Young's Modulus and Poisson's ratio of a

material is 2 × 106𝑘𝑔/𝑐𝑚2 and 0.25 respectively.

Find the Bulk Modulus.

A. 6

5× 106 𝑘𝑔/𝑐𝑚2 B.

3

4× 106 𝑘𝑔/𝑐𝑚2

C. 5

6× 106 𝑘𝑔/𝑐𝑚2 D.

4

3× 106 𝑘𝑔/𝑐𝑚2

9. Shear stress at the centre of shaft having 200 mm

radius and subjected to twisting moment of 300

N.m.is

A. 0.19 𝑁/𝑚𝑚2 B. Zero

C. 0.38 𝑁/𝑚𝑚2 D. 0.095 𝑁/𝑚𝑚2


10. Ratio of the maximum bending stress in the

flange to that in the web of an 𝐼 section at any dis-

tance along the length of a beam is always:

A. Less than one B. Equal to one

C. More than one D. No relationship exists

11. Where does maximum shear stress occur in a

rectangular shaft subjected to torsion?

A. Centre B. Corners

C. Middle of smaller side D. Middle of longer side

12. What is the greatest eccentricity which a load 'W'

can have without producing tension on the cross sec-

tion of a short column of external diameter 'D' and in-

ternal diameter 'd' ?

A. 𝐷+𝑑

8𝐷 B.

𝜋(𝐷4+𝑑4)

32𝐷3 C.

𝐷2+𝑑2

8𝐷 D.

𝐷2−𝑑2

8𝐷

13. What is the strain energy stored in a rod of length

‘L’ and axial rigidity AE due to an axial force `P'?


STRENGTH OF MATERIALSQUESTIONS




A. 𝑃2𝐿

𝐴𝐸 B.

𝑃2𝐿

2𝐴𝐸 C.

𝑃2𝐿

3𝐴𝐸 D.

𝑃2𝐿

4𝐴𝐸

14. Which allowable stress governs the permissible

bending capacity of a structural section?

A. Bending compressive stress

B. Tensile stress

C. Bending tensile stress

D. Bending compressive or bending tensile stress

15. If a circular shaft is subjected to a torque 'T' and

bending moment 'M', the ratio of: maximum bending

stress to maximum shear stress is

A. 2M/T B. M/2T C. M/T D. 2T/M

16. Where is the bending stress on a beam section

zero?

A. Depends on the shape of the beam

B. Top fibre

C. Bottom fibre

D. Centroid of the section

17. What will be the circumferential stress developed

in case of thin cylindrical pipe with diameter `d' and

thickness t, subjected to internal pressure `p'?

A. 𝑝𝑑

4𝑡 B.

𝑝𝑑

2𝑡 C.

𝑝

𝑡 D.

𝑝

2𝑡

MES Civil Gr. B Preliminary Exam 2012

18. The process of tempering is applied to steel in

hardening process for improving

A. Ductility B. Strength

C. Roughness D. All of the above

19. A cantilever beam 'AC' of uniform cross-section

carries a uniformly distributed load over the portion

'AB' of length / as shown. Slope at free end 'C' will be

l l/4

w

AB C

A. 𝑤𝑙3

6𝐸𝐼 B.

5𝑤𝑙2

96𝐸𝐼 C.

5𝑤𝑙3

48𝐸𝐼 D.

𝑤𝑙2

2𝐸𝐼

20. The at any section in a given beam is

equal to conjugate beam.

A. Slope, Shear force

B. Deflection, shear force

C. Slope, bending moment

D. Slope, Deflection

21. A cast iron beam is a T-section as shown. It is

supported and carrying a uniformly distributed load.

Which of the following is the correct bending stress

distribution diagram if the element is stressed per-

fectly within plastic limit?

A. B. C. D.

22. Which part of the beam is subjected to pure

bending in the following figure?

20 kN 30kN

2m 1m

A B C D

2m

A. AB

B. BC

C. CD

D. No part of beam is subjected to pure bending

23. Calculate the maximum stress acting on the

cross-section of following element:

Take 𝑃1 = 45 𝑘𝑁, 𝑃2 = 445 𝑘𝑁 𝑎𝑛𝑑 𝑃4 = 130 𝑘𝑁.

1m 1.2m

P1

1m

P2

P3 P4

200mm

100m

m

C/S

A. 20 𝑁/𝑚𝑚2 B. 22.5 𝑁/𝑚𝑚2

C. 28.75 𝑁/𝑚𝑚2 D. 6.5 𝑁/𝑚𝑚2

24. A rectangular timber beam (b x d) is cut out of a

cylindrical log of diameter ‘D'. The width (b) of the

strongest timber beam will be:

A. √3 𝐷 B. 𝐷

√3 C. √2 𝐷 D.

𝐷

√2

25. If AC is principal plane, then magnitude of princi-

pal tensile stresses will be

AB

C

40mpa

τ=15mpa

A. 15 MPa B. 5 MPa C. 45 MPa D. Zero




2. STRENGTH OF MATERIAL ANSWER KEY

QUE ANS QUE ANS QUE ANS QUE ANS QUE ANS

01 C 26 D 51 A 76 A 101 D

02 C 27 B 52 C 77 C 102 C

03 B 28 A 53 A 78 B 103 D

04 A 29 D 54 B 79 C 104 D

05 C 30 D 55 D 80 B 105 B

06 A 31 A 56 D 81 B 106 B

07 B 32 B 57 A 82 A 107 A

08 D 33 D 58 B 83 B 108 B

09 B 34 B 59 C 84 D 109 A

10 C 35 C 60 C 85 D 110 D

11 D 36 C 61 A 86 C 111 A

12 C 37 B 62 D 87 A 112 D

13 B 38 A 63 C 88 B 113 D

14 D 39 C 64 C 89 D

15 A 40 C 65 D 90 A

16 D 41 A 66 A 91 C

17 B 42 B 67 B 92 D

18 D 43 B 68 D 93 B

19 A 44 A 69 A 94 C

20 A 45 D 70 D 95 D

21 C 46 B 71 C 96 C

22 D 47 D 72 C 97 D

23 A 48 D 73 D 98 D

24 B 49 A 74 C 99 C

25 B 50 B 75 D 100 A


ANSWER KEYSTRENGTH OF MATERIALS




2. STRENGTH OF MATERIALS DETAILED SOLUTIONS

1. ANSWER: C

Given ⟹

K = Bulk modulus 𝑁/𝑚𝑚2

𝐺 = 𝑚𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝑟𝑖𝑔𝑖𝑑𝑖𝑡𝑦 𝑁/𝑚𝑚2

𝐸 = 2𝐺(1 + 𝜇) …………….(1)

𝐸 = 3𝑘 (1 − 2𝜇) …………….(2)

𝐸 =9𝑘𝐺

3𝑘+𝐺 ……………..(3)

From equation (1), (2) & (3)

𝜇 =3𝑘 − 2𝐺

6𝑘 + 2𝐺

2. ANSWER: C

Given 𝐸 = 𝑀𝑜𝑑𝑢𝑙𝑢𝑠 𝑜𝑓 𝐸𝑙𝑎𝑠𝑡𝑖𝑠𝑖𝑡𝑦 = 2.1 × 106𝑘𝑔/

𝑐𝑚2

Length = L = 75 cm whole Rod

1st cross section:

𝐿 = 20 𝑐𝑚, 𝑑1 = 2.5 𝑐𝑚

2nd

cross section:

𝐿 = 30 𝑐𝑚, 𝑑2 = 2 𝑐𝑚

3rd cross section:

𝐿 = 20 𝑐𝑚, 𝑑3 = 1.5 𝑐𝑚

Solution: To find elongation = 𝛿𝑙 =?

d =2.5 cm1 d =2 cm2 d =1.5 cm3

15

Tonnes

20 cm30 cm20 cm

A 𝐿1Bars are subjected to some tensile force

𝑃 = 𝑃1 = 𝑃2 = 𝑃3 = 15 × 103𝑘𝑔

Material is same

𝐸 = 2.1 × 106𝑘𝑔/𝑐𝑚2

𝛿𝑙 =𝑃1𝐿1𝐴1𝐸1

+𝑃2𝐿2𝐴2𝐸2

+𝑃3𝐿3𝐴3𝐸3

=𝑃

𝐸[𝐿1𝐴1+𝐿2𝐴2+𝐿3𝐴3]

=15 × 103

2.1 × 106[

20𝜋4× 2.52

+30

𝜋4× 22

+20

𝜋4× 1.52

]

= 0.178 𝑚

3. ANSWER: B

Given: Solid circular shaft

𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 = 𝐷 = 100 𝑚𝑚

𝑇𝑜𝑟𝑞𝑢𝑒 = 𝑇 = 25𝑘𝑁𝑚 = 25 × 106𝑁.𝑚𝑚

Angle of twist = 𝜃 = 0.09 𝑟𝑎𝑑

Length of bar = 𝑙 = 3𝑚 = 3000𝑚𝑚

Find: G = Modulus of rigidity =?

Solution:

We know that 𝐺𝜃

𝐿=𝑇

𝐽=𝜏

𝑅

𝐽 = 𝑃𝑜𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡 𝑖𝑛𝑒𝑟𝑡𝑖𝑎

𝜏 = 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠

𝑅 = 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒

From given data 𝐺𝜃

𝐿=𝑇

𝐽

𝐺 × 0.09

3000=25 × 106

𝜋32× 1004

𝐺 = 84.882 × 103𝑁/𝑚𝑚2

𝐺 = 8.482 × 104𝑁/𝑚𝑚2

4. ANSWER: A

Given:

Modulus of elasticity = 𝐸 = 205 𝐺𝑝𝑎 = 205 ×

109𝑁/𝑚2

𝐷0 = 𝐸𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑖𝑎 = 38 𝑚𝑚

𝐷𝑖 = 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑑𝑖𝑎 = 38 − 2.5

= 35.5 𝑚𝑚

Find: Euler’s crippling Load 𝑃𝐸 =?

Solution:

𝑃𝐸 =𝜋2𝐸𝐼

𝐿𝑒2

Fixed at both end then effective length 𝐿𝑒𝑓𝑓 =𝐿

2


DETAIL EXPLANATION STRENGTH OF MATERIALS

PAPER - I




𝑃𝑒 =𝜋2𝐸𝐼

(𝐿2)2 =

4𝜋2𝐸𝐼

𝐿2

Moment of inertia for Hollow cylinder

𝐼 =𝜋

64[𝐷4 − 𝑑4]

𝐼 =𝜋

64[384 − 35.54] = 24.391 × 10−9𝑚4

𝑃𝐸 =4𝜋2 × 205 × 109 × 24.391 × 10−9

(2.3)2

𝑃𝐸 = 37.315 𝑘𝑁

5. ANSWER: C

Given: Bar length = L= 400 mm 𝐶 𝑠⁄ 𝑑𝑖𝑎 𝐷 = 20 𝑚𝑚

Modulus of elasticity 𝐸 = 200 𝐺𝑝𝑎 = 2 × 105𝑚𝑝𝑎

Find: Elongation of bar =?

As we know,

𝛿𝑙 =𝑃𝐿

𝐴𝐸

𝛿𝑙 =50 × 103 × 400𝜋4× 202 × 2 × 105

𝛿𝑙 = 0.318 𝑚𝑚

6. ANSWER: A

Given:

Propped cantilever beam

𝐿𝑝 One end fix and other roller support

Find Reaction At 𝑉𝐵 =?

A

w/m

B

VB

Net deflection at B = 0

= Compatibility condition

↓ 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑈𝐷𝐿 = 𝑤𝐿4

8𝐸𝐼

↑ 𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜𝑉𝐵 =𝑉𝐵𝑙

3

3𝐸𝐼

Net deflection at B = 0

𝑉𝐵𝑙

3

3𝐸𝐼−𝑤𝑙4

8𝐸𝐼= 0

𝑉𝐵 =3𝑤𝑙

8

7. ANSWER: B

Given, Cantilever beam of end load

L

P

θ

At free end

𝑆𝑙𝑜𝑝𝑒, 𝜃 =𝑃𝐿2

2𝐸𝐼

𝐷𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛, 𝛿 =𝑃𝐿3

3𝐸𝐼

8. ANSWER: D

Given, young’s modulus = 2 × 106𝑘𝑔/𝑐𝑚2

Poisson's ratio = 𝜇 = 0.25

Find bulk modulus = k =?

Solution:

𝐸 = 3𝑘(1 − 2𝜇)

2 × 106 = 3𝑘(1 − 2 × 0.25)

2 × 106 = 3𝑘 × 0.5

𝑘 =4

3× 106𝑘𝑔/𝑐𝑚2

9. ANSWER: B

Given: Shaft radius R = 200 𝑚𝑚

Twisting moment 𝑇 = 300 𝑁.𝑚

Find shear stress (𝜏) at centre=?

Solution:

We know that 𝑇

𝐽=

𝜏

𝑅

At the centre shear stress is zero.

10. ANSWER: C

Given I – section

Find: Ratio of maximum bending stress in flange to

that in web

c/s of I section σc - stress in compression

σt - stress in tension

[Bending stress diagram]

As shown in bending stress diagram. Extreme fibre is

maximum bending stress.

So, Ratio of maximum bending stress in flange to that

in web is more than one.

11. ANSWER: D

Given: Rectangular shaft

Find: maximum shear stress=?

Solution:




b

d

b

d

In rectangular shaft maximum shear stress is devel-

oped on middle surface of longer side.

12. ANSWER: C

Given:

External dia. of Rod = D

Internal dia. = d

Load at middle beam = w

Find: what is the greatest eccentricity of column=?

d D

For No tension, minimum combined stress must be

greater or equal to zero. 𝑃

𝐴−𝑀𝑦

𝑍≥ 0

𝑃

𝐴−

𝑃𝑒 (𝐷2)

𝜋64(𝐷4 − 𝑑4)

≥ 0

𝑒 ≤𝜋

32

(𝐷4 − 𝑑4)

𝐷𝜋4(𝐷2 − 𝑑2)

𝑒 ≤𝐷2 + 𝑑4

8𝐷

13. ANSWER: B

Given:

Rod of length = L

Axial rigidity = AE

Axial Force = P

Solution:

a) Strain energy due to axial load.

𝑆. 𝐸. =1

2𝑃∆ , ∆=

𝑃𝐿

𝐴𝐸

𝑺. 𝑬. =𝟏

𝟐𝑷 ×

𝑷𝑳

𝑨𝑬=𝑷𝟐𝑳

𝟐𝑨𝑬

b) Strain energy stored due to shear force

𝑆. 𝐸. =𝜎𝑏2𝑉

2𝐸

Where,

𝜎𝑏 − 𝑠𝑡𝑟𝑒𝑠𝑠 𝑖𝑛 𝑏𝑒𝑛𝑑𝑖𝑛𝑔

𝑉 − 𝑠ℎ𝑒𝑎𝑟 𝑉𝑜𝑙𝑢𝑚𝑒

c) Strain energy stored due to torsion

𝑆. 𝐸. =1

2𝑇. 𝜃

𝐵𝑢𝑡 𝑇

𝐽=𝐺𝜃

𝐿

𝜃 =𝑇𝐿

𝐺𝐽

𝑆. 𝐸. =𝜏𝑚𝑎𝑥2 𝑉

4𝐺=𝑇2𝐿

2𝐺𝐽

Where,

𝑇 − 𝑡𝑜𝑟𝑞𝑢𝑒

𝜏𝑚𝑎𝑥 − 𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠

𝑉 − 𝑉𝑜𝑙𝑢𝑚𝑒

14. ANSWER: D

Permissible bending capacity of a structural section is

bending compressive or bending tensile stress.

15. ANSWER: A

Given:

Circular shaft

Torque (T)

Bending Moment ‘M’

Find: Ratio of maximum bending stress to maximum

shear stress.

Maximum bending stress,

𝜎𝑚𝑎𝑥 =𝑀

𝐼𝑦𝑚𝑎𝑥

𝜎𝑚𝑎𝑥 =𝑀

𝜋64 × 𝐷4

×𝐷2

𝜎𝑚𝑎𝑥 =32 𝑀

𝜋𝐷3

Maximum shear stress

𝜏𝑚𝑎𝑥 =𝑇

𝐽𝑅 =

𝑇𝜋32× 𝐷4

×𝐷

2=16𝑇

𝜋. 𝐷3

𝜎𝑚𝑎𝑥𝜏𝑚𝑎𝑥

=32 𝑀

𝜋𝐷3×𝜋. 𝐷3

16 𝑇

𝜎𝑚𝑎𝑥𝜏𝑚𝑎𝑥

=2𝑀

𝑇

16. ANSWER: D

Find: Where is bending stress is zero=?

Solution: On section stress 𝜎 =𝑀

𝐼× 𝑦

𝑦 = 𝑑𝑖𝑠𝑡. 𝑓𝑟𝑜𝑚 𝑛𝑒𝑢𝑡𝑟𝑎𝑙 𝑎𝑥𝑖𝑠

𝑓 ∝ 𝑦

N A

σmax

σmax

Bending stress is zero at neutral axis or centroid of

section and it is maximum at top and bottom fibre.





3. THEORY OF STRUCTURES AND STRUCTURAL ANALYSIS QUESTIONS


1. A propped cantilever of span ‘𝐿’ is carrying a point

load ′𝑃′ acting at midspan. The plastic moment of the

section is 𝑀𝑝. The magnitude of collapse load is

A. 6𝑀𝑃/𝐿 B. 8𝑀𝑃/𝐿 C. 2𝑀𝑃/𝐿 D. 4𝑀𝑃/𝐿

2. The curved geometry of masonry arches allows its

load carrying capability only in the form of

A. Compressive forces B. Tensile forces

C. Bending forces D. Torsional forces

3. In moment distribution method the sum of distribu-

tion factors for all the members meeting at a joint is

always

A. Equal to zero B. Equal to one

C. Greater than one D. Smaller than one

4. A continuous beam ABCDE has four simple sup-

ports A, B, C and D. DE is over hang. Hinge `F' is

provided along span BC. The degree of static indeter-

minacy is

A. 4 B. 1 C. 2 D. 3


5. Match List-I (Type of structure) with List-II (Statically

indeterminacy) and select the correct answer using

the codes given below.

List —I List — II

a. Rigid jointed plane frame 1. (𝑚 + 𝑟) — 3𝑗

b. Pin jointed space frame 2. (6 𝑚 + 𝑟) — 6𝑗

c Rigid jointed space frame 3. (6 𝑚 + 𝑟) — 3𝑗

4. (3 𝑚 + 𝑟) − 3𝑗

Where

𝑚 = no. of members

𝑗 = no. of joints

𝑟 = no. of reactions

Codes:

a b c

A 1 2 3

B 4 3 2

C 2 1 3

D 4 1 2

6. What do three hinges in an arch make it?

A. Statically unstable structure

B. Statically determinate structure

C. Geometrically unstable structure

D. Indeterminate structure

7. What is the shape of influence line diagram for the

maximum bending moment in a simply supported

beam under 𝑢𝑑𝑙?

A. Rectangular B. Triangular

C. Parabolic D. Circular

8. What is Kani's method based on?

A. Moment Distribution method

B. Column Analogy method

C. Method of Consistent Deformation

D. Strain Energy method

9. The ratio of the plastic moment capacity to the

yield moment of a section is always:

A. Less than one B. Equal to one

C. More than one D. None of the above

10. If a moment is applied to the free end of a pris-

matic propped cantilever, then the moment

A. M B. M/2 C. M/3 D. M/4

11. A single concentrated load 𝑊 rolling over the

beam of span 𝐿 will cause the maximum bending

moment and shear force on a section 𝑋 at a distance

𝑥 from left support. When the load is on the section,

its maximum bending moment will be

A. 𝑤𝑥𝐿2/(𝐿 − 𝑥) B. 𝑤𝑥(𝐿 − 𝑥)/𝐿


THEORY OF STRUCTURES ANDSTRUCTURAL ANALYSISPREVIOUS YEAR QUESTIONS

3.1PAPER - I




3. THEORY OF STRUCTURES AND STRUCTURAL ANALYSIS ANSWER KEY

QUE ANS QUE ANS QUE ANS QUE ANS QUE ANS QUE ANS

01 A 26 D 51 A 76 C 101 C 126 D

02 A 27 D 52 D 77 A 102 D 127 D

03 B 28 C 53 A 78 C 103 A 128 D

04 B 29 D 54 C 79 B 104 D 129 B

05 D 30 A 55 A 80 C 105 A 130 A

06 B 31 C 56 A 81 B 106 D 131 D

07 B 32 # 57 D 82 C 107 B 132 B

08 A 33 D 58 D 83 B 108 B 133 B

09 C 34 B 59 B 84 D 109 A 134 C

10 B 35 A 60 A 85 B 110 D 135

11 B 36 C 61 B 86 B 111 B

12 B 37 B 62 D 87 A 112 B

13 B 38 B 63 B 88 B 113 A

14 C 39 C 64 B 89 D 114 C

15 C 40 C 65 D 90 # 115 A

16 B 41 B 66 A 91 C 116 C

17 D 42 C 67 C 92 B 117 D

18 C 43 D 68 B 93 B 118 A

19 D 44 D 69 A 94 B 119 B

20 D 45 C 70 C 95 B 120 C

21 C 46 C 71 D 96 C 121 A

22 D 47 B 72 # 97 D 122 D

23 C 48 D 73 C 98 A 123 A

24 A 49 B 74 A 99 A 124 A 25 A 50 C 75 B 100 C 125 C

3.2


ANSWER KEY

THEORY OF STRUCTURES ANDSTRUCTURAL ANALYSIS

PAPER - I




3. THEORY OF STRUCTURES AND STRUCTURAL ANALYSIS DETAILED SOLUTIONS

1. ANSWER: A

P

CA B

L/2 L/2

𝐷𝑆𝐼 = 1

So number of hinges required to form collapse mech-

anism are = 𝐷𝑆𝐼 + 1 = 2

So plastic hinges will form at A and B by principle of

virtual work

P

MP

MP

θ

θθ

θ

θ

l/2

𝐸𝑥𝑡𝑟𝑒𝑛𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 = 𝐼𝑛𝑡𝑒𝑟𝑛𝑎𝑙 𝑤𝑜𝑟𝑘 𝑑𝑜𝑛𝑒

𝑃 ×𝑙

2× 𝜃 = 𝑀𝑃𝜃 +𝑀𝑃(𝜃 + 0)

𝑃 ×𝑙

2𝜃 = 3𝑀𝑃𝜃

𝑃 =6𝑀𝑃

𝑙

2. ANSWER: A

Curved geometry of arches causes the load coming

on material are compressive only.

3. ANSWER: B

At any rigid joint, the sum of distribution factor for

moments in all members is always equal to one

4. ANSWER: B

A B C D E

Number of unknown reactions = 4 (𝑉𝐴, 𝑉𝐵, 𝑉𝐶 , 𝑉𝐷)

Number of equilibrium equation = 2[Σ𝑀 = 0 Σfy =

0]

Internal hinge in span BC will also give one more

equation Σ𝑀 ℎ𝑖𝑛𝑔𝑒

∴ 𝐷𝑆𝐼 = 4 − (2 + 1)

= 1

5. ANSWER: D

Static indeterminacy

Rigid jointed plane frame (3𝑚 + 𝑟) − 3𝑗

Pin jointed space frame (𝑚 + 𝑟) − 3𝑗

Rigid jointed space frame (6𝑚 + 𝑟) − 6𝑗

6. ANSWER: B

B

CA

Number of unknown = 4

Number of equilibrium equation = 2[Σ𝑀 = 0 Σfy =

0 Σfx = 0 ]

So the 3rd hinge (i. e.) gives one more equilibrium

equation 𝑖. 𝑒. ΣMB = 0

∴ 𝐷𝑆𝐼 = 4 − (3 + 1) = 0

∴ Three hinge arch is statically determinate structure.

7. ANSWER: B

l

w/m

AB

C

a b

Maximum BM in simply supported beam subjected to

UDL is at centre i. e. at B

∴To draw BM ILD at B, assume hinge at B & lift it by

an amount 𝑎𝑏

𝑙


DETAIL EXPLANATION

THEORY OF STRUCTURES ANDSTRUCTURAL ANALYSIS3.3

PAPER - I




A B C

ab/l

ILD is triangular

8. ANSWER: A

Kani’s method is an iterative version of slope deflec-

tion and moment distribution method.

9. ANSWER: C

When a material is loaded it first reaches its yield ca-

pacity, then when it is further loaded it reaches to

plastic stage

∴ 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝑚𝑜𝑚𝑒𝑛𝑡 > 𝑦𝑖𝑒𝑙𝑑 𝑚𝑜𝑚𝑒𝑛𝑡

∴𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝑚𝑜𝑚𝑒𝑛𝑡

𝑦𝑖𝑒𝑙𝑑 𝑚𝑜𝑚𝑒𝑛𝑡> 1

10. ANSWER: B

M/2

M

When moment M is applied to one end of proper

cantilever, half of moment gets transfer to fixed end.

11. ANSWER: B

W

A B

xC

L

𝑀𝐴 = 0 𝑀𝑋𝑥 = 𝑉𝐵 × 𝐿

𝑉𝐵 =𝑉𝑥𝐿

Σ𝐹𝑦 = 0 𝑉𝐴 + 𝑉𝐵 = 𝑤

𝑉𝐴 = 𝑤 (𝐿 − 𝑥

𝐿)

𝑀𝐶 = 𝑉𝐴 × 𝑥 = 𝑤𝑥 (𝐿 − 𝑥

𝐿)

12. ANSWER: B

In concurrent force system all the forces passes

through same point. Hence at that point we have 2

equilibrium equations i.e. Σfx = Σfy = 0

∴Hence the maximum unknowns can be determined

will be 2

13. ANSWER: B

AB

C

w kN/m

l l

Fixed moments

𝑀𝐹𝐴𝐵 = −𝑤𝑙2

12 𝑀𝐹𝐵𝐴 =

𝑤𝑙2

12

𝑀𝐹𝐵𝐶 = −𝑤𝑙2

12 𝑀𝐹𝐶𝐵 =

𝑤𝑙2

12

Joint Member fk Σ 𝑘 D.f.

B

BA 3𝐸𝐼

𝐿 6𝐸𝐼

𝐿

0.5

BC 3𝐸𝐼

𝐿

0.5

Member AD BA BC CB

FEM

DF

−𝑤𝑙2

12 𝑤𝑙2

12

0.5

−𝑤𝑙2

12

0.5

𝑤𝑙2

12

Release A & C

Joint COM +𝑤𝑙2

12 𝑤𝑙2

24 −𝑤𝑙2

24 −

𝑤𝑙2

12

Final

moment

0 𝑤𝑙2

8 −

𝑤𝑙2

8

0

∴ 𝑀𝐵 =𝑤𝑙2

8

14: ANSWER: C

A

B

Cw

l l

From Q.13 we get that final moment at middle sup-

ports is 𝑤𝑙2

8⁄

AB

Cw

VA VBA VBC VC

wl /82

B

w0 0

Now ΣMA = 0 -for left span AB only.

𝑤 × 𝑙 ×𝑙

2+𝑤𝑙2

8= 𝑉𝐵𝐴 × 𝑙

𝑤𝑙2 [1

2+1

8] = 𝑉𝐵𝐴 × 𝑙

∴ 𝑉𝐵𝐴 =5𝑤𝑙

8

∴Now for span BC ΣMC = 0

𝑉𝐵𝐶 × 𝑙 = 𝑤 × 𝑙 ×𝑙

2+𝑤𝑙2

8



MPSC Civil Engg. Mains Exam – Previous Year Question Papers 2020 Paper - II Page | 173

MPSC CIVIL ENGG. [MES] MAINS EXAM Previous Year Questions, Answer key and Detailed Explanation.

Paper II

Include

1. Surveying

2. Estimating, Costing and Valuation

3. Geo-technical Engineering

4. Fluid Mechanics and machines

5. Engineering Hydrology

6. Irrigation Engineering

7. Highway Engineering

8. Bridge Engineering

9. Tunneling

10. Environmental Engineering




MPSC CIVIL ENGG. [MES] MAINS EXAM PAPER 2

INDEX 1. SURVEYING


1.2 Answer Key 182


2. ESTIMATING, COSTING AND VALUATION




3. GEO-TECHNICAL ENGINEERING




4. FLUID MECHANICS AND MACHINES




5. ENGINEERING HYDROLOGY




6. IRRIGATION ENGINEERING




7. HIGHWAY ENGINEERING




8. BRIDGE ENGINEERING




9. TUNNELING




10. ENVIRONMENTAL ENGINEERING







1. SURVEING PREVIOUS YEAR QUESTIONS


1. The imaginary line joining the intersection of cross

hairs of diaphragm to the optical centre of object glass

and its continuation is called as

A. Axis of telescope B. Line of sight

C. Axis of diaphragm D. Line of telescope

2. The least count of leveling staff is

A. 1 mm B. 1 cm C. 5 mm D. 5 cm

3. In a closed theodolite traverse ABCDA, the following

latitudes and departures were calculated

Station N S E W

A 300.50 200.25

B 200.50 299.05

C 298.50 199.50

D 199.50 300.25

If the relative error of closure 1 in 5000, then the pe-

rimeter of transverse is

A. 71560.11 m B. 1397.41 m

C. 17890.30 m D. 5000 m

4. The U fork blumb bob is used for

A. Masonry work

B. Levelling of plane table

C. Centering of plane table

D. Orientation of plane table

5. The latitude of line is given as

A. 𝑙 cos 𝜃 B. 𝑙 sin 𝜃

C. 𝑙 𝑡𝑎𝑛 𝜃 D. None of the above

6. Two tangents intersect at the chainage 1200m, the

deflection angle being 40°. The number of 30m long

normal chords for setting tangents are

A. 7 B. 4 C. 5 D. 6

7. The representation fraction of a map scale 1cm =5

km is

A. 1/500000 B. 1/500

C. 1/5000 D. 1/50000


8. A compound curve of Sun or any fixed star is

A. Celestial Survey B. Astrological Survey

C. Heaven Survey D. Astronomical Survey

9. If an equation is subtracted from a constant k, the

weight of the resulting equation will be

A. Weight of equation divided by k

B. Weight of equation multiplied by k

C. Weight of equation multiplied by 𝑘2

D. Weight of equation remains unchanged

10. In theodolite transverse computations, Gale's ta-

ble is useful for determination of:

A. Independent co – ordinates

B. Dependent co – ordinates

C. Both (1) and (2)

D. None of the above

11. The representative fraction (R. F) of scale 1 cm =

500 m is:

A. 1:500 B. 1:5000

C. 1:50000 D. 1:50

12. The magnetic bearing of sun at noon was 170°.

Hence magnetic declination is:

A. 10° E B. 10° W C. 10° s D. 10° N

13. A back sight reading on B.M. = 200m, was

2.250 m. The inverted staff reading to the bottom of

beam was 1.450m.The R.L. of bottom of beam is:

A. 200.800m B. 201.450 m

C. 201.000 m D. 203.700 m

14. A tower is situated on the far side of the river and

is inaccessible. But it is visible. It can be located by:


PREVIOUS YEAR QUESTIONSSURVEYING

PAPER II




1. SURVEYINGANSWER KEY

QUE ANS QUE ANS QUE ANS QUE ANS

01 B 26 B 51 B 76 A

02 C 27 C 52 C 77 D

03 C 28 A 53 B 78 C

04 C 29 D 54 C 79 C

05 A 30 D 55 D 80 B

06 A 31 B 56 D 81 A

07 A 32 A 57 A 82 #

08 D 33 D 58 C 83 C

09 D 34 B 59 A 84 B

10 C 35 D 60 B 85 C

11 C 36 D 61 D 86 C

12 A 37 A 62 D 87 A

13 D 38 B 63 D 88 B

14 C 39 B 64 B 89 B

15 D 40 D 65 B 90 B

16 C 41 B 66 A 91 B

17 B 42 A 67 B 92 C

18 A 43 D 68 D 93 A

19 C 44 C 69 D 94 D

20 C 45 B 70 A 95 A

21 C 46 B 71 D 96 D

22 B 47 A 72 C 97 A

23 C 48 D 73 A 98 A

24 B 49 B 74 D 99 D

25 D 50 B 75 B 100


ANSWER KEYSURVEYING

PAPER II




1. SURVEYING DETAILED SOLUTIONS

1. ANSWER. B

Line of sight ⇒ the imaginary line joining the intersec-

tion of cross hairs of diaphragm to the optical Centre

of object glass and its continuation is called line of

sight

Axis of telescope⇒should coincide with axis of the tel-

escope

2. ANSWER. C

The least count of levelling staff= 5𝑚𝑚

Ranging Rod⇒used to locate intermediate points nom-

inal dia. 30𝑚𝑚

Offset Rod⇒used to set out offset lines at right angles

to the survey lines & also measured small offsets

Length of offsets= 3𝑚.

3. ANSWER. C

Relative closure is 1in5000

Find perimeter of Travers

Solution: We know

Relative error of closure =closing error

perimeter of travers

Closing error,𝑒 = √(Σ𝐿)2 + (Σ𝐷)2

Σ𝐿 = 300.5 + 2000.5 − 298.5 − 199.5 = 3

Σ𝐷 = 299.05 + 199.5 − 200.25 − 300.25 = 1.95

𝑒 = √(3)2 + (−1.95)2

𝑒 = 3.578

1

5000=

3.578

𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑎𝑣𝑒𝑟𝑠

𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡𝑟𝑎𝑣𝑒𝑟𝑠 = 17890.29 𝑚

4. ANSWER. C

The 𝑈 fork plumb bob is used for centering of plane

table

Uses: The plumb bob is used for centering the plane

table over the station and also for transpiring the

ground point to the drawing sheet.

5. ANSWER. A

Latitude (L): Projection of a line on 𝑁 − 𝑆 axis is called

latitude

W

NA

L

D

E

S

lθ

𝐿 = 𝑙 cos 𝜃

Nature of Latitude:

North Axis +𝑣𝑒

South Axis −𝑣𝑒

Departure (D): Projection of a line 𝐸 −𝑊 axis is called

departure

𝐷 = 𝑙 sin 𝜃

Nature of departure:

East +𝑣𝑒

West −𝑣𝑒

6. ANSWER. A

Tangent length

𝑇1𝑉 = 𝑉𝑇2 = 𝑅. tanΔ

2

= 300 tan40

2

= 109.19𝑚

Chain age of 𝑇1 = 𝑐ℎ𝑎𝑖𝑛𝑎𝑔𝑒 𝑜𝑓 𝑉 − 𝑇1𝑉

= 1200 − 109.19

= 1090.81𝑚

Length of curve = 𝑅. Δ° ×π

180

= 300 × 40 ×π

180= 209.44𝑚

Chain age of 𝑇2 = 𝑐ℎ𝑎𝑖𝑛𝑎𝑔𝑒 𝑜𝑓 𝑇1 +

𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑐𝑢𝑟𝑣𝑒

= 1090.81 + 209.44

= 1300.25 𝑚

Length of 1st chord = [𝑛𝑒𝑥𝑡 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 30] −

[𝑐ℎ𝑎𝑖𝑛𝑎𝑔𝑒 𝑜𝑓 𝑇1]

= 1110 − 1090.81

= 19.19𝑚

Length of last chord = 𝐶ℎ𝑎𝑖𝑛𝑎𝑔𝑒 𝑜𝑓 𝑇2 −

𝑃𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 30

= 1300.25 − 1290


DETAIL EXPLANATIONSURVEYING

PAPER II




= 10.25𝑚

𝑁𝑜. 𝑜𝑓 𝑓𝑢𝑙𝑙 𝑐ℎ𝑜𝑟𝑑 =1290 − 1110

30= 6

No. of full 30𝑚 chord = 6

Total chord = 6 + 2 = 8

7. ANSWER. A

𝑅𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡𝑎𝑡𝑖𝑣𝑒 𝐹𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑛 𝑝𝑎𝑝𝑒𝑟

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑛 𝑔𝑟𝑜𝑢𝑛𝑑

=1𝑐𝑚

5 × 103 × 102

=1

500000

8. ANSWER. D

Many times it is not possible to provide a curve of con-

stant radius to connect the two straight line in that case

we provide more than one curve of different radii to

connect them

9. ANSWER. D

If an equation is subtracted from constant 𝐾 the weight

of the resulting equation will remains in unchanged

10. ANSWER. C

In theodolite travers computations Gale's table is use-

ful for determination of independent co-ordinates &

dependent co-ordinates

11. ANSWER. C

Representative Fraction,

𝑅𝐹 =𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑛 𝑝𝑎𝑝𝑒𝑟

𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑛 𝑔𝑟𝑜𝑢𝑛𝑑=

1𝑐𝑚

50000=

1

50000

12. ANSWER. A

100

1700

TN MH

W E

MSTS

Declination is 10 E0

True Bearing = 𝑀𝑎𝑔𝑛𝑒𝑡𝑖𝑐 𝑏𝑒𝑎𝑟𝑖𝑛𝑔 ± 𝑑𝑒𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛

+ Declination is toward east

− → Declination is toward west

180° = 170° + 𝑑𝑒𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛

𝑑𝑒𝑐𝑙𝑖𝑛𝑎𝑡𝑖𝑜𝑛 = 10°(+𝑣𝑒)

i.e. East, ∴ 10° 𝐸

13. ANSWER. D

Given:𝐵.𝑀.= 200𝑚

𝐵. 𝑆. = 2.250

Inverted staff reading = 1.450𝑚

Find: 𝑅. 𝐿.of bottom of beam

Solution:

A

2.2

5 m

1.45m

𝐻𝐼 = 𝐵𝑀 + 𝐵𝑆 = 200 + 2.250 = 202.250 𝑚

𝑅𝐿 = 𝐻𝐼 − (−𝐼𝑛𝑣𝑒𝑟𝑡𝑒𝑑 𝑠𝑡𝑎𝑓𝑓)

= 202.250 + 1.450

= 203.700 𝑚

14. ANSWER. C

Intersection: Suitable for point are inaccessible but vis-

ible

Radiation: The distance of these points are measured

and scaled of on the respective radial lines to locate

the points

Resection: It is the method of locating a station occu-

pied by the plane table when the position of that sta-

tion has not been earlier plotted on the drawing sheet

when the plane table occupied their stations

15. ANSWER. D

Astronomical survey: A survey which consists of obser-

vations of the heavenly bodies such as sun or any fixed

star

16. ANSWER. C

Given: radius of curve = 300𝑚

Intersection angle= 120°

Find: Tangent length =?

Tangent length= 𝑅 tan∆

2

= 300 tan60

2

= 173.205 𝑚

600

1200

∆ =Angle of deflection

= 180 -120 =600 0 0

17. ANSWER. B

Change Point: It is the point which denotes the shifting

of the instrument/level. Both 𝐵𝑆 & 𝐹𝑆 are taken on the

staff at the change point



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