ACE Engineering Academy · 2017. 4. 10. · 150 189 3 5 2 2 = 36.59 mm 38 mm ... For ISHB 350 @ 576.8 N/m; D = 350mm & bf = 250mm When column ends are machined for complete bearing,

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: 2 : Civil Engg. _ ESE MAINS


01. (a)

Sol:

Axial load on column P = 1800 kN

Permissible bearing stress on concrete

cc = 5 N/mm2

Permissible bending stress in slab base bs

= 189 N/mm2.

Area of slab base required A =cc

P

=5

101800 3

= 360000 mm2

Depth of steel column D = 250 + 25 + 25

= 300 mm

Width of steel column bf = 300 mm

Let L, B and ts be the length, width and

thickness of slab base respectively

To have optimum (minimum) thickness of

slab base, the bigger and lesser projections of

the slab base should be equal

The column shape with cover plates is square.

Hence the square shape slab base can be

provided to have equal projections

Side of square slab base L= B = A

L= B = 360000 = 600mm

Upward pressure from concrete pedestal

below the slab base

w =1A

P=

600600

101800 3

= 5 N/mm2 ≤ cc = 5 N/mm2

Hence concrete pedestal is safe

Projections of slab base beyond the steel

column a = b

mm1502

300600

2

DLba

Thickness of base plate

ts=

4

ba

w3 22

bs

=

4

150150

189

53 22

= 36.59 mm 38 mm

Therefore a slab base 600 × 600 × 38 may

be used at the finished end of the column.

4545

P = 1800 N

L

B

b

a300 mmab

300 mm

b

: 3 : Test – 7




(b) (i)

Ans:

1. The main difference is that the position

of the bucket is the reverse to that of

the power shovel;

2. It is designed to dig below the level of

the machine

3. It digs by pulling the load toward the

power unit

4. It is used for digging narrow pits

formation

5. Its prime mover capacity is less

compared to power shovel.

(ii)

Sol:

Net cutting blade volume capacity

=25.1

25= 20m3

Probable round-trip time:

Pushing 30m @ 2.4 km/hr =4.2

601000

30

= 0.75 min

Returning 30m@ 4.8 km/hr

=8.4

601000

30

= 0.375 min

Fixed time for gear shift = 0.88 min

Total time = 0.75 + 0.375 + 0.88

= 2.005 min

Trip per hour =005.2

50= 25 trips per hour

(rounded figure)

Bull Dozer output per hour = 25 20

= 500 m3/hr

(c)

Sol:

Factored axial compressive load

P = 1350 kN

For Fe410 grade steel, fu = 410MPa &

fy = 250MPa

For grade 6.8 bolt, fub = 600MPa

For ISHB 350 @ 576.8 N/m; D = 350mm &

bf = 250mm

When column ends are machined for

complete bearing, it is assumed that 50%

of factored axial compressive load is

transferred through bearing action and splice

should be designed for 50% factored axial

compressive load

Factored axial load on splice

P1= kN6252

1350

Factored axial load on each side of splice

P2 = kN5.3122

625

Area of splice plate

A = moy

2

/f

P

= 2

3

mm01.13751.1/250

105.312

: 5 : Test – 7


Thickness of splice plate ts =250

01.1375

= 5.5mm

Assume width of splice plate B = 250mm

Adapt size of splice plate 250mm 6mm

Assume shank diameter of bolt

d = 6.01 t = 6.01 6 = 14.72 mm

≃ 16mm

Number of bolts required on each side of

splice

db

2

V

P

boltoneofStrengthDesign

spliceofsideeachonloadaxialFactoredn

Design strength of one bolt

Vdb = Minimum of Vdsb or Vdpb

Design shear strength of one bolt in single

shear (Assuming thread of the bolt is

intercepting the shear plane)

Vdsb= )AnAn(3

fsbsnbn

mb

ub

=25.13

600

(1×10.78

π4

(16)2 + 0)

= 43.46 103 N = 43.46 kN

Minimum pitch p = 2.5 d = 2.5 16

= 40mm ≃ 50mm

Minimum end distance, emin = 1.5do

= 1.5 18 = 27mm ≃ 30mm

Design bearing strength of one Bolt

Vdpb = 2.5×Kb× d× t×mb

ubf

Where Kb is bearing factor is lesser of

74.0183

40

d3

e

o

67.025.0183

5025.0

d3

p

o

46.1410

600

f

f

u

ub

1.0

Hence bearing factor Kb = 0.67

Vdpb = 2.5 0.67× 16 6 410 /1.25

= 52.74 103 N = 52.74 kN

Design strength of one bolt, Vdb = 43.46 kN

No of bolt required

n =46.43

5.312

V

P

db

2 =7.19 ≃ 8 No’s

Provide 8 No’s of M16 bolts for connection

of each side of splice plate.

Length of splice plate L =2×[(350)+ (240)]

= 460mm

Provide two numbers of splice plates of size

460 2506 mm on each side of splice



(d)

Ans:

Seasoning of Timber: When a tree is newly

felled, it contains about 50 percent or more of

its own dry weight as water. This water is in

the form of sap and moisture. The water is to

be removed before the timber can be used for

any engineering purpose. In other words, the

timber is to be dried. This process of drying

of timber is known as the seasoning of timber

and the moisture should be extracted during

seasoning under controlled conditions as

nearly as possible at a uniform rate from all

parts of the timber.

Preservation of Timber: The art of treating

the timber with same chemical so as to

increase its life is called preservation of

timber.

Proper seasoning is the most effective means

of preservation, the objective of preservative

treatment of timber with chemicals is to

prolong it’s life and to increase its durability.

Methods for preservation of timber:

Following are the six methods adopted for

preservation of timber:

(i) Brushing (ii) Charring

(iii) Dipping and steeping

(iv) Hot and cold open tank treatment

(v) Injecting under pressure

(vi) Spraying.

Charring: This method of charring is rather

very old and as such, no preservative is used

in this method. The surface to be charred is

kept wet for about half an hour and it is then

burnt upto a depth of about 15 mm over a

wood fire. The charred portion is then cooled

with water. Due to burning, a layer of coal is

formed on the surface. This layer is not

affected by moisture and it is not attacked by

white ants, fungi, etc.

The disadvantages of this method are:

(a) The charred surface becomes black in

appearance and hence it cannot be used

for exterior work.

(b) There is some loss of strength of timber

as the cross-section is reduced due to

charring.

The process of charring is generally adopted

for lower ends of posts for fencing,

telephone, etc. Which are to be embedded in

the ground or to be inserted in moist soil.

(e)

Ans:

Composition of Good Brick Earth:

Following are the constituents of good brick

earth:

(1) Alumina: It is the chief constituent of every

kind of clay. A good brick earth should

contain about 20% to 30% of alumina. This

constituent imparts plasticity to the earth so

: 7 : Test – 7


that it can be moulded. If alumina is present

in excess, with inadequate quantity of sand,

the raw bricks shrink and warp during drying

and burning and become too hard when

burnt.

(2) Silica: It exists in clay either as free or

combined. As free sand, it is mechanically

mixed with clay and in combined form, it

exists in chemical composition with alumina.

A good brick earth should contain about 50

percent to 60 percent of silica. The presence

of this constituent prevents cracking,

shrinking and warping of raw bricks. It thus

imparts uniform shape to the bricks. The

durability of bricks depends on the proper

proportion of silica in brick earth. The excess

of silica destroys the cohesion between

particles and the bricks become brittle.

(3) Lime: A small quantity of lime not

exceeding 5 percent is desirable in good

brick earth. It should be present in a very

finely powdered state because even small

particles of the size of a pin-head cause

flaking of the bricks. The lime prevents

shrinkage of raw bricks. The sand alone is

infusible.

(4) Oxide of iron: A small quantity of oxide of

iron to the extent of about 5 to 6 percent is

desirable in good brick earth. It helps as lime

to fuse sand. It also imparts red colour to the

brick. The excess of oxide of iron makes the

bricks dark blue or blackish. If, on the other

hand, the quantity of iron oxide is

comparatively less, the bricks will be

yellowish in colour.

(5) Magnesia: A small quantity of magnesia in

brick earth imparts yellow tint to the bricks

and decreases shrinkage. But excess of

magnesia leads to the decay of bricks.



0

01

A

6

626

B30

336

76

F80

116

156

5 H14

1708170

I12

158

158

138

1386 20

G

12E

4126

126

120D

C10

J

24

194

1949

7

02. (a)

Sol:

EventsACT

MM

Duration

(d)EST LST EFT LFT TF

1 – 2 A 6 0 0 6 6 0

2 – 3 B 30 6 46 36 76 40

2 – 6 C 10 6 128 16 138 122

2 – 4 D 120 6 6 126 126 0

4 – 6 E 12 126 126 138 138 0

3 – 5 F 80 36 120 116 156 40

6 – 7 G 20 138 138 158 158 0

5 – 8 H 14 116 156 130 170 40

7 – 8 I 12 158 158 170 170 0

8 – 9 J 24 170 170 194 194 0

EFT Earliest Finish Time

EST Earliest Start Time

LFT Latest Finish Time

LST Latest Start Time

LST = LFT – Duration LFT – d

EFT = EST + Duration EST+ d

Total float = TF = LST – EST = LFT – EFT

For which ever activity total float is zero that activity is on critical path

1 – 2 – 4 – 6 7 – 8 – 9

A – D – E – G – I – J is the critical path

: 9 : Test – 7




(b)

Sol:

Allowable bearing stress, p = 187.5 N/ mm2

End reaction = 900 kN

Bearing area required end-bearing stiffener

Ae =5.187

10900 3= 4800 mm2

Try 4 ISA 130×130×10 of two on either side

of web as shown in figure.

A = 2512 mm2;

Ixx= Iyy = 405.3×104 mm4;

Cxx = Cyy=35.9mm

Bearing area provided by outstanding legs

= 4 × (130–10)×10 = 4800 mm2

Effective area of the stiffener

A = 6130225124 = 11608 mm2

Moment of inertia of bearing stiffener about

centre of web

12

3626029.3512321124103.4054

minI

= 42.25106 mm4

Minimum radius of gyration, rmin

11608

1025.42

A

I 6min = 60.33 mm

Effective length of bearing stiffener

= 0.7(2000212) = 1383.2 mm

Slenderness ratio of bearing stiffener

92.2233.60

2.1383

rmin

Permissible compressive stress

ac= 2092.22

2030

139145145

=143.25 N/mm2

Area required =25.143

10900 3

= 6282 mm2 ≯ 11608 mm2

Which is safe.

Assume nominal diameter of rivet = 20 mm

Gross (or) effective diameter of rivet,

d = 20 + 1.5 = 21.5mm

vf = 100 MPa; pf = 300 MPa

Shear strength of one rivet in double shear

1005.214

2d4

2P 2vf

2s

= 72.61 103 N = 72.61 kN

Strength of one rivet in bearing

Pb = d t pf = 21.5 6 300

= 38.7 × 103 N = 38.7 kN

Rivet value Rv = Smaller of Ps and Pb

= 38.7 kN

No of rivets required

2425.237.38

900

valueRivet

reactionEndn

Provide 12 No's- 20mm diameter of PDS

rivets @120 mm C/C on each pair of end

stiffener.

130 mm

Web plate

Filler Plate 12 mm

Bearing Stiffener

20 × 6 = 120 mm 20 × 6 = 120 mm

: 11 : Test – 7


(c)

Ans:

Soundness: The purpose of this test is to

detect the presence of uncombined lime in

cement. This test is performed with the help

of Le Chatelier apparatus as shown in fig. It

consists of a brass mould of diameter 30mm

and height 30mm. There is a split in mould

and it does not exceed 0.50mm. On either

side of split, there are two indicators with

pointed ends. The thickness of mould

cylinder is 0.50mm.

Following procedure is adopted.

(i) The cement paste is prepared. The

percentage of water is taken as

determined in the consistency test.

(ii) The mould is placed on a glass plate and

it is filled by cement paste.

(iii) It is covered at top by another glass

plate. A small weight is placed at top

and the whole assembly is submerged in

water for 24 hours the temperature of

water should be between 24oC to 34oC.

(iv) The distance between the points of

indicator is noted. The mould is again

placed in water and heat is applied in

such a way that boiling point of water is

reached in about 30 minutes. The

boiling of water is continued for one

hour.

(v) The mould is removed from water and it

is allowed to cool down.

(vi) The distance between the points of

indicator is again measured .The

difference between the two readings

indicates the expansion of cement and it

should not exceed 10 mm.

03. (a)

Sol:

DaysActivity

t0 tm tp te

variance

2

1 – 2 2 4 6 4 2/3 4/9

1 – 3 2 3 4 3 1/3 1/9

2 – 3 6 8 10 8 2/3 4/9

2 – 4 4 10 16 10 2 4

3 – 5 7 10 19 11 2 4

4 – 5 3 6 9 6 1 1

30 mm

165 mm

Split 0.5 mm max0.5mmA

A

Top view without glasssheets

30mm

Glass sheet

Glass sheet

Le chatelier apparatus



1

0

0

3

4

4

42

312

12

11 23

235

6

174

10

10

8

estimateTime6

tt4tt pm0

e

deviationdStandar6

tt 0p

ancevari6

tt2

0p2

Critical path 1–2–3–5

Duration TE = 23 days

(variance)cp = (var)1-2 + (var)2-3 + (var)3-5

= 494

94

=944

93644

Standard deviation of critical path

= pcpc var

= 9/44 = 2.211 days

Desired completion date = D = 28 days

Critical path time estimate = TE = 23 days

26.2211.2

2328TDZ

pc

e

61.982.2

93.983.2obPr26.2

dz = 0.1 dp = 0.32

dz1=0.04 dp1 = ?

dpdzdz

P93.981

1

= 128.032.01.0

04.0

P1 = Probability of completion

= 98.93 – 0.128 = 98.8

21.981.2

61.982.25.981Z

dz = 0.1; dp = 0.4

dz1 = ? dp1 = 0.11

dz1 = dzdpdp1

0275.01.04.011.0

z2.2 1

1725.20275.02.2z1

pc

e1

TDz

211.223D

1725.2

D = 23 + 2.211 ×2.1725

= 23 + 4.8 = 27.8 days

(b)

Sol:

Effective span of beam l = 6 m

Uniformly distributed load w = 21.5 kN/m

Permissible bending tensile and bending

compressive stress bt = bc = 165 MPa

Permissible average shear stress

τvf=100 MPa

: 13 : Test – 7


Maximum bending moment,

M=8

65.21

8

w 22

l= 96.75 kN-m

Maximum shear moment

V =2

65.21

2

w

l= 64.5 kN

Section modulus required Zr =btbc or

M

=165

1075.96 6

= 586.36103 mm3

Choosing ISLB 325 @422.8 N/m

Zxx = 607.7103 mm3

Ixx = 9874.6 104 mm4

tw = 8.8 mm & tf = 9.7 mm

Check for bending stress

Calculated maximum bending compressive

or bending tensile stress

bc cal = bt cal = yI

M

xx

= )2/325(106.9874

1075.964

6

= 159.21 N/mm2 <165MPa

(beam section is safe against bending

moment)

Check for shear stress

Calculated average shear stress τva cal

=wt.h

V

8.8325

105.64 3

= 22.55 N/mm2 τva =100 N/mm2

Section is safe against shear

Check for deflection

Allowable (or) Maximum deflection for a

simply supported beam

mm46.18325

6000

325max

Calculated maximum deflection =EI384

w5 4l

=45

4

106.9874102384

60005.215

= 18.37 mm < 18.46 mm

Section is safe against deflection

(c)

Ans:

Requirements of a Good Aggregate:

Following are the desirable properties or

requirements of a good aggregate.

(1) Adhesion (2) Cementation

(3) Durability (4) Hardness

(5) Shape (6) Strength

(7) Toughness

(1) Adhesion: The aggregates which are to be

used for the construction should have less

affinity with water as compared with the

binding material. If this quality is absent in

the aggregate, it will lead to the separation of

bituminous or cement coating in the presence

of water.

(2) Cementation: The binding quality of the

aggregate depends on its ability to form its

own binding material under different loading

so as to make the rough broken stone pieces

grip together to resist displacement.



(3) Durability: The durability of an aggregate

indicates its resistance to the action of

weather and is largely dependent upon its

petrological composition. The metal is

subjected to the oxidizing influence of air

and rain water. It is therefore desirable that

the aggregate should possess sufficient

soundness to resist the action of weather and

age so that the life of the structure made with

it may be prolonged.

(4) Hardness: The aggregates should be

reasonable hard to offer resistance to the

actions of abrasion and attrition. The

aggregates are always subjected to the

constant rubbing action. It is known as

abrasion and it will be increased due to the

presence of abrasive material like sand

between the exposed top surface and the

tyres of moving vehicles. The abrasive action

is very severe for roads which are used by the

steel tyred vehicles. The mutual rubbing of

stones is known as attrition and it may also

cause a little wear in the aggregates.

(5) Shape: The shape of aggregates may be

rounded, cubical, angular, flaky or elongated.

The flaky and elongated particles possess

less strength and durability and their use in

the construction should be avoided as for as

possible. The rounded particles are preferred

in cement concrete construction. But they are

unsuitable in W.B.M. construction,

bituminous construction and in granular base

course because their stability due to

interlocking in less. The angular particles are

preferred in such types of construction.

(6) Strength: The aggregates should be

sufficiently strong to withstand the stresses

developed due to the wheel loads of the

traffic. This property is especially desirable

for the road aggregates which are to be used

in top layers of the pavement. Thus, the

wearing course of road should be composed

of aggregate which possess enough strength

in addition to enough resistance to crushing.

(7) Toughness: The toughness of an aggregate is

that property which enables the aggregate to

resist fracture when struck with a hammer

and it is necessary in a metal to withstand the

impact blows caused by traffic. The

magnitude of impact is governed by the

roughness of surface, speed of the vehicle

and other vehicular characteristics. It is

desirable that the aggregate is reasonably

tough.

Fineness modulus: Fineness Modulus is a

ready index of coarseness or fineness of the

material. Fineness modulus is an empirical

factor obtained by adding the cumulative

percentages of aggregate retained on each of

the standard sieves ranging from 80 mm to

: 15 : Test – 7


150 micron and dividing this sum by an

arbitrary number 100. The larger the figure,

the coarser is the material.

Many a time, fine aggregates are designated

as coarse sand, medium sand and fine sand

may be really medium or even coarse sand.

To avoid this ambiguity fineness modulus

could be used as a yard stick to indicate the

fineness of sand.

The following limits may be taken as

guidance:

Fine sand: Fineness Modulus: 2.2 – 2.6

Medium sand: F. M. : 2.6 – 2.9

Coarse sand : F. M. : 2.9 -3.2

A sand having a fineness modulus more than

3.2 will be unsuitable for making satisfactory

concrete.

04. (a) (i)

Ans:

Properties of good Mortar Mix And Mortar:

The important properties of a good mortar

mix are mobility, placeability and water

retention.

1) Mobility: The term mobility is used to

indicate the consistency of mortar mix

which may range from stiff to fluid. The

mobility of mortar mix depends on the

composition of mortar and the mortar

mixes to be used for masonry work,

finishing work, etc. are made sufficiently

mobile.

2) Placeability: The placeability or the

ease with which the mortar mix can be

placed with minimum cost in a thick and

uniform layer over the surface depends

on the mobility of the mortar. The

placeability of mortar mix should be

such that a strong bond is developed

with the surface of the bed.

3) Water retention: A good mortar mix

should posses the ability of retaining

adequate humidity during transportation

and laying over the porous bed. If water

retention power of mortar mix is low, It

separates into layers during

transportation and when it comes into

contact with porous bed such as brick,

wood, etc., it gives away its water to that

surface. Thus the mortar becomes poor

in amount of water and the remaining

water proves to be insufficient for its

hardening. Hence the required strength

of mortar will not be achieved with such

a mortar mix.

Properties of a good mortar:

Following are the properties of a good

mortar:

(i) It should be capable of developing

good adhesion with the building units

such as bricks, stones, etc.

(ii) It should be capable of developing the

designed stresses.

(iii) It should be capable of resisting

penetration of rain water.



(iv) It should be cheap.

(v) It should be durable.

(vi) It should be easily workable.

(vii) It should not affect the durability of

materials with which it comes into

contact

(viii) It should set quickly so that speed in

construction may be achieved.

(ix) The joints formed by mortar should

not develop cracks and they should be

able to maintain their appearance for a

sufficiently long period.

Uses of Mortar:

Following are the uses of mortar:

(i) To bind the building units such as

bricks, stones, etc, into a solid mass.

(ii) To carry out pointing and plaster

work on exposed surfaces of

masonry

(iii) To form an even and soft bedding

layer for building units

(iv) To form joints of pipes

(v) To improve the general appearance

of structure,

(vi) To prepare moulds for coping,

corbels, cornice, etc.,

(vii) To serve as a matrix or cavity to

hold coarse aggregates, etc.,

(viii) To distribute uniformly the super

incumbent weight from upper layer

to lower layer of bricks or stones,

(ix) To hide the open joints of brick

work and stonework,

(x) To fill up the cracks detected in the

structure during maintenance

processes, etc.

(ii)

Ans: Schmidt’s rebound hammer developed in

1948 is one of the commonly adopted

equipments for measuring the surface

hardness.

It consists of a spring control hammer that

slides on a plunger within a tubular housing

When the plunger is pressed against the

surface of the concrete, the mass rebound

from the plunger, it retracts against the

force of the spring. The hammer impacts

against the concrete and the spring control

mass rebounds, taking the rider with it

along the guide scale. By pushing a button,

the rider can be held in position to allow the

reading to be taken. The distance traveled

by the mass, is called the rebound number.

It is indicated by the rider moving along a

graduated scale.

Plunger

Spring

LatchBody

indicator

Hammer

(a)

MassRelease Button

Concrete

: 17 : Test – 7


Limitation: Although, rebound hammer

provides a quick inexpensive means of

checking uniformity of concrete it has

serious limitations and these must be

recognized. The results are affected by:

a) Smoothness of surface under test.

b) Size, shape and rigidity of the specimen

c) Age of specimen

d) Surface and internal moisture condition

of the concrete

e) Type of coarse aggregate

f) Type of cement

(a), (b), (c) & (d) indicates the following

(a) Instrument ready for test

(b) Body pushed toward test object

(c) Hammer is released

(d) Hammer rebounds

(b)

Ans: The following common factors needs to

consider while selecting construction

equipment for a dam construction project.

1. Scope of work:The time frame within

which the construction work is required

to be carried out, and the specification

of works will be of concern.

2. Use of available: For a work where full

utilization of equipment for its entire

working life and consequently unit cost

of work. Economic should be worked

out.

3. Suitability for job conditions: The

equipment selected should suit the

demands of the job conditions. Climate

of region and working conditions

should be kept in view.

4. Uniformity in type: It is desirable to

have minimum number of types so that

there is uniformity in the type of

equipment on a project.

5. Size of equipment: While large size of

machines are capable of giving large

outputs on full load, the cost of

production is usually greater than that

of smaller units if worked out on part

loads. Large size of equipment requires

corresponding large size of matching

equipment.

plunger

Spring

LatchBody

indicator

Hammer

(a)(b)

(c) (d)



6. Use of standard Equipment: Standard

equipment are commonly manufactured

and are commonly available and are

moderately priced. The spare parts of

standard equipment are easily available

and less expensive.

7. Unit cost production: The economics

is one of the most important

considerations in selection of

equipment. While working out owing

cost all items of expenses, such as

freight, packaging and forwarding,

insurance, erection and commissioning

etc., should be included with the price

paid to supplier.

8. Availability of spare parts: Down time

for want of necessary spare parts

commonly accounts for long idle periods

during working life of equipment,

especially of imported equipment.

9. Versatility: The equipment selected

should be if possible, be capable of

performing more than one function and

should have feature of inter

convertibility as far as possible.

10. Selection of manufacturer: It is

desirable to have equipment of the same

manufacturer on a project as far as

possible and to have minimum number

of different makes of equipment.

11. Suitability of local labour: Available

operators and technicians should be able

to handle selected equipment.

12. Technical consideration: The efficient

performance of any piece of equipment

and its service life are conditioned by

factors like Strength. Rigidity, Vibration

stability, Resistance to wear etc.

13. Balance of resources: A balance

between reliability, investment cost and

operating cost should be found since a

policy of selecting the lowest priced

equipment can often lead to higher life

cycle costs etc.

14. Economical Life of Construction

Equipment: The owner of construction

equipment look for obtaining the lowest

possible cost per unit of production, for

which different costs considered are: a.

Depreciation and replacement. b.

Investment. c. Maintenance and repairs

d. Downtime. e. Obsolescence. An

analysis of the effect which hours of

usage will have on each of these costs

will establish the time at which a

equipment should be replaced.

: 19 : Test – 7




(c)

Sol: Effective span of beam L = 5.0 m

Yield stress of steel fy = 250 Mpa

` Factored U.D.L on beam is inclusive of self-

weight w = 50 kN/m

Let ‘W’ be the factored central concentrated

load in kN

Section classification

Section classification ε =250

250

f

250

y

=1.0

Outstand of flange, b = mm952

190

2

b f

Depth of web,

d = h–2×(tf+R1) = 550–2×(19.3+18)

= 475.4 mm

4.992.43.19

95

t

b

f

6744.422.11

4.475

t

d

w

Hence the section is plastic section

Since V≤ 0.6 Vd this is low shear case

Design bending strength of laterally

restrained beam

Md =mo

y

mo

ypb

fZe2.1

fZ

For plastic section b=1

10.1

2501098.27110.1M 3

d

10.1

250108.23592.1

fZe2.1 3

mo

y

= 616.35×106 N-mm ≤ 643.57×106 N-mm

= 616.35kN-m ≤ 643.57×106 kN-mm

Which is all right

Factored U.D.L on beam is inclusive of self-

weight w = 50 kN/m

Let ‘W’ be the central concentrated load in

kN

Design bending moment due to udl (w) and

Central concentrated load W

4

5W

8

)5(50

4

WL

8

wLM

22

zz

W25.125.156 kN-m

Equating Mzz = Md

156.25+1.25 W = 616.35

W = 68.08 kN ---- (1)

Design shear strength of joist,

mo

yw

mo

yvd

3

fth

3

fAV

10.13

2502.11550

kN31.808

Design shear force due to udl (w) and

Central concentrated load W

2

W

2

wLV = W5.0125W5.0

2

550

Equating V = Vd

125+0.5 W = 808.31

W = 1366.62 kN----(2)

Limiting deflection for a simply supported

beam (assuming brittle cladding)

: 21 : Test – 7


mm66.16300

5000

300itlim

Calculated maximum deflection under

service loads

Service UDL on beam ws = w/γf

= 50/1.5 = 33.33 kN/m

EI48

WL

EI

Lw

384

5 34s

caitlim

61094.648

510248

35000)5.1/1000W(

61094.648

5102

4)5000(33.33

384

567.16

16.67=2.09+0.0133W

W = 1096 kN----(3)

Factored central concentrated load W

= 368.08 kN

Design shear force due to udl (w) and

Central concentrated load

kN04.30908.3685.0125W5.0125V ≤ 0.6 Vd = 0.6 ×808.31=484.98 kN

Hence this is low shear and assumption is

correct

Factored central concentrated load W

= 368.08 kN

05. (a) (i)

Ans:

1) Masonry in foundation and plinth:

Lime mortar and cement mortar are used in

the construction of foundation and plinth

2) Masonry in superstructure: Cement

mortar and lime mortar are used in super

structure.

(3) Lime Plaster: In lime plaster mostly

fat lime should be used. Hydraulic

lime may contain particles which slake

slowly as and when they come in

contact with atmospheric moisture.

The result into blisters on the plastered

surface, known as blowing. If

hydraulic lime has to be used for

plastering, it should be ground dry

with sand. Thereafter, it is left for two

to three weeks. It is ground again

before use.

Cement Plaster: Cement plaster consists

of an intimate mixture of cement and clean

coarse angular river sand with suitable

amount of water. Proportions of cement

and fine aggregate may vary according to

the requirements of the plaster. But usual

proportions are 1:3 or 1:4. The materials in

the required proportion are stacked on a

watertight platform and thoroughly mixed

in dry condition. Before plastering, water is

added only to that such dry mix which

could be used in less than 30 minutes time.

Alternatively, dry mix is taken in pan and

water in required amount is added to the

pan itself and pan is carried to the mason

for use.

Cement plaster may be done in one coat or

two coats. Usually it is done in one coat

only.



4) Mortar for Pointing:

Pointing may be done in lime mortar or

in cement mortar. Cement mortar

pointing is mostly used as it is more

durable and weather resisting.

Lime Mortar: Lime mortar for pointing

is prepared by mixing lime and sand in

proportion of 1:1. The lime mortar

should be ground in mortar mills.

Cement Mortar: Cement mortar

consists of cement and fine sand in

proportion of 1:2 and 1:3.

Sand to be used for preparing lime or

cement mortar for pointing should be

clean, fine, and free from organic

impurities.

Method of Pointing:

Pointing is done in the following stages:

1. The mortar joints of the surface to be

pointed are raked out to a depth of about

13 mm.

2. Raked joints are cleaned from loose

mortar and thoroughly wetted.

3. Now mortar is taken in small flat

rectangular plates, made of iron. Plate

with mortar is held in one hand by the

mason, abutting the wall and with the

help of a small trowel in other hand

mortar is forced into the open spaces of

raked masonry joints. The mortar is

slightly pressed with trowel to bring it

into close contact with the masonry

mortar of the joint.

4. The pointing should be finished as per

the required finishing with the help of

specific tool. Finishing is done when

mortar is still green.

5. The pointed surface should be cured for

at least 3 days in case of lime mortar

and 10 days in case of cement mortar

pointing.

(ii)

Sol: Volume of concrete = 1 m3

Void ratio of cement, ec = 62%

Void ratio of Fine aggregate, eF.A = 41%

Void ratio of coarse aggregate, %45e A.C

Mix proportion = 1 : 2: 4

Weight of cement = Wc

Weight of Fine aggregate = WF.A

Weight of Course aggregate = WC.A

Weight of water = Ww

Concrete = Voids + water + Cement + Fine

aggregate + Coarse aggregate.

Void ratio =solidsofVolume

VoidsofVolume

Volume of voids = void ratio volume of

Solids

=solidofdensity

solidofmassratiovoid

: 23 : Test – 7


Concrete =

A.C

A.C

A.F

A.F

c

c

w

w

A.C

A.CA.C

A.F

A.FA.F

c

cc

WWWW

We

We

We

1600

W4

1700

W2

1440

W1000

W55.0

1600

W445.0

1700

W241.0

1440

W62.01

ccc

cccc

1000

W55.0

1600

W8.5

1700

W82.2

1440

W62.11 cccc

WC = 143.7 kg

WF.A = 287.4 kg

WC.A = 574.81 kg

(b)

Sol: Web stiffeners:

IS 800:1984 recommends to provide web

stiffeners as follows:

85andf

1344and

816oflesser

t

d

ycal,vaw

1

No stiffener is required

200and3200

oflessert

d

cal,vaw

2

Vertical stiffeners are provided

250and4000

oflessert

d

cal,vaw

2

Vertical stiffener + first horizontal

stiffener at a distance from compression

flange equal to two-fifth of distance from

compression flange to neutral axis

400and6400

oflessert

d

cal,vaw

2

Vertical stiffener + first horizontal

stiffener at a distance from compression

flange equal to two-fifth of distance from

compression flange to neutral axis + a

horizontal stiffener at neutral axis

d2 = 2 clear distance from compression

flange angle or plate to the neutral

axis.

In no case the greater clear dimension of a

web panel less than 270 tw and the lesser

clear dimension of the web panel less than

180 tw

(1) Transverse (stability or vertical) stiffener:

Intermediate transverse stiffener increases

buckling resistance of web against shear

Angle sections are provided for riveted or

bolted plate girders and flat or plate

section for welded plate girders

Minimum moment of inertia of transverse

stiffener

I ≥ 1.5 d13 t3 / c2

t = Minimum required thickness of web

c = the maximum permitted clear distance

between vertical stiffener for thickness tw

Maximum spacing of vertical stiffeners -

1.5d1.

Minimum spacing of vertical stiffness -

0.33d1.



If vertical stiffeners are subjected ti

external force, the moment of inertia

should be increased as

Bending moment on stiffener due to

eccentricity of vertical loading with

respect to vertical axis of web

4

w

2

cmtE

DM150IofIncrease

Lateral loading on stiffener

4

w

3

cmtE

DV3.0IofIncrease

M – Applied bending moment (kN-m)

D – Overall depth of girder (mm)

E – Young’s modulus (2 105 MPa)

tw – Thickness of web

V–Transverse force taken by stiffener

(kN)

(2) Longitudinal (Horizontal) stiffener:

Longitudinal stiffener increases buckling

resistance of web against bending

These are provided between transverse

stiffener

Minimum moment of inertia of first

longitudinal stiffener provided at 0.2d

from compression flange I ≥ 4c t3

Minimum moment of inertia of second

longitudinal stiffener provided at N.A

I ≥ d2 t3

d2 – 2 distance from compression flange

to N.A

(c) (i)

Sol:

Rolling resistance (neglect crawler weight)

= 45.5 (t) 50 (kg/t) = 2275 kgf

Grade resistance = (36+45.5) 1000 0.04

= 3260 kgf

Total resistance = 2275 + 3260 = 5532kgf

=54.27KN

Power wasted in resistance= Total Resistance

× Velocity of the equipment

= 5532×9.81×30×5/18

=452.24KW =615HP.

(ii)

Sol:

Path Duration

1–2–4 (A– D) 9

1–2–3–4(A–C–E) 18*

1–3–4(B–E) 10

ActivityCost slope =

JT

cc

CN

NC

A (300 – 200) /5–3 = 50

B (500 – 350) /4–3 = 150

C (450 – 300) /7–6 = 150

D (200 – 200) /4–4 = ----

E (400 – 300) /6–5 = 100

1

2

3

4B(4)

A(5)

C(7)

E(6)

D(4)

: 25 : Test – 7


By crashing activity ‘A’ by 2 days, the new

network is as following

Path Duration

1–2–4 (A– D) 7

1–2–3–4(A–C–E) 16*

1–3–4(B–E) 10

By crashing actions ‘E’ by 1 day, the networks is

changed as following

Path Duration

1–2–4 (A– D) 7

1–2–3–4(A–C–E) 15*

1–3–4(B–E) 9

From the network, it is observed that crashing

possible activity is ‘C’; but it cannot be

crashed due to high crash cost slope-(cost

slope of activity ‘C’ is Rs 150)

Optimum project duration = 15 days

Total project cost = normal cost + crashing

cost + Indirect cost/day project duration

= (200 + 350 + 300 + 200 + 300) + crashing

cost of A+ crashing cost of E + Indirect

cost/day project duration.

= 1350 + 250 + 1 100 + 100 15

= 1350 + 100 + 100 + 1500 = 3050/-

1

2

3

4B(4)

A(3)

C(7)

E(5)

D(4)

1

2

3

4B(4)

A(3)

C(7)

E(6)

D(4)



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