CIE_100_intro_civilengg_questionbank.pdf

PART A
CIE 100 - INTRODUCTION TO CIVIL
ENGINEERING
2

UNIT-1
SURVEY AND CONSTRUCTION MATERIALS
2 MARK QUESTION AND ANSWERS
1. Define surveying?
Surveying is the art of a determining the relative positions of points on above or beneath the
surface of the earth by means of direct or indirect measurements of distances direction and
elevation.
2. What are the different types of Surveying?
According to the instruments used the Surveying is classified as follows.
1. Chain Surveying 2. Compass Surveying
3. Theodolite Surveying 4. Plane table Surveying
5. Techometric Surveying
3. Name the two principles of Surveying?
i. To work from whole to part
ii. To fix up new points by at least two independent processes.
4. What is chain surveying?
The method of measuring a distance with a chain or tape is called chaining. In chain
surveying the area is divided into a network of triangles. No angular measurements are taken.
Chain Surveying is suitable when the area to be surveyed is small, fairly level and open and has
simple details.
5. List the suitability of the chain surveying?
i. It is suitable when the ground is fairly level.
ii. It is suitable when the area to be surveyed is comparatively small in extent.
6. Define bearing of a line?
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The horizontal angle between a survey line and a fixed line of reference or meridian is
called bearing of the line.
7. Distinguish between True Meridian and Magnetic Meridian?
The line intersection of the plane passing through a point and the geographical north and
south poles with the surface of the earth is called the ‘True Meridian’ of that point.
Magnetic meridian at a point the direction indicated by a freely floating and balanced magnetic
needle at that point, free from all other attractive forces.
8. Name the two types of compass?
Prismatic compass, Surveyor’s compass.
9. Differentiate the Whole circle bearing and reduced bearing?
Whole circle bearing (W.C.B): The bearing of line is measured with north in clock wise
direction. The value of bearing thus varies from 0 to 360.
Reduced bearing (R.B): The bearing of a line is measured, either from the north or south
clockwise.
10. Define levelling?
Levelling is the art of determining the relative positions of the points in the vertical plane.
11. What are the objects of levelling?
The following are the main objects or uses of levelling.
i. To layout grades.
ii. To determine the relative heights of different points on the surface of the earth.
iii. To find the profile of roads, railways, canals etc.
iv. To get the contours of an area for future planning and designing etc.
12. Name of the various types of levelling?
1. Simple levelling.
2. Differential levelling. (Fly levelling)
13. What are the bench mark and reduced levels?
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Bench marks (B.M): It is fixed point of reference of known elevation. This is used as a
starting point for levelling. The levelling is closed on a bench mark as a check.
Reduced level: Reduced level of a point is its height above or below the datum.
14. Distinguish between back sight (B.S) and Fore sight (F.S)?
Back sight (B.S): This is the first staff reading taken after levelling instrument is step. It is
the staff reading taken on a point of known elevation, such as a bench mark or change point.
Fore sight (F.S): it is the staff reading taken on a point whose elevation is to be determined
The last staff reading in all leveling work before shifting the instrument should be a fore sight.
15. What are the two basic principles of survey?
i) Working from the whole to the part.
ii) Fixing new points by at least two independent process
16. Compare the height of collimation method with rise and fall method
S.No Height of collimation Rise and fall
1
2
3
It is more rapid and saves time and
labour.
There is no check on the R.L of
Intermediate stations.
Error in any of the intermediate sights
are not noticed
It is laborious as the staff reading of each
station is compared to get a rise or fall.
There is complete check on the R.Ls of
intermediate stations.
Error in the intermediate sights are noticed,
these are used for finding out the rises and
falls.
17. State Simpson’s rule for determining areas.
A=d/3[h1+hn + 2(h3+h5+h7+…..hn-2) + 4(h2+h4+---+hn-1)]
18. List any four tasks of a civil Engineering?
(i) Investigation (ii) surveying (iii) Planning (iv) Design (v) Execution (vi) Research
& Development.
19. What are the constituents of bricks?
(i) Alumina (ii) Silica (iii) Lime (iv) Oxide of iron (v) Magnesia
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20. What are the classifications of bricks?
(i) Un burnt (or) sun dried bricks.
(ii) Burnt bricks. a) First class bricks. b) Second class bricks.
c) Third class bricks. d) Fourth class bricks.
21. What are the uses of bricks?
 Bricks are mainly used for the construction of walls
 Bricks are used in the construction of dams and bridges.
 Fire bricks made up of fire clay can be used as a refractory material
 Sand- lime bricks are used for ornamental work.
22. What are the major operations involved in the manufacture of bricks?
1. Preparation of bricks earth. 2. Moulding of bricks
3. Drying of bricks 4. Burning of bricks
23. What are the types of bricks?
(i) Perforated bricks. (ii) Hollow bricks
(iii) Paving bricks. (iv) Sand line bricks
(v) Pressed bricks (vi) Fire (or) Refractory bricks
(vii) Fly ash bricks.
24. What is the function of alumina in a brick and give its percentage?
It is the chief constituent of clay. A good brick should have 20-30% of alumina. This
imparts plasticity to the earth.
25. What is meant by Quarrying?
It’s defined as the process of taking out stones from natural rock beds is known as
Quarrying.
26. Give examples for igneous, sedimentary and metamorphic rocks?
Example: Igneous rocks : Basalt, Granite and Dolerite.
Sedimentary rocks : Sand stone, Lime stone, Gypsum, Gravel etc.
Metamorphic rocks : Slate, Marble, Gneisses.
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27. How are igneous rock formed?
These rocks are formed by the cooling of molten rocky material called magma, which is
inside the earth’s surface.
28. Define dressing of stone?
The surfaces of stones obtained from quarry are rough. The blocks are irregular in shape
and non-uniform in size. Hence their dressing is essential. The process of cutting the stones into
suitable size &with suitable surface is known as dressing of stones.
29. Define the following terms a) Siliceous rock b) Argillaceous rock?
Siliceous rock: In these rocks, silica is the main constituent. Eg: Granite, Quartzite’s etc….
Argillaceous rock: In these rocks, clay is the main constituent. Eg: Lime stone, Marbles etc…
30. List out any four uses of stones?
 In the construction of buildings form the very ancient times.
 For foundations, walls, columns, lintels arches, roofs, floors, damp proof courses, etc...
 For facing works in brick masonry to give a massive appearance.
 Since stones are hard, these can be used for pavements.
31. Define marble and lime stone?
Marble: It is a metamorphic rock .It is very compact and durable stone.
Example: Specific gravity- 2.65., Compressive strength- 720 kg/cm2
.
Lime stone: It is a sedimentary rock. It contains large preparation of calcium carbonate. It is
easy to work.
Example: Specific gravity-2 to 2.75., Compressive strength- 550 kg/cm2
.
32. Define cement?
Cement is obtained by burning at a very high temp a mixture of calcareous and argillaceous
materials. Calcined Product is known as clinker. A small quantity of gypsum is added to the clinker
and is pulverized into very fine powder known as cement.
33. Define Portland cement?
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Cement resembles a variety of sand stone found in Portland in England and is, therefore
called Portland.
34. What are the uses of cement?
a. Cement mortar, a mixture of cement and sand, is used for masonry work,
plastering, pointing and in joints of pipes, drains etc.
b. Cement is the binding material in concrete used for laying floors, roofs and
constructing lintels, beams, weather sheds stairs, pillars, etc.
c. Expensive cements which expands while setting, can be used in repair works of
cracks.
35. What are the types of cement?
1. Quick setting cement.
2. Low heat cement.
3. High alumina cement.
4. Expanding cement.
5. Rapid hardening cement.
6. Acid resistant cement.
7. Sulphate resisting cement.
36. List the composition of an ordinary Portland cement?
There are two main constituents in Portland cement.
(i) Argillaceous materials (ii) Calcareous materials.
In Argillaceous materials, clay is the main ingredient and in calcareous materials, calcium
Carbonate is the main ingredient.
37. Define the white cement?
This cement is white in colour and it is free from colouring ingredients such as iron oxide,
manganese oxide (or) chromium oxide.
(OR)
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The colour of this cement is white and it has the same properties of ordinary Portland
cement. This can be used for architectural purpose and for manufacturing coloured concrete,
flooring tiles etc.
38. Define water cement ratio?
The ratio of weight of water used to that of cement is termed as water cement ratio.
39. What is the need for reinforcement in RCC?
1. It should be capable of resisting tensile, compressive, bending (or) shear
stress.
2. There should be proper cover to reinforcement. So that corrosion of steel is
prevented.
40. What is meant by M15 concrete?
The letter “M’ refers to mix and the number indicates the characteristic compressive
strength of the mix at 28 days expressed in N/mm2
.
41. What are the advantages of RCC?
1. It has good resistance to fire, temperature and weathering actions.
2. RCC construction is easy and fast.
3. RCC is tough and durable.
42. What are the types of light weight concrete and polymer concrete?
Light weight concrete.
1. Light weight aggregate concrete.
2. Aerated concrete.
3. No- fine concrete.
Polymer concrete.
1. Polymer impregnated concrete.
2. Polymer cement concrete.
3. Polymer concrete.
4. Partially impregnated and surface coated polymer concrete.
43. What are the several ways in which aerated concrete is manufactured?
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There are several ways in which aerated concrete can be manufactured. One important way
is by the formation of gas (or) air bubbles using finely powdered material. (Usually aluminum
powder). Chemical reaction takes place in the concrete and finally large Quantity of hydrogen gas
is liberated which gives the cellular structure.
44. What do you mean by no- fine concrete?
Omitting sand aggregate from the concrete is called no- fine concrete.
45. What is high density concrete and where is it mainly used?
The concrete whose unit weight range from about 3360- 3840 kg/m3
and which is about
50% higher than the unit weight of normal concrete is known as high density concrete.
46. Define curing of concrete?
The finished concrete surface should be kept wet for at least 7 days to promote continued
hydration of cement. This is called curing of concrete.
47. Draw stress- strain diagram for mild steel?
 a to c is the elastic stage
 d to e is the yield stage
 f to g ductile stage
 g to h plastic yielding stage
 h is the ultimate stage
 i is the breaking point
48. What is expanded metal and where it is used?
The material is formed by cutting and expanding either plain sheets (or) ribbed sheet of
steel
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Uses: It is used as ferro cement reinforcement for concrete, plaster, pavement formation
and as partition.
49. What are the uses of channel and T-sections?
Channel: It used as structural members of the steel-framed structures.These are used in the
construction of built-in columns, crane girders, beams and steel bridges.
T – Sections: It’s widely used as members of steel roof trusses and to form built up sections.
These are also used in T-connections in steel water tanks, steel chimneys steel bridges, etc.
50. Draw the stress strain curve for the Tor steel?
51. What is the main advantage of using Tor steel?
1. It has 65% greater yield strength.
2. It has 100% greater bond strength.
3. It has highest fatigue strength.
4. It gives lesser crack width.
52. Write down the chemical composition of tor steel and what are the grades of tor steel?
Chemical composition: carbon-0.3%, sulphur-0.055%, phosphorous-0.055%.
Grades: Tor to with yield strength of 415 N/mm2
, Tor 50 with yield strength of 500 N/mm2
.
54. State the various types of steel?
Depending upon the carbon content there are three verities of steel.
 Mild steel (or) low carbon steel.
 Medium hard (or) medium carbon steel.
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 Hard steel (or) high carbon steel
1. Briefly explain the classification of survey.
Classification of Surveys
Depending on the use and purpose of the finished work, surveys are classified under the
following heads.
1. Classification based upon the nature of filed
a) Land survey b) marine (or) Navigation c) As tronomical
Surveys survey
2. Classification based upon the objective of survey
a) Engineering survey b) military or defence survey
[These are carried out for the determinations [These are carried out for the
Of quantities which is used for designing preparation of maps of important
Works.] Military areas.]
c) Geological survey d) mine survey e) archeological
[Survey carried out the [carried out exploring the [prepare maps of
Composition of Earth mineral wealth below the ancient culture.]
crust] ground.]
3. Classification based on methods employed
a) Triangulation survey b) Traverse survey
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4. Classification based on the instrument used
a) Chain b) Compass c) Plane d) Theodolite e) Theometric f) Aerial g)photo-
surveying. Surveying. Table Surveying. Surveying. Surveying. -graphic
Surveying. Surveying.
2. Explain the principle of surveying?
All surveys are based on two fundamental principals. They are
 Working form whole to part
 Fixing a point with reference to two fixed points.
Working from the whole to part
Whether it is a plane survey or a geodetic survey, the main principle adopted is to work
from whole to the part. In the case of surveying of extensive areas, such as a town or a big estate,
the survey is started from control points with high precision, the line joining these points will form
the boundary lines of the area, and otherwise, this is the main Skelton of the survey. Control points
may be established by triangulation or by running a traverse surveying the area.
Main purpose or reason to work from the whole to part is to avoid the accumulation of
errors and control any localized errors.
Fixing a point with reference to two fixed points.
Suppose point ‘A’ and ‘B’ are known on the ground and the distance between them is
measured.
Let it be measured to locate or mark a point ‘C’ the relative position of the point C is
located with reference to the two fixed points A and B by one of the following methods.
a) Linear measurement
b) Angular measurement
c) Both linear and angular measurement
Eg fixing of point.
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A and B are control point C is the New point
a) Linear measurement
The distance AC and BC may be measured and the position of C may be fixed by drawing
the arcs.
b) Angular measurement
By dropping a perpendicular from C to the base line AB. Here the distance CD and 90°
angle of intersection are the two different measurement made to locate ‘C’.
3. What are the instruments used for chain surveying. Explain briefly with neat sketches.
Various instrument needed for chain surveying
1. Chain (20m or 30m length)
2. Arrows or Marking pin (10 nos for one chain)
3. Pegs are used for marking the positions of stations
4. Ranging rods are used for marking the position of stations and for ranging the line
5. Offset rods are used to align offset line
6. Cross-Staffs are used to set out right angles.
1. Chain: A surveying chain is a device used to measure distance between two points on the
ground
The chain generally consists of 100 links for 20m chain & 150 links 30m chains.
Each link having =20cm length (or) 0.2m length
The links are formed y pieces of galvanized mild steel wire 4mm in diameter & the chain consists
of brass handle with brass eyebolt & collar and wire rings are provided at even 1m length & brass
tallies are provided at even 5m length.
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2. Arrows or Marking pin:Arrows are used with chain for marking ends chain length on the
ground. The arrow is driven into the ground at the end of each chain length is measured, the
diameter of wire and lengths are shown in fig.
3. Pegs: Wooden pegs of 15cm length and 3cm square in section are used to measuring station
points. They are driven in the ground with help of a wooden hammer and kept 4cm projecting
above the ground surface.
4. Ranging rods: It is used for ranging (2 or 3m) or aligning long lines on the ground in field
surveying. It is made by timber or mild steel of 30mm dia & 2 or 3m length. The ranging rod is
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painted rod & white or black & white in alternate bond length of 200mm each.
Ranging rod offset rod
5. Offset: Offset rod is similar to ranging rod, with two short, narrow, vertical sighting slots passing
through the centre of section. A hook is fitted or a groove is cut at the top of enable pulling or
pushing of the chain through obstruction like hedges. Offset rods are meant setting out lines
approximately at right angles to the main line.
6. Cross-Staffs: This is the instrument used for setting out right angles to a chain line. It consists of
four metal arms with vertical slits marked on a pole.
Two opposite slits are positioned along the length of a line (main line). A line perpendicular to the
main line is formed or sighted through the other two slits.
The common forms of cross staff are
a) Open cross staff
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b) French cross staff
Adjustable cross staff
7. Plumb bop: Plumb bop consists of a solid conical piece and a string attached to it at its centre.
When in use, the solid piece is at the bottom. It is used to test the verticality of the ranging rod and
to transfer the points to the ground. Plumb bop is used while chain surveying on sloping ground.
4. Explain in detail chain triangulation or principle of chain surveying?
The principle of chain surveying is to divided the area into a no of triangles of suitable
sides, chain surveying is simplest kind of surveying is simplest kind of surveying, in this case,
there is no need for measuring angles.
Chain surveying is some times called chain triangulations.
The triangle said to be well conditional or well proportioned when it contains no angle smaller than
30° and no angle greater than 120°
Total Area ‘A’= A1+A2+A3+A4
a. Linear Measurement
b. Angular measurement
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c. Both linear and Angular Measurement.
Suitability & unsuitability of chain surveying
 Suitability (Advantages)
1. Where the ground is fairly levels and offset with simple details
2. When large scale plans are needed.
3. Does not required costly equipment
 Unsuitability (Disadvantages)
1. Unsuitable for crowded area & large area
2. Unsuitable for jungle area crowded area
3. Unsuitable for undulating area
4. Chain surveying is not always assurance
5. The following perpendicular offset was taken at 10m intervals from a survey line to an
irregular boundary line.
4.2m, 4.3m, 6.50m, 5.6m, 6.85, 4.2m, 8.2m, 5.6m, 4.3m.
Calculate the area enclosed between the survey line, the irregular boundary line, and first
and last offsets, by the application of
a. Average ordinate rule
b. Trapezoidal rule
c. Simpson’s rule.
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Solution
n = no of division = 8
d = 10m (interval between offsets)
L= length of base line.
a. Average ordinate rule
A= [(h0+h1+h2+h3+------+hn) / n+1] x2
A= [(4.2+4.3+6.5+5.6+6.85+4.2+5.2+5.6+4.3) / (8+1)] x 80
A= 442.222m2
.
b. Trapezoidal rule
A= [(h0+hn/ 2) + h1+h2 +-----+ hn-1] d
d = 10m
A= [(4.2+4.3/2)+ 4.3+6.5+5.6+6.85+4.2+8.2+5.60] 10
A=455.0m2
C. Simpson’s rule.
A= [h0+hn+2(h1+h3+h5+h7)+4(h2+h4+h6)]xd/3
A= [ (4.2+4.3)+ 2( 4.3+5.6+4.2+5.6)+4(6.5+6.85+8.2)] x 10/3
A= 434.67m2
6. The following offsets in meters were measured from a survey line to an uneven boundary at
interval of 10m.
0.00, 1.20, 5.40, 6.00, 4.21, 3.88, 6.20, 0.00 compute the area
a) Trapezoidal rule
b) Simpson’s rule.
Here, the survey line meets the boundary both ends. That is the reason why the ordinates
at the two ends of the survey line happen to be zero. However, these end ordinates with
zero length also should be included in the formula.
a. Trapezoidal rule
A= [(0+0/2)+ (3.2+5.4+6+4.21+3.88+6.2)] 10
A=288.90m2
b. Simpson’s rule.
Since the no of ordinates is even 8, Simpson’s rule cannot used straight way, however, the
area up to the seventh ordinate can be calculated by using Simpson’s rule. The area between the
seventh and eighth ordinates may be calculated by trapezoidal rule.
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A1= [(0+6.2)+2(5.4+4.21)+4(3.2+6+3.88)] 10/3
A1= 259.13m2
A2= (6.2+0/2) 10
= 31.00m2
A=A1+A2
A=259.13+31
A=290.13m2
7. The following offsets were measured from a chain line to the boundary of a plot.
Distance in ‘M’ 0 10 20 30 40 60 80
Offset in ‘M ‘ 5.10 4.55 6.10 8.22 5.31 6.58 3.82
Find the area enclosed between the boundary, chain line and the end offsets.
Solution
The interval between the offset is not the same throughout the chain line, hence the chain
line is divided into sectional having the same interval. In this case, the chain line is divided into
two sections.
Up to 5th
ordinates is common interval of 10m will form first section
The second section being the plot between 5th
and 7th
offsets with interval of 20m.
The area of the two sections is calculated separately. The sum of two areas will give the total
required area.
Trapezoidal method
A1= 10/2 [ (5.1+5.31/2) + (4.53+ 6.10+8.22)] = 240.75m2
A2=20/2[((5.31+3.82)/2) + 6.58] = 222.90m2
A=A1+A2 = 240.75+ 222.90 = 463.65m2
A=0.0464 hectare [1000m2
= 1 hectare]
Simpson’s rule
A1=10/3 [5.10+5.31+2(6.1) +4(4.55+8.22)] = 245.63 m2
A2=20/3 [5.31+3.82+4(6.58)] [No odd ordinates]
= 236.33 m2
A=A1+A2 = 245.63+236.33 = 481.96 m2
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8. What are the types of compass? Explain the component of prismatic compass with sketch.
There are two types of compass survey
1. Surveyors compass
2. Prismatic compass
1. Prismatic compass
A magnetic needle is balanced over a pivot in a circular box of 85mm to 110mm in diameter. A
graduated aluminum ring is attached to the magnetic needle.
The box is covered by a glass lid. Object vane and eye vane are provided at diametrically
opposite ends. Eye vane carriers a reflecting prism which can be raised or levered as desired. A
vertical horse hair or fine wire is provided at the middle of the object vane. The graduations in the
aluminum ring are made in clockwise direction 0° at north & 180° at south marking. A triangular
prism magnification of readings to suit observer’s eye. Based on this arrangement, the compass is
named prismatic compass
Compass is fixed over a tripod with ball and socket arrangement.
9. Explain the principle of compass survey with suitable sketches.
Wherever a number of base lines are to be run for obtaining the details as in traversing, linear
measurements made by chain surveying will not be sufficient, the angles
Eg:
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1. Closed traverse - Uses: boundaries measurement, lakes, pond, wooden
land
2. Open traverse(does not form polygon)
Included between the adjacent lines should also be measured compass is one of the instruments
used to measure angles. This instrument will be supported by a tripod stand while supported by a
tripod stand while taking observation.
10. Convert the following W.C.B of lines to Quadrantal bearings,
a) OA = 54°15’ b) OB = 128°32’
c) OC = 245°12’ d) OD = 313°45’
11. Convert the following RB bearing of lines to W.C.B bearings?
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a) OA = N 48°15’ E b) OB = S 18°30’ E
c) OC = S 32°45’W d) OD = N 68°50’W
13. Find the included angle between the line PA, PB and PC given their bearings
PA = 22°45’ PB = 100°00’ PC = 162°30’
14. Explain briefly the dumpy level with sketches.
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A leveling instrument (Dumpy level)
The simplest and easy operation from of leveling instrument is dumpy level.
The different parts of a instrument is
a) Telescope
b) Eye-piece
c) Focusing knob
d) Level tube
e) Foot screw
f) Levelling read.
g) Diaphragm etc.
The dumpy level is light, stable and sufficiently accurate. The telescope is rigidly fixed with
support and therefore, can neither be rotated about the longitudinal axis, nor can it be removed
from its support. A long bubble tube is attached to the top of the telescope. The levelling head
generally consists of two parallel plates with either three foot screws of four food screws. The
upper plate is known as tribranch and the lower plate is known as trivet which can be screwed on a
tripod.
Dumpy level
15. Briefly explain the fly leveling.
Differential leveling (fly leveling)
Differential leveling is used in order to find the difference in elevation between two points
which are too for apart or if there are any obstacles between them or if the difference in elevation is
high then differential leveling is adopted. This is a simple leveling adopted in successive stages.
Hence, it is also called compound or continuous levelling.
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(Differential levelling)
Let A and E be two points whose difference in elevation is necessary. The staff reading at
‘A’ is noted as ‘a’ form station point O1, After adjusting the bubble, the staff reading at a firm point
‘B’ is noted form O1 as ‘b1’is foresight.
‘B’ is selected such that AO1 is approximated equal to O1B. Now the instrument is shifted to O2
and staff reading at B, form O2is taken and noted as b2. Another firm point C selected and the
procedure is repeated till the point E is reached. The difference in level between A and B is (a-b1).
The difference in level between B and C is (b2-c1) and so else.
The difference in level between A and E are the algebraic sum of these differences.
Diff in
Level bw A & E = (a-b1) + (b2-c1) + (c2-d1) + (d2-e1)
16. Explain the methods of reduction of levels.
These two methods for calculation the RL
1) Height of collimation (or) Height of instrument method
2) Rise and fall method
1. Height of collimation
In this method first of all the height of the instrument is calculated (HI). This is done by
adding the back sight to the elevation of BM, and the back sight is taken by keeping the instrument
at the first station. Now the fore sight also observed form the same instrument station by keeping
the staff at point.
R.L of BM + Back sight - FS
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Now the instrument is shifted to second station. The BS is taken by keeping the staff on
point ‘A’and FS is taken by keeping the staff at point ‘B’.
This procedure is repeated for all points for which RL are to be calculated.
The following gives clear idea
Eg RL at point ‘A’= RL of Bm+ Bs on Bm – FS on point A
= RL of Bm + h1-h2
Eg RL at point ‘B’= RL of point A+ Bs on A – FS on B
= RL of point A + h3-h4
Check
έBS – έFS = Last RL – First RL
Table
B.S I.S F.S Height of instrument(HI) R.L Remarks
2. Rise and fall method
The difference of level between two consecutive points for each setting of the instrument is
obtained by comparing their staff readings. A rise is indicated if the BS reading is greater then the
FS and a fall is shown if it is less than the FS
Rise: If the point is above the preceding point, the difference between the readings is known as
rise (+ve).
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Fall: It the point is blow the preceding point, then the difference between the reading is termed as
fall (-ve).
Arithmetical check
B.S F.S Rise Fall R.L Remarks
17. The following staffs reading were taken with a level, the instrument having been moved
after third and seventh readings.
2.350, 1.625, 0.700, 2.900, 1.955, 1.400, 0.900, 0.600, 1.500, 1.150
Assume BM = 100.00m, to find reduced level
(i) Height of collimation method
(ii) Rise and fall method
1. Height of Collimation
έBS – έFS = Last RL – First RL
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Instrument
station
Staff station B.S I.S F.S RL
S1
S2
S3
A
B
C
D
E
F
G
H
2.650
2.900
0.600
1.625
1.955
1.400
1.500
0.700
0.900
1.150
100.000
100.725
101.650
102.595
103.150
103.650
102.750
103.100
T.K B.S=5.850 F.S=2.750 Last RL -
First RL
= 3.100
1) R.L of height of instrument = At instrument station S1 = R.L of BM + BS
= 100+2.350
= 102.350
2) R.L of Intermediate points = R.L of H.I at S1 – I.S (or) F.S
R.L of B = 102.350- 1.625 = 100.705
R.L of C = 102.350 – 0.700 = 101.650
3) R.L of H.I of instrument station S2 = R.L of C + B.S
= 101.650 + 2.900 = 104.550
R.L of D = H.I at S2 – I.S = 104.550- 1.955
= 102.955
R.L of E= 104.55 – 1.400 = 103.150
R.L of F = 104.55 – 0.900 = 103.650
4) R.L of H.I instrument station S3 = R.L of F +BS
= 103.650 + 0.600
= 104.250
R.L of G = 104.250 – 1.50 = 102.750
R.L of H = 104.250 – 1.150 = 103.150
Check έB.S - έF.S = 5.850 – 2.750 = 3.150
Last R.L – First R.L = 103.150- 100 = 3.150
3.100 = 3.10
28

3) Rise and Fall method
Staff
station
B.S I.S F.S Rise Fall R.L Remarks
A
B
C
D
E
F
G
H
2.350
2.900
0.600
1.625
1.955
1.400
1.500
0.700
0.900
0.350
0.725
0.925
0.945
0.555
0.500
0.350
0.900
100.00
100.725
101.650
102.545
103.150
103.650
102.750
103.100
BM
CP 1
CP2
Sum of
B.S= 5.85
Sum of
F.S = 2.75
Sum of
Rise =
ssss4.00
Sum of
Fall =
0.900
Difference of level between consecutive points:
A-B = 2.350 - 1.625 = 0.725 (Rise)
B-C = 1.625 - 0.750 = 0.920 (Rise)
C-D = 2.900 - 1.955 = 0.945 (Rise)
D-E = 1.955 - 1.400= 0.555 (Rise)
E-F = 1.400 –0.900 = 0.500 (Rise)
F-G = 0.600 – 1.500 = 0.900 (Rise)
G-H = 1.500 – 1.150 = 0.350 (Rise)
[ + Rise , - Fall ]
R.L of
Station A is BM = 100
R.L of B = 100 + 0.725 = 100.725
C = 100.725 + 0.925 = 101.650
And so on
Check
[έB.S - έF.S = έ Rise - έFall]
[ 3.10 = 3.10 ] Hence ok
18. a) Explain the constituents (or) composition of bricks?
29

Constituents percentage uses
Alumina 20-30% Impact plasticity of earth
Silica 50-60% Prevent cracking
Lime Up to 50 % Prevent shrinkage if the percentage of lime is increased, it will
give irregular shape.
Oxide of iron 5-6% Give red colour to brick
Magnesia Less than 1% Impact yellow tin to brick and reduce shrinkage
b) Explain the types of bricks?
a. Hallow bricks
b. Fly ash bricks
c. Fire bricks
d. Perforated bricks
e. Paving bricks
f. Pressed brick
19. What is the requirement (quality) and uses for a stone which is to be used as a building
material?
Qualities of good stone:
 The pressing strength of the stone should be greater than 100 N/mm2
 The colour of the stone should not be easily attacked by weathering agent.
 A good building stone must be durable.
 It should not be easily affected by temperature, rain, sunlight, wind, etc.
 When the stone is immersed in water for about 24 hours the percentage of absorption of
water should not exceed 0.6%.
Uses of stone:
 Stones are uses in construction of foundation, wall, column and lintel.
 In the construction of building from the very ancient times.
 Stones are also uses as ballast for railway track.
 Stones are uses in the construction of bridge, pier, dams, abutments, retaining walls.
20. Explain the classification of Rocks?
Rocks are classified into three categories. They are,
1 Igneous rocks
30

2 Sedimentary rocks
3 Metamorphic rocks
Igneous rocks: Igneous rocks are formed by cooling of the molten material below the earth
surface. Stones of these rocks are said to be harden.
Example: granite, basalt, dolerite, etc,
Sedimentary rocks:
Sedimentary rocks are formed by deposition of weathering product on the existing rocks.
The deposits are in layer and when the load is applied along the layer, these rocks easily split.
Example: sand stone, limestone, gravel.
Metamorphic rocks:
Metamorphic rocks are formed by changes in the characteristics of igneous and the
sedimentary rocks. These will be hard if the basic rock is an igneous rock.
Example: marble, gneisses.
21. Explain properties and uses of cement?
Properties of cement
1. The colour should be uniform.
2. It should be free from lumps.
3. Cement should be uniform when touched.
4. It should be cool when felt with the hand.
5. If a small quantity of cement is thrown into a bucket of water, it should sink.
6. In cement, the ratio of percentage of alumina to that of iron oxide should not be less
0.66%.
7. Cement mortar at the age of three days should have a compressive strength of 11.5
N/mm2
and tensile strength of 2 N/mm2
also at the age of seven days the compressive
strength should not be less than 2.5 N/mm2
8. The total sulphur content of cement should not be less than 2.75%
9. The initial setting time of cement should not be less than 30 minutes and the finial
setting time shall be allowed to 10 hours.
Uses of cement:
1. Cement is used for constructing engineering structures, where great strength is required,
such as dams, bridges, storages reservoirs etc.
31

2. Cement is used for making joints for pipes, drains etc.
3. Cement is used for preparation of foundation, foot paths etc.
4. Cement is used for manufacture of precast pipes, piles etc.
22. a) Describe the various types of cement?
Types of cement:
There are two types of cement. They are,
1. Portland cement
2. Special cement
1. Portland cement:
1. Normal cement (or) ordinary Portland cement
2. Rapid hardening (or) high early setting
3. Low heat cement
2. Special cement:
 Quick setting cement
 High alumina cement
 Sulphate resisting cement
 Expanding cement
 White cement
 Colour cement
22.) b. With the help of a flow diagram briefly explain (or) list the main operation involved in
the manufacture of ordinary Portland cement?
There are three main operations involved in the manufacturing of ordinary Portland cement. They
are
 mixing of raw material
 Burning
 Grinding
Mixing of raw materials:
There are two methods in the mixing of raw materials of cement.
32

 Dry process
 Wet process
Dry process:
In this method, calcareous material such as limestone and argillaceous material such as clay
are separately reduced in size about 25 mm in crusher. After drying these materials are grinned in
boll mill or in a tube mill. This powder is then stored in hopper. Then they are mixed in correct
proportion. This raw material is then stored in the storage tank.
Wet process:
In wet process calcareous material such as limestone are crushed and stored in a storage tank.
Argillaceous material such as clay are washed and stored in basin. Now, crushed limestone from
the storage tank and wet clay from the basin are allowed to fall in a channel in correct proportion.
This material is grinned in ball mill or in a tube mill to form slurry. The chemical test of slurry
stored in slurry in storage basin, Pumping tool portion of rotary kiln.
Burning:
The burning of dry mixture or fine slurry is carried out in a long rotary kiln. The diameter
of kiln varies from 250 cm to 300 cm and its length varies from 90-120 m. It is laid with a slight
inclination of 1 in 20 to 1 in 30 towards the discharge end. The kiln is supported on roller such that
it can rotate about its longitudinal axis at the rate of one revolution per minute. Refractory lime is
provided on the inside surface of the rotary kiln. From a storage tank, the correct slurry is injected
at the upper end of the kiln. The hot gases or flames are forced through the lower end of the kiln.
The portion of the kiln near its upper zone is known as dry zone, and in this zone, the water from
slurry is evaporated. As the dried slurry descends towards the burning zone, carbon dioxide from
the slurry is evaporated. It is converted into the small lumps is called nodules. The nodules are
gradually role down and ultimately reach the burning zone when the temperature is about 1500 to
17000
c.
33

The size of clinker varies from 5 to 10mm. the hot clinker is collected in the container of suitable
size.
Grinding:
The grinding of the clinker is done in a boll mill or in tube mill. During grinding, a small
quantity of about 3-4% of gypsum is added. Gypsum controls the initial setting time of the cement.
If the gypsum were not added, cement would set as soon as the water is added. Finally, ground
cement is stored in a storage tank. It is then weighted and packed in bags. Each bag of cement
contains 50 kg of cement. These bags are carefully stored in dry place.
UNIT-2
BUILDING COMPONUNTS AND STRUCTURES - FOUNDATION
1. What are the functions of foundation?
1. Distribution of load. 2. Minimization of differential settlement.
3. Safety against sliding & overturning. 4. Provisions of level surface.
2. Distinguish: Shallow foundation and deep foundation.
When the depth of the foundation is less than (or) equal to its width, It is defined as a
shallow foundation. When the depth is more than the width, it ii termed as a deep foundation.
34

3. List out the types of shallow foundation:
Shallow foundation:
1. Isolated footing.
2. Wall footing.
3. Combined footing.
4. Cantilever footing.
5. Continuous footing.
6. Inverted arch footing.
7. Grillage foundation.
8. Raft (or) mat foundation.
Deep foundation:
1. Pile foundation. 2. Pier foundation. 3. Well foundation.
4. List the types of machine foundation?
Machine foundation can be classified into 3 categories.
1. Block foundation usually adopted for reciprocating machines and light rotary machines.
2. Block and trough foundation for impact machines, such as drop and forge hammer.
3. Frame foundation generally adopted for heavy rotary machines, such as turbo generators.
5. Define safe bearing capacity?
The load that can be safely applied on the soil is called the safe bearing capacity. It is
obtained by dividing the ultimate bearing capacity of soil by a factory of safety.
6. What are the essential requirements of a good foundation?
 The foundation should be so located that it is able to resist any unexpected future
influence which may adversely affect its performance.
 The foundation should be stable (or) safe against any possible failure.
 The foundation should not settle (or) deflect to such an extent that will impair its
usefulness.
7. What is foundation?
35

Foundations are the lowest artificially prepared of the structure which are in direct contact
with the ground and which transmit the loads of superstructure to the ground.
8. What are the different types of loads acting on the foundation of a building?
1. Live load 2. Dead load 3. Wind load
9. What is mean by bonding of bricks?
A bond is an arrangement of layers of bricks by which no continuous vertical joints are
formed. Bricks can be arranged in various forms.
10. What is masonry?
Masonry may be defined as the construction of building units bonded together with mortar.
11. Classify stone masonry based on its construction?
1. Rubble masonry.
2. Ashlar masonry.
1. Rubble masonry: a. Random rubble e masonry. (i) Coarsed (ii) Un coarsed.
b. Square Rubble masonry. --- (i) Coarsed (ii) Un coarsed
c. Polygonal Rubble masonry
d. Filnt rubble e masonry
e. Dry rubble e masonry
2. Ashlar masonry a. Ashlar fine.
b. Ashlar rough tooled.
c. Ashlar rock (or) Quarry faced.
d. Ashlar chamfered.
e. Ashlar block in course.
12. Give a list of bonds in brick work?
a. Stretcher bond. b. Header bond. c. English bond. d. Flemish bond.
13. Explain with neat sketch the English bond for one brick thick wall?
It is mostly commonly used bond for all wall thickness. This bond is considered to be the
strongest. Following are the feature of an English bond.
36

1.) This bond consists of alternate courses of header and stretchers.
2.) The queen closer is placed next to the quoin header to break the continuity of the vertical
joints.
3.) The alternate courses consist of header and stretchers.
4.) Continuous vertical joints are not formed.
5.) If the thickness of the wall is an odd number of half brick, the same course will present
stretchers on one faced header on the other.
14. Explain with neat sketch the Stretcher bond for one brick thick wall?
All the bricks are arranged in Stretcher courses. The Stretcher bond is useful for one brick
partition walls as there are no headers in such walls. As the internal bond is not proper this is not
used for wall of thickness greater than one brick.
15. Explain with neat sketch the header bond for one brick thick wall?
In this type all the bricks are arranged in header courses. It is used in curved surface since
the length will be less.
16. What are the causes of cracks in a brick masonry wall?
(i) Combining the brick work with other materials having greater deflection and strains.
(ii) Effect of deflection and shrinkage of the concrete slabs resting on walls.
37

(iii) Development of internal forces due to moisture absorption, temperature variations, etc.
17. Write down the remedial measures to prevent cracks in brick masonry wall?
(i) The foundation supporting masonry walls should be designed with sufficient stiffness.
(ii) The provision of horizontal & vertical expansion joints in the walls helps in reducing
the occurrence of cracks.
(iii) The usage of concrete with low shrinkage characteristic also prevents cracking.
(iv) It is preferable to have short spans for floor slabs.
18. What is a beam?
Beams are defined as horizontal load carrying members in a structure.
19. How are beams classified?
Depending upon the support, the beams are classified as follows:
(i) Simply supported beam. ii) Rigidly fixed beam. (iii) Cantilever beam. (iv) Over
hanging beam. v) Continuous beam.
20. What is lintel? Where do you provide it in a building?
Lintel is a horizontal member which is placed across the openings; openings are invariably
left in the wall for the provision of doors, windows, cupboards, almirahs etc
21. What is a column?
The vertical load carrying members of a structure is called column.
22. Differentiate short columns from long column?
If leff / a > 12 it is a long columns.
If leff / a < 12 it is a short columns.
leff = Effective length of column which depends on the conditions of the end support.
a = Least lateral dimensions of the column. or diameter of column, in case of circular
columns.
23. What is meant by built-up column?
38

Columns can also be made of timber and masonry, which are not often used. When two or
three steel members are connected to form a column then it is called as box or built-up column.
24. What are the factors to be considered while selecting a suitable roofing material?
A roof is the uppermost part of a building which is supported member and covered with
roofing materials to give protection to the building against rain, wind, heat, snow etc..
25. What is a truss?
Truss is a pin jointed frame made of axially loaded members
26. How are the roofs classified in general?
a) Single roofs. b) Double (or) Purlin roofs. c) Trussed roofs.
27. What are the advantages of granolithic flooring?
1. It is provided a hand surface for the floors .
2. To improve the wearing qualities, sand should be replaced by fine aggregate of crushed
granite.
28. What is stone flooring provided and why?
Stone flooring provides a hard, durable and wears resisting floor surface and hence it can be
used for go downs, stores, workshops, etc. As it dose not give a pleasing appearance it is not used
for in residential and public building. It can be easily constructed & maintained
29. What are the components of industrial flooring?
a) Base concrete. b) Wearing surface.
30. How can the method of laying cement concrete flooring be broad divided?
The concrete flooring consists of a top layer of cement concrete 25- 40mm thick laid in a
100-150mm base of suitable mix of concrete.
31. What are the disadvantages of industrial flooring?
 It cannot be satisfactorily repaired by patch work.
39

 Defects in carelessly made floor cannot be rectified and as such, it requires proper
attention while laying.
32. What are the types of asphalt flooring?
 Asphalt martic flooring.
 Asphalt tiles flooring.
 Asphalt tarroza flooring.
33. What is damp proofing? What are the materials used for it?
Damp proofing is the method adopted to prevent the entry of dampness into a building, so
as keep them dry habitable and safe.
Materials: Hot bitumen. 2. Matric asphalt 3. Bituminous felts. 4. Metal sheets of lead, copper
& aluminum
34. Define plastering?
Plastering is a process of covering rough surface of walls and with a thin plastic material
called plaster.
35. Types of plastering?
1. Lime plastering 2. Cement plasters 3. Mud plasters plasters 4. Water proof plasters
36. Define load and stress?
Load: - The external force acting on the body is called the load.
Stress: - It is defined internal force per unit area of cross section is called stress.
37. How are loads classified according to the manner of application?
Static loads are those which are applied gradually from zero to their value. Such loads
produce practically no vibration in the structure.
38. What are the types of strains?
i) Tensile strain. ii) Compressive strain. iii) Shear strain.
39. What is meant by volumetric strain?
Volumetric strain = change in volume/original volume
40

40. What is meant by superficial strain?
Superficial strain = change in area / original area
41. Define strain?
Strain= change in length / original length.
42. Define shear strain?
The ratio of the transverse displacement to the distance from the fixed face is called shear
strain.
43. Define young’s modulus?
The ratio of the axial stress to the corresponding axial strain, with in the elastic limit is
called young’s modulus (or) modulus of elasticity. It is denoted by ‘E’
44. Define Hooke’s law?
Hooke’s law states that when an elastic material is stressed with in elastic limit, the stress is
proportional to the strain. That is the ratio of axial stress to the corresponding axial stress to the
corresponding axial strain is constant.
Stress Q strain ---- Stress = k x strain , Stress / strain = k; k is the constant.
45. Define modulus of rigidity (or) shear modulus?
Modulus of rigidity = shear stress / shear strain.
Bulk modulus of elasticity:
It is denoted by ‘k’
K= Directed stress / volumetric strain.
46. Define Poisson’s ratio?
It is denoted by ‘1/m’
Poisson’s ratio = lateral strain / longitudinal strain.
The value of Poisson’s ratio for elastic materials usually lies between 0.25 to 0.33, and in no case
exceed 0.5.
47. Define factor of safety?
Factor of safety = Ultimate stress / Allowable stress.
41

48. Give the relation ship between modulus of elasticity (E), modulus of Rigidity (N), bulk
modulus (K)?
i)E = 2G (1+ 1/m) ii .E = 3k (1- 2/m) iii.E = 9kG / 3k + G
E is the young’s modulus (or) modulus of elasticity.
G is the shear modulus (or) modulus of rigidity.
K is the bulk modulus.
1/m is the Poisson’s ratio.
49. Modular ratio?
The ratio E1 / E2 is called the modular ratio of the first material to the second.
E1 = young’s modulus of bar 1.
E2 = young’s modulus of bar 2.
50. Different between gravity dams and Earth dam?
Gravity can be defined as a structural which is designed in such a way that its own
weight resists external forces. This types of dam more durable and has maximum rigidity.
Earth dam are built area where the foundation is not strong enough to bear the weight of a
gravity dam.
51. List out the various types of dams?
i) Rigid dam ii) Non- rigid dam
1. Draw sketches of two types of shallow foundations and two types of Deep foundations and
Labels the parts and indicate the materials used.
Types of foundations:
Foundation may be broadly classified into following two categories.
1. Shallow foundation
2. Deep foundation
42

Shallow foundations are those in which the depth is equal to or less than its width. When the
depth is more than the width, it is termed as a deep foundation.
1. Shallow foundation
The various types of shallow foundations are
 Isolated (or) Column footing.
 Wall footing
 Combined footing
 Cantilever footing
 Continuous footing
 Inverted arch footing
 Grillage foundation
 Raft or Mat foundation
 Stepped foundation
I. Isolated (or) Column footing:
Isolated footing is provided under column to transfer the load safely to the soil bed. If the
column is loaded lightly a spread is given under the base of the column. This spread is known as
footing.
43

II. Wall footing: If the footing is provided through out the length of the wall in the case of load
bearing walls then it is called wall footing. Wall footings can be either simple or stepped.
Depth of footing: The minimum depth of footing is given by Rankine’s formula as
D = P / W { 1- sinØ / 1 + sin Ø}2
Where
D = Minimum depth of footing
P = Safe bearing capacity of soil in Kg / m2
W = Unit weight of soil in Kg / m3
Ø = Angle of repose of soil in degrees.
The minimum depth of footing for the load bearing wall is limited to 90cm for the stability.
Width of footing: It is obtained by dividing the load including dead load, live load and wind
load etc. by the allowable bearing capacity of the soil.
B = T / P
Where
B = Width of footing in metre.
T = Total load per metre run in Kg.
P = Safe bearing capacity of the soil in Kg / m2
Deep Foundation
Deep foundation consists of pile and pier foundations .Pier foundations are rarely used for
buildings. This consists in carrying down through the soil a huge masonry cylinder which may be
supported on solid rock.
44

1. Pile Foundation:
Pile is an element of construction used as foundation. it may be driven in the ground
vertically or with some inclination to transfer the load safely. Loads are supported by in two ways,
i.e. either by the effect of friction between the soil and the pile skin or by resting the pile on a very
hard stratum. Former is called friction pile and the letter one is load bearing pile.
Friction piles may be made of cast iron, cement concrete, timber, steel, wrought iron and
composite materials. Load bearing piles are steel sheet piles, concrete piles and timber piles. Piles
may be cast-in-situ or precast. They may be cased or uncased.
2. Under –reamed piles
Structures built on expansive soil often crack due to the differential movement caused by
the alternate swelling and shrinking of soil .under –reamed piles provide a satisfactory solution to
the above problem.
The principle of this type of foundation is to transfer the load to hard strata which have
sufficient bearing capacity to take the load.
Single and double under –reamed piles may also be provided for foundations of structures
in poor soils overlaying firm soil strata .in such soils if double under –reamed piles are provided
both the under –reamed shall rest within the firm soil strata.
45

2. Explain the different methods to improve the bearing capacity of soil.
Methods of improved Bearing Capacity of Soils
In some cases, the bearing capacity of soil is so low that the dimension of the footing, work
out to be very large and uneconomical. Under such situation, it becomes necessary to improve the
bearing capacity of the soils which can be done by following methods.
1. Increasing the depth of foundation
2. Compact the soil. This can be done by using the following methods.
 Running moist soil
 Rubble compaction into the soil
 Flooding the soil
 Vibrating the soil
 Compaction by pre-loading
 Using sand piles
3. Draining the subsoil water
4. Confining the soil mass
5. Grouting with cement
6. Chemical treatment like injecting silicates etc
Loads on foundation
The foundation to be used depends upon the loads borne by it. There are three types of
loads borne by the foundation-dead line, live load and wind load.
Dead Load
This is the self weight of the various components of a building. The provision for the future
construction must also be made in the dead load calculation. In order to calculate the dead load,
knowledge of weight of the common building materials is necessary.
Live Load
This also known as superimposed load and it is the movable load on the floor. This
includes the weight of person standing on a floor, weight of materials stored temporarily on a floor,
weight of snow, etc.
Wind Load
46

Incase of tall buildings the effect due to wind should be considered. The exposed sides and
roofs of such buildings are subjected to wind pressure and its effect is to reduce the pressure on the
foundation in the windward side and to
Increase the pressure in the leeward side.
3. Compare brick masonry with stone masonry?
S
.no
Brick masonry Stone masonry
1.
Brick masonry is easy to construct. Less
skilled labour is sufficient.
Skilled labour is required for construction.
2.
It is cheaper where stones are not easily
available.
It is cheaper where stones are available in
plenty.
3.
It has reasonably good compressive
strength.
It has very good compressive strength.
4.
Proper bond can be easily provided since
bricks have regular size and shape.
Stones are to be dressed if proper bond is to
be maintained.
5.
Any type of mortar including mud mortar
can be used.
Only cement and sometimes good lime mortar
can be used.
6. Reasonably durable. Very durable and long lasting.
7. Highly fire and weather resistant.
Reasonably resistant to fire and weather
effects.
8.
Walls of small thickness can be
constructed.
Not suitable for small thicknesses.
9. Absorbs moisture from atmosphere. Water absorption is less.
10. Self-weight is less. Self-weight is more.
11. Quantity of mortar used is less. Quantity of mortar used is generally more.
12.
Openings and connection can be made
easily.
It is very difficult to make holes and
connections.
13.
Requires plastering and periodic
maintenance.
Exterior plastering is not required
maintenance.
14. Architectural effects cannot be given.
Great architectural effects can be given and
massive appearance can be obtained. For the
same reason, stone masonry is used for
monuments temples, etc.
4. Write short notes on the following:
1. Beams 2. Columns 3. Lintels
47

Beams are structural members that can carry transverse loads which produce bending moments and
shear. Beams may be horizontal(most common) or sloping(as roof beams).Horizontal beams carry
only transverse loads while sloping beams carry both axial and transverse loads.
Beams may be termed simple beams when the end connections do not carry any end
moments due to any continuity developed by the connection. A beam is continuous when it extends
across more than two supports and is called a fixed beam if the ends are rigidly attached to other
members so that a moment can be carried across the connection.
Columns are structural elements used primarily to support compressive loads. Compression
member in a truss may be known as chord members or web members, depending on their truss
location. Columns may also be called by various other names, such as braces or struts.
Failure depends on the length of the member compared to its cross sectional dimensions.
Based on their length, the columns can be classified as long or short.
If leff / a ≤ 12, it is short column and
If leff / a > 12, it is designated as a long column
A lintel is a horizontal member who is placed across an opening to support the portion of the
structure above it .The function of a lintel is same as that of a beam. Lintel may be of wood, steel,
brick, stone and reinforced cement concrete.
5. What are roofs? Explain briefly a reinforced cement concrete roof?
Now-a-days, most of the roofs constructed are of reinforced cement concrete. R.C.C. roof is
durable and fire resistant. It is economical. It is also easy to construct.
Construction: The thickness and reinforcement required for the roof slab are determined for the
given span and loading conditions. If the slab is supported by R.C.C. beams, the dimensions of the
beams and the reinforcement to be provided are also determined.
At the top level of walls, centering sheet or planks are arranged horizontally (or at a predetermined
inclination in the case of pitched R.C.C. roof) of over wooden or steel props. Steel rods are placed
in the form of grid on the top of centering sheets is given by using pre-cast cover blocks of 15 mm
thickness made of CM of mix at least 1:2.
Cement concrete of mix usually 1:2:4, mixed thoroughly with sufficient water is placed on the
centering sheets. It is well compacted preferably with mechanical vibrator to the required
48

thickness. The top is leveled. After about 12 hours of air drying, the exposed surface of slab is
cured properly for at least 7 days.
Fig. shows a section of R.C.C. slab resting on walls.
Fig. shows a section of R.C.C. slab resting on beams and walls. The arrangement of steel
reinforcement also is shown.
T-beam slab
When long sans are to be covered, they are divided into bays and the slabs are stiffened by
means of ribs. These ribs monolithically with part portion of the slab and form beams in the shape
of ‘T’. Such beams are called T beams. The slab portion of the beam is known as the flange of T-
beam. Fig.96 shows a section of a T-beam slab with steel reinforcement. The width of the rib must
be sufficient enough to accommodate the tensile reinforcement at the bottom.
T-beam slab is the most popularly adopted floor or roof slab for residential, office and
public buildings.
6. Explain the construction of the following types of floors?
a) Cement concrete flooring (b) Mosaic flooring
(c) Terrazzo flooring (d) Brick flooring
Cement concrete flooring
Cement concrete flooring is used in residential, office, school, hospital and small industries.
Construction
The ground is well compacted. A layer of 150mm thickness sand is spread over the ground
is rammed. Over the sand bed, lean cement concrete is laid upto 150mm and thoroughly
compacted. A coat of cement slurry is applied to the rough surface. Then topping cement concrete
49

of mix 1:2:5; 3:5 is applied uniformly and leveled. This surface is finally finished with a trowel and
cured for at least 10 days.
Merits
1. It is smooth, hard and pleasing in appearance.
2. Easily cleaned and washed.
3. It is durable and economical.
4. Maintenance is easy and has long life.
Demerits
1. Repair is not satisfactory
2. It is cooler in winter and warmer is summer.
Mosaic Flooring
This consists of titles available in the verity of patterns and colour. This is widely used in
theatres, temples, bathroom and superior types of building, residences, offices, hospitals and school
and colleges.
The earth is filled with sand and compacted. Over this layer a 10cm thick concrete is laid
uniformly and compacted and leveled. Now a layer of rich cement mortar is laid with thickness of
10mm and tiles are laid with hand and set properly. Dry cement is sprinkled and pressed in the
joints. After at least 2 days of laying of tiles, the flooring is ground with different grades of
carborundum stones. It is then washed with moistened oxalic acid powder. Finally, the floor is
polished using polishing machine.
These tiles make the flooring smooth. And for Decorative purpose and cool.
The initial cost is high.
50

Terrazzo flooring
This is a special type of concrete flooring in which marble ships are used as aggregates and
this concrete upon polishing with carborundum stone present a smooth surface.
Brick flooring
Brick flooring is used for go downs, stores, and platform and is subject to heavy wear and
tear and rough use.
Lean concrete is laid over well compacted sand filling, compacting properly and leveled to
a rough surface. It is 10cm thick and roughly finished to develop good bond between the base and
topping. A coat of reach cement is mortar is applied evenly over the lean concrete. Over this
cement mortar, bricks are laid in desired paten and all the joints are filled with mortar.
Merits
1. It is cheap, hard and durable
2. It is not slippery
3. Can be laid, maintained and repair easily
4. It does not crack due to the temperature differences.
Demerits
1. Cannot be used in bathroom, latrines and kitchens.
2. Cannot be washed and cleaned properly
51

7.A square steel bar of c/s area 10mm x 10mm is subjected to axial pull of 10 Kn.
Determine the stress ,strain and elongation of rof. Assume Young’s modulus 2 x 105
N/mm2
and length of rod as 1m.
Given data: A= 10mm x 10mm2
=100 mm2
Load = 10 KN = 10000N
Length = 1m = 1000mm2
E = 2x 105
N/mm2
Solution:
Formula to be used is “load / Area”
Stress (σ) = 10000/100 = 100N/ mm2
σ = 100N/ mm2
Strain =? E= Stress / Strain.
Strain = stress / e = 100 / 2x 105
Strain = 5 x 10 -4
Elongatio of rod (δl)
Strain = δl /l
5 x 10 -4
= δl / 1000 ---- δl = 0.5mm.(Ans)
8) A steel rod gets shortened by 0.75cms when subjected to compressive load of 50KN.
Determine the young’s modulus .assume dia of rod as 20mm and length of rod is 1m.
Given data:
D=20mm
L=1m=1000mm p=50KN=50x103
N
E=?
Req data:
Formula to be used E=Stress/Strain
Stress=load/area =50000/314 =159.236N/mm2
Strain = δl / l =0.75/1000 =7.5X10-4
E=Stress/Strain =159.236/7.5x105
=2.12x105
N/mm2
9) A 20m steel tape of cross section 10mmX0.8mm is subjected to a pun of 120N. Find the
actual value of the tape, if E=2x105
N/mm2
Given data
52

L=20m=20x1000mm
Size=10mmX0.8mm
E=2x105
N/mm
Req data
The tape is subjected to a pull.
Hence it will elongate
Actual length=20+change in length.
Change in length = Elengation= δl
Δl=PL/AE =120 x (20x1000)/8 x 2 x 105
Change in length=δl=1.5mm=1.5x10-3
m
Actual length=20+δl
Actual length= 20+0.0015=20.0015m
10) A load of 120KN was applied over a specimen of 20mm diameter and 120mm length in a
compression testing machine. The reduction in length due to the compressive force was
0.15mm. Determine the modulus of elasticity.
Solution:
E=PL/Aδl [δl=PL/AE]
E=120x103
x120/314x0.15
E=3.056X105
N/mm2
11) A hollow steel tube has an internal dia 0.6times as that of the external. The material
attains an ultimate stress of 420N/mm2
.If the working load is 1250KN and FS allowed is 4.
Determine the external and internal diameters.
Given:
Di=0.6 do Ultimate stress=420N/mm2
Working load=1250KN FOS=4
Solution
Working stress=ultimate stress/FS =420/4
53

Working stress=105N/mm2
Area of cross section=working load/working stress
w. stress= w. length/w. area
A= 1250x1000/105=11904.76mm
Area of hollow =3.14(Do2
-Di2
)/4
Do=153.9mm
Di=92.3mm
12) A steel wire is used to lift a load of 8KN determine the min dia of the steel wire, if the field
stress is 30N/mm2
Allow a FS of 4
Given data :
Working load=8KN
FS =4
Stress = 30N/mm2
Solution:
w.s=stress/FS =30/4 =7.5N/mm
Working .stress=Load/Area
7.5=8x103
/3.14/4x
=36.9cm
13) A square box of 75mm side is subjected to an axial pull of 240Kn. The general extension
of the box is 0.18mm even gauge length of 200mm. The change in diameter of each side is
0.010mm.Compute Young’s modulus, Poisson’s ratio, rigidity modulus and bulk modulus?
Given data :
Load = 240 KN
δl = 0.18 mm
l = 200 mm
54

δd = 0.010 mm
Solution
Area=75x75=5625mm2
Stress=24-0x1000/5625=42.67N/mm2
Lateral stress =δl/l=0.18/200=0.0009
E=Stress/Strain =42.67/0.009 =0.474x10N/mm2
Lateral strain =0.010/75=0.000133
Poisson’s ratio=lateral strain/longitudinal (linier)=0.000133/0.0009
1/m=0.148(no unit)
We know
E=2 [1+U]
0.474x10=2 [1+0.148]
Shear modulus =0.2065x10N/mm2
We know
E=3K[1-2U]
0.47x10=3k[1-2x0.148]
Bulk modulus=0.224x10N/mm2
14) The following observations were made in a tension test on a mild steel rod of dia 25mm
Original gauge length=200mm
Gauge length at a rod well below elastic=200.3mm
Load at elastic limit=25KN
Yield lead=36KN
Ultimate lead=76KN
Lead at fracture=55KN
Length of specimen after failure=212mm
Dia of specimen after failure = 16.5mm
55

Calculation:
E, yield stress, ultimate stress, % of elongation,
% of reduction area, working stress of FOS 3.
Solution
A=3.14/4x25x25=491.07mm2
Stress at elastic= 25x10/490.1=50.91N/mm2
Strain=200.3-200/200=0.30/200=0.0015
E=50.91/0.0018=0.339x10N/mm2
Yield stress=36x10/491.07=73.30N/mm2
Ultimate =76x10/491.07=154.76N/mm2
Nominal& true stress=55x10/491.07=112N/mm2
True stress at breaking point = 55x10/3.14/4x16.5=257.35N/mm2
%of elongation=(final length-original length)x100/orginal length
= (212-200) x100/200=6%
%of reduction area= (original area-final area)/original area
= (491.07-213.72)/491.07 =56.48%
W.S = Ultimate stress / FOS=154.70/3=51.59N/mm2
15. What are the various types of bridge? Explain any two in details.
Arched bridge:
Arched bridges require very stable abutment to withstand the inclined end reactions. Hence,
they are used only where it is possible to have rigid foundation.
56

Show a masonry arch bridge. It consists of arches, pier, abutments, spandrels, wearing
course, drift walls and parapets. R.C.C arch bridge used for span ranging from 35m to 200m.
Masonry arched bridges are becoming outdated due to the high cost of skilled labour involved.
Open spandrel arched bridge: Fig shows an open spandrel arched bridge. The different parts of
this are (i) Arch (ii) Piers (iii) Abutments (iv) open spandrel (v) arch flooring (vi) wearing course
and (vii) Parapet
Bow-string Girder Bridge:
Fig shows a bow-string girder bridge. It is constructed of R.C.C or steel. It consists of
a. Horizontal ties
b. Arch ribs
c. Hangers
d. Floor beam and
e. End bearing.
The roadway is supported by cross girder that rest on main beams. The main beams are
hung from the arch by hungers. Some times, lateral bracing are also constructed across the bridge
at the top of hangers, if there is enough vertical clearance. The load carrying structure consisting of
arch, hanger and the tie is known as bow- string girder. The bridge is adopted for spans varying
from 30m to 35m
Cantilever bridge: Fig shows a single R.C.C cantilever bridge. In this type, one cantilever from
each support extends to cover one span. Sometimes, a central beam is provided between the two
free ends of the cantilevers. The cantilevers are counter balanced by counter weights.
57

Girder and Slab (T- beam) bridge: Fig shows the cross section of the super-structure of a girder
and slab (T-beam) bridge made of R.C.C In the T- Beam Bridge, the longitudinal beams and deck
slab are monolithic and form a shape like T. Cross beams are provided connecting T- beam as
shown in the figure. Wearing surface of concrete is laid over the slab to arrest wearing of the slab.
The beams transmit the load from the slab to the piers and abutments the pier and abutments may
be of R.C.C or masonry. Kerb of R.C.C is provided on either side of the roadway. Parapet of R.C.C
wall or R.C.C. pillars and horizontal rails is also constructed.
T- Beam Bridge may be used up to a span of about 20m. T- Beam like any another R.C.C.
bridge is cheaper than steel bridge. Its maintenance cost is less than that of steel bridges. It is
durable. It is compacted and presents a neat appearance.
58

Continuous R.C.C Bridge: Fig shows a continuous R.C.C bridge.
Rigid Frame Bridge: Fig shows a ridged frame bridge. The entire frame is monolithic. The frames
are fixed at the bottom.
Suspension bridge: Fig shows a suspension bridge. Here, tall towers erected on piers and abutments
from the end supports two sets of cables / chains are hung in the shape of a catenary from tower to
tower. The roadway is suspended from the cables by means of suspenders. The cables pass over the
saddles at the top of towers and are anchored at the two ends.
Plate Girder Bridge: Fig shows the decking of a plate girder bridge. The deck slab is of R.C.C or
it may be constructed of steel plate. Longitudinal ribs support the deck slab / plate. These ribs are
59

carried by transverse floor beams that rest on the plate girders. Lateral bracing are provided both at
the top and bottom of the girders to take the wind pressure.
Through type steel bridge: Fig shows the superstructure of a through type steel bridge. The track
or the deck slab is carried by stringers running longitudinally. Stringers are connected to the cross
graders. The cross girder are connected to the main girders. Sway bracing and knee bracing are
provided at the top as shown in the figure.
Slab Bridge: Fig shows a sectional view of an R.C.C slab bridge. It is a single span bridge and is
suitable for spans not more than 10m. it is the simplest form of bridge and the easiest to
constructed .The bridge consists of reinforced cement concrete slab of uniform thickness. The ends
of the slab rest on abutments of brick masonry, stone masonry or R.C.C. over the slab is laid a
wearing coat of cement concrete of thickness 75mm.
Kerbs of brick masonry or R.C.C are provided on either side. Parapet wall is also provided
on both sides of the bridge
16. What are the various types of dam? Explain any two in details.
Types of dam:
60

All types of dams can generally classify into two types, Viz., rigid dams and non-rigid
dams. Rigid dams are constructed of materials such as stone, brick, and plain concrete on
reinforced cement concrete. In the case of non-rigid dams, earth and rock are used.
a) Types of Rigid dams:
i) Gravity dams
ii) Arch dams
iii) Buttress dams
iv) Steel and timber dams
b) Types of Non- Rigid dams:
i) Earth dams
ii) Homogeneous type
iii) Rock- fill dams
a)Types of Rigid dams:
1. Gravity dams:
The cross section of gravity dam is shown in figure. As the name implies, this type of dam resist
the external forces to its weight. The forces acting on the dam are
1. External water pressure
2. Weight of the dam
3. Uplift pressure from below the dam due to the seepage water
4. Wave pressure and ice pressure at the M.W.L.,
5. Pressure due to earthquake forces
6. Pressure due to silt accumulation in the reservoir
61

2. Arch dams: The arch dam is curved in plan with the convex face holding the Water. As
compared to gravity dam, the section of the requirement is the availability of strong abutments.
This types of dam economical when its length is less that the height and hence preferred for very
great height as in the case of deep narrow gorge.
3. Buttress dams: The buttresses in turn transfer the load to the base slab which Forms the
buttress Dam has thin sections and suited for construction in relatively weak foundations. The large
amount of space available between the buttresses can be used advantageously for providing the
power house and water treatment plants.
4. Steel and timber dams: These are mostly constructed for temporary purposes and when the
height of the dam is not high. A timber dam may be constructed to divert the flow of water in the
river for purpose of constructing the main dam. Timber dams are not made water tight. Steel dams
are water tight but require periodic painting and maintenance. However, steel dams are not in
common use
62

Steel and timber dam
Types of Non- Rigid dams:
I. Earth dams: These are well suited and economical when the height of the dam is medium and
can be constructed on any type of foundation. The soil is laid in this layer, added with the proper
moisture Content and compacted with rollers to develop the required strength. The upstream face
of the dam holding the water is provided with stone revetment. It is not made water right. Filters
are provided so that the seepage water does not carry any material of the dam. Earth dam’s last long
if properly designed constructed and maintained.
Homogeneous earth dam
Homogeneous type
The dam is constructed of one type of soil material only. Upstream stone pitching and a horizontal
filter provided.
Zone type
Here there are two zones the inner zone made of impervious material like clay and the outer
zones on either side made of pervious material like locally available soil. The inner impervious
zone provides more strength and reduces the seepage of water through the dam. Grater heights of
dams can be constructed by adopting this type. Filters are provided to safely lead the seepage water
through the dam in to the toe drains.
Diaphragm type:
63

In this type, at the centre of the section, a thin impervious wall of masonry or concrete is
constructed along the length f the dam. This diaphragm wall is tied to the bed rock or to the
impervious foundation material. The impervious wall very much reduces the seepage from the
dam.
ii ) Rock- fill dams: This type is
suitable for moderate heights. A
concrete slab is provided on the
upstream side. The dam section
consists of dry rubble stone masonry
on the upstream side and loose rock fill on the down stream side. If rock foundation is available,
then the settlement of the dam is minimum. This type has better resistance to earth quake forces
because of its flexible nature.
64

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