Civil Engineering Test

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Civil Engineering Test - 1

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Question 1
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Consider the following types of soil tests:

1. California bearing ratio

2. Consolidation

3. Unconfined compression

The soil tests required to be done in the case of undisturbed samples include

A
1, 2 and 3

B
1 and 2

C
1 and 3

D
2 and 3

Solution

Concept:

California bearing ratio (CBR) test

The test may be performed

(a) On undisturbed soil specimen

(b) On remoulded soil specimen

Consolidation Test:

This is used to determine the rate and magnitude of soil consolidation when the soil is restrained laterally and loaded axially. This test is conducted by an Oedometer. As disturbing specimens have a certain pre-consolidation pressure, so for conducting the consolidation test undisturbed soil specimen is required.

An odometer is used to find out the void ratio at the end of various stress levels and hence curve is plotted. This curve is useful in calculating the settlement of the soil layer.

Unconfined Compressive Strength – UCS Test:

The soil specimen to be used for the unconfined compressive strength test depends on the purpose for which it is tested. So specimens can be taken in compacted, moulded, or undisturbed forms.

Hence all the above tests can be performed in the case of an undisturbed sample.

Question 2

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Quality parameters of rainwater from different sources are given below:

Sources	pH	Turbidity	TDS (mg/l)	Chlorides (mg/l)	Fluorides (mg/l)
A	9.5	1	500	100	15
B	6.0	2	1000	200	1.5
C	8.0	5	1400	250	1.0
D	7.0	10	1600	300	0.1

Water drawn from which of the above sources can be tolerated for drinking without any treatment?

A and B

B and C

C only

D only

Solution

Concept:

The acceptable limit (AL) and cause for rejection limit (CFR) for various water quality parameters are given below.
If the value of any parameter is less than its CFR value, then it can be tolerated for drinking without any treatment.

water quality parameter	AL	CFR
TS(Total solids)	500 mg/l	2000 mg/l
Turbidity	1 mg/l	5 mg/l
Colour	5 TCU	15 TCU
PH	6.5 - 8.5	-
Alkalinity	200 mg/l	600 mg/l
Total hardness	200 mg/l	600 mg/l
chloride	250 mg/l	1000 mg/l

At least 1 mg/l of fluoride is essential to prevent dental cavities. During the formation of permanent teeth, it combines chemically with tooth enamel, resulting in harder and stronger teeth that are more resistant to decay.
Flouride content greater than 5 mg/l will lead to deformation of bones called FLOUROSIS.
Hardness in drinking water should be in the range of 75 - 115 mg/l for the proper functioning of the body.

Question 3
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Bituminous surface dressing is provided to:

A
newly constructed water bound macadam pavement to serve as load bearing coat

B
a newly constructed pavement to serve as thin wearing coat

C
an existing water bound macadam pavement to serve as load bearing coat

D
an existing pavement to serve as thin wearing coat
Solution
Explanation:

Bituminous surface treatment:
- A bituminous surface dressing or treatment is also known as a seal coat or chip seal
- It is a thin protective wearing surface that is applied to a pavement or base course
- It makes it more difficult for water to enter the base material and preventing freeze thaw damage for those locations with below-freezing temperatures
- BSTs are created using two main materials: asphalt and a cover aggregate
BSTs can provide all of the following:
- A waterproof layer to protect the underlying pavement
- Increased skid resistance
- A filler for existing cracks or raveled surfaces.
- An anti-glare surface during wet weather and an increased reflective surface for night driving.
Question 4
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A level instrument at a height of 1.320 m has been placed at a station having a RL of 115.385 m. The instrument reads - 2.835 on levelling staff held at the bottom of bridge deck. The RL of the bottom of the bridge deck is _________ m.

A
111.230

B
113.870

C
119.540

D
116.900

Solution

Concept:

According to H. I method or Collimation method,

H.I = R.L + B.S

R.L = H.I - (B.S/F.S)

Where,

H.I height of instrument, R.L = reduced = level, B.S = Backsight, F.S = foresight

Inverted staff reading

When a point for which R.L is to be determined is at a very high level above the line of sight, for example, the roof of a building, chajja, etc the leveling staff should be inverted such that the bottom of leveling staff should touch a point. It is taken as a negative reading.

Height from the base to bottom of inverted staff is kept = B.S - F.S

As staff is inverted, therefore reading of B.S is taken as negative.

Calculation

Given,

B.S = 1.32 m

RL of station = 115.385 m

F.S = -2.835

H.I = R.L +B.S

H.I = 115.385 + 1.32 = 116.705 m

Now, R.L bottom of the bridge deck,

R.L = H.I – F.S

R.L = 116.705 - (- 2.835) = 119.54 m
Question 5
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The gross commanded area for a distributary is 8000 hectares, 60 percent of which is culturable irrigable. The intensity of irrigation for Rabi season is 60 percent and the corresponding average duty at the head of the distributary is 2000 hectares/cumec, find the discharge required at the head of the distributary from average demand considerations.

A
3.45 cumec

B
1.44 cumec

C
2.45 cumec

D
1.11 cumec
Solution
Concept:

Culturable command area (CCA):
- It is that portion of gross command area (GCA) that is culturable or cultivable.
The intensity of irrigation:
- The intensity of irrigation is the percentage of CCA proposed to be irrigated annually.
Calculation:

Given,

Gross command area = 8000 hectare

Culturable command area = 60% of GCA = 0.6 × 8000 = 4800 hectare

Intensity of irrigation for Rabi season = 60%

Duty for the rabi season = 2000 hectare/cumec

So, area under rabi season = 0.6 × 4800 = 2880 hectare

Discharge required for rabi season

= area under rabi season/duty for rabi season

= 2880/2000 = 1.44 cumec
Question 6
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Cumulative frequency curve that shows the percent of time specified discharge was equalled or exceeded during a given period is known as:

A
mass duration curve

B
flow duration curve

C
hydrograph

D
power duration curve
Solution
Explanation:

Flow duration curve:
- Flow-duration curve is a cumulative frequency curve that shows the percent of time specified discharges were equaled or exceeded during a given period.
- It is a graphical method to present the discharge data of a river or a stream.
- it represents flow characteristics of a river for a particular region subjected to recorded flows in the river.
Mass duration curve:
- It is a plot between the stored volume of water in a stream and time in days.
- The curve represents the reservoir capacity.
Hydrograph:
- It is the graphical representation between discharge (rate of flow) and time in a particular river channel.
- it is used in various engineering applications such a estimation flood water and risk analysis of dam etc.
Question 7
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Find the expansion of the e^x - cos x at x = 0

A

B

C

D

Solution
Question 8
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The Fourier series expansion of x³ in the interval −1 ≤ x < 1 with periodic continuation has

A
only sine terms

B
only cosine terms

C
both sine and cosine terms

D
only sine terms and a non-zero constant

Solution

Explanation:

f(x) = x³

find f(x) is even or odd

put x = -x

f(-x) = - x³

f(x) = -f(-x) hence it is odd function

for odd function, a_o = a_n = 0

Fourier Series for odd function has only b_n term
Question 9
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Sewer is running half full. When Manning's coefficient is increased from 0.011 to 0.022, the slope of sewer to carry the same flow at the same velocity running half full will be

A
increased by 2 times

B
decreased by 2 times

C
increased by 4 times

D
decreased by 4 times

Solution

Concept:

The discharge through the sewer can be given as,

Q = A.V

By manning`s equation,
Question 10
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A consolidation test was done in the lab. A sample of 20mm thickness was consolidated at 50% in 15 minutes with a single drainage condition. how much time a 5m thick layer of the same soil will take to consolidate by 50% and 30% respectively with double drainage conditions in the field.

A
1.78 year & 0.16 year

B
0.445 year & 0.16 year

C
0.64 year & 0.445 year

D
0.64 year & 1.78 year

Solution