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BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test BPSC AE Civil Mock Test Series 2024 - BPSC AE Civil Paper 5 (Civil) Mock Test - 2

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 for Civil Engineering (CE) 2024 is part of BPSC AE Civil Mock Test Series 2024 preparation. The BPSC AE Civil Paper 5 (Civil) Mock Test - 2 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The BPSC AE Civil Paper 5 (Civil) Mock Test - 2 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 below.
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BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 1

Which of the following statements is/are correct?

1. A steel structure designer can guarantee the safety of the structure.

2. Working stress method of design of steel structures offers a safer and economical structure.

3. Strength and serviceability of a structure cannot be predicted on account of several unforeseen factors.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 1

Concept:

Statement 1:

No steel structure designer can guarantee the safety of the structure, as the design is carried out using the results based on some experiments and adopting the probabilistic approach, so the safety can not be guaranteed by any designer.

Statement 2:

  • WSM results in an uneconomical structure.
  • In WSM the design strength is calculated such that the stress in the material is restrained to its yield limit, under which the material follows Hooke's law and hence the term elastic design is used in WSM.
  • So this method yields an uneconomical design of simple beam or other structural elements where the design governing criteria is stress.

Statement 3:

  • The strength and serviceability of a structure cannot be predicted on account of several unforeseen factors.
  • So the factor of safety is used while designing of structure.

So only statement 3 is correct.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 2

A thin-walled spherical shell is subjected to an internal pressure of 10 MPa. The shell has a uniform wall thickness of 25 mm and a diameter of 500 mm. Take E = 20 GPa and μ = 0.3.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 2

Concept:

Thin spherical walls.

The hoop stress/longitudinal stress of a spherical shell is given by

The hoop strain / longitudinal strain of a spherical shell is given by

The volumetric strain is given by 3 times the hoop strain as the strain is same in all directions because sphere is same about all the axes. (Option 4)

Calculation:

Given:

P = 10 MPa; t = 25 mm; d = 500 mm; E = 20 GPa; μ = 0.3;

Now

ϵV = 3 × ϵH = 3 × 1.75 × 10-4 = 5.25 × 10-4

Due to symmetricity the strain of the sphere in all directions is same.

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BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 3

Assertion (A): Trusses comprise triangular figures.

Reason (R): A pin-jointed stable figure is a triangle.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 3
Trusses are structural frameworks made up of interconnected triangular units. In the context of trusses, pin-jointed connections allow for rotation at the joints without resistance. When a truss is in equilibrium, the triangular configuration ensures stability.
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 4
The group index for a soil, whose liquid limit is 40 percent, plasticity index is 10 percent and percentage passing 75 micron IS Sieve is 35, is :
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 4

Concept:

Group index of a soil is given by

G.I = 0.2 a + 0.005 ac + 0.01 bd

a = It is the portion of % passing through 75 μ sieve greater than 35 but not exceeding 75 expressed as whole number in between [0-40].

b = It is the portion of % passing through 75 μ sieve greater than 15 but not exceeding 55 expressed as whole number in between [0-40].

c = It is the portion of the numerical liquid limit greater than 40 but not exceeding 60 expressed as whole number in between [0-20].

d = It is the portion of the numerical plasticity index greater than to 10 but not exceeding 30 expressed as whole number in between [0-20].

Calculation:

IP = 10%, Liquid Limit = 40%

% Passing through 75 micron is 35%

a = 35 - 35 = 0 < 40

a = 0

b = 35 - 15 = 20

c = WL - 40

c = 40 - 40 = 0

d = IP - 10

d = 10 - 10 = 0

G.I = 0.2 a + 0.005 ac + 0.01 bd

G.I = 0.2 × 0 + 0.005 × 0 × 0 + 0.01 × 20 × 0

G.I = 0

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 5

Choose the correct combination for the given Table:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 5

The chart shows the relation between stress-strain in different materials.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 6
For batching 1 : 3 : 6 concrete mix by volume, the ingredients required per bag of 50 kg cement, are
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 6

Concept:

Generally in volumetric batching system the amount are measured in volume (m3).This method is not accurate as compared to weight batching.In weight batching method, materials are measured on the basis of weight. It is accurate method of batching.

Density of cement = 1440 kg/m3

Volume of one bag (50 kg) of cement = 0.03472 m3

Calculation:

Mix proportion given = 1 : 3 : 6

Given weight of cement = 50 kg

Volume of cement = 0.034m3

Volume of sand = 0.03472 × 3 = 0.1042 m3 = 104.2 ℓ = 105 ℓ

Volume of aggregate = 0.03472 × 6 = 0.20832 m3 = 210 ℓ
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 7

Which of the following failure is not a type of failure in tension members in steel structures?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 7

Concept:

Modes of failure in steel tension members:

a) Gross section yielding:

  • Generally, a tension member without bolt holes can resist loads up to the ultimate load without failure.
  • But such a member will deform in the longitudinal direction considerably (nearly 10%-15% of its original length) before fracture. At such a large deformation a structure becomes unserviceable.

b) Rupture Failure:

  • When a tension member with a hole is loaded statically, the point adjacent to the hole reaches the yield stress first.
  • On further loading, the stress at that point remains constant at yield stress and each fiber away from the hole progressively reaches the yield stress. Deformations continue with increasing load until finally rupture of the member occurs.

c) Block Shear Failure:

  • The block shear failure becomes a possible mode of failure when the material bearing strength and bolt shear strength is higher.
  • Block shear failure is the rupturing of the net tension plane(BC) and yielding on the gross shear plane(AB & CD), as shown in the figure, which results in rupturing of the shear plane as the connection length becomes shorter.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 8

A straight bar which is fixed at the ends A and B and having elastic modulus (E) and cross-sectional area (A), is subjected to a load P = 120 N at C as shown in the figure. The reactions at the ends are:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 8

Given, P = 120 N at C

Free body diagram

Let RA and RB are the reaction at the fixed end A and B.

RA + RB = P

Sign convetion: +ve(Tension), -ve(Compression)

As the Beam AB is fixed, So

ΔAB = 0

ΔAC + ΔCB = 0

After solving,

RB = P/3 = 120/3 = 40 N

RA = P - RB = 120 - 40

RA = 80 N

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 9

As per IS 875 part-2, imposed loads on balconies in dwelling houses shall be ________ Consider type of occupancy as residential building.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 9

As per CI. 3.1 of IS 875 (part 2), Imposed floor loads for different occupancies:

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 10

In the moment distribution method, the sum of distribution factors for all the members meeting at a joint is always:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 10

Concept:

Distribution factor (DF):

Or

Where M1, M2, M3, and M4 are moment induced in member OA, OB, OC, and OD respectively.

The summation of DF for all members at a joint is one.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 11

The Muller Breslar principal in structural analysis is used for

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 11

Muller-Breslau Principle:

  1. It states that "If an internal stress component or a reaction component is considered to act through some small distance and thereby to deflect or displace a structure. The curve of deflected structure will influence line stress or reaction component".
  2. It is used to draw influence line diagrams for determinate and indeterminate structures.
  3. ILD for determinate structure is linear while that for the indeterminate structure will be curvilinear.
  4. This method is based on the concept that Influence lines are deflection curves.
  5. It is a straight application of Maxwell's reciprocal theorem.
  6. ILD is a graphical representation of variation in the reactions, Shear force, and bending moment at each and every section when unit load moves from one to another end of the structure. They can be drawn for any type of structure – beams, arches, trusses, etc.

For example, to draw the influence line diagram for a vertical reaction at A in a simply supported beam as shown below.

Remove the ability to resist movement in the vertical direction at A by using the guided roller

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 12

In which of the following sections will the position of actual neutral axis (n) shift above the critical neutral axis (nc )i.e. n < nc and steel is fully stressed and concrete is under stressed?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 12

The characteristics of under-reinforced section and over a reinforced section when compared with balanced section are given below in the tabulated form.

So, For an under-reinforced beam, steel yield prior to concrete failure. Stresses in steel reach permissible value whereas stress in concrete has not reached up to its permissible value.

σs = fy and σc < fc

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 13
For a clay slope of height 20 m, the stability number is 0.05, γ = 25 kN/m3, and c = 30 kN/m2, the critical height of slope is:
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 13

Concept:

Taylor's stability number method:

(i) Taylor introduced a dimensional parameter, called Taylor's stability number, which is given by

Where, H = Effective height of slope

Hc = Critical height of slope

F = FOS

(ii) In this method, Taylor's stability number is read from Taylor's chart in order to analyze the stability of slope for a given value of strength parameters of the soil.

(iii) Taylor's method is suitable for c-ϕ soils and for ϕ = 0 (pue clays)

Calculation:

Given, C = 30 kN/m2, γ = 25 kN/m3 and Sn = 0.05, F = 1

∵ We know that,

Hc = 24 m

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 14

Match List - I with List - II and select the correct answer using the codes given below the lists:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 14

The cement of the desired color may be obtained by intimately mixing mineral pigments with ordinary cement. The amount of coloring material may vary from 5% to 10%. If the percentage exceeds 10%, the strength of the cement is affected.

The list of pigments used in cement and their corresponding colure obtained is given below in the tabulated form:

Note:

  1. The chloride does not impact the color of cement and it will produce the cement of its natural color i.e., grey. There is an error in question.
  2. Iron Oxide pigments are one of the most used pigments. These pigments protect the concrete/cement from fading or prevent the leaching out of concrete/cement.
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 15

When the air permeability method is used to determine the fineness of cement, the fineness is expressed as

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 15

The fineness of the cement is checked to test the proper grinding of the cement which significantly influences the rate of hydration. These two methods are used for the determination of the fineness of cement:

1) Air permeability method ( Blaine)

The fineness of cement is measured as the specific surface.

Specific surface is expressed as

  • The total surface area in square meters of all the cement particles in one kilogram of cement or
  • The total surface area in square cm per gram of cement.

The higher the specific surface is, the finer cement will be.

Fineness test is performed on the Blaine apparatus. It is practically a manometer in the U-tube form. One arm of the manometer is provided at the top with a conical socket to form an airtight fit with the conical surface of the cell.

2) Sieving method

  • This method serves only to demonstrate the presence of coarse cement particles. This method is primarily suited to checking and controlling the production process. The fineness of cement is measured by sieving it on standard sieves. The proportion of cement of which the grain sizes are larger than the specified mesh size is thus determined.

Additional Information

  • The soundness of cement may be determined by two methods, namely the Le-Chatelier method, and the autoclave method.
  • The Vicat Apparatus: It is used to measure the setting time and consistency of concrete.
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 16

The ratio of θA and θB for beam as shown will be:-

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 16

Solution:

The θA and θB can be calculated using conzugate beam method.

In this method,

Rotation at any point in real beam = shear force at that point in conzugate beam

Deflection at any point in Real beam = Bending moment at that point in conzugate Beam

Also, the loading on conzugate beam is the diagram of real beam.

Calculation:

Congugate Beam:

(clockwise)

Now, ∑Fy = 0

θA : θB = 1 : 2

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 17

Shear span is called the zone where:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 17

Shear span

  • It is the span between the points of application of concentrated load to its adjacent Reaction force in a beam.
  • If a simply–supported beam of span L carries a moment force at its mid-span, then the shear force diagram will be rectangular, and the bending moment diagram will be triangular.

Shear span is defined as the zone where shear force is constant.

Note:

  • Throughout Shear Span the Shear Force is constant.
  • There might be multiple shear spans for a single beam depending upon the number and position of applied force to the numbers of supports.
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 18
For painting corrugated steel sheet, surfaces shall be measured flat and the area worked out shall be increased by
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 18
Concept:
As par IS 1200 ( Part 13 ) : 1994
Corrugated surfaces shall be measured flat as fixed and not girthed Quantities so measured shall be increased by the following percentage and the results shall be included in general areas.
  • Corrugated steel sheets 14%
  • Corrugated asbestos cement sheets 20%
  • Semi–corrugated asbestos cement sheets 10%
  • Nainital pattern roofs ( plain sheeting with rolls ) 10%
  • Nainital pattern roofs with corrugated sheets 25%

Additional Information

1. Cornices and other wall features, when not picked out in a different finish/colour, shall be girthed and included in the general area.

2. The painting for building surfaces shall be kept separate and the surfaces to be painted shall be described. It shall be stated whether measurements are flat or girthed. Alternatively. different surfaces may be grouped into one general item, areas of uneven surfaces are converted into equivalent plain areas by increasing the areas as under:

  • External walls of plain brickwork faced with recessed, raised or weather stuck pointing-20 percent
  • Sand face plaster with up to 4 mm Size 50 percent
  • Roughcast plaster with stone aggregate up to 10 mm-IOO percent
  • Pebbledash finish beyond 10 mm-275 percent
  • Sponge finished plaster-25 percent
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 19

Determine the single shear stress induced in a circular pin which has an area of 115 mm2 undergoing a tension force of 12 kN.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 19

Concept:

1) Single shear case:

  • The shear force P in the shear plane is equal to tension force F.
  • The average shear stress in the plane is .
  • This joint is said to be in single shear.

2) Double shear case:

  • If the plates, which are connected by a rivet as shown in the following figure, are subjected to tension forces, shear stresses will develop in the rivet.
  • This joint is said to be in double shear.
  • Observing that the shear P in each of the planes is , the average shearing stress is

Calculation:

Given:

Force, F = 12 kN, Area, A = 115 mm2

Single shear stress-induced = τavg

τavg = 104.34 N/mm2

∴ The average shear stress, in this case, is 104.34 N/mm2

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 20
The development length in compression for a 20 mm diameter deformed bar of grade Fe415 embedded in concrete of grade M25 whose design bond stress is 1.40 N/mm2 is -
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 20

Concept:

Development length:

(i) The calculated tension or compression in any bar at any section shall be developed on each side of the section by an appropriate development length or end anchorage or a combination thereof.

(ii) Development length can be calculated as:

Where, ϕ = Diameter of bar

τbd = Design bond stress = Permissible value of average bond stress

The value of bond stress is increased by 60% for a deformed bar in tension and a further increase of 25% is made for bars in compression.

Calculation:

Given,

ϕ = 20 mm, τbd = 1.4 N/mm2 (In tension), fy = 415 MPa

∵ Bar in compression, so Increased the value of τbd by 25% and for deformed bar it is increased by 60 %

So, τbd = 1.4 × 1.25 × 1.6 = 2.8 N/mm2

Development length,

= 645 mm

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 21

If σcbc is permissible compressive stress of centre in bending, then modular ratio is given by -

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 21

Modular Ratio:

(i) In case of working stress method of analysis, the composite section of reinforced concrete is transformed into an equivalent section of single material of concrete. for this, we need to define modular ratio which is the ratio of elastic modulus of steel to concrete.

(ii) However, the short term modulus of elasticity of concrete does not take into account the long term effects of creep and shrinkage and thus it is not considered in defining the modular ratio (m). However, partly taking into account the long term effects of creep and shrinkage, CI. B-1.3 of IS 456:2000 defines modular ratio (m) as:

Where, σcbc = Permissible stress n concrete in bending compression

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 22

The utilization of concrete in tension zone of prestressed concrete member saves concrete ranging between

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 22

As per IS code:

An additional saving of 15 to 30 percent is possible in concrete due to the use of concrete in the tension zone, compared to reinforced concrete high tensile steels used in prestressed members and its ultimate strength is equal to 2100 N/mm2 and the savings include steel and is even higher, 60 to 80 percent mainly due to higher permissible stresses permissible in high tension wires.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 23

Which of the following is not one of the four distinct operations involved in bricks manufacturing?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 23

Operations involved in manufacturing of clay brick are:

Preparation of brick earth

1. Unsoiling

The soil used for making building bricks should be processed so as to be free of gravel, coarse sand (practical size more than 2 mm), lime and kankar particles, organic matter, etc. About 20 cm of the top layer of the earth, normally containing stones, pebbles, gravel, roots, etc., is removed after clearing the trees and vegetation.

2. Digging

After removing the top layer of the earth, proportions of additives such as fly ash, sandy loam, rice husk ash, stone dust, etc. should be spread over the plane ground surface on a volume basis. The soil mass is then manually excavated, puddled, watered and leftover for weathering and subsequent processing.

3. Weathering

The soil is left in heaps and exposed to the weather for at least one month in cases where such weathering is considered necessary for the soil. This is done to develop homogeneity in the mass of soil, particularly if they are from different sources, and also to eliminate the impurities which get oxidized. The plasticity and strength of the clay are improved by exposing the clay to weather.

4. Blending

The earth is then mixed with sandy-earth and calcareous-earth in suitable proportions to modify the composition of the soil. A moderate amount of water is mixed so as to obtain the right consistency for moulding.

5. Tempering

It is done in pug mill and the process is called pugging. Tempering consists of kneading the earth with feet so as to make the mass stiff and plastics (by plasticity, we mean the property which wet clay has of being permanently deformed without cracking). It should preferably be carried out by storing the soil in a cool place in layers of about 30 cm thickness for not less than 36 hours.

6. Moulding

It is a process of giving a required shape to the brick from the prepared brick earth. Moulding may be carried out by hand or by machines.

7. Drying

The object of drying is to remove the moisture to control the shrinkage and save fuel and time during burning.

8. Burning

The burning of clay may be divided into three main stages: Dehydration stage (400 - 650°C), Oxidation period (650 - 900°C), vitrification. Burning of bricks is done either in clamps or kilns.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 24
To measure the bulking of sand, a sample of sand is filled in a container and the measured sand height is 210 mm. Then the sand is taken out and the container is filled with water and sand is slowly dropped in it. Now, the height of sand is found 150 mm. The bulking of sand is ________%
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 24

Concept:

Bulking of sand:

Bulking in sand Occurs When dry sand interacts with atmospheric moisture. The presence of moisture content forms a thin layer around sand particles. This layer generates the force which makes particles move aside from each other. This results in an increase in the volume of sand.

Percentage of bulking can be determined by:

Percentage of bulking of sand =

Where

h1 = Height of the sand sample

h2 = Height of the sand in water or saturated sand

Calculation:

Given,

h1 = 210 mm and h2 = 150 mm

Percentage of bulking of sand =

Percentage of bulking of sand =

Percentage of bulking of sand = 40 %.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 25

The type of bond provided in brick masonry for carrying heavy loads is:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 25

English bond

  • In this bond, the alternate courses consist of headers and stretchers.
  • This is considered to be the strongest bond, So this bond is provided in brick masonry for carrying heavy loads.
  • Hence it is a commonly used bond for walls of all thicknesses.

Flemish Bond

  • In this type of bond, each course comprises of alternate header and stretcher.
  • Alternate courses start with stretcher and header.
  • Every header is centrally supported on the stretcher below it.
  • Construction of Flemish bond needs greater skill.
  • Used to get a good aesthetic view.

Important Point

To break the continuity of vertical joints

  • In English bond - Queen closer is used in the beginning and end of a wall after the first header.
  • In Flemish bond - Queen closers are required if a course starts with a header and in walls having their thickness equal to odd number of half bricks, bats are essentially used to achieve the bond.
  • In Header bond - 3/4 brick bat as a quoin brick in alternating courses.
  • Stretcher bond - 1/2 brick bat is provided in alternating courses.
BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 26
The strain energy stored in a member due to bending moment M is given by:
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 26

Concepts:

The strain energy is the energy stored in a body due to its elastic deformation.

The most general formula for strain energy is:

Where, F is the applied force and δ is deformation

However, When stress

; V is the volume of material.

Based on the above formals, the strain energy stored in the material due to different types of forces/moments is given as:

1. Strain energy due to bending is:

2. Strain energy due to Torsion is :

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 27

The imaginary beam of same span as the original beam loaded with M/EI diagram of the original beam, such that the shear force and bending moment at a section will represent the rotation and deflection at that section in the original beam ________.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 27

Concept:

Conjugate beam method:

Conjugate beam is defined as a imaginary beam with the same dimension as that of original beam but load at any point on the conjugate beam is equal to bending moment at that point divided by EI (Flexural Rigidity).

Conjugate beam can be made by changing the supports

Note: The shear force in conjugate beam at a point is slope at the point in real beam. The bending moment in conjugate beam at a point is the deflection at the point in real beam.

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 28

Horizontal stiffness coefficient ( K11) of a bar ‘AB’ is given by –

(Where A is cross section area and E is Young’s Modulus)

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 28

Degree of kinematic indeterminacy, DK= 1

So size of [K] = [1×1]

Calculation:

θAB = 45°

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 29

Where the horizontal stiffeners are provided 'd' in mm shall be taken as a clear distance between the horizontal stiffeners and tension flange. These vertical stiffeners shall be designed so that I is not less than

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 29

Example:

Vertical Stiffeners:

  • Vertical stiffeners are provided to prevent buckling of the web due to diagonal compression.

Some criteria regarding the design of vertical stiffness:

  • The spacing between the vertical stiffness should be between 0.33 × d and 1.5 × d, d is the distance between flange angles,s, and where there is no flange angle, the distance between flanges ignoring fillets is taken.
  • The greater unsupported clear dimension of the web panel should not be greater than 270 times the thickness of the web and the lesser unsupported clear dimension of the same web panel should not be greater than 180 times the thickness of the web.
  • The length of the outstanding leg of vertical stiffness may be taken equal to 1/30 of the clear depth of girder plus 50 mm and the outstanding of stiffener from web shall not be more than , where t is the thickness of the section )
  • The length of the connected leg of vertical stiffness should be sufficient to accommodate the rivets connecting the stiffness to the web. The amount of moment `of inertia I of the stiffness selected should not be less than

I <

Where I is the moment of inertia of pair of stiffness about the center of the web

tw = Minimum required thickness of the web

C = Maximum permitted distance between the vertical stiffeners

BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 30

When Timber is subjected to dry rot, ________ is an excellent prevention

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 2 - Question 30

Dry rot:

(i) The fungi of certain types feed on wood and during feeding, they attack on wood and convert it into powder form. This is known as the dry rot.

(ii) The dry rot occurs at places where there is no free circulation of air such as improperly ventilated basements, rooms, etc., and in damp situations like kitchen, toilets, etc.

(iii) The unseasoned softwood and sap wood are easily attacked by dry rot.

(iv) The most favorable conditions for the rapid growth of the fungus responsible for dry rot are the absence of sunlight, dampness, presence of sap, stagnant air, and warmth.

(v) When part of timber is seriously affected by dry rot, the damaged portion may be completely removed and the remaining unaffected portion should be painted with a solution of copper sulphate.

Solignum paint:

  • These paints preserve the timber from white ants as they are highly toxic in nature. They can be mixed with colour pigments and applied in hot state with the help of brush.

Coal tar:

  • In this method, timber section is applied with hot coal tar that increases its resistance against the penetration of water and fire.
  • Coal tar gives an unpleasant smell and appearance hence it is generally used for works of minor importance
  • The process of application of coal tar over timber surface is referred to as tarring.

Creosote:

  • In this case, the timber surface is coated with creosote oil. The process is known as the creosoting or Bethel's process of preservation of timber.
  • The creosote oil is obtained by the distillation of tar.
  • It is observed to have the best antiseptic properties there by is highly affected in killing fungi, insects and termite.
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Top Courses for Civil Engineering (CE)