Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  BPSC AE Civil Mock Test Series 2024  >  BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Civil Engineering (CE) MCQ

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Civil Engineering (CE) MCQ


Test Description

30 Questions MCQ Test BPSC AE Civil Mock Test Series 2024 - BPSC AE Civil Paper 5 (Civil) Mock Test - 1

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 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 - 1 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The BPSC AE Civil Paper 5 (Civil) Mock Test - 1 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 - 1 below.
Solutions of BPSC AE Civil Paper 5 (Civil) Mock Test - 1 questions in English are available as part of our BPSC AE Civil Mock Test Series 2024 for Civil Engineering (CE) & BPSC AE Civil Paper 5 (Civil) Mock Test - 1 solutions in Hindi for BPSC AE Civil Mock Test Series 2024 course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt BPSC AE Civil Paper 5 (Civil) Mock Test - 1 | 50 questions in 60 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study BPSC AE Civil Mock Test Series 2024 for Civil Engineering (CE) Exam | Download free PDF with solutions
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 1

What is the main advantage of using the double integration method over other methods for analyzing slopes and deflections?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 1
  • The main advantage of using the double integration method for analyzing slopes and deflections in structures is its simplicity, and generality, and gives more accurate results.
  • The double integration method is a classical approach that provides a straightforward and systematic way to determine the slope and deflection along the length of a structural member subjected to external loads.
  • It may not be as efficient for solving certain complex or irregular structural problems compared to more advanced numerical methods.
  • However, for introductory purposes and relatively straightforward structural analyses, the double integration method remains a valuable and widely taught technique.

Additional Information

There are other methods of finding slope deflections other than double integration method.

  1. Moment–area method.
  2. Mecaulay's method.
  3. Conjugate beam method.
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 2

As per IS: 1892 – 1979; what should be the maximum thickness of cutting edge of sampling tube of 70 mm external diameter which is required for sampling in undisturbed stiff clay soil?

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

Concept:

Area ratio:

It can be defined as the ratio of the maximum cross-sectional area of the cutting edge to the area of the soil sample.

The area ratio can be expressed as

where

D1 = inner diameter of cutting edge

D2 = outer diameter of cutting edge

Note:

For stiff formation (Ar)max = 20%

Soft sensitive clay (Ar)max = 10%

Maximum thickness of cutting edge =

Calculation:

Given, D2 = 70 mm

Soil is stiff clay, So (Ar)max = 20%

D1 = 63.9 mm

Maximum thickness of cutting edge =

= = 3.05 mm

1 Crore+ students have signed up on EduRev. Have you? Download the App
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 3

Match List-I (Type of Arch) with List-II (Indeterminacy) and select the correct answer using the codes given below the lists:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 3

Three hinge arch:

  • In the case of a three-hinged arch, we have three hinges, two at the support and one at the crown and there are four reaction components in the three-hinged arch.
  • One more equation is required in addition to three equations of static equilibrium for evaluating the four reaction components.
  • Taking moment about the hinge of all the forces acting on either side of the hinge can set up the required equation, making 3 hinge arch determinate.

Two hinge arch:

  • In the two hinge arch, we have 2 hinges at two support of the arch and have four reaction components, as in a 3 hinge arch.
  • But here we don't have an additional equation to solve the structure and hence our 2 hinge arch is indeterminate by degree 1.

​Hinge less/ Fixed arch:

  • In the fixed arch, both the support are fixed. So, in a fixed arch six reaction forces ( One moment extra on each support ).
  • But, we have only 3 equilibrium equations. So, the degree of indeterminacy of fixed arch becomes 6 - 3 = 3.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 4

The minimum cement content (kg/m3) for a pre-specified strength of concrete (using standard notations) premised on 'free water-cement ratio' will be as

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 4

Free water-cement ratio is either be selected or specified based on the slump value and size of aggregates as per IS methods of concrete mix proportioning. Once the free water-cement ratio is determined, cement content can be calculated using the following formula:

Cement Content =

Note:

1. The formula is used to find out the absolute volume of aggregates, where C and W are cement and water content respectively and Sc is specific gravity of cement.

2. The formula is used to determine the percentage of pulverized ash content in cementing material.

The above two points are based on British DOE method of mix design.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 5
Which of the following is an ODD one regarding" requirements of good brick-earth"?
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 5

Concept:

Requirements of good earth brick are

1. It should not be mixed with salty water.

2. It must be free from lumps of lime.

3. It should not contain pebbles and organic matter.

4. It should be homogenous.

Following are the constitutions of good brick earth:

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

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

(3) Lime: A small quantity of lime not exceeding 5 percent is desirable in good brick earth.

(4) Oxides of Iron: A small quantity of oxide of iron to the extent of about 5 to 6 percent is desirable in good brick earth.

(5) Magnesia: A small quantity of magnesia in brick earth imparts yellow tint to the bricks and decreases shrinkage.

So, here statement 3 is the odd one.
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 6

The cement concrete from which entrained air and excess water are removed after placing it in position is called

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 6

Vacuum concrete:

It is the type of concrete in which the entrained air and excess water is removed for improving concrete strength. The water is removed by use of vacuum mats connected to a vacuum pump.

Lightweight concrete:

It can be defined as a type of concrete which includes an expanding agent in it, so that it increases the volume of the mixture while giving additional qualities such as lessened the dead weight.

Pre-stressed concrete:

It is a form of concrete where initial compression is given in the concrete before applying the external load so that stress from external loads are counteracted in the desired way during the service period.

Air entrained concrete:

It has effects on compressive strength of concrete and its workability.

Air entrained concrete increases the workability of concrete without much increase in water-cement ratio.

To maintain the desired compressive strength and workability of concrete together, generally in the case of higher strength concrete, admixtures are used. Air entraining agent is one such concrete admixture to increase the workability without affecting much reduction in compressive strength.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 7

In plate girders, the web plate is provided with stiffness when the ratio of clear depth to thickness of web is greater than

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 7

Case 1: Web buckling due to shear will not happen (No need of stiffners)

< 85

Case 2: Transverse stiffners are provided to prevent buckling of web due to diagonal compression which is developed due to shear force

> 85

Case 3: Horizontal stiffeners are provided above NA as they prevent buckling web due to bending compressive stress

> 200

Case 4: Additional horizontal stiffners are provided at NA

> 250

Case 5: Section must be redesigned

> 400

Where tw = Thickness of web, d = Depth of the web

Additional Information

The main purpose of Stiffener in a plate girder is to prevent the buckling of web. In which different type of stiffeners are used.

Stiffeners: Stiffeners are used to make plate girder stiff or rigid.

Different type of stiffeners according to IS 800 : 2007

a) Intermediate transverse web stiffener to improve the buckling strength of a slender web due to shear.

b) Load carrying stiffener — To prevent local buckling of the web due to concentrated loading.

c) Bearing Stiffener — To prevent local crushing of the web due to concentrated loading.

d) Torsion stiffener — To provide torsional restraint to beams and girders at supports.

e) Diagonal stiffener — To provide local reinforcement to a web under shear and bearing.

f) Tension stiffener — To transmit tensile forces applied to a web through a flange.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 8

_______ is the phenomenon of a material failing under very little stress due to repeated cycles of loading.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 8

Fatigue:

(i) When a material is subjected to repeated stresses, it fails at stresses below the yield point stresses. Such type of failure of a material is known as fatigue.

(ii) In materials science, fatigue is the progressive and localized structural damage that occurs when a material is subjected to cyclic loading.

Creep:

(i) Creep is the plastic permanent deformation of a structure under constant load for a very long period of time.

(ii) It can occur as a result of long-term exposure to high levels of stress that are still below the yield strength of the material. Creep is more severe in materials that are subjected to heat for long periods and generally increases as they near their melting point.

Resilience

(i) It is the property of materials to absorb energy and to resist shock and impact loads.

(ii) It is measured by the amount of energy absorbed per unit volume within the elastic limit this property is essential for spring materials.

(iii) The resilience of material should be considered when it is subjected to shock loading.

(iv) Resilience is the total strain energy stored in a given volume of material within the elastic limit. On removal of load, this energy is released, Hence it is Recoverable strain energy. In other words, it is the area under the load-deflection curve within the elastic limit.

Plasticity:

(i) This is the ability of material to deform without any rupture by the non-returnable way. After removing the load there are staying permanent deformations.

Endurance limit (σe):

(i) It is defined as the maximum value of completely reversed bending stress which a polished standard specimen can withstand without failure, for an infinite number of cycles (usually 106 cycles).

(ii) Hence for steel in fatigue loading the endurance limit is the maximum reversed bending stress it can withstand without failure for an infinite number of cycles.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 9

The maximum lateral deflection of a gantry girder between rails:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 9

a) For Vertical Deflection of a gantry girder

b) For Lateral deflection of a gantry girder

  • The Lateral deflection Relative displacement between rails supporting 10 mm or crane is span/400
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 10

If kx and kz are are the permeabilities in the X and Z directions respectively in a two dimensional flow situation, the effective permeability k, is given by

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 10

For an anisotropic soil, if kx and kz are permeability in two directions, modified permeability in any direction is considered to be

Additional Information

Case 1: When the flow is along the bedding plane i.e. Horizontal flow

Equivalent permeability,

Case 2: When the flow is perpendicular to the bedding plane i.e. Vertical flow

Equivalent permeability,

Keq for the horizontal direction is always greater than Keq for the vertical direction

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 11

What is the stiffness factor for a beam simply supported at both ends?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 11

Stiffness (k) is defined as the force per unit deflection or moment required per unit rotation.

Stiffness factor for a beam simply supported at both ends is 3EI/L.

Important Points:

For the far end being Hinged Supported:

If a unit rotation is to be caused at an end A for the far end being hinged support. Moment of 3EI / l is to be applied at end and hence stiffness for the member is said to be 3EI / l.

For the far end being Fixed Supported:

If a unit rotation is to be caused at an end A for the far end being fixed supported. Moment of 4EI / l is to be applied at end and hence stiffness for the member is said to be 4EI / l.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 12
Identify the INCORRECT statement with respect to analysis of trusses using method of sections.
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 12

Concepts:

The method of joints and method of section both are used for analyzing the truss structures.

Method of Sections

  • In this method, we will cut the truss into two sections by passing a cutting plane through the members whose internal forces are to be determined.
  • This method permits us to solve directly any member by analyzing the left or the right section of the cutting plane.
  • Each section may constitute of non-concurrent force system from which three equilibrium equations can be written.

Σ Fx = 0 and Σ Fy = 0 and Σ M = 0

  • Because we can only solve up to three unknowns, it is important not to cut more than three members of the truss. Depending on the type of truss and which members to solve, one may have to repeat the Method of Sections more than once to determine all the desired forces.

Additional InformationMethod of Joints

  • The free-body diagram of any joint is a concurrent force system in which the summation of the moment will be of no help.
  • Recall that only two equilibrium equations can be written

Σ Fx = 0 and Σ Fy = 0

  • This means that to solve completely for the forces acting on a joint, we must select a joint with no more than two unknown forces involved. This can be started by selecting a joint acted on by only two members.
  • The method of sections is commonly used when the forces in only a few particular members of a truss are to be determined. The method of sections is always used together with the method of joints to analyze trusses.
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 13
Which of the following range of Pigment Volume Concentration Number is recommended for paint for prime coat on metal?.
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 13

PVC (Pigment Volume Concentration): It is the ratio of the volume of pigment to the volume of total non-volatile material present in a coating. This is usually expressed as a percentage.

Coatings are usually composed of a mixture of multiple pigments and binders. In this case contributions of individual components must be considered.

Where, Vp = Volume of pigment, and Vb = Volume of binder.

PVC values for different paints is tabulated below:

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 14

The coefficient of permeability of clay is not more than

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 14

Permeability is the property of soil to transmit water through it.

It is usually expressed either as a permeability rate in centimeters per hour (cm/h), millimeters per hour (mm/h), or centimeters per day (cm/d) or as a coefficient of permeability k in meters per second (m/s) or in centimeters per second (cm/s).

For the different soil types as per grain size, the orders of magnitude of permeability are as follows:

Important Points

Permeability (k) =

The factors affecting the permeability of soil can be summarised in the below-tabulated form:

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 15

Which one of the following statements is NOT correct related to the earthquake resistant design?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 15

As per IS 13920, the code for earthquake-resistant design, the following are some important provisions:

  1. The factored axial stress on the member under earthquake loading should not exceed 0.1fck; fck is the characteristic compressive strength.
  2. The width to depth ratio should be greater than 0.3.
  3. The width of a member shall not be less than 200 mm.
  4. The overall depth of a beam should not be greater than one-fourth of the clear span.
  5. The top, as well as bottom longitudinal reinforcement, shall consist of at least two bars throughout the member length.
  6. The reinforcement resisting positive moments at a joint face must be at least half the negative moment reinforcement.
  7. The percentage tensile reinforcement should not exceed 2.5 %.
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 16

Pick up the incorrect statement from the following.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 16

Retaining wall:

  • The retaining wall is used to retain earth and resist the lateral pressure of soil at a place with a sudden change in elevation.
  • Cantilever type retaining wall is used for heights up to 6m.
  • Cantilever type retaining wall has the following components:
  1. Stem
  2. Toe slab
  3. Heel slab
  • The stability of a cantilever-type retaining wall against overturning can be obtained by taking moment about the toe of the retaining wall.
  • Due to backfill pressure overturning moment is generated which causes the overturning of the retaining wall about the toe of the wall.

Resisting moment against overturning is generated due to the:

  1. Self-weight of the wall
  2. Backfill over the heel slab

  • Let Mr = Resisting moment against overturning and Mo = Overturning moment
  • If Mr > MO⇒ Retaining wall is safe against overturning
  • If the Heel slab length increases, the Resisting moment also increases as more backfill over the heel slab tends to increase the resistance of the wall against overturning thus wall becomes safe against overturning.

Hence, In the heel slab of retaining wall, reinforcement is provided at the top of the slab.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 17
Which of the following constituents present in clay is correct in regards to Bluish green colour of bricks?
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 17

Concept-

The colours of bricks, as obtained in its natural course of manufacture, depend on the following factors:

  • Degree of dryness achieved before burning.
  • Natural colour of clay and its chemical composition.
  • Nature of sand used in moulding operation.
  • Quality of fuel used in burning operation.
  • Quantity of air admitted to the kiln during burning.
  • Temperature at which bricks are burnt.

Important Points

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 18

According to IS 800, the maximum effective slenderness ratio for a member carrying compressive loads resulting from dead loads and imposed loads is:

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 18

Maximum slenderness ratios for tension members:


Maximum slenderness ratios for compression members:

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 19

Consider the following statements regarding the slabs :

  1. When the longer span to shorter span ratio is greater than or equal to two, it is a two-way slab.
  2. In one-way slab, the load transfer is chiefly by bending in the shorter direction.
  3. In two-way slabs, the load transferred by bending in both orthogonal directions.

Which of the above statements is/are correct ?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 19

Based on bending action, the slab is classified as:

  1. One-Way Slab:
  • In the case of a one-way slab, bending is comparable in one direction to another orthogonal direction.
  • The slab is classified as a one-way slab:-
  • A rectangular slab supported on opposite sides is always a one-way slab irrespective of dimension.
  • Rectangular slab supported on all edges and aspect ratio >2. (Aspect ratio = longer span/ shorter span)

Note- The aspect ratio is valid for rectangular slab supported on all sides

2. Two-way slab:

  • In the case of the two-way slab, bending is comparable in the two-orthogonal direction.
  • A rectangular slab is classified as a two-way slab based on the following two conditions:
  • Aspect ratio
  • Supporting condition
  • If the aspect ratio <=2 and the rectangular slab is supported from all four sides then it is classified as a Two-way slab.

where, Aspect Ratio = longer span/ shorter span

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 20

What largest internal pressure can be applied to a cylindrical tank of 2 m diameter and 10 mm wall thickness? (Ultimate tensile strength of steel is 480 MPa and a factor of safety is 8.)

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 20

As per elastic theory of design, the ratio of yield stress to the working stress is called factor of safety, i.e.,

Circumferential or hoop stress:

Longitudinal or axial stress:

where d is the internal diameter and t is the wall thickness of the cylinder.

where d is the internal diameter and t is the wall thickness of the cylinder.

Calculation:

Given:

d = 2 m = 2000 m, t = 10 mm, N = 8, Ultimate strength = 480 MPa

Working Strength (σh) = 60 MPa

Maximum or Circumferential or hoop stress,

P = 0.6 MPa

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 21

A well-graded sand should have

1. Uniformity coefficient greater than 6.

2. Coefficient of curvature between 1 and 3.

3. Effective size greater than 1 mm.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 21

For well-graded sand, only Cu and Cc are defined,

  1. The uniformity coefficient (Cu) is greater than 6.
  2. Coefficient of curvature (Cc) between 1 and 3.

Note:

No specific value for Effective size is given for well-graded sand.

Coefficient of Uniformity / Uniformity Coefficient:

It is defined as the ratio of D60 and D10 sieve sizes in sieve analysis of granular material.

Higher is the value of Cu larger is the range of the particle size.

For uniformly graded soil, Cu = 1

For well graded sand, Cu > 6

For well graded gravel Cu > 4

Coefficient of Curvature / Curvature Coefficient:

For well graded soil, 1 < Cc < 3

For gap graded soil, 1 > Cc or Cc > 3

where

D60 = size at 60% finer by weight

D30 = size at 30% finer by weight

D10 = size at 10% finer by weight = Effective size

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 22
According to IS-800, which of the following is NOT a limit state of durability consideration?
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 22

Concept:

Definition of Limit state:

According to IS 800:2007 clause 5.2, Limit states are the states beyond which the structure no longer satisfies the performance requirements specified. The limit states are classified

  1. Limit state of strength - Strength-related factors
  2. Limit state of serviceability - Durability related factors

Limit state of strength:

The limit states of strength are those associated with failures (or imminent failure), under the action of a probable and most unfavorable combination of loads on the structure using the appropriate partial safety factors, which may endanger the safety of life and property. The limit state of strength includes

  1. Loss of equilibrium of the structure as a whole or any of its parts or components
  2. Loss of stability of the structure (including the effect of sway where appropriate and overturning) or any of its parts including supports and foundations
  3. Failure by excessive deformation, rupture of the structure or any of its parts or components
  4. Fracture due to fatigue
  5. Brittle fracture

Limit state of serviceability

  1. Deformation and deflections, which may adversely affect the appearance or effective use of the structure or may cause improper functioning of equipment or services, or may cause damages to finishes and non-structural members
  2. Vibrations in the structure or any of its components cause discomfort to people, and damage to the structure, and its contents or which may limit its functional effectiveness. Special consideration shall be given to systems susceptible to vibration, such as large open floor areas free of partitions to ensure that such vibrations are acceptable for the intended use and occupancy
  3. Repairable damage or crack due to fatigue
  4. Corrosion, durability
  5. Fire​
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 23

Choose the correct statement.

a) The strength of fully compacted concrete is inversely proportional to the water cement ratio.

b) The strength of fully compacted concrete is directly proportional to the water cement ratio.

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 23

The compressive strength of concrete decreases with increase in the water-cement ratio of the concrete mix.

Water is added to the concrete mix for hydration of cement and workability.

Abrams water/cement ratio law states that the strength of concrete is only dependent upon water/cement ratio provided the mix is workable. According to Abram’s law, the compressive strength increases parabolically with decreasing water-cement ratio.

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 24

The difference between maximum void ratio and minimum void ratio of a sand sample is 0.30, if the relative density of this sample is 66.6% at a void ratio of 0.40. then the void ratio of this sample at its loosest state will be

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 24

Concept:

Relative density or Density index (ID):

where,

emax = Maximum voids ratio, in loosest state

emin = Minimum voids ratio, in densest state

e = Natural or In-situ void ratio

emin ≤ e ≤ emax

ID ranges from 0 ≤ ID ≤ 100%


Calculation:

Given,

Difference between maximum void ratio and minimum void ratio of a sand sample is 0.30

∴ emax - emin = 0.30

ID = 66.6%, e = 0.40

We know that

∴ emax = 0.60

Hence the void ratio of this sample at its loosest state will be 0.60

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 25

A simply supported beam AB of length 9 m, carries a uniformly distributed load of 10 N/m for a distance of 6 m from end A. What are the reaction forces at A and at B?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 25

Concept:

  • The Sum of forces in the x and y direction must be equal to zero.
    • ​∑Fx = 0,∑Fy = 0
  • The sum of the moments is equal to zero.
    • ∑MA = 0

​Calculation:

Given:

Length = 9 m, uniformly distributed load 10 N/m for 6 m from the end.

∑Fx = 0

∑Fy = 0

RA + RB = 60.....(1)

∑MA = 0

60 × 3 - RB × 9 = 0

RB = 20 N, putting RB in equation (1)

RA = 40 N

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 26

Which type of soil classification system is also called American Association of State Highway officials (AASHO)?

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 26

Highway research Board classification of soils is also called American Association of State Highway Officials (AASHO) classification of Revised Public Roads Administration soil classification system.

According to HRB classification, Soils are divided into seven groups:

A - 1, A - 2, A - 3, A - 4, A - 5, A - 6, and A - 7

Where,

A – 1, A – 2 and A – 3 soils are granular soils (percentage fines passing 0.074 mm sieve being less than 35) and

A – 4, A – 5, A – 6, A – 7 soils are fine grained or silt - clay soils (percentage fines passing 0.074 mm sieve being greater than 35).

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 27
The minimum thickness of steel of the tension members exposed to weather and not accessible is:
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 27

Concept:

Following are the specifications for steel member exposed to weather:

  • When the steelwork is directly exposed to weather and is fully accessible for cleaning and repainting the thickness shall not be less than 6 mm.
  • When the steelwork directly exposed to weather and is not accessible for cleaning and repainting the thickness shall not be less than 8 mm.
  • When the steelwork is not directly exposed to weather, the thickness of steel in main members shall not be less than 6 mm.
  • When the steelwork is not directly exposed to weather, the thickness of steel in secondary members shall not be less than 6 mm.
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 28

For a shaft transmitting power 'P' at rpm N, the diameter of shaft would be proportional to

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 28

Power Transmitted by a shaft P

From torsion equation

τ is the allowable shear stress in the shaft, J = Polar moment of inertia, T is the torque acting on the shaft, r is the radius of the shaft

Now from torsion equation

BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 29

The phenomenon for the internal transfer of forces from one leg to the other is

Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 29

Shear lag occurs when some elements of the member cross-section are not connected.

  • Shear lag is a function of the distribution of steel in the section and the length of the load transfer L. The effect of shear lag is less in the large length of the connection.
  • Consider an angle section tension member connected with one leg only. Consequently, due to this partial connection, the connected leg will be overloaded and the unconnected leg will not be fully stressed.
  • Further, since at the joint/connection more of the load is carried by the connected leg, and it takes the transition distance, for the stress to spread uniformly across the whole angle, stress distribution in the two legs of the section would be different.
  • In the transition region, the stress in the connected part of the member may even exceed fy and go into strain–hardening range; the member may fracture prematurely. Away from joint/connection, the stress distribution is more uniform. In the transition region, the shear transfer lags.
  • Since shear lag reduces the effectiveness of the component plates of the tension member that are not connected directly to a gusset plate, the outstanding legs are kept shorter in length. For this reason, unequal angles with long legs connected are preferred.
  • It is independent of the type of load and applies to both bolted and welded connections. Of course, bolted connections will be affected more than welded connections because of the reduction of the effective area due to bolt holes. The shear lag can be accounted for by using reduced net area, and is known as effective net area.
BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 30
Maximum depth of neutral axis for singly reinforced beam with Fe500 is (here, depth of the beam = d)
Detailed Solution for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 - Question 30

Concept:

Design formulas of singly reinforced Beam for limit state design,

1) Limiting depth of neutral axis

2) Actual depth of neutral axis

3) Type of section

xu,max = xu for balanced section

xu,max > xu for under reinforced section

xu,max < xu for over reinforced section

4) Moment of resistance

Mu = 0.36 fck B xu (d - 0.42xu) for compression side

Mu = 0.87 fy Ast (d - 0.42xu) for tensile side

View more questions
15 tests
Information about BPSC AE Civil Paper 5 (Civil) Mock Test - 1 Page
In this test you can find the Exam questions for BPSC AE Civil Paper 5 (Civil) Mock Test - 1 solved & explained in the simplest way possible. Besides giving Questions and answers for BPSC AE Civil Paper 5 (Civil) Mock Test - 1, EduRev gives you an ample number of Online tests for practice

Top Courses for Civil Engineering (CE)

Download as PDF

Top Courses for Civil Engineering (CE)