Previous Year Paper

BPSC AE Paper 5 (Civil Engineering) 2012 Official Paper (Previous Year Paper)

Held in 2012

50 questions60 minutes
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Technical (50 questions)

Question 1

For steel members exposed to weather and not accessible for repainting, the thickness of steel should not be less than

  1. AOption (a)

    4.5 mm

  2. BOption (b)

    6 mm

  3. COption (c)

    8 mm

  4. DOption (d)

    10 mm

View answer & explanation

Correct answer

Answer: (c)C.

8 mm

Why this answer is correct

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.

Question 2

Limiting deflection for steel becomes supporting floor and false ceiling should be

  1. AOption (a)

    span/600

  2. BOption (b)

    span/500

  3. COption (c)

    span/300

  4. DOption (d)

    span/200

View answer & explanation

Correct answer

Answer: (c)C.

span/300

Why this answer is correct

As per IS 800: 2007, Table 6 Deflection limits,

Limiting deflection for steel becomes supporting floor and roof (elements not susceptible to cracking) should not be more than span/300.

Limiting deflection for steel becomes supporting floor and roof (elements susceptible to cracking) should not be more than span/360.

Question 3

At liquid limit, the

  1. AOption (a)

    shear strength of soil is infinitesimal small

  2. BOption (b)

    shear strength is infinity

  3. COption (c)

    soil is partially saturated

  4. DOption (d)

    shear strength is considerable

View answer & explanation

Correct answer

Answer: (a)A.

shear strength of soil is infinitesimal small

Why this answer is correct

Concept:

Liquid limit:

It is the water content corresponding to the arbitrary limit between liquid and plastic state of consistency of soil.

It is defined as the minimum water content at which the soil is still in the liquid state, but has infinitesimal resistance against flow which can be measured by any standardized procedure.

At liquid limit ,shear strength is very small i.e. 2.7KN/m2 

With reference to the standard liquid limit device, it is defined as the minimum water content at which a part of soil cut by a groove of standard dimensions, will flow together for a distance of 12 mm under an impact of 25 blows in the device.

Other Important points:

Plastic limit: It is defined as the minimum water content at which a soil will just begin to crumble when rolled into a thread approximately 3 mm in diameter.

Shrinkage limit: It is defined as the maximum water content at which a reduction in water content will not cause a decrease in the volume of a soil mass. It is the lowest water content at which a soil can still be completely saturated.

Question 4

Falling head permeability test is used for

  1. AOption (a)

    sandy soil

  2. BOption (b)

    clayey soil

  3. COption (c)

    silty sand

  4. DOption (d)

    sandy gravel

View answer & explanation

Correct answer

Answer: (b)B.

clayey soil

Why this answer is correct

Concept:

Constant head permeability test is carried out to determine the permeability of coarse-grained soil only where a reasonable discharge can be collected in a given time.

However, the falling head permeability test is used for relatively less permeable soils (fine-grained) where the discharge is small.

So here in clayey soil falling head permeability test can be used.

For falling head permeability test,

K=aALt;ln(h1h2)K = \frac{a}{A}\frac{L}{t};ln\left( {\frac{{{h_1}}}{{{h_2}}}} \right)

k=2.303aA×Lt10g10(h1h2){\rm{k}} = 2.303\frac{{\rm{a}}}{{\rm{A}}} \times \frac{{\rm{L}}}{{\rm{t}}}10{{\rm{g}}_{10}}\left( {\frac{{{{\rm{h}}_1}}}{{{{\rm{h}}_2}}}} \right)

Where,

k = permeability, a = Area of tube in m2, A = Area of sample in m2, t = time in sec, L = length in m, h1 = Level of time t = 0, and h2 = Level of time t

Question 5

Westergaard’s theory is more suitable for

  1. AOption (a)

    layered soil

  2. BOption (b)

    homogenous soil

  3. COption (c)

    isotropic soil

  4. DOption (d)

    normally consolidated homogenous soil

View answer & explanation

Correct answer

Answer: (a)A.

layered soil

Why this answer is correct

Westergaard’s Theory Assumptions: Elastic medium of semi-infinite extent but containing numerous, closely spaced horizontal sheets of a negligible thickness of an infinite rigid material which permits only downward deformation as a whole without allowing it to undergo any lateral strain.

Westergaard's assumptions are more close to the field reality, especially for over-consolidated and laminated sedimentary or stratified soils, which exhibit marked anisotropy.

Based on this criterion of no lateral displacement, Westergaard derived the following equation for a point load, Q, at a depth z from the surface:

σz=QZ2π[11+2(rz)2]3/2{\sigma _z} = \frac{Q}{{{Z^2}\pi }}{\left[ {\frac{1}{{1 + 2{{\left( {\frac{r}{z}} \right)}^2}}}} \right]^{3/2}}

∴ Westergaard’s analysis for stress distribution beneath loaded areas is more suitable to stratified or layered soils.

Question 6

Expansion of soil under shear is called

  1. AOption (a)

    liquefaction

  2. BOption (b)

    volumetric deformation

  3. COption (c)

    critical expansion

  4. DOption (d)

    dilatancy

View answer & explanation

Correct answer

Answer: (d)D.

dilatancy

Why this answer is correct

Dilatancy: Volume change observed in the soil sample when subjected to shear deformation at constant pressure is called dilatancy.

Fine-grained soils can be identified in the field by dilatancy test. In dilatancy test, a wet soil sample is kept in the palm and observe the movement of water during shaking, squeezing (by closing hand)   and following criteria is used generally for classifying the fine-grained soil:

Soil TypeObservation during Dilatancy test
ClayWater slowly appears and remains on the surface during shaking or water slowly disappears upon squeezing.
SiltWater quickly appears on the surface during shaking and quickly disappears upon squeezing.

 

Liquefaction: If the rate of loading is larger and the soil is fully saturated with gravity water then positive pore water pressure will develop. This will reduce effective stress and hence shear strength of the soil. A time comes when the effective stress reduces to zero, the soil will lose all its shear strength. This phenomenon is known as liquefaction.

Question 7

Most of the shear tests are done on equipment which are

  1. AOption (a)

    stress-controlled

  2. BOption (b)

    strain-controlled

  3. COption (c)

    drainage-controlled

  4. DOption (d)

    volume-controlled

View answer & explanation

Correct answer

Answer: (b)B.

strain-controlled

Why this answer is correct

Shear Tests of soil:

Most of the shear tests are done to determine the shear strength of soil on equipments which are strain-controlled, as  because in a shear test, the focus is to apply a constant rate of strain and measure the resultant stress. This practice provides a clearer understanding of the material's behavior under shear stress.

Additional Information

Type of testType of SoilType of Shear Strength that can be measured
Direct shearSaturated sandsUndrained shear strength because there is no mechanism to measure pore water pressure
Triaxial shearAll type of soils i.e. sand, clay, siltBoth drained and undrained shear strength
Vane shearSaturated claysUndrained shear strength because there is no mechanism to measure pore water pressure
Unconfined compressionSoft clays having zero angle of internal frictionBoth drained and undrained shear strength

Question 8

The maximum size of plate for plate load test is

  1. AOption (a)

    30 cm square

  2. BOption (b)

    45 cm square

  3. COption (c)

    60 cm square

  4. DOption (d)

    75 cm square

View answer & explanation

Correct answer

Answer: (d)D.

75 cm square

Why this answer is correct

Plate load test:

Plate load test is performed to determine soil bearing capacity and settlement.

It is a semi-direct method to measure the allowable pressure of soil by inducing a given amount of settlement. The load on the plate is applied by making use of a hydraulic jack. The reaction of the jack load is taken by a cross beam or a steel truss anchored suitably at both the ends. The settlement of the plate is measured by a set of three dial gauges of sensitivity 0.02 mm placed at 120° apart. The dial gauges are fixed to independent supports which do not get disturbed during the test.

Shape and size of the plate load test are given below:

  • Square or circular in shape
  • Thickness > = 25 mm
  • Size may vary from 300 to 750 mm. therefore, the maximum size is 750 mm
  • For clayey and silty soils and for loose to medium dense sandy soils with N < 15, 450 mm square plate is used
  • For dense sandy or gravelly soils(15 < N < 30) three plates of sizes 300 mm to 750 mm is used depending  on reaction loading and maximum grain size
  • Side of plate = 4 × size of particles

Question 9

Which of the following exhibits maximum deformation on ground level?

  1. AOption (a)

    Local shear failure

  2. BOption (b)

    General shear failure

  3. COption (c)

    Punching shear failure

  4. DOption (d)

    Composite failure

View answer & explanation

Correct answer

Answer: (b)B.

General shear failure

Why this answer is correct

Concept:

1) General Shear Failure:

  • This type of failure is seen in dense and stiff soil.
  • There is a considerable bulging of sheared mass of soil adjacent to the footing, resulting in the well-defined failure surface which reaches up to the ground level hence the deformation on the ground level is maximum in this case
  • Dense or stiff soil that undergoes low compressibility experiences this failure.

2) Local Shear Failure:

  • Bulging around the soil mass is very less and hence the deformation caused by it is also less
  • This type of failure is seen in relatively loose and soft soil.
  • Failure is not sudden and there is no tilting of footing.

3) Punching Shear Failure:

 

  • No bulging of the foundation, and hence no deformation on the ground level
  • This type of failure is seen in loose and soft soil and at deeper elevations.

Confusion Points In case of punching failure the settlement is maximum, but the question is not asking for settlement, but it is asking about deformation on the ground surface, which can be seen by the bulging of the ground level.

Question 10

Which one of the following planes is not most likely to be the failure plane in Sandy soil?

  1. AOption (a)

    Maximum shear stress plane

  2. BOption (b)

    Planes carrying maximum normal stress

  3. COption (c)

    Principal plane

  4. DOption (d)

    Planes with maximum angle of obliquity

View answer & explanation

Correct answer

Answer: (d)D.

Planes with maximum angle of obliquity

Why this answer is correct

Concept of failure in soils:

Soil is a visco-elastic material therefore, it is difficult to define the point of failure in soils.

Hence, it is based on a hypothesis that material fails when the shear stresses on the failure plane at failure reaches the value which is a function of normal stress on that plane.

Where,

σ1 = major principle stress, σ2 = minor principle stress, σ = normal stress, and τ = Shear stress

Maximum shearing stresss occurs on the planes inclined at 45° to principle planes,

τ=σ1−σ32{\rm{\tau }} = \frac{{{{\rm{\sigma }}_1} - {{\rm{\sigma }}_3}}}{2}

The normal stress on plane of maximum shear are equal to each other,

σ12=σ1+σ32{{\rm{\sigma }}_{12}} = \frac{{{{\rm{\sigma }}_1} + {{\rm{\sigma }}_3}}}{2}

The resultant stress at any point is σ2+τ2\sqrt {{{\rm{\sigma }}^2} + {{\rm{\tau }}^2}} and the obliquity (β) is tan⁡−1(τσ){\tan ^{ - 1}}\left( {\frac{{\rm{\tau }}}{{\rm{\sigma }}}} \right)

The plane of maximum obliquity is most liable to failure but in case of sandy soil it is not most likely to be the failure plane due to internal friction (i.e. β = ϕ).

Question 11

In which one of the following pairs of soil types would you anticipate negative pore pressure when subjected to shearing?

  1. AOption (a)

    Normally consolidated clay and dense sand

  2. BOption (b)

    Overconsolidated clay and loose sand

  3. COption (c)

    Less consolidated clay and overconsolidated clay

  4. DOption (d)

    Dense sand and overconsolidated clay

View answer & explanation

Correct answer

Answer: (d)D.

Dense sand and overconsolidated clay

Why this answer is correct

Concept:

A pore pressure parameter greater than one is associated with a loose structure, either in sand or clay, which collapses upon load application.

On the other hand, if we have a sample of sand or clay which tends to expand upon loading, negative pore pressure can be developed.

If a sample of dense sand is subjected to shear, it tends to expand. This is due to the soil particles trying to ride on each other during sliding. The expansion under shear is termed as “Dilatancy”.

In case of saturated dense sands if there is sudden application of stress in the field, the soil does not get time to expand and negative pore water pressure develop.

In case of loose sands, they try to compress by bringing their grains closer together. This results in reduced effective stress and reduction in corresponding shear strength.

If fine sands in the saturated condition are subjected to sudden loading, the tendency towards volume reduction causes pore water pressure to increase instantaneously. This causes a large decrease in the shear resistance. 

So dense sand and over-consolidated clay would anticipate negative pore pressure when subjected to shearing.

Question 12

In a saturated clay layer consolidation with single drainage, the initial isochrones is a

  1. AOption (a)

    triangle

  2. BOption (b)

    square

  3. COption (c)

    rectangle

  4. DOption (d)

    parabola

View answer & explanation

Correct answer

Answer: (d)D.

parabola

Why this answer is correct

Concept:

 

Figure shows a clay layer, of thickness H, sandwiched between two layers of sand which serves as drainage faces.

When the layer is subjected to a pressure increment Δσ, excess hydrostatic pressure is set up in the clay layer. At the time to, the instant of pressure application, whole of the consolidating pressure Δσ is carried by the pore water so that the initial excess hydrostatic pressure u0 is equal to Δσ and is represented by a straight line u’ = Δσ on the pressure distribution diagram.

The straight line CED joining the water levels in the piezometric tubes represent this distribution. As water starts escaping into the sand, the excess hydrostatic pressure at the pervious boundaries drops to zero and remains so at all times.

After a very time tf, the whole of the excess hydrostatic pressure is dissipated so that u’=0, represented by line AGB. At an intermediate time t, the consolidating pressure Δσ is partly carried by water and partly by soil, and the following relationship is obtained.

Δσ = Δσ’ + u’

The distribution of excess hydrostatic pressure u’ at any time t is indicated by the curve AFB, joining water levels, in the piezometric tubes; this curve is known as isochrones, and number of such isochrones can be drawn at various time intervals t1,t2,t3 etc.

The slope of isochrones at any point at a given time indicates the rate of change of u’ with depth.

In a saturated clay layer consolidation with single drainage, the initial isochrones is a parabola.

Question 13

As per IS code, the frequency ratio (ratio of operating frequency of a machine to the natural frequency of soil) should not be within the range of

  1. AOption (a)

    0.5 to 1.5

  2. BOption (b)

    1.0 to 2.5

  3. COption (c)

    1.5 to 3.0

  4. DOption (d)

    3.0 to 6.0

View answer & explanation

Correct answer

Answer: (a)A.

0.5 to 1.5

Why this answer is correct

Concept:

IS code of practice (IS : 2974,Part I : 1964) gives the following criteria :

  1. The size of foundation block (in plan) should be larger than the bed plate of the machine with a minimum allround clearance of 15 cm.
  2. The width of the foundation should be at least equal to the distance of the centre of gravity of crank shaft to the bottom of foundation in all vertical machines.
  3. The depth of the foundation should be such as to rest the foundation on a good bearing strata and to ensure stability against rotation in a vertical plane.
  4. The combined centre of gravity of machine and the foundation block should be as much below the top of the foundation as possible.
  5. Whenever possible, the operating frequency should be lower than the natural frequency of the foundation soil system and the frequency ratio should be less than 0.5. When the operating frequency is higher than the natural frequency of the foundation soil system of the machine, the frequency ratio should be more than 2 for important machines and more than 1.5 for others.

Question 14

Under dynamic loading, the soil most likely to be affected is

  1. AOption (a)

    loose saturated clay

  2. BOption (b)

    loose saturated sand

  3. COption (c)

    dense saturated sand

  4. DOption (d)

    loose saturated silt

View answer & explanation

Correct answer

Answer: (b)B.

loose saturated sand

Why this answer is correct

Concept:

In a soil mass

Total stress (σ)

= Effective stress (σ’) + pore water pressure (U)

Effective stress (σ’)

= Total stress (σ) - pore water pressure (U)

In loose saturated sand, due to seismic disturbance (Earthquake) or dynamic loading volume decreases hence pore pressure change is positive.

Due to build-up of high pore pressure sudden decrease in effective stress and decrease in shear strength is recorded, consequently, large settlement of foundation suddenly occurs along with vertical upward flow of muddy water. Such a phenomenon is called liquefaction of sand.

It is generally observed near river and sea during the earthquake.

So Under dynamic loading, the soil most likely to be affected is loose saturated sand.

Question 15

Negative skin friction on a pile under vertical compressive load acts

  1. AOption (a)

    downwards and increase the load-carrying capacity

  2. BOption (b)

    downwards and reduce the load-carrying capacity

  3. COption (c)

    upwards and increase the load-carrying capacity

  4. DOption (d)

    upwards and reduce the load-carrying capacity

View answer & explanation

Correct answer

Answer: (b)B.

downwards and reduce the load-carrying capacity

Why this answer is correct

Concept:

Negative skin friction: It is usually a downward shear drag acting on a pile or pile group because of downward movement of surrounding soil relative to the piles. Such a force can exist under one of the following conditions:

  1. If a fill of clay soil is placed over a granular soil layer into which a pile is driven, the fill will exert a downward drag force on the pile due to consolidation.
  2. If a fill of granular soil is placed over a layer of soft clay, it will induce the process of consolidation in the clay layer and thus exert a downward drag on the pile.
  3. Lowering of the water table will increase the vertical effective stress on the soil at any depth, so it can induce negative skin friction on piles.
<br>

Now axial capacity of a pile is summation of upward reaction due to bearing at base and net upward skin frictional resistance. As, the negative skin friction (acting downward) lowers the net skin resistance, it in turn reduces the axial capacity of piles.

A schematic diagram of negative skin friction on piles is shown below:

∴ Negative skin friction on a pile under vertical compressive load acts downwards and reduce the load-carrying capacity.

Question 16

According to Skempton’s formula for a surface floating of square shape, the net ultimate bearing capacity on a purely cohesive soil of cohesion ‘C’ is

  1. AOption (a)

    1.4 C

  2. BOption (b)

    6.0 C

  3. COption (c)

    7.4 C

  4. DOption (d)

    9.0 C

View answer & explanation

Correct answer

Answer: (b)B.

6.0 C

Why this answer is correct

Concept:

Skempton’s formula for Bearing Capacity: Skempton had provided formulas for bearing capacity factor Nc for different depth to width ratio and for purely cohesive soil. According to Skempton for bearing capacity factor Nc of a square shaped foundation for purely cohesive soil is,

\({{\rm{N}}{\rm{c}}} = 6\left[ {1 + 0.2\frac{{{{\rm{D}}{\rm{f}}}}}{{\rm{B}}}} \right]\)where;DfB{\rm{;}}\frac{{{{\rm{D}}_{\rm{f}}}}}{{\rm{B}}} is less than 2.5  

Also, net ultimate bearing capacity is given by,

\({{\rm{q}}{{\rm{nf}}}} = {\rm{c}}{{\rm{N}}{\rm{c}}} + {\rm{;\gamma }}{{\rm{D}}{\rm{f}}}\left( {{{\rm{N}}{\rm{q}}} - 1} \right) + 0.5{\rm{\gamma B}}{{\rm{N}}_{\rm{\gamma }}}\)

Where, c is the cohesion of the soil, Df is the depth of foundation, B is the width of foundation, γ is unit weight of the soil, Nc, Nq and Nγ are the bearing capacity factors.

Calculations:

Now, for surface floating foundation depth of the foundation is zero.

∴;Nc=6[1+0.2×0B]=6\therefore {\rm{;}}{{\rm{N}}_{\rm{c}}} = 6\left[ {1 + 0.2 \times \frac{0}{{\rm{B}}}} \right] = 6

Now, for purely cohesive soils, Nq =1 and Nγ = 0.

\(\therefore {\rm{Ultimate;bearing;capacity}},{\rm{;}}{{\rm{q}}{{\rm{nf}}}} = {\rm{C}} \times 6 + {\rm{;\gamma }}{{\rm{D}}{\rm{f}}}\left( {1 - 1} \right) + 0.5{\rm{\gamma B}} \times 0 = 6{\rm{;C}}\)

Question 17

Undisturbed soil samples are required for conducting

  1. AOption (a)

    hydrometer test

  2. BOption (b)

    shrinkage limit test

  3. COption (c)

    consolidation test

  4. DOption (d)

    specific gravity test

View answer & explanation

Correct answer

Answer: (c)C.

consolidation test

Why this answer is correct

Concept:

Hydrometer Test: This is used to determine the particle size distribution of fine-grained soils passing 75 µ sieve. For this test both disturbed and undisturbed soil sample can serve the purpose.

Shrinkage Limit Test: This is one of the Atterberg Limit test to find out the shrinkage limit of the soil. For this test the soil specimen should have water content greater than liquid limit of the soil but both disturbed and undisturbed soil sample can serve the purpose.

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 disturbed specimens have a certain pre consolidation pressure, so for conducting the consolidation test undisturbed soil specimen is required.

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

Specific Gravity Test: This is used for determination of specific gravity of soil particles of both fine grained and coarse grained soils. This test is conducted by Pycnometer. For this test both disturbed and undisturbed soil sample can serve the purpose.

∴ Undisturbed soil samples are required for conducting consolidation test

Question 18

In consolidation testing curve, fitting method is used to determine

  1. AOption (a)

    compression index

  2. BOption (b)

    swelling index

  3. COption (c)

    coefficient of consolidation

  4. DOption (d)

    time factor

View answer & explanation

Correct answer

Answer: (c)C.

coefficient of consolidation

Why this answer is correct

Concept:

Time Fitting Method: It is the method for determination of the coefficient of consolidation \(({{\rm{C}}{\rm{v}}})\) in the consolidation testing in laboratory. In this method the graph for the theoretical relationship between time factor \(({{\rm{T}}{\rm{v}}})\) & degree of consolidation (U) is plotted and also the graph for the practical observed relationship between elapsed time (t) & degree of consolidation (U) of the specimen is plotted. Then those two characteristics are correlated by convenient method to find coefficient of consolidation.

Cv=T50d2t50{C_v} = \frac{{{T_{50}}{d^2}}}{{{t_{50}}}}

Where, T50 = time factor for 50% consolidation, d = length of drainage path, t50 = time required for 50% consolidation

Generally, two kinds of time fitting method are used which are the followings:

  1. Square root of time fitting method: Theoretical curve plotted between degree of consolidation (U) at Y axis and square root of time factor (√Tv)(\surd {{\rm{T}}_{\rm{v}}}) at X axis. Practical curve plotted for dial gauge reading (R) at Y axis and square root of elapsed time (t)\left( {\sqrt {\rm{t}} } \right) at X axis.
  2. Logarithmic of time fitting method: Theoretical curve plotted between degree of consolidation (U) at Y axis and logarithmic of time factor \(({\log {10}}{{\rm{T}}{\rm{v}}})\) at X axis. Practical curve plotted for dial gauge reading (R) at Y axis and square root of elapsed time (log⁡10t)\left( {{{\log }_{10}}{\rm{t}}} \right) at X axis.
<br>

∴ In consolidation testing curve, fitting method is used to determine coefficient of consolidation.

Question 19

Mohr’s circle is drawn for­

  1. a point
  2. a square block
  3. a rectangular block
  4. isotropic material

The correct answer is

  1. AOption (a)

    1, 2 and 4

  2. BOption (b)

    2 and 4

  3. COption (c)

    1 and 4

  4. DOption (d)

    2, 3 and 4

View answer & explanation

Correct answer

Answer: (c)C.

1 and 4

Why this answer is correct

Concept:

  • Mohr’s circle is drawn for representing the state of stress of the points inside a material situated at different locations with respect to external loading.
  • A point on Mohr’ circle will represent the state of stress for all points which are located on a particular plane inside the material.
  • A square or rectangular block is frequently used to depict the bi axial stress system acting on a certain point inside the material but it does not imply that Mohr’s circle is drawn for square or rectangular block.
  • Mohr’s circle is always drawn for isotropic material, because the theoretical approach for constructing Mohr’s circle does not applicable to anisotropic materials.
<br>

∴ Mohr’s circle is drawn for a point and isotropic material.

Question 20

In a plane stress problem

  1. normal stress in the third direction is zero
  2. strain in the third direction is zero
  3. strain in all directions is present
  4. normal stresses in all directions are present

The correct answer is

  1. AOption (a)

    1 and 4

  2. BOption (b)

    2 and 4

  3. COption (c)

    1 and 3

  4. DOption (d)

    2 and 3

View answer & explanation

Correct answer

Answer: (c)C.

1 and 3

Why this answer is correct

Concept:

Plane Stress Problem: The plane stress problem is characterized by very small dimensions of the structure/ member in one of the normal directions as compared to other two dimensions. In this case state of stress is defined by only three stress components which are normal stress in X direction (σxx) , normal stress in Y direction (σyy) and shear stress in XY direction (τxy) and all other stress components normal stress in Z direction (σzz) , shear stress in YZ direction (τyz) and shear stress in XZ direction (τxz)are taken as zero where Z axis is the axis through small dimension. However, all strain components are present in this case due to constitutive relationship.

Examples: Thin plate subjected to in plane loading, 2D planner element with negligible thickness subjected axial loading etc. which are shown in following diagram:

          

Related Concept:

Plane Strain Problem: Problems involving long bodies whose geometry and loading do not vary significantly in the longitudinal direction are referred to as plane strain problems. In this case the strain components \({{\rm{\varepsilon }}{{\rm{zz}}}}{\rm{;}},{\rm{;}}{{\rm{\varepsilon }}{{\rm{yz}}}}{\rm{;and;}}{{\rm{\varepsilon }}_{{\rm{xz}}}}\) are taken as zero where Z axis is the longitudinal axis.

Examples: Dam and reservoirs are subjected to water pressure loading etc. which are shown in following diagram:

Question 21

Choose the correct statements from the following:

  1. Hooke’s law is valid up to the limit of proportionality.
  2. Hooke’s law is valid up to elastic limit.
  3. Limit of proportionality is always less than the elastic limit.
  4. Limit of proportionality is either equal to or less than the elastic limit.

The correct answer is

  1. AOption (a)

    1 and 4

  2. BOption (b)

    2 and 4

  3. COption (c)

    1 and 3

  4. DOption (d)

    2 and 3

View answer & explanation

Correct answer

Answer: (a)A.

1 and 4

Why this answer is correct

Statement 1: True

Hooke’s Law states that the strain in a solid body is directly proportional to the applied stress and this condition is valid upto the limit of proportionality (i.e. point A in the figure).

Stress;∝Strain→;σ∝ε;→;σ=EεStress; \propto Strain \to ;\sigma \propto \varepsilon ; \to ;\sigma = E\varepsilon

Statement 4: True

Limit of proportionality is the stress at which the stress-strain curve ceases to be a straight line. It is the stress at which extension ceases to be proportional to strain. The proportional limit is important because all subsequent theory involving the behavior of elastic bodies is based on the stress-strain proportionality.

Elastic limit is is that point in the stress-strain curve up to which the material remains elastic, i.e. the material regains its shape after the removal of the load.

However, for many materials, elastic limit and proportional limit are almost numerically the same and the terms are sometimes used synonymously. In case where the elastic limit and proportional limit are different, the elastic limit is always greater than the proportional limit.

Question 22

A copper rod of square cross-section is fixed between two rigid supports and over which a steel rod of square cross-section is simply placed. If the temperature of the whole assembly is raised by T °C, the stresses in steel and copper respectively are 

  1. AOption (a)

    tensile and compressive

  2. BOption (b)

    zero and compressive

  3. COption (c)

    compressive and tensile

  4. DOption (d)

    compressive and zero

View answer & explanation

Correct answer

Answer: (b)B.

zero and compressive

Why this answer is correct

Concept:

Thermal Stress: This kind of stresses are induced in the material when its temperature is changed but the material is unable to accommodate its change in dimensions due to constraints. As long as the material is free to move, the material can expand or contract freely without generating thermal stresses. When temperate of the material is increased, the constraints exert a compressive stress on the material. The opposite happens while cooling i.e. the stress generated on the material will be tensile.

Now for this case,

  1. The copper rod is fixed between two rigid supports, it means it is unable to expand. So, when the temperature is raised compressive stress is generated in the copper rod.
  2. On the other hand the steel is placed freely, it means it can expand freely without any constraints. So, when the temperature is raised there will be no thermal stress generation in steel rod.

 

Note: This is not a case of composite bar.

Question 23

Choose the correct statements from the following:

  1. A stable and statically determinate real beam will have a stable and statically determinate conjugate beam.
  2. A statically indeterminate real beam will have unstable conjugate beam.
  3. A statically indeterminate conjugate beam is unstable real beam.

The correct answer is

  1. AOption (a)

    1 and 3

  2. BOption (b)

    2 and 3

  3. COption (c)

    1 and 2

  4. DOption (d)

    1, 2 and 3

View answer & explanation

Correct answer

Answer: (d)D.

1, 2 and 3

Why this answer is correct

Concept:

Conjugate Beam Method: This method is defined as the imaginary (or conjugate) beam with the same dimensions (length) as that of the original beam but load at any point on the conjugate beam is equal to the bending moment at that point divided by EI. According to this method a hinged, roller, fixed, free and intermidaite support in real beam is tranformed as hinged, roller, free, fixed and internal hinge in conjugate beam respectively.

Now consider the statements one by one with the following examples and diagrams:

Statement 1:

 

So, a stable and statically determinate real beam will have a stable and statically determinate conjugate beam.

∴ Statement 1 is correct.

Statement 2: 

So, a statically indeterminate real beam will have unstable conjugate beam.

∴ Statement 2 is correct.

Statement 3: 

So, a statically indeterminate conjugate beam is unstable real beam.

∴ Statement 3 is correct.

Question 24

A simply-supported beam of length l carries two equal unlike couples M at two ends. If the flexural rigidity EI = constant, then the central deflection of the beam is given by

  1. AOption (a)

    Ml24EI\frac{{{\rm{M}}{{\rm{l}}^2}}}{{4{\rm{EI}}}}

  2. BOption (b)

    Ml264EI\frac{{{\rm{M}}{{\rm{l}}^2}}}{{64{\rm{EI}}}}

  3. COption (c)

    Ml216EI\frac{{{\rm{M}}{{\rm{l}}^2}}}{{16{\rm{EI}}}}

  4. DOption (d)

    Ml28EI\frac{{{\rm{M}}{{\rm{l}}^2}}}{{8{\rm{EI}}}}

View answer & explanation

Correct answer

Answer: (d)D.

Ml28EI\frac{{{\rm{M}}{{\rm{l}}^2}}}{{8{\rm{EI}}}}

Why this answer is correct

Concept:

A simply supported beam subjected to couple M at the ends with its BMD and SFD is represented by:

By double integration method we get,

d2ydx2=−MEI\frac{{{{\rm{d}}^2}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^2}}} = - \frac{{\rm{M}}}{{{\rm{EI}}}}

dydx=−MEI∫;dx;\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = - \frac{{\rm{M}}}{{{\rm{EI}}}}\smallint {\rm{;dx;}}

y=−;Mx22EI+C1x+C2{\rm{y}} = - {\rm{;}}\frac{{{\rm{M}}{{\rm{x}}^2}}}{{2{\rm{EI}}}} + {{\rm{C}}_1}{\rm{x}} + {{\rm{C}}_2}

At x = 0, y =0 we get C1=;ML2EI{{\rm{C}}_1} = {\rm{;}}\frac{{{\rm{ML}}}}{{2{\rm{EI}}}}

At x = L, y =0 we get C2 = 0

So, deflection at any point at distance x from the left end is given by,

y(x)=−Mx22EI+ML2EIx=Mx2EI(L−x){\rm{y}}\left( {\rm{x}} \right) = - \frac{{{\rm{M}}{{\rm{x}}^2}}}{{2{\rm{EI}}}} + \frac{{{\rm{ML}}}}{{2{\rm{EI}}}}{\rm{x}} = \frac{{{\rm{Mx}}}}{{2{\rm{EI}}}}\left( {{\rm{L}} - {\rm{x}}} \right)

The maximum deflection will be at the midpoint of the beam. So,

∴;ymax=y(L2)=;ML28EI\therefore {\rm{;}}{{\rm{y}}_{{\rm{max}}}} = {\rm{y}}\left( {\frac{{\rm{L}}}{2}} \right) = {\rm{;}}\frac{{{\rm{M}}{{\rm{L}}^2}}}{{8{\rm{EI}}}}

Important point:

Deflection and slope of various beams is given by:

yB=PL33EI{y_B} = \frac{{P{L^3}}}{{3EI}}θB=PL22EI{\theta _B} = \frac{{P{L^2}}}{{2EI}}
yB=wL48EI{y_B} = \frac{{w{L^4}}}{{8EI}}θB=wL36EI{\theta _B} = \frac{{w{L^3}}}{{6EI}}
yB=ML22EI{y_B} = \frac{{M{L^2}}}{{2EI}}θB=MLEI{\theta _B} = \frac{{ML}}{{EI}}
yB=wL430EI{y_B} = \frac{{w{L^4}}}{{30EI}}θB=wL324EI{\theta _B} = \frac{{w{L^3}}}{{24EI}}
yc=PL348EI{y_c} = \frac{{P{L^3}}}{{48EI}}θB=wL216EI;{\theta _B} = \frac{{w{L^2}}}{{16EI;}}

yc=5384wL4EI{y_c} = \frac{5}{{384}}\frac{{w{L^4}}}{{EI}}θB=wL324EI{\theta _B} = \frac{{w{L^3}}}{{24EI}}

yc=0{y_c} = 0θB=ML24EI{\theta _B} = \frac{{ML}}{{24EI}}
yc=ML28EI{y_c} = \frac{{M{L^2}}}{{8EI}}θB=ML2EI{\theta _B} = \frac{{ML}}{{2EI}}

yc=PL3192EI{y_c} = \frac{{P{L^3}}}{{192EI}}θA=θB=θC=0{\theta _A} = {\theta _B} = {\theta _C} = 0
yc=wL4384EI{y_c} = \frac{{w{L^4}}}{{384EI}}θA=θB=θC=0{\theta _A} = {\theta _B} = {\theta _C} = 0

Where, y = Deflection of beam, θ = Slope of beam

Question 25

The ordinates of influence line diagram for bending moment always have the dimension of:

  1. AOption (a)

    force

  2. BOption (b)

    length

  3. COption (c)

    force × length

  4. DOption (d)

    force / length

View answer & explanation

Correct answer

Answer: (b)B.

length

Why this answer is correct

An influence line for a given function, such as a reaction, axial force, shear force, or bending moment, is a graph that shows the variation of that function at any given point on a structure due to the application of a unit load at any point on the structure.

The ordinates of influence line diagram for bending moment always have the dimension of length.

Question 26

Addition of pozzolana to ordinary portland cement increase

  1. AOption (a)

    Bleeding 

  2. BOption (b)

    Shrinkage

  3. COption (c)

    Permeability

  4. DOption (d)

    Heat of hydration

View answer & explanation

Correct answer

Answer: (b)B.

Shrinkage

Why this answer is correct

Addition of pozzolana:

Pozzolana is a volcanic powder. Pozzolanic materials are essentially siliceous or aluminous compounds that do not possess any cementitious property but when in finely divided form mixed in the presence of water, react with calcium hydroxide to form cementitious property.

The addition of pozzolana to ordinary Portland cement increases shrinkage but reduces the permeability of the concrete.

Advantages:

  • It increases the workability
  • It can reduce permeability
  • Reduces water evaporation
  • Honey combing will be reduced
  • Improves cohesion that means less segregation
  • Increases the elastic modulus of concrete
  • Less heat of hydration
  • Offers greater resistance to aggressive weather
  • It can reduce creep and bleeding
  • Leaching in concrete reduces
  • Offers great resistance to the expansion i.e. increases shrinkage.
  • Used in mass concreting
  • It is particularly used in marine and hydraulic structure

Disadvantages:

  • Earliest strength gain is less

Question 27

The development of strength of cement and its fineness are:

  1. AOption (a)

    Directly proportional

  2. BOption (b)

    Inversely proportional

  3. COption (c)

    Not related

  4. DOption (d)

    Randomly related

View answer & explanation

Correct answer

Answer: (a)A.

Directly proportional

Why this answer is correct

Explanation:

If the fineness of the cement is greater, surface area of the particles of the cement is also greater. And greater surface area leads to an increase in rapid hydration reaction (i.e. exothermic reaction between cement particles and water).

And hydration reaction is the backbone of the strength development.

So, increase in fineness of the cement will ultimately lead to increase in development of the strength of the cement.

Question 28

Which of the following pairs are correctly related for rapid hardening cement?

  1. Fineness: 10%
  2. Final setting time: 10 hours
  3. Initial setting time: 30 minutes

The correct answer is

  1. AOption (a)

    1, 2 and 3

  2. BOption (b)

    1 and 2

  3. COption (c)

    2 and 3

  4. DOption (d)

    1 and 3

View answer & explanation

Correct answer

Answer: (c)C.

2 and 3

Why this answer is correct

Concept:

The difference of rapid hardening cement to that of ordinary Portland cement is the quantity of Limestone (tri-calcium silicate) used as raw material, which gives the high early strength to the cement.

It attains early strength due to larger proportion of lime grounded finer than normal cement.

Properties of Rapid Hardening Cement:

  1. It gains strength faster than OPC. In 3 days it develops 7 days strength of OPC with same water cement ratio.
  2. Its initial setting times is 30 mins and final setting time is 600 mins which is same as OPC.
  3. It emits more heat during setting, therefore this cement is unsuitable for mass concreting.
  4. This cement is lighter and costlier than OPC. Its short curing period makes it economical.
  5. This cement should be stored in a dry place, or else its quality deteriorates due to premature carbonation and hydration. 6. Fineness of the rapid hardening cement should not be more than 5%.

Question 29

The inclination of lacing bars with the axis of the compressive member should be more than

  1. AOption (a)

    40°

  2. BOption (b)

    25°

  3. COption (c)

    60°

  4. DOption (d)

    70°

View answer & explanation

Correct answer

Answer: (a)A.

40°

Why this answer is correct

As per IS 800 Design specifications of lacing are:

  1. The angle of inclination of the lacing with the longitudinal axis of the column should be between 40° to 70°.

i.e inclination > 40 ° & < 70 ° 

  1. The slenderness ratio of the lacing bar should not exceed 145.
  2. Minimum width of lacing in riveted connections should be:
ϕ (mm)22201816
Width (mm)65605550

Which is approximated to 3 times the nominal diameter.

Note: The inclination of lacing bars with the axis of the compressive member should be more than 40° and less than 70°.

Question 30

According to Tresca, yield locus is a/an

  1. AOption (a)

    rectangle

  2. BOption (b)

    circle

  3. COption (c)

    hexagon

  4. DOption (d)

    ellipse

View answer & explanation

Correct answer

Answer: (c)C.

hexagon

Why this answer is correct

Explanation:

Yield loci for two yield criteria in-plane stress as per Von Mises and Tresca Criteria is given below:

∴ According to Tresca, the yield is a hexagon.

 ​

For brittle material

Theories of failureShape
Maximum Principal Stress theory (RANKINE’S THEORY)Square
Maximum Principal Strain theory (St. VENANT’S THEORY)Rhombus

For Ductile material

Theories of failureShape
Maximum Shear Stress Theory (GUEST AND TRESCA’S THEORY)Hexagon
Maximum Distortion Energy Theory (VON MISES AND HENCKY’S THEORY)Ellipse
Total Strain Energy theory (HAIGH’S THEORY)Ellipse

Question 31

Load factor in plastic design depends upon

  1. AOption (a)

    nature of loading

  2. BOption (b)

    support conditions

  3. COption (c)

    geometrical shape

  4. DOption (d)

    All of the above

View answer & explanation

Correct answer

Answer: (d)D.

All of the above

Why this answer is correct

Concept:

Load factor in plastic design:

  • Load factor is the ratio of ultimate collapse load to the working load that can be applied on the structure.
  • Load Factor = Factor of safety x Shape factor.
  • As factor of safety at member level largely depends upon the nature of loading, support conditions and mode of failure, so the load factor also depends upon those factors.
  • The shape factor of a member solely depends upon the geometry of the cross sectional area, so the load factor also depends on the geometrical shape of the member.
  • Normally a load factor of 1.7 to 2 is used in plastic design method.

Question 32

For elastic design, the redistribution of moment in a continuous beam is limited to

  1. AOption (a)

    10%

  2. BOption (b)

    15%

  3. COption (c)

    30%

  4. DOption (d)

    20%

View answer & explanation

Correct answer

Answer: (c)C.

30%

Why this answer is correct

Moment redistribution refers to the behavior of statically indeterminate structures that are not completely elastic but have some reserve plastic capacity.

As per clause 37.1.1 of IS: 456-2000:

The ultimate moment of resistance at any section should not be less than 70% of the factored moment at that section as obtained from the elastic moment envelope (considering all loading combinations).

The elastic moment at any section in a member due to a particular combination of loads shall not be reduced by more than 30% of the numerical e largest moment given anywhere by the elastic maximum moment diagram for the particular member, covering all appropriate combinations of loads.

Question 33

The most economical section for a steel column is

  1. AOption (a)

    Rectangle section

  2. BOption (b)

    Solid section

  3. COption (c)

    I-section

  4. DOption (d)

    Tubular section

View answer & explanation

Correct answer

Answer: (d)D.

Tubular section

Why this answer is correct

Tubular section is best suited and most economical for the design of smaller compression member subjected to smaller load.

Various reasons are as follows:

  1. Round tubes have the same radius of gyration in all directions and have a high local buckling strength. These are usually very economical unless moments are too large for the sizes available.
  2. Tube has excellent torsional resistance.
  3. In the case of members subjected to wind, round tubes are subjected to less force than flat sections.
  4. They have less surface area to paint or fireproof.
  5. Tubes do not have the problem of dirt collection or cleaning.
  6. The weight of tube sections usually is less than one half the weight required for open profile sections. Although tube sections cost about 25 percent more than open sections, but about 20 percent of cost savings can still be achieved.

It can be seen that only the tubular section has the same moment of inertia about any axis through its center of gravity, hence The tubular section is the ideal compression member.

Question 34

Degree of static indeterminacy of a rigid jointed plane frame having 15 members, 3 reaction components and 14 joints is

  1. AOption (a)

    2

  2. BOption (b)

    3

  3. COption (c)

    6

  4. DOption (d)

    8

View answer & explanation

Correct answer

Answer: (c)C.

6

Why this answer is correct

Concept:

Degree of static determinacy for plane frame (rigid jointed) = 3m + R – 3J

Where, m = Number of members, R = Number of support reactions, J = Number of joints

Calculation:

m = 15, R = 3, J = 14

∴ Ds = 3m + R – 3J = 3 × 15 + 3 – 3 × 14 = 6

Important Point:

Degree of static indeterminacy in a plane frame = 3m + R – 3J

Degree of kinematic indeterminacy in a plane frame = 3J – R

Question 35

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

  1. AOption (a)

    zero

  2. BOption (b)

    less than 1

  3. COption (c)

    1

  4. DOption (d)

    greater than 1

View answer & explanation

Correct answer

Answer: (c)C.

1

Why this answer is correct

Concept:

Distribution Factors in Moment Distribution Method:

  • The moment distribution method is a structural analysis method for statically indeterminate beams and frames in which every joint of the structure to be analysed is fixed so as to develop the fixed-end moments.
  • Then each fixed joint is sequentially released and the fixed-end moments are distributed to adjacent members until equilibrium is achieved.
  • When a joint is being released and begins to rotate under the unbalanced moment, resisting forces develop at each member framed together at the joint.
  • Distribution factors can be defined as the proportions of the unbalanced moments carried by each of the members. In mathematical terms, the distribution factor of member k framed at joint j is given as: \({D_{jk}} = ;\frac{{\frac{{{{\left( {EI} \right)}k}}}{{{L_k}}}}}{{\mathop \sum \nolimits{i = 1}^n \frac{{{{\left( {EI} \right)}_i}}}{{{L_i}}}}};\)where n is the number of members framed at the joint. Here, EI is the flexural rigidity and L is length of the concerned member.
  • Hence, the sum of distribution factors of all the members connecting at any joint is always 1.

Question 36

The moment required to rotate the near end of a prismatic beam through a unit angle without translation, the far end being simply supported is given by

  1. AOption (a)

    3EI/L

  2. BOption (b)

    4EI/L

  3. COption (c)

    2EI/L

  4. DOption (d)

    EI/L

View answer & explanation

Correct answer

Answer: (a)A.

3EI/L

Why this answer is correct

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 3EIL\frac{{3EI}}{L} is to be applied at end and hence stiffness for the member is said to be 3EIL.\frac{{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 4EIL\frac{{4EI}}{L} is to be applied at end and hence stiffness for the member is said to be 4EIL.\frac{{4EI}}{L}.

Question 37

The three-moment equation is applicable only, when

  1. AOption (a)

    the beam is prismatic

  2. BOption (b)

    there is no settlement of supports

  3. COption (c)

    there is no discontinuity such as hinges within the span

  4. DOption (d)

    the spans are equal

View answer & explanation

Correct answer

Answer: (c)C.

there is no discontinuity such as hinges within the span

Why this answer is correct

Concept:

Three Moment Equation:

Three moment method is a method suitable for analysing continuous beam.

According to this, if A, B and C are three consecutive points on a continuous beam and MA, MB and MC are support moments then,

\({{\rm{M}}{\rm{A}}}{{\rm{L}}{{\rm{AB}}}} + 2{{\rm{M}}{\rm{B}}}\left( {{{\rm{L}}{{\rm{AB}}}} + {{\rm{L}}{{\rm{BC}}}}} \right) + {{\rm{M}}{\rm{C}}}{{\rm{L}}_{{\rm{BC}}}} = - \frac{{6{{\rm{a}}_1}{{\rm{x}}1}}}{{{{\rm{L}}{{\rm{AB}}}}}} - \frac{{6{{\rm{a}}_2}{{\rm{x}}2}}}{{{{\rm{L}}{{\rm{BC}}}}}}\)

Where, a1 = the area of the bending moment due to loading on AB, a2 = the area of the bending moment due to loading on BC,  x1 = the distance from A to the centroid of the bending moment diagram of beam AB,  x2 = the distance from C to the centroid of the bending moment diagram of beam BC, LAB = length of span AB and LBC = length of span BC.

This method is applicable

  1. to any type of cross section not limited to prismatic cross section.
  2. in case of support settlement. Only extra moments generated due to support settlement is incorporated into the three moment equation.
  3. if there is no discontinuity within the span because discontinuity within span cannot produce a valid M/EI diagram for the span.
  4. for different span length, not necessarily the spans should be equal.

Question 38

The Muller-Breslau principle can be used to

  1. determine the shape of the influence line
  2. indicate the parts of the structure to be loaded to obtain the maximum effect
  3. calculate the ordinates of the influence lines

The correct answer is

  1. AOption (a)

    1 only

  2. BOption (b)

    Both 1 and 2

  3. COption (c)

    2 and 3

  4. DOption (d)

    1, 2 and 3

View answer & explanation

Correct answer

Answer: (d)D.

1, 2 and 3

Why this answer is correct

According to Muller Breslau Principle:

“If an internal stress component (shear force, bending moment etc) or reaction component is allowed to act through a small distance thereby causing deformation of the structure, the curve of the deformed shape represents to some scale, the influence lines for that stress or the reaction component.”

Application: Muller-Breslau's Principle is applicable to all structures, determinate or indeterminate.

Note:

  • In the case of determinate systems, the ILD can be obtained directly by allowing unit deformation corresponding to the constraint.
  • For indeterminate structure, it is applicable only when the material is within the elastic limit and obeys Hook's law so that the law of superposition holds good.

This principle gives an excellent experiment approach to determine the influence lines for indeterminate structure.

∴ The Muller-Breslau principle can be used to determine the shape of the influence line, indicate the parts of the structure to be loaded to obtain the maximum effect and calculate the ordinates of the influence lines.

Question 39

For stable structures, one of the important properties of flexibility and stiffness matrices is that the elements on the main diagonal

  1. must be positive of stiffness matrix
  2. must be negative of stiffness matrix
  3. must be positive of flexibility matrix
  4. must be negative of flexibility matrix

The correct answer is

  1. AOption (a)

    1 and 3

  2. BOption (b)

    2 and 3

  3. COption (c)

    1 and 4

  4. DOption (d)

    2 and 4

View answer & explanation

Correct answer

Answer: (a)A.

1 and 3

Why this answer is correct

Concept:

Flexibility Matrix:

  • Any element Fij;{{\rm{F}}_{{\rm{ij;}}}}in flexibility matrix represents that the displacement (measured in the direction of force) occurs in i-th node when a unit load is applied at j-th node.
  • So, for main diagonal elements Fii;{{\rm{F}}_{{\rm{ii;}}}}it essentially implies that displacement (measured in the direction of force) occurs in i-th node when a unit load is applied at i-th node itself.
  • It is evident that if a load is applied at a certain node then the displacement at that node should be in the direction of the force. That’s why the main diagonal elements in flexibility matrix is always positive and cannot be negative at all.
<br>

Stiffness Matrix:

  • Any element Kij;{{\rm{K}}_{{\rm{ij;}}}}in stiffness matrix represents that the action (force or moment) develops in i-th node when a unit displacement (deflection or rotation) is given at j-th node which is measured in the direction of given displacement.
  • So, for main diagonal elements Kii;{{\rm{K}}_{{\rm{ii;}}}}it essentially implies that the action (force or moment) develops in i-th node when a unit displacement (deflection or rotation) is given at i-th node itself.
  • It is evident that if displacement is given at a certain node then the action at that node should developed in the direction of the displacement. That’s why the main diagonal elements in stiffness matrix is always positive and cannot be negative at all.

Question 40

For limit state of collapse flexure, the maximum strain in concrete at the outermost compression fibre is taken equal to

  1. AOption (a)

    0.002

  2. BOption (b)

    0.02

  3. COption (c)

    0.0035

  4. DOption (d)

    0.035

View answer & explanation

Correct answer

Answer: (c)C.

0.0035

Why this answer is correct

As per clause No. 38.1 of IS 456:2000, the maximum strain in concrete at the outermost compression fibre is taken as 0.0035 in bending.

While, as per clause No. 39.1 of IS 456:2000, the maximum compressive strain in concrete in axial compression is taken as 0.002.

The maximum compressive strain at the highly compressed extreme fibre in concrete subjected to axial compression and bending and when there is no tension on the section shall be 0.0035 minus 0.75 times the strain at the least compressed extreme fibre.

Question 41

In an RCC beam compression failure occurs in a/an

  1. AOption (a)

    balanced section

  2. BOption (b)

    over reinforced section

  3. COption (c)

    under reinforced section

  4. DOption (d)

    None of the above

View answer & explanation

Correct answer

Answer: (b)B.

over reinforced section

Why this answer is correct

Concept:

Over reinforced Section and compression failure:

  • For a flexure resisting reinforced section, if the depth of actual neutral axis is greater than the depth of critical neutral axis then the section is called as over reinforced section.
  • In this case the concrete in compressive zone attains its maximum compressive stress limit earlier than the steel in the tensile zone reaches its tensile stress limit.
  • So, for this the failure of the section is guided by the compression failure of the concrete in the compressive zone. In this case the failure mode is brittle in nature.
<br>

Related Concept:

In case of under reinforced section the steel in the tensile zone reaches its tensile stress limit earlier than the concrete in compressive zone attains its maximum compressive stress limit. So, the failure of the section is guided by the tensile failure of the steel in the tensile zone which is ductile in nature.

Question 42

In a two-way spanning slab, if corners of the slab are not allowed to lift, torsion reinforcement is provided at corners in the form of a mesh. The size of mesh should be equal to

  1. AOption (a)

    0.125 lx × 0.125 ly

  2. BOption (b)

    0.125 lx × 0.25 ly

  3. COption (c)

    0.25 lx × 0.25 ly

  4. DOption (d)

    0.20 lx  × 0.20 ly

View answer & explanation

Correct answer

Answer: (d)D.

0.20 lx  × 0.20 ly

Why this answer is correct

Concept:

Torsional Reinforcement:

  • It is provided as per guidelines provided in annex D of IS 456: 2000.
  • In a two way spanning slab torsional reinforcement shall be provided at corners of the slab. The torsional moment are high near the corner therefore, torsional reinforcement is essential to prevent the corner slab from lifting and prevents cracks. It is provided at corners in the form of a mesh, sometimes referred as corner reinforcement.
  • It shall consist of top and bottom reinforcement, each with layers of bars placed parallel to the sides of the slab and extending from the edges a minimum distance of one-fifth of the span (0.2 times of the span).
  • The area of reinforcement in each of these four layers shall be three-quarters of the area required for the maximum mid-span moment in the slab.
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The following diagram illustrates the mesh of corner reinforcement:

Question 43

In an RCC slab, having effective depth equal to 100 mm maximum, spacing of main reinforcement should not be greater than

  1. AOption (a)

    100 mm

  2. BOption (b)

    200 mm

  3. COption (c)

    300 mm

  4. DOption (d)

    400 mm

View answer & explanation

Correct answer

Answer: (c)C.

300 mm

Why this answer is correct

Concept:

As per IS 456: 2000, clause 26.3.3,

  1. The horizontal distance between parallel main reinforcement bars shall not be more than
  • three times the effective depth of solid slab or
  • 300 mm

whichever is smaller.

<br>
  1. The horizontal distance between parallel reinforcement bars provided against shrinkage and temperature shall not be more than
  • five times the effective depth of a solid slab or
  • 450 mm

whichever is smaller.

Calculations:

The spacing of the main reinforcement should be

  1. 3 × 100 = 300 mm or
  2. 300 mm

whichever is smaller.

So, the spacing of the main reinforcement should not be greater than 300 mm.

Question 44

In a steel roof truss, if angle iron purlin is used, its depth should not be less than

  1. AOption (a)

    1/60 × span of purlin

  2. BOption (b)

    1/45 × span of purlin

  3. COption (c)

    1/60 × spacing of purlin

  4. DOption (d)

    1/45 × spacing of purlin

View answer & explanation

Correct answer

Answer: (b)B.

1/45 × span of purlin

Why this answer is correct

Concept:

Purlins and its design recommendation: Purlins are horizontal beam members which runs parallel to ridge and connects the trusses along the length of ridge.

When an iron angle section is used as purlins for steel roof trusses then generally following recommendations are required to be fulfilled. These recommendations are applicable for roof truss with pitch angle less than 30° and for a maximum imposed load of 0.75 kN/m2.

  1. The depth of the angle purlin in the plane approximate to the incidence of the maximum load should not be less than (1/45) th of the length of the purlin.
  2. The width of the angle purlin should not be less than (1/60) th of the length of the purlin.
  3. The maximum bending moment in the purlin in taken as (WL/10) where W is the total uniformly distributed load on the purlin including wind load in the normal direction of the roof and L is the center to center distance between purlin supports.

Question 45

Lateral stability of steel beam increases

  1. AOption (a)

    axial compressive stress in beam

  2. BOption (b)

    shear stress in beam

  3. COption (c)

    bending tensile stress in beam

  4. DOption (d)

    bending compressive stress in beam

View answer & explanation

Correct answer

Answer: (d)D.

bending compressive stress in beam

Why this answer is correct

Concept:

Lateral stability of steel beams:

  • A steel beam loaded predominantly in flexure would attain its full moment capacity if the local and lateral instabilities of the beam are prevented.
  • If adequate lateral restraints are not provided to the steel beams in the plane of their compression flanges, the beams would results in lateral instability and would buckle laterally resulting in a reduction of their maximum moment capacity.
  • As per clause number 8.2.2 of IS 800: 2007, The design bending strength of laterally unsupported beam as governed by lateral torsional buckling is given by, \({{\rm{M}}{\rm{d}}} = {{\rm{\beta }}{\rm{b}}}{{\rm{Z}}{\rm{p}}}{{\rm{f}}{{\rm{bd}}}}\) where  βb= 1.0 for plastic and compact sections, Zp is plastic sectional modulus and fbd is the bending compressive stress in the beam.
  • As, bending compressive stress in the beam increases the design bending strength will increase which in turn implies that the lateral stability is increasing.

Question 46

Distance between purlins on sloping roof depends on

  1. AOption (a)

    slope of truss

  2. BOption (b)

    type of truss

  3. COption (c)

    type of roof coverings

  4. DOption (d)

    spacing of trusses

View answer & explanation

Correct answer

Answer: (c)C.

type of roof coverings

Why this answer is correct

Concept:

Purlins and its spacing:

  • Purlins are the members which are spanning on the roof frames to support the roof coverings and runs parallel to ridge to connect different trusses situated in the longitudinal direction.
  • The spacing of purlins depends largely on the maximum safe span of the roof covering and glazing sheets. Hence, they should be less than or equal to their safe spans when they are directly placed on purlins.
  • In India, corrugated galvanized iron (GI) sheets are usually adopted as coverings for roofs and sides of industrial buildings. Sometimes asbestos cement (AC) sheets are also provided as roof coverings owing to their superior insulating properties.
  • Thus for corrugated GI sheets, the purlin spacing may vary from 1.5 to 1.75 m, and for corrugated AC sheets, it is limited to 1.4 m, for 6-mm thick sheets, and 1.6 m, for 7-mm thick sheets.

For economical spacing of truss,

The cost of truss should be equal to twice the cost of purlins plus the cost of roof covering.

t= 2p + r

where t= cost of truss, p= cost of purlin, r= cost of the roof.

However, the above expression is used to check the spacing of the roof truss. It can not be used to design the spacing as spacing does not occur in the equation. As a guide, the spacing of roof truss can be kept 1/4 of the span up to 15m and 1/5 of span for 15-30 m span of roof trusses.

Question 47

The strength of ‘strut’ depends on

  1. AOption (a)

    diameter of rivet used

  2. BOption (b)

    thickness of gusset plate

  3. COption (c)

    net area of strut

  4. DOption (d)

    slenderness ratio of strut

View answer & explanation

Correct answer

Answer: (d)D.

slenderness ratio of strut

Why this answer is correct

Concept:

Struts:

These are long and slender structural members in an assembly that are subjected only to an axial compressive force.

They predominantly fail by buckling but sometimes the yielding failure occurs in compression before buckling.

These members are structurally analogous to columns. So, the buckling strength of struts is similar to that of the column and can be taken as,

\({{\rm{P}}{{\rm{buck}}}} = {\rm{K}}\frac{{{{\rm{\pi }}^2}{\rm{EI}}}}{{{\rm{L}}{\rm{}}^2}},{\rm{;K;}}\)is a factor depending upon the end conditions.

\(\therefore {\rm{;}}{{\rm{P}}{{\rm{buck}}}} \propto \frac{{{\rm{EI}}}}{{{\rm{L}}{\rm{e}}^2}} \propto \frac{{{\rm{EA}}{{\rm{r}}^2}}}{{{\rm{L}}{\rm{e}}^2}} \propto \frac{{{\rm{EA}}}}{{{{\left( {\frac{{{{\rm{L}}{\rm{e}}}}}{{\rm{r}}}} \right)}^2}}} \propto \frac{{{\rm{EA}}}}{{{{\left( {\rm{\lambda }} \right)}^2}}}{\rm{;;}};;\)

∴Pbuck∝1(λ)2;\therefore {{\rm{P}}_{{\rm{buck}}}} \propto \frac{1}{{{{\left( {\rm{\lambda }} \right)}^2}}}; where λ is the slenderness ratio.

where,

Pbuck is the buckling strength of the strut,

E is the modulus of elasticity of the strut,

I is the moment of inertia, 

A is the cross-section and

Le effective length of the strut respectively.

Note: The net area of the strut is mainly the deciding factor in case for yielding failure but not for the buckling strength failure.

Question 48

When purlins are placed between panel points the principal rafter is to be designed for

  1. AOption (a)

    axial compression and tension

  2. BOption (b)

    axial compression

  3. COption (c)

    axial compression and bending moment

  4. DOption (d)

    axial tension and bending moment

View answer & explanation

Correct answer

Answer: (c)C.

axial compression and bending moment

Why this answer is correct

Concept:

Purlins: These are the members which are spanning on the roof frames to support the roof coverings and runs parallel to ridge to connect different trusses situated in the longitudinal direction.

Rafters: These are a series of sloped structural members that extend from the ridge to the downslope perimeter or to the bottom chords and are designed to support the roof deck and its associated loads.

Now, the principal rafter of a roof truss with purlins placed at intermediate parts on panel length is essentially becomes a continuous beam with several intermediate supports and with a fixed end support. So, these are subjected to external bending moment due to loading between supports. Also, axial compressive forces generated due to truss action and inclined component of dead load, live load and wind load. The whole concept in a pictorial view has shown in the following diagram:

Question 49

If diameter of reinforcement bar is d, the anchorage value of the U type hook alone will be

  1. AOption (a)

    4d

  2. BOption (b)

    8d

  3. COption (c)

    12d

  4. DOption (d)

    16d

View answer & explanation

Correct answer

Answer: (d)D.

16d

Why this answer is correct

As per IS 456: 2000,

A standard hook has an anchorage value equivalent to a straight length of 16ϕ.

The anchorage value of standard U-type hook shall be 16 times the diameter of the bar. 

The anchorage value of standard bend shall be considered as 4 times the diameter of the bar for each 45o bend subject to a maximum value of 16 times the diameter of the bar.

Anchorage Value of Bends & Hooks
Bends45°4ϕ
90°8ϕ
135°12ϕ
HooksU type Hook (180°)16ϕ

Question 50

Web crippling in beams generally occurs at the points where:

  1. AOption (a)

    Concentrated loads act.

  2. BOption (b)

    Bending moment is maximum.

  3. COption (c)

    Shear force is maximum.

  4. DOption (d)

    Deflection is maximum.

View answer & explanation

Correct answer

Answer: (a)A.

Concentrated loads act.

Why this answer is correct

Web cripling:

High concentrated loads create large bearing stress concentration at the junction of web and flange. The web near the portion of stress concentration tends to fold over the flange. This type of local buckling phenomenon is called as web crippling.

50 questions in the full paper