Limit State Method MCQ Quiz - Objective Question with Answer for Limit State Method - Download Free PDF

Last updated on Jul 10, 2025

Latest Limit State Method MCQ Objective Questions

Limit State Method Question 1:

As per IS 456:2000, what is the maximum water-cement ratio for M30 concrete with severe exposure conditions?

  1. 0.50
  2. 0.40
  3. 0.55
  4. 0.45

Answer (Detailed Solution Below)

Option 4 : 0.45

Limit State Method Question 1 Detailed Solution

Explanation:

As per IS 456: 2000, Table 5;

Environment

 Minimum grade 
of RCC

 Minimum grade 
of PCC

Nominal Cover

 1. Mid

M 20

-

20 mm

 2. Moderate

M 25

M 15

30 mm

 3. Severe

M 30

M 20

45 mm

 4. Very severe

M 35

M 20

50 mm

 5. Extreme

M 40

M 25

75 mm

 

According to IS 456: 200, the Minimum cement content and maximum water-cement ratio based on exposure conditions for plain cement concrete and reinforced cement concrete are given below:

Exposure condition Minimum cement content (kg/m3) maximum water-cement ratio Minimum cement content (kg/m3) maximum water-cement ratio
  PCC RCC
Mild 220 0.6 300 0.55
Moderate 240 0.6 300 0.5
Severe 250 0.5 320 0.45
Very severe 260 0.45 340 0.45
Extreme 280 0.40 360 0.4

Limit State Method Question 2:

For slender cylinders, as the height to lateral dimension ratio of the concrete specimen increases, the compressive strength:

  1. remains constant 
  2. increases
  3. first increases, then decreases
  4. decreases

Answer (Detailed Solution Below)

Option 4 : decreases

Limit State Method Question 2 Detailed Solution

Explanation:

  • When testing concrete specimens for compressive strength, the height-to-lateral dimension ratio (also known as slenderness ratio) plays a significant role in how the material behaves under load.
  • Shorter, stubby specimens (low height-to-diameter ratio) experience greater lateral confinement due to friction at the loading platens.

  • This confinement increases the apparent strength as it resists lateral expansion.

  • As the height increases relative to the diameter, the confining effect reduces.

  • This leads to a more uniform stress distribution but less lateral restraint, causing the measured compressive strength to drop.

  • Slender specimens are more prone to column-like failure (buckling or instability) rather than pure crushing.

  • This changes the failure behavior and reduces the load-carrying capacity under compression.

  • The standard height-to-diameter ratio used in most concrete compressive tests is 2:1.

  • Deviating significantly from this ratio (especially increasing it) may result in underestimating the actual compressive strength of the concrete.

Additional Information

  • Compressive strength is the maximum load per unit area that concrete can bear without failure when subjected to axial compression.

  • It is considered the most essential mechanical property of concrete, as concrete is primarily used to resist compression rather than tension.

  • This property helps engineers design structural elements like columns, beams, slabs, and footings safely and economically.

  • The standard test is conducted by casting cylindrical or cubical concrete specimens, typically 150 mm × 150 mm × 150 mm cubes or 150 mm diameter × 300 mm height cylinders.

  • These are cured under controlled conditions (usually 28 days) before testing.

  • The specimen is placed in a compression testing machine, and the load is applied until failure.

Limit State Method Question 3:

According to Whitney's theory, depth of stress block for a balanced section of a concrete beam is limited to

  1. 0.537 d
  2. 0.637 d
  3. 0.8 d
  4. 0.75 d

Answer (Detailed Solution Below)

Option 1 : 0.537 d

Limit State Method Question 3 Detailed Solution

Explanation:

  • According to Whitney’s stress block theory and IS 456, the depth of the neutral axis xu,maxx_{u, \text{max}} for a balanced section (where steel yields just as concrete reaches its ultimate strain) is limited to about 0.53 to 0.537 times the effective depth dd.

  • This ensures the maximum permissible depth of the stress block for safe design before concrete crushing dominates.

Additional Information

  • Whitney simplified the actual nonlinear stress distribution in concrete to an equivalent rectangular block.

  • The factor 0.53d is key in limiting the depth of compression zone in design calculations.

  • It helps distinguish between under-reinforced, balanced, and over-reinforced sections in concrete beam design.

Limit State Method Question 4:

For design of concrete structures, failure strain of concrete under direct compression and flexure respectively is ______ and ______.

  1. 0.02, 0.0035
  2. 0.002, 0.035
  3. 0.002, 0.0035
  4. 0.0035, 0.002
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 0.002, 0.0035

Limit State Method Question 4 Detailed Solution

Explanation:

Following assumptions have been made in IS 456: 200 regarding maximum strain in compression members:

1. As per, Clause 38.1, the maximum compressive strain in concrete in bending compression is taken as 0.0035.

2. As per, clause 39.1, the maximum compressive strain in concrete in axial compression is taken as 0.002.

Important Points

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.

Limit State Method Question 5:

For mild exposure, for main reinforcement steel bars up to 12 mm diameter, the nominal cover may be reduced by:

  1. 10 mm
  2. 7 mm
  3. 5 mm
  4. 6 mm
  5. 4 mm

Answer (Detailed Solution Below)

Option 3 : 5 mm

Limit State Method Question 5 Detailed Solution

Explanation:

As per IS 456 (2000), Table 16, For main reinforcement, up to 12 mm diameter bar for mild exposure the nominal cover may be reduced by 5 mm. 

Additional Information

Nominal Cover to Meet Durability Requirements:

Exposure

Nominal concrete cover 

(in mm) not less than

Mild

20

Moderate 

30

Severe

45

Very Severe

50

Extreme

75

Nominal Cover to Meet Fire Resistance:
The nominal cover is the design depth of concrete cover to all steel reinforcements.

It shall be not less than the diameter of the bar.

Minimum values of the nominal cover of normal-weight aggregate concrete to be provided to all reinforcement to meet the specified period of fire resistance shall be given in the table below 

Member

Minimum Cover

Slab

20 mm

Beam

20 mm

Column

40 mm for d > 12 mm

25 mm for d < 12 mm

Footing

50 mm

Top Limit State Method MCQ Objective Questions

In mix design for M25 concrete, the assumed standard deviation for estimation of target mean strength of concrete mix, as recommended by IS 456 ∶ 2000 is (in N/mm2):

  1. 4.5
  2. 4.0
  3. 5.0
  4. 3.5

Answer (Detailed Solution Below)

Option 2 : 4.0

Limit State Method Question 6 Detailed Solution

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The assumed value of standard deviation for initial calculations as per IS 456 – 2000 are given below in the table:

Sl. No

Grade of Concrete

Characteristic compressive strength (N/mm2)

Assumed standard deviation (N/mm2)

1.

M10

10

3.5

2.

M15

15

3.

M20

20

4.0

4.

M25

25

5.

M30

30

5.0

6.

M35

35

7.

M40

40

8.

M45

45

9.

M50

50

10.

M55

55

A reinforced concrete beam is subjected to the following bending moments.

Moment due to dead load = 50 kNm

Moment due to live load = 50 kNm

Moment due to seismic load = 20 kNm

The design bending moment for limit state of collapse is:

  1. 180 kNm
  2. 150 kNm
  3. 120 kNm
  4. 144 kNm

Answer (Detailed Solution Below)

Option 2 : 150 kNm

Limit State Method Question 7 Detailed Solution

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Given:

Moment due to Dead load (DL)= 50 kNm

Moment due to live load (LL) = 50 kNm

Moment due to Seismic load (EL) = 20 kNm

Computation:

Design bending moment is Maximum of the following

1) Mu considering moment due to dead load and live load

Mu = 1.5 (DL+LL) = 1.5 x (50+50) = 150 kN-m

2) Mu considering moment due to dead load and Seismic load

Mu = 1.5 (DL+EL) = 1.5 x (50+20) = 105 kN-m

3) Mu considering moment due to dead load, live load and seismic load together

Mu = 1.2 (DL+LL+EL) = 1.2 x (50+50+20) = 144 kN-m.

So the answer is Max of (150105, 144 kN-m) = 150 kN-m

Which of the following Indian Standard Codes provides conservatively imposed loads for building and structures?

  1. IS 875 (part 3) - 1987
  2. IS 875 (part 1) - 1987
  3. IS 875 (part 2) - 1987
  4. IS 875 (part 4) - 1987

Answer (Detailed Solution Below)

Option 3 : IS 875 (part 2) - 1987

Limit State Method Question 8 Detailed Solution

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Concept:

IS 875 (part 2) - 1987: Indian Standard Codes provides conservatively imposed loads for building and structures.

IS 875 (part 1) - 1987: Indian Standard Codes provides design dead loads(Unit weights of building material and stored materials) for buildings and structures.

IS 875 (part 3) - 1987: Indian Standard Codes provides design wind loads for buildings and structures.

IS 875 (part 4) - 1987: Indian Standard Codes provides design snow loads for buildings and structures.

IS 875 (part 5) - 1987: Indian Standard Codes provides design special loads(load combinations) for buildings and structures.

Additional InformationIS 800:2007 - Code for practice for general construction in steel.

IS 456:2000 - Code of practice for plain and reinforced concrete.

IS 1343:2012 - Code of practice for prestressed concrete.

For a structural member, dead load = 20 kN and live load = 12 kN. What will be its design load as per the limit state of collapse philosophy?

  1. 20 kN
  2. 48 kN
  3. 32 kN
  4. 8 kN

Answer (Detailed Solution Below)

Option 2 : 48 kN

Limit State Method Question 9 Detailed Solution

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Concept:

Values of the factor of safety (partial) for load combination:

Load combination

Limit state of collapse

1) Dead load & live load

1.5(DL + LL)

2) Dead seismic/wind load

a) Dead load contributes to the stability

b) Dead load assists overturning

 

0.9 DL + 1.5 (EL/WL)

1.5 (DL + EL/WL)

3) Dead, live load, and Seismic/wind load

1.2 (DL + LL + EL/WL)


Where, DL = Dead load, LL = Live load WL = Wind load EL = Earthquake load

Calculation:

Given: Dead load (DL) = 20 KN and Live load (LL) = 12 kN

Partial factor of safety = 1.5 (DL + LL) = 1.5 (20 + 12) = 48 kN

Additional Information

Values of the factor of safety (partial) for load combination:

Load combination

Serviceability limit state

1) Dead load & live load

DL + LL

2) Dead seismic/wind load

a) Dead load contributes to stability

b) Dead load assists overturning

 

DL + EQ/WL

DL + EQ/WL

3) Dead, live load and Seismic/wind load

DL + 0.8LL + 0.8EQ/WL

Calculate the target mean compressive strength at 28 days curing, for M20 grade concrete, as per IS 10262-2009 (take standard deviation as 4 N / mm2).

  1. 26.6 N / mm2
  2. 20.0 N / mm2
  3. 22.4 N / mm2
  4. 28.7 N / mm2

Answer (Detailed Solution Below)

Option 1 : 26.6 N / mm2

Limit State Method Question 10 Detailed Solution

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Concept:

The correct expression as per IS 456: 2000 to find Target mean strength is:

fcm = fck + (1.65 × σ)

fcm = Target mean strength

fck = Characteristic strength

σ = Standard deviation

1.65 is generally a risk factor

Calculations:

fck = 20 N/mm2

σ = 4.0 N/mm2

Target mean strength is:

fcm = fck + (1.65 × σ)

fcm = 20 + (1.65 × 4)

fcm = 26.6 N/mm2

As per IS 456 ∶ 2000, the minimum grade of concrete to be used for plain concrete and reinforced concrete, respectively, in concrete works exposed directly along the sea coast are:

  1. M20 and M30
  2. M15 and M25
  3. M25 and M40
  4. M20 and M25

Answer (Detailed Solution Below)

Option 1 : M20 and M30

Limit State Method Question 11 Detailed Solution

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Explanation:

Concrete in Sea-water (as per IS 456: 2000)

Concrete in sea-water or exposed directly along the sea-coast shall be at least M 20 Grade in the case of plain concrete and M 30 in case of reinforced concrete.

The use of slag or pozzolana cement is advantageous under such conditions.

Exposure

 Minimum grade of 
Plain Concrete

Minimum grade of
 reinforced concrete 

Mild

-

M20

Moderate

M15

M25

Severe

M20

M30

 Very severe 

M20

M35

Extreme

M25

M40

What is modular ratio of concrete grade M30 when effect of creep is NOT considered?

  1. 9.33
  2. 8.12
  3. 10
  4. 7.30

Answer (Detailed Solution Below)

Option 4 : 7.30

Limit State Method Question 12 Detailed Solution

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Concept:

Modular ratio:

In the elastic theory for the reinforced concrete structure, concrete and reinforcing steel are converted into one material. This is done by using the modular ratio ‘m’.

It is the ratio of modulus of elasticity of steel and concrete.

\(\rm m = \frac{{{E_s}}}{{{E_c}}}\)

However, concrete has varying moduli, as it is not a perfectly elastic material.

Short term modulus of elasticity:

The short-term modulus of elasticity is given by

 \(\left( {{E_c} = 5000\sqrt {{f_{ck}}} } \right)\) 

Long term modulus of elasticity:

The long-term modulus of elasticity is considered to take into account the effect of creep and shrinkage.

The modular ratio is taken as

\(m = \frac{{280}}{{3 \times {\sigma _{cbc}}}}\)

where σcbc is permissible compressive stress in concrete in bending.

Calculation:

Grade of concrete = M30

Modular ratio ‘m’ (without considering creep effect)

 \(m = \frac{{{E_s}}}{{{E_c}}} = \frac{{2 \times {{10}^5}}}{{5000\sqrt {{f_{ck}}} }} =\frac{{2 \times {{10}^5}}}{{5000\sqrt {{30}} }}\) = 7.30

In a certain project, the minimum specified cement content is 370 kg/cum. Due to a failure in the aggregate crushing and screening plant, the project manager is left with only 10 mm aggregate and sand. How will it affect the minimum cement content requirement?

  1. The cement content will have to be reduced by 40 kg/cum.
  2. The cement content will have to be increased by 40 kg/cum.
  3. The cement content will have to be reduced by 30 kg/cum.
  4. The cement content will have to be increased by 30 kg/cum.

Answer (Detailed Solution Below)

Option 2 : The cement content will have to be increased by 40 kg/cum.

Limit State Method Question 13 Detailed Solution

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Explanation:

As per IS 456, Clause 8.2.4.1 Table 6

Adjustment to Minimum Cement Contents or Aggregate other than 20 mm Nominal Maximum Size
SI No. Nominal Max aggregate size (mm) Adjustment to Minimum Cement Contents in Table 5 (kg/m3)
i 10 + 40
ii 20 0
iii 40 - 30

Important Points

According to IS 456: 200 Table 5, the minimum cement content and maximum water-cement ratio based on exposure conditions for plain cement concrete and reinforced cement concrete are given below:

Exposure condition Minimum cement content (kg/m3) maximum water-cement ratio Minimum cement content (kg/m3) maximum water-cement ratio
  PCC RCC
Mild 220 0.6 300 0.55
Moderate 240 0.6 300 0.5
Severe 250 0.5 320 0.45
Very severe 260 0.45 340 0.45
Extreme 280 0.40 360 0.4
 

The minimum allowable factor of safety against sliding for a cantilever retaining wall is

  1. 3.0
  2. 2.0
  3. 1.5
  4. 2.5

Answer (Detailed Solution Below)

Option 3 : 1.5

Limit State Method Question 14 Detailed Solution

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Concepts:

The factor of safety against sliding is defined as the resisting forces (friction and passive forces) divided by the driving lateral force. 

As per IS Code 1904: 1986, the minimum allowable factor of safety against sliding for a cantilever retaining wall should be 1.50 when dead load, live load, earth pressure are considered together with  either seismic forces or wind forces.  

However, in above case, if effect of wind force or earthquake forces is not considered then factor of safety is taken as 1.75 against sliding.

Note:

The minimum allowable factor of safety against overturning shall be 1.50, when dead load, live load, earth pressure are considered together with either seismic forces or wind forces. 

However, in above case, if effect of wind force or earthquake forces is not considered then factor of safety is taken as 2 against overturning.

IS 456 : 2000 specifies the modulus of elasticity of concrete (Ec) as:

  1. \(3000\sqrt {{{\rm{f}}_{{\rm{ck}}}}}\)
  2. \(5000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \)
  3. \(9000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \)
  4. \(7000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \)

Answer (Detailed Solution Below)

Option 2 : \(5000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \)

Limit State Method Question 15 Detailed Solution

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Explanation:

Modulus of elasticity of concrete:

It is the parameter that measures concrete's resistance to elastic deformation when stress is applied to it. As per IS 456: 2000, short term modulus of elasticity of concrete Ec is given by,

\({\rm{E_c}} = 5000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \)

Where fck is the characteristics strength of concrete. It means the modulus of elasticity in concrete is proportional to the compressive strength of concrete.

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Actual measured values may differ by ± 20 percent from the values obtained from the above expression, \({\rm{E_c}} = 5000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} \).

Long term elastic modulus of elasticity (EL\(\dfrac{{5000\sqrt {{{\rm{f}}_{{\rm{ck}}}}} }}{{1 + {\rm{\theta }}}}\)

Where, \({\rm{\theta }} = {\rm{Creep\;coefficient}} = \dfrac{{{\rm{Ultimate\;creep\;strain\;}}}}{{{\rm{Elastic\;strain}}}}\) and fck = characteristic compressive strength

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