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

Last updated on May 27, 2025

Latest Limit State Method MCQ Objective Questions

Limit State Method Question 1:

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 1 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 2:

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 2 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

Limit State Method Question 3:

The IS code that deals with modular coordination in prefabricated buildings is:

  1.  IS 456
  2. IS 15916
  3. IS 875
  4. IS 4926

Answer (Detailed Solution Below)

Option 2 : IS 15916

Limit State Method Question 3 Detailed Solution

Explanation: 

IS 15916 deals with modular coordination in prefabricated buildings. It provides guidelines for the dimensional coordination of building components such as walls, windows, doors, and other prefabricated elements. The purpose of modular coordination is to standardize the dimensions of building components to improve the ease and efficiency of construction, especially when using prefabricated materials.

Key aspects covered by IS 15916:

  • Standardized dimensions for building components.

  • Coordination between different components to ensure ease of assembly and minimize errors in prefabricated construction.

  • Design guidelines for maintaining uniformity across various elements in a modular building system.

Additional Information

  •  IS 456: This is the Indian Standard for Plain and Reinforced Concrete, focusing on the design and construction of concrete structures, not specifically for modular coordination.

  • IS 875: This code deals with Design Loads (Other than Earthquake) for buildings and structures, covering wind loads, dead loads, live loads, etc.

  • IS 4926: This code covers the Production and Supply of Ready-Mixed Concrete, specifying the requirements for ready-mixed concrete production and delivery, not modular coordination.

Limit State Method Question 4:

What is the depth of limiting neutral axis for beam having cross section 200 × 450 mm and reinforcement of 4 -18 mm dia bars of Fe500? Effective cover will be 50 mm.

  1. 192 mm
  2. 200 mm
  3. 126 mm
  4. 184 mm
  5. 220 mm

Answer (Detailed Solution Below)

Option 4 : 184 mm

Limit State Method Question 4 Detailed Solution

Concept:

The ratio of limiting depth of neutral axis to the effective depth of the beam is given by

\(\frac{{{x_u}}}{d} = \frac{{{\rm{Max\;strain\;concrete}}}}{{\frac{{{\rm{Grade\;of\;steel}}}}{{{\rm{Modulus\;of\;Elasticity\;of\;steel}} \times {\rm{F}}.{\rm{O}}.{\rm{S}}}} + 0.002 + {\rm{Max\;strain\;concrete}}}}\)

Note:

Without getting into the tedious calculation,

We know the standard values of the ratio of limiting depth of neutral axis to the effective depth (k) of the beam for different steel sections as following:

Grade of Steel

Fe 500

Fe 415

Fe 250

‘k’ value

0.46

0.48

0.53

Calculation:

From the above table we can say that Limiting depth xu is function of grade of steel and Effective Depth

Given, Grade of steel = Fe500, Effective cover = 50 mm,

Effective depth (d) = 450 - 50 = 400 mm

Reinforcement 4 - 18 mm diameter bars. 

∴ For Fe 500, xu ≈ 0.46× 400 = 184 mm.

Limit State Method Question 5:

The characteristics strength of concrete is defined as that compressive strength below which not more than

  1. 10% of results fail
  2. 5% of results fail
  3. 3% of results fail
  4. 1% of results fail
  5. None of the above

Answer (Detailed Solution Below)

Option 2 : 5% of results fail

Limit State Method Question 5 Detailed Solution

Concept:

As per IS 456:2000, Cl.36.1

Characteristics strength of the material is defined as the strength of material below which not more than 5% of the test results are expected to fail.

Also,

Mean strength is defined as the strength below which not more than 50% of the test results are expected to fail.

fm = fck + 1.65σ

Here,

fck – characteristics strength (MPa)

fm – mean strength (MPa)

σ – standard deviation (MPa)

The values of standard deviation for different grades of concrete are given below –

Grade

M10, M15

M20, M25

≥M30

Std. Deviation (MPa)

3.5

4.0

5.0

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

Download Solution PDF

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

Download Solution PDF

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

Download Solution PDF

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

Download Solution PDF

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

Download Solution PDF

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

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 11 Detailed Solution

Download Solution PDF

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

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 12 Detailed Solution

Download Solution PDF

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

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

Download Solution PDF

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

Download Solution PDF

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.

For moments at supports where two unequal spans meet or in cases where the spans are not equally loaded, ______ of two values for the negative moment at the support may be taken for design.

  1. addition
  2. product
  3. difference
  4. average

Answer (Detailed Solution Below)

Option 4 : average

Limit State Method Question 15 Detailed Solution

Download Solution PDF

Explanation:

AS Per IS 456 - 2000, Clause: 22.5.1

For moments at supports where two unequal spans meet or in cases where the spans are not equally loaded, the average of the two values for the negative moment at the Support may be taken for design.

Additional InformationFor Slab:

Minimum reinforcement criteria-

Mild steel- 0.15% of total C/S area

HYSD BARS- 0.12%  of total C/S area

Maximum diameter = d should not be greater than 1/8 ( thickness of slab)

For Columns:

Longitudinal reinforcement criteria-

It should not be less than 0.8%
It should not be greater than 6%

For Beams:

Tension Reinforcement criteria-

The minimum area of reinforcement  \({ Ast \over bd } = { 0.85 \over fy}\)

Maximum area of reinforcement = 0.04bD

Get Free Access Now
Hot Links: teen patti chart teen patti joy vip all teen patti master teen patti palace teen patti refer earn