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?
Answer (Detailed Solution Below)
Limit State Method Question 1 Detailed Solution
Explanation:
As per IS 456: 2000, Table 5;
Environment |
Minimum grade |
Minimum grade |
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:
Answer (Detailed Solution Below)
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.
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This confinement increases the apparent strength as it resists lateral expansion.
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As the height increases relative to the diameter, the confining effect reduces.
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This leads to a more uniform stress distribution but less lateral restraint, causing the measured compressive strength to drop.
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Slender specimens are more prone to column-like failure (buckling or instability) rather than pure crushing.
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This changes the failure behavior and reduces the load-carrying capacity under compression.
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The standard height-to-diameter ratio used in most concrete compressive tests is 2:1.
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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.
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It is considered the most essential mechanical property of concrete, as concrete is primarily used to resist compression rather than tension.
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This property helps engineers design structural elements like columns, beams, slabs, and footings safely and economically.
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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.
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These are cured under controlled conditions (usually 28 days) before testing.
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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
Answer (Detailed Solution Below)
Limit State Method Question 3 Detailed Solution
Explanation:
-
According to Whitney’s stress block theory and IS 456, the depth of the neutral axis 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 .
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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.
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The factor 0.53d is key in limiting the depth of compression zone in design calculations.
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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 ______.
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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):
Answer (Detailed Solution Below)
Limit State Method Question 6 Detailed Solution
Download Solution PDFThe 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:
Answer (Detailed Solution Below)
Limit State Method Question 7 Detailed Solution
Download Solution PDFGiven:
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 (150, 105, 144 kN-m) = 150 kN-mWhich of the following Indian Standard Codes provides conservatively imposed loads for building and structures?
Answer (Detailed Solution Below)
Limit State Method Question 8 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Limit State Method Question 9 Detailed Solution
Download Solution PDFConcept:
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).
Answer (Detailed Solution Below)
Limit State Method Question 10 Detailed Solution
Download Solution PDFConcept:
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:
Answer (Detailed Solution Below)
Limit State Method Question 11 Detailed Solution
Download Solution PDFExplanation:
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 |
Minimum grade of |
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?
Answer (Detailed Solution Below)
Limit State Method Question 12 Detailed Solution
Download Solution PDFConcept:
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?
Answer (Detailed Solution Below)
Limit State Method Question 13 Detailed Solution
Download Solution PDFExplanation:
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
Answer (Detailed Solution Below)
Limit State Method Question 14 Detailed Solution
Download Solution PDFConcepts:
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:
Answer (Detailed Solution Below)
Limit State Method Question 15 Detailed Solution
Download Solution PDFExplanation:
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.
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