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 ______.
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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:
Answer (Detailed Solution Below)
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.
Answer (Detailed Solution Below)
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
Answer (Detailed Solution Below)
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):
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
What is modular ratio of concrete grade M30 when effect of creep is NOT considered?
Answer (Detailed Solution Below)
Limit State Method Question 11 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
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 12 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 |
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.
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.
Answer (Detailed Solution Below)
Limit State Method Question 15 Detailed Solution
Download Solution PDFExplanation:
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