# Design of RC Structures Questions and Answers – Limit State Method – Singly Reinforced Sections

This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “Limit State Method – Singly Reinforced Sections”.

1. What is the value of maximum strain in concrete for limit state of collapse in flexure at the compression fiber?
a) 0.0035
b) 0.035
c) 0.35
d) 3.5

Explanation: The value of maximum strain in limit state of collapse at the outermost compression fiber is taken as 0.0035 in bending. In flexure, the plane sections normal to the axis remain plane after bending. The maximum strain in concrete is the strain at which the section reaches its maximum moment capacity.

2. The stresses in the reinforcement are derived from the stress – strain curve for the type of steel used.
a) True
b) False

Explanation: The stresses in the reinforcement are derived from representative stress – strain curve for the type of steel used.
For the design purpose, the partial safety factor for steel is applied as:
γms = 1.15.

3. For design purpose, the compressive strength of concrete in structure is assumed to be 0.70 times the characteristic strength.
a) True
b) False

Explanation: The relationship between the compressive stress distribution in concrete and the strain in concrete can be assumed to be rectangular, trapezoidal, or any other shape. For the design purpose the compressive strength of concrete in structure is assumed to be 0.67 times the characteristic strength. The partial safety factor applied given by:
γmc = 1.5.
The stress – strain curve gives the shape of compressive stress distribution in concrete. The stress distribution is known as the stress block. It is the combination of a parabola and a rectangle.

4. What is the value of maximum compressive stress in concrete for design purpose according to limit state method?
a) 0.50 fck
b) 0.45 fck
c) 0.55 fck
d) 0.67 fck

Explanation: The compressive strength of concrete in the structure is 0.67 fck. The 0.67 factor is introduced to account for difference in strength indicated by cube test and the strength of concrete in structure. Thus maximum compressive stress in concrete for design is equal to:
(0.67 / γm)fck = (0.67 / 1.5)fck = 0.45fck.

5. The stress – strain relationship for steel in tension and compression is taken same.
a) True
b) False

Explanation: The stress – strain relationship for steel in tension and compression is assumed to be the same. The modulus of elasticity for all types of reinforcing steel is:
Es = 2 × 106 N/mm2.
For mild steel, the stress is proportional to strain up to yield point and increases at constant stress.

6. Which of the following equation is used to determine the depth of neutral axis according to IS: 456 – 2000?
a) (0.36 fy.Ast) / (0.87 fck.bd )
b) (0.70 fy.Ast) / (0.36 fck.bd )
c) (0.87 fy.Ast) / (0.30 fck.bd )
d) (0.87 fy.Ast) / (0.36 fck.bd )

Explanation: While obtaining the moment of resistance of rectangular section without compression reinforcement, the depth of neutral axis is determined as: (xu / d) = (0.87 fy.Ast) / (0.36 fck.bd )
Here, (Ast) is area of steel,
b and d are the dimensions of the section.

7. Which of the following is equation for moment of resistance when the value of (xu / d) is less than the limiting value?
a) 0.87 Ast.d[1 – (Ast / bd).(fy / fck)]
b) 0.80 fy.Ast.d[1 – (Ast / bd).(fy / fck)]
c) 0.87 fy.Ast.d[1 – (Ast / bd).(fy / fck)]
d) 0.36 fy.Ast.d[1 – (Ast / bd).(fy / fck)]

Explanation: If the value of (xu / d) is less than the limiting value the moment of resistance of the section can be calculated by:
Mu = 0.87 fy.Ast.d[1 – (Ast / bd).(fy / fck)].
Also if the value of (xu / d) is equal to the limiting value the moment of resistance of the section can be calculated by:
Mulim = 0.36 fck.(xumax / d) [1 – 0.416(xumax / a)]bd2.
These values are obtained from IS: 456 – 2000.

8. If the value of (xu / d) is greater than the limiting value, the section is under – reinforced.
a) True
b) False

Explanation: If the value of (xu / d) is greater than the limiting value, the section is over – reinforced. The moment of resistance of such a section is limited to Mu(lim) given by:
Mu(lim) = 0.36 fck.(xumax / d)[1 – 0.416(xumax / a) bd2
In this case there is wastage of excess steel. The concrete reaches its ultimate capacity before steel yields resulting in sudden failure.

9. For 200 mm width and 400 mm effective depth, reinforced with bars of 16 mm diameter of Fe 415 steel. What is the value of moment for the singly reinforced concrete beam? (Take M20 concrete. Use IS code method).
a) 88.37 kN-m
b) 90.37 kN-m
c) 88.00 kN-m
d) 88.98 kN-m

Explanation: Ast = 4(π / 4)162 = 804.25 mm2;
Pst = Ast / bd = 804.25 / (200 × 400) = 0.0101
For Fe 415 steel bars, fy = 415 N/mm2,
For M 20 concrete, fck = 20 N/mm2
Hence from, (xu / d) = 2.417 pt.(fy / fck) = 2.417 × 0.0101 × (415 / 20) = 0.504
The limiting value of (xu / d) is given by:
(xumax / d) = 700 / (1100 + 0.87 × 415) = 0.479
Thus, the actual N.A. depth is more than the limiting one. Such a beam is over – reinforced and hence is undesirable, since the failure is sudden and without warning. The code recommends that such a beam should be redesigned. The limiting moment of resistance for such a beam is found as:
Mu(lim) = 0.36 fck.(xumax / d) [1 – 0.416(xumax / a) bd2
= 0.36 × 20 × 0.479 (1 – 0.416 × 0.479) 200(400)2
= 88.37 × 102 N-mm = 88.37 kN-m.

10. Design a balanced singly reinforced concrete beam section for an applied moment of 60 kN-m. The width of the beam is limited to 175 mm. Use M20 and Fe 415 steel bars. What is the value of area of steel provided?
a) 724 mm2
b) 722 mm2
c) 620 mm2
d) 622 mm2

Explanation: MD = 60 kN-m = 60 × 106 N-mm
Since, ϒf = 1.5 for both dead and live loads, MuD = 1.5 × 60 × 106 = 90 × 106 N-mm
For a balanced design,
xu, max / d = 700 / (1100 + 0.87 fy) = 700 / (1100 + 0.87 × 415) = 0.479
The moment of resistance of a balanced beam section is given by
Mu,lim = 0.36 fck.xu,max / d (1 – 0.416 × xu,max / d)bd2
Equating Mu, lim and MuD, we get
90 × 106 = 0.36 × 20 × 0.479 (1 – 0.416 × 0.479) 175d2
From which we get d = 432 mm
The reinforcement is given by:
pt,lim = (xu, max / d) . (fck / fy) × (1 / 2.417) = 0.479 × 20 / 415 × 2.417 = 9.55 × 10-3
Ast = 9.55 × 10-3 × 175 × 432 = 722 mm2

11. A beam simply supported over an effective span of 7 m carries a live load of 20 kN/m. Design the beam, using M 20 concrete and HYSD bars of grade Fe 415. Keep the width equal to half the effective depth. Assume unit weight of concrete as 25 kN/m3. Which of the following is the correct value of moment for the section?
a) 1424 mm2
b) 1422 mm2
c) 1426 mm2
d) 1464 mm2

Explanation: For the computation of self weight, let us assume that the beam has a size of 300 mm × 600 mm.
Self weight = 0.3 × 0.6 × 1 × 25 = 4.5 kN/m;
Total load, w = 4.5 + 20 = 24.5 kN/m
Ultimate design load, wuD = 1.5 × 24.5 = 36.75 kN/m
MuD = 36.75 (7)2 / 8 = 225.09 kN-m = 225.09 × 106 N-mm
For M 20 concrete and Fe 415 steel, fck = 20 N/mm2 and fy = 415 N/mm2, respectively.
xu, max / d = 700 / (1100 + 0.87 fy) = 700 / (1100 + 0.87 × 415) = 0.479
Let d = effective depth, so that b = 0.5d.
Also,
Mu, lim = 0.36 × 20 × 0.479 (1 – 0.416 × 0.479) × 0.5d × d2 = 1.381 d3

12. A reinforced concrete beam has width equal to 300 mm and total depth equal to 700 mm, with a cover of 40 mm to the centre of the reinforcement. Design the beam if it is subjected to a total bending moment of 150 kN-m. Use M 20 and HYSD bars of grade 415. Which of the following is the value of area of steel?
a) 1164 mm2
b) 1163 mm2
c) 1063 mm2
d) 1064 mm2

Explanation: b = 300 mm; d = 700 – 40 mm = 660 mm
Υf = 1.5, MuD = 150 × 1.5 = 225 kN-m = 225 × 106 N-mm
xu, max / d = 700 / (1100 + 0.87 fy) = 700 / (1100 + 0.87 × 415) = 0.479
Mu, lim = 0.36 × 20 × 0.479 (1 – 0.416 × 0.479) × bd2 = 2.761 bd2
= 2.761 × 300 × (600)2 = 360.8 × 106 > MuD
Hence compression reinforcement is not required. Also, the beam is under reinforced.
Now, MuD = 0.87 fy Ast d [ 1 – (fck / fy) . (Ast / bd)]
225 × 106 = 0.87 × 415 × Ast × 600 [1 – (415 / 20) . (Ast / 300 × 600)]
Ast [1 – 1.048 × 10-4 Ast] = 944.2 or A2st – 9542 Ast + 9.0097 × 106 = 0
From which, Ast = 1062.5 mm2
Alternatively, we have equation for Ast,
Ast = (0.5fck / fy)[1 – 1 – 4.6 Mu / fckbd2] bd
= (0.5 × 20 / 415) [1 – 1 – 4.6 × 225 × 106 / 20 × 300 × (600)2] 300 × 600
= 1063 mm2

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