This set of Design of RC Structures Multiple Choice Questions & Answers (MCQs) focuses on “Axially Loaded Columns – Limit State Method”.

1. What is the design strength of the concrete (f_{ck}=characteristic strength for particular grade of concrete)?

a) .446f_{ck}

b) .96f_{ck}

c) .78f_{ck}

d) f_{ck}

View Answer

Explanation: The compressive strength of the concrete = .87 f

_{ck}; the partial safety factor for limit state = 1.5;

The design strength of concrete is = .87 f

_{ck}/1.5=.446 f

_{ck}.

2. If we consider an axially loaded (purely compressive in nature) column, what is the value of maximum strain (compressive)?

a) .0035

b) .0045

c) .0002

d) .0008

View Answer

Explanation: According to the assumption of the limit state method, if we draw the strain diagram of an axially loaded column we can realize the fact.

3. On which factor the value of ἀ_{s} depends [where ἀ_{s}f_{y}=f_{s}(f_{s}=stress in steel in failure at a strain of .0002)]?

a) Length of the section

b) Depth of the column section

c) Grade of concrete

d) Grade of steel

View Answer

Explanation: The factor ἀ

_{s}depends on the grade of steel reinforcement. The co-efficient is multiplied with the design stress as full design stress will not develop at strain of .0002, therefore overestimated reinforcement may result in.

4. The interaction curve is plotted between __________

a) P_{u} and M_{u}

b) f_{ck} and P_{u}

c) f_{y} and M_{u}

d) f_{y} and P_{u}

View Answer

Explanation: For the design of the column section it is a compulsion to know the failure at different regions. The interaction curve represents that graphically by plotting P

_{u}and M

_{u}respectively at y and x axis.

5. A column is subjected to 50 kNm moment and 200 kN compressive load, what is the eccentricity (e) of the section?

a) 150 mm

b) 160 mm

c) 190 mm

d) 250 mm

View Answer

Explanation: In the limit state method of design, e = M

_{u}/P

_{u}

= (50*1000/200)mm = 250mm.

6. What is the shape of the stress diagram where the neutral axis is situated outside the section?

a) Rectangular

b) Partially rectangular and partially parabolic

c) Parabolic

d) Triangular

View Answer

Explanation: As we can see from the diagram that the stress diagram is rectangular upto a strain of .002 and after that the diagram takes the shape of the parabola. Therefore, the shape of the stress diagram is partially rectangular and partially parabola.

7. Bresler’s load contour method is used for analysis of_____________________

a) short column subjected to only uniaxial loading

b) short column subjected to only biaxial loading

c) short column subjected to uniaxial and biaxial loading

d) short column subjected to any loading

View Answer

Explanation: The Bresler’s load contour method is utilized for analysis short column subjected to uniaxial and biaxial loading by studying the mode of failure in the section. This method has the provision of considering both uniaxial and biaxial loading; Pu and Mu is plotted in the curve for this method.

8. For a 300*300 mm^{2} column section of length 5m, what is the value of minimum eccentricity (e_{min})?

a) 36.67mm

b) 30mm

c) 10mm

d) 20mm

View Answer

Explanation: e

_{min}= (l/500)+(D/30)

= (5000/500)+(300/30)=20mm;

Now another check criteria for this e

_{min}is e

_{min}/D > .05

20/300 = .067 > .05.

9. What is the minimum value for percentage of steel (p_{t}) for Fe500?

a) .17%

b) .27%

c) .37%

d) .47%

View Answer

Explanation: According to clause 26.5.1.1 of IS 456:2000, p

_{t}≥ 85/f

_{y}

p

_{tmin}=85/500

p

_{tmin}=.17%.

10. What is the limit state total factor of safety?

a) 2.22

b) 1.5

c) 1

d) 3.33

View Answer

Explanation: According to Indian Standard Code IS 456:2000,

The stress factor of safety = 2.22;

The load factor of safety = 1.5;

The total factor of safety = 2.2*1.5

= 3.33.

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