RCC Design Test 2

Question 1
2 / -0.33

For the working stress method, limiting depth of Neutral axis depends on ____

A
Grade of concrete

B
Grade of steel

C
Percentage of steel

D
All of the above

Question 2
2 / -0.33

While designing, for a steel column of Fe250 grade, a base plate resting on a concrete pedestal of M20 grade, the bearing strength of concrete (in N/mm²) in limit state method of design as per IS : 456 - 2000 is __________

A
9-9

B
94-91

C
97-96

D
99-92

Question 3
2 / -0.33

An axially loaded column is of 300 mm × 300 mm size. The effective length of the column is 3 m. What is the minimum eccentricity of the axial load for the column?

A
25 mm

B
10 mm

C
16 mm

D
20 mm

Question 4
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A reinforced cement concrete beam section has a size of 300 mm width and 400 mm effective depth, the grade of concrete used is M20 and grade of steel is Fe 415. If the ultimate bending moment of 150 kN-m acts at the section, the beam has to be designed as

A
Singly reinforced beam

B
Doubly reinforced beam

C
Beam cannot be designed

D
Balanced beam

Question 5
2 / -0.33

In a reinforced concrete section, the stress at the extreme fibre in compression is 12 MPa. The depth of neutral axis in the section is 60 mm and grade of concrete is M20.

Assume linear elastic behavior of the concrete.

A
Radius of curvature is 11.16 cm

B
Effective curvature is 8.96 × 10⁵mm^-1

C
Effective curvature is 8.96 × 10⁵mm

D
Radius of curvature is 11.16 m

Question 6
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If ϕ = nominal diameter of reinforcing bar, f_s= compressive stress in the bar and f_bd = design bond stress of concrete,
Which of the following statements are true for development length (L_d)

A
Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

B
Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

C
Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

D
Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

Question 7
2 / -0.33

In the design of a reinforced concrete beam the requirement for bond is not getting satisfied. The economical option to satisfy the requirement for bond is by

A
Bundling of bars

B
Providing smaller diameter bars more in number

C
Providing larger diameter bars less in number

D
Providing smaller diameter bars less in number

Question 8
2 / -0.33

Reduction coefficient taken for calculating axial load carrying capacity of long column as per working stress method, for the given column dimension is______
(Column dimension = 300 × 400 mm of L_effective = 4500 mm)

A
1.2500

B
1.0156

C
1.1161

D
0.9375

Question 9
2 / -0.33

A rectangular simply supported beam has factored shear force at the centre of the support is 200 KN. The beam has 2 bars of 16 mm dia Fe 415 steel at the centre of the support, on tension zone. M20 grade concrete with τ_bd = 1.2 MPa is used in the beam. Moment of resistance of the beam at support is 60 KN - m. What is the minimum extension of the main steel bars (in mm) with 90° bend to satisfy anchorage bond?

A
230-240

B
2304-2401

C
2307-2406

D
2309-2402

Question 10
2 / -0.33

A short column having cross section 460 × 600 mm (overall depth). It is given that the column is meeting the general requirements of the relevant code for the minimum eccentricity (e_min) as per IS 456 : 2000. If 10 bars of 20 mm diameter is provided as main bar in the column. Calculate the ultimate load carrying capacity of the column (in kN) as per IS 456 : 2000. Use M - 25 concrete and Fe - 415 steel?

A
3601-3603

B
36014-36031

C
36017-36036

D
36019-36032

Question 11
2 / -0.33

At the limit state of collapse, an RC beam is subjected to flexural moment of 200 kN-m, Shear force 20 kN, and torque 9 kN-m. The beam is 300 mm wide and gross depth of 425 mm, with an effective cover of 25 mm. The equivalent nominal shear stress (τ _ve) as calculated by using design code turns out to be lesser than the design shear strength (τ _c) of the concrete.

A
Equivalent shear force is 40 kN

B
Equivalent flexural moment required for longitudinal reinforcement is 212.79 kN

C
Equivalent flexural moment required for longitudinal reinforcement is 200 kN

D
Equivalent shear force is 68 kN

Question 12
2 / -0.33

For a RC column fixed at both ends having length of 15 m and cross-section 400 mm × 550 mm. If 4 – 25ɸ bars are used in the column. The load carrying capacity of the column is _____ kN
Take σ_cc = 5 mpa and σ_sc = 195 mpa

A
1090-1100

B
10904-11001

C
10907-11006

D
10909-11002

Question 13
2 / -0.33

The positive bending moment coefficient at the middle of the end-span of a continuous one way slab is

A
\(\left( {\frac{{{w_l}}}{{10}} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

B
\(\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{10}}} \right){L^2}\)

C
\(\left( {\frac{{{w_l}}}{{12}} + \frac{{{w_d}}}{{16}}} \right){L^2}\)

D
\(\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

Question 14
2 / -0.33

A prestressed beam of size 250 × 400 mm deep is prestressed by 14 tendons of 6 mm diameter. The cable is located at 150 mm from bottom of the beam. If the initial prestress in the beam is 1250 N/mm². The loss of prestress due to elastic deformation and creep of concrete respectively will be?
Assume:
M-40 Grade concrete
creep coefficient = 1.6
Modulus of elasticity = 2 × 10⁵ MPa.
Neglect the effect due to dead load.

A
54.72 MPa and 87.55 MPa

B
37.13 MPa and 59.42 MPa

C
25.41 MPa and 40.65 MPa

D
42.30 mPa and 55 MPa

Question 15
2 / -0.33

Consider a bar of Diameter ‘D’ of length (L) embedded in a large concrete block. If a pull out force P being applied. Let σ_band σ_st be the bond strength (between the bar and concrete) and the tensile strength of the bar respectively. If the block is held in position and material of the block does not fall, which of the following options represents the maximum value of P?

A
Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\)and (π × D × L× σ _st)

B
Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)and (π × D × L× σ_b)

duplicate options foud. English Question 1 options 1,4

C
Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) and (π × D × L× σ _b)

D
Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\) and (π × D × L× σ st)

Question 16
2 / -0.33

For a reinforced concrete beam with M 20 concrete with Fe 415 steel, the working moment corresponding to ‘balanced failure’ by the Limit Sate Method of Design is A times of the relatable magnitude of the working moment computed by the Working Stress Method. This value of A is nearly

A
1.0

B
1.5

C
2.0

D
2.5

Submit Test

RCC Design Test...

Question 1

Grade of concrete

Grade of steel

Percentage of steel

All of the above

Question 2

9-9

94-91

97-96

99-92

Question 3

25 mm

10 mm

16 mm

20 mm

Question 4

Singly reinforced beam

Doubly reinforced beam

Beam cannot be designed

Balanced beam

Question 5

Radius of curvature is 11.16 cm

Effective curvature is 8.96 × 105 mm-1

Effective curvature is 8.96 × 105 mm

Radius of curvature is 11.16 m

Question 6

Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{5 \times {f_{bd}}}}\)

Development length of deformed reinforcement bar is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

Development length of straight bar in compression is \({L_d} = \frac{{\phi \times {f_s}}}{{6.4 \times {f_{bd}}}}\)

Question 7

Bundling of bars

Providing smaller diameter bars more in number

Providing larger diameter bars less in number

Providing smaller diameter bars less in number

Question 8

1.2500

1.0156

1.1161

0.9375

Question 9

230-240

2304-2401

2307-2406

2309-2402

Question 10

3601-3603

36014-36031

36017-36036

36019-36032

Question 11

Equivalent shear force is 40 kN

Equivalent flexural moment required for longitudinal reinforcement is 212.79 kN

Equivalent flexural moment required for longitudinal reinforcement is 200 kN

Equivalent shear force is 68 kN

Question 12

1090-1100

10904-11001

10907-11006

10909-11002

Question 13

\(\left( {\frac{{{w_l}}}{{10}} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

\(\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{10}}} \right){L^2}\)

\(\left( {\frac{{{w_l}}}{{12}} + \frac{{{w_d}}}{{16}}} \right){L^2}\)

\(\left( {\frac{{{w_l}}}{9} + \frac{{{w_d}}}{{12}}} \right){L^2}\)

Question 14

54.72 MPa and 87.55 MPa

37.13 MPa and 59.42 MPa

25.41 MPa and 40.65 MPa

42.30 mPa and 55 MPa

Question 15

Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\)and (π × D × L× σ st)

Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)and (π × D × L× σb) duplicate options foud. English Question 1 options 1,4

Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) and (π × D × L× σ b)

Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\) and (π × D × L× σ st)

Question 16

1.0

1.5

2.0

Effective curvature is 8.96 × 10⁵mm^-1

Effective curvature is 8.96 × 10⁵mm

Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _b}} \right)\)and (π × D × L× σ _st)

Maximum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\)and (π × D × L× σ_b)

duplicate options foud. English Question 1 options 1,4

Minimum of \(\left( {\frac{\pi }{4} \times {D^2} \times {\sigma _{st}}} \right)\) and (π × D × L× σ _b)