Test: RCC Design- 1


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10 Questions MCQ Test GATE Civil Engineering (CE) 2023 Mock Test Series | Test: RCC Design- 1

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Test: RCC Design- 1 - Question 1

Consider the Exposure condition as "Concrete surfaces protected against weather or aggressive conditions, except those situated in coastal areas".

As per IS 456:2000, The above exposure condition will correspond to:

Detailed Solution for Test: RCC Design- 1 - Question 1

As per clause 8.2.2.1, Table No 3 of IS 456:2000,

Test: RCC Design- 1 - Question 2

As per IS 456:2000, the diagonal tension failure in RCC beams depends upon:

Detailed Solution for Test: RCC Design- 1 - Question 2

The allowable shear stress (design shear strength) depends on both percentage of longitudinal steel provided and grade of concrete used in the beam.

Test: RCC Design- 1 - Question 3

Beam sections designed in accordance with LSM as compared to sections designed in accordance with WSM will have

Detailed Solution for Test: RCC Design- 1 - Question 3

Balanced Sections designed on the basis of LSM give smaller depth or width (less concrete area) and more reinforcement than those designed with WSM.

Test: RCC Design- 1 - Question 4

Minimum shear reinforcement in beams is provided in the form of stirrups

Detailed Solution for Test: RCC Design- 1 - Question 4

The minimum shear reinforcement is provided for the following:

(i) Any sudden failure of beams is prevented if concrete cover bursts and the bond to the tension steel is lost.

(ii) Brittle shear failure is arrested which would have occurred due to principal tension without shear reinforcement.

(iii) Tension failure is prevented which would have occurred due to shrinkage, thermal stresses and internal cracking in beams.

(iv) To hold the reinforcement in place when concrete is poured.

(v) Section becomes effective with the tie effect of the compression steel.

Test: RCC Design- 1 - Question 5

Consider the following statements and Identify the correct option with regard to provisions of IS 456: 2000:

P. The minimum stripping time for props to slab spanning over 4.5 m is 7 days

Q. The minimum stripping time for props to arches spanning over 6 m is 14 days.

R. To ensure the strength of Designed concrete mix, Random sampling procedure is adopted

S. When a bar of two different diameters is to be spliced, the lap length shall be calculated on the basis of the bar of the larger diameter.

Detailed Solution for Test: RCC Design- 1 - Question 5

As per IS 456 : 2000,

(i) The minimum stripping time for props to slab spanning up to 4.5 m and spanning over 4.5 m is 7 days and 14 days respectively.

(ii) The minimum stripping time for props to aches spanning up to 6 m and spanning over 6 m is 14 days and 21 days respectively.

(iii) When bar of two different diameter are to be spliced the lap length shall be calculated on the basis of bar of smaller diameter.

(iv) A random sampling procedure shall be adopted to ensure that each concrete batch shall have a reasonable chance of being tested that is, the sampling should be spread over the entire period of concreting and cover all mixing units.

Test: RCC Design- 1 - Question 6

If after 28 days of the casting of RC beam, the creep strain is 60% more than that of Elastic strain Percentage decrease in modulus of elasticity according to IS 456 : 2000 is

Detailed Solution for Test: RCC Design- 1 - Question 6

Modulus of Elasticity concrete

θ = 1.6

% decrease = 61.5%

Test: RCC Design- 1 - Question 7

A rectangular beam of width b = 250 mm, effective depth d = 350 mm is reinforced in tension with mild steel of grade Fe250, Ast = 1800 mm2. The grade of concrete used is M30. The ultimate moment of resistance of the beam is ________

Detailed Solution for Test: RCC Design- 1 - Question 7


Xu,max = 0.53 d= 0.53 × 350 = 185.5 mm

Calculating actual neutral axis of the beam

0.36 fck b Xu = 0.87 fAst

 

= 145 mm < xu,max

∴ Under reinforced section.

∴ Moment of resistance,

Mu = 0.87 fAst (d – 0.42 xu)

= 0.87 × 250 × 1800 (350 – 0.42 × 145)

= 113.182 × 106 N – mm

= 113.182 kN-m

Test: RCC Design- 1 - Question 8

An isolated T-beam used as the walking – way is shown in the figure. The clear span is 5m and fixed at both the ends. Calculated the effective flange width of the T-beam

Detailed Solution for Test: RCC Design- 1 - Question 8

effective length, lo = 0.7 × 5000 = 3500 mm

for isolated ‘T’ beam, effective flange width, 
 

where, b = 1500 mm

bw = width of the web = 450 mm
  + 450 = 1002.63mm

Test: RCC Design- 1 - Question 9

A doubly reinforced concrete beam with an effective cover of 50 mm to center of both tension and compression reinforcement with an effective depth of 550 mm. If the maximum permissible stress in the outermost fibres in both steel and concrete reaches at the same time, what will be the maximum strain in concrete at the level of compression reinforcement satisfying codal provisions of IS 456:2000 according to limit state method.

Take Fe 500 grade of steel for both tension and compression reinforcement and M 20 grade of concrete.

Detailed Solution for Test: RCC Design- 1 - Question 9

As the maximum permissible stress in the outermost fibers in both steel and concrete reaches at the same time, therefore it is the case of balanced section:

So,

Xu = Xulim

For Fe 500

Xulim = 0.46 d

d = effective depth of the beam = 550 mm

Xulim = 0.46 × 550 = 253 mm

d' = effective cover to compression reinforcement = 50 mm

Maximum strain in concrete at outermost compression reinforcement as per IS 456:2000 = 0.0035

Now from similar triangle,

ϵsc = 0.00281

Test: RCC Design- 1 - Question 10

A simply supported beam of 300 mm wide and 500 mm deep is loaded as the shear force at support is 140 KN and throughout the beam, the SF does not exceed 70 KN. One out of six bars of 22 mm dia, Fe 415 grade is bent up at 45˚ near the support to resist the shear. Analyze the beam and choose the correct statement: - 

M20 grade concrete is used with τ= 0.48Mpa

Detailed Solution for Test: RCC Design- 1 - Question 10

Shear resisted by concrete = τc bd

= 0.48 × 300 × 500 × 10 - 3

= 72 KN > 70 KN

Now, at the support shear to be resisted by bent up bar,

Vus = Vu - τc bd = 140 - 72 = 68 KN

Shear resisted by bent up bars near supports,

Vusb = 0.87 fy Ast Sinα

Vusb = 0.87 × 415 × π/4 × 222 × sin45

= 97 KN > 68 KN

So the beam is safe in shear throughout the beam with bent up bars.

No need of transverse shear reinforcement.

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