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**Q.1 A singly-reinforced rectangular concrete beam of width 300 mm and effective depth 400 mm is to be designed using M25 grade concrete and Fe500 grade reinforcing steel. For the beam to be underreinforced, the maximum number of 16 mm diameter reinforcing bars that can be provided is [2018 : 2 Marks, Set-II](a) 3(b) 4(c) 5(d) 6Ans. **(C)

B = 300 mm

d = 400 mm (effective depth)

M25 and Fe500

Number of 16 mm φ

For A

(a) 0.50

(b) 0.65

(c) 0.70

(d) 0.80

Ans.

One end is fixed

Other end is pin jointed

Effective length of column (as per IS:456- 2000)

= 0.80 L

Solution:

Here support width = 230 mm

Effective depth of slab = 80 mm

Clear span = 3.3 - 0.23 = 3.07 m

Effective span = Min. of

Solution:

Since eccentricity effect is being neglected so column can be considered as concentrically loaded. Ultimate axial load carrying capacity of column.

(a) Section X has less flexural strength and is less ductile than section Y

(b) Section X has less flexural strength but is more ductile than section Y

(c) Section X and Y have equal flexural strength but different ductility

(d) Sections Xand Y have equal flexural strength and ductility.

Ans.

Due to presence ot more compression steel in section Y, NA of section of Xis above than as of X. It means Yis more under-reinforced than X so ductility of Y is more.

Since compression steel of Yis more so flexure resistance of X is less than as of Y.

(a) 250 and 3500

(b) 205 and 4500

(c) 270 and 2000

(d) 300 and 2500

Ans.

Minimum tension reinforcement is given by,

Maximum compression reinforcement = 0.04 bD

= 0.04 x 250 x 450

= 4500 mm

Solution:

Shear force at section X-X,

V

Depth at section X-X,

Moment at section X-X,

M

Design shear force at section X-X,

(a) 20.0 and 20.0

(b) 26.0 and 21.0

(c) 26.0 and 20.0

(d) 21.0 and 15.0

Ans.

x-x will be major axis and y-y will be minor axis.

Solution:

or

(a) 475.2

(b) 717.0

(c) 756.4

(d) 762.5

Ans.

or,

Take 717.00 kNm

(a) 200.0

(b) 223.3

(c) 236.3

(d) 273.6

Ans.

Given, X

Effective width of flange of L-beam:

Where l

D

b

Taking b

∴ Length of beam is not given

∴ Equating compressive force and tensile force

C = T

It shows our assumption was correct.

So, x

(Nearest match = 236 mm)

(a) M15

(b) M20

(c) M30

(d) M40

Ans.

Working axial load

The developed bearing pressure on concrete is given as,

The developed stress in direct compression in various grades of concrete as per IS 456 : 2000 are tabulated below = 0.25 f

The permissible stress in concrete should be more than the developed bearing pressure. Thus the minimum grade of concrete which should be used is M 20.

The moment of resistance of the section is

(a) 206.00 kN-m

(b) 209.20 kN-m

(c) 237.80 kN-m

(d) 251.90 kN-m

Ans.

Moment of resistance M

The resultant compressive force in the concrete is given by,

The resultant compressive force in the compression steel is given by,

The resultant tensile force in the tensile steel is given by,

we know that,

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