The document Shear, Torsion, Bond, Anchorage & Development Length Civil Engineering (CE) Notes | EduRev is a part of the Civil Engineering (CE) Course Topic wise GATE Past Year Papers for Civil Engineering.

All you need of Civil Engineering (CE) at this link: Civil Engineering (CE)

**Q.1 An RCC beam of rectangular cross-section has factored shear of 200 kN at its critical section. Its width b is 250 mm and effective depth d is 350 mm. Assume design shear strength Ï„ _{c} of concrete as 0.62 N/mm^{2} and maximum allowable shear stress Ï„_{c max} in concrete as 2.8 N/mm^{2}. If two legged 10 mm diameter vertical stirrups of Fe250 grade steel are used, then the required spacing (in cm, up to one decimal place) as per limit state method will be ________ [2018 : 2 Marks, Set-I]**

Solution:

Nominal shear stress,

SF taken by stirrups

= 82 mm = 8.2 cm

(a) grade of concrete and grade of steel

(b) grade of concrete only

(c) grade of steel only

(d) grade of concrete and percentage of reinforcemen

Ans.

From table 20 of IS 456 : 2000 it is evident that Ï„

(a) additional shear resistance from reinforcing steel

(b) additional shear stress that comes from accidental loading

(c) possibility of failure of concrete by diagonal tension

(d) possibility of crushing of concrete by diagonal compression

Ans.

The check for Ï„

Solution:

Bond strength of concrete

= Tensile force in steel

â‡’

For deformed bars Ï„

âˆ´

âˆ´ k = 6.4

(a) 8

(b) 10

(c) 12

(d) 16

Ans.

Factored SF = 45 kN - V

We have to calculate the dia of Fe 500 2-Legged stirrup to be used at a spacing of 325 mm c/c

âˆ´ %tensile steel

â‡’ Min shear reinforcement is required

â‡’ Min shear reinforcement is given by

Solution:

(a) Maximum of

(b) Maximum of

(c) Minimum of

(d) Minimum of

Ans.

Maximum force resisted by bond strength

Maximum tensile force resisted by bar

âˆ´ Maximum value of P = Minimum of

(a) Shear reinforcement should be designed for 175 kN for beam Pand the section for beam Q should be revised.

(b) Nominal shear reinforcement is required for beam Pand the shear reinforcement should be designed for 120 kN for beam Q.

(c) Shear reinforcement should be designed for 175 kN for beam Pand the shear reinforcement should be designed for 525 kN for beam Q.

(d) The sections for both beams Pand Q need to be revised.

Ans.

Shear strength of the concrete = Ï„c x Bd

= 0.75 x 400 x 750 x 10

Maximum shear strength of concrete

Design shear in beam P is 400 kN which is less than the maximum shear that the concrete can sustain. But the design shear is more than the nominal shear of 225 kN.

âˆ´ Shear reinforcement is designed for

= 400-225 = 175 kN

Design shear in beam Q is 750 kN which is more than the maximum shear that the concrete can sustain. Hence the beam Q should be redesigned.

(a) Corrosion causes circumferential tensile stresses in concrete and the cracks will be parallel to the corroded reinforcing bar.

(b) Corrosion causes radial tensile stresses in concrete and the cracks will be parallel to the corroded reinforcing bar.

(c) Corrosion causes circumferential tensile stresses in concrete and the cracks will be perpendicular to the direction of the corroded reinforcing bar.

(d) Corrosion causes radial tensile stresses in concrete and the cracks will be perpendicular to the direction of the corroded reinforcing bar.

Ans.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

### Footing, Columns, Beams and Slabs

- Doc | 7 pages
### Prestressed Concrete

- Doc | 3 pages

- Working Stress & Limit State Method
- Doc | 7 pages