Test: RCC Design- 1 - Civil Engineering (CE) MCQ

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

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

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

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.

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.

Modulus of Elasticity concrete

θ = 1.6

% decrease = 61.5%

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

Calculating actual neutral axis of the beam

0.36 f_ck b X_u = 0.87 f_y A_st

 

= 145 mm < x_u,max

∴ Under reinforced section.

∴ Moment of resistance,

M_u = 0.87 f_y A_st (d – 0.42 x_u)

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

= 113.182 × 10⁶ N – mm

= 113.182 kN-m

effective length, l_o = 0.7 × 5000 = 3500 mm

for isolated ‘T’ beam, effective flange width, 
 

where, b = 1500 mm

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

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,

X_u = X_ulim

For Fe 500

X_ulim = 0.46 d

d = effective depth of the beam = 550 mm

X_ulim = 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,

ϵ_sc = 0.00281

Shear resisted by concrete = τ_c bd

= 0.48 × 300 × 500 × 10 ^{- 3}

= 72 KN > 70 KN

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

V_us = V_u - τ_c bd = 140 - 72 = 68 KN

Shear resisted by bent up bars near supports,

V_usb = 0.87 f_y A_st Sinα

V_usb = 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.