Test: Beams & Slabs - 2 - Civil Engineering (CE) MCQ

for f_y = 415 N/mm² 
and E_s = 2 x 10⁵ N/mm² 
x_{u, max} = 0.48 d and 
for  f_y = 250 N/mm²
x_{u, max} = 0.53 d

The various load combinations are as follows:
(i) For Dead Load and Live Load the ultimate bending moment is given by,
M_u = 1.5 (DL + LL)
= 1.5 x (20 + 30) = 75 kN-m
(ii) For Dead Load and Earthquake (Seismic) Load the ultimate bending moment is given by,
M_u = 1.5 (DL + EL)
= 1.5 x (20 + 10) = 45 kN-m
(iii) For Dead Load, Live Load and Earthquake (Seismic) Load the ultimate bending moment is give by,
M_u = 1.2 (DL + LL + EL)
= 1.2 x (20 + 30 + 10) = 72 kN-m
So, design BM is maximum of all three combinations i.e. 75 kN-m.

A doubly reinforced beam is a type of reinforced concrete beam with reinforcement bars placed in both the tension and compression zones. These beams are used when the dimensions of the beam (both depth and breadth) are restricted due to architectural or structural considerations, but the load-carrying capacity needs to be increased.

When both the depth and breadth of a beam are restricted, it becomes challenging to provide adequate reinforcement using only singly reinforced beams. In such cases, a doubly reinforced beam is recommended to enhance the strength and load-carrying capacity of the beam without increasing its size.

For example, if a beam is used in a building with low ceiling height or limited space for structural elements, increasing the depth or breadth of the beam may not be feasible. In such cases, using a doubly reinforced beam allows the beam to carry more load without increasing its size.

In summary, doubly reinforced beams are recommended when both the depth and breadth of a beam are restricted, and there is a need to increase its load-carrying capacity without altering its dimensions.

In cantilever beam the top fibers above the neutral axis are subjected to tensile stresses. So main reinforcement is provided above the neutral axis.

∴ 

For under-reinforced beam,
(i) x_u < x_{u, lim} or x_c
(ii) Steel reaches ultimate stress before concrete reaching the ultimate stress
(iii) M_u < M_{u, lim}
(iv) Lever arm = d - 0.42 x_u will be more in the case of under-reinforced section.

To ensure ductility the maximum strain in tension reinforcement in the section at failure shall not be less than 
 

For cantilever and continuous beam the ratio of span to depth to satisfy stiffness criterion shall not exceed 7 and 26 respectively.
(clause 23.2.1 of IS : 456-2000)

As per clause 26.5.1.1 of IS : 456-2000, the maximum area of tension reinforcement in beams shall not exceed 0.04 bD.
As per clause 26.5.1.2 of IS : 456-2000, the maximum area of compression reinforcement in beams shall not exceed 0.04 bD.

Moment of resistance for a balanced section is given by

But for Fe 415,
x _{u, lim} = 0.48 d
∴ M_u.lim = 0.36 f_ck x 0.48 x (1 - 0.42 x 0.48) x bd²
⇒  M_u.lim = 0.13796 f_ck x bd²
On comparison, we get,
K = 0.13796 f_ck
= 0.13796 x 20 = 2.76

As per clause 26.5.1.5 of IS: 456-2000, the maximum spacing of shear reinforcement measured along the axis of member shall not exceed 0.75 d for vertical stirrups and ‘d ’ for inclined stirrups at 45°, where 'd' is the effective depth of the section under consideration. In no case shall the spacing exceed 300 mm.
∴ Maximum spacing =0.75 d
= 0.75 x 300
= 225 mm

As per clause 22.2 of IS:456-2000, for simply supported beam or slab, the effective span of a member that is not built integrally with its supports shall be taken as clear span plus the effective depth of slab or beam or centre to centre of supports, whichever is less.
∴ Effective span = 5 x 1000 + 400 
= 5400 mm
and effective span =  5 x 1000 + (300/2) + (300/2)
= 5300 mm

As per IS: 456-2000 modular ratio is given by,
m = (280/3σ_cbc)
For M20 concrete,
σ_cbc =  7N/mm²
∴ m = 280/(3x7) = 13.3

Total compressive force without considering the partial safety factor of material,
⇒ 0.36 f_ck b x_u x 1.5 = 0.54 f_ck b x_u

For T-beams effective width of compression flange,
⇒ b_f = (L₀/6) + b + 6t
For L-beam,
⇒ b_f = (L₀/12) + b_w + 3D_f