Question

The cross-section at mid-span of a beam at the edge of a slab is shown in the sketch. A portion of the slab is considered as the effective flange width for the beam. The grades of concrete and reinforcing steel are M 25 and Fe 415 respectively. The total area of reinforcing bars (A) is 4000 mm${}^2$. At the ultimate limit state, $x_u$ denotes the depth of the neutral axis from the top fibre. Treat the section as under-reinforced and flanged ($x_u>100$ mm).

The value of $x_u$ (in mm) computed as per the Limit State Method of IS 456: 2000 is

• A. 200.0
• B. 223.3
• C. 236.3
• D. 273.6

Answer 1

The correct option is C 236.3


Grade of concrete = M25
Grade of steel Fe415

$x_u>100$ mm
Section is under reinforced
D = 650 mm
d = 570 mm
$b_w=$325 mm

For L beam effective width of flange

$b_f=\frac{l_0}{12}+b_w+3D_f$

$D_f=$ Depth of flange
$l_0=$ Distance between points of zero moment in a beam

As the length of beam is not given. So taking effective width of the flange
= 1000 mm
For Depth of Neutral Axis $X_u$
Assuming,
$D_f>\frac{3}{7}{x}_u$


C = T
$0.36f_{ck}.b_w{x}_u+(1000-b_w)\times D_f\times 0.446f_{ck}$

$=0.87f_y.A_{st}$

$0.36\times 25\times 325x_u+675\times 100\times 0.446\times 25$

$=0.87\times 415\times 4000$

$x_u=236.43mm$

$\frac{3}{7}{x}_u=\frac{3}{7}\times 236.43$

$=101.33mm>D_f$

Hence our assumption is correct.