A T beam of span 5m has a flange 125 mm x 12.5 mm and web :(187.5 mm x 8 mm). If the maximum permissible stress is 150 MPa. Find the maximum udl the beam can carry.

Note: For calculating maximum u.d.l on beam, there should be Maximum Bending M which is governed by the values of Yt and Yb which must be minimum, here Yt < Yb, which gives maximum bending momenet and corresponding maximum u.d.l (w).

1. Calculation of $\bar y_t$ and $I_{NA}$ : -

Now, $\bar y_t=\frac{\sum_ay}{A}=\frac{169140.625}{3062.5}=55.23 mm$

$\bar y_b=(12.5+187.5)-\bar y_t=144.77mm$

Moment of inertia of section about axis (1)-(1)

$I_{1-1}=\sum I_{self}+\sum_{ay^2}=4414876.3+16.994*10^6$

$I_{1-1}=21.408*10^6 mm^4$

$I_{N-A}=I_{1-1}- Ay_t^2=21.408*10^6-3062.5*55.23^2$

$I_NA=12.066*10^6 mm^4$

Note: For calculating maximum u.d.l on beam, there should be Maximum Bending M which is governed by the values of Yt and Yb which must be minimum, here Yt < Yb, which gives maximum bending moment and corresponding maximum u.d.l (w).

2. Calculation of maximum UDL the beam carried :

As we know that,

$\frac{M}{I}=\frac{f}{y}$

Maximum permissible stress=$150MPa=150 N/mm^2$

$M=\frac{I*f}{y_t}=\frac{12.066*10^6*150}{55.23}$

$M=32.77*10^6 Nmm=32.77*10^3 Nm$

Note: But for the case of maximum UDL:

$M=\frac{wL^2}{8}$

and for the case of maximum point load,

$M=\frac{wL}{4}$

Now, $M=\frac{wL^2}{8}$=32.77 KNm

$M=\frac{w5^2}{8}=32.77$

$w=\frac{32.77}{5^2}*8$

w=10.48KN/m