UDL on C16

[Kindregg]

HI,

I've just started a new job in a sawmills and we are supplying timber for the inside of a shipping container and have been asked to work out the max UDL that can be distributed on the timber.

I remember doing these type of calculations in college but have not done them in a while now and am very unsure of myself. I am not an engineer.

The Beam is simply supported on both ends and is C16 grade under euro-codes spec.

2.46m x 75mm x 75mm is the dimension of the beam.

I have been trying to work backwards to get the max load before faliure including the k mods for medium term loading (3weeks 0.8) and partial material factor (solid timber 1.3)
These are the formula i have been working with looking for max W

sigma = (5.3 for C16 x kmod) / partial material factor
Sigma = (5.3(.8))/ 1.3
Sigma = 3.261 N/mm^2

I = (b d^2)/6
I = (0.075m x (0.075mm^2))/6
I = .0000703 m^3

M = sigma I
M = 3.261 N/mm2 x .0000703 m^3
M = 229.25 N/m

(W L^2)/8 = M
W = (8(M))/L^2
W = (8(229.25 N/m))/2.46^2
W = .303 KN/M

Thats the way I have attempted it but it can't be right.

I'm am trying to to find the maximum UDL that the C16 2.46m x 75mm x75mm simply supported beam can support

Any help would be greatly appreciated

 

[jayrod12]

In your M calculation you're missing a factor of 1e9. Change your I into mm^3 and recheck. That should give you more realistic numbers.