Kindregg
Industrial
- Jan 15, 2016
- 1
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
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