Simply supported beam

Nov 2015
10
0
Hello, I'm a bit confused on my coursework when calculating the bending moment, friends are saying different things and I don't know anymore.

We have a thin walled cylinder that is simply supported over 8m and it has a mass of 15000kg. The inner diameter is 1750mm and it has a wall thickness of 10mm. The cylinder operates at a pressure of 0.075 Mn/m^2 and during operation it is subjected to an axial force of 150kN and a torque of 800 kNm.
(i) Calculate the component stresses
(ii) Evaluate the 2D complex stress system

It was not stated if it was a udl or point load so we kinda just assumed it was a udl therefore;

F = m*g
F = 15000*9.8
F = 147000N

M = W x L /8
M = 147000 x 8/8
M = 147000 N/m
M = 147x10^6 N/mm

Is this correct? My friend says he has proof that it can be a point load but hasn't showed me the literature.

sigma d = -F/A = 150x10^3/55292.03
sigma d = -2.7 MN/m^2

sigma h = pd/2t = (0.075x10^6)x1.75/2*0.01
sigma h = 6.6MN/m^2

sigma l = pd/4t = (0.075x10^6)x1.75/4*0.01
sigma l = 3.3MN/m^2

tau = Tr/J = (800x10^3)x0.885/0.043
tau = 16.5MN/m^2

J was calculated as 0.043m^4
I was calculated as 2.141x10^10mm^4

Once I figure out the bending stress I'm going to determine the magnitude of the principal stressses analytically and graphically.
 
Last edited:

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
The weight of the beam is 147,000 N, and is simply supported at each end that means that the support reaction force at each end is half of that, or 73,500 N each. You can assume that the load of the weight is uniformly distributed along the length of the cylinder. Hence the shear as a function of position x is:

V(x) = F_r - 147,000 (x/L)

and the moment is the integral of that:

M(x) = F_r x - 147,000 x^2/(2L)
 
Nov 2015
10
0
The weight of the beam is 147,000 N, and is simply supported at each end that means that the support reaction force at each end is half of that, or 73,500 N each. You can assume that the load of the weight is uniformly distributed along the length of the cylinder. Hence the shear as a function of position x is:

V(x) = F_r - 147,000 (x/L)

and the moment is the integral of that:

M(x) = F_r x - 147,000 x^2/(2L)
Would it be wrong to assume it was a point load or is this ok as the weight acts on the centre of gravity? I uploaded a picture as well if that helps. :)
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
Mass (and hence weight) is distributed, not a point load. And yes, it makes a difference.

But now that you've uploaded the picture I see an additional compressive force of 150 KN, which I assume is in addition to the cylinder's mass of 15 Kg. You hadn't mentioned this, and it introduces a different thing to consider. So - does this new force act on the cylinder in addition to the cylinder's weight?
 
Nov 2015
10
0
Mass (and hence weight) is distributed, not a point load. And yes, it makes a difference.

But now that you've uploaded the picture I see an additional compressive force of 150 KN, which I assume is in addition to the cylinder's mass of 15 Kg. You hadn't mentioned this, and it introduces a different thing to consider. So - does this new force act on the cylinder in addition to the cylinder's weight?
Yes it does but I think I'm ok with that. So my compressive stress ie sigma d would be;

sigma d = -F/A = 150x10^3/55292.03
sigma d = -2.7 Mn/m^2

I've done most of the calculations in relation to stress I just couldn't calculate the bending stress as I was unsure about the bending moment. We were also given the torque and internal pressure but I didn't feel that those values were necessary as I only needed help with the bending moment. I've calculated the hoop stress and longitudinal stress as well though.
 
Apr 2015
1,089
252
Somerset, England
I agree with chip, you need to consider (and provide here) all the information.

1) That picture shows the supports almost at the quarter points so the basic elastic curve will have three sections.

2) The horizontal force shown is effectively prestressing, which will affect the bending stress.

3) The pressure induced stresses will subject the the tube wall material to a triaxial stress system.