Have I got this correct? Simple beams with UDL.
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Original post by TheGrinningSkull
Ahah, yea, give it a little brush over. But yea, it's much easier for these stress questions to work in Newtons and millimeters 
As the magapascal (MPa) is more recognised or one most people and books are familiar with.
Wanna go through that question again? Also, if there's any other problems give me a shout. (Don't forget to quote me at times as I may forget!)
As the magapascal (MPa) is more recognised or one most people and books are familiar with.
Wanna go through that question again? Also, if there's any other problems give me a shout. (Don't forget to quote me at times as I may forget!)
Cheers chap, it would certainly of taken me a long time otherwise to get to grips with it.
Original post by Per Ardua
Cheers chap, it would certainly of taken me a long time otherwise to get to grips with it.
It's no worries
And it's why this forum is here 
Original post by TheGrinningSkull
Wanna go through that question again? Also, if there's any other problems give me a shout. (Don't forget to quote me at times as I may forget!)
Wanna go through that question again? Also, if there's any other problems give me a shout. (Don't forget to quote me at times as I may forget!)
Next attempt. What do you think?
How would I draw the Max Stress Distribution? (It's probably nothing like I've drawn!
)(edited 10 years ago)
The next question (As pictured) asked me to reduce the max stress by increasing the cross sectional area of the beam (L x B) by 20% so instead of 20,000cm2, I have 24,000cm2, the beam was to remain a rectangle and neither the original dimensions could be less than before.
So:
I've increased 'd' to 240mm giving the following formulae:
So my proposed dimensions of the beam cross section are up from 100x200 to 100x240.
So:
I've increased 'd' to 240mm giving the following formulae:
So my proposed dimensions of the beam cross section are up from 100x200 to 100x240.
Next up was calculating the percentage reduction on the max stress by increasing the cross sectional area in such a way:
Original post by Per Ardua
Next attempt. What do you think?
How would I draw the Max Stress Distribution? (It's probably nothing like I've drawn!)
How would I draw the Max Stress Distribution? (It's probably nothing like I've drawn!
)Original post by Per Ardua
The next question (As pictured) asked me to reduce the max stress by increasing the cross sectional area of the beam (L x B) by 20% so instead of 20,000cm2, I have 24,000cm2, the beam was to remain a rectangle and neither the original dimensions could be less than before.
So:
I've increased 'd' to 240mm giving the following formulae:
So my proposed dimensions of the beam cross section are up from 100x200 to 100x240.
So:
I've increased 'd' to 240mm giving the following formulae:
So my proposed dimensions of the beam cross section are up from 100x200 to 100x240.
Original post by Per Ardua
Next up was calculating the percentage reduction on the max stress by increasing the cross sectional area in such a way:
All looking good there!! And you didn't forget that y changed to 120mm
Good work! Original post by TheGrinningSkull
All looking good there!! And you didn't forget that y changed to 120mm Good work!
Good work! Cheers. It's a combination of looking at the damn books and your advice, eventually it sticks, so hopefully soon I'll be ready to get among the modules which I know well.
Original post by Per Ardua
Cheers. It's a combination of looking at the damn books and your advice, eventually it sticks, so hopefully soon I'll be ready to get among the modules which I know well.
Good stuff, keep up the good work! Once you know what you're doing, you'll get confident with it, just watch out for silly errors
Right then, Columns....
Attached is the column that I'm getting questioned on. Now I did some sneaking around via Google and found that someone else has answered this question before (Edexcel standard by the looks of it). Rather than just cheat and use his work and hope it goes unnoticed I want to understand what he(she) has done, as it looks like hieroglyphics to me at the moment.
I've highlighted in red the work that I have found.
1.) What is the minimum length of the column at which buckling is likely to occur?
The buckling equation is: F= ((∏^2)*E*I)/(K*L)^2
E is the modulus, I is the area moment of inertia, K is the buckling coefficient, and L is the length of the column.
Looking at my books, I can't see how this formulae has been constructed.
the mass moment of inertia for a hollow tube is;
(using meters)
I = (0.08^4-0.06^4)∏/64 = 1.37 x 10^-6 m^4
Yep, get this. Standard area stuff.
for the force, F, that will cause yielding.
The equation is:
σ= F/A
The cross sectional area of the column is:
A = ∏(do^2-di^2) / 4
A = 0.002m^2
The next bit however, WTF?
Need to convert MN/m^2 to GN/m^2 by multiplying by 10^-3.
Therefore, F = Aσ = 0.02 x 141 x 10^-3 = 3.08x10-4 GN
^^^
This bit loses me, why has 141 been used as the Yield Stress and not 140? Plus how is this entered into a Calc?
The value of K for fixed-fixed end conditions is 0.5
I found this in my books so yep, got that.
Therefore, rearrange the equation and solve for L
L= (SQRT) ((∏^2)*E*I)/F*k^2 = 5.93 m
So yeah, I'm sort of following the path laid before me but I don't fully understand it (yet). My main area of confusion is just how he calculated F.
Any clues?
Attached is the column that I'm getting questioned on. Now I did some sneaking around via Google and found that someone else has answered this question before (Edexcel standard by the looks of it). Rather than just cheat and use his work and hope it goes unnoticed I want to understand what he(she) has done, as it looks like hieroglyphics to me at the moment.
I've highlighted in red the work that I have found.
1.) What is the minimum length of the column at which buckling is likely to occur?
The buckling equation is: F= ((∏^2)*E*I)/(K*L)^2
E is the modulus, I is the area moment of inertia, K is the buckling coefficient, and L is the length of the column.
Looking at my books, I can't see how this formulae has been constructed.
the mass moment of inertia for a hollow tube is;
(using meters)
I = (0.08^4-0.06^4)∏/64 = 1.37 x 10^-6 m^4
Yep, get this. Standard area stuff.
for the force, F, that will cause yielding.
The equation is:
σ= F/A
The cross sectional area of the column is:
A = ∏(do^2-di^2) / 4
A = 0.002m^2
The next bit however, WTF?
Need to convert MN/m^2 to GN/m^2 by multiplying by 10^-3.
Therefore, F = Aσ = 0.02 x 141 x 10^-3 = 3.08x10-4 GN
^^^
This bit loses me, why has 141 been used as the Yield Stress and not 140? Plus how is this entered into a Calc?
The value of K for fixed-fixed end conditions is 0.5
I found this in my books so yep, got that.
Therefore, rearrange the equation and solve for L
L= (SQRT) ((∏^2)*E*I)/F*k^2 = 5.93 m
So yeah, I'm sort of following the path laid before me but I don't fully understand it (yet). My main area of confusion is just how he calculated F.
Any clues?
Right, I've ignored what I've dug up (as it's confusing me) and gone with working out the E.S.R first.
From working out the E.S.R., I took my calculations of I and A, using them in the formula for Le which gave me 2.968, doubled that to get my effective length, so 5.94!
I'll post a pic up later as I'm meant to be busy at work but hey ho.
From working out the E.S.R., I took my calculations of I and A, using them in the formula for Le which gave me 2.968, doubled that to get my effective length, so 5.94!
I'll post a pic up later as I'm meant to be busy at work but hey ho.
Original post by Per Ardua
Next up is deciding on the mode of failure....
I've worked out the load where failure will occur:
F = (∏^2 (200e9 * 1.3745e-6)) / ((0.5 * 5.9372) ^2)
= 913,950.76804 N
= 913.95 KN
So if true, how do I work out if it buckles or bends when it reaches this critical load?
I've worked out the load where failure will occur:
F = (∏^2 (200e9 * 1.3745e-6)) / ((0.5 * 5.9372) ^2)
= 913,950.76804 N
= 913.95 KN
So if true, how do I work out if it buckles or bends when it reaches this critical load?
Hi, hmm, unfortunately I haven't learnt about columns yet in terms of buckling, so I'm not familiar with the concept yet and I don't want to give you false ideas just by internet skimming :P
I'll try to let you know as soon as I understand the concept
TheGrinningSkull
Hi, hmm, unfortunately I haven't learnt about columns yet in terms of buckling, so I'm not familiar with the concept yet and I don't want to give you false ideas just by internet skimming :P
I'll try to let you know as soon as I understand the concept
I'll try to let you know as soon as I understand the concept
No probs, I'm just going to keep staring at the books until I have a Eureka moment!
Original post by Per Ardua
No probs, I'm just going to keep staring at the books until I have a Eureka moment!
:P Good luck!
Original post by TheGrinningSkull
:P Good luck!
Right, I've found this so I'll be having a crack at it tomorrow. Thoughts?
http://www.engineersedge.com/column_buckling/column_ideal.htm
Original post by Per Ardua
Right, I've found this so I'll be having a crack at it tomorrow. Thoughts?
http://www.engineersedge.com/column_buckling/column_ideal.htm
http://www.engineersedge.com/column_buckling/column_ideal.htm
Well, I attempted to look ahead in my notes on columns, and from what I gather there are 3 modes of failure (one is the combination of the other 2), and then terms I don't understand :L
Unfortunately I don't take columns until February so I can't properly help you out.
Although if you do manage to find 2 sources of 3 that agree with each other in what you need to answer and the maths behind it, then you can probably use them.
Although my best advice would be to speak to a relevant member of staff that's teaching you or a lecturer you can email regarding your query!
Post the columns problem and I will try to help.
Original post by Smack
Post the columns problem and I will try to help.
Thank you. I've attached 4 photos.
1. The diagram accompanying the problem.
2. The initial problem.
3. A calculation I am unsure of how to apply. (This I worked on after my findings in picture 4)
4. My working out thus far.
Any advice is greatly appreciated.
(edited 10 years ago)
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