It is amazing how the sort of maths we are taught at school says "everything is linear and reversible" whereas we all know that is nonsense (try mending a broken window by "reversing the forces".
]]>I think I previously laid out the problem in a similar way, but got stuck on the "trial and improvement" step.
Glad I can close out this issue which has been left unfinished since college. Now, to find a ghost writer for my novel ... :-)
Beam
]]>angle = 0.0097313... radians
R = 271293.83203...
d = 12.84541736
LATER EDIT: I need to modify this calc. It is slightly wrong. Only a few microns out I hope. Back soon.
EVEN LATER EDIT.
Forman's problem is not the same as BeamReacher's. He adds a foot. The Op here has the beam pushed in by an inch. So the arc length is 1 mile; the straight line distance is 1 mile less one inch. The diagram below shows the re-worked version. As I suspected the difference wouldn't come to much and it doesn't. The angle is unchanged.
Bob
]]>For the sake of solving the problem we are forced to assume that the steel bar is a hypothetical one whose curve we know.
That is close to the answer for the problem in the book. But the OP's problem is different. You will have to recompute for it.
]]>Firstly, I agree with MrWhy. There are too many variables to correctly determine the nature of the curve. But let's assume it is part of a circle.
In my diagram, I've converted the distances to feet.
The following equations apply:
From these last two
I used trial and improvement to find a = 0.03370775.... radians.
This gives the radius as
From the top two equations
Using the value of R and the quadratic formula
d = 44.498509.....
(My earlier answer of 63 was because I had used 2641 rather than 2640.5 for the arc length.)
Bob
]]>≈ 44.5 ft is for a 1 foot bar, this is for 1 inch so the method is the same but the answer is different.
Also there is the difference that one problem has 1 ft added to the bar while the other has the bar pushed in by an inch.
]]>The bending moment at any position x of the bar is the sum of the axial forces times the y deflection at that point. In this we should add any forces due to the weight of the bar.
A useful formula is the deflection of the bar at point x is
Bending moment M at point x is E I d2ydx2 , where E is Young's modulus and I is the moment of area of the beam's cross-section at position x.
Sorry, I missed the bit where you had the answer. You're saying 44 ft ?
At the moment I'm getting 63 ft, so I'm thinking there's a flaw in my working. Difficult to get theta.
Bob
]]>Do you require the answer to this question? That I have on hand. If you need the working that will take some time to latex up, so please tell me now.
I guess my solution is not necessary, okay.
]]>(Otherwise, I guess I could buy/borrow the book.)
Merci beaucoup,
Beam
Numerical methods That Usually Work by Forman S. Acton.
This is available from Amazon. The problem is shown in the 'Look Inside' advert. The solution is not.
Forman has the curvature as an arc of a circle.
I think I can generate an answer if you ask me nicely.
Bob
]]>He meant a physicist. Although his English is excellent, English is not his primary language.
He also meant the solution to my rhombus problem in another thread.
Do you require the answer to this question? That I have on hand. If you need the working that will take some time to latex up, so please tell me now.
]]>