Shigleys Indeterminate Beam Derivation

  • Thread starter bugatti79
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  • #1
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Homework Statement



Folks,

I am having difficulty deriving the moment expressions for a rigidly supported beam fixed at either ends and subjected to a point load. I have two attachments, one for the expressions given in Shigleys and the other for my attempted derivation.

The problem is that I want to derive the left hand fixing moment [itex]M_1[/itex] and [itex]M_{ab}[/itex] as in Shigleys. However, I believe my attempts are not leading to these expressions.

Is anyone good at these indeterminate derivations?

Thanks
Bugatti79



Homework Equations



In attachments



The Attempt at a Solution



In attachments
NOte that I have posted this in the math help forum http://www.mathhelpforum.com/math-help/f9/shigleys-indeterminate-beam-derivation-189693.html"

I will inform both post of any updates on a daily basis.
 
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Answers and Replies

  • #2
SteamKing
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It seems your work is OK as far as it goes. Remember, the slope and deflection of the beam are both zero at the right end of the beam as well.
 
  • #3
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Hi Steamking,

Thanks for your reply.
[itex]EI \frac{dy}{dx}=M_1 x+\frac{F(x-a)^2}{2}-\frac{R_1 x^2}{2}+c1[/itex]

[itex]EIy=\frac{M_1 x^2}{2}+\frac{F(x-a)^3}{6}-\frac{R_1 x^3}{6}+c1 x+c2[/itex]

applying the BC's gives c1 and c2 both =0.

Yes, I get 2 equations and 2 unknowns as below...eliminating R1 to find M1

[itex]\frac{1}{6} M_1 x^2 =-\frac{F(x-a)^3}{6}+\frac{F(x-a)^2 x}{6}[/itex]

I dont see how this leads to M1 in shigleys because it also has a b term in it. Also, I am curious how to derive [itex]M_{ab}[/itex]...........
 
  • #4
SteamKing
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You have determined M1 in terms of F and x. You should be able to substitute for M1 in the slope equation and evaluate it at x = L. Knowing the value of the slope should allow you to solve for R1.
 
  • #5
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Dear Steam King,

I have obtained both M1 and Mab! Thanks

bugatti79
 

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