Understanding Singularity Functions and Solving Equations with Assumptions

AI Thread Summary
The discussion focuses on solving equations involving singularity functions in beam deflection analysis. It clarifies that the equation (w/24)(<x-0.5L>^4) = 0 is valid when x is less than 0.5L, leading to <x-0.5L> being zero at the boundary condition x = 0. The deflection function is derived through double integration, necessitating the determination of constants from boundary conditions, specifically at points A and C. The analysis emphasizes the need to account for statically indeterminate reactions in the beam. Understanding these principles is crucial for accurately determining the beam's deflection and reactions.
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Homework Statement


http://me.erciyes.edu.tr/mkapalak/MKA-COURSES/STRENGTH-2/EXAMS/STR2-39-2.pdf
How can (w/24)( <x-0.5L>^4 )= 0 ?

by doing so, the author assume x <0.5L ? so that <x-0.5L> = 0
How do we know that?
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Homework Equations

The Attempt at a Solution

 
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or ( <x-0.5L>^4 )= 0 is due to the author taking x less than 0.5L ?
 
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chetzread said:
or ( <x-0.5L>^4 )= 0 is due to the author taking x less than 0.5L ?
The boundary condition for the beam is evaluated at x = 0, so the singularity function evaluates to zero.
 
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SteamKing said:
The boundary condition for the beam is evaluated at x = 0, so the singularity function evaluates to zero.
if we do not consider x = 0 , then x can by any point along the beam ABC ?
 
chetzread said:
if we do not consider x = 0 , then x can by any point along the beam ABC ?
That's not the point.

The deflection function has been constructed for this beam using singularity functions. Because the deflection function was obtained using double integration, there are some unknown constants of integration which must be determined so that the particular deflection function for this beam can be determined.

At point A, the length coordinate is x = 0 and the deflection there is y = 0. This is one of the boundary conditions for this beam. By taking the general deflection function and substituting x = 0 and y = 0 into it, one can see that the constant c2 = 0. That's why this exercise was done in the first place. By using a different boundary condition, such as the fixed end at point C, then substituting x = L and y = 0 should determine the value of the constant c1.

Because this beam is statically indeterminate, the reactions RA, RC, and MC will also require solution. You should study the worked out example from the OP, since there are several pages of calculations altogether.
 

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