1. Derivation formula for b(x)

Hello,

A non prismatic figure / beam I need to derive a formula for b (x).
See the picture in the link below:

https://myalbum.com/album/Qs5BcjTUrEdt

The beam has a varying cross-section.

It is given that the dimensions at the end of the beam is determined by the relationship:
b (l) = αb (0), with α less than or equal to 1.

lead a formula for the dimension b (x) at a distance x from the left bearing.

b(x) = b(l) + ((l-x)/l)* (b(0)-b(l))

I self must step by step show the derivation of the function.
How do I do that ?

2. Re: Derivation formula for b(x)

Can someone help me?

3. Re: Derivation formula for b(x)

so if I understand this you've got a beam that has a cross section that varies as the distance from the left bearing.

at the left bearing it's width $b(0)$, at the furthest end, a distance $\ell$ from the left bearing it is width $b(\ell)$

the cross section changes linearly with the distance, $x$, from the left bearing.

We have values at two points, the left and right edges of the beam.

At the left edge we have point $(0, b(0))$

At the right edge we have point $(\ell, b(\ell))$

The slope of the width change is

$m = \dfrac{b(\ell)-b(0)}{\ell - 0} = \dfrac{b(\ell)-b(0)}{\ell }$

and using the point-slope formula for a line we have

$b(x) - b(0) = \dfrac{b(\ell)-b(0)}{\ell }\left(x - 0\right)$

$b(x) = \dfrac{b(\ell)-b(0)}{\ell }x + b(0)$

and I see for some reason your book reversed everything.

Well if we look at it in reverse

$m = \dfrac{b(0)-b(\ell)}{-\ell}$

$b(x)-b(\ell) = \dfrac{b(0)-b(\ell)}{-\ell} (x - \ell)$

$b(x) = \dfrac{b(0)-b(\ell)}{\ell}(\ell - x)+b(\ell)$

and this matches the formula in your book.

4. Re: Derivation formula for b(x)

Thank you very much for your comment!

If I understand correctly, the book has a function
given b(x) for the right bearing? Not from the left bearing.