# Inverting Shifted Laplace function

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1. Apr 10, 2017

### danmel413

1. The problem statement, all variables and given/known data
A beam is supported at one end, as shown in the diagram (PROBLEM 11 page 281 of Lea, 159 of the course pack). A block of mass M and length l is placed on the beam, as shown. Write down the known conditions at x = 0. Use the Laplace transform to solve for the beam displacement.

2. Relevant equations
I've gotten almost the entire way through the problem using the differential equations for beam deflections (most relevant ones given here: http://www.me.berkeley.edu/~lwlin/me128/BeamDeflection.pdf (8.4 and 8.50), to get to a Laplace transform I now have to invert:

L{y(x)} = Y(x) = [(Mge-sx0)/(EILs5)] * (1-e-sL) - (Mg(x0 + L/2))/(EIs3) - (Mg)/EIs4). (This is correct, I have the solutions).

The general shift of a Laplace transform: L{S(t-t0)f(t-t0)}=e-st0)F(s) for t0 > 0

The Mellin inversion integral: f(t) = 1/(2πi) ∫F(s)estds from γ-i∞ to γ+i∞
3. The attempt at a solution
So up in my first equation, the only part that can't be inverted easily is the first term which has to be shifted. The solution doesn't walk through the inversion, just gives the answer, which is annoying. It's this part:[(Mge-sx0)/(EILs5)] * (1-e-sL). I'm working in x instead of t, of course, and my F(s) would be [(Mg)/(EILs5)] * (1-e-sL), and of course the other part of the integral would be e-sx0. Can someone put me on the first step to getting through this?

2. Apr 13, 2017

### DrDu

Multiply it out and you get two terms, one with a shift X0 and one with a shift (X0+L).