When you're dropping the truck from a given height, all you are really looking to find is the force that that drop will exert on the beam. Using F=ma, to determine the Force, you need to know the mass (the weight you want to use) and the acceleration (change in velocity over a given time). It's the acceleration part that needs to be iteratively solved for a solution. As for the deflection, I'd say you can use the simply supported, centrally loaded beam equation for this one because of the (hopefully) small displacements compared to the length of the beam.
The equation is for max deflection is:
Ymax = (F(L^3))/(48EI)
Pretty much, you have to pick a starting point to get this iterative process rolling. The most logical point would be right before the beam reaches the yield point (max deflection). First you need to calculate the force required to acheive yield stress. To do this, solve the bending stress equation for the force:
Stress = Fl/4Z
F = (4*Z*(Yield Stress))/L
F = Force to reach yield stress
l = length of beam
Z = Section modulus (you have these)
Yield Stress = Yield Stress of the material you are using.
With that you can calculate the force (Force1) to reach max deflection. With that deflection (distance traveled) you know that you will slow from the freefall speed to zero in that distance. With a speed of "V", and a distance traveled of "D", the equation for acceleration is:
a = (Velocity at impact)^2/(2*distance traveled)
Where distance traveled = deflection of beam
Knowing the acceleration, you can calculate the force on the beam using F=ma. (Force 2)
Now, that Force came from an ASSUMED acceleration, with came from an ASSUMED displacement. You'll notice that if you take the final force, it will not be enough to make the beam reach a max deflection. What you need to do now is to take "Force2" and plug it back into the deflection equation, calculate a new deflection, a new acceleration, and yet another force and repeat this process until the final calculated force lines up with the inital force. (Force1~Force2). At this point, the solution will have converged and you'll notice that there will be very little difference in Force2 compared to Force1 for subsequent iterations.
Excel can do this easily using the it's built in solver. If you've never used the solver, it's easy. You just tell it what cell to compare to what cell, what their relationship should be, and which cell it can change to find a solution. Just tell it to find a solution where Force1 = Force2. Apply a couple of contraints, like they have to be positive numbers, etc and let it run.
Will's equation will be the same as mine, he just stated a height and weight on that corner of the truck. (Both sound reasonable) I'm not sure about the 36". Coming down from 3 feet and landing would be a very hard hit, but it's a good place to start. If you calculate the tubes out and they seem huge, you could maybe do some thinking about how high you'll actually drop in real life.
To calculate the velocity at impact from a certain height, you can use the potential and kinetic energy equations. I like to use the assumption that your potential energy at the top is equal to your kinetic energy at the bottom. so PE(at top) = KE (at bottom). PE = mgh KE = (mv^2)/2. Solving for velocity "v" gives you the equation:
v = sqrt(2gh)
Where:
v = velocity
g = gravitational constant
h = height off of ground
This will give you an impact velocity to use in the above iteration.
Remember to always work out the units along with the numbers to make sure you don't need any conversion factors in there. As a check, take the units that are associated with each number, put them through the equations just like you did with the numbers, cancel out as necessary, and you should come up with your desired units (i.e. "ft/s" for velocity).
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