I am not a structural engineer so there may be such a thing as an "H" beam but I am not acquainted with it. However, I suspect your "H" beam is masquerading as none other than my "I" beam. Rotate your "H" 90 degrees until one of the flanges is on top and the other on the bottom. Voila--the "H" becomes an "I".
A 8x58 I beam has flange width of 8.22 inches versus 8.75 height with flange thickness of .81" and web thickness of .51". I am pretty sure this is the beam you have. The key characteristic that determines stress and deflection is something called "moment of inertia". For the 8x58, in the strongest direction, it is 227. In the perpendicular direction, it is only 75, so it makes a three-fold difference in strength by choosing how you load the beam. In other words, using it as an H would be 1/3rd as strong as using it as an I. Having said that, it still may be PLENTY strong as you plan to use it and there may be advantages to doing it that way.
I did NOT take into account the end plates--I am not that smart. I don't think they'd make much difference anyway because their function is really just something to get the load into the beam.
It is hard to find sites or even examples in books that match exactly what you're doing. Essentially, we are looking at the stresses and deformations that occur in the throat section of a "C" clamp. The beam (the "throat" of the C-clamp) has to hold the 54k tensile stress plus the bending stress caused by the fact that the 54k is not acting through the center of the beam.
Tensile stress from the 54k is just that force divided by the cross sectional area of the beam (16.953 sq inches) So, tensile stress from the 54k load is:
54,000lb/(16.953) = 3185 psi and it acts uniformly on both flanges and the web.
Max stress from bending moment is calculated using the formula
S = (M*c)/I
where
M= bending moment (54,000 x 7" (I assumed a value of 7"))
c = max distance from neutral axis (8.75/2)
I = moment of inertia in loaded axis (227)
For the 8x58,
S = (54000*7*4.375)/227
S = 7317 psi
Total stress in the tension side of the beam adds these two stresses together:
S = 3185 + 7317 = 10,502 psi
The stress in the opposite side of the beam will be less because the bending stress is compressive and the tensile stress subtracts from it:
S = 7317-3185 =4132 psi in compression
Wait a minute---that doesn't agree with what I posted earlier. I either have a faulty memory or made a mistake. I like these values better--they are even less, which means the deflection will be less, too.
Deflection is a slightly tougher cat to skin. Even harder to explain without the benefit of a chalkboard. So without further adieu, I will just throw out the following equation as the one that I used in this case. It is NOT the right equation, but the best I could find:
v = (M * x^2)/(2xExI)
where
v= deflection in inches
x= distance along beam where deflection is measured
M= moment in inch-pounds
E= modulus of elasticity for steel (30,000,000 psi)
I= moment of inertia for this beam (227 inches^4)
Plugging in the numbers gives a max deflection of .035 for middle of the 6 ft long I beam. If you want the values for using the beam sideways, then just substitute 75 for 227 in the equations.
Bear in mind these deflection numbers are probably garbage because I'm not using the exact equation. If you search on "beam equations" or similar you will probably find all kinds of sites (usually schools) that have them. You will need to know the values of E, I, and dimensions I've given above for whatever beam you're interested in. They're called 'W' I-beams--those values ought to be out there, too.
