Simple Question, Bolt Axial Load

I trying to get an approximate value of Torque to apply to 2 bolts in a clamp to achieve 5 and 10 metric tons of clamping force and I seem to be getting an answer that is not realistic.

Here is a quick diagram. (Tube with 2 clamps and 2 bolts, with nuts)
[IMG]http://www.engineersedge.com/engineering-forum/attachment.php?attachmentid=1157&stc=1[/IMG]
Bolts are 3/8"-16 Hex Head. 304SS
Tensile Strength ~ 75000psi
Tensile Stress Area = 0.0775in^2

Therefore: Max Tensile Load (P) = 0.0775 * 75000 = 5812.5lbs.
Proof Load = 90%(P) = 5231lbs.
Typical Clamp Load = 75% Proof Load = 3923.lbs.

Torque = K x d x F
K = Nut Factor = 0.2 (Stainless steel bolt and nut with lubrication?)
d = Nominal Dia. = 3/8"
F = 3923lbs

T = 24 ft.lbs.

This torque Seems to meet recommendations I see in my industry for this type and size of fastener. But my question is, because there are 2 bolts creating this clamping force (3923lbs) would I have to apply half the torque? I.e 12ftlbs to each bolt. If torque was only applied to one bolt in this situation no axial load would be applied to the tube. Thank for all the help you can give me.

Assuming your clamp is not otherwise restrained, and ignoring contact friction with the pipe, the torque loading of one bolt will automatically be transferred to the other bolt (think of your assembly as a beam with the pipe as a center pivot). If you tighten one bolt to a given torque when you try to tighten the second bolt it will immeadiately exhibit a torque resistence equal to that of the first bolt. By a similar beam analogy with one bolt as an end pivot and the pipe in the center, the loading on the pipe will be equal to twice the torque loading of the bolts.
However, if you desire to have equal thread makeup on both bolts then you need to tighten the bolts in an alternating fashion with successively increasing torques until the desired torque is achieved.

Thanks you for the reply JAlberts. I have 1 question if I could just get a little more clarification.

If I snug up both bolts so there is no movement. Then torque the bolt on the right to 12ft.lbs, both bolts will have the same axial load (ignoring friction) and the total clamping force will be equal to that created by one bolt tightened at 24ft.lb.s? I understand this is greatly simplified.

What you say would only be true if that one bolt were directly centered over the pipe.

In an attempt to resolve your confusion let try another analogy. For a minute let's forget about the bolts. Now if you place, for example, a 50 lb weight on one end of the clamp, then (assuming the pipe is exactly centered in the clamp) how much weight do you have to place on the other end of the clamp to keep it level? The answer, I sure you immediately realize, is 50 lbs, right?
As a result, the total loading on the pipe is: 50+50+100 lbs correct?

This is what I was describing when I stated that any torque load on one bolt would immeadiately result in an equal torque load on the other bolt, ignoring friction, etc.

I hope this resolves your confusion; if not, let know. It is actually a little bit confusing for anyone. I think having to realize that there are two loadings, one on each end of a clamp, is what makes it a bit difficult.

Sorry I should have specified that the one bolt torqued to 24ft.lbs in my last message would have to be centered in the middle of the pipe. So if that was the case my last example would be shown correct? You have helped a lot and and I think I just needed to be walked through it.

I don't understand the diagram...