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Actually disregard that. I end up with zero=zero. That is not the way to go.

Let's call the bottom two supports point A and B. Do sum of moments about A. Now do sum of moments about B. And finally sum of forces in Y equal zero. You now have 3 equations with 3 unknowns.

I don't understand.

The OP is simply asking if you could have the bullet come out of the rifle at a higher velocity if you didn't waste some of its energy in making it rotate inside the barrel.

rconner, you are correct. I am often astounded at the answers here. People don't read the question.

Racook, I don't understand what you mean. I don't want to seem mean or anything..... I just don't get it. I'd like further explanation of your point. Thanks.

medeek, are you required to double check the values provided by the truss manufacturer? I buy Crosby shackles but I don't feel compelled to check that they are correct in their provided 4 or 5:1 safety factor. Just wondering if this is more of personal interest than a requirement of your...

I need to say that moment distribution by hand is more than a 10-minute exercise. Closer to an hour or more to double check the tedius algebra.

I agree with racookpe1978, partially. Use the correct reason: that you find it too complicated to calculate. Not that it will be overstressed no matter what.

Maturin, do you mean that you will refuse to do calculations on this arrangement? Just curious.

It would be the same. Force on the wall due to water is calculated using the pressure.

Another simple way to look at it is that on the discharge, the friction force is pushing on the column of water towards the pump (therefore increasing pressure at the pump). And on the suction side, the friction force pushes on the column of water away from the pump (decreases the pressure at...

Another note, when transferring your bending moment to your weld I would suggest applying an appropriate stress concentration factor due to the abrupt change in bending cross section. I'd also apply a more conservative FOS than usual to this kind of problem.

The most complicated aspect of this arrangement for me would be to figure out how the load is transferred from below to the weld. As someone suggested in the thread referenced above, maybe assume the effective weld length coming up at a 30 or 45 degree from the load? After the above is assumed...

I worked out both....the general solution taking into account the rotation then I presented the solution of the specific case where the vertical leg lenght is small/negligeable (as a check to the general solution).

Rb1957, my solution includes part of that rotation due to the L2 beam bending. I forgot the bending due to the horizontal load though.

In the 2-beam cantilever assumption with pinned ends at the vessel, I believe the reaction at the top beam should be W L2^3/(L1^3 +L2^3) = 2472 Lbs (see my posted solution above).

Zekeman, I think it's a good assumption to assume pinned at the vessel since this would yield a conservative solution at the supports 'a' and 'b'. But if pdiculous963 were to insist that it isn't pinned at the vessel then the problem can still be solved with the same approach. Concerning the...

Here's my solution inluding the bottom vertical beam. My goal was to find the sharing of the load between top beam and bottom beam. I only showed loads on my links that are necessary for this task so take it easy when you see Fx not shown in there. http://imgur.com/a/JDGtC

I understand your explanation. My resoning is that we already have 10mm of aluminum, which is heated from the underside. Therefore the temperature distribution is already adjusted somewhat on the top surface. Is 10mm enough? I'm not sure but I have a feeling that it is. I'm interested to...