Dman1801: For sizing this particular round steel tube, I guess you could use a yield factor of safety of FSy = 1.50. That seems to be what section 25.41(a) at
this[/color] link implies. (I am referring to structural steel, not forged steel parts.) However, check the yield factor of safety in the crane design code for your jurisdiction, and adjust the FSy value accordingly.
I am assuming your round steel tube is ASTM A500M, grade B, steel, which means tensile yield strength is Sty = 290 MPa, in which case you would need to use a tube thinness (n1) not exceeding 48, where n1 = d2/t1, d2 = tube outside diameter, and t1 = tube wall thickness.
It sounds like your given applied load is P1 = (4540 kg)*g = 44 537 N. Your boom fully-extended length is L1 = 8534 mm.
You must install a braking mechanism that limits braking deceleration to a maximum of 1.35 g, for both vertical and horizontal deceleration. You must install motor limits to limit load acceleration to 0.35 g, for both vertical and horizontal acceleration. Thus, when you combine vertical and horizontal acceleration or deceleration, I think a combined dynamic amplification factor of daf = 1.50 is allowed.
Therefore, using all of the above data and assumptions, and solving for d2, the necessary tube outside diameter at the base of the first tube segment would be, d2 ≥ 4.022*[FSy*daf*L1*P1/(290 MPa)]^(1/3) = 4.022*(1.50*1.50*8534*44 537/290)^(1/3) = 576.8 mm. And the tube wall thickness would be t1 ≥ d2/n1 = 576.8/48 = 12.02 mm.
Note that the above d2 equation is not a general formula, and therefore does not apply to any other scenario.