For problems 2 and 3, see figure 5.27 in order to account for various end fixity
ID: 1843821 • Letter: F
Question
For problems 2 and 3, see figure 5.27 in order to account for various end fixity conditions. The figure shows a hydraulic cylinder having a clevis mount. Because of the bearing and seal and because of the stiffness of the cylinder itself, the piston end of the rod may be regarded as fixed. The outboard end of the rod may be either a free end or rounded and guided, depending upon the application. Of course, the column length is taken as the distance/when fully extended. Consider a typical application with a hydraulic pressure of 3500 poi, a 3 -in cylinder bore, a factor of safety of 3, and a medium-carbon-steel-rod material having a yield strength of 70 kpsi. Use the recommended value for end-condition constant, based on one end fixed and the other end pinned and guided, to find a safe diameter d for piston-rod lengths of 96, 48, and 24 in. An aluminum column of length L and rectangular cross section has a fixed end B and supports a centric load at A. Two smooth and rounded fixed plates restrain end A from moving in one of the vertical planes of symmetry of the column, but allow it to move in the other plane. Determine the ratio a/b of the two sides of the cross section corresponding to the most efficient design against buckling.Explanation / Answer
Pb2:
As suggested in the problem, piston end of the rod for practical purposes may be considered fixed and the other end as pinned. Thus we have critical load Pcr= EI/(KL)2. Take E as 29.5 x106 psi
Where K is effective length factor. For one end fixed and the other pinned column, K = 1/2=0.707. thus we have Pcr= EI/(0.707L)2 = EI/2L2
We know Load Pcr= Cylinder area x 3500 psi = x32 x 3500 = 24740 lbs.
We know L= 96, 48, 24 inches.
In the above equation I is section Moments of Inertia of Rod = d4/64
For L= 96, we have ( x E x d4/64)/(2x962) = Pcr= 24740 lbs
Calculate d as 3.16 inches. The rod diameter is larger than the cylinder diameter. Impossible. Thus we cannot have rod of lengths 96” for this cylinder!
For L= 48, we have ( x E x d4/64)/(2x482) = Pcr= 24740 lbs
Calculate d as 2.24”.
For L= 24, we have ( x E x d4/64)/(2x242) = Pcr= 24740 lbs
Calculate d as 1.58”.
Pb3:
The column can buckle either in xy pane or xz plane.
The Column buckling in xz plane is considered end B fixed and end A as free. Thus Pcr ( xz plane)= EI/(KL)2 with K= 2.
I for this case is ab3/12
Pcr ( xz plane)= E ab3/(12 x (2L)2)
The Column buckling in xy plane is considered end B fixed and end A as guided. Thus Pcr (xy plane) = EI/(KL)2 with K= 1/2=0.707 and I for this case is ba3/12
Pcr (xy plane) = E ba3/(12 x 2 L2)
Most efficient design will have Pcr ( xz plane)= Pcr (xy plane)
E ab3/(12 x (2L)2) = E ba3/(12 x 2 L2)
Simplifying we get ab3/4L2 = ba3/2L2
Simplifying we get b2 = 2a2
Thus most efficient design is when b = 2a
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