Elisha M.

asked • 04/27/17

Strength of material

The piston rod of a double-acting hydraulic cylinder is 20 cm diameter and 4 m long.
The piston has a diameter of 40 cm, and is subjected to 10 MN/m2 water pressure on one
side and 3 MN/m2 on the other. On the return stroke these pressures are interchanged.
Estimate the maximum stress occurring in the piston-rod, and the change of length of the
rod between two strokes, allowing for the area of piston-rod on one side of the piston.
Take E = 200 GN/m*.

1 Expert Answer

By:

Roger N. answered • 04/27/17

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. BE in Civil Engineering . Senior Structural/Civil Engineer

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The total force in the piston,  Fp = Ap * σpmax 
 
Ap = Area of the piston = ¶D2/4 =  ¶(0.4m)2/4 = 0.126 m2
 
σpmax = Maximum stress in the piston = 10 MN/m2
 
σpmin = Minimum stress in the piston = 3 MN/m2
 
The maximum force in the piston, Fp = 0.126 m2(10x106 N/m2) = 1.26x106 N = 1.26 MN
 
The maximum force in the piston rod is the same as the force exerted by the piston
 
Area of the rod = ¶D2/4 = ¶(0.2m)2/4 = 0.031 m2
 
The maximum stress in the piston rod = σrod = Fp / Arod = 1.26 MN/ 0.031 m2 = 1.26 x 106 N / 0.031 m2
 
= 40.65 x 106 N /m2
 
The change in length of the rod between the two strokes is the difference in elongation caused by the maximum and the minimum force
 
 
The maximum elongation is due to the maximum force = Δmax = FL/AE =( 1.26 x 106 N x 4m)/(0.031m2x 200x109 N/m2)
 
Δmax = 
 
 
 

Δmax = 0.000813 m = 0.081 cm The min force in the rod = Frod = 0.126 m2(3x106 N/m2) = 0.38x106 N = 0.38 MN The min elongation in the rod is due to the min force = Δmin = FL/AE =( 0.38 x 106 N x 4 m) / (0.031 m2x 200x109 N/m2) = 0.000245 m = 0.0245 cm The difference in elongation =Δmax - Δmin = 0.0813-0.0245= 0.056 cm

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Elisha M.

But an answer is 31 MN/m2
 written in cambridge book
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04/27/17

Elisha M.

does it,have anything to do with this question ?
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04/27/17

Roger N.

tutor
Elisha I have looked this up. If we combine the water pressure in both pistons, we will get 13 MN , and the piston rod stress will be 52.6 MN greater than 40.65 MN/m2, if we subtract the pressures we will get 7 MN and the piston rod stress will be 28.5 MN/m2. Therefore  using 31 MN/m2 stress in the rod l worked the problem backward to calculate the piston pressure to be 7.63 MN, which is closer to the difference in piston pressures of 7 MN than the sum of 13 MN. I believe that the analysis is correct but the result is off. More understanding of how a double acting hydraulic cylinder is needed to get the correct water pressure acting on the piston rod...
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04/27/17

Roger N.

tutor
I have looked up the mechanism of a double acting hydraulic cylinder
It has of what is called extend and retract strokes. the extend stroke extends the piston rod and the retract stroke pulls it back. the extend stroke is usually much larger than the retract stroke and I must be the 10MN/m2 pressure in the problem, and the retract stroke is the smaller of the pressures or 3MN/m2. The extend and retract strokes do not act simultaneously, therefore I believe that the max pressure that will cause the maximum stress in the piston rod is that of the extend pressure = 10MN/m2 . I hope this helps...
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04/27/17

Elisha M.

Thanks i think that helps 
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04/27/17

Elisha M.

But piston area equals to the total piston area - rod piston area
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04/27/17

Roger N.

tutor
The piston area is used in the problem to calculate the force in the piston. The force in the piston rod will be the same and will act on the piston rod area to generate the axial stress in the piston rod.. 
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04/27/17

Roger N.

tutor
The piston and the piston rod area are different and are not the same. Normally the piston rod are has a smaller diameter than that of the piston itself and therefore a smaller cross sectional area. In this problem the piston rod diameter is 20 cm, and the piston area is 40 cm..
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04/27/17

Roger N.

tutor
Sorry I meant piston diameter of 40 cm..
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04/27/17

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