A trunnion mounted hydraulic cylinder is shown in Fig. 23.12. The internal diameter of the cylinder is 75 mm and the maximum operating pressure in the cylinder is 25 N/mm^2. The piston rod is made of steel 4OCr1 (Syt = 530 N/mm^2 and E = 207 000 N/mm^2). For buckling considerations, the effective

Question

A trunnion mounted hydraulic cylinder is shown in Fig. 23.12. The internal diameter of the cylinder is 75 mm and the maximum operating pressure in the cylinder is 25 N/mm². The piston rod is made of steel 4OCr1 [latex] \left(S_{y t}=530 N / mm ^{2} ext { and } E= ight. \left.207000 N / mm ^{2} ight) [/latex]. For buckling considerations, the effective length of the piston rod is considered as the distance between the trunnion and the clevis-mount, when the piston rod is extended to its full working stroke and this distance is 1000 mm. Determine the diameter of the piston rod, if the factor of safety is 2.5.

Accepted Answer

Given For piston rod, l = 1000 mm (fs) = 2.5

[latex] S_{y t}=530 N / mm ^{2} \quad E=207000 N / mm ^{2} [/latex].

For cylinder, D = 75 mm p = 25 N/mm².

Step I Estimation of critical load
Although one end of the piston rod is fixed in the piston, considering the complete assembly between trunnion and clevis-mount, the end fixity coefficient is taken as one (both ends hinged). The maximum force on the piston rod is given by,

[latex] P=\frac{\pi}{4} D^{2} p=\frac{\pi}{4}(75)^{2}(25)=110446.6 N [/latex].

Using a factor of safety of 2.5,

[latex] P_{c r}=2.5 P=2.5(110446.6)=276116.5 N [/latex].

Step II Diameter of piston rod
For circular cross-section,

[latex] k=\sqrt{\frac{I}{A}}=\sqrt{\left[\frac{\left(\pi d^{4} ight) / 64}{\left(\pi d^{2} / 4 ight)} ight]}=\left(\frac{d}{4} ight) mm [/latex].

At this stage, it is not clear whether one should use Euler’s or Johnson’s equation. Using Euler’s equation as a first trial,

[latex] P_{c r}=\frac{n \pi^{2} E A}{(l / k)^{2}} [/latex].

[latex] (276116.5) ext { or }=\frac{(1) \pi^{2}(207000)\left(\pi d^{2} / 4 ight)}{\left(\frac{1000}{d / 4} ight)^{2}} [/latex].

∴ d = 40.73 mm
Step III Check for design

[latex] \left(\frac{l}{k} ight)=\frac{1000}{(40.73 / 4)}=98.21 [/latex] (i).

The boundary line between Euler’s and Johnson’s equations is given by

[latex] \frac{S_{y t}}{2}=\frac{n \pi^{2} E}{(l / k)^{2}} \quad ext { or } \quad \frac{530}{2}=\frac{(1) \pi^{2}(207000)}{(l / k)^{2}} [/latex].

[latex] \left(\frac{l}{k} ight)=87.8 [/latex] (ii).

In this example, the slenderness ratio (98.21) is greater than the boundary value of (87.8). Therefore the assembly is treated as a long column and Euler’s equation used in the first trial is justified.

d = 40.73 mm or 42 mm.

Question 23.6: A trunnion mounted hydraulic cylinder is shown in Fig. 23.12...

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