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

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

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

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,

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

Using a factor of safety of 2.5,

P_{c r}=2.5 P=2.5(110446.6)=276116.5 N .

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

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

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,

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

(276116.5) \text { or }=\frac{(1) \pi^{2}(207000)\left(\pi d^{2} / 4\right)}{\left(\frac{1000}{d / 4}\right)^{2}} .

∴        d = 40.73 mm
Step III Check for design

\left(\frac{l}{k}\right)=\frac{1000}{(40.73 / 4)}=98.21            (i).

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

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

\left(\frac{l}{k}\right)=87.8                 (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.