Question 23.8: It is required to design the screw of a screw-jack by buckli...

\text { Given } S_{y t}=380 N / mm ^{2} \quad E=207000 N / mm ^{2} .

l = 500 mm P = 20 kN (fs) = 2.5
Step I Diameter of screw using Euler’s equation
The end fixity coefficient is 0.25 when one end is fixed and the other free. As a fi rst trial, using Euler’s equation,

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

(20000)(2.5)=\frac{(0.25) \pi^{2}(207000)\left(\pi d^{2} / 4\right)}{\left[\frac{500}{(d / 4)}\right]^{2}} .

or d = 26.57 mm

Step II Check for design

k = d/4 = 6.64 mm

\left(\frac{l}{k}\right)=\frac{500}{6.64}=75.3 .        (i)
The border 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{380}{2}=\frac{(0.25) \pi^{2}(207000)}{(l / k)^{2}} .

\therefore \quad\left(\frac{l}{k}\right)=51.85             (ii).

From (i) and (ii), it is concluded that the screw is a long column and Euler’s equation considered in the first trial is justified. Therefore,
d = 26.57 mm or 28 mm.