Question 23.9: A piston rod of rectangular cross-section, with both ends hi...

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

P = 15 kN (fs) = 4 l = 150 mm
Step I Ratio (b/d) by using Euler’s equation
For the purpose of convenience, the planes are called A and B as shown in the figure.
In the plane-A, the ends are hinged.

n_{A}=1 \quad \text { and } \quad I_{A}=\left(\frac{d b^{3}}{12}\right)          (i).

In the plane-B, the ends are fixed.

n_{B}=4 \quad \text { and } \quad I_{B}=\left(\frac{b d^{3}}{12}\right)             (ii).

Using Euler’s equation and equating the buckling load in two planes

\frac{n_{A} \pi^{2} E A}{\left(\frac{l}{k_{A}}\right)^{2}}=\frac{n_{B} \pi^{2} E A}{\left(\frac{l}{k_{B}}\right)^{2}} .

or        n_{A} k_{A}^{2}=n_{B} k_{B}^{2} .

n_{A} I_{A}=n_{B} I_{B} .

\text { (1) }\left(\frac{d b^{3}}{12}\right)=(4)\left(\frac{b d^{3}}{12}\right) .

or        \left(\frac{b}{d}\right)=2 .

Step II Dimensions of cross-section
As a first trial, using Johnson’s equation in the plane A,

k_{A}^{2}=\frac{I_{A}}{A}=\left(\frac{d b^{3}}{12}\right)\left(\frac{1}{b d}\right)=\left(\frac{b^{2}}{12}\right) .

\left(\frac{l}{k_{A}}\right)^{2}=\frac{(150)^{2}}{\left(\frac{b^{2}}{12}\right)}=\frac{270000}{b^{2}} .

\left(15 \times 10^{3} \times 4\right) .

=(380)\left(0.5 b^{2}\right)\left[1-\frac{380}{4(1) \pi^{2}(207000)}\left(\frac{270000}{b^{2}}\right)\right] .

b = 18.12 mm.

\left(\frac{l}{k_{A}}\right)^{2}=\frac{270000}{b^{2}}=\frac{270000}{(18.12)^{2}}=28.68 .

Since the slenderness ratio is small, Johnson’s equation is justified.
∴ b = 18.12 mm or 20 mm
a = b/2 = 20/2 = 10 mm.