A piston rod of rectangular cross-section, with both ends hinged, is shown in Fig. 23.14. It is made of steel 40C8 $\left(S_{y t}=380 N / mm ^{2}\right. \text { and } \left.E=207000 N / mm ^{2}\right)$ and subjected to an axial compressive force of 15 kN. Determine the ratio of (b/d) for equal buckling strength in either plane.
Also determine the dimensions of cross-section, if the factor of safety is 4.

Question

A piston rod of rectangular cross-section, with both ends hinged, is shown in Fig. 23.14. It is made of steel 40C8 [latex] \left(S_{y t}=380 N / mm ^{2} ight. ext { and } \left.E=207000 N / mm ^{2} ight) [/latex] and subjected to an axial compressive force of 15 kN. Determine the ratio of (b/d) for equal buckling strength in either plane.
Also determine the dimensions of cross-section, if the factor of safety is 4.

Accepted Answer

[latex] ext { Given } S_{y t}=380 N / mm ^{2} \quad E=207000 N / mm ^{2} [/latex].

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.

[latex] n_{A}=1 \quad ext { and } \quad I_{A}=\left(\frac{d b^{3}}{12} ight) [/latex] (i).

In the plane-B, the ends are fixed.

[latex] n_{B}=4 \quad ext { and } \quad I_{B}=\left(\frac{b d^{3}}{12} ight) [/latex] (ii).

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

[latex] \frac{n_{A} \pi^{2} E A}{\left(\frac{l}{k_{A}} ight)^{2}}=\frac{n_{B} \pi^{2} E A}{\left(\frac{l}{k_{B}} ight)^{2}} [/latex].

or [latex] n_{A} k_{A}^{2}=n_{B} k_{B}^{2} [/latex].

[latex] ext { Substituting } k^{2}=I / A ext {, } [/latex]

[latex] n_{A} I_{A}=n_{B} I_{B} [/latex].

[latex] ext { (1) }\left(\frac{d b^{3}}{12} ight)=(4)\left(\frac{b d^{3}}{12} ight) [/latex].

or [latex] \left(\frac{b}{d} ight)=2 [/latex].

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

[latex] k_{A}^{2}=\frac{I_{A}}{A}=\left(\frac{d b^{3}}{12} ight)\left(\frac{1}{b d} ight)=\left(\frac{b^{2}}{12} ight) [/latex].

[latex] \left(\frac{l}{k_{A}} ight)^{2}=\frac{(150)^{2}}{\left(\frac{b^{2}}{12} ight)}=\frac{270000}{b^{2}} [/latex].

[latex] P_{c r}=S_{y t} A\left[1-\frac{S_{y t}}{4 n \pi^{2} E}\left(\frac{l}{k} ight)^{2} ight] [/latex]

[latex] \left(15 imes 10^{3} imes 4 ight) [/latex].

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

b = 18.12 mm.

[latex] \left(\frac{l}{k_{A}} ight)^{2}=\frac{270000}{b^{2}}=\frac{270000}{(18.12)^{2}}=28.68 [/latex].

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.

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

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