Let us consider the following problean. A rod having roundings of radius $R$ at its ends (Fig 114) is compressed withoat friction between two solid slabs. It is necessary to determine the critical load.
Taking the line of compressive forces action as the $x$ axis and designating the lateral deflection of the beam axis as $y$ we obtain as usual:
$$
\begin{aligned}
& E J y^{\prime \prime}+P y=0, \quad y^{\prime \prime}+\alpha^2 y=0 \\
& y=A \sin \alpha x+B \cos \alpha x \quad\left(a^2=\frac{P}{E J}\right)
\end{aligned}
$$
The deflection $y$ at the rod ends is proportional to the angle of rotation $y^{\prime}$, i.e. $y=-R y^{\prime}$ under $x=0$. As the negative deflection $y$ corresponds to the positive angle of rotation $y^{\prime}$ the sign before $R y^{\prime}$ is negative. Thus we get
$$
B=-\alpha R A, \quad y=A(\sin \alpha x-\alpha R \cos \alpha x)
$$
In addition, $y=+R y^{\prime}$ under $x=l$; here $y$ is posit ive for positive $y^{\prime}$. Hence
$$
A(\sin \alpha-a R \cos \alpha l)=A\left(\alpha R \cos \alpha l+\alpha^2 R^2 \sin \alpha l\right)
$$
As $A \neq 0$ we obtain
$$
\tan \alpha l=\frac{2 a \frac{R}{l}}{1-(\alpha l)^2 \frac{R^2}{\ell^2}}
$$
Solving this transcendental equation we determine the least non-zero value of $\alpha l$ dependent on the relation $R / L$ and then find the critical load.
As $a^2=P /(E J)$ the critical load is equal to
$$
P_{c r}=\frac{(a l)^2 E J}{l^2}
$$
The first (the least) root of equation (1) must be substituted here instead of al .
Now let us find the least root od of equation (1) as a function of $R /$ l. The most convenjent way to ascertain this relationship is specifying values of $a l$ and then determining $R / l$ from equation (1).
The calculation results are shown as a curve (Fig. 115). We reveal that $\alpha t=\pi$ in case of $R / l=0$, and therefore according to (2) :
$$
P_{c r}=\frac{\pi^2 E J}{l^2}
$$
The critical bad is equal to the ordinary Euler force as one would expect.
As $R / l$ increases the value of $a l$ grows too. Hence the critical load also increases. It also seems to be sufficiently obvious. But when $R / l=0.5$ the critical load vanishes (falls to zero value) as it follows from the curve and then while further ascending of radius $R$ the critical load begins to grow again tending under $R / t=\infty$ to a limit equal once more to Euler's force.
Interpret the result obtained above.