Question 21.P.8: In the experimental determination of the buckling loads for ...
In the experimental determination of the buckling loads for 12.5 mm diameter, mild steel,
pin-ended columns, two of the values obtained were:
(i) length 500 mm, load 9800 N,
(ii) length 200 mm, load 26 400 N.
(a) Determine whether either of these values conforms to the Euler theory for buckling load.
(b) Assuming that both values are in agreement with the Rankine formula, find the constants \sigma_{\mathrm{s}} and k. Take E = 200 000 N/mm².
a. The Euler buckling load for a pin-ended column is given by Eq. (21.5) P_{\mathrm{CR}}=\frac{\pi^{2} E I}{L^{2}}. The second moment of area of a circular section column is \pi D^{4}/64 which, in this case, is given by I = π × 12.5^{4}/64 = 1198.4 mm{4}
The area of cross section is
A=\pi \times \frac{12.5^{2}}{4}=122.72 \mathrm{~mm}^{2}
i.
P=\pi^{2} \times 200000 \times \frac{1198.4}{(500)^{2}}
i.e.
P = 9462.2 N
Therefore the test result conforms to the Euler theory.
ii.
P=\pi^{2} \times 200000 \times \frac{1198.4}{(200)^{2}}
i.e.
P = 59 137.5 N
Therefore the test result does not conform to the Euler theory.
b. From Eq. (21.27) P=\frac{\sigma_{S} A}{1+k\left(L_{e} / r\right)^{2}} and considering the first test result From Eq. (21.27) and considering the first test result
9800=\frac{122.72 \sigma_{\mathrm{s}}}{\left[1+\frac{k L^{2}}{\left(\frac{11984}{122 / 2}\right)}\right]}
which simplifies to
79.86+2.05 \times 10^{6} k=\sigma_{\mathrm{s}}
Similarly, from the second test result
215.12+0.88 \times 10^{6} k=\sigma_{\mathrm{s}}
Solving gives
k=1.16 \times 10^{-4}, \quad \sigma_{\mathrm{s}}=317.2 \mathrm{~N} / \mathrm{mm}^{2}