Question 23.7: A column of hollow rectangular cross-section and made of ste...

A column of hollow rectangular cross-section and made of steel plates is shown in Fig. 23.13. The thickness of the section is 2.5 mm throughout. The plate material is steel 30C8 \left(S_{y t}=\right. \left.400 N / mm ^{2} \text { and } E=207000 N / mm ^{2}\right) . The end fixity coefficients can be taken as 1.5 and 1 for bending about long and short axes respectively. The effective length of the column is 1 m. Determine the load capacity of the column from buckling consideration.

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

\text { Given } S_{y t}=400 N / mm ^{2} \quad E=207000 N / mm ^{2} .

l = 1 m n = 1.5 (for bending about long axis)
n = 1 (for bending about short axis)
Step I Slenderness ratio about XX and YY axis
A = 30 × 20 – 25 × 15 = 225 mm².

I_{x x}=\frac{1}{12}\left[30(20)^{3}-25(15)^{3}\right]=12968.75 mm ^{4} .

I_{y y}=\frac{1}{12}\left[20(30)^{3}-15(25)^{3}\right]=25468.75 mm ^{4} .

k_{x x}=\sqrt{\left(\frac{I_{x x}}{A}\right)}=\sqrt{\frac{12968.75}{225}}=7.59 mm .

k_{y y}=\sqrt{\left(\frac{I_{y y}}{A}\right)}=\sqrt{\frac{25468.75}{225}}=10.64 mm .

\left(\frac{l}{k_{x x}}\right)=\frac{1000}{7.59}=131.75            (i).

\left(\frac{l}{k_{y y}}\right)=\frac{1000}{10.64}=93.98          (ii).

Step II Critical load along XX-axis

\frac{S_{y t}}{2}=\frac{n \pi^{2} E}{(l / k)^{2}} \text { or } \frac{400}{2}=\frac{(1.5) \pi^{2}(20701}{\left(\frac{l}{k_{x x}}\right)^{2}} .

\therefore \quad\left(\frac{l}{k_{x x}}\right)=123.78                   (iii).

From (i) and (iii), the column is treated as a long column. Using Euler’s equation,

P_{c r}=\frac{n \pi^{2} E A}{\left(l / k_{x x}\right)^{2}}=\frac{(1.5) \pi^{2}(207000)(225)}{(131.75)^{2}} .

=39723.05 N               (a)
Step III Critical load along YY-axis

\frac{S_{y t}}{2}=\frac{n \pi^{2} E}{\left(l / k_{y y}\right)^{2}} \quad \text { or } \quad \frac{400}{2}=\frac{(1) \pi^{2}(207000)}{\left(\frac{l}{k_{y y}}\right)^{2}} .

\therefore \quad\left(\frac{l}{k_{y y}}\right)=101.07               ( iv)
From (ii) and (iv), the column is treated as a short column. Using Johnson’s equation,

P_{c r}=S_{y t} A\left[1-\frac{S_{y t}}{4 n \pi^{2} E}\left(\frac{l}{k_{y y}}\right)^{2}\right] .

=(400)(225)\left[1-\frac{400}{4(1) \pi^{2}(207000)}(93.98)^{2}\right] .

or                P_{c r}=51091.61 N               (b)
Step IV Load carrying capacity
From (a) and (b), the load carrying capacity of the column is 39 723.05 N.

Related Answered Questions