A mild steel column of height 5 m is hollow and circular in cross section with an external diameter of 400 mm and a wall thickness of 20 mm. If the column carries a compressive load of 2500 kN calculate the direct stress in the column and also the shortening of the column. Take E = 200 000 N/mm^2

Question

A mild steel column of height 5 m is hollow and circular in cross section with an external diameter of 400 mm and a wall thickness of 20 mm. If the column carries a compressive load of 2500 kN calculate the direct stress in the column and also the shortening of the column. Take E = 200 000 N/mm² . If the ends of the column are then fixed so that no further axial movement is possible calculate the total direct stress in the column when it is subjected to a temperature rise of 20 K. The coefficient of linear expansion of the material of the column is 0.00005/K and with the usual notation [latex] L=L_{0}(1+\alpha T) [/latex].

Accepted Answer

The cross sectional area, A, of the column is given by

[latex] A=\pi\left(400^{2}-360^{2} ight) / 4=23876.1 mm ^{2} [/latex]

Then, from Eq. (7.1)

[latex] \sigma=\frac{P}{A} [/latex] (7.1)

[latex] \sigma=2500 imes 10^{3} / 23876.1=104.7 N / mm ^{2} ext { (compression) } [/latex]

The direct strain in the column is obtained from either of Eqs (7.7) or (7.8) and is

[latex] \sigma=E \varepsilon[/latex] (7.7)

[latex] \frac{\sigma}{\varepsilon}=E [/latex] (7.8)

[latex] \varepsilon=\sigma / E=104.7 / 200000=0.00052 [/latex]

so that the shortening, δ, of the column is, from Eq. (7.4), given by

[latex]\varepsilon =\frac{\delta }{L_0} [/latex] (7.4)

[latex] \delta=0.00052 imes 5 imes 10^{3}=2.62 mm [/latex]

Due to the temperature rise the column would, if not prevented, increase in height with no corresponding change in stress. However, this change in height is prevented thereby causing, in effect, a strain and an accompanying stress. Now [latex] L=L_{0}(1+\alpha T) [/latex] so that the change in height due to a temperature increase would be [latex] L-L_{0} [/latex] which is equal to [latex] L_{0} \alpha T [/latex]. The strain corresponding to the suppression of this change in height is then [latex] \left(L-L_{0} ight) / L_{0}=\alpha T [/latex]. The accompanying direct stress is then, from Eq. (7.7), given by

[latex] \sigma=0.00005 imes 20 imes 200000=200 N / mm ^{2} [/latex]

which is compressive since the increase in height is prevented. The total stress in the column produced by the load and the temperature rise is then

[latex] \sigma( ext { total })=200+104.7=304.7 N / mm ^{2} [/latex]

Question 7.7: A mild steel column of height 5 m is hollow and circular in ...

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