Question

a solid round bar with 80 mm diameter is used as a column. find the safe compressive load for the column by euler's formula (a) one end fixed and other end hinged, (b) both ends fixed. take e = 2x10; n/mm and factor of safety 3.​

Answer 1

Given:

A solid round bar with 80 mm diameter is used as a column. Take

E = 2x10 N/mm^2 and factor of safety 3.​

To Find:

Find the safe compressive load for the column by Euler's formula:

(a) one end fixed and other end hinged,

(b) both ends fixed.

Solution:

let, solid round bar be a unit column.

$l=1m$ =$1000 mm$ , $d=80mm$ , and factor of safety =3

$\[\begin{gathered} E = 2 \times 10N/m{m^2} \hfill \\ I = \frac{\pi }{{64}} \times {d^4} = \frac{\pi }{{64}} \times {80^4} = 2009600m{m^4} \hfill \\ \end{gathered} \]$

(a) one end fixed and other end hinged

$\[\begin{gathered} {P_{cr}} = \frac{{{\pi ^2}EI}}{{{l^2}}} = \frac{{{\pi ^2} \times 2 \times 10 \times 2009600}}{{{{1000}^2}}} = 396.277N \hfill \\ = 0.396kN \hfill \\ \end{gathered} \]$

so, safe compressive load for column=${P_{cr}$/factor of safety.

$=\dfrac{0.396}{3}=0.132kN$

(b) both ends fixed.

$\[\begin{gathered} {P_{cr}} = \frac{{4{\pi ^2}EI}}{{{l^2}}} = \frac{{4 \times {\pi ^2} \times 2 \times 10 \times 2009600}}{{{{1000}^2}}} = 1585.11N \hfill \\ = 1.585kN \hfill \\ \end{gathered} \]$

so, safe compressive load for column=${P_{cr}$/factor of safety.

$=\dfrac{1.585}{3}=0.52kN$