Question

The slenderness ratio of a 4 m column with fixed ends having a square cross-sectional area of side 40 mm is:

• A. 173
• B. 17.3
• C. 1.73
• D. 100

Answer 1

Answer:

A. 173

Given Length of the column (L) = 4m Side of a square = 40mm = 0.04m both ends are fixed, so, $L_{e} =L/2=2m$ , And $I_{min} = \frac{a^ 4}{12} = \frac{(0.04)^ 4}{12} =2.13\times 10^ {-7} m^ 4$ minimum radius of gyration, $K_{min} = \sqrt{\frac{I_{min}}{ A}} \Rightarrow K_{min} = \sqrt{\frac{ 2.13\times 10^ {-7}}{1.6\times 10^ {-3}}}$ =$0.0115 m$ And slenderness ratio, $S=\frac{L_e}{K_{min}} =\frac{ 2}{ 0.0115} =173.9$