Slenderness ratio of a 5 m long column hinged at both ends and having a circular cross section with diameter 16 cm is ______.

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SSC JE CE Previous Year Paper 17 (Held On: 3rd March 2017 Evening)
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  1. 31.25
  2. 62.5
  3. 100
  4. 125

Answer (Detailed Solution Below)

Option 4 : 125
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Concept:

Slenderness ratio is defined as the ratio of effective length of column to the least radius of gyration.

\({\rm{\lambda }} = \frac{{{{\rm{L}}_{{\rm{eff}}}}}}{{{{\rm{r}}_{{\rm{min}}}}}} \)

where,

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{\rm{I}}}{{\rm{A}}}} \)

\({\rm{I}} = \frac{{{\rm{\pi }} \times {{\rm{D}}^4}}}{{64}} \)

\({\rm{A}} = \frac{{\rm{\pi }}}{4} \times {{\rm{D}}^2} \)

Where,

Leff = Effective length of column

rmin = Minimum radius of gyration

I = Moment of inertia of the centroidal axis.

A = Area of the cross section of the column

Calculation:

For column hinged at both ends, Leff = Length of the column itself = 5 m = 500 cm

\({{\rm{r}}_{{\rm{min}}}} = \sqrt {\frac{{{{\rm{D}}^2}}}{{16}}} = \frac{{\rm{D}}}{4} \)

D = 16 cm

rmin = 4 cm = 0.04 m

\({\rm{\lambda }} = \frac{{500}}{4} = 125 \)
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