The maximum stress intensity at the base of square column of area

The maximum stress intensity at the base of square column of area ‘A’ and side ‘b’ subjected to load W’ at an eccentricity ‘e’ equals to

A. \(\frac{W}{A}\left( {1 + \frac{{2e}}{b}} \right)\)

B. \(\frac{W}{A}\left( {1 - \frac{{4e}}{b}} \right)\)

C. \(\frac{W}{A}\left( {1 + \frac{{6e}}{b}} \right)\)

D. \(\frac{W}{A}\left( {1 + \frac{{8e}}{b}} \right)\)

Please scroll down to see the correct answer and solution guide.

Right Answer is: C

SOLUTION

Concept:

Bending equation

\(\frac{M}{I} = \frac{\sigma }{y} = \frac{E}{R}\)

M = bending moment, I = moment of inertia of the area of cross-section, σ = bending stress, y = distance of extreme fibre from the neutral axis, E = young’s modulus for the material of the beam, R = radius of curvature

Calculation:

Given:

Side of square column = b, Load at an eccentricity (e) = W, Area = b²

The column will be under compression and bending.

Point B is heavily stressed.

Now, stresses at B,

We know,

\(\frac{M}{I} = \frac{\sigma }{y}\)

\(I = \frac{{{b^4}}}{{12}},\;y = \frac{b}{2}\)

The maximum stress intensity at the base of the square column,

σ = σ_compr. + σ_bending

\(\sigma = \frac{W}{A} + \frac{{W \times e \times \frac{b}{2}}}{{\frac{{{b^4}}}{{12}}}} \Rightarrow \frac{W}{A} + \frac{{W \times e}}{{{b^2} \times \left( {\frac{b}{6}} \right)}} = \frac{W}{A} + \frac{{W \times e \times 6}}{{A \times b}}\)

\( \sigma= \;\frac{W}{A}\left( {1 + \frac{{6e}}{b}} \right)\)