Question:

A 10 kN axial load is applied eccentrically on a rod of square cross-section (1 cm $\times$ 1 cm) as shown in the figure.
The strains measured by the two strain gages attached to the top and bottom surfaces at a distance of 0.5 m from the tip are \( \varepsilon_1 = 0.0016 \) and \( \varepsilon_2 = 0.0004 \), respectively.
The eccentricity in loading, \( e \), is _________ mm.

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The eccentricity in loading can be calculated using strain measurements and the bending stress distribution. Ensure to account for the geometry of the section when calculating bending moments.
Updated On: Nov 27, 2025
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Correct Answer: 0.95

Solution and Explanation

The strain at a distance from the neutral axis in bending is given by the formula: \[ \varepsilon = \frac{M y}{E I}, \] where \( M \) is the bending moment, \( y \) is the distance from the neutral axis, \( E \) is the modulus of elasticity, and \( I \) is the second moment of area.
For the square cross-section, the moment of inertia \( I \) is: \[ I = \frac{b^4}{12} = \frac{(1\, \text{cm})^4}{12} = 0.0000833 \, \text{cm}^4. \] The bending moment at the point of interest is given by: \[ M = F e = 10000 \, \text{N} \times e \, \text{m}. \] From the strain difference equation: \[ \varepsilon_1 - \varepsilon_2 = \frac{M (y_1 - y_2)}{E I}. \] Solving for \( e \), we get: \[ e \approx 0.95 \, \text{mm}. \] Thus, the eccentricity in loading is: \[ \boxed{0.95 \, \text{mm}}. \]
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