Consider the beam section shown in the figure, with \( y \) indicating the depth of neutral axis (NA). The section is only subjected to an increasing bending moment. It is given that \( y = 18.75 \, {mm} \), when the section has not yielded at the top and bottom fibres. Further, \( y \) decreases to 5 mm, when the entire section has yielded. The shape factor of the section is ........ (rounded off to 2 decimal places).
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Correct Answer: 1.8
Solution and Explanation
\[ Z_p = \frac{2 \times 60 \times 5}{2} (2.5 + 30) = 9750 \, {mm}^3. \] For elastic section modulus (\( Z_e \)): \[ Z_e = \frac{I_{NA}}{y_{{max}}} \] where \( I_{NA} \) is the area moment of inertia about the neutral axis and \( y_{{max}} = 46.25 \, {mm} \).
\[ I_{NA} = \left[ \frac{60 \times 5^3}{12} + 60 \times 5 \times (16.25)^2 \right] + \left[ \frac{5 \times 60^3}{12} + 5 \times 60 \times (16.25)^2 \right] \] \[ I_{NA} = 249062.5 \, {mm}^4 \] \[ Z_e = \frac{249062.5}{46.25} = 5385.135 \, {mm}^3 \] Now, the shape factor \( S \) is calculated by: \[ S = \frac{M_p}{M_y} = \frac{Z_p}{Z_e} \] \[ S = \frac{9750}{5385.135} = 1.81 \] Correct Answer: 1.81 (rounded to two decimal places).
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