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GATE 2025 CE (CE1) – Question 53

Solid Mechanics · Simple bending theory, flexural and shear stresses, shear centre · 2 marks · Numerical answer

Consider the beam section shown in the figure, with $\bar{y}$ indicating the depth of neutral axis (NA). The section is only subjected to an increasing bending moment. It is given that $\bar{y} = 18.75$ mm, when the section has not yielded at the top and bottom fibres. Further, $\bar{y}$ decreases to 5 mm, when the entire section has yielded. The shape factor of the section is _____________ (*rounded off to 2 decimal places*).

A T-section with a flange 60 mm wide and 5 mm thick on top of a web 5 mm wide and 60 mm deep. The neutral axis is marked at a depth $\bar{y}$ below the top. All dimensions are in mm.

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Correct answer: 1.80 to 1.82

Explanation

The flange area is $60 \times 5 = 300$ mm$^2$ and the web area is $5 \times 60 = 300$ mm$^2$. The elastic neutral axis is at $\frac{300 \times 2.5 + 300 \times 35}{600} = 18.75$ mm from the top, as given, and the section is 65 mm deep. The moment of inertia is $I = \left[\frac{60 \times 5^3}{12} + 300 \times 16.25^2\right] + \left[\frac{5 \times 60^3}{12} + 300 \times 16.25^2\right] = 249062$ mm$^4$. The bottom fibre is farther from the axis, at $65 - 18.75 = 46.25$ mm, so $Z = \frac{249062}{46.25} = 5385$ mm$^3$. When all of the section yields the plastic neutral axis divides the area equally, which is at 5 mm (the base of the flange), so $Z_p = 300 \times (35 - 2.5) = 9750$ mm$^3$. The shape factor is $\frac{9750}{5385} = 1.81$.