GATE 2025 CE (CE1) – Question 54

Steel Structures · Connections: simple, eccentric, beam-column · 2 marks · Numerical answer

Consider the built-up column made of two I-sections as shown in the figure, with each batten plate bolted to a component I-section of the column through 6 black bolts. Each connection of the batten plate with the component section is to be designed for a longitudinal shear of 70 kN and moment of 10 kN.m. The minimum bolt value required (in kN) is _____________ (*rounded off to the nearest integer*).

Two I-sections side by side, 70 mm wide each with a 100 mm gap between them, joined by batten plates. Each connection has 6 bolts in two columns 70 mm apart and three rows 140 mm apart, with 30 mm edge distances at the top and bottom of the plate, which is 340 mm tall. The clear gap between the two battens is 210 mm. All dimensions are in mm.

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Correct answer: 23

Explanation

The six bolts are at $(\pm 35, 0)$ and $(\pm 35, \pm 140)$ mm from the centre of the group. The shear of 70 kN is shared equally, so each bolt takes $\frac{70}{6} = 11.67$ kN along the column. For the moment, $\sum r^2 = 2(35^2) + 4(35^2 + 140^2) = 85750$ mm$^2$. A corner bolt gets a component $\frac{10000 \times 140}{85750} = 16.33$ kN across the column and $\frac{10000 \times 35}{85750} = 4.08$ kN along it. At the worst corner the along-column parts add up to $11.67 + 4.08 = 15.75$ kN, so the resultant is $\sqrt{15.75^2 + 16.33^2} = 22.7$ kN. The bolt value must be at least this, about 23 kN.