Question

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).

Hint:

When calculating the minimum bolt value, always consider both direct shear and torsional shear. The maximum resultant shear force on the bolt can be found using vector addition.

Answer 1

Correct Answer: 21

\[ r_{{max}} = \sqrt{140^2 + 35^2} = 144.31 \, {mm} \quad {(Radius of gyration)} \] \[ V_b = 70 \, {kN} \quad {(Longitudinal shear)} \] \[ M_b = 10 \, {kN.m} \quad {(Moment)} \] % Concept Concept: The bolt value \( F_{{max}} \) is the maximum resultant shear force on the bolt. Step 1: Direct shear force on bolt, \( F_1 \) \[ F_1 = \frac{V_L}{n} = \frac{70}{6} = 11.67 \, {kN}. \] Step 2: Maximum torsional shear on bolt, \( F_2 \) \[ F_2 = \frac{(T \cdot M)_{{max}}}{r_{{max}}} = \frac{10 \times 10^3 \times 144.31}{4 \times 144.31^2 + 2 \times 35^2} = 16.82 \, {kN}. \] Step 3: Minimum angle of inclination between \( F_1 \) and \( F_2 \), \( \theta \) \[ \theta = \tan^{-1} \left( \frac{140}{35} \right) = 75.96^\circ. \] Step 4: Maximum resultant shear force on bolt, \( F_R \) \[ F_R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta} \] \[ F_R = \sqrt{11.67^2 + 16.82^2 + 2 \times 11.67 \times 16.82 \times \cos75.96^\circ} = 22.67 \, {kN}. \] Thus, the minimum bolt value required is approximately \( \boxed{23} \, {kN} \) (rounded to the nearest integer).