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

Given Data:

• Radius of gyration: \[ r_{{max}} = \sqrt{140^2 + 35^2} = 144.31 \, \text{mm} \]
• Longitudinal shear: \[ V_b = 70 \, \text{kN} \]
• Moment: \[ M_b = 10 \, \text{kN} \cdot \text{m} \]

Concept:

Bolt Value: The bolt value \( F_{{max}} \) is the maximum resultant shear force on the bolt.

Step 1: Direct Shear Force on Bolt, \( F_1 \)

Using the formula:

\[ F_1 = \frac{V_L}{n} = \frac{70}{6} = 11.67 \, \text{kN} \]

Step 2: Maximum Torsional Shear on Bolt, \( F_2 \)

Using the equation:

\[ F_2 = \frac{(T \cdot M)_{{max}}}{r_{{max}}} \]

Substituting values:

\[ F_2 = \frac{10 \times 10^3 \times 144.31}{4 \times 144.31^2 + 2 \times 35^2} = 16.82 \, \text{kN} \]

Step 3: Minimum Angle of Inclination Between \( F_1 \) and \( F_2 \), \( \theta \)

Using the formula:

\[ \theta = \tan^{-1} \left( \frac{140}{35} \right) \]

Solving:

\[ \theta = 75.96^\circ \]

Step 4: Maximum Resultant Shear Force on Bolt, \( F_R \)

Using the resultant force equation:

\[ F_R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos\theta} \]

Substituting values:

\[ F_R = \sqrt{11.67^2 + 16.82^2 + 2 \times 11.67 \times 16.82 \times \cos75.96^\circ} \]

Solving:

\[ F_R = 22.67 \, \text{kN} \]

Final Answer:

Minimum Bolt Value Required: \( \boxed{23} \, \text{kN} \) (rounded to the nearest integer).