Bolt Strength Calculation

The question asks us to calculate the strength of a 20 mm diameter bolt of grade 4.6, given its net tensile stress area and a partial safety factor. Based on the options and the provided data, it appears the question requires the calculation of the shear strength of the bolt, using the provided net area for the threaded portion in shear.

Given Data

• Bolt Diameter (\(\Phi\)) = 20 mm
• Bolt Grade = 4.6
• Net tensile stress area (\(A_{nb}\) or \(A_n\)) = 245 mm\(^2\)
• Partial safety factor for bolt material (\(\gamma_{mb}\)) = 1.25

Bolt Material Properties

For a bolt of grade 4.6:

• Ultimate tensile strength (\(f_u\)) = 400 MPa = 400 N/mm\(^2\)
• Yield strength (\(f_y\)) = 0.6 \(\times\) \(f_u\) = 0.6 \(\times\) 400 = 240 MPa = 240 N/mm\(^2\)

Shear Strength Calculation

We will calculate the design shear strength of the bolt. The formula for the nominal shear strength (\(V_{nsb}\)) of a bolt, as per relevant codes (like IS 800:2007), considering shear planes passing through the threaded portion, is:

\(V_{nsb} = \frac{f_u}{\sqrt{3}} (n_n A_n + n_s A_s)\)

Where:

• \(f_u\) is the ultimate tensile strength of the bolt (400 N/mm\(^2\))
• \(n_n\) is the number of shear planes intersecting the threaded portion
• \(A_n\) is the net tensile stress area of the bolt (245 mm\(^2\)), which is \(A_{nb}\) provided
• \(n_s\) is the number of shear planes intersecting the shank (unthreaded) portion
• \(A_s\) is the shank area (gross cross-sectional area)

Assuming there is one shear plane and it passes through the threaded portion (\(n_n = 1\), \(n_s = 0\)), the formula simplifies to:

\(V_{nsb} = \frac{f_u}{\sqrt{3}} (1 \times A_n + 0 \times A_s) = \frac{f_u A_n}{\sqrt{3}}\)

Substitute the given values into the formula for \(V_{nsb}\):

\(V_{nsb} = \frac{400 \, \text{N/mm}^2 \times 245 \, \text{mm}^2}{\sqrt{3}}\)

\(V_{nsb} = \frac{98000}{\sqrt{3}} \, \text{N}\)

\(V_{nsb} \approx 56568.54 \, \text{N}\)

The design shear strength (\(V_{dsb}\)) is obtained by dividing the nominal shear strength by the partial safety factor (\(\gamma_{mb}\)):

\(V_{dsb} = \frac{V_{nsb}}{\gamma_{mb}}\)

Substitute the calculated \(V_{nsb}\) and the given \(\gamma_{mb}\) (1.25):

\(V_{dsb} = \frac{56568.54 \, \text{N}}{1.25}\)

\(V_{dsb} \approx 45254.83 \, \text{N}\)

Convert the strength from Newtons to kilonewtons (kN) by dividing by 1000:

\(V_{dsb} \approx \frac{45254.83}{1000} \, \text{kN}\)

\(V_{dsb} \approx 45.255 \, \text{kN}\)

Rounding to two decimal places, the design shear strength is approximately 45.26 kN.

Result Summary

The calculated shear strength of the bolt is approximately 45.26 kN.

Comparing this value with the given options, 45.26 kN matches one of the options.