The breaking strength of a certain kind of mild steel in tension is 389 MPa . A tie-bar of angle steel supports an axial pull of \( 167,7 \mathrm{kN} \). Find suitable dimensions if the mean thickness of metal is \( 12,7 \mathrm{~mm} \). The factor of safety is 6 .

To ensure the tie-bar can safely handle the axial pull, we first need to determine the allowable stress using the given factor of safety. The allowable stress is calculated by dividing the breaking strength by the factor of safety: \[ \text{Allowable stress} = \frac{\text{Breaking Strength}}{\text{Factor of Safety}} = \frac{389 \, \text{MPa}}{6} = 64.83 \, \text{MPa} \] Next, we convert the axial pull to the same units: \[ 167.7 \, \text{kN} = 167700 \, \text{N} \] Now, using this allowable stress, we can find the required cross-sectional area (A) of the tie-bar to ensure it can support the load: \[ A = \frac{F}{\sigma_{\text{allowable}}} = \frac{167700 \, \text{N}}{64.83 \, \text{MPa}} \approx 2583.49 \, \text{mm}^2 \] Given that the mean thickness of the angle steel is 12.7 mm, we can denote the width of the angle as \( b \). For an angle section of thickness \( t \), the area of the cross-section can be approximated assuming equal legs: \[ A = 2(b \cdot t) - t^2 \] Substituting \( t = 12.7 \, \text{mm} \): \[ 2583.49 \, \text{mm}^2 = 2(b \cdot 12.7) - (12.7^2) \] This gives: \[ 2583.49 = 25.4b - 161.29 \] \[ 25.4b = 2583.49 + 161.29 = 2744.78 \] Thus, \[ b = \frac{2744.78}{25.4} \approx 108.63 \, \text{mm} \] So, suitable dimensions for the angles can be approximately 12.7 mm in thickness and 108.63 mm in width.