Two plates are joined with bolts and subjected to shear with a force of $P = 44$ kN. If the bolt material has a tensile fracture stress of $\sigma_{\theta p} = 3700$ N/cm$^2$ and a safety factor of $v = 5$, calculate the minimum bolt diameter.

1. Problem Description

Two plates are joined with bolts and subjected to shear with a force of $P = 44$ kN. If the bolt material has a tensile fracture stress of $\sigma_{\theta p} = 3700$ N/cm$^2$ and a safety factor of $v = 5$, calculate the minimum bolt diameter.

2. Solution Steps

First, we need to find the allowable tensile stress, $\sigma_{allow}$, by dividing the tensile fracture stress by the safety factor:

$\sigma_{allow} = \frac{\sigma_{\theta p}}{v}$

$\sigma_{allow} = \frac{3700 \text{ N/cm}^2}{5} = 740 \text{ N/cm}^2$

The force $P$ is supported by two bolts in shear, as illustrated in the diagram. Therefore, each bolt is subjected to a shear force equal to $P/2$. The problem states that the bolts are subjected to *shear*, but it provides a tensile fracture stress. We need to make an assumption that the shear stress at failure is proportional to the tensile stress at failure, perhaps using a Tresca or von Mises yield criterion. Lacking any other information, let us *assume* the permissible shear stress $\tau_{allow}$ is $\sigma_{allow}$.

Therefore we have:

$\tau_{allow} = \sigma_{allow} = 740 \text{ N/cm}^2$

The shear force acting on each bolt is:

$F = \frac{P}{2} = \frac{44 \text{ kN}}{2} = 22 \text{ kN} = 22000 \text{ N}$

The area of each bolt subjected to shear is:

$A = \frac{\pi d^2}{4}$

The shear stress is:

$\tau = \frac{F}{A}$

We want to find the minimum diameter $d$, so we need to ensure that $\tau \le \tau_{allow}$. Let us set $\tau = \tau_{allow}$:

$\tau_{allow} = \frac{F}{A}$

$740 \text{ N/cm}^2 = \frac{22000 \text{ N}}{\frac{\pi d^2}{4}}$

Rearrange the equation to solve for $d^2$:

$d^2 = \frac{4 \times 22000 \text{ N}}{\pi \times 740 \text{ N/cm}^2} = \frac{88000}{\pi \times 740} \text{ cm}^2$

$d^2 \approx 37.97 \text{ cm}^2$

Take the square root to find the diameter:

$d = \sqrt{37.97} \text{ cm} \approx 6.16 \text{ cm}$

3. Final Answer

The minimum bolt diameter is approximately 6.16 cm.