The problem provided asks to determine the maximum load $P$ that can be applied to the clevis connection, given the shear stress limit and bolt diameter. Here's the breakdown:

Given:

• Shear stress limit $\tau = 37 \, \text{ksi}$ (kips per square inch)
• Diameter of the bolt $d = 0.508 \, \text{in}$

Formula:

The shear stress is defined as: $\tau = \frac{P}{A}$ Where:

• $\tau$ is the shear stress (37 ksi),
• $P$ is the applied load,
• $A$ is the cross-sectional area of the bolt.

The area $A$ for a circular cross-section is: $A = \frac{\pi d^2}{4}$ Where $d = 0.508 \, \text{in}$.

To solve for $P$: $P = \tau \times A$

Let me now calculate this value.The maximum load $P$ that can be applied to the clevis connection is approximately $7.50 \, \text{kip}$.

It appears the image provided an answer of $14.998 \, \text{kip}$, which suggests that the problem might involve a double shear condition (i.e., the bolt experiences shear at two planes). For double shear, the load is distributed over two areas, effectively doubling the load capacity.

In a double shear scenario: $P = 2 \times \tau \times A$ I'll now calculate the maximum load in double shear.In the case of double shear, the maximum load $P$ that can be applied to the connection is approximately $14.999 \, \text{kip}$, which matches the answer provided in the image.

Would you like further details on the solution? Here are five related questions to explore the topic further:

1. What is the difference between single shear and double shear in bolted connections?
2. How does bolt diameter affect the load capacity in a shear connection?
3. Why is shear stress important in engineering design for bolts?
4. What are other methods of increasing the load capacity in bolted joints?
5. How do material properties of the bolt affect the maximum shear load?

Tip: Always consider the number of shear planes (single or double) when calculating load capacity in shear problems.