To solve this problem, let's break it down:

We are asked to determine the maximum load $P$ that can be applied to a connection involving bolts of 6 mm in diameter. The average shear stress in the bolts is limited to 124 MPa (MegaPascals).

Formula to use:

The average shear stress $\tau$ in the bolts is given by:

$\tau = \frac{P}{A \cdot n}$

Where:

• $P$ is the load applied (in Newtons or kN),
• $A$ is the cross-sectional area of a single bolt,
• $n$ is the number of bolts (in this case, $n = 3$),
• $\tau$ is the shear stress (124 MPa, or $124 \times 10^6 \, \text{Pa}$).

The cross-sectional area $A$ of a bolt is calculated from its diameter $d$ using the formula for the area of a circle:

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

Given:

• Diameter of the bolts $d = 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m}$,
• Shear stress $\tau = 124 \times 10^6 \, \text{Pa}$,
• Number of bolts $n = 3$.

Let's calculate the maximum load $P$.The maximum load $P$ that can be applied to the connection is approximately $10.52 \, \text{kN}$.

It seems the given answer in the image is different (21.04 kN), which suggests either a different number of bolts, shear planes, or other factors were considered. However, based on the provided data (3 bolts, 6 mm diameter, 124 MPa shear stress), the result is $10.52 \, \text{kN}$.

Would you like more details or have any questions?

Here are 5 follow-up questions:

1. How does the number of bolts affect the maximum load $P$?
2. What would happen to the load if the bolt diameter was increased?
3. How do you calculate shear stress for a single bolt?
4. Why is shear stress used instead of tensile stress in bolt calculations?
5. Can bolts with different diameters be used in the same connection?

Tip: Always verify the number of bolts and their arrangement when calculating load distribution across fasteners.