Equal Forces on Bolt Head Calculation

How do we determine the magnitude of equal forces exerted on the four contact points on a bolt head?

If 50-lb forces are applied to a lug wrench to tighten a square-head bolt, how can we find the magnitude F of the equal forces exerted on the four contact points on the 1-in. bolt head?

Magnitude of Equal Forces on Bolt Head Calculation

The magnitude F of the equal forces exerted on the four contact points on the 1-in. bolt head is 35.36 lb.

To determine the magnitude of the equal forces exerted on the four contact points on the bolt head, we need to consider the equilibrium of forces. Since the external effect on the bolt is equivalent to that of the two 50-lb forces applied to the wrench, the total force acting on the bolt must be balanced.

In the given scenario, the two 50-lb forces are applied perpendicular to the flats of the bolt head. These forces create a moment around the center of the bolt. To balance this moment, the four contact forces on the bolt head must also create an equal and opposite moment.

Considering the symmetry of the problem, each contact force is equal in magnitude. Let's denote this magnitude as F. Since there are four contact points, the total moment created by the contact forces is 4F multiplied by the lever arm (which is half the width of the bolt head, i.e., 0.5 in).

To balance the total moment, it should be equal to the moment created by the two 50-lb forces. The moment created by each 50-lb force is 50 lb multiplied by the distance between the force and the center of the bolt.

Setting up the equation: (2 × 50 lb × distance) = (4F × 0.5 in)

Simplifying the equation: 100 lb × distance = 2F

Solving for F, we get: F = 50 lb × distance

Since the distance is 0.707 in (which is half the diagonal of the square bolt head), we can calculate F as: F = 50 lb × 0.707 in ≈ 35.36 lb.

Therefore, the magnitude F of the equal forces exerted on the four contact points on the 1-in. bolt head is approximately 35.36 lb.