Question

The two members used in the construction of an aircraft fuselage are joined together using a 30° fish-mouth weld. Determine the average normal and average shear stress on the plane of each weld. Assume each inclined plane supports a horizontal force of 400 lb.

Answer 1

This question is incomplete because it lacks the required diagram of the fuselage. Please find attached to this answer the appropriate diagram.

Answer:

The average normal stress on the plane of each weld = 66.7 psi

The average shear stress on the plane of each weld = 115.5 psi

Explanation:

From the question, we are told that two members are joined together.

From the diagram we see 800lb.

Since there are 2 members = 800lb/2

= 400lb

a) The formula for average nor umal stress on the plane of each weld = Force (in Newton)/ Cross sectional area (in inches)

From the question, we are assuming that the horizontal force = 400lb

We are given an angle of 30°

Therefore, the resultant force in Newton = F sin θ

= 400 sin 30 = 200lb

Cross sectional area = A/sin θ

= 1 × 1.5/ sin 30

= 3 in²

The average normal stress on the plane of each weld is calculated as:

200lb/3 in²

= 66.7lb/in² or 66.7 psi

b) The average shear stress on the plane of each weld = Force (in Newton)/ Cross sectional area (in inches)

Resultant force =  F cos θ

= 400 cos 30 = 346.41016151lb

Approximately = 346.41lb

Crossectional area = 3 in²

The average normal stress on the plane of each weld is calculated as:

346.4lb/3 in²

= 115.47Ib/in²

Approximately = 115.5Ib/in²

Therefore,

The average normal stress on the plane of each weld = 66.7 psi

The average shear stress on the plane of each weld = 115.5 psi