### Answers

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

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

the correct answer, according to e2020, is the second choice there is an occluded front in region a and a cold front in region b.

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