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.
1 answer:
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
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