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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Here is the complete question
The seeds in bush bean pods are each the product of an independent fertilization event. Green seed color is dominant to white seed color in bush beans.
If a heterozygous plant with green seeds self-fertilizes, what is the probability that 6 seeds in a single pod of the progeny plant will consist of:
A) 4 green and 2 white seeds?
B) all white seeds?
C) at least 5 white seeds?
Answer:
A) 0.3
B) 0.0002
C) 0.001
Explanation:
Let the allele for Green plant be = P
Let the allele for white plant be = p
Let the heterozygous plant be Pp, if self fertilization occurs we have:
P p
P PP Pp
p Pp pp
From the punnet square above, we have the following cross;
PP, Pp, Pp, pp
Since P is dominant to p; we have a phenotypic ratio of 3:1 between the dominant phenotype ( PP, Pp, Pp ) and recessive phenotype ( pp )
NOW, the probability for the green plant(i.e PP or Pp) will be:
the probability for the white plant (pp) will be:
A)
what is the probability that 6 seeds in a single pod of the progeny plant will consist of 4 green and 2 white seeds?
=
The probability includes the use of permutation:
=
= 0.2966
≅ 0.3
B)
What is the probability that 6 seeds in a single pod of the progeny plant will consist of all white seeds?
Since probability for white seed =
The probability that 6 seeds will consist of all white seeds will be =
= 2.44 × 10⁻⁴
= 0.000244
≅ 0.0002
C)
What is the probability that 6 seeds in a single pod of the progeny plant will consist of at least 5 white seeds?
For at least 5 white seeds, the probability will consist of 5 white seeds and 6 white seeds.
For 5 white seeds; we have:
5 White and 1 Green Since there are 6 permutations.
The probability For 5 white seeds will therefore be :
= 7.32 × 10⁻⁴
= 0.000732
For 6 white seeds; we have:
Only one permutation for 6 whites since there will no be any presence of any green seed.
= 2.44 × 10⁻⁴
= 0.000244
∴
The total probability for at least 5 white seeds = 0.000732 + 0.000244
= 9.76 × 10⁻⁴
= 0.000976
≅ 0.001