Explain the difference, on the basis of the test results, between the ultimate strength and the "true" stress at fracture.
1 answer:
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Answer:
P = 4.745 kips
Explanation:
Given
ΔL = 0.01 in
E = 29000 KSI
D = 1/2 in
LAB = LAC = L = 12 in
We get the area as follows
A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²
Then we use the formula
ΔL = P*L/(A*E)
For AB:
ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB
For AC:
ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)
⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC
Now, we use the condition
ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in
⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in
Knowing that PAB*Cos 30°+PAC*Cos 30° = P
we have
(2.107*10⁻⁶ in/lbf)*P = 0.01 in
⇒ P = 4745.11 lb = 4.745 kips
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Answer:
T = 20.42 N
Explanation:
given data
standard altitude = 30,000 ft
velocity Ca = 500 mph = 0.4 m/s
inlet areas Aa = 7 ft² = 0.65 m²
exit areas Aj = 4.5 ft² = 0.42 m²
velocity at exit Cj = 1600 ft/s = 487.68 m/s
pressure exit j = 640 lb/ft² = 0.3 bar
solution
we get here thrust of the turbojet that is express as
thrust of the turbojet T = Mg × Cj - Ma × Ca + ( j Aj -
a Ag ) .............1
here Ma = Mg
Ma = a × Ca Aa = 0.042 kg/s
put value in equation 1 we get
T = 0.042 × (487.68 -0.14) + ( 0.3 × - 0.3 × 0.65 )
T = 20.42 N