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
The pic shown can help to understand the question.
Answer:
Creating vacancies in ceramics. Consider doping ZrO₂ with small concentrations of Nb205. The valence of Nb is 5. Assume that Nb ions sit in Zr ion sites
a. A substitutional defect will be produced.
b. With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.
Explanation:
(a) The defect produced by dopping a little concentration of Nb₂O5 with Nb in the +5 charge state is known as a substitutional defect.
(b) With this dopping, the Nb increases electron band conduction and decreases oxygen anion conduction in ZrO₂.
Moreover, if oxygen vacancies are rate-limiting defect, the corrosion of ZrO₂ decreases and if electrons are rate-limiting then the corrosion of ZrO₂ is accelerated.
Answer:
What is one of the “don’ts” in drawing dimension lines? they should never be labeled they should never be stacked they should never cross each other they should never have only one measurement value
