Answer:
square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.
Required:
a. What is the nominal maximum tensile stress on the bar?
b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?
Answer:
The maximum length is 3.897×10^-5 mm
Explanation:
Extension = surface energy/elastic modulus
surface energy = 1.05 J/m^2
elastic modulus = 198 GPa = 198×10^9 Pa
Extension = 1.05/198×10^9 = 5.3×10^-12 m
Strain = stress/elastic modulus = 27×10^6/198×10^9 = 1.36×10^-4
Length = extension/strain = 5.3×10^-12/1.36×10^-4 = 3.897×10^-8 m = 3.897×10^-8 × 1000 = 3.897×10^-5 mm
Answer:
see pictures
Explanation:
In the pictures you can see the different processes.
In Menlo Park, California, Hwan Rhee is debating whether to launch ChargeAll, a software firm that will create full-room wireless chargers for any kind of mobile electronic device.
In Menlo Park, California, Hwan Rhee is debating whether to launch ChargeAll, a software firm that will create full-room wireless chargers for any kind of mobile electronic device. In order to examine the financial information for a startup loan that would pay for the components and product manufacture, Hwan is using an Excel worksheet. He requests your assistance in fixing mistakes and performing financial computations in the worksheet.
Visit the worksheet for loan analysis. Hwan requests that you fix the mistakes in the worksheet before he can compute the principal and interest payments on the loan.
To know more about software click here:
brainly.com/question/985406
#SPJ4
Answer:
a. Solid length Ls = 2.6 in
b. Force necessary for deflection Fs = 67.2Ibf
Factor of safety FOS = 2.04
Explanation:
Given details
Oil-tempered wire,
d = 0.2 in,
D = 2 in,
n = 12 coils,
Lo = 5 in
(a) Find the solid length
Ls = d (n + 1)
= 0.2(12 + 1) = 2.6 in Ans
(b) Find the force necessary to deflect the spring to its solid length.
N = n - 2 = 12 - 2 = 10 coils
Take G = 11.2 Mpsi
K = (d^4*G)/(8D^3N)
K = (0.2^4*11.2)/(8*2^3*10) = 28Ibf/in
Fs = k*Ys = k (Lo - Ls )
= 28(5 - 2.6) = 67.2 lbf Ans.
c) Find the factor of safety guarding against yielding when the spring is compressed to its solid length.
For C = D/d = 2/0.2 = 10
Kb = (4C + 2)/(4C - 3)
= (4*10 + 2)/(4*10 - 3) = 1.135
Tau ts = Kb {(8FD)/(Πd^3)}
= 1.135 {(8*67.2*2)/(Π*2^3)}
= 48.56 * 10^6 psi
Let m = 0.187,
A = 147 kpsi.inm^3
Sut = A/d^3 = 147/0.2^3 = 198.6 kpsi
Ssy = 0.50 Sut
= 0.50(198.6) = 99.3 kpsi
FOS = Ssy/ts
= 99.3/48.56 = 2.04 Ans.