Answer: preventive maintenance
Explanation:
Answer:
Maximum Normal Stress σ = 8.16 Ksi
Maximum Shearing Stress τ = 4.08 Ksi
Explanation:
Outer diameter of spherical container D = 17 ft
Convert feet to inches D = 17 x 12 in = 204 inches
Wall thickness t = 0.375 in
Internal Pressure P = 60 Psi
Maximum Normal Stress σ = PD / 4t
σ = PD / 4t
σ = (60 psi x 204 in) / (4 x 0.375 in)
σ = 12,240 / 1.5
σ = 8,160 P/in
σ = 8.16 Ksi
Maximum Shearing Stress τ = PD / 8t
τ = PD / 8t
τ = (60 psi x 204 in) / (8 x 0.375 in)
τ = 12,240 / 3
τ = 4,080 P/in
τ = 4.08 Ksi
Houses the CYLINDERS, Water Jacket & Crankcase
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.
