Answer:
Explanation:
Soil bearing pressure=
Since we're given pressure of 2500 psf and load of 45000 pounds
The area=
Therefore, the smallest area of safe footings should not be less than
A crate with dimension 12 x 10 x 4in and a mass of 120 lbs slides on an inclined plane of dimension 80 x 10 x 3in. Answer the fo
llowing questions:
1. Model the system in ADAMS without friction and compare your result with the analytical solution.
2. Model the system in ADAMS with friction coefficients, μs=0.4 and μd= 0.3 and find the minimum inclination that will ensure that a crate slides off an inclined plane. Find the value of the crate's constant acceleration. compare and verify your result with the analytical solution.
3. For the previous part, simulate the model for an end time of 0.5 seconds and plot the create acceleration vs. time. Show all the calculations and results clearly.
1. Model the system in ADAMS without friction and compare your result with the analytical solution.
2. Model the system in ADAMS with friction coefficients, μs=0.4 and μd= 0.3 and find the minimum inclination that will ensure that a crate slides off an inclined plane. Find the value of the crate's constant acceleration. compare and verify your result with the analytical solution.
3. For the previous part, simulate the model for an end time of 0.5 seconds and plot the create acceleration vs. time. Show all the calculations and results clearly.
1 answer:
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Answer:
LAOD = 6669.86 N
Explanation:
Given data:
width
thickness
crack length 2c = 0.5 mm at centre of specimen
stress intensity factor = k will be
we know that
[c =0.5/2 = 2.5*10^{-4}]
K = 0.1724 Mpa m^{1/2} for 1000 load
if then load will be
LAOD = 6669.86 N
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
