Answer:
the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Explanation:
Given the data in the question;
yield strength σ
= 690 Mpa
plane strain fracture toughness K
= 32 MPa-
minimum component thickness for which the condition of plane strain is valid = ?
Now, for plane strain conditions, the minimum thickness required is expressed as;
t ≥ 2.5( K / σ )²
so we substitute our values into the formula
t ≥ 2.5( 32 / 690 )²
t ≥ 2.5( 0.0463768 )²
t ≥ 2.5 × 0.0021508
t ≥ 0.005377 m or 5.38 mm
Therefore, the minimum component thickness for which the condition of plane strain is valid is 0.005377 m or 5.38 mm
Answer:
diagram of an electrical curcuit
an sketch of an HVAC system
Also 3D image of a hydrualic piston
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Answer:
a) 152000 slugs
b) 2220000 kg or 2220 metric tons
Explanation:
A body with a weight of 4.9*10^6 lbf has a mass of
4.9*10^6 lbm * 1 lbf/lbm = 4.9*10^6 lbm
This mass value can then be converted to other mass values.
1 slug is 32.17 lbm
Therefore:
4.9*10^6 lbm * 1 slug / (32.17 lbm) = 152000 slugs
1 lb is 0.453 kg
Therefore:
4.9*10^6 lbm / (1/0.453) * kg/lbm = 2220000 kg
Answer:
Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s
Explanation:
du/u² = dt = ∫du/2.3183 = ∫dt
0.4313 u = t + c
(a) t = 0, u= 15m/s, c = 0.647
u = t+c/0.4313 = t + 0.647/0.4313
(a) when t= 1 u = 1+ 0.647/0.4313 = 3.8187m/s
(b) when t= 10 u = 10 + 0.647/0.4313 = 24.0858m/s
(c)when t= 1000 u = 1000 + 0.647/0.4313 = 3220.071m/s
