Answer:
a) 0.0663 m³/s
b) 3.312 N/(m/s)²
c) 16.665 m/s
d) 0.1105 m³/s
Explanation:
See attached pictures.
Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.
1 answer:
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Answer:
The amplitude of the absorbed mass can be found
for ka:
now
Answer:
The required size of column is length = 15 ft and diameter = 4.04 inches
Explanation:
Given;
Length of the column, L = 15 ft
Applied load, P = 10 kips = 10 × 10³ Psi
End condition as fixed at the base and free at the top
thus,
Effective length of the column, = 2L = 30 ft = 360 inches
now, for aluminium
Elastic modulus, E = 1.0 × 10⁷ Psi
Now, from the Euler's critical load, we have
where, I is the moment of inertia
on substituting the respective values, we get
or
I = 13.13 in⁴
also for circular cross-section
I =
thus,
13.13 =
or
d = 4.04 inches
The required size of column is length = 15 ft and diameter = 4.04 inches
Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws
Explanation:
<u>Plane -Strain Fracture toughness is calculated as</u>
=
б
F=geometry factor of the flaw
б=Stress applied
=Fracture toughness
a=Flaw size
<u>Given that </u>
Internal Flaw,a=0.001cm
Fracture Toughness =45MPa
Tensile Strength б=550 MPa
Geometry Factor,=1
<u>Calculation</u>
An internal Flaw i s 0.001 cm
2a=0.001cm
a=0