What are the minimum and maximum required footing projections for a concrete footing having a thickness of 8 inches
1 answer:
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Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂
The heat supplied = × Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied = ·
= 20 kg/s
The heat supplied = 20* = 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K
T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust = ×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.
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
Answer:
Explanation:
Given that
Lattice constant = a
Radius of unit cell cell =R
Atom is in BCC structure.
In BCC unit cell (Body centered cube)
1.Eight atoms at eight corner of cube which have 1/8 part in each cube.
2.One complete atom at the body center of the cube
So the total number of atoms in the BCC
Z= 1/8 x 8 + 1 x 1
Z=2
In triangle ABD
In triangle ABC
AC=4R
BC=a
So
So the relationship between lattice constant and radius of unit cell
