A rigid vessel with a volume of 10 m3 contains a water-vapor mixture at 400 kPa. If the quality is 60 percent, find the mass. Th
1 answer:
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Answer:
A.) 0.3088
B.) 0.0017
C.) part A
Explanation:
A.)
B.)
C.) Since the seat performance for an individual pilot is more important than 39 different pilots.
Answer:
a) Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
b) principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Explanation:
Combined normal stress and shear stress sketches attached below
The terms in the sketch are :
бx = tensile stress in x direction
бy = tensile stress in y direction
Txy = y component of shear stress acting on the perpendicular plane to x axis
бn = Normal stress acting on the inclined plane EF
Tn = shear stress acting on the inclined plane EF
A) Normal and shear stresses on inclined plane
Normal stress :
бn =[ ( бx + бy ) / 2 + ( бx - бy ) / 2 ] cos2∅ + Txysin2∅
shear stress
Tn = ( - бx - бy ) / 2 sin2∅ + Txy cos2∅
B) principal and maximum shear stresses
principal stress :
б1 = ( бx + бy ) / 2 - ( ( бx - бy ) / 2 )^2 + T^2xy
maximum shear stress:
Tmax = ( б1 - б2) / 2 = √ (( бx - бy ) / 2 )^2 + T^2xy
Answer:
a. = 40 psf
b. L ≈ 30.80 psf
c. The uniformly distributed total load for the beam = 812.8 ft./lb
d. The alternate concentrated load is more critical to bending , shear and deflection
Explanation:
The given parameters of the beam the beam are;
The span of the beam = 26 ft.
The width of the tributary, b = 16 ft.
The dead load, D = 20 psf.
a. The basic floor live load is given as follows;
The uniform floor live load, = 40 psf
The floor area, A = The span × The width = 26 ft. × 16 ft. = 416 ft.²
Therefore, the uniform live load, = 40 psf
b. The reduced floor live load, L in psf. is given as follows;
For the school, = 2
Therefore, we have;
The reduced floor live load, L ≈ 30.80 psf
c. The uniformly distributed total load for the beam, = b ×
=
∴ = = 16 × (20 + 30.80) ≈ 812.8 ft./lb
The uniformly distributed total load for the beam, = 812.8 ft./lb
d. For the uniformly distributed load, we have;
= 812.8 × 26/2 = 10566.4 lbs
= 812.8 × 26²/8 = 68,681.6 ft-lbs
= 5×812.8×26⁴/348/EI = 4,836,329.333/EI
For the alternate concentrated load, we have;
= 1000 lb
= 20 × 16 = 320 lb/ft.
= 1,000 + 320 × 26/2 = 5,160 lbs
= 1,000 × 26/4 + 320 × 26²/8 = 33,540 ft-lbs
= 1,000 × 26³/(48·EI) + 5×320×26⁴/348/EI = 2,467,205.74713/EI
Therefore, the loading more critical to bending , shear and deflection, is the alternate concentrated load
Answer:
So the bouy does not float with its axis vertical
Explanation:
From the question we are told that:
Diameter
Length
Weight
Specific weight of sea water
Generally the equation for weight of cylinder is mathematically given by
Weight of cylinder = buoyancy Force
Where
Therefore
Therefore
Center of Bouyance B
Center of Gravity
Generally the equation for\BM is mathematically given by
Therefore
Therefore
Therefore
So the bouy does not float with its axis vertical