5.You are designing the fit for two bearings. For each case, specify the maximum and minimum hole and shaft diameter:a.A journal
bearing and bushing with basic size is 1.25 in that requires a close running fit.b.A ball bearing assembly with basic size 35 mm that requires a locational interference fit.
1 answer:
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To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.
The stagnation temperature can be defined as
Where
T = Static temperature
V = Velocity of Fluid
Specific Heat
Re-arrange to find the static temperature we have that
Now the pressure of helium by using the Adiabatic pressure temperature is
Where,
= Stagnation pressure of the fluid
k = Specific heat ratio
Replacing we have that
Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa
<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>
Answer:
1.176
Explanation:
When the bullets impact the mass they become embedded on it, it is a plastic collision, therefore momentum is conserved.
v2 * (M + mb) = v1 * mb
Where
v1: muzzle velocity of the bullet
M: mass of the bob
mb: mass of the bullet
v2: mass of the bob with the bullet after being hit
v2 = v1 * mb / (M + mb)
Upon being impacted the bob will acquire speed v2, this implies a kinetic energy. The bob will then move and raise a height h. Upon acheiving the maximum height it will have a speed of zero. At that point all kinetic energy will be converted into potential energy.
Ek = 1/2 (M + mb) * v2^2
Ep = (M + mb) * g * h
Ek = Ep
1/2 (M + mb) * v2^2 = (M + mb) * g * h
1/2 * (v1 * mb / (M + mb))^2 = g * h
1/2 * v1^2 * mb^2 / (M + mb)^2 = g * h
v1^2 = g *h * (M+ mb)^2 / (1/2 * mb^2)
The height h that it reaches is related to the length L of the pendulum arm and the angle it forms with the vertical.
h = L * (1 - cos(a))
For the 9 mm:
For the 0.44 caliber:
The ratio is 460 / 391 = 1.176
Answer:
M10 × 1.5
least number of threads is 4.3
Explanation:
given data
static tensile load = 4 kN = 4000 N
safety factor = 5
coarse‐thread metric = 5.8
solution
we know that proof load for 5.8 class is Sp = 380 MPa
so area of cross section is express as
area = \frac{force \times FOS}{Sp} ......................1
area = \frac{4000 \times }{380}
area = 52.6 mm²
so by table at 58 mm² we can say we resist M10 × 1.5
and
bolt tensile strength is express as
bolt tensile strength = nut shear strength ......2
so here bolt tensile strength = At × Sy (bolt) ...............3
and nut shear strength = πd ( 0.75 t ) Sys
nut shear strength = π × 10 × ( 0.75 t ) × 0.58 × × Sy(bolt) ............4
so from equation 3 and 4 we get
t = 6.37 mm
and
for the pitch is = 1.5 mm
least number of threads is 4.3