Answer:
Explanation: Kindly refer to the attached files for a detailed solution process.
Take note that the diagram file is the first one and is positioned accordingly.
A solid circular shaft has a uniform diameter of 5 cm and is 4 m long. At its midpoint 65 hp is delivered to the shaft by means
1 answer:
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Answer:
r=0.31
Ф=18.03°
Explanation:
Provided:
Original diameter of bar = 75 mm
Diameter post-cutting = 73 mm
Average diameter of the bar d= (75+73)/2=74 mm
Average length of uncut chip = πd
Average length of uncut chip = π x 74 =232.45 mm
Thus, cutting ratio r
r=0.31
Therefore, the cutting ratio equals 0.31.
Now, the shearing angle is given as
Next by substituting the values
\
Ф=18.03°
Concluding, the shearing angle is 18.03°.
Answer:
a) The phase before eutectoid is commonly referred to as cementite, with the chemical formula Fe₃C.
b) The total mass of ferrite obtained is 0.8311 kg.
The total cementite mass equals 0.1689 kg.
c) The total cementite mass accounts for 0.9343 kg.
Explanation:
Provided:
1 kg of austenite
a carbon content of 1.15 wt%
Cooled below 727°C
Questions:
a) Identify the proeutectoid phase.
b) Calculate the mass of total ferrite and cementite, Wf =?, Wc =?
c) Determine the mass of both pearlite and the proeutectoid phase, Wp =?
d) Create a schematic to illustrate the resulting microstructure.
a) The proeutectoid phase is referred to as cementite with the formula Fe₃C.
b) To find the total mass of formed ferrite:
With:
Ccementite = composition of cementite = 6.7 wt%
C₁ = composition of phase 1 = 0.022 wt%
C₂ = overall composition = 1.15 wt%
Inserting the values yields:
For the total mass of cementite:
c) The mass of pearlite:
d) The diagram illustrates the different compositions: (pearlite, proeutectoid cementite, ferrite, eutectoid cementite)
Answer:
Change in length = 0.0913 in
Explanation:
Given data:
Length = 6 ft
Diameter = 0.2 in
Load w = 200 lb/ft
Solution:
We start by applying the equilibrium moment about point C, expressed as
∑M(c) = 0.............1
This can be used to find the force in AB.
10× 200 × ( 5) - (T cos(30)) × 10 = 0
Solving gives us
Tension in wire T(AB) = 1154.7 lb
We also know the modulus of elasticity for A992 is
E = 29000 ksi
And the area will be
Area =
The change in length is expressed as
Change in length = .........2
Substituting values results in
Change in length =
Change in length = 0.0913 in