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fredd [130]
3 years ago
8

In a turning operation, spindle speed is set to provide a cutting speed of 1.8 m/s. The feed and depth of cut are 0.30 mm and 2.

6 mm, respectively. The tool rake angle is 8°. After the cut, the deformed chip thickness is measured to be 0.49 mm.
Determine:
(a) shear plane angle, (b) shear strain, and (c) material removal rate.
Use the orthogonal cutting model as an approximation of the turning process.
Engineering
1 answer:
love history [14]3 years ago
6 0

Answer:

(a) shear plane angle (∅) = 33.53°

(b) shear strain (y) = 1.989

(c) material removal rate (MRR) = 1404mm³/s

Explanation:

Cutting speed = 1.8m/s

Feed = 0.3mm

Depth = 2.6mm

Angle = 8°

Thickness = 0.49

(a) calculating the chip thickness ratio using the formula;

r = t₀/tc

 = 0.3/0.49

 = 0.6122

Calculating the shear angle using the formula;

∅ = tan⁻¹[rcosα/1-rsinα]

  = tan⁻¹[(0.6122*cos8)/(1-0.6122sin8)]

  = 33.53°

(b) Calculating the shear strain using the formula;

y = cot∅ + tan(∅-α)

   = cot 33.53 + tan(33.53-8)

   = 1.509 + 0.477

  = 1.989

(c) Calculating the material removal rate using the formula;

MRR =f*d*V

        = 0.3 * 2.6 *1.8 *1000mm

        = 1404mm³/s

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For a copper-silver alloy of composition 25 wt% Ag-75 wt% Cu and at 775C (1425F) do the following: (a) Determine the mass frac
allochka39001 [22]

Answer:

a)

mass fraction of α = 0.796

mass fraction of β = 0.204

b)

mass fraction of primary α (Wα) = 0.734

mass fraction of  eutectic micro-constituents (We) = 0.266

c)

α in eutectic mixture = 0.062

Explanation:

Assumptions:

(i) the  system is in thermal equilibrium with its surroundings

(ii) There are no impurities or other alloying elements present

(a) Determine the mass fractions of α and β phases

From the Cu - Ag phase diagram, Cα = 8.0 wt% Ag, Cβ = 91.2 wt% Ag, and C0 = 25 wt% Ag .Using the lever-rule:

Total mass fraction of α = (Cβ - C0) / (Cβ - Cα) = (91.2 - 25) / (91.2 - 8) = 0.796

Total mass fraction of β = (C0 - Cα) / (Cβ - Cα) = (25 - 8) / (91.2 - 8) = 0.204

(b) Determine the mass fractions of primary α and eutectic microconstituents

Ceutetic = 71.9 wt.%Ag

mass fraction of primary α (Wα) = (Ceutetic - C0) / (Ceutetic - Cα) = (71.9 - 25) / (71.9 - 8) = 0.734

mass fraction of  eutectic micro-constituents (We) = (C0 - Cα) / (Ceutetic - Cα) = (25 - 8) / (71.9 - 8) = 0.266

(c) Determine the mass fraction of eutectic α.

From the eutetic reaction, L  ↔    α + β

Total α = Primary α + α in eutectic mixture

Therefore: α in eutectic mixture = Total α - Primary α = 0.796 - 0.734 = 0.062

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Answer:

V_t=6 = 32 m/s

Explanation:

The origin is at point 0 with the positive motion to the right  

The instantaneous acceleration is change of velocity measured at infinitesimal interval of time, so the expression for instantaneous acceleration is:  

a=dv/dt

From here we can express dv as:

dv = a dt

Replace a by 2t — 1

dv = (2t — 1) dt

Integrate both sides of equation  

\int\limits^v_a  {2t-1} \, dv

v=t

a=t_0

putting these value in integral

<em>v-v_0=(t^2-t)-(t_0^2-t_0)</em>

We know that v_0 = 2 at t_0 = 0, so we'll replace t_0 and v_0 by their values

v — 2 = (t^2 — t) — (0^2 — 0)

From here we can write the expression for v as:  

v_t=6=6^2-6+2                             (1)  

So the velocity at t = 6 s is:

v_t=6 = 32 m/s

V_t=6 = 32 m/s

In order to determine the total distance travelled, we must check how maw times the particle has changed its direction, i.e. how many times its speed was equal to zero  

To do that, we'll just replace v by 0 in expression (1)

0 = t^2 — t + 2

The roots of the quadratic equation are:

t_1/2=1±  √(1^2-4*2*1)/2

Since 1^2-4*2*1 < 0, the quadratic equation have no real roots, so we can say that the velocity is always positive, i.e. to the right  

Now that we have all the details, we can correctly draw the path of the particle  

We can see from the sketch that the total distance traveled is:  

s^T=Δs_0-1

s^T=| s_1 - s_0 |

Replace s_0 by its value  

s^T=| s_1 - 1 |                                        (2)  

In order to determine the position of particle at t = 6 s, we'll need to determine the expression for s as function of time  

Since we have already wrote expression for v as function of time (step 2), we'll use expression  to get the expression for s

v= ds/dt  

Multiply both sides of equation by dt

v dt = ds

Replace v by expression (1)

(t^2 — t + 2) dt = ds

Integrate both sides of equation  

\int\limits^t_b {x} \, dx

t=s

b=(s=0)

x=(t^2 — t + 2)

dx=ds

putting these value in integral

(t^3/3-t^2/2+t)-(t_0^3/3-t_0^2/2+t_0)= s-s_0

Since s = 1 m at t = 0, and we want to determine the position s at t = 6, we'll replace so by 1, t_0 by 0 and t by 6  

(6^3/3-6^2/2+6)-(0^3/3-0^2/2+0)=s_t=6-1

4 0
4 years ago
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