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dexar [7]
4 years ago
6

The shaft is hollow from A to B and solid from B to C. The shaft has an outer diameter of 79 mm, and the thickness of the wall o

f the hallow segment is 10 mm.
Determine the maximum shear stress developed in the shaft. Express your answer to three significant figures and include appropriate units.

Engineering
1 answer:
SVETLANKA909090 [29]4 years ago
7 0

Answer:

<u><em>Maximum Shear Stress is 62.0 MPA </em></u>

Explanation:

<h2><u /></h2><h3><u>Please refer to the attachment below for explanation.</u></h3>

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Gear A has a mass of 1 kg and a radius of gyration of 30 mm; gear B has a mass of 4 kg and a radius of gyration of 75 mm; gear C
Kruka [31]

Answer:

(4.5125 * 10^-3 kg.m^2)ω_A^2

Explanation:

solution:

Moments of inertia:

I = mk^2

Gear A: I_A = (1)(0.030 m)^2 = 0.9*10^-3 kg.m^2

Gear B: I_B = (4)(0.075 m)^2 = 22.5*10^-3 kg.m^2

Gear C: I_C = (9)(0.100 m)^2 = 90*10^-3 kg.m^2

Let r_A be the radius of gear A, r_1 the outer radius of gear B, r_2 the inner radius of gear B, and r_C the radius of gear C.

r_A=50 mm

r_1 =100 mm

r_2 =50 mm

r_C=150 mm

At the contact point between gears A and B,

r_1*ω_b = r_A*ω_A

ω_b = r_A/r_1*ω_A

       = 0.5ω_A

At the contact point between gear B and C.

At the contact point between gears A and B,

r_C*ω_C = r_2*ω_B

ω_C = r_2/r_C*ω_B

       = 0.1667ω_A

kinetic energy T = 1/2*I_A*ω_A^2+1/2*I_B*ω_B^2+1/2*I_C*ω_C^2

                           =(4.5125 * 10^-3 kg.m^2)ω_A^2

6 0
4 years ago
A sample of cast iron with .35 wt%C is slow cooled from 1500C to room temperature. What is the fraction of proeutectoid Cementit
Marizza181 [45]

Answer:

So the fraction of proeutectoid cementite is 44.3%

Explanation:

Given that

cast iron with 0.35 % wtC and we have to find out fraction of proeutectoid cementite phase when cooled from 1500 C to room temperature.

We know that

Fraction of  proeutectoid cementite phase gievn as

{w_p}'=\dfrac{{C_o}'-0.022}{0.74}

Now by putting the values

{w_p}'=\dfrac{0.35-0.022}{0.74}

{w_p}'=0.443

So the fraction of proeutectoid cementite is 44.3%

6 0
3 years ago
The constant A in Equation 17.2 is 12π4 R/5(θD) 3 where R is the gas constant and θD is the Debye temperature (K). Estimate θD f
kirill115 [55]

Answer:

The Debye temperature for aluminum is 375.2361 K

Explanation:

Molecular weight of aluminum=26.98 g/mol

T=15 K

The mathematical equation for the specific heat and the absolute temperature is:

C_{v} =AT^{3}

Substituting in the expression of the question:

C_{v} =(\frac{12\pi ^{4}R }{5\theta _{D}^{3}  } )T^{3}

\theta _{D} =(\frac{12\pi ^{4}RT^{3}  }{5C_{v}   } ) ^{1/3}

Here

C_{v} =4.6\frac{J}{kg-K} *\frac{1kg}{1000g} *\frac{26.98g}{1mol} =0.1241J/mol-K

Replacing:

\theta _{D}  =(\frac{12\pi ^{4}*8.31*15^{3}  }{5*0.1241} )^{1/3} =375.2361K

3 0
3 years ago
A 8.0-m-diameter parachute of a new design is to be used to transport a load from flight altitude to the ground with an average
S_A_V [24]

Answer:

drag coefficient for the parachute is Cd = 1.84177

Explanation:

given data

diameter = 8.0 m

average vertical speed = 3 m/s

total weight of load = 200 N

to find out

drag coefficient for the parachute

solution

we will apply here drag coefficient formula that is express as here  

drag coefficient Cd = \frac{2w}{\rho v^2A}    ......................1

here w is total load and A is area and v is speed

and here ρ = 1.22 kg/cu

put here value we get  

Cd = \frac{2*4*200}{1.229*5^2*\pi * 3^2}  

Cd = 1.84177

drag coefficient for the parachute is Cd = 1.84177

3 0
3 years ago
Free pts was my day I was a dog but it would need to be
Ne4ueva [31]

Answer:

hi

Explanation:

8 0
3 years ago
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