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3 years ago

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The driveshaft of an automobile is being designed to transmit 134 hp at 2900 rpm. Determine the minimum diameter d required for

a solid steel shaft if the allowable shear stress in the shaft is not to exceed 6100 psi.

1 answer:

Misha Larkins [42]3 years ago

4 0

Answer:

The minimum diameter is 1.344 in

Explanation:

The angular speed of the driveshaft is equal to:

$w=\frac{2\pi N}{60}$

Where

N = rotational speed of the driveshaft = 2900 rpm

$w=\frac{2\pi *2900}{60} =303.69rad/s$

The torque in the driveshaft is equal to:

$\tau=\frac{P}{w}$

Where

P = power transmitted by the driveshaft = 134 hp = 73700 lb*ft/s

$\tau=\frac{73700}{303.69} =242.68lb*ft$

The minimum diameter is equal to:

$d_{min} =(\frac{16T}{\pi *\tau } )^{1/3}$

Where

T = shear stress = 6100 psi

τ = 242.68 lb*ft = 2912.16 lb*in

$d_{min} =(\frac{16*2912.16}{\pi *6100} )^{1/3} =1.344in$

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A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w

Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( $\frac{M1}{100}$ )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

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3 years ago

Many manufacturers currently offer SAE level _________ equipped vehicles

Pepsi [2]

Answer should be B.3 hopefully

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3 years ago

In 1945, the United States tested the world’s first atomic bomb in what was called the Trinity test. Following the test, images

Zarrin [17]

Answer:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

$A= \frac{r^5 \rho}{t^2}$

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

Explanation:

Notation

In order to do the dimensional analysis we need to take in count that we need to conditions:

a) The energy A is released in a small place

b) The shock follows a spherical pattern

We can assume that the size of the explosion r is a function of the time t, and depends of A (energy), the time (t) and the density of the air is constant $\rho_{air}$ .

And now we can solve the dimensional problem. We assume that L is for the distance T for the time and M for the mass.

[r]=L with r representing the radius

[A]= $\frac{ML^2}{T^2}$ A represent the energy and is defined as the mass times the velocity square, and the velocity is defined as $\frac{L}{T}$

[t]=T represent the time

$[\rho]=\frac{M}{L^3}$ represent the density.

Solution to the problem

And if we analyze the function for r we got this:

$[r]=L=[A]^x [\rho]^y [t]^z$

And if we replpace the formulas for each on we got:

$[r]=L =(\frac{ML^2}{T^2})^x (\frac{M}{L^3})^y (T)^z$

And using algebra properties we can express this like that:

$[r]=L=M^{x+y} L^{2x-3y} T^{-2x+z}$

And on this case we can use the exponents to solve the values of x, y and z. We have the following system.

$x+y =0 , 2x-3y=1, -2x+z=0$

We can solve for x like this x=-y and replacing into quation 2 we got:

$2(-y)-3y = 1$

$-5y = 1$

$y= -\frac{1}{5}$

And then we can solve for x and we got:

$x = -y = -(-\frac{1}{5})=\frac{1}{5}$

And if we solve for z we got:

$z=2x =2 \frac{1}{5}=\frac{2}{5}$

And now we can express the radius in terms of the dimensional analysis like this:

$r=K A^{1/5} \rho^{-1/5} t^{2/5}$

And K represent a constant in order to make the porportional relation and equality.

The problem says that we can assume the constant K=1.

And if we solve for the energy we got:

$A^{1/5}=\frac{r}{t^{2/5} \rho^{-1/5}}$

$A= \frac{r^5 \rho}{t^2}$

And now we can replace the values given. On this case t =0.025 s, the radius r =140 m, and the density is a constant assumed $\rho =1.2 kg/m^2$ , and replacing we got:

$A=\frac{140^5 1.2 kg/m^3}{(0.025 s)^2}=1.033x10^{14} \frac{kg m^2}{s^2}$

And we can convert this into ergs we got:

$A= 1.033x10^{14} \frac{kgm^2}{s^2} * \frac{1 x10^7 egrs}{1 \frac{kgm^2}{s^2}}=1.033x10^{21} ergs$

And then we know that 1 g of TNT have $4x10^4 erg$

And we got:

$A=1.033x10^{21} ergs *\frac{Kg TNT}{4x10^{10} erg}=2.58x10^{10} Kg TNT$

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3 years ago

For some metal alloy it is known that the kinetics of recrystallization obey the Avrami equation, and that the value of k in the

JulsSmile [24]

Answer:

t = 1456.8 sec

Explanation:

given data:

contant k = 2.60*10^{-6}

rate of crystallization is 0.0013 s-1

rate of transformation is given by

$r = \frac{1}{t_0.5}$

use specifies value to solve $t_0.5$

it is ime required for 50% tranformation

$r = \frac{1}{.0013}=769.2 sec$

Avrami equation is given by

$y = 1 - e^{-kt^n}$

$0.5 = 1 - e^{-kt_0.5^n}$

$1-0.5 = e^{-kt_0.5^n}$

$ln (1 - 0.5) = -kt_0.5^n$

$ln \frac{ln (1 - 0.5)}{-k} = nln t_0.5$

$n = \frac{ ln \frac{ln (1 - 0.5)}{-k}}{ln t_0.5}$

$n = \frac{ ln \frac{ln (1 - 0.5)}{-2.60*10^{-6}}}{ln 769.2}$

n = 1.88

second degree of recrystalization may be determine by rearranging original avrami equation

$t = [\frac{-ln(1-y)}{k}]^{1/n}$

for 90%completion

$t = [\frac{-ln(1-0.9)}{2.60*10^{-6}}]^{1/1.88}$

t = 1456.8 sec

5 0

3 years ago

A wood pipe having an inner diameter of 3 ft. is bound together using steel hoops having a cross sectional area of 0.2 in^2. The

Minchanka [31]

Answers:

31.7 inches

Explanation:

Given:

Diameter = 3ft

Let D = Diameter

So, D = 3ft. (Convert to inches)

D = 3 * 12in = 36 inches

Coss-sectional area of the steel = 0.2in²

Gauge Pressure (P) = 4psi

Stress in Steel (σ)= 11.4ksi

Force in steel = ½ (Pressure * Projected Area)

Area (A) = 2 * Force/Pressure

Also, Area (A) = Spacing (S) * Wood Pipe Diameter

Area = Area

2*Force/Pressure = Spacing * Diameter

Substitute values I to the above expression

2 * Force / 4psi = S * 36 inches

Also

Force in steel (F) = Stress in steel (σ) × Cross-sectional area of the steel

So, F = 11.4ksi * 0.2in²

F = 11.4 * 10³psi * 0.2in²

F = 2.28 * 10³ psi.in²

So, 2 * Force / 4psi = S * 36 becomes

2 * 2.28 * 10³/4 = S * 36

S = 2 * 2.28 * 10³ / (4 * 36)

S = 4560/144

S = 31.66667 inches

S = 31.7 inches (approximated)

5 0

4 years ago