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2 years ago

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The steel shaft has a radius of 15 mm. Determine the torque Tin the shaft if the two strain gages, attached to the surface of th

e shaft, report strains of elementof_x = -80(10^-6) and elementof_y = 80(10^-6). Also, determine the strains acting in the x and y directions. E_st = 200 GPa. v_st = 0.3. The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x' and y direction a torque T = 2 kN middot m is applied to the shaft.

1 answer:

GalinKa [24]2 years ago

7 0

Answer:

Part A.

T = 65.248 N-m

Part B.

εx' = 0.00245

εy' = - 0.00245

Explanation:

Part A.

Given

R = 15 mm = 0.015 m

εx' = - 80*10⁻⁶

εy' = 80*10⁻⁶

Est = E = 200 GPa

υst = υ = 0.3

T = ?

In order to understand the question we can see the pic shown.

Pure shear.

εx = εy = 0

We apply the formula

εx' = εx*Cos²θ + εy*Sin²θ + γxy*Sin θ*Cos θ

If

θ = 45° we have

- 80*10⁻⁶ = 0 + 0 + γxy*Sin 45*Cos 45°

⇒ γxy = - 160*10⁻⁶

Also,

θ = 135° we have

80*10⁻⁶ = 0 + 0 + γxy*Sin 135*Cos 135°

⇒ γxy = - 160*10⁻⁶

Then we get G as follows:

G = E/(2*(1 + υ))

⇒ G = 200 GPa/(2*(1 + 0.3)) = 76.923 GPa = 76.923*10⁹Pa

we can use the equation

τ = G*γxy ⇒ τ = (76.923*10⁹Pa)(160*10⁻⁶) = 12.308*10⁶Pa

Finally, we can obtain T as follows

T = τ*Jz/R

where

Jz = π*R⁴/2 ⇒ Jz = π*(0.015 m)⁴/2 = 7.95*10⁻⁸m⁴

⇒ T = 12.308*10⁶Pa*7.95*10⁻⁸m⁴/0.015 m

⇒ T = 65.248 N-m

Part B.

The shaft has a radius of 15 mm and is made of L2 tool steel. Determine the strains in the x' and y' direction if a torque T = 2 kN-m is applied to the shaft.

Given:

R = 15 mm = 0.015 m

Jz = 7.95*10⁻⁸m⁴

E = 200 GPa

υ = 0.3

G = 76.923*10⁹Pa

T = 2 kN-m = 2000 N-m

We can apply the equation

T = τ*Jz/R ⇒ τ = T*R/Jz

⇒ τ = 2000 N-m*(0.015 m)/(7.95*10⁻⁸m⁴)

⇒ τ = 3.77*10⁸ Pa

We use the formula

τ = G*γxy ⇒ γxy = τ/G

⇒ γxy = 3.77*10⁸ Pa/76.923*10⁹Pa

⇒ γxy = 0.0049

Now, we apply the equations

εx' = εx*Cos²θ + εy*Sin²θ + γxy*Sin θ*Cos θ

If

θ = 45° we have

εx' = 0 + 0 + 0.0049*Sin 45°*Cos 45°

⇒ εx' = 0.00245

If

θ = 135° we have

εy' = 0 + 0 + 0.0049*Sin 135°*Cos 135°

⇒ εy' = - 0.00245

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1) Which of the following statements are true?

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Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

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$g$ - Gravitational acceleration, measured in meters per square second.

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The maximum height of the ball is 1.417 meters.

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2 years ago