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4 years ago

What tightening torque should be used for a hexagonal head screw (not split-bolt) on a 250 kcmil conductor?

Physics

1 answer:

Rudiy274 years ago

3 0

Answer:

325 inches-pounds

Explanation:

Tightening technique guide suggests that;

4/0 kcmil ---requires---→250 in-pounds tightening torque with 500 split-bolt

250kcmil ---requires---→325 in-pounds tightening torque with 650 split-bolt

300kcmil ---requires---→325 in-pounds tightening torque with 650 split-bolt

350kcmil ---requires---→325 in-pounds tightening torque with 650 split-bolt

400kcmil ---requires---→325 in-pounds tightening torque with 825 split-bolt

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In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle =100 . The chip thickness ratio

Rudik [331]

Answer:

The percentage of the total energy dissipated into shear plane is 89.46%.

Explanation:

Given that,

Rake angle = 10°

Thickness ratio= 0.5

Cutting Force = 400 N

Thrust force = 200 N

Speed =3 m/s

Suppose the shear force is 345.21 N.

We need to calculate the shear plane angle

Using formula shear angle

$\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}$

Put the value in to the formula

$\tan\phi=\dfrac{0.5\cos10}{1-0.5\sin10}$

$\tan\phi=0.539$

$\phi=\tan^{-1}(0.539)$

$\phi=28.32^{\circ}$

We need to calculate the shear velocity

Using formula of shear velocity

$v_{2}=\dfrac{v\cos\alpha}{\cos(\phi-\alpha)}$

Put the value into the formula

$v_{2}=\dfrac{3\times\cos10}{\cos(28.32-10)}$

$v_{2}=3.11\ m/s$

We need to calculate the percentage of the total energy dissipated into shear plane

Using formula of energy dissipated

$\%d=\dfrac{P_{s}}{P}\times100$

$\%d=\dfrac{F_{s}\times v_{c}}{F_{c}\times v}\times100$

Put the value into the formula

$\%d=\dfrac{345.21\times3.11}{400\times3}\times100$

$d=89.46\%$

Hence, The percentage of the total energy dissipated into shear plane is 89.46%.

6 0

3 years ago

A sled that has a mass of 8 kg is pulled at a 50 degree angle with a force of 20 N. The force of friction acting on the sled is

Alik [6]

1) Acceleration of the sled

The acceleration of the sled is given by the net force acting in the direction parallel to the incline. There are two forces acting along this direction: the component of the weight parallel to the ramp (downward) and the friction (upward). Therefore, the net force acting in this direction is

$F=mg sin \theta- F_f =(8 kg)(9.8 m/s^2)(sin 50^{\circ})-2.4 N=57.7 N$