contestada

A 42 mm thick plate made of low carbon steel is to be reduced to 34 mm in one pass in a rolling operation. As the thickness is reduced, the plate widens by 4%. The yield strength of the steel plate is 174 MPa and the tensile strength is 290 MPa. The entrance speed of the plate is 15 m/min. The roll radius is 325 mm and the rotational speed is 49 rev/min. Determine: i) the minimum required coefficient of friction that would make this rolling operation possible ii) the exit velocity of the plate iii) the forward slip

Respuesta :

Answer:

minimum required coefficient of friction  = 0.157

exit velocity = 17.8 m/min

forward slip  = 0.0947

Explanation:

given data

thick plate = 42 mm

reduced in one pass  = 34 mm

plate widens = 4%

yield strength of steel plate = 174 MPa

tensile strength = 290 MPa

entrance speed = 15 m/min

roll radius = 325 mm

rotational speed = 49 rev/min

solution

we know that maximum draft is express as here

d(max) = [tex]\mu ^2[/tex] × R    ............1

and here

d is = 42 - 34

d = 8 mm

so put value in equation 1 we  get

[tex]\mu ^2[/tex] = [tex]\frac{8}{325}[/tex]

[tex]\mu[/tex]  = 0.157

and

here plate wide by 4 percent so we can exit velocity as

t × w × v = t1 × w1 × v1   ..........................2

here w2 = 1.04 w

so put here value

42 × w × 15 = 34 × 1.04 w × v2

solve it we get

v2 = 17.8 m/min

and

now we get roll speed that is

v = π × r² × N   ...............3

v = π × 0.325² × 49

v = 16.26 m/min

so forward slip is

s = \frac{17.8 - 16.26}{16.26}

s = 0.0947

Otras preguntas