Respuesta :
Answer:
a. 20.265 MN
b. 0.555 MNm
c. 403.44 KW
Explanation:
Given:-
- The width ( w ) = 1000 mm
- Original thickness ( to ) = 10 mm
- Final thickness ( t ) = 5 mm
- The radius of the rollers ( R ) = 600 mm
- The peripheral speed of the roller ( v ) = 0.12
- Deformation efficiency ( ε ) = 55%
Find:-
a) the roller force ( F )
b) the roller torque ( T )
c) the performance on the pair of rollers. ( P )
Solution:-
- The process of flat rolling entails a pair of compressive forces ( F ) exerted by the rollers on the steel sheet that permanently deforms.
- The permanent deformation of sheet metal is seen as reduced thickness.
- We will assume that the compressive force ( F ) acts normal to the point of contact between rollers and metal sheet.
- The roll force ( F ) is defined as:
[tex]F =L*w*Y_a_v_g[/tex]
Where,
L: The projected length of strip under compression
Y_avg: The yielding stress of the material = 370 MPa
- The projected length of strip under compression is approximated by the following relation:
[tex]L = \sqrt{R*( t_o - t_f )} \\\\L = \sqrt{0.6*( 0.01 - 0.005 )} \\\\L = 0.05477 m[/tex]
- The Roll force ( F ) can be determined as follows:
[tex]F = (0.05477)*(1 )*(370*10^6 )\\\\F = 20.265 MN[/tex]
- The roll torque ( T ) is given by the following relation as follows:
[tex]T = \frac{L}{2} * F\\\\T = \frac{0.05477}{2} * 20.265\\\\T = 0.555 MNm[/tex]
- The rotational speed of the rollers ( N ) is determined by the following procedure:
[tex]f = \frac{v}{2\pi* R} = \frac{0.12}{2*\pi 0.6} = 0.03181818 \frac{rev}{s} \\\\N = f*60 = 1.9090 rpm[/tex]
- The power consumed by the pair of rollers ( P ) is given by:
[tex]P = \frac{2\pi * F * L * N}{e*60,000} KW \\\\P = \frac{2\pi * ( 20.265*10^6) * (0.05477) * (1.90909 ) }{60,000*0.55} KW\\\\P = 403.44 KW[/tex]
