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# TORQUE CALCULATION ON A CIRCULAR DISC

A disc of dia 300 mm and thickness of 5 mm made of mild steel.It has to be rotated by a motor at a rpm of 100.The motor shaft is connected to the disc by a gear mechanism.The outer periphery of the disc is having a profile of gear teeth.Now my question is,

a)What is the required torque to rotate the disc?

b)What will be the calculation to determine the horse power of the said motor?

a)What is the required torque to rotate the disc?

b)What will be the calculation to determine the horse power of the said motor?

Dear,

According to me find the following calculations.

Diameter(d)=300 mm

Thick(t)= 5 mm

RPM(N)=100

Let, Y.S. of mild steel(Y)=250 N/mm^2

Now ,Allowable Shear Stress(S)=Y*0.5 =250*0.5 =125 N/mm^2

Let, F.O.S.(F)=2.5

Therefore, Allowable Shear Stress=S/F =125/2.5 = 50 N/mm^2

Also, Let Polar moment of Inertia is "J" and the Radius of shaft "R"(also the maximum distance from center)

Now, the formula for torque is,

Torque(T)= (J*S)/R

Now, J= (3.14*d^4)/32

= (3.14*300^4)/32

= 794812500 mm^4

Now, substitute all the values,

T= (794812500*50)/150

= 264937500 N-mm

Now, The formula for power required,

Power(P) = T*W

where, W= angular velocity

Now, W= (2*3.14*N)/60

= (2*3.14*100)/60

= 10.5 RPS

now substitute the values,

P= (265*10.5)

= 2774 KN-m/s

Kindly note, If you don't consider the calculations correct then please don't use them.

According to me find the following calculations.

Diameter(d)=300 mm

Thick(t)= 5 mm

RPM(N)=100

Let, Y.S. of mild steel(Y)=250 N/mm^2

Now ,Allowable Shear Stress(S)=Y*0.5 =250*0.5 =125 N/mm^2

Let, F.O.S.(F)=2.5

Therefore, Allowable Shear Stress=S/F =125/2.5 = 50 N/mm^2

Also, Let Polar moment of Inertia is "J" and the Radius of shaft "R"(also the maximum distance from center)

Now, the formula for torque is,

Torque(T)= (J*S)/R

Now, J= (3.14*d^4)/32

= (3.14*300^4)/32

= 794812500 mm^4

Now, substitute all the values,

T= (794812500*50)/150

= 264937500 N-mm

**T = 265 KN-M**

Now, The formula for power required,

Power(P) = T*W

where, W= angular velocity

Now, W= (2*3.14*N)/60

= (2*3.14*100)/60

= 10.5 RPS

now substitute the values,

P= (265*10.5)

= 2774 KN-m/s

**P= 2774 KW**Kindly note, If you don't consider the calculations correct then please don't use them.

---Sir,PRAD CHAKRABORTYA disc of dia 300 mm and thickness of 5 mm made of mild steel.It has to be rotated by a motor at a rpm of 100.The motor shaft is connected to the disc by a gear mechanism.The outer periphery of the disc is having a profile of gear teeth.Now my question is,

a)What is the required torque to rotate the disc?

b)What will be the calculation to determine the horse power of the said motor?

Please let me know how 150 has come while calculating the torque

it might be the radius of the disc?

PRAD CHAKRABORTYit might be the radius of the disc?

Dear,PRAD CHAKRABORTY---Sir,

Please let me know how 150 has come while calculating the torque

150 is the radius(R) of disc.

As your diameter is 300 mm, hence the radius is 150 mm.

Also, Note that the value of the torque(265 KN-m) i get is actually the maximum torque you can apply on this particular disc and the value of power(2774 KW) i get the power required to produce 265 KN-m torque or in other words you can also say that 2774 KW power can be safely transmitted through this disc with some obvious deflection(by safely i means the shaft will not break but can deflect). SO, If you want the torque required to start the rotation then the calculations will be different.

ok si

rPRAD CHAKRABORTYit might be the radius of the disc?