# Homework Help: A solid steel shaft is subjected to a torque of 45 KN/m.

1. Jun 28, 2012

### manal950

A solid steel shaft is subjected to a torque of 45 KN/m. If the angle of twist is 0.5 degree per meter length of the shaft and the shear stress is not exceed 90 MN/m^2
(I) find the suitable diameter of the shaft . Take C = 80 GN/m^2
(II) Maximum shear strain

we can solve the question by torsion equation
T/Ip = e/R = CQ/l

T maximum twisting torque in Nm
Ip polar moment of inertia = pid^4/32 in m^4
e shear stress in N/m^2
Q the angle of the shaft
l length if the shaft in m

my question know why in solving fined two diameter ?

2. Jun 28, 2012

### rock.freak667

Well from your formula, which two should you equate with what you are given in order to get the diameter? (You know one quantity as one of the terms must be related to the diameter).

3. Jun 29, 2012

### manal950

yes why ?

and here I have the solution

Diameter on the basic of twist
= 0.16 m

and then diameter on the basis of shear stress :
D = 0.1365 m

Now my question why in answer give us two diameter ?

and which one is correct ?

4. Jun 29, 2012

### rock.freak667

Using shear stress, you got d=0.1365 m, so if you put d = 0.16 m will your shaft be able to withstand the shear stress or will it fail?

(Hint: You don't want your shaft to fail)

5. Jun 29, 2012

### manal950

How I can know if the shaft It will withstand the shear stress or not ... ?

for finding the second diameter is depend on first diameter ?

6. Jun 29, 2012

### rock.freak667

At the maximum shear stress value the diameter is 0.1356 m. (Meaning that this is the minimum diameter the shaft should be)

For the angle of twist, the diameter is 0.16 m. Since you want your shaft to be subjected to the specified angle of twist, then d = 0.16 m will be sufficient.

7. Jun 29, 2012

### manal950

thanks so much ..

8. Jun 29, 2012

### rock.freak667

In essence, what you did was to get the diameter for the angle of twist.

Then found the minimum diameter of the shaft to prove the first diameter would not exceed the shear stress.

Alternatively, you could have used the first diameter (found using the angle of twist) to get the shear stress using

$$\frac{T}{I_p} = \frac{e}{R}$$

9. Jun 29, 2012

### manal950

thanks .. I forget ... how I can figure out the equation from above equation to solve for maximum strain ?? there is no simple for strain ?? ...

10. Jun 29, 2012

### rock.freak667

How is shear stress and shear strain related to the modulus of rigidity (your value for C)?

11. Jun 29, 2012

### manal950

module of rigidity = steers / strain .

12. Jun 29, 2012

### rock.freak667

so from e/R = CQ/l, what do you get when you replace e with C*strain ?