## Chapter 7

## Q. 7.10

Design of a Rotating Shaft

A solid shaft for a cement kiln produced from the tool steel in Figure 7-18 must be 96 in. long and must survive continuous operation for one year with an applied load of 12,500 lb. The shaft makes one revolution per minute during operation. Design a shaft that will satisfy these requirements.

## Step-by-Step

## Verified Solution

The fatigue life required for our design is the total number of cycles N that the shaft will experience in one year:

N = (1 cycle/min)(60 min/h)(24 h/day)(365 days/yr)

N = 5.256 × 10^5 cycles/yr

From Figure 7-18, the applied stress therefore, must be less than about 72,000 psi. Using Equation 7-13, the diameter of the shaft is given by

\pm \sigma = \frac{32 F L}{\pi d^3} (7-13)

72,000 \ psi = \frac{(32) (12,500 \ lb)(96 \ in.)}{\pi d^3}

d = 5.54 in.

A shaft with a diameter of 5.54 in. should operate for one year under these conditions; however, a significant margin of safety probably should be incorporated in the design. In addition, we might consider producing a shaft that would never fail.

Let us assume the factor of safety to be 2 (i.e., we will assume that the maximum allowed stress level will be 72,000/2 = 36,000 psi). The minimum diameter required to prevent failure would now be

36,000 \ psi = \frac{32 (12,500 \ lb)(96 \ in.)}{\pi d^3}

d = 6.98 in.

Selection of a larger shaft reduces the stress level and makes fatigue less likely to occur or delays the failure. Other considerations might, of course, be important. High temperatures and corrosive conditions are inherent in producing cement. If the shaft is heated or attacked by the corrosive environment, fatigue is accelerated. Thus, for applications involving fatigue of components, regular inspections of the components go a long way toward avoiding a catastrophic failure.