Question 10.4: Specify a suitable material for the shaft shown in Figure 10...

Objective    Specify a suitable material for the shaft.

Given          Shaft and loading in Figure 10–21

Power = P = 3.75 kW; rotational speed = n = 120 rpm

Shaft diameter = D = 55 mm

Sled-runner key seats at B and C

Simple supports at A and D

Analysis   The several steps to be used to solve this problem are outlined here:

1. The torque in the shaft will be computed for the known power and rotational speed from T = P/n, as developed in Chapter 4.

2. The tensions in the chains for sprockets B and C will be computed. These are the forces that produce bending in the shaft.

3. Considering the shaft to be a beam, the shearing force and bending moment diagrams will be drawn for it.

4. At the section where the maximum bending moment occurs, the equivalent torque T_{e} will be computed from Equation (10–4).

T_{e} = \sqrt{M^{2} + T^{2}}

5. The polar section modulus Z_{p} and the stress concentration factor Kt will be determined.

6. The maximum shear stress will be computed from Equation (10–7).

\tau_{max} = \frac{T_{e}}{K_{t}}{Z_{p}}

7. The required yield strength of the shaft material will be computed by letting τ_{max} = \tau_{d} in  Equation (10–6) and solving for s_{y} . Remember, let N = 4 or more.

\tau_{d} = \frac{s_y}{2N}

8. A steel that has a sufficient yield strength will be selected from Appendix A–11.

Results       Step 1. The desirable unit for torque is N · m. Then it is most convenient to observe that the power unit of kilowatts is equivalent to the units of kN · m/s. Also, rotational speed must be expressed in rad/s:

n = \frac{120  rev}{min} \times \frac{2 \pi  rad}{rev} \times \frac{1  min}{60  s} = 12.57 rad/s

We can now compute torque:

T = \frac{P}{n} =  \frac{3.75  kN·m}{s} \times \frac{1}{12.57  rad/s} = 0.298 kN·m

Step 2. The tensions in the chains are indicated in Figure 10–21 by the forces F_{1} and F_{2} .
For the shaft to be in equilibrium, the torque on both sprockets must be the same in magnitude but opposite in direction. On either sprocket, the torque is the product of the chain force times the radius of the pulley. That is,

 T = F_{1}R_{1} = F_{2}R_{2}

The forces can now be computed:

F_{1} = \frac{T}{R_{1}} = \frac{0.298  kN·m}{75  mm} \times \frac{10^{3}  mm}{m} = 3.97 kN

F_{2} = \frac{T}{R_{2}} = \frac{0.298  kN·m}{50  mm} \times \frac{10^{3}  mm}{m} = 5.96 kN

Step 3. Figure 10–24 shows the complete shearing force and bending moment diagrams found by the methods of Chapter 5. The maximum bending moment is 1.06 kN · m at section B, where one of the sprockets is located.

Step 4. At section B, the torque in the shaft is 0.298 kN · m and the bending moment is 1.06 kN · m. Then,

T_{e} = \sqrt{M^{2}+T^{2}} = \sqrt{(1.06)^{2} + (0.298)^{2}} = 1.10 kN · m

Step 5. Z_{p} = \frac{\pi D^{3}}{16} = \frac{ \pi (55  mm)^{3}}{16} = 32.67 ×10³  mm³

For the key seat at section B securing the sprocket to the shaft, we will use K_{t} reported in Figure 4–16.

Step 6. \tau_{max} = \frac{T_{e}K_{t}}{Z_{p}} = \frac{1.10 \times 10^{3} N · m(1.6)}{32.67 \times 10^{3}  mm^{3}} \times \frac{10^{3}  mm}{m} 53.9 MPa

Step 7. Let \tau_{max} = \tau_{d} = \frac{s_{y}}{2N}

Then s_{y} = 2Nτ_{max} = (2)(4)(53.9 MPa) = 431 MPa

Step 8. Referring to Appendix A–11, we find that several alloys could be used. For example, SAE 1040, cold-drawn steel, has a yield strength of 490 MPa. Alloy SAE 1141 OQT 1300 has a yield strength of 469 MPa and also a very good ductility, as indicated by the 28% elongation. Either of these would be a reasonable choice.