Question 18.CS.5: Gearbox Shafting Design Figure 18.5 shows the input shaft of...
Gearbox Shafting Design
Figure 18.5 shows the input shaft of the crane gearbox, supported in the gearbox by bearings A and B and driven by electric motor. Determine
a. The factor of safety n for the shaft using the maximum energy of distortion theory incorporated with the Goodman criterion
b. The rotational displacements or slopes at the bearings
c. The stresses in the shaft key
Given: The geometry and dimensions ofthe hollow shaft and square shaft key are known.
Data: Refer to Figure 18.5a and Case Study 18.4.
F_t=206 N , \quad F_r=75 N , \quad T=2.06 N \cdot m , \\ a=66 mm , \quad b=84 mm , \quad L=150 N \cdot m , \\ d=6 mm , \quad d_p=20 mm , \quad D=12 mm \\ w=2.4 mm , \quad L_k=25 mm , \\ I=\frac{\pi}{64}\left(D^4-d^4\right)=954.3 mm ^4
The operating environment is room air at a maximum temperature of 50°C.
Assumption: Bearings act as simple supports.
Design Decisions:
1. The shaft and shaft key are made of 1030 CD steel with machined surfaces:
S_u=520 MPa , \quad S_y =440 MPa (from Table B.3)
E =210 GPa
2. At the keyway, K_{f} = 2.
3. The shaft rotates and carries steady loading at normal temperature.
4. The factor of safety is n = 3 against shear of shaft key.
5. A survival rate of 99.9% is used.
See Figure 18.5; Table A.9, Section 9.5.
a. The reactions at A and B, as determined by the conditions of equilibrium, are indicated in Figure 18.5b. The moment and torque diagrams are obtained in the usual manner and drawn in Figure 18.5c. Observe that the critical section is at point C. We have
\begin{array}{c} M_C=\left[(2.77)^2+(7.62)^2\right]^{1 / 2}=8.11 N \cdot m \\ T_C=2.06 N \cdot m \end{array}
The mean and alternating moments and torques, by Equation 9.10, are then
\begin{array}{c} M_m=1 / 2\left(M_{\max }+M_{\min }\right) \\ T_m=1 / 2\left(T_{\max }+T_{\min }\right) \end{array} and \begin{array}{c} M_a=1 / 2\left(M_{\max }-M_{\min }\right) \\ T_a=1 / 2\left(T_{\max }-T_{\min }\right) \end{array} (9.10)
\begin{array}{c} M_m=0 \quad M_a=8.11 N \cdot m \\ T_m=2.06 N \cdot m \quad T_a=0 \end{array}
The modified endurance limit, through the use of Equations 7.1 and 7.6 and
referring to Section 7.7, is
S_e^{\prime}=0.5 S_u (7.1)
S_e=C_f C_r C_s C_t\left(1 / K_f\right) S_e^{\prime} (7.6)
C_f=A S_u^b (7.7)
where
C_f=4.51\left(520^{-0.265}\right)=0.86
C_{r} = 0.76 (by Table 7.3)
C_{s} ≈ 0.85 (using Equation 7.9)
C_{1} = 1
K_{f} = 2
S_e^{\prime} = 0.5(520) = 260 MPa ( from Equation 7.1 )
C_s=\left\{\begin{array}{ll} 0.85 \quad (13 mm <D \leq 50 mm ) \quad \left\lgroup \frac{1}{2}<D \leq 2 in . \right\rgroup \\ 0.70 \quad(D>50 mm ) \quad(D>2 in .) \end{array}\right. (7.9)
Hence,
S_e=(0.86)(0.75)(0.85)(1)\left\lgroup \frac{1}{2} \right\rgroup(260)=71.27 MPa
Since the loading is steady, the shock factors K_{sb} = K_{st} = 1 by Table 9.1.
Substituting the numerical values into Equation 9.12 and replacing D³ with D^3\left[1-(d / D)^4\right] , we obtain
\frac{S_u}{n}=\frac{32}{\pi D^3}\left[\left\lgroup M_m+\frac{S_u}{S_e} M_a \right\rgroup^2+\frac{3}{4}\left\lgroup T_m+\frac{S_u}{S_e} T_a \right\rgroup^2\right]^{1 / 2} (9.12)
\begin{aligned} \frac{520\left(10^6\right)}{n} &=\frac{32}{\pi(0.012)^3\left[1-(6 / 12)^4\right]} \\ & \times\left[\left\lgroup 0+\frac{520 \times 8.11}{71.27} \right\rgroup^2+\frac{3}{4}(2.06)^2\right]^{1 / 2} \end{aligned}
from which
n = 1.4
b. The slopes at ends A and B (Figure 18.5b) are given by Case 6 of Table A.9. Note that L^2-b^2=(L+b)(L-b)=(L+b) a and similarity L^2-a^2=(L+a) b . Introducing the given data, the results are
\begin{aligned} \theta_A &=-\frac{F_t a b(L+b)}{6 E I L} \\ &=-\frac{206(66)(84)(150+84)}{6\left(210 \times 10^3\right)(954.3)(150)} \\ &=-1.482\left(10^{-3}\right) rad =-0.085^{\circ} \end{aligned}
\begin{aligned} \theta_B &=\frac{F_t a b(L+a)}{6 E I L} \\ &=\frac{206(66)(84)(150+66)}{6\left(210 \times 10^3\right)(954.3)(150)} \\ &=1.368\left(10^{-3}\right) rad =0.078^{\circ} \end{aligned}
where a minus sign means a clockwise rotation.
Comments: Inasmuch as the bearing and gear stiffnesses are ignored, the negligibly small values of \theta _{A} and \theta _{B} estimated by the preceding equations represent higher angles than the true slopes. Therefore, self-aligning bearings are not necessary.
c. The compressive forces acting on the sides of the shaft key equal F_{1} = 206 N (Figure 18.5). The shear stress in the shaft key is
\tau=\frac{F_t}{w L_k}=\frac{206}{(0.0024)(0.025)}=3.433 MPa
We have, from Equation 6.20,
\text {Maximum shearing stress theory:} S_{yx} = 0.50S_{y} \\ \text {Energy of distortion theory, or its equivalent,} \\ \text {the octahedral shear stress theory:} S_{yx} = 0.577S_{y} (6.20)
S_{y s}=0.577 S_y=(0.577)(440)=253.9 MPa
The allowable shear stress in the shaft key is
\tau_{\text {all }}=\frac{S_{y s}}{n}=\frac{253.9}{3}=84.63 MPa
Since \tau _{all} ≫ \tau, shear should not occur at shaft key. We obtain the same result on the basis of compression or bearing on key (see Section 9.9).
Comments: On following a procedure similar to that in the preceding solution, the design of the remaining three shafts and the associated keys in the gearbox of the winch crane can be analyzed in a like manner.
| Table 7.3 Reliability Factors |
|
| Survival Rate (%) | C_{r} |
| 50 | 1.00 |
| 90 | 0.89 |
| 95 | 0.87 |
| 98 | 0.84 |
| 99 | 0.81 |
| 99.9 | 0.75 |
| 99.99 | 0.70 |
| Table 9.1 Shock Factors in Bending and Torsion |
|
| Nature of Loading | K_{sb}, K_{st} |
| Gradually applied or steady | l.0 |
| Minor shocks | l.5 |
| Heavy shocks | 2 |