The steel step shaft shown in Fig. 4–7a is mounted in bearings at A and F.
A pulley is centered at C where a total radial force of 600 lbf is applied. Using singularity functions evaluate the shaft displacements at{1}/{2}-in increments. Assume the shaft is simply supported.
The reactions are found to be R_1= 360 \ lbf \ and \ R_2= 240 \ lbf.
Ignoring R_2 , using singularity functions, the moment equation is
M=360x-600\left(x-8\right) ^1 (1)
This is plotted in Fig. 4–7b.
For simplification, we will consider only the step at D. That is, we will assume section AB has the same diameter as BC and section EF has the same diameter as DE.
Since these sections are short and at the supports, the size reduction will not add much to the deformation. We will examine this simplification later. The second area moments for BC and DE are
l_{BC}=\frac{\pi }{64} =1.5^4=0.2485 \ in^4 i_{DE}=\frac{\pi }{64} =1.75^4=0.4604 \ in^4
A plot of {M}/{l}[ is shown in Fig. 4–7c. The values at points b and c, and the step change are
\left(\frac{M}{l} \right)_b=\frac{2760}{0.2485}=11 106.6 \ {lbf}/{in^3}\left(\frac{M}{l} \right)_c=\frac{2760}{0.4604}=5 994.8\ {lbf}/{in^3}
\Delta \left(\frac{M}{l} \right) =5 994.8-11 106.6=-5 111.8 \ {lbf}/{in^3}
The slopes for ab and cd, and the change are
m_{ab} =\frac{2760-2880}{0.2485\left(0.5\right) } =-965.8 \ {lbf}/{in^4}m_{cd} =\frac{-5 994.8}{11.5} =-521.3\ {lbf}/{in^4}
\Delta m=-521.3-\left(-965.8\right) =444.5 \ {lbf}/{in^4}
Dividing Eq. (1) by I_BC \ and \ , \ at \ x= 8.5 \ in , adding a step of −5111.8 \ {lbf}/{in^3} and a ramp of slope, adding a step of 444.5\ {lbf}/{in^4} , gives
\frac{M}{l} =1448.7x-2414.5\left\langle x-8\right\rangle ^1-5111.8\left\langle x-8.5\right\rangle ^0+444.5\left\langle x-8.5\right\rangle ^1 (2)
Integration gives
E\frac{dy}{dx} =724.35x^2-1207.3\left\langle x-8\right\rangle ^2-5111.8\left\langle x-8.5\right\rangle ^1+222.3\left\langle x-8.5\right\rangle ^2+C_1 (3)
Integrating again yields
Ey=241.5x^3-402.4\left\langle x-8\right\rangle ^3-2555.9\left\langle x-8.5\right\rangle ^2+74.08\left\langle x-8.5\right\rangle ^3+C_1x+C_2 (4)
At x = 0 \ , \ y = 0. This gives C_2= 0 (remember, singularity functions do not exist until
the argument is positive).
At x = 20 \in, \ y = 0 , and
0=241\left(20\right) ^3-402.4\left(20-8\right) ^3-2555.9\left(20-8.8\right) ^2+74.08\left(20-8.5\right) ^3+C_1\left(20\right)Solving, gives C_1= −50 565 \ {lbf}/{in^2}. Thus, Eq. (4) becomes, with E 30\left(10\right) ^6 \ psi ,
y=\frac{1}{30\left(10^6\right) } \left(241.5x^3-402.4\left\langle x-8\right\rangle^3-2555.9\left\langle x-8.5\right\rangle^2 +74.08\left\langle x-8.5\right\rangle^3-50565x \right) (5)
When using a spreadsheet, program the following equations:
y=\frac{1}{60\left(10^6\right) } \left(241.5x^3-50565x\right) \ 0\leq x\leq 8 \ in
y=\frac{1}{30\left(10^6\right) } \left[241.5x^3-402.4\left(x-8\right)^3- 50565x\right] \ 8\leq x\leq 8.5 \ in
y=\frac{1}{30\left(10^6\right) } \left[241.5x^3-402.4\left(x-8\right)^3- 2555.9\left(x-8.5\right)^2-74.08\left(x-8.5\right)^3-50565x\right] \ 8.5\leq x\leq 20 \ in
The following table results.
| x | y | x | y | x | y | x | y | x | y |
| 0 | 0.000000 | 4.5 | -0.006851 | 9 | -0.009335 | 13.5 | -0.007001 | 18 | -0.002377 |
| 0.5 | -0.000842 | 5 | -0.007421 | 9.5 | -0.009238 | 14 | -0.006571 | 18.5 | -0.001790 |
| 1 | -0.001677 | 5.5 | -0.007931 | 10 | -0.009096 | 14.5 | -0.006116 | 19 | -0.001197 |
| 1.5 | -0.002501 | 6 | -0.008374 | 10.5 | -0.008909 | 15 | -0.005636 | 19.5 | 0.000600 |
| 2 | -0.003307 | 6.5 | -0.008745 | 11 | -0.008682 | 15.5 | -0.005134 | 20 | 0.000000 |
| 2.5 | -0.004088 | 7 | -0.009037 | 11.5 | -0.008415 | 16 | -0.004613 | ||
| 3 | -0.004839 | 7.5 | -245.000000 | 12 | -0.008112 | 16.5 | -0.004075 | ||
| 3.5 | -0.005554 | 8 | -0.009362 | 12.5 | -0.007773 | 17 | -0.003521 | ||
| 4 | -0.006227 | 8.5 | -0.009385 | 13 | -0.007403 | 17.5 | -0.002954 |
where x and y are in inches. We see that the greatest deflection is at x = 8.5 \ in \ where \ y = −0.009385 in .
Substituting C_1 into Eq. (3) the slopes at the supports are found to be \theta _A=1.686\left(10^{-3} \right) \ rad = 0.09657 \ deg, \ and \ theta _F= 1.198(10^{−3}) \ rad = 0.06864 \ deg.
You might think these to be insignificant deflections, but as you will see in Chap. 7, on shafts, they are not.
A finite-element analysis was performed for the same model and resulted in
y|_{x=8.5 \ in}= −0.009380 \ in \ \theta _A= −0.09653◦ \ \theta _F= 0.06868◦◦
Virtually the same answer save some round-off error in the equations.
If the steps of the bearings were incorporated into the model, more equations result,but the process is the same.
The solution to this model is
y|_{x=8.5 \ in}= −0.009387 \ in \ \theta _A= −0.09763◦ \ \theta _F= 0.06873◦
The largest difference between the models is of the order of 1.5 percent.
Thus the simplification was justified.
