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.

Question

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[latex]{1}/{2}[/latex]-in increments. Assume the shaft is simply supported.

Accepted Answer

The reactions are found to be [latex]R_1= 360 \ lbf \ and \ R_2= 240 \ lbf[/latex].

Ignoring [latex]R_2[/latex] , using singularity functions, the moment equation is

[latex]M=360x-600\left(x-8 ight) ^1[/latex] (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

[latex]l_{BC}=\frac{\pi }{64} =1.5^4=0.2485 \ in^4[/latex] [latex]i_{DE}=\frac{\pi }{64} =1.75^4=0.4604 \ in^4[/latex]

A plot of [latex]{M}/{l}[[/latex] is shown in Fig. 4–7c. The values at points b and c, and the step change are

[latex]\left(\frac{M}{l} ight)_b=\frac{2760}{0.2485}=11 106.6 \ {lbf}/{in^3}[/latex]

[latex]\left(\frac{M}{l} ight)_c=\frac{2760}{0.4604}=5 994.8\ {lbf}/{in^3}[/latex]

[latex]\Delta \left(\frac{M}{l} ight) =5 994.8-11 106.6=-5 111.8 \ {lbf}/{in^3}[/latex]

The slopes for ab and cd, and the change are

[latex]m_{ab} =\frac{2760-2880}{0.2485\left(0.5 ight) } =-965.8 \ {lbf}/{in^4}[/latex]

[latex]m_{cd} =\frac{-5 994.8}{11.5} =-521.3\ {lbf}/{in^4}[/latex]

[latex]\Delta m=-521.3-\left(-965.8 ight) =444.5 \ {lbf}/{in^4}[/latex]

Dividing Eq. (1) by [latex]I_BC \ and \ , \ at \ x= 8.5 \ in [/latex], adding a step of [latex]−5111.8 \ {lbf}/{in^3}[/latex] and a ramp of slope, adding a step of [latex]444.5\ {lbf}/{in^4}[/latex] , gives

[latex]\frac{M}{l} =1448.7x-2414.5\left\langle x-8 ight angle ^1-5111.8\left\langle x-8.5 ight angle ^0+444.5\left\langle x-8.5 ight angle ^1[/latex] (2)

Integration gives

[latex]E\frac{dy}{dx} =724.35x^2-1207.3\left\langle x-8 ight angle ^2-5111.8\left\langle x-8.5 ight angle ^1+222.3\left\langle x-8.5 ight angle ^2+C_1[/latex] (3)

Integrating again yields

[latex]Ey=241.5x^3-402.4\left\langle x-8 ight angle ^3-2555.9\left\langle x-8.5 ight angle ^2+74.08\left\langle x-8.5 ight angle ^3+C_1x+C_2[/latex] (4)

At [latex] x = 0 \ , \ y = 0[/latex]. This gives [latex]C_2= 0[/latex] (remember, singularity functions do not exist until
the argument is positive).

At [latex]x = 20 \in, \ y = 0[/latex] , and

[latex]0=241\left(20 ight) ^3-402.4\left(20-8 ight) ^3-2555.9\left(20-8.8 ight) ^2+74.08\left(20-8.5 ight) ^3+C_1\left(20 ight) [/latex]

Solving, gives [latex]C_1= −50 565 \ {lbf}/{in^2}[/latex]. Thus, Eq. (4) becomes, with [latex]E 30\left(10 ight) ^6 \ psi [/latex],

[latex]y=\frac{1}{30\left(10^6 ight) } \left(241.5x^3-402.4\left\langle x-8 ight angle^3-2555.9\left\langle x-8.5 ight angle^2 +74.08\left\langle x-8.5 ight angle^3-50565x ight) [/latex] (5)

When using a spreadsheet, program the following equations:

[latex]y=\frac{1}{60\left(10^6 ight) } \left(241.5x^3-50565x ight) \ 0\leq x\leq 8 \ in[/latex]

[latex]y=\frac{1}{30\left(10^6 ight) } \left[241.5x^3-402.4\left(x-8 ight)^3- 50565x ight] \ 8\leq x\leq 8.5 \ in [/latex]

[latex]y=\frac{1}{30\left(10^6 ight) } \left[241.5x^3-402.4\left(x-8 ight)^3- 2555.9\left(x-8.5 ight)^2-74.08\left(x-8.5 ight)^3-50565x ight] \ 8.5\leq x\leq 20 \ in[/latex]

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 [latex] x = 8.5 \ in \ where \ y = −0.009385 in [/latex].

Substituting [latex]C_1[/latex] into Eq. (3) the slopes at the supports are found to be [latex] heta _A=1.686\left(10^{-3} ight) \ rad = 0.09657 \ deg, \ and \ theta _F= 1.198(10^{−3}) \ rad = 0.06864 \ deg[/latex].

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
[latex]y|_{x=8.5 \ in}= −0.009380 \ in \ heta _A= −0.09653◦ \ heta _F= 0.06868◦◦[/latex]
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

[latex]y|_{x=8.5 \ in}= −0.009387 \ in \ heta _A= −0.09763◦ \ heta _F= 0.06873◦[/latex]
The largest difference between the models is of the order of 1.5 percent.

Thus the simplification was justified.

Question 1.4.7: The steel step shaft shown in Fig. 4–7a is mounted in bearin...

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