Question 7.15: Design the stepped shaft to transmit 5 kW at 1725 rpm of the......

Torque transmitted is obtained from

T=\frac{\mathrm{kW} \times 9550}{\mathrm{rpm}}=\frac{5 \times 9550}{1725}=27.68 \mathrm{~N} \cdot \mathrm{m}

Length of each segment is

l_{A B}=40 \mathrm{~mm}

l_{B C}=160 \mathrm{~mm}

l_{C D}=50 \mathrm{~mm}

l_{D E}=20 \mathrm{~mm}

The polar moments of inertia in terms of diameter d are

J_{D E}=\frac{\pi}{32} d^4=0.0982 d^4

J_{C D}=\frac{\pi}{32}(1.1 d)^4=0.1437 d^4

J_{B C}=\frac{\pi}{32}(1.25 d)^4=0.2397 d^4

J_{A B}=\frac{\pi}{32}(1.5 d)^4=0.497 d^4

Total angle of rotation between gear and pulley is

\theta=\frac{T}{G}\left(\frac{l_{B C}}{J_{B C}}+\frac{l_{C D}}{J_{C D}}\right)

=\frac{27.68 \times 1000}{84 \times 10^3}\left(\frac{160}{0.2397 d^4}+\frac{50}{0.1437 d^4}\right) \leq 0.25

\frac{27.68 \times 1015.5}{84 \times d^4} \leq 0.25

d^4=\frac{27.68 \times 1015.5}{84 \times 0.25}

d = 6.05 mm

Taking a higher value in order to keep the angular deflection below the specified limit d = 7 mm

Now the calculated diameter can be used to verify the safe criterion on lateral deflection and strength basis.

Lateral deflection can be obtained from the deflection equation. So it is necessary to obtain the moment diagram. The forces acting on gear and pulley and the reactions at the bearing supports can be obtained as follows

Gear forces

The tangential force on the gear tooth is obtained from the relation

F_t=\frac{T}{R_G}=\frac{27.68}{15 \times 10^{-3}}=1845.33 \mathrm{~N}

The radial force at the gear tooth is

F_r=F_t \tan \phi=1845.33 \times \tan 20^{\circ}=671.65 \mathrm{~N}

Pulley forces

Assuming 0.3 as coefficient of friction between pulley and belt, and 180° as angle of lap, we can use the following equations to obtain the belt tensions

\frac{T_1}{T_2}=e^{\mu \theta}=e^{0.3 \times \pi}=2.57

T_1=2.57 T_2 (7.46)

Total force due to pulley tensions that will try to bend the shaft is

F_s=T_1+T_2

The net force that is associated with the driving torque is

F_n=T_1-T_2=\frac{T}{R_p}

R_p is the pulley radius = 12.5 mm

F_n=T_1-T_2=\frac{27.68 \times 1000}{12.5}=2214.4 \mathrm{~N}

2.57 T_2-T_2=2214.4 \mathrm{~N}

T_2=1410.45 \mathrm{~N}

T_1=2.57 \times 1410.45=3624.85 \mathrm{~N}

F_s=T_1+T_2=3624.85+1410.45=5032.3 \mathrm{~N}

Assuming the gear and pulley forces acting as shown in Figure 7.16(a) reactions at support, bending moment and deflections are calculated.

The horizontal and vertical forces acting on the shaft with their position are shown below. It is assumed that forces are acting on the mid plane of pulley, gear and bearings.

Bearing reactions R_{A H} \text { and } R_{C H} are calculated from force and moment equilibrium. From horizontal force equilibrium,

\sum F_H=0=R_{A H}+R_{C H}=5032.3+1845.33=6877.63 \mathrm{~N} (7.47)

Taking moment of the forces about point A

1845.33 \times 0.06-R_{C H} \times 0.22+5032.3 \times 0.27=0 (7.48)

From (7.47) and (7.48)

R_{A H}=6877.63-6679.28=198.35 \mathrm{~N}

The horizontal bending moment diagram is shown in Figure 7.16(c).

Bending moment equation can be written as

M=R_{A H} \times[x]-1845.33[x-0.06]+R_{C H}[x-0.22]-5032.3[x-0.27]

Evaluating the bending equation for each length segments

At point D (x = 0.27 m)

M_D=198.35[0.27]-1845.33[0.27-0.06]+6679.28[0.27-0.22]=0

At point C (x = 0.22 m)

M_C=198.35[0.22]-1845.33[0.22-0.06]=-251.616 \mathrm{~N} \cdot \mathrm{m}

At point B (x = 0.06 m)

M_B=198.35[0.06]=11.901 \mathrm{~N} \cdot \mathrm{m}

Vertical forces are shown in Figure 7.16(d).

Bearing reactions R_{A V} \text { and } R_{C V} are calculated from force and moment equilibrium as follows.

From horizontal force equilibrium,

\sum F_V=0=R_{A V}+R_{C V}=671.65 (7.49)

Taking moment of the forces about point A

671.65 \times 0.06-R_{C V} \times 0.22=0

R_{C V}=183.18 \mathrm{~N} (7.50)

From Eqs. (7.49) and (7.50)

R_{A V}=6.71 .65-183.18=488.47 \mathrm{~N}

The vertical bending moment diagram is shown in Figure 7.16(e).

Bending moment equation is

M=R_{A V} \times[y]-671.65[y-0.06]+R_{C V}[y-0.22]

Evaluating the bending equation for each length segments, the bending moment diagram is shown in Figure 7.16(e).

The combined moment diagram using Pythagorean Theorem is shown in Figure 7.16(f).

Deflection can be obtained from the M/EI diagram. The following section describes about the M/EI diagram.

The bending deflection is given by

\delta=\iint \frac{M}{E I} d x+C x+D

The first function of the above equation is written as

or the moment equation for horizontal loads is written as

M=R_{A H} \times[x]-1845.33[x-0.06]+R_{C H}[x-0.22]-5032.3[x-0.27] (7.51)

The moment equation for vertical loads is written as

M_V=R_{A V} \times[y]-671.65[y-0.06]+R_{C V}[y-0.22]+Q[y-0.27] (7.52)

where Q is a dummy load in the vertical direction applied in order to use the Castigliano’s theorem.

According to Castigliano’s theorem

\delta_{\max }=\frac{\partial U}{\partial F}

where U=\int_0^l \frac{M^2 d x}{2 E I} \text { or } \quad U=\int_0^l \frac{M^2 d y}{2 E I}

To determine the deflection at each location or the maximum deflection, deflection produced due to horizontal and vertical loads are determined individually and they are combined vectorially to obtain the resultant deflection diagram.

Now to minimize the calculations it can be assumed on the basis of combined moment diagram that the maximum deflection may be at point D.

Vertical deflection

\frac{\partial M_V}{\partial Q}=[y-0.27] from (7.52)

U_V=\int_0^{0.06} \frac{\left(M_V\right)^2}{2 E I_{A B}} d y+\int_{0.06}^{0.22} \frac{\left(M_V\right)^2}{2 E I_{B C}} d y+\int_{0.22}^{0.27} \frac{\left(M_V\right)^2}{2 E I_{C D}} d y

\frac{\partial U_V}{\partial M_V}=\int_0^{0.06} \frac{\left(M_V\right)}{E I_{A B}} d y+\int_{0.06}^{0.22} \frac{\left(M_V\right)}{E I_{B C}} d y+\int_{0.22}^{0.27} \frac{\left(M_V\right)}{E I_{C D}} d y

Calculating the derivatives individually

\int_0^{0.06} \frac{\left(M_V\right)}{E I_{A B}} d y=\int_0^{0.06} \frac{(488.47 \times[y])}{E I_{A B}} d y=\left.\frac{488.47 \times y^2}{2 E I_{A B}}\right|_0 ^{0.06}=\frac{488.47 \times 0.06^2}{2 E I_{A B}}=\frac{1.7585}{2 E I_{A B}}

\int_{0.06}^{0.22} \frac{\left(M_V\right)}{E I_{B C}} d y=\int_{0.06}^{0.22} \frac{R_{A V} \times[y]-671.65[y-0.06]}{E I_{B C}} d y

=\left.\frac{R_{A V} \times y^2-671.65(y-0.06)^2}{2 E I_{B C}}\right|_{0.06} ^{0.22}

=\frac{488.47 \times\left(0.22^2-0.06^2\right)-671.65(0.22-0.06)^2}{2 E I_{B C}}=\frac{4.6892}{2 E I_{B C}}

\int_{0.22}^{0.27} \frac{\left(M_V\right)}{E I_{C D}} d y=\int_{0.22}^{0.27} \frac{\left(R_{A V} \times[y]-671.65[y-0.06]+R_{C V}[y-0.22]\right)}{E I_{C D}} d y

=\frac{\left(488.47 \times[0.0245]-671.65[0.0185]+183.18\left[2.5 \times 10^{-3}\right]\right)}{2 E I_{C D}}

=\frac{-6 \times 10^{-5}}{2 E I_{C D}}

The deflection at D in vertical direction (y-direction) is

\delta_{D y}=\frac{\partial U_V}{\partial M_V} \frac{\partial M_V}{\partial Q}

where Q is a dummy load at location D in the vertical direction.

=\left(\frac{1.7585}{2 E I_{A B}}+\frac{4.6892}{2 E I_{B C}}-\frac{6 \times 10^{-5}}{2 E I_{C D}}\right)(y-0.27)

Similarly it can be written for vertical direction.

As the given shaft is stepped one, it is difficult to obtain the deflection using Castigliano’s theorem, hence M/El diagram is used to obtain the deflections.

To determine M/El diagram, the moment of inertia for each step part of the shaft is determined.

Portion AB

Shaft diameter = 1.5d = 1.5 × 7 = 10.5 mm

I=\frac{\pi}{64}(10.5)^4=117.86 \mathrm{~mm}^4

Portion BC

Shaft diameter = 1.25d = 1.25 × 7 = 8.75 mm

I=\frac{\pi}{64}(8.75)^4=287.74 \mathrm{~mm}^4

Portion CD

Shaft diameter = 1.10d = 1.10 × 7 = 7.7 mm

I=\frac{\pi}{64}(7.7)^4=172.57 \mathrm{~mm}^4

Portion DE

Shaft diameter = d = 7 mm

I=\frac{\pi}{64}(7.0)^4=117.86 \mathrm{~mm}^4

The combined M/El diagram is shown in Figure 7 .17.

Section AB

At location B, \frac{M}{E I}=\frac{31.63}{(1.5)^4 E I}=\frac{31.63 \times 1000}{207 \times 10^3 \times 596.66}=256 \times 10^{-4} / \mathrm{mm}

Section BC of the shaft

At location B, \frac{M}{E I}=\frac{31.63}{(1.25)^4 E I}=\frac{31.63 \times 1000}{207 \times 10^3 \times 287.74}=5.31 \times 10^{-4} / \mathrm{mm}

At location C, \frac{M}{E I}=\frac{251.616}{(1.25)^4 E I}=\frac{251.616 \times 1000}{207 \times 10^3 \times 287.74}=42.24 \times 10^{-4} / \mathrm{mm}

Section CD of the shaft

At location C, \frac{M}{E I}=\frac{251.616}{(1.10)^4 E I}=\frac{251.616 \times 1000}{(1.10)^4 \times 207 \times 10^3 \times 117.86}=70.44 \times 10^{-4} / \mathrm{mm}

At location D, \frac{M}{E I}=0

Let V_{D, C}=\text { moment of area of } \frac{M}{E I} diagram between D and C about C

V_{D, C}=\left(\frac{1}{2} \times 50 \times 70.44 \times 10^{-4}\right) \times\left(\frac{1}{3} \times 50\right)

= 2.935 mm

V_{A, C}=\text { moment of area of } \frac{M}{E I} diagram between A and C about C

V_{A, C}=\left(\frac{1}{2} \times 60 \times 2.56 \times 10^{-4}\right) \times\left(\frac{1}{3} \times 60+160\right)+\left(\frac{[5.31+42.24]}{2} \times 10^{-4} \times 160 \times \frac{160}{2}\right)

= 1.3824 + 30.426

 31.808 mm

\theta_A=\frac{V_{A, C}}{220}=\frac{31.808}{220}=0.145

Deflection at C

y_C=\delta_C=\theta_A \times 220-V_{A, C}

= 0.145 × 220 – 31.808

= 0.092 mm

Deflection at B

y_B=\delta_B=\theta_A \times 60-V_{A, B}

=0.145 \times 60-\frac{1}{2} \times 60 \times 2.56 \times 10^{-4} \times \frac{60}{3}

= 8.5464 mm

Similarly,

\theta_C=\frac{V_{C, A}}{220}

V_{C, A}=\left(\frac{1}{2} \times 2.56 \times 10^{-4} \times 60\right)\left(\frac{2}{3} \times 60\right)+\left(\frac{5.31+42.24}{2}\right) \times 10^{-4} \times 160 \times\left(60+\frac{160}{2}\right)

= 0.3072 + 53.256

= 53.563

\theta_C=\frac{V_{C, A}}{220}=\frac{53.563}{220}=0.244

y_D=\delta_D=\theta_C \times 50-V_{C, D}

=0.244 \times 50-\frac{1}{2} \times 50 \times 70.44 \times 10^{-4} \times \frac{2 \times 50}{3}

= 12.2 – 5.87 = 6.33 mm

The maximum deflection is at B and it is more than the given value. Hence all diameters must be modified by multiplying with the factor obtained as follows.

The ratio of the maximum and permissible deflection is

\frac{8.5464}{1.5}=5.6976

The amount to be multiplied to each diameter is obtained as

(5.6976)^{1 / 4}=1.55

The new diameter is

d = 1.55 × 7 = 10.85 mm

d = 11.0 mm