Question 7.3: A machine shaft is transmitting 50 kW power at a speed of 11......
A machine shaft is transmitting 50 kW power at a speed of 1100 rpm with mild shock. The maximum bending moment due to weight of the shaft is 1200 N·m. The shaft is also subjected to axial thrust of 65 kN. The shaft is supported at an interval of 1.5 m. If inner diameter of the shaft is 50 mm, determine the outer diameter of the hollow shaft. The allowable strength of the shaft material is 56 MPa and the yield strength in tension is 320 MPa, E = 205 GPa.
T=\frac{9550 \times \mathrm{kW}}{\mathrm{rev} / \mathrm{min}}=\frac{9550 \times 50}{1100}=434.09 \mathrm{~N} \cdot \mathrm{m}
M = 1200 N·m
F = 65 kN
Take k_b=1.5, k_t=1.5
d_o=\left[\frac{16}{\pi \tau_y\left(1-k^4\right)} \sqrt{\left[\left(k_b M+\frac{\alpha F d_o\left(1+k^2\right)}{8}\right)\right]^2+\left(k_t T\right)^2}\right]^{1 / 3}
This problem requires an iterative or trial & error method to find out the diameter.
d_o=\left[\frac{16}{\pi \times 56 \times 10^6\left(1-k^4\right)} \sqrt{\left[\left(1.5 \times 1200+\frac{\alpha \times 65000 d_o\left(1+k^2\right)}{8}\right]\right]^2+(1.5 \times 434.09)^2}\right]^{1 / 3}
With the inner diameter as the initial trial value, the following calculations are made.
A=\pi \times 50^2 / 4=1963.5 \mathrm{~mm}^2
I=\frac{\pi}{64} \times 50^4=306796.15 \mathrm{~mm}^4
\kappa=\sqrt{306796.15 / 1963.5}=12.5 \mathrm{~mm}
L / \kappa=1500 / 12.5=120>115
\alpha=\frac{\tau_y}{\pi^2 n E}\left(\frac{L}{\kappa}\right)^2=\frac{56.0}{\pi^2 \times 1.6 \times 205 \times 10^3} \times(120)^2=0.249
The trial estimate of outer diameter is
d_o=\left[\frac{16}{\pi \times 56 \times 10^6} \sqrt{\left[\left(1800+\frac{0.249 \times 65000 \times 0.05}{8}\right)\right]^2+(434.09 \times 1.5)^2}\right]^{1 / 3}
d_o=0.05675 \mathrm{~m}
= 56.75 mm
Second trial d_o=57.0 \mathrm{~mm}
A=\pi \times \frac{57^2-50^2}{4}=588.26 \mathrm{~mm}^2
I=\frac{\pi}{64}\left(d_o^4-d^4\right)=\frac{\pi}{64}\left(57^4-50^4\right)=211370.3 \mathrm{~mm}^4
\kappa=\sqrt{I / A}=\sqrt{211370.3 / 588.26}=18.96 \mathrm{~mm}
L / \kappa=1500 / 18.96=79.11<115
\alpha=\frac{1}{1-0.0044 \times 79.11}=1.54
k=\frac{50}{57}=0.88
New diameter is
d_o=\left[\frac{16}{\pi \times 56 \times 10^6\left(1-0.88^4\right)}\right.
\left.\sqrt{\left[\left(1800+\frac{1.54 \times 65000 \times 0.057\left(1+0.88^2\right)}{8}\right)\right]^2+(651.135)^2}\right]^{1 / 3}
= 0.0893 m
= 89.3 mm
which is not close to the trial value. Considering the other values, the results are inner diameter = 50 mm and outer diameter = 71.5 mm. Reader may try with these results.
MATLAB Program
Script File
% computation of shaft diameter, Example 6.3
% di inner diameter of the shaft, mm
% d0 outer diameter of the shaft, mm
% f axial load, N
% moment
% t
bending moment, N mm
torque, N mm
% kt, km fatigue factors for torque and bending moment
clear all;clc;
di= 50;
taus = 56;
kt= 1.5;
km= 1.5;
moment = 1200000;
f = 56000;
t = 434000;
out dia(l) = 70;
for i = 2:100
k = 50./out_dia(i - l);
MI = (pi/64)*out_dia(i - l)A4;
area = (pi/4)*out_dia(i - l)A2;
kappa = sqrt(Ml/area);
lk = 1500/kappa;
alpha = l/(l-0.0044*lk);
al = 16/(pi*taus*(l-kA4));
a2 = (km*moment+(alpha*f*out_dia(i-1)*(1 +kA2)/8))A2;
a3 = (kt*t)A2;
out_dia(i) = (al *(a2+a3)A0.5)A0.3333333;
error= abs(out_dia(i)-out_dia(i-1));
if error < 0.5e-3
d0 = out_dia(i-1);
break;
end
end
Results = 'inner diameter (mm)=',di,'outer diameter (mm)=',d0
Output of the MATLAB program
Results =
inner diameter (mm)=
\mathrm{d}_{\mathrm{i}}=50
outer diameter (mm)=
\mathrm{d}_{\mathrm{o}}=71.4665
Program through Excel Spreadsheet of Example 7.3
Note: Data entries start from 7th row of the spreadsheet. Some of the commands to obtain the value in the appropriate cell are shown in Table 7.3. Commands are written for 7th row of the spreadsheet. This is an illustration only. Result is taken when the output and input diameters differ by a permissible amount.