Question 7.4: A hollow shaft having outside diameter of 45 mm and inner di......
A hollow shaft having outside diameter of 45 mm and inner diameter of 30 mm is operating without failure when driven by a torque of 345 N·m. Now the same shaft is to be operated under an additional internal pressure of 110 MPa to produce a new product. If the shaft is made of steel having ultimate strength of 670 MPa and yield strength of 560 MPa, do you recommend for use?
Case I When the shaft is under the action of torque T = 345 N·m.
Neglecting weight of the shaft and no other force is acting on it, the shear stress is obtained as follows.
d_o=45 \mathrm{~mm} ~d_i=30 \mathrm{~mm} ~\mathrm{k}=30 / 45=0.667
\tau_y=0.3 \sigma_y=0.3 \times 560=168 \mathrm{~MPa}
=0.18 \sigma_{\mathrm{ult}}=0.18 \times 670=120.6 \mathrm{~MPa}
J=\frac{\pi\left(d_o^4-d_i^4\right)}{32}=\frac{\pi\left(45^4-30^4\right)}{32}=323056.35 \mathrm{~mm}^4
\tau=\frac{T d_o}{2 J} \frac{345000 \times 45}{2 \times 323056.35}=24.03 \mathrm{~N} / \mathrm{mm}^2
Principal stresses are
\sigma_{1,2}=\tau=\pm 24.03 \mathrm{~MPa}
Case II When the shaft is to work under T and internal pressure p
The stresses develop due to internal pressure can be obtained assuming it as thick cylinder subjected to internal pressure. The stresses are
\sigma_{r r}=-p=-110 \mathrm{~MPa}
\sigma_{\theta \theta}=\frac{p\left(r_o^2+r_i^2\right)}{\left(r_o^2-r_i^2\right)}=\frac{110\left(22.5^2+15^2\right)}{\left(22.5^2-15^2\right)}=286 \mathrm{~MPa}
\sigma_{z \theta}=\tau=24.03 \mathrm{~MPa}
The state of stress is
\sigma=\left[\begin{array}{ccc}\sigma_{r r} & \sigma_{r \theta} & \sigma_{r z} \\\sigma_{\theta r} & \sigma_{\theta \theta} & \sigma_{\theta z} \\\sigma_{z r} & \sigma_{z \theta} & \sigma_{z z}\end{array}\right]=\left[\begin{array}{llc}-110 & 0 & 0 \\0 & 286 & 24.03 \\0 & 24.03 & 0\end{array}\right]
Principal stresses are obtained from the following characteristic equation
\sigma^3-I_1 \sigma+I_2 \sigma-I_3=0
where
I_1= sum of normal stresses = -110 + 286 + 0 = 176.0
I_2=\left|\begin{array}{lc}-110 & 0 \\0 & 286\end{array}\right|+\left|\begin{array}{cc}286 & 24.03 \\24.03 & 0\end{array}\right|+\left|\begin{array}{cc}-110 & 0 \\0 & 0\end{array}\right|=31460-577.4409+0
=-3.203 \times 10^4
I_3=\left|\begin{array}{lcc}-110 & 0 & 0 \\0 & 286 & 24.03 \\0 & 24.03 & 0\end{array}\right|=6.3518 \times 10^4
The characteristic equation is
\sigma^3-176 \sigma-3.20 \sigma+6.352=0
Solving the equation, we get principal stresses as:
\sigma_1=-110 \quad \sigma_2=-2.005 \text { and } \sigma_3=288.005~(\mathrm{MPa})
According to von Mises theory
\sigma_{\mathrm{von}}=\left[\frac{\left(\sigma_1-\sigma_2\right)^2+\left(\sigma_2-\sigma_3\right)^2+\left(\sigma_3-\sigma_1\right)^2}{2}\right]^{1 / 2}
\sigma_{\mathrm{von}}=\left[\frac{(-110+2.005)^2+(-2.005-288.005)^2+(288.005+110)^2}{2}\right]^{1 / 2}
= 356.5 MPa which is less than yield strength.
Alternative
von Mises stress is given by
\sigma_{\mathrm{von}}=\frac{1}{\sqrt{2}}\left[\left(\sigma_{r r}-\sigma_{\theta \theta}\right)^2+\left(\sigma_{\theta \theta}-\sigma_{z z}\right)^2+\left(\sigma_{z z}-\sigma_{r r}\right)^2+6\left(\sigma_{r \theta}^2+\sigma_{\theta z}^2+\sigma_{z r}^2\right)\right]^{1 / 2}
\sigma_{\mathrm{von}}=\frac{1}{\sqrt{2}}\left[(-110-286)^2+(286-0)^2+(0+110)^2+6\left(0+24.03^2+0\right)\right]^{1 / 2}
=356.5 \mathrm{~MPa}<\sigma_y,
Hence yielding will not occur. Hence it can be used for the given condition of loading.
Matlab Programming and output of sample problem 7.4 is presented below.
Script File
% Solution of Example 7.4
% shaft design
sig_yield = 560;
sig_uts = 670;
sig_shear = 0.5*sig_yield;
sig_ von = 0.577*sig_yield;
a = [-110 0 0; 0 286 24.03; 0 24.03 0];
il = a(l, l)+a(2,2)+a(3,3);
xl = [a(l,l) a(l,2);a(2,l) a(2,2)];
x2 = [a(2,2) a(2,3);a(3,2) a(3,3)];
x3 = [a(l, 1) a(l ,3);a(3, 1) a(3,3)];
xxl = det(xl);
xx2 = det(x2);
xx3 = det(x3);
i2 = xxl + xx2 + xx3;
i3 = det(a);
pr = eig(a);
prl = pr(l, l);
pr2 = pr(2, l);
pr3 = pr(3, l);
vonl = sqrt((prl-pr2)A2+(pr2-pr3)A2+(pr3-prl)A2);
von = (l/sqrt(2))*vonl;
maxssl_2 = 0.5*abs(prl-pr2);
maxss2_3 = 0.5*abs(pr2-pr3);
maxss3_1 = 0.5*abs(pr3-prl);
ss = [maxssl_2, maxss2_3, maxss3_1];
max_shear_stress = max(ss)
von_mises_stress = von
fos_ss = sig_shear/max_shear_stress
if fos_ss>l
'safe to use as per maximum shear stress theory'
else
'unsafe as per maximum shear stress theory'
end
fos_ von_mises = sig_yield/sig_ von
if fos_ von_mises> 1
'safe to use as per von Mises theory'
else
'unsafe as per von Mises theory'
end
Output of the program is as follows
>>
max_shear_stress =
199.0025
von_mises_stress =
356.4945
fos_ss =
1.4070
ans=
safe to use as per maximum shear stress theory
fos_ von_mises =
1.7331
ans=
safe to use as per von Mises theory