Question 9.4: The layout of an intermediate shaft of a gear box supporting...

\text { Given } S_{u t}=770 N / mm ^{2} \quad S_{y t}=580 N / mm ^{2} .

k_{b}=1.5 \quad k_{t}=2.0 .

\text { For gears, }\left(d_{p}^{\prime}\right)_{B}=900 mm \quad\left(d_{p}^{\prime}\right)_{C}=600 mm .

Step I Permissible shear stress

0.30 S_{y t}=0.30(580)=174 N / mm ^{2} .

0.18 S_{u t}=0.18(770)=138.6 N / mm ^{2} .

The lower of the two values is 138.6 N/mm² and there are keyways on the shaft.

\therefore \quad \tau_{\max .}=0.75(138.6)=103.95 N / mm ^{2} .

Step II Bending and torsional moment
The forces and bending moments in vertical and horizontal planes are shown in Fig. 9.7. The maximum bending moment is at C. The resultant bending moment at C is given by,

M_{b}=\sqrt{(3496203)^{2}+(121905)^{2}} .

=3498327.4 N – mm .

M_{t}=4421(450)=1989450 N – mm .

Step III Shaft diameter
From Eq. (9.15),

\tau_{\max .}=\frac{16}{\pi d^{3}} \sqrt{\left(k_{b} M_{b}\right)^{2}+\left(k_{t} M_{t}\right)^{2}}                  (9.15).

d^{3}=\frac{16}{\pi \tau_{\max }} \sqrt{\left(k_{b} M_{b}\right)^{2}+\left(k_{t} M_{t}\right)^{2}} .

=\frac{16}{\pi(103.95)} \sqrt{(1.5 \times 3498327.4)^{2}+(2.0 \times 1989450)^{2}} .

or d = 68.59 mm.