Question 3.4: A solid steel shaft ABCDE (Fig. 3-17) having diameter d = 30...
A solid steel shaft ABCDE (Fig. 3-17) having diameter d = 30 mm turns freely in bearings at points A and E. The shaft is driven by a gear at C, which applies a torque T_{2} = 450 N⋅m in the direction shown in the figure. Gears at B and D are driven by the shaft and have resisting torques T_{1} = 275 N⋅m and T_{3} = 175 N⋅m,respectively, acting in the opposite direction to the torque T_{2} . Segments BC and CD have lengths L _{BC} = 500 mm and L _{CD} = 400 mm, respectively, and the shear modulus G = 80 GPa.
Determine the maximum shear stress in each part of the shaft and the angle of twist between gears B and D.
Each segment of the bar is prismatic and subjected to a constant torque (Case 1). Therefore, the first step in the analysis is to determine the torques acting in the segments, after which we can find the shear stresses and angles of twist.
Torques acting in the segments. The torques in the end segments (AB and DE) are zero since we are disregarding any friction in the bearings at the supports.
Therefore, the end segments have no stresses and no angles of twist.
The torque T_{CD} in segment CD is found by cutting a section through the segment and constructing a free-body diagram, as in Fig. 3-18a. The torque is assumed to be positive, and therefore its vector points away from the cut section.
From equilibrium of the free body, we obtain
T _{CD}=T_{2} – T_{1}=450 N⋅m – 275 N⋅m=175 N⋅mThe positive sign in the result means that T_{CD} acts in the assumed positive direction.
The torque in segment BC is found in a similar manner, using the free-body diagram of Fig. 3-18b:
T _{BC}= – T_{1}= – 275 N⋅m
Note that this torque has a negative sign, which means that its direction is opposite to the direction shown in the figure.
Shear stresses. The maximum shear stresses in segments BC and CD are found from the modified form of the torsion formula (Eq. 3-12); thus,
\tau _{max}=\frac{16T}{\pi d^{3} } (3-12)
\tau _{BC}=\frac{16T_{BC} }{\pi d^{3} } =\frac{16\left(275 N⋅m\right) }{\pi \left(30 mm\right)^{3} }=51.9 MPa
\tau _{CD}=\frac{16T_{CD} }{\pi d^{3} } =\frac{16\left(175 N⋅m\right) }{\pi \left(30 mm\right)^{3} }=33.0 MPa
Since the directions of the shear stresses are not of interest in this example, only absolute values of the torques are used in the preceding calculations.
Angles of twist. The angle of twist \phi _{BD} between gears B and D is the algebraic sum of the angles of twist for the intervening segments of the bar, as given by Eq. (3-19); thus,
\phi = \phi _{1} + \phi _{2} + … + \phi _{n} (3-19)
\phi _{BD}=\phi _{BC} + \phi _{CD}When calculating the individual angles of twist, we need the moment of inertia of the cross section:
I_{P} =\frac{\pi d^{4} }{32} =\frac{\pi \left(30 mm\right)^{4} }{32} =79,520 mm^{4}Now we can determine the angles of twist, as follows:
\phi _{BC}=\frac{T_{BC}L_{BC} }{GI_{P} } =\frac{\left(-275 N⋅m\right) \left(500 mm\right) }{\left(80 GPa\right)\left(79,520 mm^{4} \right) } =- 0.0216 rad\phi _{CD}=\frac{T_{CD}L_{CD} }{GI_{P} } =\frac{\left(175 N⋅m\right) \left(400 mm\right) }{\left(80 GPa\right)\left(79,520 mm^{4} \right) } =- 0.0110 rad
Note that in this example the angles of twist have opposite directions. Adding algebraically, we obtain the total angle of twist:
\phi _{BD} = \phi _{BC} + \phi _{CD}= – 0.0216 + 0.0110 = – 0.0106 rad= – 0.61^{°}The minus sign means that gear D rotates clockwise (when viewed from the righthand end of the shaft) with respect to gear B. However, for most purposes only the absolute value of the angle of twist is needed, and therefore it is sufficient to say that the angle of twist between gears B and D is 0.61°. The angle of twist between the two ends of a shaft is sometimes called the wind-up.
Notes: The procedures illustrated in this examplecan be used for shafts having segments of different diameters or of different materials, as long as the dimensions and properties remain constant within each segment.
Only the effects of torsion are considered in this example and in the problems at the end of the chapter. Bending effects are considered later, beginning with Chapter 4.
