Mechanics of Materials

T/ Mohammed. A. Mustafa Introduction Consider a bar rigidly clamped at one end and twisted at the other end by a torque (twisting moment) T = Fd applied in a plane perpendicular to the axis of the bar, as shown in Fig. 1. Such a bar is in torsion. An alternative representation of the torque is the curved arrow shown in the figure.

Fig. 1 Torque applied to a circular shaft Polar Moment of Inertia

A mathematical property of the geometry of the cross section which occurs in the study of the stresses set up in a circular shaft subject to torsion is the polar moment of inertia J, defined in a statics course

For a hollow circular shaft of outer 휋 diameter 푫 with a concentric circular J = (퐷4 − 푑4) hole of diameter 풅 the polar moment of 32 inertia of the cross-sectional area is given by

For a solid circular shaft of diameter 풅 휋 the polar moment of inertia of the J = 푑4 cross-sectional area is given by 32 Polar Moment of Inertia Torsional Shearing Stress So, for either a solid or a hollow circular shaft subject to a twisting moment 푻 the torsional shearing stress τ at a distance r from the center of the shaft is written as

Distribution of shear stress along the radius of a circular shaft. The maximum shear stress is found by replacing ρ by the radius r of the shaft: Shearing Strain

The ratio of the shear stress 휏 to the shear strain 훾 is called the shear modulus 휏 퐺 = 훾 휏 = shear stress 훾 = shear strain G = shear modulus 푟휃 훾 = 훾 = tan ∝ ≈ ∝ 퐿

Again the units of G are the same as those of shear stress, since the shear strain is dimensionless 푇퐿 휃 = angle of twist 휃 = 퐿 = length of the shaft 퐺퐽 SOLVED PROBLEMS Example (1): If a twisting moment of 1100 N.m is impressed upon a 4.4 cm diameter shaft, what is the maximum shearing stress developed? Also, what is the angle of twist in a 150 cm length of the shaft? The material is steel for which G = 85 GPa. SOLUTION Step one: The polar moment of inertia of the cross-sectional area is 휋 휋 J = (푑4) = × 0.0444 = ퟑ. ퟔퟖ × ퟏퟎ−ퟕ푚4 32 32 Step two: The maximum shear stress is developed at the outer fibers where r = 0.022 m: 푇푟 1100 × 0.022 휏 = = = ퟔퟓ. ퟖ × ퟏퟎퟔ 푷풂 풐풓 ퟔퟓ. ퟖ푴푷풂 푚푎푥 퐽 3.68 × 10−7 Step three: The angle of twist q in a 1.5 m length of the shaft is 푇퐿 1100 × 1.5 휃 = = = ퟎ. ퟎퟓퟐퟕ 퐫퐚퐝 퐨퐫 퐺퐽 85 × 109(3.68 × 10−7) Example (2): A hollow 3 m long steel shaft must transmit a torque of 25 kN.m. The total angle of twist in this length is not to exceed 2.5° and the allowable shearing stress is 90 MPa. Determine the inside and outside diameters of the shaft if G = 85 GPa. SOLUTION

Step one: The angle of twist is θ = TL/GJ. Thus, in the 3-m length we have 휋 푟푎푑 (25000)(3) 2.5° = ∴ 퐷4 − 푑4 = 206 × 10−6 푚4 → (ퟏ) 180° 85 × 109(휋/32)(퐷4 − 푑4)

Step two: The maximum shearing stress occurs at the outer fibers where r = D/2 (25000)(퐷/2) 90 × 106 = ∴ 퐷4 − 푑4 = (1415퐷 × 10−6)푚4 → (ퟐ) (휋/32)(퐷4 − 푑4) Comparison of the right-hand sides of equations (1) and (2) indicates that 206 × 10−6 = (1415퐷 × 10−6) and thus D = 0.146 m or 146 mm. Substitution of this value into either of the equations then gives d = 0.126 m or 126 mm. Example (4): Consider two solid circular shafts connected by 5-cm- and 25-cm-pitch-diameter gears as in Fig. (a). Find the angular rotation of D, the right end of one shaft, with respect to A, the left end of the other, caused by the torque of 280 N.m applied at D. The left shaft is steel for which G = 80 Gpa and the right is brass for which G = 33 GPa. SOLUTION Step one: A free-body diagram of the right shaft CD [Fig. (b)] reveals that a tangential force F must act on the smaller gear. For equilibrium, Fr = T 0.025 F = 280 ∴ F = 11200 N Step two: The angle of twist of the right shaft is 푇퐿 280 × 1.00 휃 = = = ퟎ. ퟏퟎퟔퟕ퐫퐚퐝 1 퐺퐽 (33 × 109)휋 × 0.034/32 Step three: A free-body diagram of the left shaft AB is shown in Fig. (c). The force F is equal and opposite to that acting on the small gear C. This force F acts 12.5 cm from the center line of the left shaft; hence it imparts a torque of 0.125(11200) = 1400 N.m to the shaft AB. Because of this torque there is a rotation of end B with respect to end A given by the angle 휃2, where 푇퐿 1400 × 1.20 휃 = = = ퟎ. ퟎퟏퟔퟓ퐫퐚퐝 2 퐺퐽 (80 × 109)휋 × 0.064/32

Step four: This angle of rotation 휃2 induces a rigid-body rotation of the entire shaft CD because of the gears. In fact, the rotation of CD will be in the same ratio to that of AB as the ratio of the pitch diameters, or 5:1. Thus a rigid-body rotation of 5(0.0165) rad is imparted to shaft CD. Superposed on this rigid body movement of CD is the angular displacement of D with respect to C, previously denoted by 휃1. Hence the resultant angle of twist of D with respect to A is ° 휃 = 5휃2 + 휃1 = 5 × 0.0165 + 0.1067 = ퟎ. ퟏퟖퟗ 풓풂풅 풂풓 ퟏퟎ. ퟖ Thank You