Question

CASE STUDY:

Torsion in Solid and Hallow Shaft Application

DIRECTIONS:

Select an application on Torsion Axel Problem, research the problem and submit a report according to

Part 1:

- Present your example of Torsion Application for solid and hallow shaft

- Identify the concepts of the problem

Part 2:

- Calculate each process parameters.

- Select a case possible to your solution

Part 3:

- Calculations

Answer 1

TORSIONAL DEFORMATION OF A CIRCULAR SHAFT Torsion is a moment that twists/deforms a member about its longitudinal axis * By observation, if angle of rotation is small, length of shaft and its radius remain unchanged Circles remain circular Longitudinal lines become twisted Before deformation Radial lines remain straight After deformation (b) 3

Torsional Deformation of Circular Bars • Assumptions Plane sections remain plane and perpendicular to the AD torsional axis Material of the shaft is uniform Twist along the shaft is uniform. Axis remains straight and inextensible (a) Before deformation CO al Asteld with circular section lo 41 4 1 | | | | | @ (b) After deformation FIGURE 4.3 The deforma- tion of a circular cylinder. FIGURE 4.2 Examples of torsional deformation

Torsional Deformation L 0 = angle of twist B F ө IL R O = shear strain is the shear strain, also remember that tano = 0, thus: FF RO 0 L L Note that shear strain does not only change with the amount of twist, but also, it varies along the radial direction such that it is zero at the center and increases linearly towards the outer periphery (see next slide) 5

Torsional Deformation τ Сө 9 RL r Shear stress at any point in the shaft is proportional to the distance of the point from the axis of the shaft. dA Tmax Tmax Shear stress varies linearly along cach radial line of the cross section Fig. 3-7 Longitudinal and transverse shear stresses in a circular bar

Torque transmitted by shaft(solid) T I total turning moment due to turning force = total force on the ring x Distance of the ring from the axis x 2ur dr R Total turning moment (or total torque) is obtained by integrating the above equation between the limits and R T=S”T - SE *2#rºdr R IR τ = < 2.0 x 2л -141 R R JT =TX XR л -D 16

0 1 Torque transmitted by shaft(hollow) r total turning moment due to turning force = total force on the ring x Distance of the ring from the axis x 2.nr dr R Total turning moment (or total torque) is obtained by integrating the above equation between the limits 0 and R T= SKATS 2rdr R T - x 2053*rd - x 2r R. 4 =TXX R. -R R. 4 л 16 T D.

Power transmitted by shaft Power transmitted by the shafts N = r.p.m of the shaft T = Mean torque transmitted Angular speed of shaft 20NT Power = 60 = 0 XT 0 =

Power transmitted by shaft Power transmitted by the shafts N = r.p.m of the shaft T = Mean torque transmitted Angular speed of shaft 20NT Power = 60 = 0 XT 0 =

Torque in terms of polar moment of inertia T r R Moment dT on the circular ring dr = 250'dr xrºx 2.trdr = (dA = 2.rdr) R R xr? x dA Total Torque = SAT T= S" - L'Exr’da Srda r’dA = moment of elemnetary ring about an axis perpendicular to the plane and passing though the center of the circle Srda = moment of the circle about an axis perpendicular to the plane and passing though the center of the circle T T Polar moment of inertia = xD 32

Torque in terms of polar moment of inertia τ T= - XJ R T T I R J R τ СӨ = RL T τ CO C = Modulus of rigidity J R L e = Angle of twist L = Length of the shaft

Polar Modulus Polar modulus is defined as ration of polar moment of inertia to the radius of the shaft. Z J R л For solid shaft => J= D4 32 л D4 32 D/2 z л -D² 16 For hollow shaft => J= 3 [Di-D"] 3 [D:*-D] JT Z [D." - Dt] P D./2 16D

Torsional rigidity Torsional rigidity is also called strength of the shaft. It is defined as product of modulus of rigidity (C) and polar moment of inertia C*J

Shaft in combined bending and Torsion stresses м. Г X == JT D Shear stress at any point due to torque T 9 Т Txr =9= J J D Shear Stress at a point on the surface of the shaft ftra 2 Txr T D 16T T = J 2 AD 32 Bending stress at any point due to bending moment M o Mху I 1 D Bending Stress at a point on the surface of the shaft r=- 2 Mху M 32M 0. 7 D' 2 64 167 tane лD т 0 32M M лр == X JT AD 2x 21

Shaft in combined bending and Torsion stresses Major principal Stress 0 +T 2 32M 2 x D + 32M 2 xD + 167 D 16 D (M+ VM + T2 SOLID SHAFT Minor principal Stress 16 (M - VM +T) D Max shear Stress Max principal Stress - Min principal Stress 2 16 VM+T2 RD (VM+7)

Shaft in combined bending and Torsion stresses Major principal Stress 16D M+M+12 [D.* -D" Minor principal Stress (M-/M² + T² [D.* -D'] Max shear Stress 16D M +12 HOLLOW SHAFT 16D - *ma7(w+r)

Application to a Bar Normal Force: Fo Fn Bending Moment: ME M Shear Force: Ft Ft Torque or Twisting Moment: MA G 16 M.

a solid bar of circular cross section d 15 in. d=40 mm, L= 1.3 m, G = 80 GPa (a) T= 340 N-m, Tmax 0 = ? (b) Tall = 42 MPa, Pall = 2.5º, T = ? T (a) 16 T 16 x 340 N-M Tmar 27.1 MPa πα 1, a d'/32 TL a (0.04 m 2.51 x 10-2 m4 340 N-m x 1.3 m 0.02198 rad = 1.26° 4 GI, 80 GPa x 2.51 x 10 m

(b) due to Tall 42 MPa T=rd Tall/ 16 = 1 (0.04 m) x 42 MPa / 16 = 528 N-m due to all 2.5° 2.5 x a rad / 180° 0.04363 rad T 80 GPa x 2.51 x 10-7 m m* x 0.04363 / 1.3 m G I all/L 674 N-m thus Tall min [T, T2] . 528 N-m

A hollow shaft dia ratio 3/5 is required to transmit 450Kw at 1200pm, the shearing stress in the shaft must not exceed 60N/mm2 and the twist in a length of 2.5m is not to exceed 1°. Calculate the minimum external of the shaft. Take, C=8.0KN/mmº. Solution Dia 3/5 d=3. D=5 P - 450Kw= 450x10 W N 1200pm TO 60N/mm - 1 Il 2.5m 25x102 = 1ºx 180 radi 0201745 C 80KN/mm

D min T = 194) 16 2NT P Il 60 P60 T 35.82x10 N/mm ZNT

D- TI 35.82x101x16 = TT X 60 3,0427x10 N/mm D-3/5D) 3.0427x10 D p'-(1-1296) = 3.0427x10 D 3.0427X10 0972 D 9.106mm