Structural Analysis I Chapter 4  Torsion TORSION


 Asher Gardner
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1 ORSION orsional stress results from the action of torsional or twisting moments acting about the longitudinal axis of a shaft. he effect of the application of a torsional moment, combined with appropriate fixity at the supports, is to cause torsional stresses within the shaft. hese stresses, which are effectively shear stresses, are greatest on the outer surface of the shaft. Page 1
2 Internal Resisting orsional Moment () When a steel shaft (a circular section straight bar) with one end fixed and the other end free is subject to a concentrated moment M x at the free end. his M x is rotating about the longitudinal axis, xaxis of the shaft. x M x xaxis he action of this moment M x which is sometime called torque, will then cause a twist (torsional rotation) of the shaft. he internal torsional moment in the member under torsional load can be found by using the equation of equilibrium. M x = 0, = M x xaxis x M x It is seen in this example that the internal torsional moment in all crosssections of this member remains constant. Page 
3 M x xaxis M x x = M x Diagram In the case when more than one external torsional moment (load) is acting along the member as shown, for example A m 1= 5 knm m = 7 knm 1.m B 0.5m C m = knm 3 1.5m D xaxis Page 3
4 Between C & D xaxis m 3 CD CD = m 3 = knm Between B & C xaxis m 3 m BC BC = 7 = 3 knm Between A & B xaxis m 3 m AB m 1 AB = = 8 knm Page 
5 A B C D 8 knm 3 knm  knm Page 5
6 orsional Shear Stress in a Circular Section Assumptions: i) he material is homogeneous. ii) he material is elastic and obeys Hooke s aw. iii) he stress does not exceed the elastic limit or limit of proportionality. iv) Circular sections remain circular. v) Plane sections remain plane. vi) All diameters of the crosssection which are original straight remain straight and their magnitudes do not change. he above assumptions are based on the observation of the behaviour of circular section shafts under torsion. he observation also confirms that during the twist, a longitudinal straight line OA on the surface takes up new position as OA while the section at the free end rotates an angle of twist relative to the fixed end which does not rotate. O A' A M x Page 6
7 Geometrically, A R a r c B B' d b b' d d' xaxis C D D' It is observed that the element ABCD on the surface has been distorted to AB CD due to the torsional rotation d over a distance. his kind of deformation must be accompanied by shear stresses around the element. A B B' C D D' BB = shear displacement of B BB / = shear angle at the surface. Page 7
8 Somewhere between the centre (xaxis) & the surface, an element abcd also has the similar deformation. a r b b' r () c d d' bb = r = shear displacement of b r = shear angle On the other hand, Physically, bb = rd d r r G herefore, G d r Gr d In which is the torsional rotation of the shaft per unit length. Equation of equilibrium, r da r Page 8
9 d r r da G r da d G & are the constants in this integration and r da = Polar moment of inertia of the circular section about its R center & d G d G d From r Gr r herefore, r and max R max max It can be seen that torsional shear stress varies linearly from zero at the r centre to maximum at the outside fibre. he formula r is also valid for hollow circular sections. max max Page 9
10 When a circular shaft is subjected to torsion, the elements between two adjacent crosssections are under pure shear. 1 From Mohr s circle, it can be seen that the principal stresses, he occurrence of 1 max at 5 o to longitudinal axis causes cracking of brittle shaft because brittle material is very weak in tension, like cast iron & chalk. m x 1 1 m x Page 10
11 If this shaft is made of mild steel which is ductile material strong in tension & compression but is comparatively weak in shear. his shaft will eventually fail in shear when the external torsional load keeps on increasing. he failure section is the crosssection of the shaft perpendicular to its longitudinal axis. Ductile Material Page 11
12 orsional Rotation O A' A M x he deformation of a shaft under pure torsion is the torsional rotation of a crosssection. From d G x0 G therefore = G he is then the torsional rotation of the free end as the fixed end does not rotate. Page 1
13 Summary of Formulae for orsion of Circular Section Maximum Shear Stress: max R Shear Stress at any radial position r: r r Polar Moment of Inertia for Solid Circular Section: D D 3 R Polar Moment of Inertia for Hollow Circular Section: D o D D R R o 3 i o i D i Angle of wist for Circular Section: G Page 13
14 Example 1 A solid steel shaft 50 mm in diameter is subjected to a torsional moment of knm. Determine the maximum shear stress in the shaft. Solution: he maximum shear stress on the periphery of the shaft is max R 6 x10 x N/mm Example Determine the angle of twist between two sections 500 mm apart in a steel rod having a diameter of 5 mm with a torque of 300 Nm is applied. Shear modulus G of steel is 80 GPa. Solution: he angle of twist of the steel rod is G 300x10 80x rad Page 1
15 Example 3 Find the torque that a solid circular shaft with a diameter of 150 mm can transmit if the maximum shearing stress is 50 MPa. What is the angle of twist per meter of length if the shear modulus G is 80 GPa? Solution: he polar moment of inertia is given by, D x10 mm R max 50(.97x10 ) 6 max 33.13x10 R 75 = knm Nmm he angle of twist is given by, x x G 80x10.97x10 rad Page 15
16 Example What torque should be applied to the end of the steel shaft to produce a twist of 3 o? Use the value G = 80 GPa for the shear modulus of steel shaft. he outside diameter D o = 80 mm and the inside diameter D i = 70 mm. 1.5m Solution: G = 80x10 3 N/mm o o 360 = 1.5 m = 1500 mm rad Nmm x10 mm G G x x x =.67 knm Page 16
17 Example 5 he preliminary design of a large shaft connecting a motor to a generator calls for the use of a hollow shaft with inner and outer diameters of 15 mm and 175 mm respectively. Knowing that the allowable shearing stress is 100 MPa, determine the maximum torque which may be transmitted (a) by the shaft as designed, (b) by a solid shaft of the same weight, (c) by a hollow shaft of the same weight and 50 mm outside diameter. Solution: d1 d 50 (b) (a) (c) (a) Hollow shaft as Designed x10 mm 3 R 175/ max x x10 Nmm 77.8 knm Page 17
18 (b) Solid shaft Area of hollow shaft = Area of solid shaft d1 d1 1.5 d mm.11x R 1.5/ max x x10 Nmm 36.1kNm mm (c) Enlarged hollow shaft Area of hollow shaft = Area of enlarged hollow shaft d d x10 mm 3 R 50 / max x x10 Nmm knm mm Page 18
19 Example 6 A solid alloy shaft of 100 mm diameter is to be frictionwelded concentrically to the end of a hollow steel shaft of the same external diameter. Find the internal diameter of the steel shaft if the angle of twist per unit length is to be 70% of that of the alloy shaft. What is the maximum torque that can be transmitted if the limiting shear stresses in the alloy and the steel are 75 MPa and 100 MPa respectively? Given that G steel = G alloy. D = D a alloy a s steel s d s Solution: alloy = steel = s a and 0. 7 s s Hence, 0. 7 Gs s Since s = a and a a Ga a G s = G a a = 0.7** s Da Ds d 0.7xx = 1.*(100 d s ) d s = 73.1 mm s Page 19
20 he torque that can be carried by the alloy is max x10 Nmm 1.73 R 50 knm he torque that can be carried by the steel is max x10 R 50 Hence the maximum allowable torque is 1.03 knm 6 Nmm 1.03 knm Page 0
21 BC Structural Analysis I Example 7 M 1 = 10 knm, M = 8 knm and G = 100 knm A is a fixed end which does not rotate. Find the torsional rotations of sections B & C. Solution: A M =10 knm B 1 1.5m 1.m M =8 knm C BC =8 knm = knm AB B C A 1 Plan View of the Shaft Since A does not rotate, section B will rotate an angle 1 and it is a rotation in a absolute sense knmx1.5 m 1 100kNm 0.03rad Page 1
22 If section B did not rotate, then section C rotates relative to B an angle of 8x 1. BC rad 100 Since section B does not rotate, therefore section C rotates, relative to A, an angle of = 1  BC = = rad. Page 
23 Example 8 A B R A 1 C R B A circular shaft with two ends fixed is subjected to a torsional moment in its span as shown. Find the torsional reactions R A & R B and draw a torsional moment diagram for the shaft. Solution: A C C B M x = 0, R A + R B = 0 Geometrical compatibility CA = CB = C Physically, CA 1 G 1 and BC G Equation of compatibility, herefore, G G Page 3
24 Page  Note that 1 + =, R A = 1 and R B = R B 1 R A 1 A C B 1
25 Strain Energy in orsion O A' A M x d 1 W ext M x E W strain ext W int Within elastic limit, d G dw int 1 d W int G G x0 therefore, E strain G Page 5
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