1. CHAPTER OBJECTIVES
Discuss effects of applying torsional loading to a long straight member
Determine stress distribution within the member under torsional load
Determine angle of twist when material behaves in a linear-elastic and inelastic manner
Discuss statically indeterminate analysis of shafts and tubes
Discuss stress distributions and residual stress caused by torsional loadings

2. CHAPTER OUTLINE
Torsional Deformation of a Circular Shaft
The Torsion Formula
Power Transmission
Angle of Twist
Statically Indeterminate Torque-Loaded Members
*Solid Noncircular Shafts
*Thin-Walled Tubes Having Closed Cross Sections
Stress Concentration
*Inelastic Torsion
*Residual Stress

3. 5.1 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

4. 5.1 TORSIONAL DEFORMATION OF A CIRCULAR SHAFT
By definition, shear strain is
 = (/2)  lim ’
CA along CA
BA along BA
Let x  dx and   = d
BD =  d = dx 
 =  d dx
Since d / dx =  / = max /c
 = max  c
( )
Equation 5-2

5. ( ) ∫A 2 dA 5.2 THE TORSION FORMULA
For solid shaft, shear stress varies from zero at shaft’s longitudinal axis to maximum value at its outer surface.
Due to proportionality of triangles, or using Hooke’s law and Eqn 5-2,
 =  max  c
( )
...
 =
 max c
∫A 2 dA

6. 5.2 THE TORSION FORMULA
The integral in the equation can be represented as the polar moment of inertia J, of shaft’s x-sectional area computed about its longitudinal axis
 max = Tc J
 max = max. shear stress in shaft, at the outer surface
T = resultant internal torque acting at x-section, from method of sections & equation of moment equilibrium applied about longitudinal axis
J = polar moment of inertia at x-sectional area
c = outer radius pf the shaft

7. 5.2 THE TORSION FORMULA
Shear stress at intermediate distance, 
 = T J
The above two equations are referred to as the torsion formula
Used only if shaft is circular, its material homogenous, and it behaves in an linear-elastic manner

8. 5.2 THE TORSION FORMULA
Solid shaft
J can be determined using area element in the form of a differential ring or annulus having thickness d and circumference 2 .
For this ring, dA = 2 d
J = c4  2
J is a geometric property of the circular area and is always positive. Common units used for its measurement are mm4 and m4.

9. 5.2 THE TORSION FORMULA
Tubular shaft
J = (co4  ci4)  2

10. 5.2 THE TORSION FORMULA
Absolute maximum torsional stress
Need to find location where ratio Tc/J is maximum
Draw a torque diagram (internal torque  vs. x along shaft)
Sign Convention: T is positive, by right-hand rule, is directed outward from the shaft
Once internal torque throughout shaft is determined, maximum ratio of Tc/J can be identified

11. 5.2 THE TORSION FORMULA
Procedure for analysis
Internal loading
Section shaft perpendicular to its axis at point where shear stress is to be determined
Use free-body diagram and equations of equilibrium to obtain internal torque at section
Section property
Compute polar moment of inertia and x-sectional area
For solid section, J = c4/2
For tube, J = (co4  ci2)/2

12. 5.2 THE TORSION FORMULA
Procedure for analysis
Shear stress
Specify radial distance , measured from centre of x-section to point where shear stress is to be found
Apply torsion formula,  = T /J or max = Tc/J
Shear stress acts on x-section in direction that is always perpendicular to 

13. EXAMPLE 5.3
Shaft shown supported by two bearings and subjected to three torques.
Determine shear stress developed at points A and B, located at section a-a of the shaft.

14. EXAMPLE 5.3 (SOLN)
Internal torque
Bearing reactions on shaft = 0, if shaft weight assumed to be negligible. Applied torques satisfy moment equilibrium about shaft’s axis.
Internal torque at section a-a determined from free-body diagram of left segment.

15. EXAMPLE 5.3 (SOLN)
Internal torque
 Mx = 0;
4250 kN·mm  3000 kN·mm  T = 0
T = 1250 kN·mm
Section property
J = /2(75 mm)4 = 4.97 107 mm4
Shear stress
Since point A is at  = c = 75 mm
B = Tc/J = ... = 1.89 MPa

16. EXAMPLE 5.3 (SOLN)
Shear stress
Likewise for point B, at  = 15 mm
B = T /J = ... = MPa
Directions of the stresses on elements A and B established from direction of resultant internal torque T.

17. 5.3 POWER TRANSMISSION
Power is defined as work performed per unit of time
Instantaneous power is
Since shaft’s angular velocity  = d/dt, we can also express power as
P = T (d/dt)
P = T
Frequency f of a shaft’s rotation is often reported. It measures the number of cycles per second and since 1 cycle = 2 radians, and  = 2f T, then power
P = 2f T
Equation 5-11

18. 5.3 POWER TRANSMISSION
Shaft Design
If power transmitted by shaft and its frequency of rotation is known, torque is determined from Eqn 5-11
Knowing T and allowable shear stress for material, allow and applying torsion formula, J c T
allow =

19. 5.3 POWER TRANSMISSION
Shaft Design
For solid shaft, substitute J = (/2)c4 to determine c
For tubular shaft, substitute J = (/2)(co2  ci2) to determine co and ci

20. EXAMPLE 5.5
Solid steel shaft shown used to transmit 3750 W from attached motor M. Shaft rotates at  = 175 rpm and the steel allow = 100 MPa.
Determine required diameter of shaft to nearest mm.

21. ( ) ( ) EXAMPLE 5.5 (SOLN) Torque on shaft determined from P = T,
Thus, P = 3750 N·m/s
( )
 = = rad/s
175 rev
min
2 rad
1 rev
1 min
60 s
( )
Thus, P = T, T = N·m
= = J c
 c4
2 c2 T
allow
. . .
c = mm
Since 2c = mm, select shaft with diameter of d = 22 mm

22. 5.4 ANGLE OF TWIST
Angle of twist is important when analyzing reactions on statically indeterminate shafts
 =
T(x) dx
J(x) G ∫0 L
 = angle of twist, in radians
T(x) = internal torque at arbitrary position x, found from method of sections and equation of moment equilibrium applied about shaft’s axis
J(x) = polar moment of inertia as a function of x
G = shear modulus of elasticity for material

23.  5.4 ANGLE OF TWIST Constant torque and x-sectional area TL JG  =
 = TL JG
If shaft is subjected to several different torques, or x-sectional area or shear modulus changes suddenly from one region of the shaft to the next, then apply Eqn 5-15 to each segment before vectorially adding each segment’s angle of twist:
 = TL JG 

24. 5.4 ANGLE OF TWIST
Sign convention
Use right-hand rule: torque and angle of twist are positive when thumb is directed outward from the shaft

25. 5.4 ANGLE OF TWIST
Procedure for analysis
Internal torque
Use method of sections and equation of moment equilibrium applied along shaft’s axis
If torque varies along shaft’s length, section made at arbitrary position x along shaft is represented as T(x)
If several constant external torques act on shaft between its ends, internal torque in each segment must be determined and shown as a torque diagram

26. 5.4 ANGLE OF TWIST
Procedure for analysis
Angle of twist
When circular x-sectional area varies along shaft’s axis, polar moment of inertia expressed as a function of its position x along its axis, J(x)
If J or internal torque suddenly changes between ends of shaft,  = ∫ (T(x)/J(x)G) dx or  = TL/JG must be applied to each segment for which J, T and G are continuous or constant
Use consistent sign convention for internal torque and also the set of units

27. EXAMPLE 5.9
50-mm-diameter solid cast-iron post shown is buried 600 mm in soil. Determine maximum shear stress in the post and angle of twist at its top. Assume torque about to turn the post, and soil exerts uniform torsional resistance of t N·mm/mm along its 600 mm buried length. G = 40(103) GPa

28. EXAMPLE 5.9 (SOLN)
Internal torque
From free-body diagram
 Mz = 0; TAB = 100 N(300 mm) = 30  103 N·mm

29. EXAMPLE 5.9 (SOLN)
Internal torque
Magnitude of the uniform distribution of torque along buried segment BC can be determined from equilibrium of the entire post.
 Mz = 0;
100 N(300 mm)  t(600 mm) = 0
t = 50 N·mm

30. EXAMPLE 5.9 (SOLN)
Internal torque
Hence, from free-body diagram of a section of the post located at position x within region BC, we have
 Mz = 0;
TBC  50x = 0
TBC = 50x

31. EXAMPLE 5.9 (SOLN)
Maximum shear stress
Largest shear stress occurs in region AB, since torque largest there and J is constant for the post. Applying torsion formula
max = = ... = 1.22 N/mm2
TAB c J

32. ∫ EXAMPLE 5.9 (SOLN) Angle of twist
Angle of twist at the top can be determined relative to the bottom of the post, since it is fixed and yet is about to turn. Both segments AB and BC twist, so
A = +
TAB LAB JG
TBC dx ∫
LBC
. . .
A = rad

33. 5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
A torsionally loaded shaft is statically indeterminate if moment equation of equilibrium, applied about axis of shaft, is not enough to determine unknown torques acting on the shaft

34. 5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
From free-body diagram, reactive torques at supports A and B are unknown, Thus,
 Mx = 0; T  TA  TB = 0
Since problem is statically indeterminate, formulate the condition of compatibility; end supports are fixed, thus angle of twist of both ends should sum to zero
A/B = 0

35. 5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
Assume linear-elastic behavior, and using load-displacement relationship,  = TL/JG, thus compatibility equation can be written as
TA LAC JG
TB LBC
 = 0
Solving the equations simultaneously, and realizing that L = LAC + LBC, we get
TA = T
LBC L
( )
TB = T
LAC L
( )

36. 5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
Procedure for analysis
Equilibrium
Draw a free-body diagram
Write equations of equilibrium about axis of shaft
Compatibility
Express compatibility conditions in terms of rotational displacement caused by reactive torques
Use torque-displacement relationship, such as  = TL/JG
Solve equilibrium and compatibility equations for unknown torques

37. EXAMPLE 5.11
Solid steel shaft shown has a diameter of 20 mm. If it is subjected to two torques, determine reactions at fixed supports A and B.

38. EXAMPLE 5.11 (SOLN)
Equilibrium
From free-body diagram, problem is statically indeterminate.
 Mx = 0;
 TB N·m  500 N·m  TA = 0
Compatibility
Since ends of shaft are fixed, sum of angles of twist for both ends equal to zero. Hence,
A/B = 0

39. EXAMPLE 5.11 (SOLN)
Compatibility
The condition is expressed using the load-displacement relationship,  = TL/JG.
. . .
1.8TA  0.2TB = 750
Solving simultaneously, we get
TA = 345 N·m TB = 645 N·m

40. *5.6 SOLID NONCIRCULAR SHAFTS
Shafts with noncircular x-sections are not axisymmetric, as such, their x-sections will bulge or warp when it is twisted
Torsional analysis is complicated and thus is not considered for this text.

41. *5.6 SOLID NONCIRCULAR SHAFTS
Results of analysis for square, triangular and elliptical x-sections are shown in table

42. EXAMPLE 5.13
6061-T6 aluminum shaft shown has x-sectional area in the shape of equilateral triangle. Determine largest torque T that can be applied to end of shaft if allow = 56 MPa, allow = 0.02 rad, Gal = 26 GPa.
How much torque can be applied to a shaft of circular x-section made from same amount of material?

43. EXAMPLE 5.13 (SOLN)
By inspection, resultant internal torque at any x-section along shaft’s axis is also T. Using formulas from Table 5-1,
allow = 20T/a3; ... T = N·m
allow = 46TL/a3Gal; ... T = N·m
By comparison, torque is limited due to angle of twist.

44. EXAMPLE 5.13 (SOLN)
Circular x-section
We need to calculate radius of the x-section.
Acircle = Atriangle; ... c = mm
Limitations of stress and angle of twist require
allow = Tc/J; T = N·m
allow = TL/JGal; ... T = N·m
Again, torque is limited by angle of twist.
Comparing both results, we can see that a shaft of circular x-section can support 37% more torque than a triangular one

45. *5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
Thin-walled tubes of noncircular shape are used to construct lightweight frameworks such as those in aircraft
This section will analyze such shafts with a closed x-section
As walls are thin, we assume stress is uniformly distributed across the thickness of the tube

46. *5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
Shear flow
Force equilibrium requires the forces shown to be of equal magnitude but opposite direction, thus AtA = BtB
This product is called shear flow q, and can be expressed as
q = avgt
Shear flow measures force per unit length along tube’s x-sectional area

47. *5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
Average shear stress
avg = T
2tAm
avg = average shear stress acting over thickness of tube
T = resultant internal torque at x-section
t = thickness of tube where avg is to be determined
Am = mean area enclosed within boundary of centerline of tube’s thickness

48. *5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
Average shear stress
Since q = avgt, the shear flow throughout the x-section is
q = T
2Am
Angle of twist
Can be determined using energy methods
 = ∫ TL
4Am2G ds t O

49. *5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
IMPORTANT
Shear flow q is a product of tube’s thickness and average shear stress. This value is constant at all points along tube’s x-section. Thus, largest average shear stress occurs where tube’s thickness is smallest
Both shear flow and average shear stress act tangent to wall of tube at all points in a direction to contribute to resultant torque

50. EXAMPLE 5.16
Square aluminum tube as shown.
Determine average shear stress in the tube at point A if it is subjected to a torque of 85 N·m. Also, compute angle of twist due to this loading.
Take Gal = 26 GPa.

51. EXAMPLE 5.16 (SOLN)
Average shear stress
Am = (50 mm)(50 mm) = 2500 mm2
avg = = ... = 1.7 N/mm2 T
2tAm
Since t is constant except at corners, average shear stress is same at all points on x-section.
Note that avg acts upward on darker-shaded face, since it contributes to internal resultant torque T at the section

52.  = 0.196(10-4) mm-1[4(50 mm)] = 3.92 (10-3) rad
EXAMPLE 5.16 (SOLN)
Angle of twist
 = ∫ TL
4Am2G ds t O
= ... = 0.196(10-4) mm-1 ∫ ds O
Here, integral represents length around centerline boundary of tube, thus
 = 0.196(10-4) mm-1[4(50 mm)] = 3.92 (10-3) rad

53. 5.8 STRESS CONCENTRATION
Three common discontinuities of the x-section are:
is a coupling, for connecting 2 collinear shafts together
is a keyway used to connect gears or pulleys to a shaft
is a shoulder fillet used to fabricate a single collinear shaft from 2 shafts with different diameters

54. 5.8 STRESS CONCENTRATION
Dots on x-section indicate where maximum shear stress will occur
This maximum shear stress can be determined from torsional stress-concentration factor, K

55. 5.8 STRESS CONCENTRATION
K, can be obtained from a graph as shown
Find geometric ratio D/d for appropriate curve
Once abscissa r/d calculated, value of K found along ordinate
Maximum shear stress is then determined from
max = K(Tc/J)

56. 5.8 STRESS CONCENTRATION
IMPORTANT
Stress concentrations in shafts occur at points of sudden x-sectional change. The more severe the change, the larger the stress concentration
For design/analysis, not necessary to know exact shear-stress distribution on x-section. Instead, obtain maximum shear stress using stress concentration factor K
If material is brittle, or subjected to fatigue loadings, then stress concentrations need to be considered in design/analysis.

57. EXAMPLE 5.18
Stepped shaft shown is supported at bearings at A and B. Determine maximum stress in the shaft due to applied torques. Fillet at junction of each shaft has radius r = 6 mm.

58. EXAMPLE 5.18 (SOLN)
Internal torque
By inspection, moment equilibrium about axis of shaft is satisfied. Since maximum shear stress occurs at rooted ends of smaller diameter shafts, internal torque (30 N·m) can be found by applying method of sections

59. EXAMPLE 5.18 (SOLN)
Maximum shear stress
From shaft geometry, we have D d r
2(40 mm)
2(20 mm)
6 mm)
= = 2
= = 0.15
Thus, from the graph, K = 1.3
max = K(Tc/J) = ... = 3.10 MPa

60. EXAMPLE 5.18 (SOLN)
Maximum shear stress
From experimental evidence, actual stress distribution along radial line of x-section at critical section looks similar to:

61. *5.9 INELASTIC TORSION
To perform a “plastic analysis” for a material that has yielded, the following conditions must be met:
Shear strains in material must vary linearly from zero at center of shaft to its maximum at outer boundary (geometry)
Resultant torque at section must be equivalent to torque caused by entire shear-stress distribution over the x-section (loading)

62. *5.9 INELASTIC TORSION
Expressing the loading condition mathematically, we get:
T = 2∫A  2 d
Equation 5-23

63. *5.9 INELASTIC TORSION
A. Maximum elastic torque
For maximum elastic shear strain Y, at outer boundary of the shaft, shear-strain distribution along radial line will look like diagram (b)
Based on Eqn 5-23,
TY = (/2) Yc3
From Eqn 5-13,
d =  (dx/)

64. *5.9 INELASTIC TORSION
B. Elastic-plastic torque
Used when material starts yielding, and the yield boundary moves inward toward the shaft’s centre, producing an elastic core.
Also, outer portion of shaft forms a plastic annulus or ring
General formula for elastic-plastic material behavior,
T = (Y /6) (4c3  Y3)

65. *5.9 INELASTIC TORSION
B. Elastic-plastic torque
Plastic torque
Further increases in T will shrink the radius of elastic core till all the material has yielded
Thus, largest possible plastic torque is
TP = (2/3)Y c3
Comparing with maximum elastic torque,
TP = 4TY / 3
Angle of twist cannot be uniquely defined.

66. *5.9 INELASTIC TORSION
C. Ultimate torque
Magnitude of Tu can be determined “graphically” by integrating Eqn 5-23

67. *5.9 INELASTIC TORSION
C. Ultimate torque
Segment shaft into finite number of rings
Area of ring is multiplied by shear stress to obtain force
Determine torque with the product of the force and 
Addition of all torques for entire x-section results in the ultimate torque, Tu ≈ 2 2

68. *5.9 INELASTIC TORSION
IMPORTANT
Shear-strain distribution over radial line on shaft based on geometric considerations and is always remain linear
Shear-stress distribution must be determined from material behavior or shear stress-strain diagram
Once shear-stress distribution established, the torque about the axis is equivalent to resultant torque acting on x-section
Perfectly plastic behavior assumes shear-stress distribution is constant and the torque is called plastic torque

69. EXAMPLE 5.19
Tubular shaft made of aluminum alloy with elastic - diagram as shown.
Determine (a) maximum torque that can be applied without causing material to yield,
(b) maximum torque or plastic torque that can be applied to the shaft.
What should the minimum shear strain at outer radius be in order to develop a plastic torque?

70. EXAMPLE 5.19 (SOLN)
Maximum elastic torque
Shear stress at outer fiber to be 20 MPa. Using torsion formula
Y = (TY c/J); TY = 3.42 kN·m
Values at tube’s inner wall are obtained by proportion.

71. EXAMPLE 5.19 (SOLN)
Plastic torque
Shear-stress distribution shown below. Applying  = Y into Eqn 5-23:
TP = ... = 4.10 kN·m
For this tube, TP represents a 20% increase in torque capacity compared to elastic torque TY.

72. EXAMPLE 5.19 (SOLN)
Outer radius shear strain
Tube becomes fully plastic when shear strain at inner wall becomes 0.286(10-3) rad. Since shear strain remains linear over x-section, plastic strain at outer fibers determined by proportion:
o = ... = 0.477(10-3) rad

73. *5.10 RESIDUAL STRESS
Residual stress distribution is calculated using principles of superposition and elastic recovery

74. EXAMPLE 5.21
Tube made from brass alloy with length of 1.5 m and x-sectional area shown. Material has elastic-plastic - diagram shown. G = 42 GPa.
Determine plastic torque TP. What are the residual-shear-stress distribution and permanent twist of the tube that remain if TP is removed just after tube becomes fully plastic?

75. EXAMPLE 5.21 (SOLN)
Plastic torque
Applying Eqn 5-23,
TP = ... = 19.24(106) N·mm
When tube is fully plastic, yielding started at inner radius, ci = 25 mm and Y = rad, thus angle of twist for entire tube is
P = Y (L/ci) = ... = rad

76. EXAMPLE 5.21 (SOLN)
Plastic torque
Then TP is removed, then “fictitious” linear shear-stress distribution in figure (c) must be superimposed on figure (b). Thus, maximum shear stress or modulus of rupture computed from torsion formula,
r = (TPco)/J = ... = MPa
i = ( MPa)(25 mm/50 mm) = MPa

77. EXAMPLE 5.21 (SOLN)
Plastic torque
Angle of twist ’P upon removal of TP is
’P = (TP L)/(JG) = ... = rad
Residual-shear-stress distribution is shown.
Permanent rotation of tube after TP is removed,
+  =  = rad

78. CHAPTER REVIEW
Torque causes a shaft with circular x-section to twist, such that shear strain in shaft is proportional to its radial distance from its centre
Provided that material is homogeneous and Hooke’s law applies, shear stress determined from torsion formula,  = (Tc)/J
Design of shaft requires finding the geometric parameter, (J/C) = (T/allow)
Power generated by rotating shaft is reported, from which torque is derived; P = T

79. ∫0  CHAPTER REVIEW Angle of twist of circular shaft determined from
If torque and JG are constant, then
For application, use a sign convention for internal torque and be sure material does not yield, but remains linear elastic
 =
T(x) dx JG ∫0 L
 = TL JG 

80. CHAPTER REVIEW
If shaft is statically indeterminate, reactive torques determined from equilibrium, compatibility of twist, and torque-twist relationships, such as  = TL/JG
Solid noncircular shafts tend to warp out of plane when subjected to torque. Formulas are available to determine elastic shear stress and twist for these cases
Shear stress in tubes determined by considering shear flow. Assumes that shear stress across each thickness of tube is constant

81. CHAPTER REVIEW
Shear stress in tubes determined from  = T/2tAm
Stress concentrations occur in shafts when x-section suddenly changes. Maximum shear stress determined using stress concentration factor, K (found by experiment and represented in graphical form). max = K(Tc/J)
If applied torque causes material to exceed elastic limit, then stress distribution is not proportional to radial distance from centerline of shaft

82. CHAPTER REVIEW
Instead, such applied torque is related to stress distribution using the shear-stress-shear-strain diagram and equilibrium
If a shaft is subjected to plastic torque, and then released, it will cause material to respond elastically, causing residual shear stress to be developed in the shaft