1. *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

2. *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

3. *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

4. *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

5. *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

6. 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.

7. 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

8.  = 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

9. 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

10. 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

11. 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)

12. 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.

13. 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.

14. 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

15. 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

16. 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:

17. *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)

18. *5.9 INELASTIC TORSION
Expressing the loading condition mathematically, we get:
T = 2∫A  2 d
Equation 5-23

19. *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/)

20. *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)

21. *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.

22. *5.9 INELASTIC TORSION
C. Ultimate torque
Magnitude of Tu can be determined “graphically” by integrating Eqn 5-23

23. *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

24. *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

25. 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?

26. 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.

27. 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.

28. 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

29. *5.10 RESIDUAL STRESS
Residual stress distribution is calculated using principles of superposition and elastic recovery

30. 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?

31. 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

32. 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

33. 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

34. 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

35. ∫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 

36. 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

37. 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

38. 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