Engineering Mechanics of Solids by Popov Chapter 4: Torsion

1. Chapter 4
Torsion
Mechanics of Solids

2. Application of the Method of Sections
• In analyzing members for torque, the basic method of sections is
employed.
• Equation of statics. If the x-axis is directed along a member, 𝑀 𝑥 = 0
• After determining this torque, an analysis begins by separating a
member of a section perpendicular to the axis of a member. Then
either side of a member can be isolated and the ‘internal torque’
found.
• The external and the internal torques are equal, but have opposite
sense.
• A section where the largest internal torque is developed is the critical
section.

3. Part A – Torsion of Circular Elastic Bars
• Assumptions along with homogeneity of the material:
1. A plane section of material perpendicular to the axis of a circular
members remains plane after the torques are applied, i.e., no
warpage or distortion of parallel planes normal to the axis of a
member take place.
2. In a circular member subjected to torque, shear strains 𝛾 vary
linearly from the central axis reaching 𝛾 𝑚𝑎𝑥 at the periphery.
Fig. 1: Variation of strain in circular
Member subjected to torque

4. • As shown in Fig.1, this assumption means that an imaginary plane
such as DO1O3C moves to D’O1O3C when the torque is applied.
Alternatively, if an imaginary radius O3C is considered fixed in
direction, similar radii initially at O2B and O1D rotate to their
respective new positions O2B’ and O1D’. These radii remain straight.
• These assumptions hold only for circular solid and tubular members.
3. If attention is confined to the linearly elastic material, Hooke’s law
applies, shear stress is proportional to shear strain.
• Stresses vary linearly from the central axis of a circular member.

5. Fig. 2: Shear strain assumption leading to elastic shear stress distribution in a circular member
• The maximum shear stress occur at points most remote from the
center O and is designated 𝜏 𝑚𝑎𝑥, such as points C and D in Fig. 2, lie
at the periphery of a section at a distance c from the center.
• For linear shear stress variation, at any arbitrary point at a distance 𝜌
from O, the shear stress is 𝜌 𝑐 𝜏 𝑚𝑎𝑥.
• For equilibrium, internal resisting torque must equal the externally
applied torque T.

6. • Hence,
• At a given section, 𝜏 𝑚𝑎𝑥 and c are constant; hence,
𝜏 𝑚𝑎𝑥
𝑐
𝜌2 𝑑𝐴 = 𝑇
• However, J = 𝜌2 𝑑𝐴, the polar moment of inertia of a cross
sectional area, is also a constant for a particular cross-section area.
• For a circular section, 𝑑𝐴 = 2𝜋𝜌 𝑑𝜌, where 2𝜋𝜌 is the circumference
of an annulus with a radius 𝜌 of width d𝜌. Hence,
𝐽 = 𝜌2
𝑑𝐴 =
0
𝑐
2𝜋𝜌3
𝑑𝜌 =
𝜋𝑐4
2
=
𝜋𝑑4
32

7. • Where, d is the diameter of a solid circular shaft. By using the symbol
J for the polar moment of inertia of a circular area,
𝜏 𝑚𝑎𝑥 =
𝑇𝑐
𝐽
𝜏 =
𝜌
𝑐
𝜏 𝑚𝑎𝑥 =
𝑇𝜌
𝐽
• The units of the torsional shear stress Pa in SI units or psi in U.S.
customary units.
Fig.3: Variation of stress in an elastic Fig. 4: Elastic behavior of a circular member in torsion
circular tube Having an inner core of soft material

8. • For a tube as shown in Fig. 3, the limits of integration is from c to b.
• Hence, for a circular tube,
𝐽 = 𝜌2 𝑑𝐴 =
𝑏
𝑐
2𝜋𝜌3 𝑑𝜌 =
𝜋
2
𝑐4 − 𝑏4
• J for a circular tube equals +J for a solid shaft using the outer
diameter and –J for a solid shaft using the inner diameter.
• For, very thin tubes, c – b = t, the thickness of the tube
𝐽 ≈ 2𝜋𝑅 𝑎𝑣
3 𝑡
Where, 𝑅 𝑎𝑣 = 𝑏 + 𝑐 2
• If a circular bar is made from two different materials bonded together,
as shown in Fig. 4(a), the same strain assumption applies as for a solid
member. Through, Hooke’s law, the shear-stress distribution becomes
as in Fig. 4(b)

9. Procedure Summary
• For the torsion problems of circular shafts:
1. Equilibrium conditions are used for determining the internal
resisting torques at a section.
2. Geometry of deformation (kinematics) is postulated such that shear
strain varies linearly from the axis of a shaft.
3. Material properties (constitutive relations) are used to relate shear
strains to shear stresses and permit calculation of shear stresses at
a section.
Fig. 5: Existence of shear stresses on mutually perpendicular planes in a circular shaft subjected to torque

10. • Shear stresses are acting to form a couple resisting the externally
applied torques. Such as an infinitesimal cylindrical element, shown in
Fig. 5(b).
• The variation of the shear stresses on the mutually perpendicular
planes is shown in Fig. 5(c).
• Such shear stresses can be transformed into an equivalent system of
normal stresses acting at angles of 45° with the shear stresses.
𝜏 = 𝜎1 = −𝜎2
• If the shear strength of a material is less than its strength in tension, a
shear failure takes place on a plane perpendicular to the axis of a bar.
Tis kind of failure occurs gradually and exhibits ductile behavior.
• Alternatively, if the converse is true, i.e., 𝜎1 < 𝜏, a brittle fracture is
caused by the tensile stresses along a helix forming an angle of 45°
with the bar axis.

11. Design of Circular Members in Torsion
• Allowable shear stress must be selected depending upon the
information available from experiments and on the intended
applications.
• Typically, the shear strength of ductile materials is only about half as
large as their tensile strength.
• The ASME (American Society of Mechanical Engineers) code of
recommended practice for transmission shafting gives an allowable
value in shear stress of 8000 psi for unspecified steel and 0.3 of yield,
or 0.18 of ultimate, shear strength, whichever is smaller.
• After the torque to be transmitted by a shaft is determined and the
maximum allowable shear stress is selected. The proportions of a
member are given as
𝐽
𝑐
=
𝑇
𝜏 𝑚𝑎𝑥

12. • Where J/c is the parameter on which the elastic strength of a shaft
depends. For an axially loaded rod, such a parameter is the cross-
sectional area of a member. For a solid shaft, J/c = 𝜋c3/2, where c is
the outside radius.
• Large local stresses generally develop at changes in cross-sections and
at splines and keyways, where the torque is actually transmitted.
• 1 hp = 175.7 watt (W)
• For a shaft rotating with a frequency of f Hz, the angle is 2𝜋𝑓 rad/s. If
a shaft were transmitting a constant torque T measured in Nm, It
would do 2𝜋𝑓T Nm of work per second.

13. • In the U.S. customary system of units, 1 hp does work of 550 ft-lb/sec,
or 550 × 12 × 60 in-lb per minute. If the shaft rotates at N rpm,
𝑇 =
63,000 × ℎ𝑝
𝑁
𝑖𝑛 − 𝑙𝑏
Stress Concentrations
• For stepped shafts where the diameters of the adjoining portions
change abruptly, large perturbations of shear stresses take place.
• Stress-concentration factors depend only on the geometry of a
member.
• Corresponding to the given r/(d/2) ratio, the stress-concentration
factor K is read from the curve.

14. Fig. 6: Torsional stress-concentration factors in circular
shafts of two diameters
• From the definition of K, the actual maximum shear stress
𝜏 𝑚𝑎𝑥 = 𝐾
𝑇𝑐
𝐽
• Where, the shear stress Tc/J is determined for the smaller shaft.
• A study of stress-concentration factors shown in Fig. 6 suggests the
need for a fillet radius r at all sections where transition occurs.

15. Fig. 7: Circular shaft with a keyway
• A shaft prepared for a key, Fig. 7 is no longer a circular member.
• According to the procedures suggested by the ASME, the allowable
shear stress must be reduced by 25%. This presumably compensates
for the stress-concentration and reduction in cross-sectional area.
• Because of some inelastic or non-linear response in real materials,
the theoretical stress concentrations based on the behavior of linearly
elastic material tend to be somewhat high.

16. Angle-of-Twist of Circular Members
Fig. 8: Deformation of a circular bar element due to torque
• A typical element of length dx of a shaft is shown isolated in Fig. 8
• In the element shown, a line on its surface such as CD initially parallel
to the axis of the shaft. After the torque is applied, it assumes a new
position CD’. At the same time, by virtue of assumption 2, radius OD
remains straight and rotates through a small angle dφ to a new
position OD’.

17. • Denoting the small angle DCD’ by 𝛾 𝑚𝑎𝑥, from geometry,
𝑎𝑟𝑐 𝐷𝐷′ = 𝛾 𝑚𝑎𝑥 𝑑𝑥
or 𝑎𝑟𝑐 𝐷𝐷′ = 𝑑𝜙 𝑐
𝛾 𝑚𝑎𝑥 𝑑𝑥 = 𝑑𝜙 𝑐
• 𝛾 𝑚𝑎𝑥 applies only in the zone of an infinitesimal tube of constant
maximum shear stress, 𝜏 𝑚𝑎𝑥 = G 𝛾 𝑚𝑎𝑥 = 𝑇𝑐 𝐽.
• Hence, 𝛾 𝑚𝑎𝑥 = 𝑇𝑐 𝐺𝐽.
𝑑𝜙
𝑑𝑥
=
𝑇
𝐺𝐽
𝜙 = 𝜙 𝐵 − 𝜙 𝐴 =
𝐴
𝐵
𝑑𝜙 =
𝐴
𝐵
𝑇𝑥
𝐺𝐽 𝑥
𝑑𝑥
• Where, 𝜙 𝐵 and 𝜙 𝐴 are, respectively, the global rotations at ends B
and A.

18. • For a circular shaft having length L,
𝜙 =
𝑇𝐿
𝐺𝐽
• This can be compared to,
Δ =
𝑃𝐿
𝐴𝐸
• For axially loaded bars, 𝜙 ⟺ Δ, T ⟺ P, J ⟺ A, and G ⟺ E.
• Torsional stiffness, 𝑘 𝑡 =
𝑇
𝜙
=
𝐺𝐽
𝐿
𝑖𝑛−𝑙𝑏
𝑟𝑎𝑑
or
𝑁𝑚
𝑟𝑎𝑑
• This constant represents the torque required to cause a rotation of 1
rad, i.e. 𝜙 = 1.
• The reciprocal of 𝑘 𝑡 defines the torsional flexibility 𝑓𝑡. This defines the
rotation resulting from application of a unit torque, i.e. T = 1.
• In a torsion test, 𝜙, T, L, and J can be obtained from the dimensions.

19. Statically Indeterminate Problems
• In considering linearly elastic problems with one degree of external
indeterminacy, i.e., cases where there are two reactions, the force
(flexibility) method is particularly advantageous.
• Such problems are reduced to statical determinacy by removing one
of the redundant reactions and calculating the rotation 𝜙0 at the
released support.
• The required boundary conditions are restored by twisting the
member at the released end through an angle 𝜙1 such that
𝜙0 + 𝜙1 = 0
• Torsion problem also occur with internal statical indeterminacy in
composite shafts built up from two or more materials, such as shown
in Fig. 9

20. Fig. 9: Externally statically indeterminate bar in torsion
• The angle-of-twist 𝜙 is the same for each
Constituent part of the member. Total
Applied torque is the sum of its parts.
𝑇 =
𝑖
𝑘 𝑡 𝑖 𝜙
• For global equilibrium:
𝑇1 + 𝑇2 + 𝑇 = 0
• For geometric compatibility:
𝜙 𝐴𝐵 = 𝜙 𝐵𝐶
𝑇1 𝐿1
𝐽1 𝐺1
=
𝑇2 𝐿2
𝐽2 𝐺2

21. Alternative Differential Equation Approach for
Torsion Problems
Fig. 10: Infinitesimal element of a circular bar subjected to torque
• Consider an element, shown in Fig. 10, subjected to the end torques T
and T + dT and to an applied disturbed torque tx . By using the right-
hand screw rule for the torques, all the quantities are shown in the
figure as having a positive sense. For equilibrium,
𝑡 𝑥 𝑑𝑥 + 𝑑𝑇 = 0 or
𝑑𝑇
𝑑𝑥
= −𝑡 𝑥
As,
𝑑𝜙
𝑑𝑥
=
𝑇
𝐺𝐽

22. 𝐺𝐽
𝑑2 𝜙
𝑑𝑥2
=
𝑑𝑇
𝑑𝑥
= −𝑡 𝑥
• The constants appearing in the solution of this differential equations
are determined from the boundary conditions at the ends of a shaft.
Energy and Impact Loads
• The deflection of a member can be determined by equating the
internal shear stress energy Ush for a member to the external work,
We due to the applied force.
Fig. 11: Cross section of an elastic shaft subjected to a
constant torque

23. • The shear stress in an elastic circular shaft subjected to a torque
varies linearly from the longitudinal axis. Hence, the shear stress
acting on an element at a distance 𝜌 from the center of the cross-
section is 𝜏 𝑚𝑎𝑥 𝜌 𝑐.
• Integrating over the volume V of the rod L inches long.
𝑈𝑠ℎ =
𝜏2
2𝐺
𝑑𝑉 =
𝜏 𝑚𝑎𝑥
2 𝜌2
2𝐺𝑐2
2𝜋𝜌 𝑑𝜌 𝐿
=
𝜏 𝑚𝑎𝑥
2
2𝐺
2𝜋𝐿
𝑐2
0
𝑐
𝜌3 𝑑𝜌 =
𝜏 𝑚𝑎𝑥
2
2𝐺
2𝜋𝐿
𝑐2
𝑐4
4
=
𝜏 𝑚𝑎𝑥
2
2𝐺
1
2
𝑣𝑜𝑙

24. • If torque T is gradually applied to the shaft, the external work 𝑊𝑒 =
1
2
𝑇𝜙, where 𝜙 is the angular rotation of the free end in radians.
• 𝜏 𝑚𝑎𝑥 = 𝑇𝑐 𝐽, the volume of the rod 𝜋𝑐2 𝐿, and 𝐽 = 𝜋𝑐4 2. Thus,
𝑈𝑠ℎ =
𝜏 𝑚𝑎𝑥
2
2𝐺
1
2
𝑣𝑜𝑙 =
𝑇2 𝑐2
2𝐽2 𝐺
1
2
𝜋𝑐2
𝐿 =
𝑇2 𝐿
2𝐽𝐺
• Then, from 𝑊𝑒 = 𝑈𝑠ℎ
𝑡𝜙
2
=
𝑇2 𝐿
2𝐽𝐺
and 𝜙 =
𝑇𝐿
𝐽𝐺

25. Shaft Couplings
• For maintenance or assembly reasons, it is often desirable to make up
a long shaft from several pieces. To join the pieces of a shaft together,
flange-shaft couplings as shown in Fig. 12 are used.
Fig. 12: Flanged shaft coupling
• When bolted together, such couplings are termed ‘rigid’, another type
‘flexible’ provides for misalignment of adjoining shafts.

26. • For rigid couplings, it is customary to assume that shear strains in the
bolts vary linearly as their distance from the axis of the shaft.
• The shear stress in any one bolt is assumed to be uniform and is
governed by the distance from its center to the center of the
coupling. Multiplying the shear stress with bolt’s cross-section area,
the force in a bolt is found.
• For bolts of equal size in two ‘bolt circles’, the forces on the bolts
located by the respective radii 𝑎 and 𝑏 are as shown in Fig. 12(c).
• The moment of the forces developed by the bolts around the axis of a
shaft gives the torque capacity of a coupling.
• Instead of a continuous cross-section, a discrete number of points is
considered. Stress concentrations are present at the points of contact
of the bolts with the flanges of a coupling.

27. • The outlined method of analysis is valid only for the case of a coupling
in which the bolts act primarily in shear.
• However, in some couplings bolts are tightened so much that the
initial tension in the bolts is great enough to cause the entire coupling
to act in friction.
• Under these circumstances, the suggested analysis is valid only as a
measure of the ultimate strength of the coupling should the stresses
in the bolts be reduced.
• However, if high tensile strength bolts are used, there is little danger
of this happening, and the strength of the coupling may be greater
than it would be if the bolts had to act in shear.

28. Torsion of Inelastic Circular Bars
• Shear Stresses and Deformations in Circular Shafts in the Inelastic
Range:
• The equilibrium requirements at a section must be met. The
deformation assumption of linear strain variation from the axis
remains applicable. Only the difference in material properties affect
the solution.
• A section through a shaft is shown in Fig. 13(a), the linear strain
variation is shown schematically in the same figure.
• Some possible mechanical properties of materials in shear, for
example in experiments with thin tubes in torsion, are shown in Fig.
13 (b),(c) and (d). The corresponding shear stress distribution is
shown. The stresses are determined from the strain

29. Fig. 13: Stresses in circular members due to torque

30. • After the stress distribution is known, torque 𝑇 carried by these
stresses is found by integration over the cross-sectional area of the
shaft.
𝑇 = 𝜏 𝑑𝐴 𝜌
• Although the shear-stress distribution after the elastic limit is
exceeded is nonlinear and the elastic torsion formula, 𝜏 𝑚𝑎𝑥 = 𝑇𝑐 𝐽
does not apply, it is sometimes used to calculate ‘fictitious stress’ for
the ultimate torque.
• The computed stress is called the ‘modulus of rupture’; the largest
ordinates of the dashed lines on Fig. 13(f) and (g)
• For a thin-walled tube, the stress distribution is very nearly the same
regardless of the mechanical properties of a material; Fig. 14

31. Fig. 14: For thin-walled tubes the difference between elastic and inelastic
stresses is small
• If a shaft is strained into the inelastic range and the applied torque is
then removed, every imaginary ‘annulus’ rebounds elastically.
• Because of the differences in the strain paths causing permanent
strain in the material, residual stresses develop.
𝑑Φ
𝑑𝑥
=
𝛾 𝑚𝑎𝑥
𝑐
=
𝛾 𝑎
𝜌 𝑎
• To determine the rate of twist of a circular shaft or tube either the
maximum shear strain at 𝑐 or the strain at 𝜌 𝑎 determined from the
stress-strain diagram must be used.

32. Part C- Torsion of Solid Non-Circular Members
• Sections perpendicular to the axis of a member warp when a torque
is applied.
• The nature of the distortions that take place in a rectangular section
can be surmised form Fig. 15
Fig. 15: Rectangular bar (a) before and (b) after the torque is applied

33. • For a rectangular member, the corner elements do not distort. Shear
stresses are zero at the corners and maximum at the midpoints of the
long slides.
• Fig. 16 shows the shear-stress distribution along three radial lines
emanating from the center.
• For a circular section, the stress is a maximum at the most remote
point, but for the rectangular, the stress is zero at the most remote
point.
Fig. 16: Shear stress distribution in a Fig.17 : The shear stress shown at
rectangular shaft subjected the corner cannot exist
to a torque

34. • If a shear stress 𝜏 existed at the corner, it could be resolved into two
components parallel to the edges of the bar.
• However, as shears always occur in pairs acting on two mutually
perpendicular planes, these components would have to be met by
shears lying in the planes of the outside surfaces. Which is impossible
as outside surfaces are free of all stresses. At corner, 𝜏 = 0 (Fig. 17)
• Analytical solutions for torsion of rectangular, elastic members:
𝜏 𝑚𝑎𝑥 =
𝑇
𝛼𝑏𝑡2 and Φ =
𝑇𝐿
𝛽𝑏𝑡3 𝐺
Where, 𝑇 = applied torque, 𝑏 = length of the long side, 𝑡 = thickness
or width of the short side of a rectangular section.
• The values of parameters 𝛼 and 𝛽 depend upon the ratio 𝑏 𝑡 as
shown in table below.
• For thin section, where b ≫ 𝑡, the values of 𝛼 and 𝛽 approaches 1/3.

35. • The torsional stiffness 𝑘 𝑡 for a rectangular section,
𝑘 𝑡 =
𝑇
Φ
= 𝛽𝑏𝑡3
𝐺
𝐿
• The solution of the partial differential equation that must be solved in
the elastic torsion problem is mathematically identical to that for a
thin membrane, such as a soap film, lightly stretched over a hole.
• This hole must be geometrically similar to the cross section of the
shaft being studied. Light air pressure must be kept on one side of the
membrane.

36. 1. The shear stress at any point is proportional to the slope of the
stretched membrane at the same point, Fig. 18 (a)
2. The direction of a particular shear stress at a point is at right angles
to the slope of the membrane at the same point, Fig. 18(a)
3. Twice the volume enclosed by the membrane is proportional to the
torque carried by the section.
Fig. 18: Membrane analogy: (a) simply connected region, and (b) multiply connected (tubular) region

37. • ‘membrane analogy’ is a very useful mental aid for visualizing stresses
and torque capacities of members.
• Consider a narrow rectangular bar subjected to torque 𝑇 as shown in
Fig. 19 A stretched membrane for this member is shown in Fig. 19 (a)
• If such a membrane is lightly stretched by internal pressure, a section
through the membrane is a parabola, Fig. 19 (b). For this surface, the
maximum slope, hence maximum shear stress, occurs along the
edges, Fig. 19 (c)
Fig. 19: Illustration of the membrane analogy for a rectangular bar in torsion

38. • No shear stress develops along a line bisecting the bar thickness 𝑡.
The maximum shear stresses along the short sides are small.
• The volume enclosed by the membrane is directly proportional to the
torque the member can carry at a given maximum stress. For this
reason sections shown in Fig. 20 can carry approximately the same
torque at the same maximum shear stress since the volumes are
same. (for all these shapes, 𝑏 = 𝐿 and the 𝑡’s are equal)
• The contour lines of a soap film will ‘pile up’ at points 𝑎 of re-entrant
corners. High local stresses will occur at those points.
Fig. 20: Members of equal cross-section areas of the same thickness carrying the same torque

39. • The ‘sand-heap analogy’ is developed for plastic torsion.
• Dry sand is poured onto a raised flat surface having the shape of the
cross section of a member.
• The surface of the sand heap is assumes a constant slope. For
example, a cone is formed on a circular disk, or a pyramid on a square
base.
• The constant maximum slope of the sand corresponds to the limiting
surface of the membrane in the previous analogy.
• The volume of the sand heap, hence its weight, is proportional to the
fully plastic torque carried by a section.
• The other items in connection with the sand surface have the same
interpretation as those in the membrane analogy.

40. Warpage of Thin-Walled Open Sections
• No in-plane deformation can take place along the entire width and
length of the bar’s middle surface. The same holds true for middle
surfaces of curved bars, as well as for an assembly of bars.
• An 𝐼 section, shown in Fig. 21, consists of three flat bars, and during
twisting, during twisting, the three middle surfaces of these bars do
not develop in-plane deformations.
Fig. 21: Cross-sectional warpage due to applied torque

41. • Due to symmetry, this 𝐼 section twists around its centroidal axis,
which is in this case is also the center of twist.
• During twisting, as the beam flanges displace laterally, the
undeformed middle surface 𝑎𝑏𝑐𝑑 rotates about point A, Fig. 21(a).
• Plane sections of an 𝐼 beam warp, i.e., cease to be plane, during
twisting. By contrast, for circular members, the section perpendicular
to the axis remain plane during twisting. For thick sections, including
rectangular bars warpage effect is negligible.
• For thin-walled torsion-members, commonly employed in aircraft,
automobiles, ships, bridges, etc., the cross-sectional warpage, or its
restraint may have an important effect on member strength and its
stiffness.
• To maintain required compatibility of deformations, in-plane flange
moments 𝑀, as shown in Fig. 21(b), must develop.

42. Torsion of Thin-Walled Tubular Members
• Consider a tube of an arbitrary shape with varying wall thickness,
such as shown in Fig. 22 (a), subjected to torque 𝑇. Isolate an element
from these tube, as shown enlarged in Fig. 22 (b)
• These element must be in equilibrium under the action of forces 𝐹1,
𝐹2, 𝐹3, and 𝐹4.
Fig. 22: Thin-wall tubular member of variable thickness

43. • From 𝐹𝑥 = 0, 𝐹1 = 𝐹3, 𝜏2 𝑡2 𝑑𝑥 = 𝜏1 𝑡1 𝑑𝑥
• The product of the shear stress and the wall thickness is the same,
i.e., constant, on any such planes. This constant is denoted by 𝑞,
which is measured in the units of force per unit distance along the
perimeter. Unit = N/m or lb/in
• Shear stresses on mutually perpendicular planes are equal at a corner
of an element. Hence, at a corner such as A in Fig. 22 (b), 𝜏2 = 𝜏3;
similarly 𝜏1 = 𝜏4. therefore, 𝜏4 𝑡1 = 𝜏3 𝑡2, or in general 𝑞 is constant in
the plane of a section perpendicular to the axis of a member.
• Consider the cross section of a tube as shown in Fig. 22 (c). The force
per unit distance of the perimeter of this tube, is constant and is the
‘shear flow’ 𝑞.
• This shear flow multiplied by the length by 𝑑𝑠 of the perimeter gives a
force by 𝑞 𝑑𝑠 per differential length.

44. • The product of this infinitesimal force 𝑞 𝑑𝑠 and 𝑟 around some
convenient point such as 𝑂, Fig. 22 (c) gives the contribution of an
element to the resistance of applied torque 𝑇.
𝑇 = 𝑟𝑞 𝑑𝑠
• The integration process is carried around the tube along the center
line of the perimeter. For a tube, 𝑞 is a constant.
• From Fig. 22 (c), it can be seen that 𝑞 𝑑𝑠 is twice the value of the
shaded area of an infinitesimal triangle of altitude 𝑟 and base
𝑑𝑠.hence, the complete integral is twice the whole area bounded by
the center line of the perimeter of the tube.
𝑞 =
𝑇
2𝐴

45. • This applies only to thin-walled tubes. The area 𝐴 is approximately an
average of the two areas enclosed by the inside and the outside
surfaces of a tube.
• The shear stress at any point of a tube where the wall thickness is 𝑡 is,
𝜏 =
𝑞
𝑡
• For linearly elastic materials, the angle of twist for a hollow tube can
be found by applying the principle of conservation of energy.
𝑈𝑠ℎ =
𝜏2
2𝐺
𝑑𝑉 =
𝑇2
8𝐴2 𝐺𝑡
𝑑𝑠 =
𝑇2
8𝐴2 𝐺
𝑑𝑠
𝑡
• Here, 𝑑𝑉 = 1 × 𝑡 𝑑𝑠
• Equating this relation to the external work per unit length of member
expressed as 𝑊𝑒 = 𝑇𝜃 2. 𝜃 is angle-of-twist per unit length of the
tube.

46. • The governing differential equation becomes:
𝜃 =
𝑑Φ
𝑑𝑥
=
𝑇
4𝐴2 𝐺
𝑑𝑠
𝑡
• For a prismatic tube subjected to a constant torque, Φ = 𝜃𝐿.
• Torsional stiffness for a thin walled hollow tube,
𝑘 𝑡 =
𝑇
Φ
=
4𝐴2
𝑑𝑠 𝑡
𝐺
𝐿