Engineering science

1. Chapter 1- Static engineering systems
1.3 Torsion in circular shafts
1.3.1 theory of torsion and its assumptions (eg
determination of shear stress, shear strain,
shear modulus)
1.3.2 distribution of shear stress and angle of
twist in solid and hollow circular section
shafts
1

2. Torsional loads on circular shafts
• Interested in stresses and strains of
circular shafts subjected to twisting
couples or torques
• Turbine exerts torque T on the shaft
• Shaft transmits the torque to the
generator
133 2

3. Assumptions in torsion

5. Shear strain
• The value of shear stress τ is expressed by
d
Tr T
Tr 16 T
τ= τ=
J
= 24 =
πd π d3
J
32
• The shear strain γ is given by
where θ=angle of twist or degree of rotation
L= length of the specimen

6. Torque twist diagram
• The elastic properties in torsion can be obtained by using
the torque at the proportional limit where the shear stress
is calculated corresponding to the twisting moment
• The torsional elastic limit or yield strength can be
obtained from testing a tubular specimen since the stress
gradient are practically eliminated.

7. Twisting moment
• Consider a cylindrical bar subjected to a torsional moment
at one end.
• The twisting moment is resisted by shear stresses set up
in the cross section of the bar. (zero at centre, max at
surface)
• Twisting moment is the sum of shear torques over the
cross section
• Since is the polar moment of inertia of the area
with respect to the axis of the bar

8. • The maximum shear stress at the surface of the
bar is
• For a tubular specimen, the shear stress on the
outer surface is
Where D1=Outside diameter of tube
D2=Inside diameter of tube
• The modulus of elasticity in shear G or the
modulus of rigidity is as follows:

9. Problem
SOLUTION:
• Apply a static equilibrium analysis on
the two shafts to find a relationship
between TCD and T0
• Apply a kinematic analysis to relate
the angular rotations of the gears
• Find the maximum allowable torque
Two solid steel shafts are connected on each shaft – choose the smallest
by gears. Knowing that for each
shaft G = 11.2 x 106 psi and that the • Find the corresponding angle of twist
allowable shearing stress is 8 ksi, for each shaft and the net angular
determine (a) the largest torque T0 rotation of end A
that may be applied to the end of
shaft AB, (b) the corresponding angle
through which end A of shaft AB
rotates.
156 9

10. SOLUTION:
• Apply a static equilibrium analysis on • Apply a kinematic analysis to relate
the two shafts to find a relationship the angular rotations of the gears
between TCD and T0
rBφ B = rCφC
∑ M B = 0 = F ( 0.875 in.) − T0 rC 2.45 in.
φB = φC = φC
∑ M C = 0 = F ( 2.45 in.) − TCD rB 0.875 in.
TCD = 2.8 T0 φ B = 2.8φC
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11. • Find the T0 for the maximum • Find the corresponding angle of twist for each
allowable torque on each shaft – shaft and the net angular rotation of end A
choose the smallest
T L
φ A / B = AB =
( 561lb ⋅ in.)( 24in.)
(
J ABG π ( 0.375 in.) 4 11.2 × 106 psi
2
)
= 0.387 rad = 2.22o
TAB c T ( 0.375 in.)
τ max = 8000 psi = 0
π ( 0.375 in.) 4 T L 2.8 ( 561lb ⋅ in.)( 24in.)
φC / D = CD =
( )
J AB
2 J CDG π ( 0.5 in.) 4 11.2 × 106 psi
2
T0 = 663 lb ⋅ in.
TCD c 2.8 T0 ( 0.5 in.)
= 0.514 rad = 2.95o
( )
τ max = 8000 psi =
π ( 0.5 in.) 4
J CD
2 φ B = 2.8φC = 2.8 2.95o = 8.26o
T0 = 561lb ⋅ in. T0 = 561lb ⋅ in φ A = φ B + φ A / B = 8.26o + 2.22o φ A = 10.48o
156 11

12. Design of Transmission Shafts
• Principal transmission shaft • Determine torque applied to shaft at
performance specifications are: specified power and speed,
­ power P = Tω = 2πfT
­ speed P P
T= =
ω 2πf
• Designer must select shaft
material and cross­section to • Find shaft cross­section which will not
meet performance specifications exceed the maximum allowable
without exceeding allowable shearing stress,
shearing stress. Ta
τ max =
J
J π 3 T
= a = ( solid shafts )
a 2 τ max
π
J
=
a2 2 a2
( a24 − a14 ) = τ T ( hollow shafts )
max
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13. Stress concentrations
• The derivation of the torsion formula,
Ta
τ max =
J
assumed a circular shaft with uniform
cross­section loaded through rigid end
plates.
• The use of flange couplings, gears and
pulleys attached to shafts by keys in
keyways, and cross­section discontinuities
can cause stress concentrations
• Experimental or numerically determined
concentration factors are applied as
Ta
τ max = K
J
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14. Stress distribution of a solid shaft
T*r
T τ= ; T (N-m)
τ (N/m ) 2
J
J = π d4 / 32 ; (m4)
r=d/2
T’
Shear stress
Circular shafts subjected to torsion
G
τmax = T * a / J τ max
τa = T * r / J τ=G*γ
r
γ
T

15. • Multiplying the previous equation by the
shear modulus,
r
Gγ = Gγ max
a
From Hooke’s Law, τ = Gγ , so
a r r
τ = τ max
a
The shearing stress varies linearly with the
J = 1 π a4
2
radial position in the section.
a1 a2 r
(
J = 1 π a2 − a14
2
4
)

16. Angle of twist
• From observation, the angle of twist of the
shaft is proportional to the applied torque and
to the shaft length.
θ ∝T
θ ∝L
• When subjected to torsion, every cross­section
of a circular shaft remains plane and
θ undistorted.
• Cross­sections for hollow and solid circular
shafts remain plain and undistorted because a
circular shaft is axisymmetric.
• Cross­sections of noncircular (non­
axisymmetric) shafts are distorted when
subjected to torsion.
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