t/ mohammed. a. mustafa - national university
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T/ Mohammed. A. Mustafa
Consider a bar rigidly clamped at one end and twisted at the other
end by a torque (twisting moment) T = Fd applied in a plane
perpendicular to the axis of the bar, as shown in Fig. 1. Such a bar is
in torsion. An alternative representation of the torque is the curved
arrow shown in the figure.
Fig. 1 Torque applied to a circular shaft
Introduction
Polar Moment of Inertia
J =π
32(π·4 β π4)
J =π
32π4
For a hollow circular shaft of outer
diameter π« with a concentric circular
hole of diameter π the polar moment of
inertia of the cross-sectional area is
given by
For a solid circular shaft of diameter π the polar moment of inertia of the
cross-sectional area is given by
A mathematical property of the geometry of the cross section which occurs in the study of
the stresses set up in a circular shaft subject to torsion is the polar moment of inertia J,
defined in a statics course
Polar Moment of Inertia
Torsional Shearing Stress
So, for either a solid or a hollow circular shaft subject to a twisting moment
π» the torsional shearing stress Ο at a distance r from the center of the
shaft is written as
The maximum shear stress is found by replacing Ο by the radius r of the shaft:
Distribution of shear stress
along the radius of a circular
shaft.
Shearing Strain
The ratio of the shear stress π to the shear strain πΎ is called the shear modulus
Again the units of G are the same as those of shear stress, since the shear strain is dimensionless
πΊ =π
πΎπ = shear stress
πΎ = shear strain
G = shear modulus
π =ππΏ
πΊπ½π = angle of twist
πΏ = length of the shaft
πΎ = tan β β β πΎ =ππ
πΏ
SOLVED PROBLEMS
If a twisting moment of 1100 N.m is impressed upon a 4.4 cm diameter shaft, what is
the maximum shearing stress developed? Also, what is the angle of twist in a 150
cm length of the shaft? The material is steel for which G = 85 GPa.
SOLUTION
Step one: The polar moment of inertia of the cross-sectional area is
J =π
32(π4) =
π
32Γ 0.0444 = π. ππ Γ ππβππ4
Step two: The maximum shear stress is developed at the outer fibers where r = 0.022 m:
ππππ₯ =ππ
π½=
1100 Γ 0.022
3.68 Γ 10β7= ππ. π Γ πππ π·π ππ ππ. ππ΄π·π
Step three: The angle of twist q in a 1.5 m length of the shaft is
π =ππΏ
πΊπ½=
1100 Γ 1.5
85 Γ 109(3.68 Γ 10β7)= π. ππππ π«ππ π¨π«
Example (1):
A hollow 3 m long steel shaft must transmit a torque of 25 kN.m. The total angle of
twist in this length is not to exceed 2.5Β° and the allowable shearing stress is 90
MPa. Determine the inside and outside diameters of the shaft if G = 85 GPa.
Example (2):
SOLUTION
Step one: The angle of twist is ΞΈ = TL/GJ. Thus, in the 3-m length we have
2.5Β°π πππ
180Β°=
(25000)(3)
85 Γ 109(π/32)(π·4 β π4)β΄ π·4 β π4 = 206 Γ 10β6 π4 β (π)
Step two: The maximum shearing stress occurs at the outer fibers where r = D/2
90 Γ 106 =(25000)(π·/2)
(π/32)(π·4 β π4)β΄ π·4 β π4 = (1415π· Γ 10β6)π4 β (π)
Comparison of the right-hand sides of equations (1) and (2) indicates that
206 Γ 10β6 = (1415π· Γ 10β6)
and thus D = 0.146 m or 146 mm. Substitution of this value into either of the equations then gives d = 0.126 m or 126 mm.
Consider two solid circular shafts connected by 5-cm- and 25-cm-pitch-diameter
gears as in Fig. (a). Find the angular rotation of D, the right end of one shaft, with
respect to A, the left end of the other, caused by the torque of 280 N.m applied at
D. The left shaft is steel for which G = 80 Gpa and the right is brass for which G =
33 GPa.
Example (4):
SOLUTION
Step one: A free-body diagram of the right shaft CD [Fig. (b)] reveals that a tangential force F
must act on the smaller gear. For equilibrium,
Fr = T 0.025 F = 280 β΄ F = 11200 NStep two: The angle of twist of the right shaft is
π1 =ππΏ
πΊπ½=
280 Γ 1.00
(33 Γ 109)π Γ 0.034/32= π. πππππ«ππ
π2 =ππΏ
πΊπ½=
1400 Γ 1.20
(80 Γ 109)π Γ 0.064/32= π. πππππ«ππ
π = 5π2 + π1 = 5 Γ 0.0165 + 0.1067 = π. πππ πππ ππ ππ. πΒ°
Step three: A free-body diagram of the left shaft AB is shown in Fig. (c). The force F is equal
and opposite to that acting on the small gear C. This force F acts 12.5 cm from the center line
of the left shaft; hence it imparts a torque of 0.125(11200) = 1400 N.m to the shaft AB.
Because of this torque there is a rotation of end B with respect to end A given by the angle
π2, where
Step four: This angle of rotation π2 induces a rigid-body rotation of the entire shaft CD
because of the gears. In fact, the rotation of CD will be in the same ratio to that of AB as the
ratio of the pitch diameters, or 5:1. Thus a rigid-body rotation of 5(0.0165) rad is imparted to
shaft CD. Superposed on this rigid body movement of CD is the angular displacement of D
with respect to C, previously denoted by π1. Hence the resultant angle of twist of D with
respect to A is
Thank You
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