bfc 20903 (mechanics of materials) chapter 6: torsion shahrul niza mokhatar [email protected]
TRANSCRIPT
Chapter Learning Outcome
1. Defined the relationship between stress and strain2. Understand torsion theory and their applications 3. Analysed and calculate torsion in solid and hollow
circular bar 4. Analysed and calculate the torsion with end-
restraints 5. Analysed and calculate the torsion with combined
bar
BFC 20903 (Mechanics of Materials) Shahrul Niza Mokhatar ([email protected]
Introduction• Torsion commonly found in mechanical engineering.• application, for example machinery structures that has twisting member. • In civil engineering applications;
secondary beam will distributes load to main beam as point load and moment at connection part of the beams.
water retaining in a channel will produce moment that are distributed to the beam as torsion
advert post will experience torsion due to wind load that acting on advert planks.
Torsion theory• When a bar or shaft of circular section is twisted by moment, its called pure
tension & the deformed element are in a state of pure shear.
Few assumptions are taken into account in torsion analysis,• The longitudinal axis of the shaft
remains straight.• The shaft does not increase or
decrease in length.• Radial lines remain straight and radial
as the cross section rotates.• Cross sections rotate about the axis of
the member.
Using Right Hand Law, torsion vector can be determined• Planar cross-sectional parallel with
member axis will remains planar after subjected to torsion.
• Shear strain, γ is changing linearly along the bar.
Shear stress due to torsion• Shear stress in circular section, τ (tau). From Hooke’s Law,
τ = TR J
τ : Shear stress in the shaft if τmax=maximum shear stress occurs at the outer surface @ tegasan ricih. (N/m2)J : Polar moment of inertia of the cross sectional area @ moment sifat tekun kutub (m4)T : Applied torque acting at the cross section (Nm)R : Radius of the shaft (m)
Angles of twist
= TL JG
= angle of twist (radians)T = applied torqueL = length of memberG = shear modulus of material/ modulus of rigidity@modulus ketegaran (N/m2)J = polar moment of inertia
Twisting angle is angle (in radian) produced when a bar is subjected to torsion.
Power transmission• Circular bars or shafts are commonly used for transmission of
power. • Power is defined as the work performed per unit of time. The work
transmitted by a rotating shaft equals to the torque applied times the angle of rotation.
• If torque is not a function of time, then the equation for power simply becomes:
• Use consistent units for P, T, and ω. Power is commonly specified in horsepower, HP. Angular velocity is usually given in revolutions per minute or RPM. It should then be converted to rad/sec. To do this multiply the value in RPM by 2π and divide by 60.
Work = Torque x Angular Displacement Power = d/dt (Work)
P = T ω
ω is the angular velocity of the shaft (rad/s) = 2π ff : frequency (Hz @ hertz) (1 Hz = 1 cycle/s)T : Applied torque acting at the cross section (Nm)P : Power (W)(1W = 1 Nm/s)
Example 1
Example 2• Determine the maximum torque of a hollow circular shaft with inside diameter
of 60mm and an outside diameter of 100mm without exceeding the maximum shearing stress of 70MPa.
Solution;• Given;
di = 60mm, do = 100mm, τmax = 70MPa
• Remember: τmax=maximum shear stress occurs at the outer surface/radius.
4644 1055.8)60100(32
mmxJ
kNmxx
r
JT
J
rT
97.1105.0
)1055.8)(1070( 66
maxmax
maxmax
Quiz
A hollow steel shaft has an outside diameter of 150mm and an inside diameter 100mm. The shaft is subjected to a torque of 35kNm. The modulus of rigidity for the steel is 80GPa. Determine;
a) the shearing stress at the outside surface of the shaft.b) the shearing stress at the inside surface of the shaft.c) the magnitude of the angle of twist in a 2.5m length.
Quiz: SolutionSolution;
• the shearing stress at the outside surface of the shaft.
• the shearing stress at the inside surface of the shaft
4644 1089.39)100150(32
mmxJ
MPax
x
J
Trout
81.651089.39
)075.0(10356
3
MPax
x
J
Trin
9.431089.39
)05.0(10356
3
JG
TL
radxx
x027.0
)1089.39(1080
)5.2(103569
3
• the magnitude of the angle of twist in a 2.5m length.
• Combined bar consists of two or more materials to form a structure. An example is shown in Figure. Superposition principle is used to solve this problem.
• The concept to solve combined bar are: (a) Imposed external torsion is equal to total torsion formed in the bar, i.e.,
(b) Twisting angle first material is equal to twisting angle of second material at connection part, i.e.,
(c) Total twisting angle can be calculated from formula,
Composite Bars
Group Exercise• The composite bars with the different material is subjected to the torque is shown
in figure. Determine the maximum shear stress and the position. Determine the angle of twist at C.
Solution
4624
1082.932
)100(
32mmx
dJ AB
4624
1062.032
)50(
32mmx
dJ BC
• Maximum shear stress; Angle of twist at the end of C;
The maximum shear stress occurs in the bar of BC.
26
6
/55.301082.9
)50(106mmN
x
x
J
rT
AB
ABABAB
26
6
/3.1611062.0
)25(104mmN
x
x
J
rT
BC
BCBCBC
BCAB JG
TL
JG
TL
JG
TL
radxx
x
xx
x0158.0
)108(1062.0
)300)(104(
)103(1082.9
)2000)(106(46
6
46
6
Torsion of non-cylindrical member
• Generally, we deal with axisymmetric bodies and the shear strain is linear through the entire body. However, non-circular cross-sections are not axisymmetric causing complex behaviors, which may cause bulging or warping when the shaft is twisted.
Condition of bulging of non-circular shaft
Empirical formulas for various shapes
Example 3
The aluminum shaft shown in figure has a cross sectional area in the shape of an equilateral triangle. Determine the largest torque, T that can be applied to the end of the shaft if the allowable shear stress, is = 56MPa and the angle of twist at its end is restricted to = 0.02 rad. Given Gal = 26GPa.
allow
allow
Example 3: Solution
Thin-walled Having Closed Cross Sections
• Thin walled of noncircular shape are often used to construct lightweight frameworks which is used in aircraft.
• Due the applied torque, T, shear stress is developed on the front face of the element. Shear flow in a solid body is the gradient of a shear stress through the body. Shear flow is the product of the tube’s thickness and the average shear stress. This value is constant at all points along the tube’s cross section. As a result, the largest average shear stress on the cross section occurs where the tube’s thickness is small.
tq avg
Thin-walled Having Closed Cross Sections
• In non-circular thin walled shafts for closed segments. We assume that the stress is uniformly distributed across the thickness and that we can assume an average shear stress. The average shear stress in the body is;
where, - average shear stresst - the thickness of the shaft at the point of interestAm - mean area enclosed within the boundary of the
centerline of theshaft thickness.T - the applied torque
mave tA
T
2
ave
ds
t
T
h
dF
τave
τave
Thin-walled Having Closed Cross Sections
mave
mave
ave
ave
At
Adt
dsht
dsth
2
2
)(
mave tA
T
2
Since, we can determine the shear flow throughout the cross section using the equation;
tq avg
mA
Tq
2
• Angle of twist, • This angle can be determined by using the energy method. The
angle given in radians, can be expressed as;
• Here the integration must be performed around the entire boundary of the tube’s cross sectional area.where, t - thickness of the interior segmentL - length of the section G - modulus of rigidity of the section/shear modulus
t
ds
GA
TL
m24
Example 4A square aluminum tube has the dimensions as shown in figure. a) determine the average shear stress in the tube at point A if it is subjected to a torque of 85Nm. b) compute the angle of twist due to this loading. Given Gal = 26GPa.
Example 4: Solutiona) The area, Am;
Since t is a constant because of the square tube, the average shear stress is the same at all points on the cross section.
22500)50)(50( mmAm
50mm
50mm
The shaded area = Am
23
/7.1)2500)(10(2
1085
2mmN
x
tA
T
mavg
Example 4: Solutionb) Angle of twist;
Here, the integral represents the length around the centerline boundary of the tube.
t
ds
GA
TL
m24
mm
mm
x
x
10
504
)1026()2500(4
)1500)(1085(32
3
radx
x3
4
1092.3
)20(10962.1
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