Transcript Chapter 1

5. Torsion
*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
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5. Torsion
*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
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*5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
Average shear stress
T
avg =
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
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5. Torsion
*5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
Average shear stress
Since q = avgt, the shear flow throughout the xsection is
T
q=
2Am
Angle of twist
Can be determined using energy methods
∫
TL
ds
O
=
4Am2G
t
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5. Torsion
*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
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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.
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EXAMPLE 5.16 (SOLN)
Average shear stress
Am = (50 mm)(50 mm) = 2500 mm2
T
avg =
= ... = 1.7 N/mm2
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 darkershaded face, since it contributes to
internal resultant torque T at the
section
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EXAMPLE 5.16 (SOLN)
Angle of twist
∫
TL
ds
-4) mm-1 O
O
=
=
...
=
0.196(10
∫ ds
2
4Am G
t
Here, integral represents length around
centerline boundary of tube, thus
 = 0.196(10-4) mm-1[4(50 mm)] = 3.92 (10-3) rad
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5.8 STRESS CONCENTRATION
• Three common discontinuities
of the x-section are:
a) is a coupling, for connecting
2 collinear shafts together
b) is a keyway used to connect
gears or pulleys to a shaft
c) is a shoulder fillet used to
fabricate a single collinear
shaft from 2 shafts with
different diameters
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5. Torsion
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
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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)
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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.
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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.
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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
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EXAMPLE 5.18 (SOLN)
Maximum shear stress
From shaft geometry, we have
D 2(40 mm)
=
=2
d 2(20 mm)
6 mm)
r
=
= 0.15
d 2(20 mm)
Thus, from the graph, K = 1.3
max = K(Tc/J) = ... = 3.10 MPa
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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:
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*5.9 INELASTIC TORSION
•
To perform a “plastic analysis”
for a material that has yielded,
the following conditions must
be met:
1. Shear strains in material must
vary linearly from zero at center of
shaft to its maximum at outer
boundary (geometry)
2. Resultant torque at section must
be equivalent to torque caused by
entire shear-stress distribution
over the x-section (loading)
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*5.9 INELASTIC TORSION
•
Expressing the loading
condition mathematically,
we get:
2 d
T
=
2∫

Equation 5-23
A
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*5.9 INELASTIC TORSION
A. Maximum elastic torque
• For maximum elastic shear strain Y,
at outer boundary of the shaft, shearstrain distribution along radial line will
look like diagram (b)
• Based on Eqn 5-23,
TY = (/2) Yc3
• From Eqn 5-13,
d =  (dx/)
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*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)
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*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.
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*5.9 INELASTIC TORSION
C. Ultimate torque
• Magnitude of Tu can be determined “graphically”
by integrating Eqn 5-23
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*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
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*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
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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?
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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.
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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.
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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
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*5.10 RESIDUAL STRESS
• Residual stress distribution is calculated using
principles of superposition and elastic recovery
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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 residualshear-stress distribution
and permanent twist of the
tube that remain if TP is
removed just after tube
becomes fully plastic?
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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 = 0.002 rad, thus angle of
twist for entire tube is
P = Y (L/ci) = ... = 0.120 rad
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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 = ... = 104.52 MPa
i = (104.52 MPa)(25 mm/50 mm) = 52.26 MPa
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EXAMPLE 5.21 (SOLN)
Plastic torque
Angle of twist ’P upon removal of TP is
’P = (TP L)/(JG) = ... = 0.0747 rad
Residual-shear-stress distribution is shown.
Permanent rotation of tube after TP is removed,
+  = 0.120  0.0747 = 0.0453 rad
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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
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CHAPTER REVIEW
•
Angle of twist of circular shaft determined from
L T(x) dx
 =∫
0 JG
•
If torque and JG are constant, then
TL
 =
JG
•
For application, use a sign convention for
internal torque and be sure material does not
yield, but remains linear elastic
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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
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CHAPTER REVIEW
•
•
•
Shear stress in tubes determined from
 = T/2tAm
Stress concentrations occur in shafts when xsection 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
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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
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