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Unformatted text preview: LECTURE Third Edition SHAFTS: POWER, STRESS
CONCENTRATION, THINWALLED
• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering 8
Chapter
3.7 – 3.8
3.13 by
Dr. Ibrahim A. Assakkaf
SPRING 2003
ENES 220 – Mechanics of Materials
Department of Civil and Environmental Engineering
University of Maryland, College Park LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 1
ENES 220 ©Assakkaf Power Transmission
Work of a Force
– A force does work only when the particle
to which the force is applied moves. Work = U = Pd (1) P
d 1 Slide No. 2 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission ENES 220 ©Assakkaf Work in Two and Three Dimension
U = P· d = (P cos φ) d
= Px dx + Py dy P P cos φ φ (2) d Slide No. 3 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission ENES 220 ©Assakkaf Work of a Couple
– The work of a couple is defined as the
magnitude of the couple C times the
angular movement of the body. U1→2 = C∆θ
rr
dU = C ⋅ dθ (3) 2 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 4
ENES 220 ©Assakkaf Power Transmission by Torsional Shaft
The power is defined as the time rate of
doing work, that is dU
dθ
dθ
=C⋅
=T
dt
t
dt
= Tω (4) ω = angular velocity of the shaft in radians per minute LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 5
ENES 220 ©Assakkaf Power Transmission by Torsional Shaft
– But ω = 2π f, where f = frequency. The unit
of frequency is 1/s and is called hertz (Hz).
– If this is the case, then the power is given
by
P = 2πfT or
T= (5)
P
2πf 3 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 6
ENES 220 ©Assakkaf Power Transmission by Torsional Shaft
– Units of Power
SI US Customary watt (1 N·m/s) hp (33,000 ft·lb/min) LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 7
ENES 220 ©Assakkaf Power Transmission by Torsional Shaft
– Some useful relations 1 −1 1
s=
Hz
60
60
1 hp = 550 ft ⋅ lb/s = 6600 in ⋅ lb/s
1 rpm = rpm = revolution per minute 4 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 8
ENES 220 ©Assakkaf Example 5
What size of shaft should be used for a
rotor of 5hp motor operating at 3600 rpm if
the shearing stress is not to exceed 8500
psi in the shaft? 1 hp = 6600 in
5 hp = P P = 5(6600) = 33,000 in ⋅ lb/s LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 9
ENES 220 ©Assakkaf Example 5 (cont’d)
1 Hz
f = (3600 rpm ) = 60 Hz = 60/s
60 rpm
33,000
P
=
= 87.54 lb ⋅ in
T=
2πf 2π (60 ) Let J denotes the polar moment of area, and c
the maximum radius, therefore, τ= Tc
JT
⇒=
J
cτ (6) 5 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Power Transmission Slide No. 10
ENES 220 ©Assakkaf Example 5 (cont’d)
– Evaluating the term J/c in Eq. 6, yields
14
πc
1
J2
=
= π 2c 3
2
c
c
– Therefore,
1 2 3 J T 87.54
= 0.001030 in 3
πc= = =
2
c τ 8500
⇒ c = 0.1872 in ⇒ Shaft size (dia) = 2c = 0.375 in LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 11
ENES 220 ©Assakkaf The derivation of the torsion formula,
τ max = Tc
J (7) assumed a circular shaft with uniform
crosssection loaded through rigid end
plates. 6 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 12
ENES 220 ©Assakkaf Stress Concentrations in Circular
Shafts 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
τ max = K Tc
J (8) LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 13
ENES 220 ©Assakkaf 7 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 14
ENES 220 ©Assakkaf Example 6
The stepped shaft shown is to rotate at 900
rpm as it transmits power from a turbine to a
generator. The grade of steel specified in the
design has an allowable shearing stress of 8
ksi. (a) For preliminary design shown,
determine the maximum power that can be
transmitted. (b) If in the final design the radius
of the fillet is increased so that r = 15/16 in.,
what will be the percent change , relative to
the preliminary design in the power? LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 15
ENES 220 ©Assakkaf Example 6 (cont’d) 8 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 16
ENES 220 ©Assakkaf Example 6 (cont’d)
(a) Preliminary Design:
Using Fig. 3.32, and
knowing that the following are given:
D = 7.50 in., d = 3.75 in., r = 9/16 in. =
0.5625 in.
Therefore, D 7.50
=
=2
d 3.75 and r 0.5625
=
= 0.15
3.75
d • Hence, from Fig. 3.32 K = 1.33 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 17
ENES 220 ©Assakkaf Example 6 (cont’d) K = 1.33 9 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 18
ENES 220 ©Assakkaf Example 6 (cont’d)
Using Eq. 40,
τ max = K Tc
J or T = J τ max
cK 4 1
1 3.75 4
J = πc 4 = π = 19.414 in
2
22
J τ max
19.414(8)
∴T =
=
= 62.3 kip  in
(3.75 / 2)1.33
cK
From Eq. 37, the power is given by 900 6
P = 2πfT = 2π (62.3) = 5,869.9 in  kip/s = 5.87 × 10 in  lb/s 60 5.87 × 106
Pa =
= 889.3 hp
6600 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 19
ENES 220 ©Assakkaf Example 6 (cont’d)
– Final Design:
Using Fig. 3.32, and
knowing that the following are given:
D = 7.50 in., d = 3.75 in., r = 15/16 in. = 0.9375 in.
Therefore,
D 7.50
=
=2
d 3.75 and r 0.9375
=
= 0.25
3.75
d • Hence, from Fig. 3.32
K = 1.20 10 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 20
ENES 220 ©Assakkaf Example 6 (cont’d)
K = 1.20 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Stress Concentrations in Circular
Shafts Slide No. 21
ENES 220 ©Assakkaf Example 6 (cont’d)
Using Eq. 40,
τ max = K Tc
J or T = J τ max
cK 4 1
1 3.75 4
J = πc 4 = π = 19.414 in
2
22
19.414(8)
J τ max
=
= 69 kip  in
∴T =
(3.75 / 2)1.20
cK
From Eq. 37, the power is given by 900 6
P = 2πfT = 2π (69.0 ) = 6,503.1 in  kip/s = 6.50 ×10 in  lb/s 60 6.50 × 106
985 − 889.3
= 985 hp
Change in P = 100
= 11%
Pb =
6600
889.3 11 Slide No. 22 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members
Bar of Rectangular Cross Section The maximum shearing stress and the angle of
twist for a uniform bar of rectangular cross
section, and subjected to pure torsion T are
given by τ max = T τmax a
b T
L φ= T
k1ab 2 (9) TL
k 2 ab 3G (10) The coefficients k1 and k2 can be obtained from Table 1. LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 23
ENES 220 ©Assakkaf Torsion of Noncircular Members Table 1. Coefficients for Rectangular Bars in Torsion
a/b k2 1.0
Beer and
Johnston,
2002 k1
0.208 0.1406 1.2 0.219 0.1661 1.5 0.231 0.1958 2.0 0.246 0.229 2.5 0.258 0.249 3.0 0.267 0.263 4.0 0.282 0.281 5.0 0.291 0.291 10.0 0.312 0.312 ∞ 0.333 0.333 12 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 24
ENES 220 ©Assakkaf Torsion of Noncircular Members
ThinWalled Hollow Shafts – It was indicated earlier that the
determination of the stresses in noncircular
members generally requires the use of
advanced mathematical methods.
– In the case of thinwalled hollow
noncircular shaft (Fig. 16), however, a
good approximation of the distribution of
stresses can be obtained. LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 25
ENES 220 ©Assakkaf Torsion of Noncircular Members
ThinWalled Hollow Shafts
Figure 16 Center line or
Mean Perimeter t t
τ τ Ab 13 Slide No. 26 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members
ThinWalled Hollow Shafts – The shearing stress τ at any given point of
the wall may be expressed in terms of the
torque T as t τ τ=
Ab T
2tAb (11) Ab = area bounded by center line LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 27
ENES 220 ©Assakkaf Torsion of Noncircular Members
ThinWalled Hollow Shafts – The shearing stress τ of Eq. 11 represents
the average value of the shearing stress
across the wall.
– However, for elastic deformations the
distribution of the stress across the wall
may be assumed uniform, and Eq. 11 will
give the actual value of the shearing stress
at a given point of the wall. 14 Slide No. 28 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members
ThinWalled Hollow Shafts The angle of twist of a thinwalled shaft of
length L and modulus of rigidity G is given by
t τ
L φ= TL
ds
4 Ab2G ∫ t (12) Where the integral is computed along the
center line of the wall section. LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 29
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 7 Using τall = 40 MPa, determine the largest
torque which may be applied to each of the
brass bars and to the brass tube shown.
Note that the two solid bars have the same
crosssectional area, and that the square
bar and square tube have the same
outside dimensions 15 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 30
ENES 220 ©Assakkaf Torsion of Noncircular Members
Figure 17 T1 T2 m
m m
m 64 25 mm (1) t = 6 mm 40 mm 40 40 mm 40 m
m T3 (3) (2) LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 31
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 7 (cont’d)
1. Bar with Square Cross Section: For a solid bar of rectangular cross section,
the maximum shearing stress is given by Eq.
9:
T τ max = k1ab 2 where the coefficient k1 is obtained from
Table 1, therefore
a = b = 0.040 m a
= 1.00
b k1 = 0.208 16 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 32
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 7 (cont’d)
For τmax = τall = 40 MPa, we have
τ max = T1
k1ab 2 40 = T1
⇒ T1 = 532 N ⋅ m
2
0.208(0.04 )(0.04 ) 2. Bar with Rectangular Cross Section:
a 0.064
=
= 2.56
b 0.025
By interpolation, Table 1 gives : k1 = 0.259
a = 0.064 m b = 0.025 m LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 33
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 7 (cont’d)
τ max = T2
k1ab 2 40 = T2
⇒ T2 = 414 N ⋅ m
2
0.259(0.064)(0,025) 3. Square Tube:
For a tube of thickness t, the shearing stress
is given by Eq. 11 as τ= T
2tAb 17 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 34
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 7 (cont’d) where Ab is the area bounded by the center line
of the cross section, therefore, Ab = (0.034 )(0.034) = 1.156 × 10 −3 m 2 40 mm 34 mm t = 6 mm 34 mm
40 mm LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 35
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 7 (cont’d)
τ = τall = 40 MPa and t = 0.006 m.
Substituting these value into Eq. 11 gives τ= T
2tAb 40 = T3
2(0.006 ) 1.156 × 10 −3 ( ) ∴ T3 = 555 N ⋅ m 18 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 36
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 8 Structural aluminum tubing of 2.5 × 4in.
rectangular cross section was fabricated by
extrusion. Determine the shearing stress
in each of the four walls of a portion of
such tubing when it is subjected to a torque
of 24 kip·in., assuming (a) a uniform 0.160in. wall thickness (Figure 18a), (b) that, as
a result of defective fabrication, walls AB LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 37
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 8 (cont’d) and AC are 0.120in thick, and walls BD
and CD are 0.200in thick (Fig. 18b)
A
2.5 in. 4 in. A B
0.160 in. 2.5 in. 0.160 in. C D (a) 4 in. B
0.120 in.
0.200 in. C Figure 18 D (b) 19 Slide No. 38 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 8 (cont’d)
(a) Tubing of Uniform Wall Thickness:
Figure 19
A The area bounded by the center line (Fig. 19) is
given by
3.84 in. B
0.160 in. 2.34 in. Ab = (3.84)(2.34 ) = 8.986 in 2 0.160 in. C D LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) Slide No. 39
ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 8 (cont’d) Since the thickness of each of the four walls is t
= 0.160 in., we find from Eq. 11 that the
shearing stress in each wall is
τ= T
24
=
= 8.35 ksi
2tAb 2(0.160 )(8.986) (b) Tubing with Variable Wall Thickness:
Observing that the area Ab bounded by the
center line is the same as in Part a, and
substituting t = 0.120 in. and t = 0.200 in. into 20 Slide No. 40 LECTURE 8. SHAFTS: POWER, STRESS CONCENTRATION, THINWALLED (3.7 – 3.8, 3.13) ENES 220 ©Assakkaf Torsion of Noncircular Members
Example 8 (cont’d)
• Eq. 11, the following values for the shearing
stresses are obtained: τ AB = τ AC = 24
= 11.13 ksi
2(0.120 )(8.986 ) τ BD = τ CD = 24
= 6.68 ksi
2(0.200)(8.986 ) and • Note that the stress in a given wall depends
only upon its thickness t. 21 ...
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