Unformatted text preview: SECTION 3.9 Solution 3.92 T (A) STRAIN ENERGY 0.75 m d 40 mm max tmax IP ¢ L
2G ¢ 32
d4
(Eq. 2) Substitute numerical values:
U 32 MPa
16T
d3 2 d2Lt2
max
16G 45 GPa L d 3tmax
16 T 2L
2GIP U L G 237 Copper bar d T Strain Energy in Torsion (b) ANGLE OF TWIST d 3tmax
16 T 5.36 J (Eq. 1) Tf
2 U d4
32 f 2U
T Substitute for T and U from Eqs. (1) and (2):
2Ltmax
Gd f (Eq. 3) Substitute numerical values:
f 0.026667 rad d2 Problem 3.93 A stepped shaft of solid circular cross sections
T
(see figure) has length L 45 in., diameter d2 1.2 in., and
6 psi.
diameter d1 1.0 in. The material is brass with G 5.6 10
Determine the strain energy U of the shaft if the angle of twist
is 3.0°. Solution 3.93 d1 T L
—
2 1.0 in. 45 in.
5.6 L
—
2 T 2L
a 2GI
P 16 T 2( L 2 )
4
Gd2
1
4
d1 1.2 in. G 3.0 L
—
2 L
—
2 8T 2L 1
¢4
G d2 U L T STRAIN ENERGY
T d2 d1 Stepped shaft d2 d1 1.53 Also, U 16 T 2( L 2 )
4
Gd1
(Eq. 1) Tf
2 (Eq. 2) Equate U from Eqs. (1) and (2) and solve for T:
106 psi (brass) T 44
Gd1 d2 f
4
4
16L ( d1 d2 ) U Tf
2 0.0523599 rad 44
Gf2 d1 d2
¢4
4
32L d1 d2 f SUBSTITUTE NUMERICAL VALUES:
U 22.6 in.lb radians 238 CHAPTER 3 Torsion Problem 3.94 A stepped shaft of solid circular cross sections (see figure)
has length L 0.80 m, diameter d2 40 mm, and diameter d1 30 mm.
The material is steel with G 80 GPa.
Determine the strain energy U of the shaft if the angle of twist is 1.0°. Soluton 3.94 Stepped shaft
d2 Equate U from Eqs. (1) and (2) and solve for T: d1 T T T d1 30 mm d2 L 0.80 m G 1.0 U L
—
2 L
—
2 44
G d1 d2 f
4
4
16L ( d1 d2 ) Tf
2 44
Gf2 d1 d2
¢4
4
32L d1 d2 f 40 mm SUBSTITUTE NUMERICAL VALUES: 80 GPa (steel) U radians 1.84 J 0.0174533 rad STRAIN ENERGY
T 2L
a 2GI
P 16T 2( L 2 )
4
Gd2 8T 2L 1
¢4
G d2 U 1
4
d1 Also, U 16T 2( L 2 )
4
Gd1
(Eq. 1) Tf
2 (Eq. 2) Problem 3.95 A cantilever bar of circular cross section and length L is
fixed at one end and free at the other (see figure). The bar is loaded by a
torque T at the free end and by a distributed torque of constant intensity
t per unit distance along the length of the bar.
(a) What is the strain energy U1 of the bar when the load T acts alone?
(b) What is the strain energy U2 when the load t acts alone?
(c) What is the strain energy U3 when both loads act simultaneously?
Solution 3.95 Cantilever bar with distributed torque G
IP L T polar moment of inertia T t shear modulus torque acting at free end t torque per unit distance t L T SECTION 3.9 239 Strain Energy in Torsion (c) BOTH LOADS ACT SIMULTANEOUSLY (a) LOAD T ACTS ALONE (EQ. 351a)
T 2L
2GIP t (b) LOAD t ACTS ALONE dx U1 T At distance x from the free end: From Eq. (356) of Example 311:
U2 x T(x) t 2L3
6GIP T tx
L U3
0 [ T(x) ] 2
dx
2GIP T 2L
2GIP L 1
2GIP TtL2
2GIP tx ) 2dx (T
0 t 2L3
6GIP NOTE: U3 is not the sum of U1 and U2. 2T0 Problem 3.96 Obtain a formula for the strain energy U of the statically
indeterminate circular bar shown in the figure. The bar has fixed supports
at ends A and B and is loaded by torques 2T0 and T0 at points C and D,
respectively.
Hint: Use Eqs. 346a and b of Example 39, Section 3.8, to obtain the
reactive torques. Solution 3.96 L
—
4 L
—
2 3L
4 U
L
4
L T0 ¢ L 7T0
4 L
—
4 TB L
—
4 7T0
1
B¢
2GIP
4
T0
4 TDB i Pi 1
L
BT 2 ¢
2GIp AC 4 5T0
4 TCD n
Ti2Li
a 2G I i1 7T0
4 INTERNAL TORQUES
TAC L
—
2 L
—
4 D From Eq. (346a): TA D STRAIN ENERGY (from EQ. 353) REACTIVE TORQUES 3T0 C B
C TB B T0 A TA TA A Statically indeterminate bar
2T0 ( 2T0 ) ¢ T0 5T0
4 U 2
19T0L
32GIP 2 ¢ L
4 2
TCD¢ T0
4 2 ¢ L
2 L
2 2
TDB¢ 5T0
4 2 ¢ L
R
4 L
R
4 240 CHAPTER 3 Torsion Problem 3.97 A statically indeterminate stepped shaft ACB is fixed at
ends A and B and loaded by a torque T0 at point C (see figure). The two
segments of the bar are made of the same material, have lengths LA and LB,
and have polar moments of inertia IPA and IPB.
Determine the angle of rotation of the cross section at C by using
strain energy.
Hint: Use Eq. 351b to determine the strain energy U in terms of the
angle . Then equate the strain energy to the work done by the torque T0.
Compare your result with Eq. 348 of Example 39, Section 3.8. Solution 3.97 A A IPA T0
C IPB LA
LB Statically indeterminate bar IPA
C
LA T0 IPB WORK DONE BY THE TORQUE T0
B W T0f
2 LB EQUATE U AND W AND SOLVE FOR
Gf2 IPA
¢
2 LA STRAIN ENERGY (FROM EQ. 351B)
U n
GIPif2
i
a 2L
i1
i Gf2 IPA
¢
2 LA GIPAf2
2LA GIPBf2
2LB f IPB
LB IPB
LB T0f
2 T0LALB
G ( LBIPA LAIPB ) (This result agrees with Eq. (348) of Example 39,
Section 3.8.) Problem 3.98 Derive a formula for the strain energy U of the cantilever
bar shown in the figure.
The bar has circular cross sections and length L. It is subjected
to a distributed torque of intensity t per unit distance. The intensity
varies linearly from t 0 at the free end to a maximum value t t 0
at the support.
t0 t L B SECTION 3.9 Solution 3.98 Strain Energy in Torsion Cantilever bar with distributed torque t0 x
L t(x) = t0 dx x distance from righthand end of the bar d
L x STRAIN ENERGY OF ELEMENT dx ELEMENT d
Consider a differential element d at distance from
the righthand end. dU [ T ( x ) ] 2dx
2GIP t0 2 4
1
¢
x dx
2GIP 2L
2
t0
x4 dx
8L GIP
2 dT STRAIN ENERGY OF ENTIRE BAR
L d U dU
0 dT t( )d
j
t0 ¢ dj
L U ELEMENT dx AT DISTANCE x T(x) T(x) dx T(x) internal torque acting on this element T(x) total torque from x
x T(x) x dT
0 t0x2
2L t0 ¢
0 j
dj
L 0 to x x L x4 dx
0 2
t0
L5
¢
8L2GIP 5 external torque acting on this element dT 2
t0
8L2GIP t 2L3
0
40GIP 241 242 CHAPTER 3 Torsion Problem 3.99 A thinwalled hollow tube AB of conical shape has
constant thickness t and average diameters dA and dB at the ends
(see figure). B A T T (a) Determine the strain energy U of the tube when it is subjected
to pure torsion by torques T.
(b) Determine the angle of twist of the tube. L
t
t d 3t /4 for a thin circular Note: Use the approximate formula IP
ring; see Case 22 of Appendix D. dB dA Solution 3.99 Thinwalled, hollow tube
B A T Therefore, T (x) L x dx B dA L 0 dx
3
dB dA
xR
L
L t thickness 1 dA
dB 2( dB average diameter at end A
average diameter at end B d(x) L average diameter at distance x from end A d(x) dB dA dA
L dA
L 2( dB t
BdA
4 dB dA
L U 3 xR 2T 2
Gt L ( dA dB )
22
2dAdB U
0 Work of the torque T: W 2 T dx
2GIP ( x ) W
L 2T2
Gt dx
3
dB dA x
R
L B dA
0 From Appendix C:
dx
( a bx ) 3 4 xR
0 L
dA )( dA ) 2 T 2L dA dB
¢ 22
Gt dA dB (b) ANGLE OF TWIST
(a) STRAIN ENERGY (FROM EQ. 354)
L 2 Substitute this expression for the integral into the
equation for U (Eq. 1): dt
4
[ d ( x ) ] 3t
4 dB L ( dA dB )
22
2dA dB 3 IP ( x ) B dA L
dA )( dB ) 2 2( dB x POLAR MOMENT OF INERTIA
IP dA ) 1
2b ( a bx ) 2 (Eq. 1) U Tf
2 T 2L ( dA dB )
22
Gt dAdB Solve for :
f Tf
2 2TL ( dA dB )
22
Gt dAdB SECTION 3.9 **Problem 3.910 A hollow circular tube A fits over the end of
a solid circular bar B, as shown in the figure. The far ends of both
bars are fixed. Initially, a hole through bar B makes an angle with
a line through two holes in tube A. Then bar B is twisted until the
holes are aligned, and a pin is placed through the holes.
When bar B is released and the system returns to equilibrium,
what is the total strain energy U of the two bars? (Let IPA and IPB represent the polar moments of inertia of bars A and B, respectively. The
length L and shear modulus of elasticity G are the same for both
bars.) Strain Energy in Torsion IPA IPB Tube A Bar B L L Tube A
Bar B Solution 3.910 Circular tube and bar
IPA IPB Tube A Tube A Bar B
Bar B L L COMPATIBILITY TUBE A A A B FORCEDISPLACEMENT RELATIONS
T fA TL
GIPA TL
GIPB fB Substitute into the equation of compatibility and
solve for T:
T torque acting on the tube A angle of twist BAR B
T T bG IPAIPB
¢
L IPA IPB STRAIN ENERGY
B U T 2L
a 2GI
P
T 2L 1
¢
2G IPA T 2L
2GIPA T 2L
2GIPB 1
IPB Substitute for T and simplify:
U
T torque acting on the bar B angle of twist b2G IPA IPB
¢
2L IPA IPB 243 244 CHAPTER 3 Torsion **Problem 3.911 A heavy flywheel rotating at n revolutions per
minute is rigidly attached to the end of a shaft of diameter d (see figure).
If the bearing at A suddenly freezes, what will be the maximum angle of
twist of the shaft? What is the corresponding maximum shear stress in
the shaft?
(Let L length of the shaft, G shear modulus of elasticity, and
Im mass moment of inertia of the flywheel about the axis of the shaft.
Also, disregard friction in the bearings at B and C and disregard the mass
of the shaft.)
Hint: Equate the kinetic energy of the rotating flywheel to the strain
energy of the shaft. A d n (rpm) B
C Solution 3.911 Rotating flywheel
Shaft d diameter of shaft Flywheel Gd 4f2
64L U
UNITS:
G
IP
d (length)4
radians diameter n (force)/(length)2 rpm L length U (length)(force) KINETIC ENERGY OF FLYWHEEL
EQUATE KINETIC ENERGY AND STRAIN ENERGY 1
I v2
2m K.E.
v 2n
60 n 22 K.E. rpm Gd 4f2
64L n Im
1800 2 ImL
2n
2
15d B G Solve for :
f 1
2n
Im ¢
2
60 K.E. U 2 MAXIMUM SHEAR STRESS 22 n Im
1800
UNITS:
Im T(d 2)
IP t radians per second
(length)(force) t
tmax STRAIN ENERGY OF SHAFT (FROM EQ. 351b)
2 U
IP GIPf
2L
32 TL
GIP Eliminate T: (force)(length)(second)2 K.E. f d4 tmax Gdf
2L
2 ImL
Gd 2n
2L 15d 2 B G
2 GIm
n
15d B L SECTION 3.10 ThinWalled Tubes 245 ThinWalled Tubes
Problem 3.101 A hollow circular tube having an inside diameter of 10.0 in.
and a wall thickness of 1.0 in. (see figure) is subjected to a torque T 1200 kin.
Determine the maximum shear stress in the tube using (a) the approximate
theory of thinwalled tubes, and (b) the exact torsion theory. Does the approximate
theory give conservative or nonconservative results?
Solution 3.101 10.0 in.
1.0 in. Hollow circular tube
APPROXIMATE THEORY (EQ. 363)
T
2 r 2t t1 1200 kin.
2 ( 5.5 in.) 2 (1.0 in.) 6314 psi 10.0 in. tapprox
1.0 in. 6310 psi EXACT THEORY (EQ. 311)
T ( d2 2 )
IP t2
T 1200 kin. t 1.0 in. r radius to median line r 5.5 in. Td2
2¢ 32 4
d2 4
d1 16(1200 kin.)(12.0 in.)
[ (12.0 in.) 4 ( 10.0 in.) 4 ]
6831 psi d2 outside diameter d1 inside diameter texact 12.0 in. 6830 psi Because the approximate theory gives stresses that
are too low, it is nonconservative. Therefore, the
approximate theory should only be used for very thin
tubes. 10.0 in. t Problem 3.102 A solid circular bar having diameter d is to be replaced
by a rectangular tube having crosssectional dimensions d 2d to the
median line of the cross section (see figure).
Determine the required thickness tmin of the tube so that the maximum shear stress in the tube will not exceed the maximum shear stress in
the solid bar.
Solution 3.102 d d 2d Bar and tube SOLID BAR Am
16T
d3 tmax d t (Eq. 312) (d )(2d ) tmax T
2tAm 2d 2
T
4td 2 (Eq. 364)
(Eq. 361) EQUATE THE MAXIMUM SHEAR STRESSES AND SOLVE FOR t
RECTANGULAR TUBE 16T
d3 t tmin d 2d T
4td 2
d
64 If t tmin, the shear stress in the tube is less than the
shear stress in the bar. 246 CHAPTER 3 Torsion Problem 3.103 A thinwalled aluminum tube of rectangular
cross section (see figure) has a centerline dimensions b 6.0 in.
and h 4.0 in. The wall thickness t is constant and equal to
0.25 in. t
h (a) Determine the shear stress in the tube due to a torque
T 15 kin.
(b) Determine the angle of twist (in degrees) if the length L of
the tube is 50 in. and the shear modulus G is 4.0 106 psi. Solution 3.103 b Thinwalled tube t Eq. ( 371) with t1 h t 6.0 in. 0.25 in. T 15 kin. L 50 in. G 4.0 T
2tAm 1250 psi 4.0 in. t (b) ANGLE OF TWIST (EQ. 372)
f TL
0.0065104 rad
GJ
0.373 106 psi Problem 3.104 A thinwalled steel tube of rectangular cross section
(see figure) has centerline dimensions b 150 mm and h 100 mm.
The wall thickness t is constant and equal to 6.0 mm.
(a) Determine the shear stress in the tube due to a torque T 1650 N m.
(b) Determine the angle of twist (in degrees) if the length L of the tube is
1.2 m and the shear modulus G is 75 GPa.
Solution 3.104 Thinwalled tube
b 150 mm h t 100 mm ht Eq. (364): Am Eq. ( 371) with t1
10.8 10 6 m4 1.2 m G
bh 1650 N m L
b 75 GPa t: J t T
2tAm 2b2h2t
bh 9.17 MPa (b) ANGLE OF TWIST (EQ. 372)
f TL
GJ
0.140 0.015 m2
t2 (a) SHEAR STRESS (EQ. 361) 6.0 mm T J t: (a) SHEAR STRESS (EQ. 361) b h t2 28.8 in.4 J b 24.0 in.2 bh Eq. (364): Am 0.002444 rad J 2b2h2t
bh SECTION 3.10 Problem 3.105 A thinwalled circular tube and a solid circular bar of
the same material (see figure) are subjected to torsion. The tube and bar
have the same crosssectional area and the same length.
What is the ratio of the strain energy U1 in the tube to the strain energy
U2 in the solid bar if the maximum shear stresses are the same in both cases?
(For the tube, use the approximate theory for thinwalled bars.)
Solution 3.105 r2 tmax T
2tAm r T
TL
2GJ 2 2 r 2t 2 r3t A J 2 rt U1 r2 A r2
2 tmax Tr2
IP T
2 r2t
max 2 ( 2 r ttmax ) L
2G (2 r3t ) T 2L
2GIP U2 rtt2 L
max
G
But rt Bar (2) SOLID BAR (2)
Am U1 Tube (1) Thinwalled tube (1) t 2 247 ThinWalled Tubes But r2
2 A
2
At2 L
max
2G A IP
2T
r3
2 ( r3 tmax ) 2L
2 r4
2
T r3tmax
2
2 r2t2 L
2 max
4G 8G ¢ r4
22
∴ U2 2 At2 L
max
4G RATIO
U1
U2 2 t = 8 mm Problem 3.106 Calculate the shear stress and the angle of twist (in
degrees) for a steel tube (G 76 GPa) having the cross section shown
in the figure. The tube has length L 1.5 m and is subjected to a torque
T 10 kN m. r = 50 mm r = 50 mm b = 100 mm Solution 3.106 Steel tube
t = 8 mm r = 50 mm SHEAR STRESS r = 50 mm G
L (50 mm)2 2(100 mm)(50 mm)
17,850 mm2
Lm 1.5 m T
b = 100 mm 76 GPa Am r2 2br T
2tAm 10 kN . m t 10 kN . m
2(8 mm)(17,850 mm2 )
35.0 MPa ANGLE OF TWIST
f TL
GJ ( 10 kN . m ) (1.5 m )
(76 GPa)(19.83 106 mm4 ) 0.00995 rad 2(100 mm) 2 (50 mm)
514.2 mm
J 2b 2 r 0.570 4tA2
m
Lm
19.83 4 (8 mm)(17,850 mm2 ) 2
514.2 mm
6
10 mm4 248 CHAPTER 3 Torsion Problem 3.107 A thinwalled steel tube having an elliptical cross
section with constant thickness t (see figure) is subjected to a torque
T 18 kin.
Determine the shear stress and the rate of twist (in degrees
per inch) if G 12 106 psi, t 0.2 in., a 3 in., and b 2 in.
(Note: See Appendix D, Case 16, for the properties of an ellipse.) t 2b 2a Solution 3.107 Elliptical tube t FROM APPENDIX D, CASE 16:
ab Am
2b p [1.5( a Lm b) ab ]
6.0 in.2 ] [ 1.5(5.0 in.) T 18 kin. G 12 106 psi t 0.2 in a SHEAR STRESS constant t 15.867 in. 4tA2 4 (0.2 in.)(18.850 in.2 ) 2
m
Lm
15.867 in.
4
17.92 in. J 2a 18.850 in.2 (3.0 in.)(2.0 in.) 3.0 in. b t 2.0 in. T
2tAm 18 k in.
2(0.2 in.)(18.850 in.2 ) 2390 psi
ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
u f
L u 83.73 Problem 3.108 A torque T is applied to a thinwalled tube having
a cross section in the shape of a regular hexagon with constant wall
thickness t and side length b (see figure).
Obtain formulas for the shear stress and the rate of twist . T
GJ 18 kin.
10 psi)(17.92 in.4 )
6 (12
10 6 rad in. 0.0048 in. t b SECTION 3.10 Solution 3.108 ThinWalled Tubes Regular hexagon
SHEAR STRESS t t T
2tAm T3
9b2t ANGLE OF TWIST PER UNIT LENGTH (RATE OF TWIST)
J Lm b
0 b
t
Lm Length of side
u Thickness 4A2
mt
Lm 4A2
m T
GJ dS
t 9b3t
2 2T
G (9b3t ) 2T
9Gb3t (radians per unit length) 6b FROM APPENDIX D, CASE 25:
60 n 6 2 Am b
nb
cot
4
2
3 6b2
cot 30
4 3b2
2 Problem 3.109 Compare the angle of twist 1 for a thinwalled circular tube
(see figure) calculated from the approximate theory for thinwalled bars with the
angle of twist 2 calculated from the exact theory of torsion for circular bars. t
r
C (a) Express the ratio 1/ 2 in terms of the nondimensional ratio
r/t.
(b) Calculate the ratio of angles of twist for
5, 10, and 20. What conclusion
about the accuracy of the approximate theory do you draw from these results?
Solution 3.109 Thinwalled tube
RATIO t f1
f2 r
C Let b 4r2 t2
4r2 1 t2
4r2 r f1
t f2 1 1
4b2 APPROXIMATE THEORY
f1 TL
GJ 2 r3t f1 J 1/ 2 TL
2 Gr3t 5
10 f2 TL
GIP f2 TL
GIP From Eq. (317): Ip
2TL
Grt (4r2 t2 ) rt
(4r2
2 t2 ) 1.0025 20 EXACT THEORY 1.0100 1.0006 As the tube becomes thinner and becomes larger,
the ratio 1/ 2 approaches unity. Thus, the thinner
the tube, the more accurate the approximate theory
becomes. 249 250 CHAPTER 3 Torsion *Problem 3.1010 A thinwalled rectangular tube has uniform thickness t
and dimensions a b to the median line of the cross section (see figure).
How does the shear stress in the tube vary with the ratio
a/b if
the total length Lm of the median line of the cross section and the torque T
remain constant?
From your results, show that the shear stress is smallest when the
tube is square (
1). t b a Solution 3.1010 Rectangular tube
t T, t, and Lm are constants.
Let k 2T
tL2
m t
k ¢ tmin a, b
b
Lm
T thickness (constant) a
b
constant 4
2
0 T
2tAm
2b (1 b Am
b) Lm
2(1 b ) ab bb2 t 1 2 3 From the graph, we see that is minimum when
1 and the tube is square. constant
Am bB 2
Lm
R
2(1 b ) bL2
m
4(1 b ) 2
T
2tAm 8T
tL2
m 6 b) SHEAR STRESS Lm b 8 constant t k a
b b Dimensions of the tube 2(a t min a t b)2 (1 4 b constant T (4)(1 b ) 2
2tbL2
m 2T (1 b ) 2
tL2 b
m ALTERNATE SOLUTION
t
dt
db 2T (1 b ) 2
B
R
b
tL2
m
2T b (2)(1
B
tL2
m or 2 (1 ) (1 b) (1 b ) 2 (1) b2
)2 0 R 0 1 Thus, the tube is square and is either a minimum or
a maximum. From the graph, we see that is a
minimum. SECTION 3.10 *Problem 3.1011 A tubular aluminum bar (G 4 106 psi) of square
cross section (see figure) with outer dimensions 2 in. 2 in. must resist a
torque T 3000 lbin.
Calculate the minimum required wall thickness tmin if the allowable
shear stress is 4500 psi and the allowable rate of twist is 0.01 rad/ft. ThinWalled Tubes t
2 in. 2 in. Solution 3.1011 Square aluminum tube
THICKNESS t BASED UPON SHEAR STRESS t T
2tAm t
2 in. T
2t tAm t(b
T lbin. UNITS: t 4 3000 lbin.
2(4500 psi) 3000 lbin. u 4500 psi uallow 0.01 rad ft T
GJ 0.01
rad in.
12 Centerline dimension J 0.0915 in. T
Gt ( b t ) 3 t (2.0 in. G t)3 4tA2
m
Lm t)2
4t ( b
4( b Lm
t)4
t) t(b
psi (4 b
4(b 10t (2 t t) 9 0 0.140 in. 3 ANGLE OF TWIST GOVERNS
tmin 0.140 in. T
Gu 3000 lbin
10 psi)(0.01 12 rad in.)
6 Solve for t: t) t
t(b t)3 t)3 rad/in. 9
10 outer dimension (b 13
in.
3 0 in. 2.0 in. Am 1 UNITS: t allow Let b t)2 psi THICKNESS t BASED UPON RATE OF TWIST 106 psi T t)2 Solve for t: t 2.0 in. G b 3t (2 Outer dimensions:
2.0 in. in. t (2.0 in.
2 in. in. T
2t t)2 251 252 CHAPTER 3 Torsion *Problem 3.1012 A thin tubular shaft of circular cross section
(see figure) with inside diameter 100 mm is subjected to a torque of
5000 N m.
If the allowable shear stress is 42 MPa, determine the required
wall thickness t by using (a) the approximate theory for a thinwalled
tube, and (b) the exact torsion theory for a circular bar. 100 mm
t Solution 3.1012 Thin tube
(b) EXACT THEORY
t Tr2
Ip Ip 2 ( r4
2 r4 )
1 100 mm 2 [ (50 t 42 MPa
T 5,000 N m d1 inner diameter 42 MPa allow 2 t is in millimeters.
r t Outer radius T
2tAm 42 MPa Am 2 2t ( r ) r T
2 r2t 5,000 N . m
t2
2 ¢ 50
t
2 or
t2
2
Solve for t: t 6.66 mm 106 mm3 5,000 N . m
2 ( 42 MPa) 7.02 mm The approximate result is 5% less than the
exact result. Thus, the approximate theory is
nonconservative and should only be used for
thinwalled tubes. 2 t T t ¢ 50 ( 50) 4 ] Solve for t: (a) APPROXIMATE THEORY
t t) ( 5000 N . m )(2)
( ) (42 MPa)
21 50 mm 50 mm t)4 5 t
50 mm
2
Inner radius r2 [ (50 t ) 4 ( 50) 4
50 t Average radius r1 ( 50) 4 ] ( 5,000 N . m )(50 100 mm
(50 t)4 106 5
84 mm3 SECTION 3.10 ••Problem 3.1013 A long, thinwalled tapered tube AB of circular cross
section (see figure) is subjected to a torque T. The tube has length L and
constant wall thickness t. The diameter to the median lines of the cross
sections at the ends A and B are dA and dB, respectively.
Derive the following formula for the angle of twist of the tube: L
t t Hint: If the angle of taper is small, we may obtain approximate
results by applying the formulas for a thinwalled prismatic tube to a
differential element of the tapered tube and then integrating along the
axis of the tube. dB dA Thinwalled tapered tube
B A
d(z) dA
x For entire tube:
L dB 4T
GT f dx dx
3
dB dA
xR
L B dA
0 L t T dA
dB
2
dA d 2
B 2 TL
Gt Solution 3.1013 B A T From table of integrals (see Appendix C): thickness dA average diameter at end A dB (a dx
bx ) 3 1
2b ( a bx ) 2 average diameter at end B T torque d(x) dA 2 r3t J(x) dB dA
L 4T
B
Gt d 3t
4 t
[ d(x) ] 3
4 Tdx
GJ ( x ) L 1
2¢ dB dA
L ¢ dA dB dA
L 2 x t
BdA
4 dB 3 dA xR L G t B dA 4Tdx
dB dA
L 3 xR L
2( dB f 2TL dA dB
¢
22
Gt dAdB L
2
dA ) dB 2( dB S
0 x For element of length dx:
df 4T
Gt C average diameter at distance x from end A. d(x)
J f 253 ThinWalled Tubes 2
dA ) dA R 254 CHAPTER 3 Torsion Stress Concentrations in Torsion
The problems for Section 3.11 are to be solved by considering the
stressconcentration factors. D2 R
D1 Problem 3.111 A stepped shaft consisting of solid circular segments T
having diameters D1 2.0 in. and D2 2.4 in. (see figure) is subjected
to torques T. The radius of the fillet is R 0.1 in.
If the allowable shear stress at the stress concentration is 6000 psi,
what is the maximum permissible torque Tmax?
Solution 3.111 T Stepped shaft in torsion
D2 R USE FIG. 348 FOR THE STRESSCONCENTRATION D1 T FACTOR T R
D1
K
D1 2.4 in. R tmax 1.52 0.1 in. D2
D1 0.05
K 2.4 in.
2.0 in. tmom K¢ 2.0 in. D2 0.1 in.
2.0 in. allow 1.2 16 Tmax
D3
1 D3tmax
1
16K Tmax ( 2.0 in.) 3 (6000 psi)
16(1.52) 6000 psi ∴ Tmax 6200 lbin. 6200 lbin. Problem 3.112 A stepped shaft with diameters D1 40 mm and
D2 60 mm is loaded by torques T 1100 N m (see figure).
If the allowable shear stress at the stress concentration is 120 MPa,
what is the smallest radius Rmin that may be used for the fillet?
Solution 3.112 Stepped shaft in torsion
D2 T USE FIG. 348 FOR THE STRESSCONCENTRATION FACTOR R
D1
T tmax Ktnom allow 40 mm
60 mm
1100 N m
120 MPa D2
D1 16T
D3
1
( 40 mm) 3 (120 MPa)
16(1100 N # m ) D3tmax
1
16T K D1
D2
T K¢ 60 mm
40 mm 1.5 From Fig. (348) with
we get R
D1 ∴ Rmin 1.37 D2
D1 1.5 and K 0.10
0.10(40 mm) 4.0 mm 1.37, SECTION 3.11 Stess Concentrations in Torsion Problem 3.113 A full quartercircular fillet is used at the shoulder
of a stepped shaft having diameter D2 1.0 in. (see figure). A torque
T 500 lbin. acts on the shaft.
Determine the shear stress max at the stress concentration for
values as follows: D1 0.7, 0.8, and 0.9 in. Plot a graph showing
max versus D1.
Solution 3.113 Stepped shaft in torsion
D2 R D1 (in.) D2/D1 D1 T T R (in.) R/D1 K max (psi) D2 D1
2 0.5 in. 0.214 1.20 8900 1.25 0.10 0.125 1.29 6400 1.11 0.05 0.056 1.41 4900 10,000 Full quartercircular fillet (D2
R 0.15 0.9
1.0 in.
500 lbin.
0.7, 0.8, and 0.9 in. 1.43 0.8 D2
T
D1 0.7 D1 2R) max (psi) D1
2 5000 USE FIG. 348 FOR THE STRESSCONCENTRATION FACTOR
tmax Ktnom
K K¢ 16T
D3
1 16(500 lbin.)
D3
1 0 2546 K
D3
1 0.6 0.8 1 D1 (in.) Note that max gets smaller as D1 gets larger, even
though K is increasing. 255 256 CHAPTER 3 Torsion Problem 3.114 The stepped shaft shown in the figure is required to
transmit 600 kW of power at 400 rpm. The shaft has a full quartercircular
fillet, and the smaller diameter D1 100 mm.
If the allowable shear stress at the stress concentration is 100 MPa, at
what diameter D2 will this stress be reached? Is this diameter an upper or
a lower limit on the value of D2?
Solution 3.114 Stepped shaft in torsion
D2 R
D1 T P
n 600 kW
400 rpm D1 T 100 mm
100 MPa allow Full quartercircular fillet
POWER P
P 2 nT
(Eq. 342 of section 3.7)
60 watts
60P
2n
T n rpm T Newton meters 3 60(600 10 W )
2 ( 400 rpm) 14,320 N # m USE FIG. 348 FOR THE STRESSCONCENTRATION FACTOR
tmax
K Ktnom K¢ 16T
D3
1 tmax ( D3 )
1
16T
( 100 MPa)( ) (100 mm) 3
16(14,320 N # m ) 1.37 Use the dashed line for a full quartercircular fillet.
R
D1 0.075 R 0.075 D1 0.075 ( 100 mm) 7.5 mm
D2 D1 ∴ D2 2R 100 mm 2(7.5 mm) 115 mm 115 mm This value of D2 is a lower limit
(If D2 is less than 115 mm, R/D1 is smaller, K is
larger, and max is larger, which means that the
allowable stress is exceeded.) SECTION 3.11 Stess Concentrations in Torsion Problem 3.115 A stepped shaft (see figure) has diameter D2 1.5 in.
and a full quartercircular fillet. The allowable shear stress is 15,000 psi
and the load T 4800 lbin.
What is the smallest permissible diameter D1?
Solution 3.115 Stepped shaft in torsion
D2 R
D1 T D2 T 1.5 in. Use trialanderror. Select trial values of D1 allow 15,000 psi D1 (in.) R (in.) R/D1 K T 4800 lbin. 1.30
1.35
1.40 0.100
0.075
0.050 0.077
0.056
0.036 1.38
1.41
1.46 Full quartercircular fillet D2
R D2 D1
2 0.75 in. D1 2R Ktnom K¢ 16T
D3
1 K 16(4800 lbin.)
B
D3
1
24,450 K
D3
1 15,400
14,000
13,000 D1
2 USE FIG. 348 FOR THE STRESSCONCENTRATION FACTOR
tmax max(psi) Tmax (psi) 16,000
Tallow
15,000
D1=1.31in.
14,000 13,000 1.30 From the graph, minimum D1 1.40 1.31 in. 257 ...
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This note was uploaded on 10/01/2009 for the course MEGR 2144 taught by Professor Sharpe during the Fall '08 term at UNC Charlotte.
 Fall '08
 Sharpe

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