TORSION
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Torsion in a Circular Shaft
The Torsion Formula
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Stress
Strain
Torque Diagram
Angle of Twist
Power Transmission
Statically Indeterminate Torque-Loaded
Members
1
Torsion in a Circular Shaft
y
φ(x)
x
x
deformed
plane
undeformed
plane
z
The angle of twist φ(x) increases as x increases.
2
The Torsion formula
τ
τ max
=
ρ
c
τ =ρ
τ
ρ
c
T
τmax
τ max
c
τ max
c
= constant
dA
τmax
τ
c
ρ
3
T
τ
τ
dT = ρ (dF ) = ρ (τdA)
dF
ρ
τ
dT = ρ [ ρ (
dA
c
T = ∫(
T =(
τmax
τ
c
ρ
τ max
τ max
c
τ max
c
c
)dA]
constant
) ρ 2 dA
J
) ∫ ρ 2 dA
Tc
)
J
Tρ
τ =
J
τ max = (
4
• Solid Shaft
τmax
T
dρ
c
c
1
J = ∫ ρ dA = ∫ ρ (2πρdρ ) = 2π ∫ ρ 3 dρ = 2π ( ) ρ 4
4
0
0
A
2
c
2
ρ
J=
π
2
c
0
c4
• Tubular Shaft
τmax
T
τmin
co
ci
J=
π
2
4
4
(co − ci )
5
• Stress strain relationship
τ (MPa)
G=
τ
γ
γ (rad)
6
• Equilibrium and compatibility condition
T1
TA
B
A
LAB
T
TA
di
T2
TB
C
LBC
LCD
D
do
TA + T1
TA + T1 - T2 = TB
x
Equilibrium equation:
TA + T1 - T2 - TB = 0
Compatibility equation:
φD/A = φB/A + φC/B + φD/C = 0
7
• Torsion and shear stress diagram
T1
TA
B
A
C
LAB
T
TA
di = 2ci
T2
LBC
LCD
T3
D
do = 2co
TA + T1
TA + T1 - T2 = T3
+
x
τ
τ AB =
(TA )(co )
4
(π / 2)(co )
τ BC =
+
(TA + T1 )(co )
4
(π / 2)(co )
τ CD =
(T3 )(co )
4
4
(π / 2)(co − ci )
x
8
Angle of twist
T
dφ
cdφ
γmax
Tc
)
J
Tρ
τ =
J
τ max = (
c
dx
τ
cdφ = γ max dx
dφ =
γ max dx
dφ =
τ max dx
G=
c
Gc
Tdx
dφ =
JG
φ=∫
Tdx
JG
=
Tcdx
JGc
φ=
TL
JG
τ
τ
, γ =
G
γ
γ
9
• Sign Convention
φ
x
x
+φ(x)
+Τ(x)
+φ(x)
+Τ(x)
Positive sign convention
for T and φ
10
• Torsion and angle of twist diagram
TA
B
A
LAB
T
TA
di = 2ci
T2
T1
C
LBC
LCD
T3
D
do = 2co
TA + T1
+
TA + T1 - T2 = T3
x
φC / A = φ B / A +
φx/A
φB / A
(TA )( LAB )
=
4
(π / 2)(co )G
(TA + T1 )( LBC )
4
(π / 2)(co )G
φ D / A = φC / A +
(T3 )( LCD )
4
4
(π / 2)(co − ci )G
x
11
Example 1
The shaft shown is supported by two bearings and is subjected to three torques.
Determine the shear stress developed at points A and B, located at section a-a of
the shaft.
a
5 kN•m
3 kN•m
2 kN•m
a
A
B
z
30 mm
y
A
B
10 mm
Section a-a
x
12
a
5 kN•m
3 kN•m
2 kN•m
0
0
T (kN•m)
a
A
-2
z
30 mm
y
A
-5
-5
2 kN•m
B
A
B
10 mm
Section a-a
47.17 MPa
B
• Section Property
x
47.17 MPa
A
Top view A
15.72 MPa
J =
π 4 π
c = ( 0 . 03 m ) 4 = 1 . 272 (10 −6 ) m 4
2
2
• Shear Stress
τA
( 2 × 10 3 N • m )( 0 . 03 m )
Tc
=
=
= 47 . 17 MPa
J
1 . 272 (10 −6 ) m 4
τB
Tρ ( 2 × 10 3 N • m )( 0 . 01 m )
=
=
= 15 . 72 MPa
−6
4
J
1 . 272 (10 ) m
13
Example 2
The pipe shown in has an inner diameter of 30 mm and an outer diameter of 60
mm. If its end is tightened against the support at A using a torque wrench at B,
determine the shear stress developed in the material at the inner and outer walls
along the central portion of the pipe when the 80-N forces are applied to the
wrench.
A
C
80 N
di = 30 mm
80 N
B
do = 60 mm
200 mm
300 mm
14
di = 30 mm
y
T
80 N
do = 60 mm
C
z
∑M
A
x
= 0;
80 N (0.5 m) − T = 0
T = 40 N•m
80 N
• Section Property
B
200 mm
300 mm
x
D
1.006 MPa
τD = 1.006 MPa
τE = 0.503 MPa
E
• Internal Torque
40 N•m Section C
0.503 MPa
J =
π
[( 0 . 03 m ) 4 − ( 0 . 015 m ) 4 ] = 1 . 193 (10 −6 ) m 4
2
• Shear Stress
- For any point lying on the outside surface of the
pipe, ρ = co = 0.03 m
τo =
Tco 40 N • m(0.03 m)
=
= 1.006 MPa
1.193(10 −6 ) m 4
J
- For any point located on the inside surface,
ρ = ci = 0.015 m
τi =
Tc i
40 N • m ( 0 . 015 m )
=
= 0 . 503 MPa
J
1 . 193 (10 −6 ) m 4
15
Example 3
A solid steel shaft shown in the figure (G = 80 GPa), having an inner diameter of
30 mm and outer diameter of 60 mm. Draw the quantitative torsion, shear stress
and the angle of twist diagram of the shaft.
1.0 m
0.6 m
1200 N•m
C
di = 30 mm
A
800 N•m
B
do = 60 mm
16
1.0 m
0.6 m
1200 N•m
C
A
B
di = 30 mm
do = 60 mm
T (N•m)
JAB = (π/2)(0.034 )
800 N•m
= 1.272 (10-6) m4
JBC = (π/2)(0.034 - 0.0154 )
= 1.193 (10-6) m4
400
400
x (m)
+
-800
-800
Tc
J
τ AB =
9.43
+
τ (MPa)
τ =
J = (π/2)(co4 - ci4)
20.12
τ BC =
( 400 N • m )( 0 . 03 m )
= 9 . 43 MPa
(1 . 272 × 10 −6 m 4 )
x (m)
( −800 N • m )( 0 . 03 m )
= −20 . 12 MPa
(1 . 193 × 10 −6 m 4 )
17
1.0 m
0.6 m
1200 N•m
C
A
J = (π/2)(co4 - ci4)
B
di = 30 mm
do = 60 mm
T (N•m)
JAB = (π/2)(0.034 )
800 N•m
= 1.272 (10-6) m4
JBC = (π/2)(0.034 - 0.0154 )
= 1.193 (10-6) m4
400
400
x (m)
+
φB/A =
-800
-800
0.00393
( 400 )(1 . 0 )
(1 . 272 × 10 −6 )( 80 × 10 9 )
= 0 . 00393 rad = 0 . 23 o
φ (rad)
φ =
TL
JG
( −800 )( 0 . 6 )
x (m)
φC /B =
= −0 . 00503 rad = −0 . 29 o
−6
9
(1 . 193 × 10 )( 80 × 10 )
0.00393 - 0.00503 = -0.0011
18
Example 4
A solid steel shaft shown in the figure(G = 80 GPa), having an inner diameter of
30 mm and outer diameter of 60 mm. Draw the quantitative torque,
shear stress and the angle of twist diagrams of the shaft.
400 N•m
A
B
0.5 m
2800 N•m
C
0.4 m
di = 30 mm
1500 N•m
D
0.6 m
do = 60 mm
19
400 N•m
TA
B
A
T (N•m)
TA
2800 N•m
C
0.5 m
0.4 m
TA + 400
di = 30 mm
1500 N•m
D
do = 60 mm
0.6 m
TA + 400 + 2800 = 1500
+
x (m)
Equilibrium equation:
TA + 400 + 2800 -1500 = 0
TA = -1700 N•m
20
400 N•m
1700 N•m
A
T (N•m)
B
0.5 m
2800 N•m
C
0.4 m
JAC = (π/2)(0.034 )
= 1.272 (10-6) m4
di = 30 mm
1500 N•m
D
do = 60 mm
0.6 m
JCD = (π/2)(0.034 - 0.0154 )
= 1.193 (10-6) m4
1500
+
-
( −1700 N • m)( 0 . 03 m )
(π / 2 )( 0 . 03 4 ) m 4
= −40 . 08 MPa
τ AB =
x (m)
-1300
-1700
τ (MPa)
37.73
+
-40.08
-30.65
( −1300 N • m )( 0 . 03 m )
(π / 2 )( 0 . 03 4 ) m 4
= −30 . 65 MPa
τ BC =
x (m)
τ CD =
(1500 N • m )( 0 . 03 m )
( π / 2 )( 0 . 03 4 − 0 . 015 4 ) m 4
= 37 . 73 MPa
21
400 N•m
1700 N•m
A
T (N•m)
B
2800 N•m
C
0.5 m
0.4 m
D
do = 60 mm
JCD = (π/2)(0.034 - 0.0154 )
= 1.193 (10-6) m4
1500
+
-
φx/A (rad)
1500 N•m
0.6 m
JAC = (π/2)(0.034 )
= 1.272 (10-6) m4
-1700
di = 30 mm
( −1700 )( 0 . 5 )
( π / 2 )( 0 . 03 4 )( 80 × 10 9 )
= −0 . 00835 rad
φB/A =
x (m)
-1300
( −1300)(0.4)
(π / 2)(0.034 )(80 × 109 )
= −0.00511 rad
φC / B =
-0.00835
-0.00835 -0.00511 = -0.0135
x (m)
-0.00403
(1500)(0.6)
(π / 2)(0.034 − 0.0154 )(80 × 109 )
= 0.00943 rad
φD / C =
22
The meshed gears
T
A
D
L
d CD
E
CD
L AB
d AB
F
C
rB
B
rC
23
T
T
TD
A
TD = T rC
D rB
Fz
A
D
E
Ez
L
CD
Ex
F
L AB
C
rB
B
F =
Fx
T
rB
C
B
rC
F´ =
rB
T
rB
rC
• Internal torque
+ Σ MAB = 0: T - F(rB) = 0
T
F=
rB
F'= F =
+ Σ MCD = 0:
TD −
T
rB
T
rC = 0
rB
TD =
rC
T
rB
24
T
TD
A
D
E
L
CD
F
L AB
•
C
rB
B
rB
P
φB
P´
φC
P´´
•
rC
rC
Direction of motion
• Angle of twist
arc PP´ = arc PP´´
φB rB = φC rC
φB =
rC
φC
rB
25
T
Ez
TD = T rC
rB
D
Fz
A
Start at D → C → B → A
rC
T ) LCD
rB
=
J CD GCD
(
Ex
φC / D
T
F= r
B
Fx
B
C
T
F´ = r
B
rB
rC
φ C = φC / D
φA
rC
T ) LCD
rC
rC rB
φ B = φC = [
]
rB
rB J CD GCD
TLAB
φ A/ B =
J AB G AB
(
φ A = φ A/ B + φB =
TLAB
J AB G AB
rC rC
)( T ) LCD
rB rB
+[
]
J CDGCD
(
26
Example 5a
The two solid steel shafts are coupled together using the meshed gears.
Determine the shearing stress in two solid and the angle of twist of end A of shaft
AB when the torque T = 45 N•m is applied. The upper shaft has a diameter of
17.5 mm and the lower shaft has a diameter of 22.5 mm. Take G = 80 GPa. Shaft
AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D.
T = 45 N•m
A
D
E
1. 5
m
F
2
m
C
B
150 mm
75 mm
27
• Internal Torque
G = 80 MPa.
T = 45 N•m
Ez
T = 45 N•m
D
A
φ
=1
7.
5
φ
F
AB
=2
2
m
2.
5
m
TD
A
CD
E
Fz
1. 5
m
m
m
B
150 mm
E
Fx
C
m
Ex
75 mm
D
F
rC =75 mm
FB=300 N
C
FC = 300 N
rB =150 mm
B
r
75 mm
( 45 N • m)
TD = C T A =
150 mm
rB
= 22 . 5 N • m
• Shearing Stress
τ AB =
T AB c AB ( 45 N • m )( 0 . 01125 m )
=
= 20 . 12 MPa ⇐
4
J AB
( π / 2 )( 0 . 01125 m )
τ CD =
T CDc CD ( 22 . 5 N • m )( 0 . 00875 m )
=
= 21 . 38 MPa ⇐
4
J CD
(π / 2 )( 0 . 00875 m )
28
• Internal Torque
G = 80 MPa.
T = 45 N•m
Ez
T = 45 N•m
D
A
CD
2
m
φ F
AB
=2
2.
5
m
Fz
TD
D
A
φ
E
TD = T A =
=1
7.
5
1. 5
m
m
m
Fx
C
m
B
150 mm
Ex
E
FB
F
rB =150 mm
75 mm
• Angle of twist of end A
75 mm
( 45 N • m )
150 mm
= 22 . 5 N • m
rC =75 mm
φB φC
B
φB =
C
FC
rC
φ C = 0 . 5φ C
rB
φA = φA/B + φB = φA/B + 0.5φC
T LAB
rC rC T LCD
TAB LAB
rC TCD LCD
φA =
+ ( )[
] =
+ ( )[( )
]
J AB G rB rB J CD G
J AB G
rB J CD G
φA =
( 45 N • m )( 2 m )
(. 5 )( 45 N • m )(1 . 5 m )
0
.
5
[
]
+
4
9
4
9
( π / 2 )( 0 . 01125 m ) ( 80 × 10 Pa )
( π / 2 )( 0 . 00875 m ) ( 80 × 10 Pa )
φA = 0.0447 + 0.0229 = 0.0676 rad = 0.0676(180/π) = 3.87o
29
Example 5b
The two solid steel shafts are coupled together using the meshed gears.
Shaft AB is free to rotate within bearings E and F, whereas shaft DC
is fixed at D. The upper shaft has a diameter of 17.5 mm and the lower
shaft has a diameter of 22.5 mm. Determine the maximum torque the shaft
can have if it is restricted not to twist more than 10o and the allowable shear
stress is τallow = 50 MPa. Take G = 80 GPa.
T
A
D
E
1. 5
m
F
2
m
C
B
150 mm
75 mm
30
• Internal Torque
τallow = 50 MPa.
T
T
Fz
D
A
φ
CD
E
2
m
2.
5
m
TD
rC
75 mm
T =
= 0 . 5T
rB
150 mm
D
A
=1
7.
5
φ
F
AB
=2
TD =
Ez
1. 5
m
m
m
Fx
C
m
B
150 mm
Ex
E
F
rB =150 mm
75 mm
FB
B
rC =75 mm
C
FC
• Allowable Shearing Stress: τ = Tc/J
(τAB)allow = 50x106 Pa =
(T)(0.01125 m)
(π/2)(0.01125 m)4
,
T1 = 111.8 N•m
(τCD)allow = 50x106 Pa =
(0.5T)(0.00875)
(π/2)(0.00875 m)4
,
T2 = 105.2 N•m
31
G = 80 MPa.
Angle of Twist (φ) = 10o
T
T
Fz
D
A
φ
CD
E
φ
F
AB
=2
2
m
2.
5
m
TD =
Ez
TB
rC
rB
T =
D
A
=1
7.
5
1. 5
m
m
m
Ex
C
m
E
Fx
B 75 mm
150 mm
FB
F
rB =150 mm
75 mm
= 0 . 5T
150 mm
rC = 75 mm
φB φC
B
ϕB =
C
FC
rC
ϕ C = 0.5ϕ C
rB
• Angle of twist
φA = φA/B + φB = φA/B + 0.5φC
10 (
10 (
T L
π
T L
) rad = A AB + 0 . 5 [ D CD ]
180
J AB G
J CDG
( 0 . 5T )(1 . 5 m)
T ( 2 m)
π
) rad =
0
.
5
[
]
+
4
9
4
9
180
(π / 2 )( 0 . 01125 m) ( 80 × 10 Pa )
(π / 2 )( 0 . 00875 m) ( 80 × 10 Pa )
T3= 116.15 N•m
The maximum torque at A is 105.2 N•m
32
Power transmission
T
Power = Work performed per unit of time
dF
dθ
ρ
P = Power in N•m/s, or W or watts
1 W = 1 N•m/s
τ
dA
c
dP = (dF) ρ (dθ)/(dt)
=(dT) ω
P=Tω
ω = Angular velocity, in rad/s
P = 2π f T
f = Frequency, in Hz
Note
1 hp = 745.7 N•m/s
33
Example 6a
A solid steel shaft AB is to be used to transmit 5 hp from the motor M to which
it is attached. If the shaft rotates at ω = 175 rpm and the steel has an allowable
shear stress of τallow = 100 MPa, determine the required diameter of the shaft.
34
Solution
5 hp
ω = 175 rpm
τallow = 100 MPa
φ =?
⇒ τ allow =
⇒ P = Tω
175 rev 2π rad 1 min 
745 . 7 N • m/s
(5 hp )(
) =T 
×
×
1 hp
min
1
rev
60 s 

T = 203.4 N • m
Tc
Tc
2T
=
=
J (π / 2)(c 4 ) π (c 3 )
 2T 
c =

 πτ allow 
1/3
 2 ( 203 . 41 N • m ) 
=
6
2 
 π (100 × 10 N/m ) 
= 10 . 90 mm
1/ 3
d = 2c = 2(10.90 mm) = 21.8 mm
Select a shaft having a diameter of 22 mm
35
Example 6b
A tubular shaft, having an inner diameter of 30 mm and an outer diameter
of 42 mm, is to be used to transmit 90 kW of power. Determine the frequency
of rotation of the shaft so that the shear stress will not exceed 50 MPa.
Solution
τ max =
50 × 10 6 N/m 2 =
Tc
J
T ( 0 . 021 m)
(π / 2) ( 0 . 021 m) 4 − ( 0 . 015 m ) 4
[
]
T = 538 N • m
P = 2π fT
90 × 10 3 N • m/s = 2π f (538 N • m )
f = 26 . 6 Hz
36
Example 7
The two solid steel shaft are coupled together using the meshed gears. Shaft CD
connected to the machine which required power of 35 kW at frequency 10 Hz,
whereas shaft AB connected to motor to rotate within bearing E.
(a) Determine the required power and frequency of motor.
(b) If the allowable stress in both shafts must be limited to 200 MPa, determine
the size of the diameter required for each shaft.
Motor (Input)
A
E
rC = 150 mm
B
C
rB = 75 mm
D
Machine (Output)
Pout = 35 kW
fout = 10 Hz
37
Motor (Input)
A
E
rC = 150 mm
B
C
D
Machine (Output)
rB = 75 mm
Pout = 35 kW
fout = 10 Hz
• Internal torque
-Output (Machine)
-Input (Motor)
FB = FC
P = Tw = T (2πf )
TB TC
=
rB rC
35 × 103 = TD ( 2π × 10)
T C= T D = 557 N • m
TB =
rB
75
TC =
(557 N • m ) = 278 . 5 N • m
rC
150
38
Motor (Input)
A
T A = T B = 278 . 5 N • m
E
rC = 150 mm
B
C
rB = 75 mm
allowable stress in both shaft : τallow = 200 MPa
D
Machine (Output)
Pout = 35 kW
fout = 10 Hz
T C= T D = 557 N • m
• (a) The power and frequency of motor
fB =
rC
fC
rB
= 2f C
= 2(10 Hz) = 20 Hz
Pinput = T B ( 2π f B )
= ( 278 . 5 N • m)( 2π × 20 Hz)
= 35 kW
Comment : Pinput = TB (2πf B )
T
= ( C )(2π × 2 f C ) = TC ( 2πf )
2
39
Motor (Input)
T A = T B = 278 . 5 N • m
A
E
rC = 150 mm
B
C
D
Machine (Output)
rB = 75 mm
Pout = 35 kW
fout = 10 Hz
allowable stress in both shaft : τallow = 200 MPa
T C= T D = 557 N • m
• (b) Allowable diameter for each shaft
- Shaft AB
(τ AB ) allow =
200 × 10 6 Pa =
c AB = [
- Shaft CD
TB c AB
J AB
( 278 . 5 N • m )c AB
4
( π / 2 )c AB
( 278 . 5 N • m)
1/ 3
]
(π / 2)( 200 × 10 6 Pa )
TC cCD
J CD
( 557 N • m )c CD
200 × 10 6 Pa =
4
( π / 2 )c CD
(τ CD ) allow =
c CD = [
= 9 . 6 × 10 −3 m
d AB = 19 . 2 × 10 −3 m ≈ 20 mm ⇐
d CD
( 557 N • m )
1/ 3
]
(π / 2 )( 200 × 10 6 Pa )
= 0 . 0121 m
= 0 . 0242 m ≈ 25 mm ⇐
40
Statically Indeterminate Torque-Loaded Members
A
TA
C
LAC
L
B
T
TB
LBC
• Equilibrium equation
+ Σ Mx = 0:
- TA + T + TB = 0
----------(1)
• Compatibility equation:
φA/B = φA/C + φB/C = 0
TA LAC TB LBC
−
=0
JG
JG
− − − − − −( 2)
41
Example 8
A solid steel shaft shown in the figure(G = 80 GPa), having an inner diameter of
30 mm and outer diameter of 60 mm.
(a) Determine the reactions at the fixed supports A and D.
(b) Draw the quantitative torsion,shear stress and the angle of twist
diagram of the shaft.
500 N•m
B
di = 30 mm
800 N•m
C
A
D
0.5 m
0.4 m
0.6 m
do = 60 mm
42
500 N•m
TA
B
800 N•m
di = 30 mm
TD
C
A
D
0.5 m
T (N•m)
TA
0.4 m
0.6 m
TA + 500
do = 60 mm
TA + 500 + 800 = TD
+
x (m)
Equilibrium equation:
TA + 500 + 800 = TD
Compatibility equation:
φ D / A = φ B / A + φC / B + φ D / C = 0
TA (0.5)
(TA + 500)(0.4)
(TA + 1300)(0.6)
+
=0
+
4
9
4
9
4
4
9
(π / 2)(0.03 )(80 × 10 ) (π / 2)(0.03 )(80 ×10 ) (π / 2)(0.03 − 0.015 )(80 ×10 )
TA = -670.17 N•m, TD = -670.17+500+800 = 629.8 N•m
43
500 N•m
670.17 N•m
B
800 N•m
629.83 N•m
C
A
D
0.5 m
T (N•m)
di = 30 mm
0.4 m
JAC = (π/2)(0.034 )
= 1.272 (10-6) m4
0.6 m
JCD = (π/2)(0.034 - 0.0154 )
= 1.193 (10-6) m4
629.83
+
-
τ (MPa)
15.84
+
-15.80
( −670 . 17 N • m)( 0 . 03 m)
(π / 2 )( 0 . 03 4 ) m 4
= −15 . 80 MPa
τ AB =
x (m)
-170.17
-670.17
-
do = 60 mm
-4.01
( −170 N • m)( 0 . 03 m )
(π / 2 )( 0 . 03 4 ) m 4
= −4 . 01 MPa
τ BC =
x (m)
τ CD =
( 629 . 83 N • m )( 0 . 03 m )
( π / 2 )( 0 . 03 4 − 0 . 015 4 ) m 4
= 15 . 84 MPa
44
500 N•m
670.17 N•m
B
800 N•m
629.83 N•m
C
A
D
0.5 m
T (N•m)
di = 30 mm
0.4 m
JAC = (π/2)(0.034 )
= 1.272 (10-6) m4
0.6 m
JCD = (π/2)(0.034 - 0.0154 )
= 1.193 (10-6) m4
629.83
+
-
do = 60 mm
φB/A =
( −670 . 17 )( 0 . 5 )
(π / 2 )( 0 . 03 4 )( 80 × 10 9 )
x (m) = −0 . 00329 rad = −0 . 593 o
-170.17
-670.17
φC /B =
φx/A (rad)
( −170 . 17 )( 0 . 4 )
(π / 2 )( 0 . 03 4 )( 80 × 10 9 )
o
0
.
000669
rad
0
.
12
=
−
=
−
x (m)
- 0.00329
(629.83)(0.6)
(π / 2)(0.034 − 0.0154 )(80 ×109 )
- 0.00329 - 0.000669 = -0.00396 = 0.00396 rad = 0.713°
φD / C =
45
Example 9
The shaft shown is made form a steel tube, which is bonded to a brass core. If a
torque of T = 340 N•m is applied at its end, plot the shear-stress distribution
along a radial line of its cross sectional area. Take Gst = 80 GPa, Gbr = 35 Gpa.
15 mm
1m
30 mm
T = 340 N•m
46
Tbr
• Section Properties
Tst
π
π
4
c br = ( 0 . 015 m ) 4 = 7 . 952 × 10 −8 m 4
2
2
π
π
4
4
= (c st − c br ) = [( 0 . 03 m ) 4 − ( 0 . 015 m ) 4 ]
2
2
J br =
15 mm
1m
= 1.193(10-6) m4
30 mm
x
J st
T = 340 N•m
• Equilibrium
Tst + Tbr = 340
----------(1)
• Compatibility
φ = φst = φbr
Tst L
T L
= br
J st Gst J br Gst
T st L
T br L
=
(1 . 193 × 10 −6 m 4 )( 80 × 10 9 Pa ) ( 7 . 952 × 10 −8 m 4 )( 35 × 10 9 Pa )
Tst = 34.29 Tbr
Solving (1) and (2) , Tst = 330.4 N•m,
----------(2)
Tbr = 9.63 N•m
47
1.82 MPa
Tst = 330.4 N•m,
4.15 MPa
8.31 MPa
Tbr = 9.63 N•m
Jbr= 7.952(10-8) m4 , Jst= 1.193(10-6) m4
• Shear Stress
Tρ
J
( 9 . 63 N • m )( 0 . 015 m )
=
= 1 . 82 MPa
( 7 . 952 × 10 −8 m 4 )
τ=
0.015 m
A
B
0.03 m
Shear stress distribution
γmax = γB
= 1.039(10-4) rad
A
B
(τ br ) max
(τ st ) min =
(330.4 N • m)(0.015 m)
= 4.15 MPa
−6
4
(1.193 × 10 m )
(τ st ) max =
(330.4 N • m)(0.03 m)
= 8.31 MPa
(1.193 × 10 −6 m 4 )
• Shear Strain
γA =
γA = 5.2(10-5) rad
γA
rA
Shear strain distribution
=
γB
rB
(τ br ) A (τ st ) A
=
G br
G st
=(
1 . 82 MPa
−5
)
=
5
.
2
×
10
rad
br
3
35 × 10 MPa
γ max = γ B =
8 . 31 MPa
= 1 . 039 × 10 −4 rad
3
80 × 10 MPa
48