1. No category

AMME2301/AMME5301
Torsion
• Lecturer: Dr Li Chang
• Room s503, Building J07
• Tel: 9351-5572
• e-mail: [email protected]
Torsion of Circular Section: Maximum Shear Stress
Engineer's Theory of Torsion ( ETT )
where
T = the internal torque at the analyzed cross-section;
J = the shaft’s polar moment of inertia;
G = shear modulus of elasticity for the material
ρ = radial distance from the axis (centre).
The Maximum Shear Stress
The maximum shear stress can be computed as where Ro is the radius of the outer surface
of circular shaft
Torsion of Circular Section: Angle of Twist
Single Constant Torque and Uniform Cross-Section Area
ϕ
T
x
T’
φ = the angle of twist of one end with respect to other end, measured in radian
T(x) = the internal torque at arbitrary position x,
J(x) = the shaft’s polar moment of inertia expressed as a function of position x.
G = shear modulus of elasticity for the material
Torsion of Circular Section: Sign Convention
Sign Convention of Internal Torque
We will use the right-hand rule, whereby both the torque and angle will be positive, provided
the thumb is directed outward from the shaft when the fingers curl to give the tendency for
Rotation.
Torsion of Circular Section: Statically Indeterminate
5-88 Determine the maximum shear stress in the shaft. G = 75 GPa (p. 223)
1. Equilibrium Equation:
TA + TB – 750 = 0
ΣT=0
𝜑𝐵/𝐴 = 0
2. Compatibility:
𝑇𝐴𝐶 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
𝑇𝐶𝐷 𝐿𝐶𝐷
=
𝐽𝐶𝐷 𝐺𝐶𝐷
Segment CD:
𝜑𝐷/𝐶
Segment DB:
𝜑𝐵/𝐷 =
𝑇𝐴𝐶 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
+
𝑇𝑖 𝐿𝑖
𝐽𝑖 𝐺𝑖
𝑇𝐶𝐷 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
+
d=12 mm
+
d=25 mm
𝑇𝐴 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
𝑇 𝐿
− 𝐵 𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
=0
d=25 mm
TA
=0
TB
T = 750 Nm
T [Nm]
TAD
B
A
𝑇𝐴 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
300 mm
T = 750 Nm
𝑇𝐷𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
𝑇𝐷𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
B
200 mm
𝜑𝐶/𝐴 =
Segment AC:
𝜑𝐵/𝐴 =
125 mm
𝜑=
D
C
A
C
D
TDB
TAD = TA ; TDB = -TB
Torsion of Circular Section: Statically Indeterminate
5-88 Determine the maximum shear stress in the shaft. G = 75 GPa (p. 223)
1. Equilibrium Equation:
2. Compatibility:
𝑇𝐴 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
TA + TB – 750 = 0
+
𝑇𝐴 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
−
𝑇𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
=0
D
C
A
125 mm
d=12 mm
B
200 mm
300 mm
d=25 mm
d=25 mm
T = 750 Nm
𝑇𝐵
TA
𝑇𝐴 = 8.5
TA = 78.8 Nm;
TB
TB = 671.2 Nm
3. The Maximum Shear Stress
1. the outer surface of AC segment
T [Nm]
T = 750 Nm
TAD
B
2. the outer surface of DB segment
A
C
D
TDB
TAD = TA ; TDB = -TB
Torsion of Circular Section
5-88 Determine the maximum shear stress in the shaft. G = 75 GPa (p. 223)
1. Equilibrium Equation:
TA + TB – 750 = 0
ΣT=0
125 mm
d=12 mm
2. Compatibility:
𝜑𝐷/𝐴 = −𝜑𝐷/𝐵
𝜑𝐷/𝐴 = 𝜑𝐶/𝐴 + 𝜑𝐷/𝐶 = −𝜑𝐷/𝐵
𝑇𝐴𝐶 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
𝑇𝐴 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
+
+
𝑇𝐶𝐷 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
𝑇𝐴 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
=−
=
D
C
A
200 mm
B
300 mm
d=25 mm
d=25 mm
T = 750 Nm
TA
TB
𝑇𝐷𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
T = 750 Nm
𝑇𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
-T
D
Torsion of Circular Section: Statically Indeterminate
5-88 Determine the maximum shear stress in the shaft. G = 75 GPa (p. 223)
1. Equilibrium Equation:
TA + TB – 750 = 0
ΣT=0
𝜑𝐵/𝐴 = 0
2. Compatibility:
Segment AC:
𝑇𝐴𝐶 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
𝑇𝐶𝐷 𝐿𝐶𝐷
=
𝐽𝐶𝐷 𝐺𝐶𝐷
𝜑𝐷/𝐶
Segment DB:
𝜑𝐵/𝐷 =
𝑇𝐴𝐶 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
+
𝑇𝑖 𝐿𝑖
𝐽𝑖 𝐺𝑖
𝑇𝐶𝐷 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
+
d=12 mm
+
d=25 mm
𝑇𝐴𝐷 𝐿𝐶𝐷 (𝑇𝐴𝐷 −750)𝐿𝐷𝐵
+
𝐽𝐶𝐷 𝐺𝐶𝐷
𝐽𝐷𝐵 𝐺𝐷𝐵
d=25 mm
TA
=0
=0
TB
T = 750 Nm
T [Nm]
TAD
B
A
𝑇𝐴𝐷 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
300 mm
T = 750 Nm
𝑇𝐷𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
𝑇𝐷𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
B
200 mm
𝜑𝐶/𝐴 =
Segment CD:
𝜑𝐵/𝐴 =
125 mm
𝜑=
D
C
A
C
D
TDB
TAD = TA ; TDB = -TB
Torsion of Circular Section: Statically Indeterminate
5-88 Determine the maximum shear stress in the shaft. G = 75 GPa (p. 223)
𝜑𝐵/𝐴 = 0
Compatibility:
Segment AC:
Segment CD:
Segment DB:
𝜑𝐵/𝐴 =
𝑇𝐴𝐶 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
𝑇𝐴𝐷 𝐿𝐴𝐶
𝐽𝐴𝐶 𝐺𝐴𝐶
+
+
𝜑=
𝑇𝐴𝐶 𝐿𝐴𝐶
𝜑𝐶/𝐴 =
𝐽𝐴𝐶 𝐺𝐴𝐶
𝑇𝐶𝐷 𝐿𝐶𝐷
𝜑𝐷/𝐶 =
𝐽𝐶𝐷 𝐺𝐶𝐷
𝜑𝐵/𝐷
𝑇𝐶𝐷 𝐿𝐶𝐷
𝐽𝐶𝐷 𝐺𝐶𝐷
𝑇𝐷𝐵 𝐿𝐷𝐵
𝐽𝐷𝐵 𝐺𝐷𝐵
𝑇𝐴𝐷 𝐿𝐶𝐷 (𝑇𝐴𝐷 −750)𝐿𝐷𝐵
+
𝐽𝐶𝐷 𝐺𝐶𝐷
𝐽𝐷𝐵 𝐺𝐷𝐵
125 mm
B
200 mm
300 mm
d=25 mm
d=25 mm
T = 750 Nm
TA
TB
=0
=0
TDB = - 671.2 Nm
T = 750 Nm
T [Nm]
TAD
B
A
TAD = 78.8 Nm;
D
C
A
d=12 mm
𝑇𝐷𝐵 𝐿𝐷𝐵
=
𝐽𝐷𝐵 𝐺𝐷𝐵
+
𝑇𝑖 𝐿𝑖
𝐽𝑖 𝐺𝑖
C
D
TDB
TAD = TA ; TDB = -TB
Torsion & Axial Load
Loading Types
Deformation
Torsion & Axial Load
Loading Types
Deformation
Torsion: Power Transmission
P=Tω
A
B
FA = - FB
TA /TB = RA/RB
FBA
< τallow
B
A
RA
RB
FAB = - FBA
FAB
TB= FAB RB; TA = FBA RA
Torsion: Power Transmission
P=Tω
A
B
FA = - FB
TA /TB = RA/RB
B
ϕA/ϕB = RB/R A
A
ϕA
RB
ϕB
RA
lA = lB
ϕARA = ϕB RB
Torsion: Power Transmission
5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by
bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of
twist of ends B and A. G = 75 GPa (p. 215)
1. Internal Torque:
A
T [Nm]
G
F
60 Nm
TF = 60 Nm
H
B
E
G
B
A
G
F
L
F
-60 Nm
D
H
C
E
120 Nm
FFE
E
TE = -90 Nm F
TD = 30 Nm
T [Nm]
E
30 Nm
D
-90 Nm
H
C
FEF = - FFE
L
FEF
TF /TE = RF/RE
Torsion: Power Transmission
5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by
bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of
twist of ends B and A. G = 75 GPa (p. 211)
1. Internal Torque:
A
T [Nm]
G
F
B
60 Nm
TF = 60 Nm
H
E
B
A
G
F
G
L
-60 Nm
F
D
H
C
E
120 Nm
TE = -90 Nm
TD = 30 Nm
Shaft DE Segment DH: TDH = 30 Nm
T [Nm]
Segment HE: THE = -90 Nm
D
-90 Nm
Shaft AB
E
30 Nm
H
C
L
Segments AG and FB: T = 0
Segment GF: TGF = -60 Nm
Torsion: Power Transmission
5-62 The two shafts are made of A-36 steel. Each has a diameter of 25 mm, and they are supported by
bearings at A, B and C, which allow free rotation. If the support at D is fixed, Determine the angel of
twist of ends B and A. G = 75 GPa (p. 211)
1. Internal Torque:
Shaft DE Segment DH: TDH = 30 Nm
H
Segment HE: THE = -90 Nm
Shaft AB Segments AG and FB: T = 0
2. Angle of twist:
E
Segment GF: TGF = -60 Nm
G
F
/D
𝜑𝐸/𝐷 =𝜑𝐻/𝐷 + 𝜑𝐸/𝐻
𝜑𝐵/𝐷 = 𝜑𝐹/𝐷
= - 0.02086
E
150
= −𝜑𝐸/𝐷
100
𝜑𝐴/𝐷 =𝜑𝐺/𝐷 = 𝜑𝐹/𝐷 + 𝜑𝐺/𝐹
Ans.
/D
Ans.
Torsion
External Loading Types
T
0
Internal Loading Diagram
x
Stress
τmax ≤ τallow
Shear Strain
Strain
Deformation
γ = π/2 – θ’
τ = Gγ
(Compatibility)
Axial Load
External Loading Types
N
F
0
Internal Loading Diagram
x
σ ≤ σallow
Stress
Average Normal Strain
Strain
Deformation
σ = Eε
(Compatibility)
AMME2301/AMME5301
Bending
• Lecturer: Dr Li Chang
• Room s503, Building J07
• Tel: 9351-5572
• e-mail: [email protected]
Bending
Distributed Force
Concentrated Force
Internal Loadings - Bending Moments and Shear Force
SIGN CONVENTION:
• Distributed load is positive when it acts upward on
the beam;
• Shear is positive when internal shear force causes a
clockwise rotation of the beam segment on which it
acts;
• Moment is positive when it causes compression in
the top beams of the segment.
Internal Loadings - Bending Moments and Shear Force
= w(x)×dx
w(x)
M(x)
x
dx/2
V(x)
M(x) + dM(x)
dx
𝑉(𝑥) + 𝑤 𝑥 𝑑𝑥 − 𝑉(𝑥) + 𝑑𝑉(𝑥) = 0
dx
V(x) + dV(x)
𝑤 𝑥 𝑑𝑥 = 𝑑𝑉(𝑥) or,
𝑑𝑥
−𝑀 𝑥 − 𝑉 𝑥 𝑑𝑥 − 𝑤 𝑥 𝑑𝑥
+ (𝑀 𝑥 + 𝑑𝑀 𝑥 ) = 0
2
𝑑𝑀(𝑥) = 𝑉(𝑥)𝑑𝑥 or,
Graphical Method: Shear and Moments Diagrams
The slope of shear diagram = distributed loading intensity
Slope of moment diagram = shear force
Graphical Method: Shear and Moments Diagrams
A
B
1. Determine the ground reactions:
; MA = 0
; MB = 0
2. Plotting the shear and moment diagrams:
A
A
B
B
Graphical Method: Shear and Moments Diagrams
A
B
1. Determine the ground reactions:
; MA = 0
; MB = 0
2. Plotting the shear and moment diagrams:
M/L
B
A
M/L· a
A
B
M/L · (L-a)
Graphical Method: Shear and Moments Diagrams
o
1. Determine the ground reactions:
; MA = 0
; MB = 0
2. Plotting the shear and moment diagrams:
wo L/2
A
A
B
-wo L/2
B
Graphical Method: Shear and Moments Diagrams
A
1. Determine the ground reactions: RMA =wo L/2 MA =-wo L2/3;
wo
B
RB = 0
MB = 0
2. Plotting the shear and moment diagrams:
wo L/2
A
A
-wo L2/3
B
B
Graphical Method: Shear and Moments Diagrams
6-19 Draw the shear and moment diagrams for the beam. (p. 281)
30 kN/m
45 kN m
B
A
1.5 m
1.5 m
1.5 m