Chp.5 Torsion

State of Stress
τ α α τ τ n x
types of failure
Ductile
Brittle
Strength Condition
τ max ≤ [τ ]
τ max
TR T = = ≤ [τ ] I p Wp
1. Check the shear stress strength; 2. Design a reasonable size of cross section; 3. Determine the maximum loads.
Use the proportional triangle
τρ ρ = τ max R
EXAMPLE 4 The diameters of shaft AB and BC are D1 and D2 separately. Known D1=2D2 ， determine the maximum shear stressτmax.
dA=2πρdρ
Tubular shaft
π D 4 (1 α 4 )
32
Ip =
d α= D Wp =
π D 3 (1 α 4 )
16
EXAMPLE 3 The solid shaft of diameter D=10cm is subjected to torque T=19kNm. Determine the shear stress at point K (ρ=4cm) and the maximum shear stress of the area.
T—internal torque，IP— polar second moment， GIP —stiffness of torsion .
τρ is linearly increase with respect to the distance ρ
from the axis of the shaft. IP/R=WP 抗扭截面系数，then the maximum shear stress
T T1 T2 D=2d
d DLeabharlann 1 T1 = T 16 15 T2 = T 16
Isolated volume element 单元体
Complementary property of shear stress(切应力互 等定律): On any adjacent perpendicular faces, the shear stresses are the same, which point to (or depart from） the intersecting line of the two faces.
T O
ρK
D
SOLUTION （1）The shear stress of point K
T ρ 19 × 103 × 4 × 102 τK = = = 77.4 MPa 4 8 Ip π × 10 × 10 32
（2）The maximum shear stress
τ max
T 19 × 103 = = = 96.9 MPa 6 3 W p π × 10 × 10 16
Me
R x L
γ L =φ R
Shear strain is defined as the change of the right angle.
γ dx = d φ R
5.2 The Torsion Formula
Internal torque (扭矩 T) :Cutting section method
τ = τ
Not only does the internal torque T develop a linear distribution of shear stress along each radial line in the plane of cross sectional area, but also an associated shear-stress distribution is developed along an axial plane.
The shear stress in the thin-walled circular tube Approximately calculating shear stress: T= Me= ∫A τ dAR0
Me
τ = Me /（ 2πR0t R0）
A0=πR02
τ = Me / （2A0t）
d T 180o ) max × =( ≤ [θ ] dx GI p π
unit: radians , degree
Strain Energy of Torsion
1 1 dW = (τ dydz )( γ dx ) = τγ ( dxdydz ) 2 2
Shape
Solid Shaft
Tubular Shaft
Thin-Walled Tube
EXAMPLE 5 The revolution of the shaft is n=100rpm. Transmission power is P=10kw. The allowable shear stress is [τ]=80MPa. Design the diameter of a solid shaft and tubular shaft. Providedα=d/D=0.5, compare the weight of two shafts in the case of equal length.
Chapter 5 Torsion
A long straight member with circular cross section is subjected to twisting. Torsional deformation: Circles remain circles, and the length of the shaft remains unchanged.
τ max
4. Sign of shear stress
TR T = = I p Wp
τ + τ
Polar Second Moment The diameter of a circular cross section is D. IP and WP is calculated as following:
∑Mx=0
T=Me
Sign convention：right-hand rule (右手螺旋定则 )
Torque diagram
EXAMPLE 1 Refer to figure with MA=70Nm， MC=130Nm， MB=200Nm . Plot the torque diagram of this member.
SOLUTION
Torsional Stress on solid circular shaft
Deformable assumption Only shear stress on the cross section τ
dx
dφ γ =R dx
dφ γρ = ρ dx
Hooke’s law in shear stress
Statics
∫ ρτ ρ dA = T
A
dφ τ ρ = Gρ dx
dφ T = dx GI p
Tρ τρ = Ip
I p = ∫ ρ dA
2 A
is noted the polar moment of inertia or polar moment (极惯性矩 )
1. 2. 3.
note
Tρ τρ = Ip
SOLUTION For segment AB
2M n 4M n τ AB max = = 3 3 π (2 D2 ) π D2 16 For segment BC
τ BC max =
Mn
π D2
16
3
16 M n = 3 π D2
∴
τ max = τ BC
Question
Can you tell me the fraction of T that is resisted by the shaded area.
TL T and stiffness are constant along L, gives = GI p
Quantities are constant in each segment. For this case
Ti Li =∑ Gi I pi
Stiffness Condition：the unit angle of twist is smaller than the allowable one.
SOLUTION
P
5.4 Angle of Twist
Angle of twist per unit length Angle of twist of one end of shaft with respect to the other end
T d = dx GI p
T ( x) =∫ dx GI p L
MA
MB
MC
SOLUTION （1）Internal Torque
（2） Torque Diagram
EXAMPLE 2 Draw the torque diagram of the following shaft.
SOLUTION T（x）= - mnx Tmax= - mnL
EXERCISE The torque shown in the figure are Μn=920kNm，mn=160kNm/m. Plot the internal torque diagram.(L=1m)
Substitute