axi symmetric shafts in torsion
DESCRIPTION
torsion materialTRANSCRIPT
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S. Socrate 2013 K. Qian
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Axisymmetric Shafts in Torsion
Loading Conditions on each Section (x) Applied loading only around the axis (x) of the shaft. The only internal resultant at any sections ┴ x is the axial torque T(x)
Find T(x)along the bar (axial torque diagram) by cutting the bar at x and imposing moment_x equilibrium.
For the example shown, equilibrium at x gives: for x<xB :Σ Mx = 0 = – T(x)+QC +QB à T(x)= QC + QB
for x>xB :Σ Mx = 0 = – T(x) +QC à T(x)= QC
And the entire axial torque diagram is:
For distributed loading t (x), with t (x) [in (N*m)/m =N] + with right hand rule along x, obtain the torque T(x) by integrating t (x) along the shaft. For the shaft shown:
T ( x ) = t ( x )x
L
∫ dxThe differential relationship between the distributed torque t (x) and the axial resultant T(x) is
dT ( x )dx
= −t ( x )
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Structural response
Angle of twist Rotation field : with ϕ (x0)=ϕ0 determined by BC (e.g., ϕ0=0 at wall)
Section deformation Section at x has rotation ϕ(x) Section at x+dx has rotation ϕ(x+dx)= ϕ(x)+dϕ
Local measure of deformation at section x: twist rate
Φ =dϕdx0
L
∫ dx
ϕ( x )=ϕ( x0 )+dϕdxdx
x0
x
∫
sections rigidly rotateà strain γ increases with radius r: γ( r ,x )= rdϕdx( x )
Kinematics constraint (geometry of deformation) Cross sections ┴ x : stay flat, rotate by ϕ(x)
Strain ßà section deformation
dϕdx
Sign Conventions
“with x” Rotation field : ϕ (x) x-component of reaction torque : TA
x
“out” Angle of twist: Φ Internal Torque resultant : T
A
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T (x )= r ⋅τ ( r ,x )⋅2πrA∫ dr
Section equilibrium The Axial Torque T(x) at section x is obtained by integrating the contributions of each elemental area dA=2πrdr, at distance r from the axis, which carries a shear stress τ(r,x)
Constitutive Properties If the material is linear elastic and the shear modulus of elemental area dA is G
the stress can be obtained as Section Response
Constant over cross section
Effective Section Stiffness:
If only 1 material, G(x)à (GIp)eff =G(x) Ip(x);
If 2 materials (G1, G2) à (GIp)eff= G1 Ip1 + G2 Ip2
τ ( r ,x )=G ( r ,x )⋅γ( r ,x )
T (x )= r ⋅τ ( r ,x )⋅2πrA∫ dr = r ⋅G ( r ,x )γ( r ,x )
A∫ ⋅dA = r ⋅G ( r ,x ) r dϕ
dxA∫ ⋅dA
T (x )= dϕdx
r 2G ( r ,x )A∫ dA = dϕ
dx(GI p )eff ( x ) invert← →%%
dϕdx( x )= T (x )
(GI p )eff ( x )
(GI p )eff ( x )= r 2G ( x ,r )Ax
∫ dAx
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Special case: homogeneous shaft (shear modulus G); constant axial torque; constant cross section with polar moment of inertia
KT =TΦ=QΦ
: torsional stiffness of the bar
τ ( r )= r TIp
: shear stress (at radius r)
γ( r )= r ΦL
: shear strain (at radius r)
G =τγ
: Shear Modulus of the material
KT = G IpL
material ,geometry
Φ =LGIp1/KT
Q
Q =G IpL
KT
Φ
Equilibrium (moments x)
ΣMx=0 à T =Q (constant along shaft)
T
Φ
1 KT
γ
1 G
τ
structure
material
∫∫=A
p dArI 2
I p = r 2 dAA∫∫ =
π R 4
2
2πR 3t
π RO4
2−π RI
4
2
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