S. Socrate 2013 K. Qian

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 )

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

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

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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