2.001 - MECHANICS AND MATERIALS I
Lecture #24
12/4/2006
Prof. Carol Livermore
Torsion (Twisting)
Mxx = Mt
Examples where torsion is important:
- Screwdriver
- Drills
- Propellers
Use superposition
Example: Unifrom along length, has circular symmetry
1
ϕ(x) = Angle of Twist (Total) at Point x (Built Up)
Compatibility:
d = ϕ(x + dx)R − ϕ(x)R
d = γdx
So:
γdx = ϕ(x + dx)R − ϕ(x)R
γdx = dϕR
dγ
γ=R
dx
What is γ?
So:
γ = γrθ Shear Strain
Thus:
γrθ = R
2
dϕ
dx
And:
γ
R dϕ
=
2
2 dx
θx =
rr = θθ = xx = rθ = rx = 0
Constitutive Relations:
σθx = 2Gθx = Gγxθ = Gr
dϕ
, where G is the shear modulus.
dx
E
2(1 + ν)
G=
σxx = σθθ = σrr = σrθ = σrx = 0
σθx = Gr
Recall beam bending σxx =
dϕ
dx
−Ey
ρ .
Equilibrium:
Mx = 0
rdF = 0
−Mt +
A
dF = σθx dA
rσθx dA = 0
−Mt +
A
Mt =
rGr
A
3
dϕ
dA
dx
Mt =
dϕ
dx
Gr2 dA
A
Recall beam bending:
M=
1
ρ
Ey 2 dA
A
Note:
dϕ
happens
dx : what
2
Gr
dA:
effective stiffness
A
Mt : what we apply
If G is constant ”Special Case”
Mt =
J≡
dϕ
G
dx
r2 dA
A
r2 dA Polar Moment of Inertia
A
So for ”special case” G is constant.
Mt = GJ
dϕ
dx
Recall beam bending:
M=
EI
ρ
If G is not constant:
(GJ)ef f =
Gr2 dA
A
When G is constant:
Mt = GJ
dϕ
dϕ
and generally σθx = Gr
dx
dx
Mt
σθx
=
GJ
Gr
So:
σθx =
4
Mt r
J
Example for J
1. Cicular solid shaft
r2 dA
J=
2π
R
=
0
A
2π
=
0
0
4
r2 rdrdθ
R
dθ
4
π 4
R
2
J=
2. Hollow circular shaft
J=
2π
0
R2
r3 drdθ
R1
π
4
4
J=
R2 − R1
2
(GJ)composite
=
=
(GJ)inside + (GJ)outside
G1 πR14
G1 π 4
R2 − R14
+
2
2
Or:
(GJ)ef f =
0
2π R1
0
3
R2
G1 r dr +
G2 r dr dθ
3
R1
Note: Due to powers of order 4, material buys you more stiffness on the outside
than the inside.
5
So why don’t we make R huge?
Now look at it at 45◦
It will fail in compression and it will buckle.
What is the total angle of twist?
For one material (constant G)
Mt = GJ
So:
6
dϕ
dx
dϕ
Mt
=
dx
GJ
ϕ(x) =
Mt
x + c1
GJ
Boundary Condition:
At x = 0, ϕ(0) = 0 ⇒ c1 = 0.
So:
ϕ(x) =
Mt
x
GJ
φ(L) =
Mt L
GJ
The total angle of twist φ(L)
7