Post on 09-Jan-2016
description
1/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
CONTINUUM MECHANICS(TORSION
as BOUNDARY VALUE PROBLEM - BVP)
2/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
Body shape: straight prismatic bar with end surfaces perpendicular to the bar axis, cross-section of arbitrary shape.
Loading: distributed loading over end surfaces yielding torque as only cross-sectional force, side surface free of loading, no volume forces.
Kinematics boundary conditions: Bar fixed at one end (all displacements and their derivatives vanish there).
MS
Problem formulation
In a further analysis we shall adopt the assumption of replacing kinematics conditions by statics ones (reaction torque); the bar is considered as being in the equilibrium but free to be twisted (free torsion).
M
We will make also use of de Sain-Venant principle replacing distributed loading with a torque
MS=M
3/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
x3
x1
x2
cossinsincoscoscoscos1 rrru
x1
x2
A
A’
r
r’
Assume: r’ = r sincossincossinsinsin2 rrru
Assume: 3x
0,, 21
sin1cos
21 sin xru
12 cos xru
Twist angle per unit length (unit angle)
1,0,0
321 xxu 312 xxu
213 , xxu ?
AA’
1
1
Total twist angle
Distortion function
u
4/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
i
j
j
iij x
u
x
u
2
1321 xxu 312 xxu
213 , xxu
2
12
3
3
223 2
1
2
1
xx
x
u
x
u
02
222
x
u 03
333
x
u01
111
x
u
02
1
2
133
1
2
2
112
xxx
u
x
u
1
21
3
3
113 2
1
2
1
xx
x
u
x
u
0
00
00
T
ijkkijij G 20kk
0
00
00
2GT
132123 ' xG
311213 ' xG
- distortion function
5/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
0
j
iji x
P
311213 ' xG
0
0
0
3
33
2
32
1
31
3
23
2
22
1
21
3
13
2
12
1
11
xxx
xxx
xxx
00 0
0 0 0
022
2
21
2
xx
0
0''0''0 21 GG
322123 ' xG
The governing equation of torsion boundary value
problem
02 or:
Laplacian
6/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
Statics boundary conditions
On a bar surface 0,0,0q 0,, 21 jijiq
0'' 221112 xGxG
On bar ends: 1,0,0 0,, 21 qqq
121 ' xGq 212 ' xGq
x1
x2
qν1
qν2
S
A
MdAxqxq 2112
S
A
MdAxxxxG 212121 '' +
MS
S
S
GJ
M
JS Torsion inertia moment
This is boundary value condition for distortion function differential equation
By de Saint Venant hypothesis
7/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
21
,x
f
x
fn
022
2
21
2
xx
2R
x1
x2
n 0, 22
22121 Rxxxxf
21
21 44 xxn
21 2,2 xxn
RxRxn
n21 ,
02
21
1
12
R
x
xx
R
x
xx
02
2
1
1
R
x
xR
x
x
0Governing equation and boundary condition are
homogeneous
Solid circular shaft
Contour equation:
No distortion!
Rxxn 22 21
21
022
111
2
x
xx
x
8/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
2
123 2 xx
GJ
M
S
S
GTT 2
dAxxxxJA
S 212121 ''
S
S
GJ
M
2
123 2 xx
1
213 2 xx
1
213 2 xx
GJ
M
S
S
2
123 xx
J
M
S
S
1
213 xx
J
M
S
S
321 xxu
312 xxu
213 , xxu
022
2
21
2
xx
02
211
12
x
xx
x 00,0,0
9/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
dAxxxxJA
S 212121 ''
S
S
GJ
M
2
123 xx
J
M
S
S
1
213 xx
J
M
S
S
0 0
222
21 JdArdAxxJ
AA
S 2
013 x
J
M S
10
23 xJ
M S
rJ
Mxx
J
M SS
0
21
21
0
223
213
r
00max W
MR
J
M SS 3
03 x
GJ
Mx S
Twist angle
Unit twist angle
Total twist angle for a shaft of length llGJ
M S
0max
)(r
Solid circular shaft
1x
2x
Torsion section modulus
Polar inertia moment
10/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
S
S
GJ
M0
S
S
W
Mmax
ss JJ ss WW
For hbb
h/2 max
hbJ s3 hbWs
2
h/b 1 1,5 2 2,5 3 4 6 8
0,208 0,231 0,246 0,258 0,267 0,282 0.299 0,307 0,333
0,141 0,196 0,229 0,249 0,263 0,281 0.299 0,307 0,333
Rectangular bar
),( bh h/2
11/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
Bars of open cross-sectionsBars of closed cross-sections
The behaviour of the above types of bars differs significantly when subjected to the action of a torque. One can make a simple experiment cutting a tube:
Thin-walled bars
12/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
1
2
1
Assume: constant distribution of shear stress across of tube thickness
2
Assume: prismatic tube of varying wall thickness
From equilibrium condition:
const 2211
02211
minmax const
Closed thin-walled cross-sections
13/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
ds
dA
r(s)
s
dssrdA )(2
1
dAdssrMss
s 2)(
const
SdAMs
s 22
S – area of the figure embedded
within central curve s
S
M s
2
s
s
W
Mmax SW s min2
minmax 2
S
M s
Closed thin-walled cross-sections
14/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
b1
h1
b2
h2
b3
h3
n
isis MM
1
i
ii bh
Solutions for torsion of rectangular bars obey:
s
si
GJ
M
si
sii W
Mmax
iiisi hbJ 3 iiisi hbW 2
A2
iii
Sii hbG
M3 iiisi hbGM 3
A3
n
iiiis hbGM
1
3 ss
n
iiiis JGMhbGM
1
3
Assumptions:Cross-section partitioning:
A1
A2
A3
A1
Open thin-walled cross-sections
15/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
si
sii W
Mmax
iiisi hbW 2
iiisi hbGM 3
ss
n
iiiis JGMhbGM
1
3
iiis
ssi hb
GJ
MGM 3
ii
i
s
s
iii
iii
s
si b
J
M
hb
hb
J
M
2
3
max
is
si b
J
Mmaxmax max
i
i
i
s
si b
J
M
maxmax max
For hi/bi >6 i =i
Open thin-walled cross-sections
16/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
is
s bJ
Mmaxmax
maxh/2
h/2
max
Open thin-walled cross-sections
17/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
ds
dA
r(s)
s
S – area of the figure embedded
within central curve s
minmax 2
S
M s
Closed thin-walled cross-sections
18/17M.Chrzanowski: Strength of Materials
SM1-12: Continuum Mechanics: Torsion as BVP
stop