Torsion Torsion of circular bar, Transmission of power by ...
Prandtl Torsion
of 14
-
Upload
eulerintegral -
Category
Documents
-
view
261 -
download
0
Transcript of Prandtl Torsion
-
8/12/2019 Prandtl Torsion
1/14
Prandtl Stress Function Summary
( )
( )
( )
,x x
y
z
u u y z
u z x
u y x
=
=
=
xxz
xxy
uG y
z
uG z
y
= +
=
satisfy equilibrium equation
by taking
0xy xz
y z
+ =
xy
xz
Gz
Gy
=
=
Prandtl stresss function ( l2 )
(1)
From (1)2 2 2
2 2
xu
G G G G Gy z y z
= = +
2 2
2 2 2y z
+ = Poissons equation
"compatibility"
-
8/12/2019 Prandtl Torsion
2/14
2 2
2 22
y z
+ =
in A
A
Cn
ts
0xn xy y xz z
dn n G
ds
= + = =
constant 0 = = on C
Torque-twist
effT G J=
2effJ dA=
( )nt on C eff on C
T
J n
=
( )maxmax
on C
eff on C
T
J n
=
bar cross section
T
-
8/12/2019 Prandtl Torsion
3/14
For the warping displacement
x
x
uy
z y
u zy z
=
= +
xu yz dzy
yz dyz
= +
= +
Thin rectangular cross section (neglect ends)
y
z
t
22
4
ty
=
/ 2
3
/ 2
12
3
y t
eff
y t
J bdy bt
=+
=
= =
max 2
/ 2
3eff y t
T TJ y bt
=
= =
xu yz=
-
8/12/2019 Prandtl Torsion
4/14
Membrane Analogy
p
w
s s
2 2
2 2
w w p
y z s + = 0w= on the boundary
2sw
p
=
22
4
ty
=
-
8/12/2019 Prandtl Torsion
5/14
For cross sections with holes we also need to satisfy
0 =
K =
0xhole
du =
2 holehole
ds G An
=
or 2 hole
hole
ds G A =
If one has multiple holes, this additional condition is applied at
each hole to solve for the multiple unknown constants
additional
unknown
supplementary condition to determine K
-
8/12/2019 Prandtl Torsion
6/14
Torsion of Thin, Closed Sections
K1 K2
a b
c
c c
a
a
b b
a a
ta
= 0
= K1
b b
tb
= K= K
1
2 = K
t c
2
c c
a= K1
ta, b=
K1 K2tb
, c = K2
tc
-
8/12/2019 Prandtl Torsion
7/14
K1
K2
q1 =
=q2
q1
q2
1 1 1 2 1 2 2 2, ,a a b b c cq t K q q q t K K q t K = = = = = = =
shear flows
q1
-q
2q - q
2 1
shear flows into or out of a junction are conserved
0outq =
-
8/12/2019 Prandtl Torsion
8/14
K1
K2
q1 =
=q2
q1
q2
1 2
i area enclosed by centerline of ith "cell'
Torque-shear flow 2 i iiT q= for each cell
1
2i ithcell
qds
G t=
max
max
qK
t
=
warping is generally small for closed sections
2 holehole
ds G A =
-
8/12/2019 Prandtl Torsion
9/14
Torsion of a Thin Closed Section
(multiple cells)
1q 2q
cell 1 cell 2
1
2
1 1 2 2
1
2
2 2
1
2
1
2
C
C
T q q
qds
G t
qds
G t
= +
=
=
1. If the torque T is known, then q1 and q2 are first found in terms
of the unknown ' from Eqs. (2) and (3). These qm 's are then
placed into Eq.(1) which is solved for the unknown ' . Once 'is known in this manner, the qm 's are completely determined.
2. If ' is known, Eqs.(2) and ( 3) can be solved directly for theqm 's and then Eq.(1) can be used to find the torque, T
(1)
(2)
(3)
( the q in Eqs.(2) and (3) is the total q flowing in
a given cross section, i.e it is q1 q2 flowingin the vertical section)
-
8/12/2019 Prandtl Torsion
10/14
P
r Tq
= area contained within thecenterline of the cross section
For a single cell, we can write these more explicitly
2
C
C
T qr ds
q r ds q
=
= =
2
Tq=
2
12
4
C
C
q dsG t
T ds
G t
=
=
effT GJ =
where
24eff
C
Jds
t
=
max
max min2 2
T T
t t
= =
(no stress concentrations)
-
8/12/2019 Prandtl Torsion
11/14
Torsion of a Thin Closed Section
(single cell)
P
r
2
Tq=
T
effT GJ =
where24
eff
C
Jds
t
=
q
= area contained within thecenterline of the cross section
1. If T is known, q follows directly from Eq. (1),
' is found from Eq.(2)
(1)
(2)
2. If ' is known, T follows from Eq.(2),and q is then found from Eq. (1)
-
8/12/2019 Prandtl Torsion
12/14
Torsion of a Thin Closed Section
(single cell)
The shear stress is not quite uniform across the thickness for thin
closed sections
t
yields uniform stress
The difference looks much like that for an open section
t
so as a small correction factor: ( )2
34 1
3eff
C
C
J t s dsds
t
= +
-
8/12/2019 Prandtl Torsion
13/14
Torsion of closed sections with fins
23 34 1 1
3 3c fT T T G t ds t dsds
t
= + = + +
In the closed sectionmax
min2
cT
t =
JeffclosedJefffor a fin (allows varaiable
thickness)
( )c eff closedT G J=
-
8/12/2019 Prandtl Torsion
14/14
In a fin
( )maxmax
f
eff fin
T t
J
=
( )f eff finT G J=
We can write this also in terms of the values since
total f others
total f others
J J J
T T T
= +
= +
f f
others others
total total
T G J
T G J
T G J
=
=
=
so
( ) ( )f total
eff eff fin total
T TG
J J= =
( )maxmax
total
eff total
T t
J
=
