Prismatic Bar Subjected to End Loadings
y
Semi-Inverse Method
Assume :  x   y   xy  0
S
x
Equilibriu m Equations
  xz

z
ℓ
R
P
Compatibil
 z
2
x
z
2
  yz
z
 z
y
0
ity Equations
2


2
 z
2

z
2

 z
2

xy
0
Integratin g 
M
 z  C 1 x  C 2 y  C 3 z  C 4 xz  C 5 yz  C 6
Extension of Cylinders
y
S
x
ℓ
R
z
Pz
Assumptions
- Load Pz is applied at centroid of crosssection so no bending effects
- Using Saint-Venant Principle, exact end
tractions are replaced by statically
equivalent uniform loading
- Thus assume stress z is uniform over any
cross-section throughout the solid
z 
Pz
,  xz   yz  0
A
and  x   y   xy  0
Using stress results into Hooke’s law and combining with the straindisplacement relations gives
u
 Pz
v
 Pz
w
Pz
Integrating and dropping

,

,

x
AE
y
AE
z
AE
rigid-body motion terms
such that displacements
u v
v w
w u

 0,

 0,

 0 vanish at origin
y x
z y
x
z
u 
 Pz
x
AE
v
 Pz
AE
w
Pz
AE
z
y
Torsion of Cylinders
y
S
x
ℓ
R
z
T
Guided by Observations from Mechanics of Materials
• projection of each section on x,y-plane rotates as
rigid-body about central axis
• amount of projected section rotation is linear
function of axial coordinate
• plane cross-sections will not remain plane after
deformation thus leading to a warping displacement
Torsional Deformations
y
u   r  sin     y
v  r  cos    x
P'
P
S
r

  z

O
x
R
 = angle of twist per unit length
u    yz
v   xz
w  w( x, y )
w = warping displacement
Now must show assumed displacement form
will satisfy all elasticity field equations!
Stress Function Formulation
e x  e y  e z  e xy  0
u    yz
v   xz
e xz 
1  w

 y 

2  x

 w

 xz   
 y 
 x

e yz 

1  w

  x 
2  y

 w

 yz   
  x 
 y

w  w( x, y )
Equilibrium Equations
  xz
x

  yz
y
 x   y   z   xy  0
Compatibility Relation
  xz
 0
y

Introduce Prandtl Stress Function  = (x,y) :
  yz
x
  2 
 xz 

y
,  yz  

x
Equilibrium will be identically satisfied and compatibility relation gives
 
2
 
2
x
2
 
2

y
2
  2 
a Poisson equation that is amenable to several analytical solution techniques
Boundary Conditions
Stress Function Formulation
On Lateral Side: S
y
 
 n  n
 

 n 0  00

 n 0 

T x   x n x   yx n y   zx n z  0  0  0
n
T
S
x
n
y
xy
n

 x ds
ℓ
y
y
T z   xz n x   yz n y
  dx
n
x
  dy
zy
z
z
z
0 
 y ds
d
 0    constant
0
ds
R
On End: R (z = constant)
Unit Normal
z
T
nx 
dy

ds
ny  
dx
ds
dx
Px 

R
Py 

R
Pz 

dn

dy
dn
M
M
M
x

y

z

T x dxdy  0
n
T y dxdy  0
n
T z dxdy  0
n
R



yT z dxdy  0
n
R
xT z dxdy  0
n
R
( xT y  yT x ) dxdy  T  T  2   dxdy
n
R
n
R
Displacement Formulation
  xz
x

  yz
y
 w
2
 0
x
2
 w
2

y
2
0
Displacement component satisfies Laplace’s equation
On Lateral Side: S
T z   xz n x   yz n y   z n z  0 
n
 w

dw
 w

 y   n x  
 x   n y  0 or
  ( yn x  xn y )


y
dn
 x



On End: R
M
z


( xT y  yT x ) dxdy  T 
n
R
n

w
w 
2
2
 dxdy
T      ( x  y )  x
 y
R
y
x 

T  J
 2
x w y w 
2
 dxdy . . . Torsional Rigidity
J     x  y 

R
 y  x 

Formulation Comparison
y
S
x
O
R
Stress Function Formulation
Displacement Formulation
 w
2
 
2
 
2
x
2
 
2

y
2
  2   R
0  S
Relatively Simple Governing Equation
Very Simple Boundary Condition
x
2
 w
2

y
2
0  R
 w

 w


 n y  0  S

y

n


x


 x 
 x

 y

Very Simple Governing Equation
Complicated Boundary Condition
Multiply Connected Cross-Sections
y
Boundary conditions of zero tractions on all lateral surfaces
apply to external boundary So and all internal boundaries
S1, . . . Stress function will be a constant and displacement
be specified as per (9.3.20) or (9.3.21) on each boundary Si,
i = 0, 1, . . .
So
C
S1
x
R
 w

 w

   i  S i or 
 y   n x  
 x   n y  0  S i
 x

 y

where i are constants. Value of i may be arbitrarily chosen
only on one boundary, commonly taken as zero on So .
Constant stress function values on each interior boundary are found by
requiring displacements w to be single-valued, expressed by

  ds
dw ( x , y )  0
S1
S1
 2  A1 where A1 is area enclosed by S1
Value of 1 on inner boundary S1 must therefore be chosen so that relation is satisfied. If crosssection has more than one hole, relation must be satisfied for each hole.
Boundary conditions on cylinder ends will be satisfied, and resultant torque condition will give
T  2   dxdy  2  1 A1
R
Membrane Analogy
Stress function equations are identical to those governing static deflection of an elastic membrane
under uniform pressure. This creates an analogy between the two problems, and enables particular
features from membrane problem to be used to aid solution of torsion problem. Generally used to
providing insight into qualitative features and to aid in developing approximate solutions.
z
Deflected Membrane
p
z
Ndy
Ndy
Ndx
dy
R
dx
Ndx
z
z
y
Ndy
pdxdy
x
x
2
dx
Ndy
Membrane Element
S
x
 z
2

x
x
Static Deflection of a Stretched Membrane
Membrane Equations

 z
2
Fz  0 
x
2
 z
2

y
z  0 on S
V 

zdxdy
R
2

p
N
Equilibrium of Membrane Element
Torsion Equations
 
2
x
2
 
Contour Line : z  constant
2

y
2
  2 
  0 on S
T 2

 dxdy
R
Equations are same with:  = z , p/N = 2 , T = 2V
t
n

z
s
 0   zn  0
    zt  
dz
dn
Torsion Solutions Derived from
Boundary Equation
y
f ( x, y )  0
S
x
R
Boundary
 
2
x
2
- Value Problem
 
2

y
2
  2   R
0S
If boundary is expressed by relation f(x,y) = 0, this
suggests possible simple solution scheme of
expressing stress function as  = Kf(x,y) where K is
an arbitrary constant. Form satisfies boundary
condition on S, and for some simple geometric
shapes it will also satisfy the governing equation
with appropriate choice of K. Unfortunately this is
not a general solution method and works only for
special cross-sections of simple geometry.
Example 9.1 Elliptical Section
y
x
2
a
2

y
2
b
2
1
b
a
Look for Stress Function Solution
x
2
 x2

y
  K  2  2  1 
b
a

 satisfies boundary condition and will satisfy governing governing if K  
a b 
2
2
a b
2
2
Since governing equation and boundary condition are satisfied, we have found solution
Elliptical Section Results
(Displacement Contours)
(Stress Function Contours)
Stress Field
2 a 
Displacement Field
2
 xz  
a b
2
2 b 
2
y  
2
 yz 

a b
2
2
x 
 xz   yz 
2
2
2 Ty
 ab
2
w
3
2
a b 
3
3
xy
2 Tx
 ba
3
x
2
a
4
2T
 ab

y
2
b
4
Loading Carrying Capacity
Angle of Twist
2 a b   1
T  2
 2
2
a b a
2
 max   ( 0 ,  b ) 
T (b  a )
2T
 ab
2
2

2
3
a b
2
1
b
R
 a b 
3
T 
x dxdy 
2
2

y dxdy 
2
R
T (a  b )
2
or

2
a b 
3
3


dxdy 
R

Elliptical Section Results
3-D Warping Displacement Contours
T
Example 9.2 Equilateral Triangular Section
y
2a
a
x
For stress function try product form of each boundary line equation
  K (x 
3 y  2 a )( x 
3 y  2 a )( x  a )
 satisfies boundary condition and will satisfy governing governing if
K  

6a
Since governing equation and boundary condition are satisfied, we have found solution
Equilateral Triangular Section Results
(Stress Function Contours)
(Displacement Contours)
Stress Field
 xz 
 yz 

Displacement Field
w 
(x  a) y
2
2
Loading Carrying Capacity
Angle of Twist
( x  2 ax  y )
2
2
2a
 max   yz ( a , 0 ) 
y (3 x  y )
6a
a


3
2
 a 
5 3T
18 a
3
T 
27
5 3
 a
4

3
5
 I p
Additional Examples That Allow Simple
Solution Using Boundary Equation Scheme
y 
a
2
 cx
2
y
y
x 
a
a  cy
2
r = 2acos
.
2
r
r=b
a

x
x
x   a  cy
2
2
y  a
2
 cx
2
Section with Higher Order
Polynomial Boundary (Example 9-4)
  K ( a  x  cy )( a  cx
2
K 
2
2
4 a (1 
2
 y )
2


( b  r )(1 
2
2
2 a cos 
2

2
2
Circular Shaft with Circular
Keyway (Exercise 9-19)
, c 3
8
2)
 max   (  a , 0 )   ( 0 ,  a ) 
2  a
As b / a  0
)
r
(  max ) keyway
(  max ) solid shaft
 Stress Concentrat

2  a
 a
ion of 2
2
Examples That Do Not Allow Simple
Solution Using Boundary Equation Scheme
y
y
y = m1x
x=a
x
b
a
x
y = -m2x
General Triangular Section
Rectangular Section
Example 9.4 Rectangular Section
Fourier Method Solution
y
Previous boundary equation scheme will not create a
stress function that satisfies the governing equation.
Thus we must use a more fundamental solution
technique - Fourier method. Thus look for stress
function solution of the standard form
b
a
x
  h   p
with
 p ( x , y )   ( a  x )
2
2
homogeneous solution must then satisfy
  h  0 ,  h (  a , y )  0 ,  h ( x ,  b )    ( a  x )
2
2
 
2
 h ( x , y )  X ( x )Y ( y )
Separation of Variables Method

 h ( x, y ) 

B n cos
nx
n 1
   ( a  x ) 
2
2
ny
cosh
2a
2a
32  a

3
2


n  1 , 3 , 5
x
B n   32  a (  1)
2
(  1)
3
( n 1 ) / 2
n cosh
nb
2a
cos
2

( n 1 ) / 2
nx
2a
2
 
2
cosh
y
2
  2 
nb 
 3 3
/  n  cosh

2a 

ny
2a
Rectangular Section Results
Stress Field
 xz 

y
 

16  a

(  1)

2
n  1 , 3 , 5
( n 1 ) / 2
nb
2
n cosh
nx
cos
sinh
ny
2a
2a
2a
 yz  

x

16  a
 2  x 

(  1)

2
n  1 , 3 , 5
 max   yz ( a , 0 )  2  a 
( n 1 ) / 2
nb
2
n cosh

16  a

2
sin
nx
cosh
2a
2a
2a
1

n  1 , 3 , 5
2
n cosh
nb
2a
Loading Carrying Capacity/Angle of Twist
16  a b
3
T 

1024  a

3

4

5
n  1 , 3 , 5
1
n
5
tanh
nb
2a
Displacement Field
w   xy 
32  a

3
2


n  1 , 3 , 5
(  1)
3
( n 1 ) / 2
n cosh
nb
2a
sin
nx
2a
sinh
ny
ny
2a
Rectangular Section Results
(Stress Function Contours)
(Displacement Contours, a/b = 0.9)
(Displacement Contours, a/b = 1.0)
(Displacement Contours, a/b = 0.5)
Torsion of Thin Rectangular Sections (a<<b)
y
Investigate results for special case of a very thin
rectangle with a << b. Under conditions of b/a >> 1
b
cosh
nb
  and tanh
nb
2a
a
1
2a
   ( a  x )
2
x
2
 max  2  a
T 
16
 a b
3
3
Composite Sections
y
Torsion of sections composed of thin
rectangles. Neglecting local regions where
rectangles are joined, we can use thin
rectangular solution over each section.
Stress function contours shown justify these
assumptions. Thus load carrying torque for
such composite section will be given by
3
1
x
2
T 
16
3
(Composite Section)
(Stress Function Contours)
N
  a i b i
3
i 1
Example 9.5 Hollow Elliptical Section
x
2
( ka )
2

y
2
( kb )
2
1
y
x
2
a
2

y
2
b
2
For this case lines of constant shear stress
coincide with both inner and outer boundaries,
and so no stress will act on these lateral
surfaces. Therefore, hollow section solution is
found by simply removing inner core from solid
solution. This gives same stress function and
stress distribution in remaining material.
1
x
2 2
2
2

a b   x
y


 2


1
2
2
2

a  b  a
b

a b 
2
Constant value of stress function on inner boundary is
i 
2
a b
2
2
k
2
1
Load carrying capacity is determined by subtracting load carried by the
removed inner cylinder from the torque relation for solid section
 a b 
3
T 
3
a b
2
2
 ( ka ) ( kb ) 
3

3
( ka )  ( kb )
2
2


a b
2
a b (1  k )
3
2
Maximum stress still occurs at x = 0 and y = b
3
 max 
4
2T
 ab
1
2
1 k
4

Hollow Thin-Walled Tube Sections
C Membrane
Tube Centerline
a
a
A
o
B
A
B
(Section aa)
t
With t<<1 implies little variation in membrane slope, and BC can be approximated
by a straight line. Since membrane slope equals resultant shear stress

Load carrying relation: T  2   dxdy  2  o Ai  2  A

R

o
t
o 
  2  o Ai  2  o A c
2 
where A = section area, Ai = area enclosed by inner boundary, Ac = area enclosed by centerline

Combining relations
Angle of twist:

 ds  2  A c 
Sc
 
T
2 Ac t
TS c
4 Ac  t
2
where Sc = length of tube centerline
Cut Thin-Walled Tube Sections
Cut
Cut creates an open tube and produces significant changes to stress function,
stress field and load carrying capacity. Open tube solution can be
approximately determined using results from thin rectangular solution.
Stresses for open and closed tubes can be compared and for identical applied
torques, the following relation can be established (see Exercise 9-21)
3 T
τ OpenTube
τ ClosedTube

2 aA s
T
6
Ac
As
, but since A c  A s 
τ OpenTube
τ ClosedTube
 1  τ OpenTube  τ ClosedTube
2 Ac t
 Stresses are higher
in open tube
and thus closed tube is stronger
Torsion of Circular Shafts
of Variable Diameter
x
.

r
z
y
e r  e   e z  e rz  0
e r 
 r      z   rz  0
u 
1  u 
1 u 
   , e z 

2  r
r 
2 z
Equilibrium Equations

Stress Function Approach
Boundary Condition
Displacement Assumption
ur = uz = 0
u = u (r,z)
u 
u 
 u 
 r   
   ,  z  
r 
z
 r
  3   u  
  3   u  
r




  0


r
r  r  r  z  z  r 
  u 
r
 r
 r

 
z
r  r 

 

r
2
3
r
3   u 
 r
 z


r
z  r 

    dr
  dz 
d


0

 0    constant


2
 z ds 
ds
r   r ds
2
2
2

3 
r r
 
2

z
2
0
Load Carrying Torque
T  2  [  ( R ( z ), z )   ( 0 , z )]
Conical Shaft Example 9-7
z
r z
2
 cos   constant
on boundary
2
r
z
2
Stress Function Solution

  C


z
r  z
2

2
1
z
3
3 (r  z )
2
2
3/2




T
C  
2  (
2
 cos  
3
 r  
 z  
2
(r  z )
2
2
5/2
C  rz
(r  z )
2
cos  )
3
3
Displacement
Stresses
C r
1
2
5/2
u  
Cr
3( r  z )
2
2
3/2
 r
r is rigid-body rotation about z-axis and
 can be determined by specifying shaft
rotation at specific z-location
Conical Shaft Example 9-7
Exercise 9-23  = 30o
Comparison with Mechanics of Materials
Max Shear Stress Comparison
0.06
Mechanics of Materials
Elasticity Theory
0.05
( )
z max
/T
0.04
0.03
0.02
0.01
0
4
5
6
7
z
8
9
10
Numerical FEA Torsion Solutions
A
(4224 Elements, 2193 Nodes)

(Stress Function Contours)
(4928 Elements, 2561 Nodes)
(Stress Function Contours)
(4624 Elements, 2430 Nodes)
(Stress Function Contours)
Flexure of Cylinders
y
S
Py
R
z
(xo,yo)
.
Px
Consider flexure of cantilever beam of
arbitrary section with fixed end at z = 0
and transverse end loadings Px and Py at
x
z = ℓ. Problem is solved in Saint-Venant
sense, so only resultant end loadings Px
and Py will be used to formulate boundary
ℓ
conditions at z = ℓ.
From general formulation  x   y   xy  0 , and motivated
from strength of materials choose  z  ( Bx  Cy )( l  z ),
where B and C are constants. Stresses xz and yz will be
determined to satisfy equilibrium and compatibility
relations and all boundary conditions.
Remaining equilibrium equation
  xz
x

  yz
y
 ( Bx  Cy )  0
satisfied if we introduce stress function F(x,y) such that
will be identically
 xz 
F
y
 yz  

F
x
1
Bx
2
2

1
2
Cy
2
Flexure Formulation
Remaining Beltrami-Michell Compatibility Relations

2
y

( F ) 

B
1 
( F ) 
2
x
 0
C
1 
 F 

2
 0
( Cx  By )  2 
1 
Zero Loading Boundary Condition on Lateral Surface S
 xz n x   yz n y  0
dF

ds
1
dy
2
( Bx
2
 Cy
ds
2
dx
)
ds
Separate Stress Function F into Torsional Part  and Flexural Part 
F ( x, y )  ( x, y )   ( x, y )
    2  in R
  
d
d
2
ds
 0 on S

2
ds
1 
 
1
2
( Cx  By ) in R
( Bx
2
dy
ds
 Cy
2
dx
ds
) on S
Flexure Formulation
General solution to  2  
 ( x, y )  f ( x, y ) 
1

6 1 
( Cx

1 
3
( Cx  By )
 By )
3
where 2f = 0
Boundary Conditions on end z = ℓ

R

R
 xz dxdy  Px
BI
y
 CI
 yz dxdy  Py
BI
xy
 CI x   P y
xy
  Px
B  
C  
Px I x  P y I xy
I x I y  I xy
2
P y I y  Px I xy
I x I y  I xy
2
where x, y and  xy are the area moments of inertia of section R

J 
R
[ x  yz  y  xz ] dxdy  x o Py  y o Px
1

 
2
2

(
Cxy

Bx
y
)

(
x

y
)  dxdy  x o Py  y o Px
 R  2
x
y 

where J is the torsional rigidity – final relation determines angle of twist 
Flexure Example - Circular Section with No Twist
a
z
P
x
ℓ
y
Polar Coordinate Formulation
   

2
1  Ix
Solution:  
 xz  
Stress Solution:
 yz 
P
P 1  2
4I x 1  
I x 8 (1   )
P
Ix
a 

1 P
2
a sin
3
 on r  a
2 Ix
P  3  2 2
1  2
1  2
2
3

a
x

xy

x


I x  8 (1   )
8 (1   )
24 (1   )

P 3  2
z  
1 
r cos 
xy
[a  y 
y (l  z )
2
2
1  2
3  2
2
x ]
 max   yz ( 0 , 0 ) 
3  2
P
 a 2 (1   )
Strength of Materials:  max 
2
4
P
3 a
2