Modified Torsion Equation of S. Vlassov to represent the Warping and the Shear Slide for thin walls.

1. MODIFIED TWIST ROTATION
(Copy Right to Simon Bechara) +961 79 170009. e-mail: matrix.FAQ@gmail.com
1 SYMBOLS
T total applied twist moment, Kip-in. (N-mm)
tT twist moment resisted by the thin walls under rotation, Kip-in. (N-mm)
wT twist moment resisted by pure warping of the cross-section, Kip-in. (N-mm)
E modulus of elasticity, ksi (MPa)
G shear modulus of elasticity of steel, ksi (MPa)
J torsional constant for the thin walls of the cross section, in4
(mm4
)
wJ 
dd 2
, in4
(mm4
)
wC warping constant of the cross section, in6
(mm6
)
 final angle of rotation of the member around its own axis, (radian)
w angle of rotation due to pure warping of the member around its own axis, (radian)
s angle of rotation due to pure shear of the member around its own axis, (radian)
 reduced area, in2
(mm2
)
ws shear stress at point s due to warping, ksi (MPa)
wsS warping statical moment at point s, in4
(mm4
)
wsW normalized warping function at point s, in2
(mm2
)
d distance from the shear stress due to bi-shear to the rotation centre perpendicular to
the direction of the stress, in (mm)
v shear displacement, in (mm)
V shear force, lbs (N)

2. 2 VLASSOV ORIGINAL EQUATION
2.1 Review
Vlassov original differential equation wtw TTECGJT  
Where  GJTt and   ww ECT
2.1 Stress Diagrams
An applied twist moment, will generate lateral and longitudinal stress diagrams within the
section.
We will show the stresses for an I or H section as a direct application.

3. 2.2.1 Rotation Convention:

Positive Rotation (Figure 1)
2.2.2 Thin Walls Shear Diagram:
The shear stress diagram due to thin walls rotation (According to Timoshinko) :
  Gtt  GJTt (Figure 2)

4. 1.2.3 Longitudinal Stress Diagram due to Warping:
  nsws EW   ww ECT (Figure 3)
1.2.4 Shear Stress Diagram Due to Warping:
dV 1w
0w dt 
t
dVV
d
cs  
t
S
E ws
ws   ww ECT (Figure 4)
ds
b

5. 3 PRESENTATION OF THE PROBLEM
3.1 Total Twist Moment Distribution
The applied twist moment T to a thin walls section will be absorbed by two different rigidities
offered by the cross-section:
- The rigidity of the thin walls under rotation
According to Timoshinko, thin walls resist the torsion by pure shear slide rotation (Figure 1)
following the equation  GJTt
- The rigidity of the warping ability of the cross-section
The warping of a cross-section results in longitudinal stresses representing the diagram of a
bending moment at each level. The twist moment resisted by the warping effect, will be the
balance between the total applied twist moment and the resisted twist moment by the thin walls
shear tww TTECT  

 dVdECT ww  twist moment generated by the shear forces at each location multiplied
by the arm to the cs accordingly.
Direct conclusion, the shear forces (stresses) resulting in wT shall generate shear displacement
parallel to the shear forceV .
The shear stress diagram (Figure 4) results in a shear slide movement due to the pure shear
stresses that is not taken into consideration by the original equation of Vlassov.
The final twist rotation  , shall include the bi-shear twist rotation due to Shear Slide.

6. 4 MODIFICATION OF THE ORIGINAL EQUATION OF VLASSOV
If:
 final angle of rotation of the member around its own axis
w angle of rotation due to warping of the member around its own axis, neglecting the bi-shear
Effect.
s angle of rotation of the member around its own axis due to the bi-shear slide (Figure 4)
ws  
s w
We will start by the equilibrium equation: wt TTT 
The twist moment resisted by the thin wall will be wt GJT  since  is the final thin wall
rotation ws   and ws  
Assume the portion dtbd  subject to a shear force dV (Figure 4)
The area dtbd  will be subject to shear displacement following the equation:

7. )( dtbG
dV
Gd
dV
dx
dv




And the generated twist moment will be dVd 
)( dtbG
dVd
Gd
dVd
dx
dvd







Knowing that sdddv 
)(
2
dtbG
dVd
Gd
dVd
dx
dd s






 
)( 22
dtdbG
dVd
dGd
dVd
s






On the other hand, sconstant along the depth of the member.
Then  

dtdbGddGdVdT ssw
22

Let us define 
 ddJw
2
, in4
(mm4
)
wwsww ECGJT  
Where:
 reduced area of the of the cross-section under warping, in2
(mm2
)
d distance to the rotation centre perpendicular to the direction of the stress, in (mm)
cs rotation centre
The Vlassov Equation:
wwwwt ECGJTTT  
s and w are related by the equation:

8. The Solution of the differential equation, will result in w
On the other hand:
swwww GJECT  
w
ww
s
GJ
EC 



The final rotation will be:
ws  
4 CONCLUSION
After solving the equation of Vlassov, the final rotation will be:
ws   Simon Torsion Theory
Where:
w is Resulting from Solving Vlassov Equation. wwwwt ECGJTTT  
s is Resulting from Solving the Equation.
w
ww
s
GJ
EC 



Simon Bechara
+961 79 170009
e-mail: matrix.FAQ@gmail.com