Lecture #1
Torsion rigidity and shear center
of closed contour
SHEAR STRESSES RELATED QUESTIONS
- shear flows due to the shear force, with no torsion;
- shear center;
- torsion of closed contour;
- torsion of opened contour, restrained torsion and
deplanation;
- shear flows in the closed contour under combined action
of bending and torsion;
- twisting angles;
- shear flows in multiple-closed contours.
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CALCULATION OF TWIST ANGLE
FOR CLOSED CROSS SECTION
To find the twist angle for a portion of thin-walled beam
subjected to external loads we use Mohr’s formula under
virtual displacement method:
Df  M   t  g  dV
V
At the left side we have real twist angle Df (from external
loads) multiplied by ̅M (unity moment). The right side
represents the energy stored in the volume V and
calculated using unity shear stresses ̅t and real strains g.
The relation between unit moment ̅M and unit shear
flow ̅q could be found using Bredt’s formula:
M  q
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CALCULATION OF TWIST ANGLE
FOR CLOSED CROSS SECTION
From the formula from previous slide we can derive that
Dz q  q
Df 
dt

M t G 
Here Dz is the length of portion of thin-walled beam
having the twist angle Df ; t is tangential coordinate.
Taking ̅M=1, we get the relative twist angle
qq

dt ;
G 
t
q 1

The absolute angle of rotation relation f
is calculated by integration and f z  f
0
using the starting value f0 :
 
.
z
    z  dz
0
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EXAMPLE OF TWIST ANGLE CALCULATIONS – GIVEN DATA
EQUIVALENT DISCRETE CROSS SECTION
G  27GPa
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EXAMPLE – SHEAR FLOW DIAGRAMS AND TWIST ANGLE
CONTINUOUS APPROACH
 = - 0.473 °/m
q  t  , kN m
DISCRETE APPROACH
 = - 0.474 °/m
(0.1% error)
qi  Li
1
  
 i Gi   i
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EXAMPLE – TWIST ANGLES FOR THREE DIFFERENT
SHEAR FORCE POSITIONS
 = - 0.988 °/m
 = - 0.473 °/m
 = + 1.327 °/m
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TORSION RIGIDITY
In Mechanics of Materials, the formula for relative twist
angle was
MT

G  Ir
The denominator G ∙ Ir is referred as torsion rigidity.
Taking the formula for  for thin-walled beam and using
Bredt’s formula, we can derive
G  Ir  
2
dt
t G  
Example: Taking the hollow circular cross section we get
Ir = 2pR3 which is same with Mechanics of Materials.
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PROPERTIES OF SHEAR CENTER
1. The transverse force applied at shear center does not
lead to the torsion of thin-walled beam.
2. The shear center is a center of rotation for a section of
thin-walled beam subjected to pure torsion.
3. The shear center is a position of shear flows resultant
force, if the thin-walled beam is subjected to pure shear.
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CALCULATION OF SHEAR CENTER POSITION
FOR CLOSED CROSS SECTION
For closed cross section, we use equilibrium
equations for moments to find q0. Thus, moment of
shear flows MC (q) would depend on the position of
force Qy and would not be useful to find XSC.
We find the shear center position from the condition
that the twist angle should correspond to the
torsional moment which is calculated as Qy
multiplied by the distance to shear center:
df
MT  Q y   X Q  X SC  
 G  Ir
dz
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CALCULATION OF SHEAR CENTER POSITION
FOR CLOSED CROSS SECTION
Thus, we need to calculate the torsion rigidity of
wingbox G·Ir and relative twist angle  :
G  Ir  
2
qq

dt ;
G 
t
dt
t G  
q 1

.
Finally, we can get shear center position XSC by
formula
 G  I
X SC  X Q 
r
Qy
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CALCULATION OF SHEAR CENTER POSITION
FOR CLOSED CROSS SECTION - EXAMPLE
For the abovementioned example, we get:
- torsion rigidity G·Ir = 1.1·106 N·m2 ;
- relative twist angle  = - 0.473 °/m ;
- torsional moment MT = - 9.2 kN·m ;
- shear center coordinate XSC = 0.192 m .
shear center
q  t  , kN m
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WHERE TO FIND MORE INFORMATION?
Megson. An Introduction to Aircraft Structural Analysis. 2010
Chapter 17.1 for torsion of closed cross sections
Paragraph 16.3.2 for shear center in closed cross sections
… Internet is boundless …
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TOPIC OF THE NEXT LECTURE
Calculation of shear flows in multiple closed contours
All materials of our course are available
at department website k102.khai.edu
1. Go to the page “Библиотека”
2. Press “Structural Mechanics (lecturer Vakulenko S.V.)”
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