Lecture #2
Shear stresses and shear center
in multiple 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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SHEAR FLOW IN MULTIPLE-CLOSED CONTOUR.
STATIC INDETERMINACY
The total shear flow is represented as a sum of
variable part qf for an opened contour and shear
flows in separate cells q0i taken with certain sign ̅qi :
q  q f   q i  q0 i
i
Here ̅qi = ±1 and is determined according to positive or
negative tangential coordinate direction.
If the contour has i closed contours (cells), the
problem has i unknown cell flows q0i but it is statically
indeterminate only i-1 times because one unknown
could be evaluated from equilibrium equation.
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EQUILIBRIUM EQUATION
An equilibrium equation includes i unknown flows in
cells q0i :
M q 0  M qf  M Q
q
0i
 W i  M qf  M Q
i
Here Mq0 , Mqf – moments from constant and variable
parts of shear flow, respectively;
MQ – moment from resultant shear force;
Wi – double area of separate i-th contour.
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DETERMINATION OF RELATIVE TWIST ANGLES
We use the formula derived at last lecture:
qq

dt ;
G 
t
q 1
W
.
Substituting the sum for shear flow, we get

qf  q j
 G   W
t
or
j
dt   
i
t
qi  q j
G   W j
q0 i
   iF    ij  q0 i
i
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THE SYSTEM OF EQUATIONS
The system of equations includes one equilibrium
equation and j = i relative twist angles equations:
  q0 i  W i  M qf  M Q ;
 i
 n
   ij  q0 i   iF   , j  1, n .
 i 1
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EXAMPLE – GIVEN DATA
EQUIVALENT DISCRETE CROSS SECTION
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q f  t  , kN m
EXAMPLE –
DISCRETE
APPROACH
11 = 6.59·10-6 °/N;
12 = - 9.65·10-7 °/N;
21 = - 7.07·10-7 °/N;
q1
22 = 5.84·10-6 °/N;
1F = - 0.87 °/m;
2F = - 3.13 °/m.
q2
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EXAMPLE – DISTRIBUTED APPROACH
For equilibrium equation, we have:
- moment from shear flows qf : Mqf = -47.3 kN·m ;
- moment from resultant shear force Qy :
MQ = -15 kN·m .
Solving the system of equations, we get
- shear flows in contours q01 = 114.2 kN/m and
q02 = 455.0 kN/m ;
- relative twist angle  = - 0.556 °/m (compare to
 = - 0.473 °/m which we calculated for similar singleclosed contour)
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q f  t  , kN m
q1  q01 , kN m
q 2  q02 , kN m
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EXAMPLE – FINAL DIAGRAM
FOR DISCRETE APPROACH
q , kN m
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EXAMPLE – CONTINUOUS APPROACH
q f  t  , kN m
q , kN m
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EXAMPLE – CONTINUOUS AND DISCRETE
APPROACHES BEING COMPARED
q , kN m
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EXAMPLE – SINGLE-CLOSED AND DOUBLE-CLOSED
CONTOURS BEING COMPARED
q , kN m
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SHEAR CENTER CALCULATION
Firstly, we solve the following system with arbitrary
chosen torsional moment MT :
  q0 i  W i  M T ;
 i
 n
   ij  q0 i  T , j  1, n .
 i 1
Next, we find torsion rigidity G·Ir = MT / T and
X SC
  G  Ir
 XQ 
Qy
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EXAMPLE – SHEAR CENTER CALCULATION
shear center
shear center
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TOPIC OF THE NEXT LECTURE
Torsion of opened cross sections
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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