1. Shear stresses and shear center in multiple closed contour
Lecture #2
Shear stresses and shear center
in multiple closed contour

2. 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. 2

3. 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 :
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. 3

4. MQ – moment from resultant shear force;
EQUILIBRIUM EQUATION
An equilibrium equation includes i unknown flows in cells q0i :
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. 4

5. DETERMINATION OF RELATIVE TWIST ANGLES
We use the formula derived at last lecture:
Substituting the sum for shear flow, we get or 5

6. THE SYSTEM OF EQUATIONS
The system of equations includes one equilibrium equation and j = i relative twist angles equations: 6

7. EQUIVALENT DISCRETE CROSS SECTION
EXAMPLE – GIVEN DATA
EQUIVALENT DISCRETE CROSS SECTION 7

8. EXAMPLE – DISCRETE
APPROACH
q11 = 6.59·10-6 °/N;
q12 = ·10-7 °/N;
q21 = ·10-7 °/N; q22 = 5.84·10-6 °/N;
Q1F = °/m;
Q2F = °/m. 8

9. Q = - 0.473 °/m which we calculated for similar single-
EXAMPLE – DISTRIBUTED APPROACH
For equilibrium equation, we have:
- moment from shear flows qf : Mqf = kN·m ;
- moment from resultant shear force Qy :
MQ = -15 kN·m .
Solving the system of equations, we get
- shear flows in contours q01 = kN/m and
q02 = kN/m ;
- relative twist angle Q = °/m (compare to
Q = °/m which we calculated for similar single-
closed contour) 9

10. 10

11. EXAMPLE – FINAL DIAGRAM FOR DISCRETE APPROACH11

12. EXAMPLE – CONTINUOUS APPROACH12

13. EXAMPLE – CONTINUOUS AND DISCRETE APPROACHES BEING COMPARED13

14. EXAMPLE – SINGLE-CLOSED AND DOUBLE-CLOSED CONTOURS BEING COMPARED14

15. SHEAR CENTER CALCULATION
Firstly, we solve the following system with arbitrary chosen torsional moment MT :
Next, we find torsion rigidity G·Ir = MT / QT and 15

16. EXAMPLE – SHEAR CENTER CALCULATION16

17. Torsion of opened cross sections
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.)” 17