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

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;

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

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  1. 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 sign̅qi: 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-1times because one unknown could be evaluated from equilibrium equation. 3

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

  8. EXAMPLE – DISCRETE APPROACH q11= 6.59·10-6 °/N; q12= - 9.65·10-7 °/N; q21= - 7.07·10-7 °/N; q22= 5.84·10-6 °/N; Q1F= - 0.87 °/m; Q2F= - 3.13 °/m. 8

  9. 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 Q= - 0.556 °/m (compare to • Q= - 0.473 °/m which we calculated for similar single- • closed contour) 9

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  11. EXAMPLE – FINAL DIAGRAM FOR DISCRETE APPROACH 11

  12. EXAMPLE – CONTINUOUS APPROACH 12

  13. EXAMPLE – CONTINUOUS AND DISCRETE APPROACHES BEING COMPARED 13

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

  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 CALCULATION shear center shear center 16

  17. 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

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