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4.9 Statically indeterminate systems

Statically indeterminate systems can be solved from conditions of equal or zero displacement of chosen points. If both parts of the structure have the same length, the fraction of the external load accepted by each part of the structure is proportional to its rigidity - the product of modulus of elasticity by cross-sectional area. Stress is proportional to the corresponding fraction of external load. At the same time the stress is inversely proportional to the cross-sectional area. This means that absolute values of the stresses in both parts are equal. For an angle of 60 degrees, length of cable 1 is twice as long as cable 2. Displacement d2 is twice as large as d1. The fraction of external force depends on rigidity, cable length and the displacement. Cables 1 and 3 accept only 20% of the external vertical load. The shaft parts have different torsional stiffness and a common part in the center. The common part is rotated at the same angle for both parts, the angle of twist. Absolute values of the angle of twist for the left and right parts are equal. The gap d1 can be deleted by elastic deformation of the cables. The equilibrium equation for moments relative to point O shows that forces in the cables are inversely proportional to the distance from point O. Both beams have equal deflection that is proportional to L3. Cutting the length in half increases the fraction of external force by 23=8. Shapes of the the moment diagram are similar, but for the second beam there are reaction moments in embedded ends. They shift the curve down and decrease the maximum moment. The second beam is more rigid. The bending moment is at a maximum in the support zones. The maximum value of the bending moment M=FL over the right support does not depend on the number of supports. To solve the system it is necessary to find the shown unknown inner forces and reactions from conditions of equal displacements. There are no transverse forces for symmetrical system. Number of equations n depends on chosen equivalent scheme. The smaller the number the better.

4.10 Three-dimensional structures

For three-dimensional structures there are six components of moments and forces in a point. If we cut the curved beam perpendicular to the beam axis the torque is equal to the product of Force by Arm : F*r1 and the bending moment is equal to F*r2. Internal forces are equal to the sum of corresponding projections of external forces. The figure shows examples of bending moment and torque diagrams for three-dimensional structures. The diagrams are built by the above-mentioned rules. There are tensile as well as compressive stresses at external surface of the arc. The bending moment changes its sign. For a thin-walled sphere, absolute value of radial stress at the external surface is equal to external pressure. It is smaller than meridian and tangential stress components. The first step of the analysis includes determination of reactions in the supports. There is symmetry in the structure. There are no torque or bending moments in part of the structure. The conclusion is purely theoretical, the structure is not stable. Loss of stability causes additional forces and moments for all parts of the structure.

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