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The plastic section modulus is used to calculate a cross-section's capacity to resist bending after yielding has occurred across the entire section. It is used for determining the plastic, or full moment, strength and is larger than the elastic section modulus, reflecting the section's strength beyond the elastic range.
In structural engineering, the plastic moment (M p) is a property of a structural section. It is defined as the moment at which the entire cross section has reached its yield stress . This is theoretically the maximum bending moment that the section can resist – when this point is reached a plastic hinge is formed and any load beyond this ...
A filled rectangular area as above but with respect to an axis collinear with the base = = [4] This is a result from the parallel axis theorem: A hollow rectangle with an inner rectangle whose width is b 1 and whose height is h 1
The root-3 rectangle is also called sixton, [6] and its short and longer sides are proportionally equivalent to the side and diameter of a hexagon. [7] Since 2 is the square root of 4, the root-4 rectangle has a proportion 1:2, which means that it is equivalent to two squares side-by-side. [7] The root-5 rectangle is related to the golden ratio ...
Quantities used in the definition of the section modulus of a beam. The maximum tensile stress at a cross-section is at the location z = c 1 {\displaystyle z=c_{1}} and the maximum compressive stress is at the location z = − c 2 {\displaystyle z=-c_{2}} where the height of the cross-section is h = c 1 + c 2 {\displaystyle h=c_{1}+c_{2}} .
In 1820, the French engineer A. Duleau derived analytically that the torsion constant of a beam is identical to the second moment of area normal to the section J zz, which has an exact analytic equation, by assuming that a plane section before twisting remains planar after twisting, and a diameter remains a straight line. Unfortunately, that ...
The radius is r=0.200 m = 200 mm, or a diameter of 400 mm. If one adds a factor of safety of 5 and re-calculates the radius with the admissible stress equal to the τ adm =τ yield /5 the result is a radius of 0.343 m, or a diameter of 690 mm, the approximate size of a turboset shaft in a nuclear power plant.
where is the volume fraction of the fibers in the composite (and is the volume fraction of the matrix).. If it is assumed that the composite material behaves as a linear-elastic material, i.e., abiding Hooke's law = for some elastic modulus of the composite and some strain of the composite , then equations 1 and 2 can be combined to give