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Torsion of a square section bar Example of torsion mechanics. In the field of solid mechanics, torsion is the twisting of an object due to an applied torque [1] [2].Torsion could be defined as strain [3] [4] or angular deformation [5], and is measured by the angle a chosen section is rotated from its equilibrium position [6].
The formula to calculate average shear stress τ or force per unit area is: [1] =, where F is the force applied and A is the cross-sectional area.. The area involved corresponds to the material face parallel to the applied force vector, i.e., with surface normal vector perpendicular to the force.
Strength depends upon material properties. The strength of a material depends on its capacity to withstand axial stress, shear stress, bending, and torsion.The strength of a material is measured in force per unit area (newtons per square millimetre or N/mm², or the equivalent megapascals or MPa in the SI system and often pounds per square inch psi in the United States Customary Units system).
Depending on the type of material, size and geometry of the object, and the forces applied, various types of deformation may result. The image to the right shows the engineering stress vs. strain diagram for a typical ductile material such as steel.
The (infinitesimal) strain tensor (symbol ) is defined in the International System of Quantities (ISQ), more specifically in ISO 80000-4 (Mechanics), as a "tensor quantity representing the deformation of matter caused by stress. Strain tensor is symmetric and has three linear strain and three shear strain (Cartesian) components."
This is only the average stress, actual stress distribution is not uniform. In real world applications, this equation only gives an approximation and the maximum shear stress would be higher. Stress is not often equally distributed across a part so the shear strength would need to be higher to account for the estimate. [2]
This deformation is differentiated from a pure shear by virtue of the presence of a rigid rotation of the material. [2] [3] When rubber deforms under simple shear, its stress-strain behavior is approximately linear. [4] A rod under torsion is a practical example for a body under simple shear. [5]
The modulus of elasticity can be used to determine the stress–strain relationship in the linear-elastic portion of the stress–strain curve. The linear-elastic region is either below the yield point, or if a yield point is not easily identified on the stress–strain plot it is defined to be between 0 and 0.2% strain, and is defined as the ...