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The ultimate tensile strength of a material is an intensive property; therefore its value does not depend on the size of the test specimen.However, depending on the material, it may be dependent on other factors, such as the preparation of the specimen, the presence or otherwise of surface defects, and the temperature of the test environment and material.
For example, the ultimate tensile strength (UTS) of AISI 1018 Steel is 440 MPa. In Imperial units, the unit of stress is given as lbf/in 2 or pounds-force per square inch. This unit is often abbreviated as psi. One thousand psi is abbreviated ksi. A factor of safety is a design criteria that an engineered component or structure must achieve.
For very small strains in isotropic materials – like glass, metal or polymer – flexural or bending modulus of elasticity is equivalent to the tensile modulus (Young's modulus) or compressive modulus of elasticity. However, in anisotropic materials, for example wood, these values may not be equivalent.
1940s flexural test machinery working on a sample of concrete Test fixture on universal testing machine for three-point flex test. The three-point bending flexural test provides values for the modulus of elasticity in bending, flexural stress, flexural strain and the flexural stress–strain response of the material.
The flexural strength is stress at failure in bending. It is equal to or slightly larger than the failure stress in tension. Flexural strength, also known as modulus of rupture, or bend strength, or transverse rupture strength is a material property, defined as the stress in a material just before it yields in a flexure test. [1]
Fig. 1: Critical stress vs slenderness ratio for steel, for E = 200 GPa, yield strength = 240 MPa. Euler's critical load or Euler's buckling load is the compressive load at which a slender column will suddenly bend or buckle. It is given by the formula: [1] = where
The relevant information is the area of the material being sheared, i.e. the area across which the shearing action takes place, and the shear strength of the material. A round bar of steel is used as an example. The shear strength is calculated from the tensile strength using a factor which relates the two strengths.
If HV is first expressed in N/mm 2 (MPa), or otherwise by converting from kgf/mm 2, then the tensile strength (in MPa) of the material can be approximated as σ u ≈ HV/ c, where c is a constant determined by yield strength, Poisson's ratio, work-hardening exponent and geometrical factors – usually ranging between 2 and 4. [9]