"Both tension and bending testing applies tensile stresses to the sample which motivates yielding and/or fracture.  Accordingly, both types of tests can be, and are routinely, used to evaluate the strength and/or fracture strengths of materials.  The difference, of course, is that the nature of the stress and strain states in tension and bending are not the same. With a tensile test, the maximum tensile stresses are experienced throughout the entire volume (and surface area) of the test piece; in bending (where the sample sees tensile stress above the neutral axis and compressive stresses below), the maximum tensile stresses are conversely concentrated in a small region on the top surface above the neutral axis.  Accordingly for similar sized test pieces, the tensile sample sees the maximum stresses throughout its entire gauge length, i.e., over a much larger volume than the corresponding bend sample,   In brittle materials which are highly sensitive to the defect population, this change in statistical sampling volume means that strength and fracture properties measured in tensile tests are likely to be somewhat lower than the corresponding properties measured in bending, because there is a higher statistical probability of finding a larger defect. "(answered by Robert O Ritchie)

Hello Mohsin Abdullah Al-Shammari

The tensile modulus (Young's modulus) ,or elastic modulus, is a measure of  the stiffness of an elastic material and is a quantity used to characterize materials but whereas the flexural modulus or bending modulus is the ratio of stress to strain in flexural deformation, or the tendency for a material to bend.

From both the tests one can get modulus of elasticity, Infact, the Tensile and Flexural modulus would be the same since these tells us the materials ability to resist deformation under loads but with different load resisting phenomenon. However the results are usually different because measurements are not made in an ideal state.

Best wishes

PVR

https://www.sciencedirect.com/search?qs=differences%20between%20the%20values%20of%20the%20mechanical%20properties%20if%20they%20are%20estimated%20using%20three%20point%20bending%20test%20or%20tensile%20test&show=25&sortBy=relevance

Mechanical properties tested in different loading modes, will result in different values- these properties as name suggest should be unique for the same mode of deformation or loading ( as they are 'ratio' of one over other). Meaning, a flexural modulus is distinct and different from youngs modulus, both of them can be used to characterize the material- the point is what sort of application you are after!!

Hi everyone,

I agree with Vikram Shenoy and P Venkateshwar Reddy

I would like to add this:

It should be noted that both tests can be used to describe the mechanical behavior of a material, but however; the choice between the two tests depends mainly upon the type of the material being tested.

If you are working with brittle materials (such as ceramics), then you will need to test them in a flexural bending configuration, since pure tension would likely break the specimen very easily. In tension the entire cross section of the specimen will be subjected to the same maximum stress at failure, but in flexural the volume subjected to the maximum stress is small and the probability of a large flaw laying in it is small as well. In simple tension, all flaws see the same stress, and that's why the flexural properties obtained for the same material are higher than their tensile properties.

If you are working with metal-based materials or polymers (i.e. ductile materials) then there shouldn't be a much difference between the two tests, it will be your choice.

Unfortunately, there is no general relationship or a direct equation correlating the properties obtained from both tests. However, an empirical equation has been used for brittle material (i.e. ceramics), in which the Flex_strength = 1.3 x tensile_strength. This equation can be proven by calculating the effective volume subjected to stress in each test method and then correlating the strength values using Weibull statistics.

I hope this helps.

- Nadim S. Hmeidat (NSH)

I agree with Mr. Nadim S. Hmeidat

I recommended P Venkateshwar Reddy answer

Thanks for your cooperation

I agree with Mr. Vikram Shenoy

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