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Dear all, I'd like to know about F 3
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Flexural ultimate strength, Sfu, sometimes ambiguously (and confusingly) given the nebulous term rupture modulus or modulus of rupture, is the strength of a cross section in bending at the instant of failure, expressed as an equivalent extreme fiber tensile stress corresponding to an equivalent linear bending stress distribution, thereby allowing you to use the elastic (linear) flexure formula, sigma = Mc/I, when analyzing the ultimate bending stress on the cross section. In other words, Sfu allows you to compute bending stress on the cross section using the simplified formula sigma = Mc/I and compare sigma to Sfu.
You asked for formulas. The remainder of this post elaborates on the above, in case you are interested in more formulas regarding flexural ultimate strength. Let's define the following nomenclature. Even though not explicitly mentioned throughout this post, if your material(s) have a different stress-strain curve in compression than in tension, use the correct compressive stress from the stress-strain curve for fibers in compression.
Sty = tensile yield strength (stress).
Stu = tensile ultimate strength.
Sfy = flexural yield strength.
Sfu = flexural ultimate strength.
M = applied moment.
My = yield moment.
Mp = plastic moment.
Mu = ultimate moment.
I = moment of inertia.
c = distance from neutral axis to cross section extreme fiber.
y = distance from neutral axis to differential element.
S = elastic section modulus = I/c.
Z = plastic section modulus = (SF)S.
SF = plasticity shape factor = Mp/My = Z/S, a number between 1 and 2.5.
eps1 = strain at onset of yield.
eps2 = strain at end of yield plateau (if one exists).
eps3 = strain at ultimate strength (peak of stress-strain curve).
sigma = extreme fiber stress (or equivalent extreme fiber stress).
sigma(eps) = stress-strain curve ordinate at abscissa "eps."
As is often customary (for simplicity and conservatism), let us assume the proportional limit coincides with the yield point. Assume cross section is homogeneous and extreme fiber is not stressed to yield point. Then, sigma = Mc/I, and sigma(eps1) = Sfy = Sty = (My)c/I.
Now let's say you have a homogeneous cross section consisting of a material that has a remarkably long plateau at the yield point. (Mild steel is one of the few materials exhibiting this remarkable property.) Then let's say you strain the outer fiber to eps2. Most of the cross section would then be at stress Sty, so it is customary to say 100% of the cross section is at Sty, as this very closely (and conservatively) approximates the stress distribution. When 100% of the cross section is at Sty, the corresponding resisting moment is called the plastic moment Mp. Furthermore, because statically, monotonically-loaded, ductile materials in bending generally do not fail by rupture but instead by a very large, excessive deflection, it is customary (for ductile materials) to say Mu = Mp, as this very closely (and conservatively) approximates the ultimate bending stress distribution. (Notice I said statically and monotonically loaded. We're not talking about fatigue here. In a fatigue problem, ductile members can indeed end up rupturing.) So we thus have, in the case of this long yield plateau, sigma = Mc/I, and Sfu = (SF)Sty = (Mu)c/I = (Mp)c/I = (c/I)integral[(Sty)
(dA)]. Note that, because you are only approximating Mu here with Mp, Sfu here will often be below Stu.
Now let's say you have a homogeneous cross section consisting of a material that does not exhibit a yield plateau. Then, sigma = Mc/I, and Sfu = (Mu)c/I = (c/I)integral[sigma(eps3*y/c)(dA)]. Sfu will exceed Stu if the cross section is homogeneous. (By the way, it is perhaps optimistic to expect theoretical calculations to match published test values.)
To compute the second integral listed above (usually done numerically), you must have or create an analytic regression of the stress-strain curve for each material in your cross section. (Perform polynomial regression of each region before and after any stress-strain curve "kink" separately, then write your final function with arithmetic conditional statements. You will never fit it well with a single polynomial regression if the curve contains a "kink" or localized bend.) Alternately, you could get Sfu from some material testing folks or from a product specification sheet.
Also, if the material and cross section are symmetric about the axis of bending, you know neutral axis is at cross section centroid. If asymmetric, it is an iterative process to solve both of the integrals listed above, meaning you assume a neutral axis location, solve problem, and repeat the process until resisting moments above and below neutral axis are equal.
Now let's graduate to nonhomogeneous cross sections. The same procedure described above is employed, using each material's respective stress-strain curve during the solution. In general, Sfu will often exceed Stu (unless the cross section proves very weak in bending). Also, as mentioned earlier, as the cross section materials or lamina approach homogeneity in the direction of bending stresses, Sfu will typically exceed Stu, unless perhaps the cross section exhibits excessive elastic or plastic deflection instead of rupture.
The term "flexural strength," though ambiguous and therefore nebulous, means "flexural ultimate strength." Though this post is not limited to plastics, in the case of plastics, erroneously leads some to believe ASTM D 790 is a test for "flexural yield strength," but Sect. 1.1 of ASTM D 790 clearly states it is a test for flexural ultimate strength. Just because certain specimens exhibit excessive deflection without rupturing doesn't mean ASTM D 790 is a test for "flexural yield strength." (Flexural yield strength would be the equivalent extreme fiber stress when the specimen in bending has an extreme fiber strain at, e.g., the 0.2% offset strain, not 5% strain.)
I hope this helps answer some of your questions about flexural strength. If not, post a reply. You also asked about flexural modulus. Answering that question is better deferred as a separate reply if time permits. Good luck.
You asked for formulas. The remainder of this post elaborates on the above, in case you are interested in more formulas regarding flexural ultimate strength. Let's define the following nomenclature. Even though not explicitly mentioned throughout this post, if your material(s) have a different stress-strain curve in compression than in tension, use the correct compressive stress from the stress-strain curve for fibers in compression.
Sty = tensile yield strength (stress).
Stu = tensile ultimate strength.
Sfy = flexural yield strength.
Sfu = flexural ultimate strength.
M = applied moment.
My = yield moment.
Mp = plastic moment.
Mu = ultimate moment.
I = moment of inertia.
c = distance from neutral axis to cross section extreme fiber.
y = distance from neutral axis to differential element.
S = elastic section modulus = I/c.
Z = plastic section modulus = (SF)S.
SF = plasticity shape factor = Mp/My = Z/S, a number between 1 and 2.5.
eps1 = strain at onset of yield.
eps2 = strain at end of yield plateau (if one exists).
eps3 = strain at ultimate strength (peak of stress-strain curve).
sigma = extreme fiber stress (or equivalent extreme fiber stress).
sigma(eps) = stress-strain curve ordinate at abscissa "eps."
As is often customary (for simplicity and conservatism), let us assume the proportional limit coincides with the yield point. Assume cross section is homogeneous and extreme fiber is not stressed to yield point. Then, sigma = Mc/I, and sigma(eps1) = Sfy = Sty = (My)c/I.
Now let's say you have a homogeneous cross section consisting of a material that has a remarkably long plateau at the yield point. (Mild steel is one of the few materials exhibiting this remarkable property.) Then let's say you strain the outer fiber to eps2. Most of the cross section would then be at stress Sty, so it is customary to say 100% of the cross section is at Sty, as this very closely (and conservatively) approximates the stress distribution. When 100% of the cross section is at Sty, the corresponding resisting moment is called the plastic moment Mp. Furthermore, because statically, monotonically-loaded, ductile materials in bending generally do not fail by rupture but instead by a very large, excessive deflection, it is customary (for ductile materials) to say Mu = Mp, as this very closely (and conservatively) approximates the ultimate bending stress distribution. (Notice I said statically and monotonically loaded. We're not talking about fatigue here. In a fatigue problem, ductile members can indeed end up rupturing.) So we thus have, in the case of this long yield plateau, sigma = Mc/I, and Sfu = (SF)Sty = (Mu)c/I = (Mp)c/I = (c/I)integral[(Sty)
(dA)]. Note that, because you are only approximating Mu here with Mp, Sfu here will often be below Stu.Now let's say you have a homogeneous cross section consisting of a material that does not exhibit a yield plateau. Then, sigma = Mc/I, and Sfu = (Mu)c/I = (c/I)integral[sigma(eps3*y/c)
(dA)]. Sfu will exceed Stu if the cross section is homogeneous. (By the way, it is perhaps optimistic to expect theoretical calculations to match published test values.)To compute the second integral listed above (usually done numerically), you must have or create an analytic regression of the stress-strain curve for each material in your cross section. (Perform polynomial regression of each region before and after any stress-strain curve "kink" separately, then write your final function with arithmetic conditional statements. You will never fit it well with a single polynomial regression if the curve contains a "kink" or localized bend.) Alternately, you could get Sfu from some material testing folks or from a product specification sheet.
Also, if the material and cross section are symmetric about the axis of bending, you know neutral axis is at cross section centroid. If asymmetric, it is an iterative process to solve both of the integrals listed above, meaning you assume a neutral axis location, solve problem, and repeat the process until resisting moments above and below neutral axis are equal.
Now let's graduate to nonhomogeneous cross sections. The same procedure described above is employed, using each material's respective stress-strain curve during the solution. In general, Sfu will often exceed Stu (unless the cross section proves very weak in bending). Also, as mentioned earlier, as the cross section materials or lamina approach homogeneity in the direction of bending stresses, Sfu will typically exceed Stu, unless perhaps the cross section exhibits excessive elastic or plastic deflection instead of rupture.
The term "flexural strength," though ambiguous and therefore nebulous, means "flexural ultimate strength." Though this post is not limited to plastics, in the case of plastics, erroneously leads some to believe ASTM D 790 is a test for "flexural yield strength," but Sect. 1.1 of ASTM D 790 clearly states it is a test for flexural ultimate strength. Just because certain specimens exhibit excessive deflection without rupturing doesn't mean ASTM D 790 is a test for "flexural yield strength." (Flexural yield strength would be the equivalent extreme fiber stress when the specimen in bending has an extreme fiber strain at, e.g., the 0.2% offset strain, not 5% strain.)
I hope this helps answer some of your questions about flexural strength. If not, post a reply. You also asked about flexural modulus. Answering that question is better deferred as a separate reply if time permits. Good luck.
"Flexural modulus" is an alternate term for "flexural modulus of elasticity," in contrast to "tensile modulus of elasticity" and "shear modulus of elasticity."
Flexural modulus, Eb, is the slope of the stress-strain curve, within the elastic limit, obtained during a flexure (bending) test. I.e., flexural modulus is the ratio of [equivalent] extreme (outermost) fiber bending stress [(corresponding to an equivalent linear stress distribution)] to extreme fiber strain. The above-mentioned flexure test consists of a simply-supported strip of material (beam) with a concentrated transverse load applied at midspan.
As you know, "bending stress," above, means axial tensile stress on tension side, and axial compressive stress on compressive side, of a cross section in bending.
If test specimen cross section is homogeneous (i.e., consistent, one material), then strike out the words I have placed within brackets, above. For homogeneous cross sections, flexural modulus of elasticity, Eb, should theoretically equal tensile modulus of elasticity, E!
Notice that, even if cross section is nonhomogeneous, because flexural modulus is given as the corresponding equivalent linear-stress-distribution stress per unit strain, it enables you to still use the linear flexure formula (Eb)(eps) = sigma = (M)/I, or rather deflection equations derived from it, to compute deflection.
As we discussed in the previous post, flexural strength, Sfu, will (depending on the lamina, of course) usually be higher than tensile strength. Might the same trend frequently apply between flexural modulus and tensile modulus? Unfortunately, I'm not sure at the moment. Hope this helps answer some of your questions about flexural modulus of elasticity. Good luck.
Flexural modulus, Eb, is the slope of the stress-strain curve, within the elastic limit, obtained during a flexure (bending) test. I.e., flexural modulus is the ratio of [equivalent] extreme (outermost) fiber bending stress [(corresponding to an equivalent linear stress distribution)] to extreme fiber strain. The above-mentioned flexure test consists of a simply-supported strip of material (beam) with a concentrated transverse load applied at midspan.
As you know, "bending stress," above, means axial tensile stress on tension side, and axial compressive stress on compressive side, of a cross section in bending.
If test specimen cross section is homogeneous (i.e., consistent, one material), then strike out the words I have placed within brackets, above. For homogeneous cross sections, flexural modulus of elasticity, Eb, should theoretically equal tensile modulus of elasticity, E!
Notice that, even if cross section is nonhomogeneous, because flexural modulus is given as the corresponding equivalent linear-stress-distribution stress per unit strain, it enables you to still use the linear flexure formula (Eb)(eps) = sigma = (M)
/I, or rather deflection equations derived from it, to compute deflection.As we discussed in the previous post, flexural strength, Sfu, will (depending on the lamina, of course) usually be higher than tensile strength. Might the same trend frequently apply between flexural modulus and tensile modulus? Unfortunately, I'm not sure at the moment. Hope this helps answer some of your questions about flexural modulus of elasticity. Good luck.
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