Plastic section modulus, however, is used when a material is allowed to yield and plastically deform. This is the most common usage, because it deals with materials that are within their elastic limit, or stresses less than the yield strength. When the term section modulus is used, it is typically referring to the elastic modulus. What is the difference between section modulus and plastic modulus? For that reason, it’s common to use specialized software to calculate the section modulus in these instances. This can be a very difficult integration to perform with a high level of accuracy for an irregular shape. Therefore, the section modulus of an irregular shape can be defined by L = the perpendicular distance from the element to the neutral axis passing through the centroid The definition of moment of inertia isĭA = the area of an element of the cross-sectional area of the irregular shape Recall that the section modulus is equal to I/y, where I is the area moment of inertia. How do you find the section modulus of an irregular shape?Įven if a shape does not have a pre-defined section modulus equation, it’s still possible to calculate its section modulus. Therefore, the required section modulus to achieve a safety factor of 2 in bending is calculated as shown below:įor this example problem, the required section modulus is 6.67 in3. Rearrange the equation from the beginning of this post into the following form:Ī36 steel is equal to the yield stress of 36,000 psi. Calculate the required section modulus with a factor of safety of 2. Consider the following example:Ī beam made from A36 steel is to be subjected to a load of 120,000 lbf-in. The required section modulus can be calculated if the bending moment and yield stress of the material are known. Y = the distance from the neutral axis to the outside edge of a beam What is the required section modulus? I = the area moment of inertia (or second moment of area) The general formula for elastic section modulus of a cross section is: Σ = F b d is larger than one.The elastic section modulus of C-channel is calculated from the following equation: Both of these forces will induce the same failure stress, whose value depends on the strength of the material.įor a rectangular sample, the resulting stress under an axial force is given by the following formula: If we don't take into account defects of any kind, it is clear that the material will fail under a bending force which is smaller than the corresponding tensile force. Conversely, a homogeneous material with defects only on its surfaces (e.g., due to scratches) might have a higher tensile strength than flexural strength. Therefore, it is common for flexural strengths to be higher than tensile strengths for the same material. However, if the same material was subjected to only tensile forces then all the fibers in the material are at the same stress and failure will initiate when the weakest fiber reaches its limiting tensile stress. When a material is bent only the extreme fibers are at the largest stress so, if those fibers are free from defects, the flexural strength will be controlled by the strength of those intact 'fibers'. In fact, most materials have small or large defects in them which act to concentrate the stresses locally, effectively causing a localized weakness. The flexural strength would be the same as the tensile strength if the material were homogeneous. Most materials generally fail under tensile stress before they fail under compressive stress Flexural versus tensile strength These inner and outer edges of the beam or rod are known as the 'extreme fibers'. At the outside of the bend (convex face) the stress will be at its maximum tensile value. At the edge of the object on the inside of the bend (concave face) the stress will be at its maximum compressive stress value. 1), it experiences a range of stresses across its depth (Fig. When an object is formed of a single material, like a wooden beam or a steel rod, is bent (Fig. 2 - Stress distribution through beam thickness
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