E is generally taken to be the elastic constant known as Young's modulus which describes the relationship between axial stress and axial strain where Hooke's law still applies (i.e. linear elasticity). Nu is Poisson's ratio which is the relationship between axial strain and radial or transverse strain. For more information, please see the related link.
There are a number of assumptions required to answer this question!1) E = Young's modulus. This assumption is based on the reference to axial forces and a request to calculate elongation.2) t = tonnes (1,000 kg) = 9,810 NewtonsFirst of all, Young's modulus is normally expressed in Pascals (N/m2). So it is sensible to convert your value of Young's modulus into SI units.Modulus (N/cm2) = (80 * 1000 * 9.81) / 10Calculation of modulus with SI units:Area = 0.1 * 0.1 = 0.01 m2Modulus = (80 * 1000 * 9.81) / 0.01= 78,480,000 N/m2 = 78.48 MN/m2= 78.48 MPaNow Young's modulus (E) can be used to find the axial strain (ΣA) that occurs in a material at a given axial stress (σA).You have not told us what the axial stress is, however:E = σA / ΣAΣA = σA / EIn order to calculate the elongation (change in length) from the above we would require the original length of the bar (which again you haven't given us), however:ΣA = change in length / original lengthSo the change in length (m) = (σA / E) * original lengthBelow is an example using the above based on assumed values for axial stress and the original length of the bar:Bar undeformed length = 0.3 mMass applied = 10 tonnes (10,000 kg)Axial Stress (σA) = Force / AreaAxial Stress (σA) = (10,000 * 9.81) / 0.01Axial Stress (σA) = 9.81x106 PaChange in length = (9.81x106 / 7.848x107) * 0.3Change in length = 0.0375 m (3.75 cm or 37.5 mm)
Yes, indeed. Sometimes tensile modulus is different from flexural modulus, especially for composites. But tensile modulus and elastic modulus and Young's modulus are equivalent terms.
The elastic modulus, also called Young's modulus, is identical to the tensile modulus. It relates stress to strain when loaded in tension.
Young's modulus
When we talk about deformatation, we are referring to two properties, namely Elasticity and Plasticity. These properties are measured using constants known as " Moduli of Elasticity". There are 4 such moduli: Young's Modulus Axial Modulus Rigidity Modulus Bulk Modulus The larger the value of the Bulk Modulus, the harder it is to compress the material.
We knew from Hook's law- "stress is proportional to strain." So, stress = k * strain [here, k is a constant] or, stress/strain= k Now, if the stress and strain occurs due to axial force then k is known as modulus of elasticity and it is denoted by E. if the stress and strain occurs due to shear force then k is known as modulus of rigidity and it is denoted by G.
E is generally taken to be the elastic constant known as Young's modulus which describes the relationship between axial stress and axial strain where Hooke's law still applies (i.e. linear elasticity). Nu is Poisson's ratio which is the relationship between axial strain and radial or transverse strain. For more information, please see the related link.
There are a number of assumptions required to answer this question!1) E = Young's modulus. This assumption is based on the reference to axial forces and a request to calculate elongation.2) t = tonnes (1,000 kg) = 9,810 NewtonsFirst of all, Young's modulus is normally expressed in Pascals (N/m2). So it is sensible to convert your value of Young's modulus into SI units.Modulus (N/cm2) = (80 * 1000 * 9.81) / 10Calculation of modulus with SI units:Area = 0.1 * 0.1 = 0.01 m2Modulus = (80 * 1000 * 9.81) / 0.01= 78,480,000 N/m2 = 78.48 MN/m2= 78.48 MPaNow Young's modulus (E) can be used to find the axial strain (ΣA) that occurs in a material at a given axial stress (σA).You have not told us what the axial stress is, however:E = σA / ΣAΣA = σA / EIn order to calculate the elongation (change in length) from the above we would require the original length of the bar (which again you haven't given us), however:ΣA = change in length / original lengthSo the change in length (m) = (σA / E) * original lengthBelow is an example using the above based on assumed values for axial stress and the original length of the bar:Bar undeformed length = 0.3 mMass applied = 10 tonnes (10,000 kg)Axial Stress (σA) = Force / AreaAxial Stress (σA) = (10,000 * 9.81) / 0.01Axial Stress (σA) = 9.81x106 PaChange in length = (9.81x106 / 7.848x107) * 0.3Change in length = 0.0375 m (3.75 cm or 37.5 mm)
there are different types of modulus it depends on what types of stress is acting on the material if its direct stress then then there is modulus of elasticity,if tis shear stress then its modulus of rigidity and when its volumetric stress it is bulk modulus and so on
Yes, indeed. Sometimes tensile modulus is different from flexural modulus, especially for composites. But tensile modulus and elastic modulus and Young's modulus are equivalent terms.
The elastic modulus, also called Young's modulus, is identical to the tensile modulus. It relates stress to strain when loaded in tension.
The Young modulus and storage modulus measure two different things and use different formulas. A storage modulus measures the stored energy in a vibrating elastic material. The Young modulus measures the stress to in still elastic, and it is an elastic modulus.
Young's modulus
1. Young's modulus of elasticity, E, also called elastic modulus in tension 2. Flexural modulus, usually the same as the elastic modulus for uniform isotropic materials 3. Shear modulus, also known as modulus of rigidity, G ; G = E/2/(1 + u) for isotropic materials, where u = poisson ratio 4. Dynamic modulus 5. Storage modulus 6. Bulk modulus The first three are most commonly used; the last three are for more specialized use
Modulus robot was created in 1984.
Young's modulus