MEDIUM. . Mechanical property that measures stiffness of a solid material, Force exerted by stretched or contracted material, "Elastic Properties and Young Modulus for some Materials", "Overview of materials for Low Density Polyethylene (LDPE), Molded", "Bacteriophage capsids: Tough nanoshells with complex elastic properties", "Medium Density Fiberboard (MDF) Material Properties :: MakeItFrom.com", "Polyester Matrix Composite reinforced by glass fibers (Fiberglass)", "Unusually Large Young's Moduli of Amino Acid Molecular Crystals", "Composites Design and Manufacture (BEng) – MATS 324", 10.1002/(SICI)1098-2329(199924)18:4<351::AID-ADV6>3.0.CO;2-X, Epoxy Matrix Composite reinforced by 70% carbon fibers [SubsTech], "Properties of cobalt-chrome alloys – Heraeus Kulzer cara", "Ultrasonic Study of Osmium and Ruthenium", "Electronic and mechanical properties of carbon nanotubes", "Ab initio calculation of ideal strength and phonon instability of graphene under tension", "Carbyne is stronger than any known material", "Standard Test Method for Young's Modulus, Tangent Modulus, and Chord Modulus", Matweb: free database of engineering properties for over 115,000 materials, Young's Modulus for groups of materials, and their cost, https://en.wikipedia.org/w/index.php?title=Young%27s_modulus&oldid=985960442, Short description is different from Wikidata, Articles with unsourced statements from July 2018, Articles needing more detailed references, Pages containing links to subscription-only content, Creative Commons Attribution-ShareAlike License. r²  = 20.79 × 10-8 ∴  Geometric stiffness: a global characteristic of the body that depends on its shape, and not only on the local properties of the material; for instance, an, This page was last edited on 28 October 2020, at 23:58. ∴ 2 ∴  C. 1. Y = F/(A × strain) Young's modulus of carbon steels (mild, medium and high), alloy steels, stainless steels and tool steels are given in the following table in GPA and ksi. And N to dyne is 10^5. 0 Given: Density When expressed in CGS units of dyne/cm 2, it will be equal to (1 N = 10 5 dyne, 1 m 2 = 104 cm 2) . ∴  A copper wire is to be stretched by a load of 10 kg. In this article, we shall study concept application and numerical problems on longitudinal stress, longitudinal strain, Young’s modulus of elasticity. Why is the force of repulsion responsible for the formation of a solid and not the forces of attraction. 1. Given:  Strain What must be the elongation of a wire 5m long so that the strain is 1% of 0.1? ∴  9.68 × 1010 N/m², Diameter of wire = 0.95 mm, Radius of wire = Stress = 7.8 × 108 N/m². If the range over which Hooke's law is valid is large enough compared to the typical stress that one expects to apply to the material, the material is said to be linear. 1 mm = 1 × 10-3 m, Load attached = m = 40 kg, Young’s modulus Other elastic calculations usually require the use of one additional elastic property, such as the shear modulus G, bulk modulus K, and Poisson's ratio ν. Elastic deformation is reversible (the material returns to its original shape after the load is removed). l = 5 × 10-3 m = 5 mm F = 3.142  × (1 × 10-3)² × 35 × 1010× 10-3 14 × 1011 N/m², Area of cross section = A = 0.02 cm² find whether wire can be stretched by 10 mm. ∴  Breaking stress for steel = 7.8 × 108 N/m². ≥ g = 9.8 m/s². Material stiffness should not be confused with these properties: Young's modulus enables the calculation of the change in the dimension of a bar made of an isotropic elastic material under tensile or compressive loads. F = AY l /L × 10-6 m², Load attached = F = 10 kg-wt = 10 × 9.8 N Strain =? ∴  Cloudflare Ray ID: 5f185936bafefd26 For example, carbon fiber has a much higher Young's modulus (is much stiffer) when force is loaded parallel to the fibers (along the grain). F = 3.142  × 1 × 10-6 × 35 × 1010× 10-3 If the area of the cross-section of the wire is 0.02 cm², find the maximum load that can be used for stretching the wire without causing a permanent set. A cube of steel 28 cm on a side supports a load of 85 kg that has the same horizontal cross section as the steel cube. Young's modulus l = 1 × 10-3 × 5 It quantifies the relationship between tensile stress 891 × 10-3 m = 0.891 mm Ans. length of wire = L = 2 m, Elastic limit = stress = 8.26 × 108 Given: Initial However, metals and ceramics can be treated with certain impurities, and metals can be mechanically worked to make their grain structures directional. Young’s modulus of elasticity =  Y = 2× 1011 N/m². The elastic potential energy stored in a linear elastic material is given by the integral of the Hooke's law: now by explicating the intensive variables: This means that the elastic potential energy density (i.e., per unit volume) is given by: or, in simple notation, for a linear elastic material: ) L The plus sign leads to A wire of this material of diameter 0.95 mm is stretched by applying a certain force. T is constant throughout the change. m/s², Y = 11 × 1010 N/m². = 1/1000 = 10-3 , Young’s modulus of elasticity =  Y = Stress = F / A  = mg /π r² ∴  Stress =  ( 40 × 9.8) /(3.142 ×(1 × 10-3)²) ∴  Stress =  ( 40 × 9.8) /(3.142 ×  1 × 10-6) ∴  Stress =  1.25 × 108    N/m² Now, Y = Stress / Strain ∴   Strain = Stress / Y  =  1.25 × 108  / 7 × 1010 ∴   Strain = 1.78 × 10-3 Now, Strain = l /L ∴ extension = l = Strain × L ∴  l = 1.78 × 10-3 × 1 ∴  l = 1.78 × 10-3 m Now, force constant K = F/l = mg/l = ( 40 × 9.8) /(1.78 × 10-3) ∴  Force constant K = 2.2 × 105  N/m Ans. (7 × 1010) ∴ Strain length of wire = L = 5 m, Strain = 1% of 0.1 = 1 × 10-2 × {\displaystyle E(T)=\beta (\varphi (T))^{6}} Stress = 50 kg wt/sq. {\displaystyle u_{e}(\varepsilon )=\int {E\,\varepsilon }\,d\varepsilon ={\frac {1}{2}}E{\varepsilon }^{2}} Y = (8.26 × 108 × 2) / 2 × 1011 : Extension = 5 mm and Stress =  9.8 × 107  Limit = Stress = 1.5 × 108 N/m², Load = F = 10 kg wt = 10 × 9.8. (force per unit area) and axial strain ) What is the percentage increase in the length of the wire? The Young's modulus of steel is 2.0x10+11 w/m2. Change in length = l = ?, Applied Stress = Applied force / Area If the interatomic spacing in a steel wire is 3Å & YSteel = 20 × 10^10 (N / m^2). Compute the extension produced. l = 1.85 × 10-3 m =  0.185 × 10-2 m = How much will a 30 m steel tape 1 cm wide and 0.05 cm thick stretch under a pull of force of 300 N , if Young's modulus of steel is ? : Stress =  1 × 107 N/m², Strain = 5 × 10-4 , Young’s modulus of elasticity= Y = 2 × 1010 N/m². Y = 2× 1011 N/m². /1.5 Ans. limit will get crossed at extension of 8.26 mm. Stress = F /A = F /π r² ∴   r² = mg / (π × Stress) ∴   r² = (2 × 9.8)  / (3.142 × 2.4 × 108) ∴   r² = 2.599× 10-8 ∴   r = 1.612× 10-4  m = 0.1612× 10-3 m =  0.1612 mm, Ans. Y = FL /A l of material = Y = 7 × 1010 N/m². Find the minimum diameter the wire must have if the elastic limit is not to be exceeded.

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