Physics (101) Properties of Matter Chapter (2) Elasticity Mansoura University PDF

Physics Department Faculty of Science Mansoura University Elasticity: Elasticity: is that property of a body by which it experiences a change in size or shape whenever a deforming force acts on the body. Elasticity: Elongation...

Physics Department Faculty of Science Mansoura University Elasticity: Elasticity: is that property of a body by which it experiences a change in size or shape whenever a deforming force acts on the body. Elasticity: Elongation Contraction Solid Material Solid Material Original length Pulling Original size Compressing Molecules away Molecules pushed Attractive molecular Repulsive molecular from their closer together from forces pull all the forces causes the equilibrium their equilibrium molecules back atoms to return position position Applied force is Change in length Compressive Reduction in size removed of the material force is removed Elasticity: Elastic materials: The material that restores its original shape after removing stress (load) from it. Plastic materials: The material that fails to restore its original dimensions after removing the applied stress, it's said to plastic. Hooke’s Law — Stress and Strain: Stress: the force (F) per unit area (A) Stress = F/A  Its units are dyne/cm2 or N/m2. Strain: the change in material dimensions relative to the original dimensions. It’s dimensionless. ΔL, is directly proportional to the magnitude of the applied force F. ∆𝑙 𝛼 𝐹 Hooke’s Law — Stress and Strain: ΔL is inversely proportional to the cross-sectional area A of the wire 1 ∆𝑙 𝛼 𝐴 ΔL is directly proportional to the original length of the wire Lo. ∆𝑙 𝛼 𝑙𝑜 So, we can write 𝑭 𝒍𝒐 ∆𝒍 𝜶 𝑨 Hooke’s Law — Stress and Strain: which can be rewritten as 𝑭 𝒍𝒐 ∆𝒍 𝜶 𝑨 ∆𝒍 𝑭 Strain 𝜶 Stress 𝒍𝒐 𝑨 Hooke’s law of elasticity: Hooke’s law state that (The applied stress on material is proportional to the resultant strain only if the material in its elastic region). Y : This constant, called Young’s modulus 𝑭 ∆𝒍 𝑭 ∆𝒍 𝜶 =𝒀 (Elastic modulus) and depends on the type 𝑨 𝒍𝒐 𝑨 𝒍𝒐 of material. Its unit (N/m2). Example 2.1: Stretching a wire. A steel wire 1.00 m long with a diameter d = 1.00 mm has a10.0-kg mass hung from it. (a) How much will the wire stretch? (b) What is the stress on the wire? (c) What is the strain? Knowing that Young’s modulus for steel is 𝐘 = 𝟐𝟏 𝒙𝟏𝟎𝟏𝟎 / 𝑵/𝒎𝟐. Solution The cross-sectional area of the wire is given by 𝒅 𝟐 𝒅𝟐 𝝅 (𝟏 𝒙 𝟏𝟎−𝟑 )𝟐 𝑨= 𝝅 𝒓𝟐 = 𝝅( ) = 𝝅 = = 𝟕. 𝟖𝟓 𝒙 𝟏𝟎−𝟕 𝒎𝟐 𝟐 𝟒 𝟒 The force acting on the wire 𝒎 𝑭 = 𝒎𝒈 = 𝟏𝟎 𝒌𝒈 𝟗. 𝟖 𝟐 = 𝟗𝟖 𝑵 𝒔 Example 2.1: 𝑭 ∆𝒍 𝟏 𝑭 𝒍𝒐 The elongation of the wire =𝒀 so, ∆𝒍 = 𝑨 𝒍𝒐 𝒀 𝑨 𝟗𝟖 𝑵 (𝟏 𝒎) ∆𝒍 = = 𝟎. 𝟓𝟗𝟒 𝒙 𝟏𝟎−𝟑 𝒎 𝑵 𝟐𝟏 𝒙 𝟏𝟎𝟏𝟎 𝟐 (𝟕. 𝟖𝟓 𝒙 𝟏𝟎−𝟕 𝒎𝟐 ) 𝒎 The stress acting on the wire is 𝑭 𝟗𝟖 𝑵 𝑺𝒕𝒓𝒆𝒔𝒔 = = −𝟕 𝟐 = 𝟏. 𝟐𝟓 𝒙 𝟏𝟎𝟖 𝑵/𝒎𝟐 𝑨 𝟕. 𝟖𝟓 𝒙 𝟏𝟎 𝒎 The strain of the wire is ∆𝒍 𝟎. 𝟓𝟗𝟒 𝒙 𝟏𝟎−𝟑 𝒎 𝑺𝒕𝒓𝒂𝒊𝒏 = = = 𝟎. 𝟓𝟗𝟒 𝒙 𝟏𝟎−𝟑 𝒍𝒐 𝟏𝒎 Stress-Strain Curve : Materials can be essentially classified into two groups according to their stress-strain curves, ductile materials and brittle materials. 1. Ductile materials curve: OA: the material obeys Hooke’s law which means that it retains or recovers its original shape whenever it is freed from stress. AB: The material in this region doesn’t obey Hooke’s law. Point A is called the proportional limit because beyond this point, stress and strain terminate their linear relationship. If we removed the stress on the rod in AB, the material would still can recover its original shape. Point B is therefore known as the elastic point (elastic limit). Stress-Strain Curve : BC: Point B is also known as the upper-yield point, as beyond it the material surrenders to stress and begins to deform. The stress decreases until point C, which is known as the lower-yield point, but the material continues to elongate. CD: The material tries to resist this change and tends to harden. This is known as strain hardening. The material growing longer and thinner, until point D, which represents the material’s maximum strength (ultimate strength). DE: Now, only the neck experiences any further deformation. The neck grows thinner and weaker until point E, where it breaks;. Point E is called the fracture or rupture point. Necking and Fracture in ductal materials. Stress-Strain Curve : 2. Brittle materials curve: Brittle materials share the same relationship except they don’t exhibit a yielding phenomenon. These materials skip the plasticity region and undergo a fracture directly after the elasticity region. It survives until the elastic limit, after which it fractures when subjected to excess stress. Elastomers: materials that can be stretched to cause large strains. Example 2.2 Compressing a steel column. A 445,000-N load is placed on top of a steel column 3.05 m long and 10.2 cm in diameter. By how much is the column compressed? Y= 21×1010 N/m2. Example 2.2: Solution The cross-sectional area of the column is 𝒅 𝟐 𝟐 𝒅𝟐 𝝅 (𝟏𝟎. 𝟐 𝒙 𝟏𝟎−𝟐 𝒎)𝟐 𝑨 = 𝝅 𝒓 = 𝝅( ) = 𝝅 = = 𝟖𝟏. 𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝟐 𝟐 𝟒 𝟒 The change in length of the column, 1 𝐹 𝑙𝑜 ∆𝑙 = 𝑌 𝐴 𝟒𝟒𝟓𝟎𝟎𝟎 𝑵 (𝟑. 𝟎𝟓 𝒎) ∆𝒍 = = 𝟕. 𝟗𝟏 𝒙 𝟏𝟎−𝟒 𝒎 𝑵 𝟐𝟏 𝒙 𝟏𝟎𝟏𝟎 𝟐 (𝟖𝟏. 𝟕 𝒙 𝟏𝟎−𝟐 𝒎𝟐 ) 𝒎 Example 2.3: Exceeding the ultimate compressive strength. A human bone is subjected to a compressive force of 5.00×105 N. The bone is 25.0 cm long and has an approximate area of 4.00 cm2. If the ultimate compressive strength for a bone is 1.70 × 108 N/m2 , will the bone be compressed, or will it break under this force? Solution 𝑨 = 𝟒 𝒄𝒎𝟐 = 𝟒 𝒙 𝟏𝟎−𝟒 𝒎𝟐 The stress acting on the bone is found from 𝐹 𝟓 𝒙 𝟏𝟎𝟓 𝑁 𝟖 𝑵/𝒎𝟐 𝑺𝒕𝒓𝒆𝒔𝒔 = = = 𝟏𝟐. 𝟓 𝒙 𝟏𝟎 𝐴 𝟒 𝒙 𝟏𝟎−𝟒 𝒎𝟐 Since this stress exceeds the ultimate compressive stress of a bone, 1.70 × 108 N/m2 , the bone will break. Hooke’s Law for a Spring: Hooke’s Law state that 𝐹 ∆𝑙 =𝑌 𝐴 𝑙𝑜 rearrangement of the equation 𝐴 𝐹 = 𝑌 ∆𝑙 𝑙𝑜 A, Y, and Lo are all constants, So, 𝐴 𝑘=𝑌 𝑙𝑜 Hooke’s Law for a Spring: We call k a force constant or a spring constant and its unit is (N/m). Then, 𝐹 = 𝑘 ∆𝑙 In general we can say, 𝐹=𝑘𝑥 Example 2.4 A spring with a force constant of 50.0 N/m is loaded with a 0.500-kg mass. Find the elongation of the spring. Solution 𝒎 𝑭 𝒎𝒈 𝟎.𝟓 𝒌𝒈 𝒙 (𝟗.𝟖 𝟐 ) 𝒔 𝒙= = = = 𝟎. 𝟎𝟗𝟖 𝒎 𝑲 𝒌 𝟓𝟎 𝑵/𝒎 Elasticity of Shape – Shear: Consider a fixed cube with a height h above the bottom. A tangential force 𝑭𝑡 is applied at the top of the cube. The tangential force applied to the body causes the layers of atoms to be displaced sideways. The tangential force thus causes a change in the shape of the body that is measured by the angle, called the angle of shear Φ. ∆𝒙 𝐭𝐚𝐧 𝝋 = 𝒉 Because the deformations are usually quite small, so the 𝑡𝑎𝑛 𝜑 can be replaced by the angle 𝜑 itself, expressed in radians. Thus, ∆𝒙 𝝋= 𝒉 Which represents the shearing strain of the body. Elasticity of Shape – Shear: we find experimentally that 𝟏 𝑭𝒕 𝝋 𝜶 𝑭𝒕 & 𝝋𝜶 𝝋𝜶 𝑨 𝑨 𝑭𝒕 Note that F/A has the dimensions of a stress and it is now defined as the shearing stress: 𝑠ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑠𝑡𝑟𝑒𝑠𝑠 = 𝑨 𝑭𝒕 𝜶𝝋 𝑨 𝑭𝒕 =𝑺𝝋 𝑨 𝑆 , called the shear modulus (torsion modulus and the modulus of rigidity). The larger the value of S, the greater the resistance to shear. Note that the shear modulus is smaller than Young’s modulus Y. Example 2.5: A sheet of copper 0.750 m long, 1.00 m high, and 0.500 cm thick is acted on by a tangential force of 50,000 N, as shown in figure. The value of S for copper is 4.20 × 1010 N/m2. Find (a) the shearing stress, (b) the shearing strain, and (c) the linear displacement ∆ x. Solution (a) The area that the tangential force is acting over is 𝑨 = 𝒃 𝒕 = 𝟎. 𝟕𝟓 𝒎 𝒙 𝟎. 𝟓 𝒙 𝟏𝟎−𝟐 𝒎 = 𝟑. 𝟕𝟓 𝒙 𝟏𝟎−𝟑 𝒎𝟐 The shearing stress is 𝑭 𝟓𝟎, 𝟎𝟎𝟎 𝑵 𝟕 𝑵/𝒎𝟐 𝒔𝒉𝒆𝒂𝒓𝒊𝒏𝒈 𝒔𝒕𝒓𝒆𝒔𝒔 = = = 𝟏. 𝟑𝟑 𝒙 𝟏𝟎 𝑨 𝟑. 𝟕𝟓 𝒙 𝟏𝟎−𝟑 𝒎𝟐 (b) The shearing strain 𝑭𝒕 =𝑺𝝋 𝑨 Example 2.5: 𝟏 𝑭𝒕 𝝋= 𝑺 𝑨 𝟏 𝟕 𝑵 −𝟒 𝒓𝒂𝒅 𝝋= 𝟏. 𝟑𝟑 𝒙 𝟏𝟎 = 𝟑. 𝟏𝟕 𝒙 𝟏𝟎 𝟒. 𝟐 𝒙 𝟏𝟎𝟏𝟎 𝑵/𝒎𝟐 𝒎𝟐 (c) The linear displacement ∆ x ∆𝒙 𝝋= 𝒉 ∆ 𝒙 = 𝝋 𝒉 = 𝟑. 𝟏𝟕 𝒙 𝟏𝟎−𝟒 𝒙 𝟏 𝒎 = 𝟑. 𝟏𝟕 𝒙 𝟏𝟎−𝟒 𝒎 = 𝟎. 𝟑𝟏𝟕 𝒎𝒎 Elasticity of Volume: If a uniform force is exerted on all sides of an object, each side of the block is compressed. Thus, the entire volume of the block decreases. The compressional stress is defined as 𝑭 𝒔𝒕𝒓𝒆𝒔𝒔 = 𝑨 ∆𝑽 𝒔𝒕𝒓𝒂𝒊𝒏 = 𝑽𝒐 From Hooke’s law, we have 𝑭 ∆𝑉 𝑨 𝛼 𝑉𝑜 So, 𝑭 ∆𝑉 = −𝐵 𝑨 𝑉𝑜 Elasticity of Volume: Constant of proportionality B, called the bulk modulus. The minus sign in the equation because an increase in the stress (F/A) causes a decrease in the volume, leaving ∆V negative. Compressibility: The reciprocal of the bulk modulus B, called the compressibility k, is a measure of how easy it is to compress the substance. The bulk modulus B is used for solids, while the compressibility k is usually used for liquids 1 𝑲= 𝐵 ‫ فى حالة غمر جسم فى سائل فإن اإلجهاد الواقع على الجسم يساوى ضغط السائل على الجسم وبالتالى‬ 𝑭 ∆𝑉 𝑷 = = −𝐵 𝑨 𝑉𝑜 Example 2.6: A solid copper sphere of 0.500 m3 volume is placed 30.5 m below the ocean surface where the pressure is 3.00 ×105 N/m2. What is the change in volume of the sphere? The bulk modulus for copper is 14 × 1010 N/m2. Solution 𝑭 ∆𝑽 ∆𝑽 = −𝑩 𝑷 = −𝑩 𝑨 𝑽𝒐 𝑽𝒐 𝑷𝑽𝒐 The change in volume ∆𝑽 = − 𝑩 𝑵 𝟑 𝒙 𝟏𝟎𝟓 𝟎. 𝟓 𝒎𝟑 ∆𝑉 = − 𝒎𝟐 = −𝟏. 𝟏 𝒙 𝟏𝟎−𝟔 𝒎𝟑 𝑵 𝟏𝟒 𝒙 𝟏𝟎𝟏𝟎 𝒎𝟐 The minus sign indicates that the volume has decreased.