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Dr. Ahmed M. Maghraby General Physics Chapter (2): Mechanical Properties of Materials - When applying an external force on a material, some materials response is the deformation. - Deformation of a body involves the change in its size, or shape, or both of them. - Deformation is of two types: 1- Elastic deformation: A temporary change in the size or shape of a body and disappears after stopping the external force. 2- Plastic deformation: A permanent change in the size or shape of a body even after stopping the external force. - Materials can be classified into two categories: • Elastic material: Is the material which is able to restore its original size and shape after removing the external force. Examples: Rubber, Steel • Plastic material: Is the material which fails to restore its original size and shape after removing the external force. Examples: Clay, Mud, Dough - Elasticity of materials can be compared in terms of “elasticity moduli”. Elasticity modulus is a property for each material. 1|Page Dr. Ahmed M. Maghraby General Physics Elasticity modulus is the ratio of stress to strain. Stress (σ): It is the instantaneous perpendicular force (F) per unit cross sectional area (A). Strain (ε): is a measure of deformation as a result of stress. - Types of Elasticity Moduli: 1 – Young’s modulus (Y): for the change in length of a linear body. 2- Shear modulus (S): for the shear effect in a body. 3- Bulk modulus (B): for the change in the volume of a bulk body. =========================== 1- Young’s modulus (Y): - It is the ratio of tensile stress to tensile strain. - Tensile stress (σ): it is the ratio of perpendicular applied force (F) to the area (A). (σ) = F A - σ has the unit of N/m2. - Tensile strain (ε): is the ratio of the change in length (ΔL) to the original length (Lo). 2|Page Dr. Ahmed M. Maghraby General Physics ε= - L Lo Tensile strain is a ratio of similar quantities and has no unit. N/m2 is called “Pascal” or “Pa” - Strain can express elongation (increase in length) or compression (decrease in length). - Hooke’s law: - Hooke’s law represents the relation between the stress and strain. Hooke’s law states that “The Stress is directly proportional to the Strain” σ=Yε - If a rod is stretched by a force F1 , change in length is ∆L, then Since Y, A and L is constant for each material, 3|Page Dr. Ahmed M. Maghraby - So we can write: So.. - General Physics F1 = K ΔL According to Hooke’s law, “the change in length (ΔL) is directly proportional to the applied force (F)”. 4|Page Dr. Ahmed M. Maghraby General Physics The figure shows the Hooks law where the relation between the applied force and the extension is linear with slope of force constant, k. - when an elastic body is stretched by external agent, the body exerts force proportional to the distance and in opposite direction to external force: F2 = - F1 - Stress – Strain Relationship: - Hook’s law does not describe the whole relationship between the stress and strain; it only describes the first linear part. The above figure shows the stress-strain relationship. 5|Page Dr. Ahmed M. Maghraby General Physics 1. Point A is called “Proportional Limit”: it represents the limit of linear range between stress and strain. 2. Point B is called “elasticity limit”: represents the limit of elasticity region, beyond which the material deformation is plastic. 3. Point C is called “Yield point”. 4. Point D is called “Maximum stress point” represents the maximum value of applied stress. 5. Point E is called “Fracture point”: is the point at which the body is fractured (broken). In region (OA) - There is a linear relationship between stress and strain and this region called Hookean behavior because the material obeys Hook’s law. - The slope of straight line give modulus of elasticity define by Modulus of elasticity= Stress Strain (Pascal or N/m2). In region (AB) - The stress increases in proportional to strain but not linearly. 6|Page Dr. Ahmed M. Maghraby - General Physics The point b is the elastic limit ( it is maximum stress, which a material can withstand without undergoing some permanent deformation. In region (BDE) - This region, Indicates the degree of permanent deformation in which a material up to the point of fracture (E). - Example (2-1): The bar shown has a square cross section for which the depth and thickness are 40 mm. If an axial force of 800 N is applied along the centroidal axis of the bar’s cross sectional area, determine the average normal stress acting on the bar? Solution: Stress is - F 800 500 * 10 3 Pa 3 2 A (40 * 10 ) Example (2-2): A 80 Kg mass is hung on a steel wire having 18m long and 3mm diameter. What is the elongation of the wire, knowing Young's modulus for steel is 21 x 1010 N/m2? Solution: m=80 kg 2r=3 mm= 0.003m L 7|Page Lo= 18 m r=0.0015m Y= 21 x 1010 N/m2 80x9.8 18 x 0.0095m 9.5mm 2 (0.0015) 21x1010 Dr. Ahmed M. Maghraby - General Physics Example (2-3): A piece of copper originally 305 mm long is pulled in tension with a stress of 276 MPa. If the deformation is entirely elastic, what will be the resultant elongation? Solution Young's modulus is given by: Then, L - (F A )L L (276x 106 )(305x 103 ) 0.76x 103 m 11 Y Y (11x 10 ) Example (2-4): A person carries a 21 kg suitcase in one hand. Assuming the humorous bone supports the entire weight, determine how much it stretches, assume the humorous is 33 cm in length and has an effective cross-sectional area of 5.2 x 10-4 m2. Where Y= 1.6x 1010 N/m2 Solution Young,s modulus is Y F /A FL mgL (21)(9.8)(0.33) L 8.17x 106 m 10 4 L / L YA YA (1.6x 10 )(5.2x 10 ) Exemple (2.5) A telephone wire 120 m long and 2.2 mm in diameter is stretched by a force of 380 N. What is the longitudinal stress? If the length after stretching is 120.10 m, what is the longitudinal strain? Determine Young’s modulus for the wire? Solution A = 𝜋 r2 = 3.14 * (1.1*10-3)2 = 3.8 * 10-6 m2 F 380 100*106 -6 A 3.8*10 ΔL 120.1 - 120 0.1 m σ 8|Page N / m 2 100 MPa Dr. Ahmed M. Maghraby ε General Physics ΔL 0.1 8.33*10-4 Lo 120 σ 100*106 Y 12*1010 N / m 2 12*10 4 MPa -4 ε 8.33*10 Exemple (2.7) A load of 102 kg is supported by a wire of length 2 m and cross sectional area 0.1 cm2. The wire is stretched by 0.22 cm . Find the tensile stress, tensile strain, and Young’s modulus of the wire ? m = 102 kg L=2m A = 0.1 cm2 ∆L = 0.22 cm Tensile Stress () F m g 102 * 9.8 999.6 *10 5 N/m 2 4 A A 0.1 *10 Tensile Strain () L 0.22 11 * 10 4 L o 2 * 100 Young's modulus (Y) 999.6 * 105 90.87 * 109 N/m2 4 11 * 10 Exemple (2.8) A structure steel rod has a radius R of 9.5 mm and a length L of 81 cm. A force F of 6.2 * 104 N stretches it axially. (Esteel = 2 * 1011 N/m2) (a) What is the stress in the rod ? (b) What is the elongation of the rod under this load ? (c) What is the strain? Radius = 9.5 mm L = 81 cm F = 6.2 * 104 N F F 6.2 * 10 4 Tensile Stress () 2 2.19 * 10 8 N/m 2 3 2 A r (9.5 * 10 ) 9|Page Dr. Ahmed M. Maghraby Tensile Strain ( ) General Physics L Lo ; E 2.19 * 108 L * Lo * Lo * 81 * 102 8.86 * 104 m 11 Y 2 * 10 Tensile Strain ( ) Y 2.19 * 108 1.1 * 103 11 2 * 10 2- Shear modulus (S): Shear stress: Shear strain: Shear Strain Tan =∆x/h but is small so Shear strain = = ∆x/h 10 | P a g e x h tan so Dr. Ahmed M. Maghraby General Physics Exemple (2.9) A horizontal force of 1.2 N is applied to the top of a pile of compact discs 13 cm in diameter and 9 cm high. The result is a 2.5 cm shear. Find the shear modulus? Solution: Shear Modulus (S) F / A / S F h F h (1.2)(0.09) = 2 = A x r x (0.13 / 2) 2 (0.025) 3- Bulk modulus (B): 11 | P a g e F / A / (x / h ) N/m 2 Dr. Ahmed M. Maghraby General Physics Note: Bulk modulus is negative if the stress results in the decrease in volume and positive if the stress results in the increase in volume. Bulk stress (Volume stress): Bulk strain (Volume strain): Bulk strain is A measure for the change in Volume Volume Strain The compressibility factor, K V V 1 B Note: Bulk modulus applies for both solids and liquids. 12 | P a g e Dr. Ahmed M. Maghraby General Physics Energy Stored in a stretched wire (Elastic potential energy) 1-The stored potential energy (U): 1 U= 𝜎𝑥𝜀 2 2-The work done = 𝜎 𝑥 𝜀 Exemple (2.10) A wire of length 120cm and diameter 0.82mm, supported from one end, A 5.3kg in the other end. Find: a) The stress in the wire b) The strain in the wire c) The strain energy where Y = 1.2x 1012 N/m2 Solution r 0.82 0.041 cm and m = 5.3kg 20 𝐹 𝑚𝑔 𝐴 𝐴 The stress = = = (5.3 𝑥 9.8) 𝜋(0.41 𝑥 10−3 ) = 40344 𝑁/𝑚2 40344 The strain Stress =3.36 x 10-8 1.2 𝑥 1012 Y 1 Strain Energy = (Stress)(Strain) = 0.5 x 40344 x 3.36 x 10-8 = 6.7 x 10-4 J 2 Exemple (2.11) A uniform wire of length 20cm density 0.78g/cm3 and mass 16gm stretched by a distance 1.2mm when 8kg is supported on it, Find: 13 | P a g e Dr. Ahmed M. Maghraby General Physics a) The stress in the wire b) Young's modulus c) The strain energy Solution Volume (V) V= 𝑚 𝜌 = 16 𝑥 10−3 0.78 𝑥 103 V = 20.5 x 10-6 m3 But V=A . L A = V/L = 20.5 x 10-6 / 0.16 =128 x 10-6 m2 Stress= 𝐹 𝐴 = 𝑚𝑔 𝐴 = 8 𝑥 9.8 128 𝑥 10−6 = 0.612 x 106 N/m2 Y = Stress/Strain = 0.612 x 106 / (1.2 x 10-3/20 x 10-2) Y = 1.02 x 1012 N/m2 Strain energy = ½ (Stress) x (Strain) Strain energy = ½ x 0.612 x 106 x (1.2 x 10-3/20 x 10-2) Strain energy = 12.75 x 108 J 14 | P a g e