Phy 1 BMS.pdf

Lecture 17,18 &19 Basic Physics 1 (FAD1020) 2024/2025 Subtopic: Basic Materials Science Dr. Salmiah Ibrahim [email protected]...

Lecture 17,18 &19 Basic Physics 1 (FAD1020) 2024/2025 Subtopic: Basic Materials Science Dr. Salmiah Ibrahim [email protected] Room No 14 Level 3 Complex PASUM Centre for Foundation Studies in Science Universiti Malaya 1  Contents 1. Introduction 2. Solids – Intermolecular forces 3. Mechanical Properties of Matter (Stress and Strain) Basic - Young Modulus, Shear Modulus and Bulk Modulus 4. Crystalline and Non-crystalline Materials Materials 5. Classification of Materials: - Metal Alloys Science - Polymers - Ceramics - Composites - Biomaterials 7. Properties of Materials - Mechanical - Thermal - Electrical 2 Lecture 17 Introduction What is material science? Definition 1: A branch of science that focuses on materials; interdisciplinary field composed of physics and chemistry. Definition 2: Relationship of material properties to its composition and structure. What is a material scientist? A person who uses his/her combined knowledge of physics, chemistry and metallurgy to exploit property-structure combinations for practical use. 3  Solid States of Matter Liquid Gas Plasma All matter consists of some distribution of atoms or molecules 4  Solids Has definite shape and volume. Molecules are held in specific locations by electrical forces Vibrate about equilibrium positions - Can be modeled as springs connecting molecules. A model of a portion of a solid. The atoms (spheres) are imagined as being attached to each other by springs, which represent the elastic nature of the interatomic forces 5  Liquid Has a definite volume No definite shape Exist at a higher temperature than the solid state. The intermolecular forces are not strong enough to keep the molecules in a fixed position. The molecules wander through the liquid in a random fashion Erratic motion of a molecule in a liquid 6  Gas Has no definite volume Has no definite shape Molecules are in constant random motion The molecules exert only weak forces on each other Average distance between molecules is large compared to the size of the molecules. These balloons are filled with helium gas. 7  Plasma Matter heated to a very high temperature - Many of the electrons are freed from the nucleus - Result is a collection of free, electrically charged ions Plasmas exist inside the Sun and other stars Plasmas also exist inside neon light and laser tubes 8  Molecular Bonding Solid A solid has definite shape, whereas liquids and gas takes the shape of the container that holds it. This shows that molecules are bound to each other. It is means that a group of two or more atoms that are strongly held together so as to function as a single unit. When atoms make such an attachment, we say that a chemical bond has been formed. In general, there are basically four kinds of molecular/chemical bonding: (I) Ionic bond (II) Covalent bond (III) Van der Waals bond (IV) Hydrogen bond 9  Example of intermolecular and intramolecular forces Intermolecular Forces the forces which mediate interaction between molecules including forces of attraction or repulsion which act between molecules and other types of neighbouring particles, e.g: atoms or ions. Ø Intermolecular Ø Intramolecular forces forces are the are the forces that forces that exist hold atoms between together within a molecules. molecule. 10  Intermolecular Forces The forces that bind the atoms or molecules are electrostatic in nature so there are repulsive and attractive forces that acts between two molecules/atoms. Originally, the two atoms are separated by a distance of r0. r = r0 11 Continued… At one instant, one atom might moved slightly closer to the neighbouring atom through distance r where r < r0 When this happens, a repulsive force will act on it. This force increases when the separation decreases. The closer the atom is to its neighbour, the greater will be the repulsion. Hence, it will be pushed back by the strong repulsive force to prevent it from crashing into its neighbouring atoms. Frepulsive Frepulsive r 12 Continued… ✓The repulsive force between two atom can be represented by 𝒂 𝑭𝒓𝒆𝒑𝒖𝒍𝒔𝒊𝒗𝒆 = 𝒑 𝒓 where a and p are positive constant and r is the separation between the atoms ✓The repulsive force is assumed positive because it is in the direction of increasing r. 13 Continued… The variation of repulsive force Frepulsive with the separation r between two atoms Figure 1 14 Continued… At one instant, one atom might have moved slightly further away from the neighbouring atom through a distance r. [r > r0 ] When this happens, an attractive force acts on it. This force increases with distance so the further the atom is from its neighbour, the greater will be the attraction. It will be pulled back by the strong attractive force to prevent it from escaping. r = r0 15 Continued… The attractive force between two atoms can be represented by; 𝒃 𝑭𝒂𝒕𝒕𝒓𝒂𝒄𝒕𝒊𝒗𝒆 =− 𝒒 𝒓 where b and q are positive constants. The negative sign indicates that the directions of the attractive force is opposite to the direction of increasing r Fattractive Fattractive r1 16 Continued… The variation of attractive force Fattractive with the separation r between two atoms. Fattractive Figure 2 17 Continued… ❖ The intermolecular forces, repulsive and attractive, vary with the separation distance between the atoms or molecules. ❖ The vector sum of the repulsive force and attractive force produces a resultant force F and is given by expression 𝑎 𝑏 𝐹= !− " 𝑟 𝑟 18 Forces between two atoms for different separation distances r0 (equilibrium separation) Frepulsive r < r0 Frepulsive Fattractive Fattractive r > r0 19 Continued… Strong attractive and repulsive atomic forces will act on an atom if it moves too far from or too close to a neighbouring atom. Because of this, each atom in a solid can only vibrate constantly at its equilibrium position, not being able to move freely within the whole volume of the solid. The attractive and repulsive atomic forces have ‘locked’ it in its equilibrium position. The fact that forces are needed to be applied to stretch or to compress a solid is evidence of the existence of the attractive and repulsive forces between atoms in matter. 20  Graph of resultant force F against separation r When r = r0 , F = 0 where IFattractive I = I Frepulsive I (r0 = equilibrium separation of the 2 atoms] r < r0 , F = positive [represents repulsive force] [Repulsive force increases as r decreases] r > r0 , F = negative [represents attractive force] [Between r0 and r1 magnitude of attractive force increases. - atoms are still bonded to each other] Figure 3 21 Continued… When r = r1 , the attractive force is maximum and this represents the breaking force. When r > r1 , the bond between the atoms has been broken and the attractive force between the atoms decreases as the separation r between the atom increases. *Solid transform to a liquid. When r is large [r = ¥ ] when compared to the diameter of the atom, F = 0. Under such conditions the materials is in the gaseous state. Figure 3 22 The relationship between Hooke’s Law and the F-r graph The F-r graph can also be used to explain Hooke’s Law Within the elastic limit of a solid material, the deformation produced by a force is proportional to the force. If the elastic limit is not exceeded, the material returns to its original shape and size after the force is removed. Hooke’s Law For values of r close to r0 , the curve F against r is a straight line. The equation for the straight line is given as F = -k (r - r0) where -k = gradient of the graph (r – r0 ) = Dr (change in the separation between atoms) 23  Example From the F-r graph, Fresultant (´ 10-9 N) (a) Determine the value of r0 34 (b) For a small distance from r0, the F-r 23 graph exhibits a linear relationship which obeys Hooke’s Law. Find the value of k. 11 r (´ 10-12 m) What is k ? 0 0 100 200 300 400 500 -11 -23 Answer: (a) r0 is the separation distance between atoms at where resultant force is zero. From the graph, when F = 0, r0 = 1.5 ´ 10-10 m (a) –k = the gradient of the F-r graph 4.0 ´ 102 N m-1 25  Potential energy between molecules Consider that F is the repulsive force acting on atom B. The work done to bring atom B to a r distance dr closer to atom A against the repulsive force F is W = F (-dr) A B F The work done causes a change of dU in - dr the potential energy, U of the system which can be expressed as dU = F (-dr) -. 0. F=- -/. =- 0/ 26 Continued… The potential energy of the system of two atoms with separation r is given by /! U = − ∫1 𝐹 𝑑𝑟 The change in potential energy of the system when the separation between atoms changes from r1 to r2 is /! /" U2 – U1 = − ∫/ 𝐹 𝑑𝑟 = ∫/ 𝐹 𝑑𝑟 " ! The equation above is equal to the area under the F-r graph between r2 and r1 27  Graph of Potential energy between molecules r < r0 : gradient of U-r graph is negative The repulsive force is positive r > r0 : gradient of U-r graph is positive The attractive force is negative r = r0 : gradient of the U-r graph is zero The force is zero,(equilibrium position) Potential energy at this point is minimum r = r1 : Positive gradient of U-r graph starts decreasing with increasing r Point P is a point of inflection where attractive force is maximum 28 Continued… When r = r0 , the potential energy U is minimum and the atoms in the solid have no thermal energy or kinetic energy because the atoms does not move. When r1 < r < ¥ , the material is in the liquid state. When r = ¥ , the material is in the gaseous state. In this state, the forces between molecules becomes negligible. 29  Quick Quiz 1. The force curve of an atom in a diatomic molecule shown in the graph below The equilibrium position of the atom is represented by the point A. P B. Q C. R D. S 30 2. The graph shows the variation of force F between two atoms with the separation r between the atoms F Which of the following statements is true? A. From P to Q the resultant force is attractive. B. From Q to R, the attractive force decreases. C. At R, the attractive force is maximum. D. From R to S, the resultant force is repulsive. 31 3. Graph (a) and (b) shows the interatomic force F and potential energy U plotted against separation r between a pair of of atoms. The graph have different scales on the x-axis. (a) (b) The equilibrium separation of the atoms is represented by the points A. X and P B. Y and P C. X and Q D. Y and Q 32 4. The graph shows the variation of potential energy, U between two molecules with separation r, At which point on the graph does a change from the solid state to liquid state take place? A. P B. Q C. R D. S 33 Lecture 18 &19 Mechanical properties of matter - Stress and strain Prepared by: Dr. Salmiah Ibrahim [email protected] PASUM 34  Deformation of Solids Deformation: Changes in shape or size (or both) of an object through the application of external forces. The behaviour of different materials when subjected to a force varies. The degree of deformation produced by a force depends on the force per unit area. Example: A force causes greater elongation on a thin wire than a thicker wire because the cross-sectional area of a thin wire is smaller. when the forces are removed, the object tends to its original shape. This is a deformation that exhibits elastic behavior. 35  There are two types of deformations: Elastic deformation - Deformation where a material returns to its original shape or size when the force acting on it is removed. Plastic deformation - Deformation where a material does not return to its original shape when the force acting on it is removed. 36  Elasticity of solids Elasticity: The property of a material which enables it to return to its original dimension once the force acting on it is removed. Elastic Limit: The maximum amounts by which an object or a material can be stretched and still regain its original shape after the distorting forces are removed. Beyond the elastic limit, permanent deformation occurs, and the material does not return completely to its original dimensions. Plastic deformation occurs when a material is deformed beyond its elastic limit. 37  Stress and strain. The elastic properties of solid are discussed in terms of stress and strain. Stress (s) is a measure of force causing a deformation and is Figure below shows a wire of defined as force per unit cross length L, that is stretched with sectional area a force F, so that it experiences an extension ∆L. SI unit for stress 𝐹 𝑆𝑡𝑟𝑒𝑠𝑠, s = is N m-2 or Pa 𝐴 Force applied perpendicular to the ∆L cross- sectional area 38 Continued… The extension produced depends on the original length of wire; extension per unit original length or strain. Strain (e) is a measure of the degree of deformation of a solid when subjected to a stress. 𝐸𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛 ∆𝐿 e = Δx = AL 𝑆𝑡𝑟𝑎𝑖𝑛, e = = 𝑂𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝐿𝑒𝑛𝑔𝑡ℎ 𝐿 - Strain has no unit (dimensionless) 39  Elastic Modulus or Young’s Modulus The elastic modulus is the constant of proportionality between stress and strain For sufficiently small stresses, the stress is directly proportional to the strain The constant of proportionality depends on the material being deformed and the nature of the deformation 𝑆𝑡𝑟𝑒𝑠𝑠, s 𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑚𝑜𝑑𝑢𝑙𝑢𝑠, = 𝑆𝑡𝑟𝑎𝑖𝑛, 𝜺 The elastic modulus can be thought of as the stiffness of the material. A material with a large elastic modulus is very stiff and difficult to deform. 40  Extension-Force and Stress – Strain Graphs The Extension-Force and Stress –Strain Graph for a steel wire (ductile material) Stress Strain Figure A Graph of Extension-Force Figure B Graph of Stress- Strain 41  Extension-Force Graph and Stress – Strain Graph A : Limit of proportionality (maximum stress where stress and strain are proportional, Hooke’s law is valid) B : Elastic limit (a point slightly above point A) Along OB (elastic deformation), wire return to its original length when the force is removed C : Yield point (plastic deformation starts) Point C to Point D (plastic deformation), the extension OX (or strain) becomes permanent i.e sliding between atomic planes occurs. D : Breaking Stress (force is maximum) After point D, the wire extent drastically with a little increase in force (or stress) E : Wire breaks 42 Continued… From point O to point A, the force is directly proportional to the extension. This is in accordance with Hooke’s law which states that the force applied is directly proportional to the extension if the proportional limit is not exceeded. That is F ∝ ∆x F = k ∆x where k is force constant (force required to produce a unit of extension in the wire) k depends on the type of materials. 43  Elastic and Plastic Deformation ❖Ductile - The property of materials that undergo plastic deformation before breaking. A ductile material can be worked out into a shape or can be permanently deformed without breaking. Example: gold, steel, copper and aluminum ❖Brittle – The property of material that do not show plastic behaviour. This characteristic is reflected in the graph as a straight line up to the breaking point. Example: glass, bone ❖A brittle material cannot be permanently deformed and fractures once the elastic limit is exceeded. 44  Rigidity or stiffness of an object is the extent to which the object can resist deformation such as stretching, bending or compressing. A material that requires a greater stress to produce the same strain than another material is said to be more rigid. 45 Stress Extension Strain Force Stress- strain graph for various Extension-force graph for materials various materials 46  Polymer ❖ Property of rubber departs from Hooke’s law and behaves elastically, regaining its original length when the load applied is removed. However, polythene undergoes plastic deformation producing permanent deformation. ❖ Vulcanized rubber undergoes elastic hysteresis since its follow different curve for increasing load and decreasing load. ❖ The area enclosed by the curve represent the energy lost per unit volume in the form of heat dissipation. Stress Stress Strain Strain Stress- strain curve for rubber Stress- strain curve for polythene 47  Strain energy The energy stored in a body due to deformation is called strain energy. If the proportional limit is not exceeded, the strain energy is given by 2 W= 3 F𝑥 F = kx 2 or W= 3 k𝑥2 This is equal to the area under the force-extension graph Force Strain energy = work done 𝟏 𝑭𝒙 𝟐 x Extension Force-extension graph 48  Strain energy per unit volume 2 Strain energy per unit volume = 3 stress × strain 𝑉𝑜𝑙𝑢𝑚𝑒, 𝑉 = 𝐴𝐿 # % ∆( = $ & ) 2 2 𝜎 = 𝜀𝑌 =3 𝜎𝜀 = 3 𝜀 3Y = area under the stress-strain graph Stress Strain Stress-strain graph 49  Young’s modulus Three Types of Elastic Moduli: Shear modulus Bulk modulus 50  (a)Young’s Modulus : Elasticity in length ❖Young’s Modulus, Y is a quantity that represents the elasticity in length of a material. ❖It is defined as the ratio of tensile stress to tensile strain Tensile stress is the ratio of the external force to the cross- sectional area. Tensile strain is the ratio of the change in length, ΔL to the original length, L. Strain is dimensionless 51  %&'()*& (+,&(( Young’s Modulus, Y = %&'()*& (+,-)' !"#$% &#"''('%$)*"+,- ,#%, =.-"+/,)*"+ "#*/*+,- -%+/)0 !. 1 = ∆3. 34 / 04 = 1 ∆0 The SI unit for Young’s Modulus is N m -2 or Pa 52  To find the value of Young’s modulus, Y from the graphs of Force-Extension or Extension-Force find k from the gradient of Force-Extension graph, 𝑭 k = ∆𝑳 or 1/k from the gradient of Extension-Force graph ∆𝑳 1/k = 𝑭 𝑭! 𝑭𝑳 𝑳 Y= 𝑨 ∆𝑳! = =𝒌 𝑳 𝑨∆𝑳 𝑨 53  Table 1: Typical Values for Elastic Moduli A material having a large value of Young’s Modulus is very stiff and difficult to be stretched or compressed. Young’s modulus applies to a stress of either tension or compression. 54  Ultimate Strength of Materials The ultimate strength of a material is the maximum stress (force per unit area) the material can withstand before it breaks or fractures. Some materials are stronger in compression than in tension Table 2: 55 Example 1: A cylindrical brass rod with Young’s Modulus 9.7 × 1010 Pa and original diameter 10.0 mm experiences elastic deformation when a tensile load of 200 N is applied. (a) Calculate the stress that produces the deformation (b) If the original length of the rod is 0.25 m, calculate the change in the length of the rod. Solution e = ∆𝑳 56 Example 2: A vertical steel beam in a building supports a load of 6.0 × 104 N. (a) If the length of the beam is 4.0 m and its cross-sectional area is 8.0 × 10-3 m2, find the distance the beam is compressed along its length. (b) What maximum load (in newtons) could the steel beam support before failing? Solution (a) Compressive strength of steel (b) (refer Table 2) 57 Exercise : Two wires X and Y of the same materials and the same length are subjects to the same load respectively. The diameter of wire X is half that of Y. If the strain in wire X is 5.0 × 10-4, then the strain in wire Y is ….. Solution *+,-** Young’s Modulus, Y = *+,./0 εy = 1.25 × 10-4 58  (b) Shear Modulus: Elasticity of Shape Shear Modulus (S) - Type of deformation that occurs when an object is subjected to a force F parallel to one of its faces while the opposite face is held fixed. The stress is called a shear stress. The shear stress displaces the top face of the block to the right relative the bottom. 59  Shear stress is the ratio of magnitude of parallel force, F to the area, A of the face being sheared. 𝑺𝒉𝒆𝒂𝒓 𝒇𝒐𝒓𝒄𝒆 𝑭 Shear stress = = 𝒂𝒓𝒆𝒂 𝒐𝒇 𝒔𝒖𝒓𝒇𝒂𝒄𝒆 𝑨 Shear strain is the lateral (horizontal) displacement in the direction of the shear force per unit perpendicular distance between the shear forces. or Shear strain is the ratio of the horizontal displacement and the height of the object 𝒉𝒐𝒓𝒊𝒛𝒐𝒏𝒕𝒂𝒍 𝒅𝒊𝒔𝒑𝒍𝒂𝒄𝒆𝒎𝒆𝒏𝒕 ∆𝒙 Shear strain = = 𝒉𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒐𝒃𝒋𝒆𝒄𝒕 𝒉 60 Example Shear Modulus When the book on the table is pushed by a pair of equal and opposite forces F to the plane of the area A, a shear deformation is produced. 61 The shear stress displaces the front cover of the book to the right relative the back cover  The shear modulus, S for a certain material is given as the ratio of shear stress to shear strain. !"#$% !&%#!! !. 1 Shear Modulus, S = !"#$% !&%$'( = ∆7. 0 For small shears within the elastic limit, the shear strain is directly proportional to the shear stress. Therefore, shear stress is related to the shear strain according to C ∆E =S Similar to Hooke’s law D F A material having a large shear modulus is difficult to bend. No change in volume in this deformation. 63 Example 3: A 500 N shear force is applied to upper face of an aluminum cube measuring 15 cm on each side. Calculate the displacement of the upper face relative to the opposite face? [Shear modulus for aluminum is 2.5 ´ 1010 N m-2] 15 cm Solution: /0123 /431// 15 cm Shear Modulus, S = F = 500 N /0123 /43256 15 cm 1G 2 S= ∆3G 4 CF ∆𝑥 = DH 64  (c) Bulk Modulus: Volume Elasticity ❑ Bulk modulus characterizes the response of an object to uniform squeezing. o Suppose the forces are perpendicular to, and acts on all the surfaces. Example: when an object is immersed in a fluid. ❑ The object undergoes a change in volume without a change in shape. 65  Bulk Modulus Example: (Vi + ∆V) When a solid is under uniform pressure, it undergoes a change in volume but no change in shape. This cube is compress on all sides by forces normal to its six faces. 66 ❑ Volume stress, ΔP, is the ratio of the force to the surface area. This is also the Pressure 𝑭 Volume stress = 𝑨 ❑ The volume strain is equal to the ratio of the change in volume to the original volume. also called as bulk strain 𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒐𝒍𝒖𝒎𝒆 ∆𝑽 Volume strain = = 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒗𝒐𝒍𝒖𝒎𝒆 𝑽 qUnder uniform bulk stress, the cube shrinks in size without changing shape. 67 For small volume changes within the elastic limit,, volume strain is directly proportional to the change in pressure, ∆𝑃. ∆5 ∆5 ∆𝑃 ∝ ∆𝑃 = B 5 5 𝑽𝒐𝒍𝒖𝒎𝒆 𝒔𝒕𝒓𝒆𝒔𝒔 Therefore, Bulk modulus, B = 𝑽𝒐𝒍𝒖𝒎𝒆 𝒔𝒕𝒓𝒂𝒊𝒏 ∆𝑭7 ∆𝑷 B=− ∆𝑽7 𝑨 =− ∆𝑽7 𝑽 𝑽 -ve sign - *increase in pressure causes a decrease in volume and vice versa. 68  A material with a large bulk modulus is difficult to compress. The compressibility is the reciprocal of the bulk modulus. 69 Example : A solid sphere of volume 0.50 m3 is initially surrounded by air, and the air pressure is exerted on it is 1.0 × 105 N m-2 (normal atmospheric pressure) is dropped in the ocean to a depth of about 2000 m (about 1 mile) where the pressure is 2.0 × 107 N m-2. Lead has a bulk modulus of 7.7 × 109 Pa. What is the change in volume of the sphere? ∆𝑭G Solution: B=− 𝑨 =− ∆𝑷 ∆𝑽G ∆𝑽G 𝑽 𝑽 𝑽∆𝑷 ∆𝑽 = − 𝑩 L.N O 9 (3.L ×2L 2L:; R2.L × 2L< S O :! ∆𝑽 = − =−𝟏. 𝟑×1089 𝒎9 7.7 × 109 Pa The negative sign indicates that the volume of the sphere is decreases. 70  QUICK QUIZ 1. Which of the following is the stress-strain graph for porcelain which is stretched until it is broken? 71 2. The behaviour of a wire when it is extended by a tensile force is represented by the graph of stress against strain as show in figure below Which of the following deduction from the graph above is not true A. Young’s modulus for the materials of the wire is the same as the gradient of the straight line OW B. The shaded area below the straight line OW is the same as strain energy C. The atomic plane in the wire slide over one another when the wire is extended from X to Y D. A permanent deformation Ɛo is produced when the stress exerted on the wire at point X is removed. 72 3. Which of the following best relates to the stiffness of solids A. Strain B. Elastic limit C. Deformation points D. Young’s modulus 4. The thigh bone of a man with an effective cross-sectional area of 2.5 × 10-4 m2 can withstand stress of 1.5 × 108 Pa. The maximum compressive force that can act on the bone is A. 32 kN B. 38 kN C.42 kN D.46 kN 73  Summary Young’s modulus Shear modulus Bulk modulus 74 Thank you 75 https://www.youtube.com/watch?v=DLE-ieOVFjI&list=PPSV 76