Tiêu đề MECHANICAL PROPERTIES OF SOLID.pdf ✅

 Digital www.allendigital.in [ 217 ] Introduction to types of forces and types of bodies Types of forces Deforming Forces External force which tends to bring about a change in the length, volume or shape of a body is called deforming force. Restoring Forces When an external force acts at any object then an internal resistance produced in the material due to the intermolecular forces which is called restoring force. Elasticity Elasticity is that property of a material of a body by virtue of which it opposes any change in its shape or size when deforming forces are applied on it, and recover its original state as soon as the deforming force are removed Types of Bodies Rigid Body A body is said to be rigid if the relative positions of its constituent particles remain unchanged when external deforming forces are applied to it. The nearest approach to a rigid body is diamond or carborundum. Perfectly Elastic Body A body which perfectly regains its original form on removing the external deforming force, is defined as a perfectly elastic body. Example : quartz. – It is quite close to a perfect elastic body. Plastic Body A body which does not have the property of opposing the deforming forces, is known as a plastic body. All bodies which remain in the deformed state even after the removal of the deforming forces are known as plastic bodies. Example : clay, wax, putty. Stress and its types, breaking stress Stress The restoring force acting per unit area of the deformed body is called stress. internal restoring forces stress area of cross sec tion = external F (at equilibrium) stress A = SI Unit: N/m2 Dimensions: [M1L–1T–2] Mechanical Properties of Solids Types of Stress Longitudinal Stress Volumetric Stress Shear Stress Tensile Stress Compressive Stress Properties of Matter and 05 Fluid Mechanics Fr F
NEET : Physics [ 218 ] www.allendigital.in  Digital Types of stress (i) Longitudinal Stress When the stress is normal to the cross sectional area, then it is known as longitudinal stress. Longitudinal Stress = F A ⊥ There are two types of longitudinal stress: (a) Tensile Stress The longitudinal stress, produced due to increase in the length of a body, is defined as tensile stress. (b) Compressive Stress The longitudinal stress, produced due to the decrease in the length of a body, is defined as compressive stress. (ii) Volume Stress/Hydraulic Stress If equal normal forces are applied over every unit surface of a body, then it undergoes a certain change in volume. The force opposing this change in volume per unit area is defined as volume stress. Volume Stess/Hydraulic Stress = F P A ⊥ = (iii) Shear Stress/Tangential Stress When the stress is tangential or parallel to the surface of a body then it is known as shear stress. Due to this stress, the shape of the body changes or it gets twisted but not its volume. Shear stress = Tangential t F F A A = (iv) Breaking Stress The minimum stress required to cause the actual fracture of a material is called the breaking stress or ultimate strength. or The maximum stress that a material can bear is called breaking stress. Breaking stress = max F A Fmax = force required to break the body. Dependence of breaking stress: (i) Nature of material (ii) Temperature (iii) Impurities. Independence of breaking stress : (i) Cross sectional area or thickness (ii) Applied force (Fmax = B.S. × Area). Maximum load (force) which can applied on the wire depends on (i) Cross sectional area or thickness (ii) Nature of material (iii) Temperature (iv) Impurities Fmax Fmax F F F F
Properties of Matter and Fluid Mechanics  Digital www.allendigital.in [ 219 ] Illustration 1: The ratio of radii of two wires of same materials is 2 : 1. Find if they are stretched by the same force, the ratio of stress: Solution: Stress = force area = 2 F r  1 2 (stress) (stress) = 2 1 F r × 2 2 r F  = 2 2 1 r r       = 2 1 2       = 1 4 Illustration 2: A body of mass 10 kg is attached to a 30 cm long wire whose breaking stress is 4.8 × 107 N/m2. The area of cross section of the wire is 10-6 m2. What is the maximum angular velocity with which it can be rotated in a horizontal circle? Solution: 2 m L A  = breaking stress (BS)   = (BS)A mL = 7 6 4.8 10 10 10 0.3 −    = 4 rad/s Illustration 3: The breaking stress of aluminium is 7.5 × 108 dyne/cm2. Find the maximum length of aluminium wire that can hang vertically without getting broken under its own weight. Density of aluminium is 2.7 g/cm3. Given : g = 980 cm/s2. Solution: Let L be the maximum length of the wire that can hang vertically without getting broken. Mass of the wire, m = cross-sectional area (A) × length (L) × density () Weight of the wire = mg = ALg This is equal to the maximum force that the wire can withstand.  Breaking stress = LA g A  = Lg Or 7.5 × 108 = L × 2.7 × 980  L = 8 7.5 10 2.7 980   = 2.834 × 105 cm = 2.3 km Strain and its types Strain change in the dimension of the body Strain original dimension of the body = It is a unitless and dimensionless quantity. Types of Strain Longitudinal Strain Volumetric Strain Shear Strain Tensile Strain Compressive Strain
NEET : Physics [ 220 ] www.allendigital.in  Digital (i) Longitudinal strain: When applied force is ⊥ to cross-section, length changes Longitudinal strain = change in length of the body initial length of the body = L L  (ii) Volumetric strain: When pressure is applied on body, volume changes Volume strain = change in volume of the body V original volume of the body V  = (iii) Shear strain: When applied force is ∥ to cross-section, shape changes tan = x L (Here  is very small) displacement of upper face relative x to the lower face L distance between two faces  = = φ = shear strain OR angle of shear Relation between angle of twist and angle of shear When a cylinder of length L and radius r is fixed at one end and a tangential force is applied at the other end, then the cylinder gets twisted. Figure shows the angle of shear  and angle of twist . From diagram r = L r L  =  = angle of twist  = angle of shear Illustration 4: A wire of length 1m and radius 2mm is clamped from it’s upper end. The lower end is twisted through an angle of 45°. The angle of shear is? Solution:  = r L  = 3 2 10 1 −  × 45° = 0.09° F⊥ F⊥ F⊥ F⊥ L L F x L F  L   r