REVIEW OF MECHANICAL CONCEPTS || EC 465 MEMS || S7 ECE KTU || LECT 7 || MODULE 1

hi students, today we are discussing about some mechanical concepts we have heard about stress and strain, right? today we are discussing about some mechanical concepts first, let's see about stress stress is developed in a response of mechanical loading what does stress develop as a part of mechanical loading? stress is developed in response to mechanical loading it all depends on Newton's 3 laws of motion Newton's 3 laws of motion are the basis for all these everyone knows Newton's laws of motion Newton's first law is every body continues a state of rest or of uniform motion unless an external force is applied to it Newton's second law is F is equal to ma force is equal to mass into acceleration Newton's second law of motion is this equation first law is moment of inertia second law is force Newton's third law is every action there is equal and opposite reaction based on these laws MEMS devices analyze static and dynamic behaviours during loading these laws are used to analyze force distribution inside a material which gives rise to stress and strain force distribution inside a material how does it happen? how does it turn into stress and strain? these laws of motion are used to analyze this this figure is a bar this bar is embedded in a brick wall an axial force F is applied to it this force is transmitted through the bar the wall produces an equal and opposite reaction according to Newton's third law this is what is shown in the figure the next diagram is we are removing the wall we are removing the wall to analyze it instead of removing it, we have drawn some actions that is the free body diagram this free body diagram of bar clearly reveals that the wall must provide an axial force with equal magnitude but opposite direction to the applied force you can see that it produces an axial force opposite direction to the applied force there is an equal force why is it considered as equal force? because the bar is stationary stationary means the total force on the bar is zero that is why it is considered as equal and opposite force that is why the axial force is equal to the applied force the next step is to find the hidden forces we have cut it into a section since the bar is in equilibrium condition all the parts of the bar will be in equilibrium this is what is shown in the diagram this is what we said consider a bar firmly embedded in a brick wall with an axial force applied at the end an axial force is applied at the end of the bar since the force is transmitted through the bar to the wall the wall must produce a reaction according to Newton's third law the wall would act on the left end of the bar with an unknown force to expose and quantify this force we imaginarily remove the wall and replace it with the actions it imparts on the bar this free body diagram of the bar clearly reveals that the wall must provide an axial force with equal magnitude but opposite direction to the applied force so that the total force on the bar is zero to maintain its stationary status we can use this technique to expose and quantify hidden forces and stress at any section since the bar is in equilibrium any part of it must be in equilibrium as well we can pick an arbitrary section of interest and imaginarily cut the bar into two halves since the force applied at the free end of the bar is equal but opposite force must develop at the cross section we applied an axial direction force earlier in this figure, we are applying a transverse direction force we have isolated a bar here the sum of forces and moments acting on the isolated bar must be zero for the net force to be zero a force of same magnitude but opposite side must act on the end of the bar attached to the wall the net force here must be zero since we are applying a transverse force we are applying a downward force to maintain its stationary status the upward force must develop in the wall this is what is shown in this diagram when we apply a downward force what will the upward force produce? it will produce a wall we have a pair of forces opposite to each other what will this pair of forces create? it will create a torque you can see the equation for torque what will this pair of forces produce? it will produce a torque there will be a reactive torque the reactive torque will be F times L but the sign will be opposite it must act on the end of the bar attached to the wall the end of the bar attached to the wall will produce a torque opposite to it we have cut this as an imaginary section after cutting if we look at the left hand piece the net force and net torque will be zero this is the transverse direction we are applying a force the bar we have considered is the isolate bar sum of forces and moments acting on isolate bar will be zero this is the transverse direction we are applying a downward force opposite to it, a force will produce a wall what will this pair of forces produce? it will produce a torque the equation for torque is length of the force length of the bar is called torque the net force and net torque on the left hand side will be zero now we have some definitions for stress and strain we can classify mechanical stress in two ways normal stress and shear stress you can see this diagram this diagram shows normal stress this is the normal stress in loaded and unloaded condition this is the normal shear in loaded and unloaded condition you can get the idea of normal stress and shear stress from this figure let us see what is normal stress we are pulling a rod if we give a longitudinal pull, what will be the length of the rod? it will increase if we increase the length of the rod, what will happen? tension will develop if we cut the rod, we can get the idea of internal stress the intensity of the force is called stress if the stress acts in the direction perpendicular to the cross section, then it is called normal stress if the stress acts perpendicular to the cross section, then it is called normal stress normal stress is commonly denoted as the force applied on a given area SI unit is known as Newton per meter square or Pascal unit is known as force per area force per area is known as Newton per meter square normal stress is denoted as sigma SI unit is known as Newton per meter square or Pascal we took a rod and pulled it what happened to the rod? normal stress developed if we push the rod, what will happen? stress will develop normal stress can be tensile or compressive tensile stress will develop when we pull the rod compressive stress will develop when we push the rod if the volume is pulled in one particular axis, the stress is tensile if the volume is pushed, the stress is compressive the unit of elongation of the rod represents the strain in this case, we can call the strain as normal strain suppose a steel bar, its original length is L0 when we applied normal stress, its length is L the resultant strain developed in the bar is S is equal to L minus L0 divided by L0 or delta L divided by L0 delta L is the change in length divided by the original length that is the strain the unit of elongation of the rod represents the strain we said that we pulled a rod to apply normal stress when we pull the rod, the length of the rod increases not only length increases, but also reduces the cross sectional area so we can say that the relative dimensional change occurs in Y and Z direction and it can be expressed as EY and EZ the general material characteristic is captured by a term called Poisson's ratio the Poisson's ratio is defined as the ratio between transverse and longitudinal elongation denoted by the letter V V can be expressed as SY divided by SX sorry, I wrote EY it is correct, XY relative dimensional change in Y is denoted by SY correct it XY divided by SX or XZ divided by SX that is the Poisson's ratio Poisson's ratio is the ratio between transverse and longitudinal elongation what is stress and strain? stress and strain are closely related stress is proportional to strain there is a law called Hooke's law what is Hooke's law? stress is proportional to strain under small deformation, the stress and strain terms are proportional to each other Hooke's law equation is sigma is equal to E into S sigma is stress, S is strain and E is modulus of elasticity modulus of elasticity is called Young's modulus it is an intrinsic property of a material modulus of elasticity is a constant value each material has a different value it does not depend on its shape, dimension or mechanical element modulus of elasticity is an inter-atomic force when a force changes its shape, it returns to its original state it regains its original shape or restore its previous value modulus of elasticity is a process that returns to its original state stress and strain are categorized as two normal stress or strain and shear stress or strain so far we have seen what is normal stress and normal strain what is shear stress? shear stress can be developed under different force loading conditions shear stress is developed under different force loading conditions this figure is a diagram of normal stress and shear stress it is the same diagram we learned earlier when shear stress is applied, two different forces are applied when two different forces are applied in a cube, the shape of the cube varies the magnitude of shear stress is defined as tau is equal to F divided by A the force divided by area is called stress shear stress is denoted by tau tau is equal to force divided by area its unit is newton per meter square shear stress has no tendency to elongate or shorten the element in x, y and z directions shear stress does not have any tendency to elongate or shorten the element what does shear stress do? it changes the shape of the element the original element we saw in the figure was a rectangular shape when shear stress is applied, it changes to an oblique parallel pipe the shape of the pipe changes shear strain is the extent of rotational displacement when shear stress is applied, it has a rotational displacement shear strain is the extent of that shear strain is defined as the extent of rotational displacement what is shear stress? sorry, there is a correction shear strain is unitless because shear strain is the extent of rotational displacement shear strain is unitless it represents the angular displacement expressed in the unit of radians shear stress does not elongate or shorten the element it changes the shape of the element shear strain is the extent of rotational displacement shear stress and shear strain are related we learned about the modulus of elasticity in Hooke's law similarly, shear modulus of elasticity is also related shear modulus of elasticity is denoted by g so g is equal to tau divided by gamma the unit of g is newton per meter square because gamma is unitless so unit of g is newton per meter square the value of g depends only on the material not the shape and dimension of the object the value of g depends only on the material not the shape and dimension of the object the relation between modulus of elasticity and shear modulus of elasticity and Poisson ratio is g is equal to E divided by 2 into 1 plus V this is the relation between g, E and V so today we learned about stress and strain that is, some mechanical, review of mechanical concept that is what we discussed today I hope everyone understood see you in the next lecture, thank you