# Introduction, Method of Sections

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1 Lecture #1 Introduction, Method of Sections Reading: 1:1-2 Mechanics of Materials is the study of the relationship between external, applied forces and internal effects (stress & deformation). An understanding of statics is essential. At least one half of every mechanics problem is a statics problem. Free body diagrams, equations of equilibrium, centroids & area moments of inertia, structural analysis (trusses & frames) and shear and bending moment diagrams are topics that are used extensively. To extend statics to the determination of internal forces we use the method of sections. If a body is in equilibrium, any part of the body must also be in equilibrium. This concept means that we can cut a body in two in order to expose a cross section on which we want to determine internal forces, draw a free body diagram of one of the parts (putting appropriate forces on the cut face, i.e. internal reactions), and apply equations of equilibrium. In general, for a three dimensional problem, there are six reactions on any cross section (3 forces & 3 moments). For two dimensional problems, this can be reduced to three reactions (2 forces & 1 moment). Note that the normal force is perpendicular to the cross section while shear forces are in the plane of the cross section. A torsional moment vector is perpendicular to the cross section and the bending moment vectors are in the plane of the cross section.

2 Lecture #2 Introduction to Normal and Shear Stress Reading: 1:3-5 Stress is the intensity with which a force affects a point in a body. Stress is broken into components in the same way that we break forces into components. There is a normal stress that we represent by Normal stress is the part of the normal force carried by an infinitesimal area at a point on a cross section divided by the differential area. There is also a shear stress that we represent by This is the part of the shear force carried by an infinitesimal area at a point on a cross section in a particular coordinate direction divided by the differential area. Recognize that there are two components of shear stress at a point on a cross section. From this we see the units on stress are force per unit area. In SI units, we would use a Pascal (Pa) which is a N/m 2, or a Mega Pascal (MPa) which is 10 6 Pa or a N/mm 2. You will notice that when I calculate stress in SI units, I will normally use N and mm. This way the stress always has units of MPa. In US units, we use psi which is a lb/in 2, or a ksi which is a kip/in 2 or 10 3 psi. You will notice that when I calculate a stress in US units, I will normally use kips and inches. We never calculate a stress in terms of feet in mechanics (always use inches).

3 The state of stress of a point is a picture showing all stresses that act on an infinitesimal element at the point. The general three dimensional state of stress is shown in the figure below. There are equal and opposite stresses on the back faces which are not shown for clarity. The subscripts represent the direction of the normal stress or the plane of the shear stress. Note that there are six components of stress for the general three dimensional state of stress. If we consider the general two dimensional state of stress, we see three components of stress. There are two normal stresses and one shear stress. A normal stress may be tensile or compressive. It is important to note the difference by either writing ( T ) or ( C ) after the stress magnitude or putting a sign on the magnitude with tension being positive and compression being negative. Do not use both conventions together since a negative sign negates the word. In other words, -10 ksi ( C ) is that same as +10 ksi ( T ). Later in the semester, we will be concerned about the direction of a shear stress, but for now we will only calculate a magnitude without a sign. If the author puts a sign on shear stress in an example or solution at this point, I want you to ignore it.

4 Let s consider a simple example of each type of stress. The Axially Loaded Bar If we pass a cutting plane through the bar perpendicular to its axis, we see that the cross section must carry a normal force equal to P. The force is not actually concentrated at a point, but rather is distributed over the area. The average normal stress can be calculated by σ Avg = P/A If the bar is homogeneous and isotropic, and if the resultant force acting on the cross section passes through the centroid of the cross section, then the actual stress is uniformly distributed and equal to the average σ = P/A A truss is an example of a structure that is composed of axially loaded bars. Direct Shear Stress An example of direct shear is the lap joint. Here, we have two plates held together by a bolt or a rivet. If there is an axial force applied to the plates and friction between the plates is negligible, then the bolt or rivet must transfer the force from one plate to the other. Cutting the bolt or rivet between the plates and looking at a free body diagram of one plate, we see that the force is transferred between the plates via a shear force on the bolt or rivet cross section.

5 The average shear stress on the cross section can be calculated Avg = V / A The shear stress is never uniformly distributed, but we often use this equation when dealing with small areas such as the area of a bolt or rivet. The lap joint described above is a single lap joint. One cross section of the bolt or rivet carries the force. A double lap joint creates a situation called double shear. You will note that the force is transferred between the plates via two cross sections of the bolt or rivet. If we cut the bolt or rivet between the plates and consider a free body diagram of the center plate, we see that each cross section of the bolt or rivet carries half of the force. When calculating the average shear stress, we can consider one area with half of the force or two areas with the entire force. Either way, the stress is halved. Avg = V / A = (P/2) / A = P / (2A)

6 Lecture #3 Introduction to Design & Factor of Safety Reading: 1:6-7 Here we introduce the idea of design or the sizing of a member based on a comparison of the largest stress experienced and the allowable stress that the material can safely carry. We must also introduce here the concept of factor of safety. The factor of safety is defined as the ratio of the maximum stress a material can carry before some form of failure occurs and the allowable stress, i.e., the stress that we are willing to accept in the material. F.S. = fail / allow or F.S. = fail / allow The failure stress is defined by the type of failure which might occur. Most often, we will define the failure stress to be the yield stress, i.e., that stress that causes the material to yield. Fail = yield or fail = yield We always allow a stress that is below the failure stress, i.e., the factor of safety is always greater that 1. The factor of safety is sometimes called the factor of ignorance because we introduce the concept to deal with uncertainty. We don t allow stresses right up to the failure stress because of uncertainties that exist. The failure stress is a statistical quantity (an average with some standard deviation). We are not certain of the exact value for any specimen of the material. The actual stress is based on applied forces which may be approximations (not known with certainty). Anyway, a comparison between the maximum stress experienced and the allowable stress for the material can be used to determine the minimum required size of the cross section. max = P / A = allow A req = P / allow max = V / A = allow A req = V / allow

7 Another stress introduced in these sections is called a bearing stress. This is a compressive stress on the surfaces of two contacting bodies. Often this stress has a complex distribution so a nominal value is calculated. For example, consider the bearing stress created by the contact between the bolt and the inside of the hole in the plate of a lap joint. Cutting the bolt between the plates and then cutting one of the plates through the hole, we see in the figure on the next page that the force P acts on half of the inside of the hole. The nominal bearing stress is calculated from a projected area - the diameter of the hole times the thickness of the plate. bearing = P / ( t d )

8 Lecture #4 Introduction to Normal & Shear Strain Reading: 2:1-2 Applied forces create deformation, i.e., changes in size and shape. Changes in size come from normal strain. We represent the normal strain using and define it as a change in length per unit length. The assumption here is that if the material is homogeneous, then the change in length will be equally distributed along the length. = L / L o Changes in shape come from shear strain. We represent shear strain using and define it as the change in a right angle. Note that strain is unitless. Therefore, must be in radians. There are 6 components of strain. There are three normal strains - one associated with each coordinate direction ( x, y, z ). There are also three shear strains - one associated with each plane ( xy, xz, yz ). We will generally assume that strains are small, i.e., << 1.

9 Lecture #5 Material Behavior & Properties Reading: 3:1-4,6-7 Material properties and strengths are vitally important in mechanics. To learn about a material, we run standardized tests. One of these tests is a tension test. In a tension test, we pull on a specimen and measure the force and the change in a prescribed gauge length. If we plot P vs L, we don t actually learn anything about the material because a specimen that has a different cross section area or a different gauge length will behave differently. To eliminate the dependence on the cross section, we plot stress instead of force. To eliminate the dependence on length, we plot strain instead of the change in length. A plot of vs will then be substantially the same for multiple test specimens of the same material but different sizes and lengths. Most material vs diagrams share some common characteristics. To discuss these characteristics, lets consider the diagram for a mild steel. Generally, there is an initial region which is linear. This is called the proportional region. In this region, stress and strain are related to one another by the slope of the curve which is a property called the modulus of elasticity or E.

11 The maximum stress a material can withstand is called the ultimate strength, u. Once the material reaches this stress, a local region of the member will begin to experience necking. Necking is a narrowing of the cross section at the place of ultimate failure. Because of the reduced cross section, the applied load can be decreased and the member continues to stretch and the neck narrow until ultimate rupture. Earlier, it was mentioned that not all materials exhibit a well defined yield point. In this event, it is customary to use the 0.2% offset stress as the yield strength. Here, we start from a strain value of and draw a line parallel to the proportional region until it intersects the stress vs strain curve. When we pull on a specimen, we also note that it gets smaller in the directions perpendicular to the load. This is called the Poisson effect. Another material property is the Poisson Ratio, = - (transverse strain)/(axial strain) = - t / a The negative sign is included to make a positive quantity since t will be negative when a is positive.

12 We have only discussed normal stress and strain. There is also a standard test for determining a shear stress vs shear strain diagram. The diagram for shear will be different, but the diagram will have analogous attributes to the normal stress vs normal strain diagram. There is a proportional region in which we can write the shear form of Hooke s Law, = G. G is the shear modulus or slope of the proportional region. There is a yield strength for shear, y, an ultimate strength, u, and other similar behaviors. The shear modulus is a third material property, but it is not independent of E and. It has been shown that G = E / (2(1+ )) There is more that we could discuss about material behavior, but this discussion is sufficient for our purposes. The back of the textbook has a table of properties and strength values for some typical engineering materials.

14 = d = ( P / (AE)) dx If P and/or A is a function of x, we will need to evaluate an integral.. If P, A and E are all constant, then = ( P / (AE)) dx = (PL) / (AE) If P, A and E are all constant over sections of the length, then = ( (PL) / (AE) ) i

15 Lecture #7 Statically Indeterminant Axially Loaded Members Reading: 4:4 A member is statically indeterminant if the number of unknown reactions is greater than the number of equilibrium equations. In an axially loaded member, any reactions will also be axial forces. First, we draw a free body diagram of the member and then apply our one equilibrium equation, F x = 0. In a structure that contains axially loaded members we may have additional equilibrium equations. Next, we look to write constraint equations which involve displacements and are based on how the problem is supported. If a bar is supported at each end and not allowed to change length, then our constraint equation would be δ = 0 We will work other problems where displacements of two members are the same or related to one another in some way.

16 Lecture #8 Temperature Effects on Axially Loaded Members Reading: 4:6 Now, we would like to add temperature effects. If we have an unconstrained bar, and we increase the temperature, the bar will change length by d T dx Where is the coefficient of thermal expansion. Within a reasonable temperature range, may be considered a constant. The material property table in the back of the textbook gives values for If T is a function of length, then T dx If T is constant, then T L If we have a member that is exposed to an axial force and a temperature change, then ( ( PL ) / ( AE) ) + T L For a member with multiple sections with constant P, A, E, and T, then PL ) / ( AE) ) + T L ] i The change in temperature does not affect the stress in the member unless the member is statically indeterminant and then the effect can be drastic. When a member is constrained and the temperature is increased, significant reaction forces may be developed in order to keep the material from expanding due to the temperature change.

17 Lecture #9 Torsion of Cylindrical Members Reading: 5:1-3 Torsion of members having a circular cross section Consider a bar with a circular cross section. First, we make a grid on the member of lines along the length and around the circumference. Now, we place a torsional moment on the member and watch how the grid deforms. We note a couple of things from the deformation. First, the vertical lines which represent a cross section in the two dimensional view remain vertical. This means that the cross section remains planar. In other words, a cross section rotates as a rigid disk. Secondly, if we consider the shape of a single element of the grid, we see the deformation gives the shape that we saw earlier caused by shear stress..

18 Now, consider a slice of the torsion member. We will consider the left side of the slice as being fixed and the right side as having a torque applied The radial line CA will rotate to a new position CA. The angle d is the angle of twist of the cross section and the deformation angle of the line PA is the max shear strain, max. The arc length AA can be written two ways. AA dx max = R d If we consider a point B a distance r from the center, it moves to position B. Because the angle d if the same, the shear strain of an element on this smaller cylinder will be smaller. We will call this angle The arc length BB can be written as BB dx r d From these equations, we see two expressions for d /dx. d /dx = max / R = / r So, max ( r / R ) From this we see that shear strain varies linearly with the distance r measured from the center of the cylindrical member. Now, assume that the material remains within the linear, elastic range of its behavior. We can apply Hooke s Law. = G G max ( r / R ) = max ( r / R ) We see then that the shear stress varies linearly from the center of the cross section. Now, lets remember that the stress distribution on the cross section must be statically equivalent to the net torque, T.

19 Considering an arbitrary element of area, da, on the cross section a distance r from the center which has a stress da = a net force df df r = a net torque around the center dt dt = r df = r da = r max ( r / R ) da dt = max / R ) r 2 da If we integrate over the cross section area, we get an equation for max. T = dt = ( max / R ) r 2 da = ( max / R ) r 2 da Note that r 2 da = J ( Polar Moment of Inertia). Now, we have J = ( / 2 ) R 4 for a solid circular cross section J = ( / 2 ) ( R o 4 - R i 4 ) for a circular tube max J / R or max = T R / J Substituting back into the equation for, we get a general equation for the shear stress. = max r / R = ( T R / J ) ( r / R ) = T r / J A torsion member may carry a torque which varies along its length. It may be a function of length, or constant in sections. We recognize that if the applied torques are concentrated, the internal torque will be constant between the applied torques. The internal torques are determined using the method of sections. An arbitrary cut must be made in each section of the member, a free body diagram must be drawn, and an equilibrium equation written to determine the internal torque distribution. The maximum shear stress in the member can be found by calculating the maximum shear stress in each section of the member and comparing. If the cross section is constant, we recognize that the maximum will come from the maximum T. But if the cross section changes, the maximum will come from where T r / J is maximum which could be any section.

20 Torsion members are often used to transfer power from a power source to a machine which will do work. Power is the rate of doing work and is related to the torque carried and the angular velocity of the shaft. P = T Units are important when using this equation. In SI units, P is in Watts which is a ( N-m / s ) so T will be in ( N-m ) and in ( rad / s ). In English units P is typically given in horsepower (hp). One hp is equal to 550 ft-lb / s. After this conversion, T will be in ( lb ft ) and in ( rad / s ). Make sure you use the correct units in this equation. We are also able to design a shaft at this point. If we know the allowable shear stress, allow, then So, from this allow = ( T R / J ) max J req / R req = T / allow For a solid cross section, From which R 4 / R = T / allow or R 3 = T / allow R = ( T / ( allow 2 Keep in mind that this is a minimum required size. Sign convention note: When finding the internal torque, we generally assume positive as a vector pointing away from the cross section. However, the sign of the torque is usually ignored in a stress calculation. The material does not distinguish between a shear stress in one direction or the other.

21 Lecture #10 Angle of Twist for Cylindrical Torsion Members Reading: 5:4 From the previous lecture, we learned that d dx = / r where is the angle of twist. Assuming the material to be linearly elastic, we can say So = / G and = T r / J d /dx = T r / ( G J r ) = T / ( G J ) For a general torsion member, we can find the angle of twist by integrating this equation. d = ( T / ( G J )) dx = ( T / ( G J )) dx If T, G or J are functions of length, we need to integrate. If these quantities are constant over the length, then = ( T / ( G J )) dx = T L / ( G J ) If T, G and J are constant in sections, then we can add together the twist from each section. ( T L / ( G J )) i When calculating the angle of twist, the sign convention for T is important. We must carry the sign into the equation for because the angle of twist for two sections could be in opposite directions, canceling one another to some degree over the entire length.

22 Lecture #11 Statically Indeterminant Torsion Members Reading: 5:5 As in the axial load problem, if we have more reactions than equations of equilibrium, the problem is statically indeterminant. First, we draw a free body diagram of the member. In this case, our one equation of equilibrium is a moment equation. T = 0 Then, we write constraint equations as needed involving the angle of twist based on how the member is constrained. If for example, a member is held at each end, we would say that φ = 0. We will solve other problems where the angle of twist of two members are the same or related to one another.

23 Lecture #12 Torsion of Noncylindrical Members Reading: 5:6-7 When the cross section was circular, we noted that cross sections remained planar after a torque was applied. This does not happen when the cross section is noncircular. The cross section will warp out of plane. For solid noncircular cross sections, the theory is beyond the scope of this course. The textbook presents some equations for 3 common solid cross sections. See the table in the textbook for these equations. Although the theory is beyond this course, there is no reason why we cannot use the results of that theory to solve problems. Another type of noncircular cross section which we can analyze is the thin walled tube. Although the wall must be thin (generally an order of magnitude smaller than the major inplane dimensions), the thickness of the wall does not have to be constant. Since the wall is thin, we will assume that the shear stress is constant through the thickness and tangent to the wall centerline. Consider a piece of the tube wall Assume a shear stress 1 at the corner of the thickness t 1 and a shear stress of 2 at the corner of thickness t 2. Because shear stress always occurs in a set of four stresses, we know the stress in the wall along the length of our piece. Where the thickness is t 1, the shear stress is 1, and where the thickness is t 2, the shear stress is 2.

24 Writing F x = 0, 1 t 1 L - 2 t 2 L = 0 or 1 t 1 = 2 t 2 We made no assumptions about the size of the piece so L and s are arbitrary. This means that t = constant We define this product as a new quantity called shear flow, q. Note this is a force per unit length. Looking back at the cross section, the shear flow around the circumference of the wall must be statically equivalent to the applied torque. Over a small piece of the circumference, q creates a net force df = q ds which in turn creates a torque around the axis of the shaft dt = q ds h The total torque is then found by integrating around the circumference. T = dt = q h ds But q is constant, so T = q h ds

25 The integral is geometry dependent only and is equal to twice the area enclosed by the wall centerline. T = q 2 A m where A m = area enclosed by wall centerline Now, using t = q, we get T = t 2 A m or = T / ( 2 A m t ) The maximum stress obviously occurs where the thickness is smallest. The angle of twist can be derived as well to find = (( T L ) / ( 4 A 2 m G ) ( ds / t ) if T is constant over the entire length. ( ds / t ) is evaluated knowing how t varies with the circumference variable s. If t is constant, ( ds / t ) = s / t If t is constant in sections of the wall, ( ds / t ) = ( s / t ) i

26 Lecture #13 Shear & Bending Moment Diagrams Reading: 6:1 A beam is a structure that carries transverse loads. This means that a cross section will carry a shear force and a bending moment. We want to be able to plot shear and bending moment over the length of the beam in order to determine the maximum value of these quantities. So that we all have the same diagrams, we introduce a sign convention. Positive shear rotates the beam clockwise and positive bending moment points to the top of the cross section. A positive bending moment creates compression at the top of the beam and tension at the bottem of the beam. One way we can plot these quantities is to use the method of sections to find V and M as functions of length. First, we must recognize that we may not be able to represent V and M as a single functions over the entire length. The beam must first be divided into sections where V and M can be represented by single functions. This occurs between points of concentrated force or moment, supports, or a change in the functional representation of a distributed load. Once we know how many sections we have, we can draw a free body diagram of the whole beam and use equilibrium to calculate the reaction forces. Then, we use the method of sections to make a cut in the interior of each section at an arbitrary distance x measured from the left hand end of the beam. At this point, we draw a free body diagram of one part of the beam assuming V and M in their positive directions and apply equilibrium equations to determine functions for V and M in terms of x. Then we simply plot the functions.

27 Lecture #14 Shear & Bending Moment Diagrams Reading: 6:2 If we consider a small piece of beam with a distributed load and draw a free body diagram with positive internal forces, we get The shear and moment on the right side will be different from the left because of the distributed load, but not much different because dx is small. Let V 2 = V 1 + dv and M 2 = M 1 + dm, then apply equilibrium equations. F y = 0 : V 1 + w dx - ( V 1 + dv ) = 0 w dx - dv = 0 dv = w dx or dv/dx = w M = 0 : ( M 1 + dm ) - M 1 - V 1 dx - w dx ( dx / 2 ) = 0 dm - V 1 dx - w dx 2 / 2 = 0 Since dm and dx are small, dx 2 is very small and can be neglected. We can also drop the subscript on V at this time. dm = V dx or dm/dx = V These equations are the basis of what I call the summation method. dv/dx = w says that the slope of the shear diagram is equal to the distributed load. Integrating dv = w dx, we get dv = w dx V 2 - V 1 = w dx or V 2 = V 1 + w dx This says that the shear at some point x 2 equals the shear at the starting point x 1 plus the area under the distributed load curve between the two points. dm/dx = V says that the slope of the moment diagram is equal to the shear force. Also, since the maximum moment occurs where dm/dx = 0, we know that the maximum moment will be where the shear force is zero.

28 Integrating dm = V dx, we get dm = V dx M 2 - M 1 = V dx or M 2 = M 1 + V dx So, the moment at a point x 2 equals the moment at the starting point x 1 plus the area under the shear diagram between the two points. These equations are valid between concentrated forces and moments. We also know that at points of concentrated forces we get a discontinuity or a jump in the shear diagram in the direction of and equal in magnitude of the force. At points of concentrated moments, we get a discontinuity or a jump iin the moment diagram equal in magnitude to the concentrated moment. The direction of the jump is up when the moment is clockwise and down when counterclockwise. Let s state a procedure for the summation method. 1. Draw a free body diagram of the beam and use equilibrium to determine the reaction forces. 2. Draw the shear diagram using the following rules. a. Start at the left end at zero b. When you encounter a concentrated force, follow the force. That is jump in the direction of the force the magnitude of the force. c. Between concentrated forces, apply V 2 = V 1 + w dx to find the shear at the end of the section and dv/dx = w to draw the line over the section. 3. Draw the moment diagram using the following rules. a. Start at the left end at zero. b. When you encounter a concentrated moment, jump up if the moment is CW or down if the moment is CCW a magnitude equal to the magnitude of the moment. c. Between concentrated moments, apply M 2 = M 1 + V dx to find the moment at the end of a section and dm/dx = V to draw the line over the section. Notes: 1. Both diagrams must return to zero at the right end of the beam. If one of them does not, then a mistake has been made. An area may have been calculated incorrectly or the reactions may be wrong. 2. The maximum moment will occur at an end or where dm/dx = 0. Therefore, you must find M at all places where V = This method breaks down if the distributed load is higher order than constant. For these sections, we go back to the method of sections to find the functions of x.

29 Lecture #15 Centroids and Area Moments of Inertia Reading: A:1-3 You were introduced to these topics in statics so we will only review quickly. However, centroids and moments of inertia are very important concepts in mechanics of materials. In general, the centroid of an area is found by x c = x da / A y c = y da / A where ( x c, y c ) are the coordinates of the centroid and A = da. In practice, we do not often need to integrate because most areas we encounter can be treated as a composite area - an area made up of simple shapes. Given that we have tables that locate the centroid of simple shapes for us, we can simplify the equations to x c = A i x ci / A y c = A i y ci / A Where ( x ci, y ci ) are the coordinates of the centroid of the simple shape A i and A = i. We know that if there is an axis of symmetry, the centroid must be located on that axis. In general, area moments of inertia are found from I x = y 2 da I y = x 2 da I xy = x y da J = r 2 da where I xy is the product of inertia and J is the polar moment of inertia. Since r 2 = x 2 + y 2, we can show that J = I x + I y. We know that I x, I y, and J are always positive, but I xy can be positive or negative.

30 Again, in practice, we don t often need to integrate because most areas we encounter are composite areas. Therefore, we can simplify the equations to I x = I xi I y = I yi I xy = I xyi J = J i Where the subscript i represents that the moment of inertia is for the i th simple shape. Although we have tables that give information about moments of inertia for simple shapes, the information given is generally for specific axes through the simple shape. Most often, the given equations are with respect to axes that pass through the centroid. The parallel axis theorem allows us to determine the moments of inertia for any set of axes parallel to the axes that pass through the centroid. I x = I cx + A d y 2 I y = I cy + A d x 2 I xy = I cxy + A d x d y J = J c + A d 2 Where ( d x, d y ) are the coordinates of the centroid, the bar above the moment of inertia designates that the moment of inertia is with respect to a centroidal axis and d 2 = d x 2 + d y 2. Often, information about the product of inertia is not given in the tables provided. If there is an axis of symmetry, I xy = 0. Therefore, we only need equations for unsymmetric shapes. For these shapes, the sign will depend on the orientation. For example, consider the right triangle. Using the parallel axis theorem, we can write I x = ( I cxi + A i d yi 2 ) I y = ( I cyi + A i d xi 2 ) I xy = ( I cxyi + A i d xi d yi ) J = ( J ci + A i d i 2 ) In statics, you probably tabulated the information when finding a centroid or moment of inertia. I will no longer do that, but feel free to make a table if that helps you.

31 Lecture #16 Principal Axes and Principal Moments of Inertia Reading: A:4 Consider an area in a coordinate system x,y which is rotated an angle to x,y. If we want to find the moments of inertia with respect to the x,y axes, we could integrate. I x = ( y ) 2 da I y = ( x ) 2 da I x y = ( x )( y ) da These integrals can be written in terms of the x,y coordinate system if we relate x and y to x and y. Note that x = x cos + y sin y = y cos - x sin Now, we can substitute these equations into our integrals. Consider I x. I x = ( y ) 2 da = ( y cos - x sin ) 2 da I x = ( y 2 cos 2 + x 2 sin 2 x y sin cos ) da Now, we can separate the integral into three integrals. Recognizing that respect to the integration, we get is a constant with I x = y 2 da cos 2 + x 2 da sin 2 x y da sin cos Recognizing the definitions of I x, I y and I xy, I x = I x cos 2 + I y sin 2 - I xy 2 sin cos

32 So we found I x in terms of I x, I y and I xy. Assuming we can find the moments of inertia for one orientation, we can now find them for any other orientation. This equation is typically used in a different form. Knowing the trig identities cos 2 = 1/2 ( 1 + cos 2 ) sin 2 cos 2 ) and 2 sin cos sin 2 we can substitute to find I x = [ ( I x + I y )/2 ] + [ ( I x - I y )/2 ] cos 2 - I xy sin 2 Following the same procedure, we also find I y = [ ( I x + I y )/2 ] - [ ( I x - I y )/2 ] cos 2 + I xy sin 2 and I xy = [ ( I x - I y )/2 ] sin 2 + I xy cos 2 The importance of these equations is that we can determine the maximum or minimum values of the moments of inertia. Consider finding the extreme values of I x by taking a derivative and setting to zero. di x /d = 0 = - [ ( I x - I y )/2 ] 2 sin 2 - I xy 2 cos 2 [ ( I x - I y )/2 ] sin 2 = I xy cos 2 tan 2-2 I xy / ( I x - I y ) This locates the extreme values of I x. There are two solutions for 2 which are different by 180 o or. Your calculator always gives the value between +90 o and - 90 o or between + and - /2. Solving for, our two roots differ by 90 o. This means that we only need one of the roots, because the second root is the opposite axis. One of the roots gives a maximum value while the other gives a minimum value. Therefore, the maximum and minimum values of moment of inertia occur for the same coordinate system - one being I x and the other I y.

33 If you compare the equations for di x /d and I xy, you will notice that they are the same except for a factor or 2. Therefore, when we have the max and min values for I x and I y, the product of inertia is zero ( I x y = 0 ). This coordinate system is called the principal axes and the moments of inertia are called the principal moments of inertia. This will become important very soon. The maximum and minimum values can also be determined from I max,min = [ ( I x + I y ) / 2 ] ± { [ ( I x - I y )/2 ] 2 + I xy 2 } 1/2 The problem with this equation is we don t know which is I x and which is I y so we still need to substitute back into one of the equations so that we know which is which. In many circumstances, we can tell which is which by looking at the area in relation to the axes x, y. If you can see that the area is concentrated closer to one axis and spread farther away from the other, then the axis which has the area farther from it has the maximum moment of inertia.

34 Lecture #17 Stress from Pure Bending Reading: 6:3-4 Consider the behavior of a beam which has a cross section with a vertical axis of symmetry under the action of pure bending. If a grid of lines is marked on the beam along the length and vertically before the moment is applied, the response of the beam to the moment can be noted. After the moment is applied, the beam takes a curved shape. We note that the lines along the length become arcs and the vertical lines become radial, intersecting at a common center of curvature. We note that the top of the beam is in compression while the bottom is in tension. Somewhere in between, there must be a point that has no stress or strain. If there were a line along the length there, it would not have changed length. Because vertical lines remain straight, we know that the cross section remains planar. Now, consider an infinitesimal length of the beam. Let the length be dx. The line that does not change length represents a plane through the thickness called the neutral plane or neutral surface. The intersection of the neutral surface with the cross section is called the neutral axis and is represented by a point in the diagram. The radius of curvature,, is defined as the distance from the center of curvature to the neutral surface. The angle made by the radial lines will be called d. We know then that dx = d. We define a distance, y, as the distance measured from the neutral surface upward. The change in the length of an arbitrary fiber, y above the neutral surface, can be seen to be du = - y d

35 The strain along this fiber would be du/dx = (-y d d -y / We see that the strain varies linearly from the neutral surface. If we assume that the material stays within the linear, elastic region of its behavior, then E E y So stress varies linearly. The maximum stress or strain occurs at the point farthest from the neutral surface. Imagine this distance is to the top and call it c. Now looking at a plot of versus y, max = - E c / So we could write max y / c This stress distribution must be statically equivalent to the forces on the beam cross section. Therefore, P = da = 0 or ( max / c) y da = 0 Since max and c are constants y da = 0 We remember that this integral is used to locate the centroid. Therefore, y da = A y = 0 This tells us that the neutral surface must pass through the centroid of the cross section. So the neutral axis is a centroidal axis. Also, the stress distribution must be equivalent to the moment on the cross section. M = - y da

36 The negative sign is included because M goes in the opposite direction caused by a tensile stress. Substituting the equation for stress, M = - ( max / c) y 2 da = - ( max / c) y 2 da Note that y 2 da = I, the moment of inertia of the cross section about the neutral axis. So now, M = - max I / c or max = - M c / I Substituting back into the equation for, we get max y / c = - M y / I This equation assumes that y is measured upward from the neutral axis and that M is pointed toward the top of the cross section. It will be to our advantage later to eliminate the need for a sign on each of these terms. Instead, from this point on the equation will be written as = M y / I where M and y are always taken as positive (magnitudes without signs) and then a sign is placed on the final stress value based on whether the stress is tensile(+) or compressive(-). This can be determined from the point where you want to calculate the stress and the direction of the bending moment. Remember that the moment always points to the side of the cross section that is in compression.

37 Lecture #18 Unsymmetric Bending Reading: 6:5 When we derived the equation for bending stress, we assumed that the cross section was symmetric, i.e. I xy = 0. Actually, = - M y / I can be applied to any cross section with a bending moment as long as the moment is around a principal axis and I is a principal moment of inertia, i.e. I xy = 0. We could consider two different situations. 1. What if the cross section is symmetric, so we know where the principal axes are, but the moment vector is at an angle to the principal coordinate system? Consider a rectangular cross section with a moment vector at an angle from the principal x axis in the plane of the cross section. If we break the moment vector into components about the x and y axes, we can apply the bending stress equation to each moment and use superposition to add the stresses together. In the equation = M y / I, we must recognize that each letter is a placeholder or a dummy variable. The moment must be about a principal axis, the moment of inertia must be with respect to the same axis as the moment, and y represents a distance measured perpendicular to the axis the moment is around. Consider the stress due to M x. The equation used will be = M x y / I x. There must be a sign placed on the stress based on whether the stress is tensile or compressive. If we want, we can write a general equation where we allow y to be positive or negative based on the coordinate of the point where we want to calculate the stress. If you consider a point where y is positive, note that the moment M x causes tension based on the given direction of the moment. So in general, = + M x y / I x.

38 Now repeat this process for M y to find = - M y x / I y. Using superposition, we write = + M x y / I x - M y x / I y. Now, we have an equation which is good for any point in the cross section. We only need to substitute the appropriate coordinate values of the point where we want to find the stress. This general equation allows for the determination of the neutral axis, the line of zero stress in the cross section. By setting our general stress equation equal to zero, we get the equation of a line. 0 = + M x y / I x - M y x / I y or y = (( M y I x )/( M x I y )) x We can plot the line or determine the angle it makes with the coordinate axes. We can also see where the maximum stress occurs from the general equation by finding the point farthest away from the neutral axis. In this case, the maximum occurs at points A and B where the stress at A is tensile and the stress at B is compressive. If we are only interested in finding the stress at one point, we do not need to develop a general expression. We can instead take all placeholders in the equation = M y / I to be positive and place a sign on the stress depending on whether it is tensile or compressive at the point. Again, the sign of the stress can be determined from the direction of the moment and the location of the point relative to the axis the moment goes around. For example, the stress at B could be written as = - M x (h/2) / I x - M y (b/2) / I y The difference is that we did not write a general equation and take the x and y values to the be coordinates of the point of interest.

39 The other type of problem we could be faced with is one where we have a general cross section which does not have an axis of symmetry with a moment applied. Consider, for example, a cross section as shown with a moment around the x axis. Let s assume that we can calculate the moments of inertia I x, I y, I xy and the product of inertia is not zero. We could use equations developed in an earlier lesson to find the principal axes and the principal moments of inertia. Imagine the principal axes to be located by the angle p and that we can determine I x and I y. Now, the moment can be broken into components in the x and y directions and we can write a general equation for stress in terms of x and y or determine the stress at a specific point as before. The difficult part of this second problem type is determining the coordinate values to be substituted into the equation. Some geometry may be required. The point where the maximum stress occurs may not be obvious. The easiest way to find the maximum stress for a general cross section is to locate the neutral axis as before, then determine which point in the cross section is furthest away from that axis. Then we can perform the necessary geometry and trigonometric manipulations required to find the coordinates.

40 Lecture #19 Shear Stress from Shear Forces Reading: 7:1-3 Consider a piece of a beam with constant shear force. Again, let s assume there is a vertical axis of symmetry. From equilibrium, (M+dM) - M - Vdx = 0 dm = V dx V = dm/dx So the moments on the two sides are not the same. If we consider the stress caused by the bending moments (note this assumes that the material is linearly elastic), the stresses must be larger on the right hand side. Now, let s measure an arbitrary distance y from the neutral axis and slice the piece horizontally. If we look at the piece we have cut off, we see that because the net horizontal force on each side is different, a shear force develops on the cut horizontal face. F 1 = da = ( M y / I ) da F 1 = ( M / I ) y da In the same way, F 2 = ( (M+dM) / I ) y da We recognize that y da = A y c Let s define this quantity as Q = y da = A y c = y c

41 Now, from equilibrium, F x = 0 : df + F 1 = F 2 df = F 2 - F 1 = ((M + dm) Q / I) - (M Q / I) = dm Q / I = V dx Q / I df/dx = VQ/I This is a force per unit length of the beam. If we divide by the thickness of the bottom cut surface, we get stress. = (df/dx)/t = VQ/(I t) Note that by looking at an infinitesimal element at the corners of the bottom surface, we find the stress on the cross section is equal to that found on the bottom surface. Also note that the shear stress on the cross section is in the same direction as the shear force on the cross section. Let s consider the shear stress variation on a rectangular cross section. To calculate the stress at an arbitrary location y above the neutral axis, first draw a line perpendicular to V through the cross section at the point of interest. Q is found from the area above this line. The thickness t is the length of the line that cuts through material. Q = A y = b (h/2 y) (h/2 + y)/2 = b/2 ((h/2) 2 - y 2 ) t = b = VQ/(I t) = [ V (b/2)((h/2) 2 - y 2 )]/[(bh 3 /12) b]

42 We see immediately that the stress varies parabolically or quadradically with y. If y = ± h/2, we see that = 0. The maximum value of shear stress occurs when y = 0. max = [V b/2 (h/2) 2 ]/[(bh 3 /12) b] = [12 V b h 2 ]/[8 b 2 h 3 ] = 3V/[2 bh] = 3V/(2A) We have just shown that the maximum shear stress for a rectangular cross section is 50% larger than the average shear stress. In general, the maximum shear stress will occur where Q/t is maximum since V and I are constants. We should note that Q is always maximum at the centroid. The thickness t of the material along the line drawn through the cross section, however, can change the location away from the centroid if it is small enough elsewhere. Keep in mind that t represents the amount of material along the line drawn through the cross section perpendicular to the shear force at the point of interest. So a cross section with a cutout would lower the value of t. This equation for shear stress is a troubled equation. There are times when it does not give an accurate stress value. In other words, it has limitations. Consider, for example, a circular cross section. If we consider a point a distance y above the neutral axis, we could calculate the shear stress from = VQ/(I t) Which acts parallel to V along the line through the cross section at y. Now, consider an infinitesimal volume element at the edge of the cross section. The boundary of the cylindrical member is called a free surface if a stress is not applied to it. The free surface has no stress or is free from stress. This means that on the cross section, there can be no stress perpendicular to the boundary. The calculated stress which is in the direction of V and was calculated to be non zero has a component in the direction perpendicular to the boundary. So there is a contradiction.

43 The actual stress will be zero at the boundary and there will be a varaiation in stress along the line with = VQ/(I t) being the average stress along the line. The shear stress equation will give the actual stress where the boundary is vertical since the free surface then is parallel to the shear force and the shear stress from the equation has no component perpendicular to the free surface. So, we can find the stress at the centroid of the circular cross section which also turns out to be the maximum stress. = VQ/(I t) Q = A y = r 2 / 2 [ 4 r / (3 r 3 / 3 = V [2 r 3 / 3] / [( r 4 / 4)(2 r)] = 4 V / [3 r 2 ] = 4V / (3A) The maximum shear stress for a circular cross section is 33% above the average value. Another limitation occurs with thin wall open sections like wide flange beams. The top and bottom of the flange are free surfaces. This says that the stress at A is zero. However, we can calculate a nonzero stress value at A using our equation. For these types of cross sections, we apply the equation to the web only which works out fine since the max shear stress is at the centroid which is in the web. Special note: When finding Q, we can take the area above or below the cut. Remember that Q=A y for area above the cut where we want to calculate the stress. However, A y = A y = 0 for the entire cross section since we are measuring y from the centroid already. Therefore, A y above = - A y below. Taking the area below the cut gives the same Q except with the opposite sign. If we take the absolute value, we get the correct value. This can work to our advantage when the area below the cut is simpler than the area above as shown.

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