ME 208 Class No. 11876 S. F. Felszeghy 
Fall 2005 

STATICS AND STRENGTH OF MATERIALS 
WEEK  DATE  TOPICS  PROBLEMS 
1  Sep. 26  1.1  1.6; 2.1  2.3  110, 113, 115, 118 
Sep. 28  2.4  2.7  23, 213, 221, 249  
2  Oct. 3  2.8, 2.9; 3.1  3.4  274, 291, 311, 333 
Oct. 5  3.4  3.9  349, 366, 375, 390  
3  Oct. 10  3.9; 4.1  4.3  397, 399, 42, 49 
Oct. 12  4.3  4.6  414, 418, 434, 437  
4  Oct. 17  5.1  5.5  51, 59, 517, 533 
Oct. 19  MIDTERM  
5  Oct. 24  5.5; 6.1, 6.2  545, 558, 567, 65 
Oct. 26  6.2  6.7  622, 637, 653, 673  
6  Oct. 31  7.1; 8.1, 8.2  723, 730, 87, 822 
Nov. 2  8.3  8.5  825, 837, 844, 853  
7  Nov. 7  8.6  8.8  858, 863, 865, 867 
Nov. 9  MIDTERM  
8  Nov. 14  9.1  9.6  92, 95, 913, 922 
Nov. 16  9.7; 10.1  10.3; 11.1  *920, 105, 1015, 1021  
9  Nov. 21  11.2  11.4  115, 1117, 1130, 1137 
Nov. 23  11.4; 7.2, 7.3  733, 741, 759, 765  
10  Nov. 28  12.1, 12.2  121, 122, 129, 1223 
Nov. 30  Review 
FINAL EXAM: Monday, December 5, 4:30  7:00 p.m.
Lecture 1
Notebook containing solutions to problems is available in CSULA Library. Call No. is 989.
Concepts and Key Words
Introduction
Statics and Strength of Materials (the latter also known as Mechanics of Materials) are branches of (classical) mechanics. Mechanics deals with the response of physical bodies to the action of applied forces. By response what is meant is the stresses (or the stress resultants, also known as internal forces) in a body, and the deformation and motion of a body .
Mechanics is oldest subject of engineering.
Principles and theories of classical mechanics apply to relatively massive and slowly moving bodies compared to atomic sized particles and the speed of light.
Classical mechanics encompasses (a) mechanics of particles and rigid bodies: statics and dynamics, and (b) mechanics of continua (deformable extended bodies): solids  strength of materials, elasticity, plasticity, viscoelasticity, etc.; fluids  ideal, viscous, compressible, etc. Statics deals with the action of forces on bodies at rest. Strength of materials deals with the stresses in a body and deformation of a body that result from applied forces.
STATICS
Some basic concepts and definitions. Mass (inertial and gravitational  they are the same). Force (vector quantity). Particle. Rigid body. Time.
Newton's laws of motion. First: A particle remains at rest or continues to move with a uniform velocity in a straight line if there is no unbalanced force acting on it. Second: The acceleration of a particle is proportional to the resultant force acting on it and is in the direction of this force. Third: The forces of action and reaction between interacting bodies are equal in magnitude, opposite in direction, and collinear.
In statics, we apply the first and third laws. The first law is a statement of the condition when forces are in state of balance or equilibrium. Third law states that forces always occur in pairs. To distinguish between action and reaction, we isolate bodies in freebody diagrams. More about this later.
Law of gravitation. Weight (W). Gravitational acceleration at surface of earth: g = 9.81 m/s^2 or 32.2 ft/sec^2.
Units. SI Units (Systè International). Fundamental quantities and units: length (m), time (s), mass (kg). Force (N) derived. USCU (U. S. Customary Units). Fundamental quantities and units: length (ft), time (sec), force (lb). Mass (slug) derived. W = mg in both systems (m is mass).
Vectors
Quantities that can be represented by directed line segments and which combine according to the parallelogram law of addition (or triangle rule of addition) are called vectors, or geometric vectors to be precise. Prototype is the displacement of a particle.
Lecture 2
Vector notation: boldface in print, e.g. A, underlined letters when handwritten. Vector algebra. Magnitude or length of vector denoted by italics in print, e.g. A. Equality of two vectors. Sum or resultant of two vectors. Properties of vector addition. Multiplication of a vector by a scalar. Properties of this multiplication. Zero vector. Unit vector.
Resultant of coplanar forces concurrent at a point. To find overall resultant F_R of set of forces F_i, apply repeatedly triangle rule or parallelogram law of addition to pairs of vectors. Thus, F_R = F. Graphical, and analytical approach using trigonometry to find resultant. Example 1: law of cosines and law of sines. The resultant F_R must be stated as a magnitude and as an arrow with an angle of inclination measured from some reference direction.
Resolution of a vector into vector components. Reverse application of parallelogram law of addition.
Rectangular vector components and rectangular scalar components of a vector. Righthanded rectangular coordinate system: x, y, z. Orthogonal projections of vector A, with tail at origin O, onto x, y, z axes produces vector components A_x, A_y, A_z. Rectangular coordinates of head of A are scalar components A_x, A_y, A_z. Rectangular base vectors i, j, k. Hence, A_x = A_x i, A_y = A_y j, A_z = A_z k. Direction angles that A makes with the x, y, z axes: , and . Hence A_x = A cos(), A_y = A cos(), A_z = A cos(), A = sqrt[(A_x)^2 + (A_y)^2 + (A_z)^2]. Example: finding rectangular scalar components of a force when its magnitude and angular orientation are given.
Lecture 3
Addition of forces by summing rectangular scalar components. Resultant is F_R. The scalar components are F_Rx = F_x, F_Ry = F_y, F_Rz = F_z. Example: The twodimensional problem in Example 1.
Relative position vectors. The relative position vector r_MN is a vector that starts at point M and ends at point N. Thus, r_MN = (x_N  x_M)i + (y_N  y_M)j + (z_N  z_M)k. Form unit vector u_MN = r_MN/r_MN. A force of magnitude F directed from point M to point N can be written as F = F u_MN.
The dot (scalar, inner) product. A*B = A B cos(), where is the angle measured from A to B. Geometric interpretation and algebraic properties of dot product. The result of A*u, where u is a unit vector, is the scalar component of A in the direction of u. Because i, j, k are orthogonal, A*B = A_x B_x + A_y B_y + A_z B_z.
The cross (vector) product. A x B = C, where C = A B abs[sin()], where is the angle from A to B, and C is so directed that A, B, and C form a righthanded triad. Geometric interpretation and algebraic properties of the cross product. Because i, j, k are orthogonal, A x B = det [ i, j, k; A_x, A_y, A_z; B_x, B_y, B_z], where "det" stands for "determinant."
Moment of a force about a point. Consider a force F and a point O not on the line of action of force F. The force and point define a plane. The moment M_O of F about point O is defined to be M_O = Fd, where d (called the moment arm) is the perpendicular distance from point O to the line of action of force F. If force F were to act on a rod one end of which is supported at O by a ballandsocket joint, then the rod would tend to rotate about an axis passing through O and oriented perpendicular to the plane defined by F and point O. This motivates the representation of the moment of F about point O as a vector M_O whose magnitude equals M_O and whose direction represents the axis and sense of the tendency of rotation according to the righthand rule, that is, when the fingers of the right hand curl in the direction of rotation, the thumb points in the direction of the moment vector M_O.
Lecture 4
Let r be a relative position vector from point O to any point on the line of action of F. Then, it can be verified that M_ O = r x F. Example: In threedimensions, moment of a force with given magnitude and pointing from one point to another point, about a third point.
Resultant moment of a system of forces. M_R_O = (r x F).
Varignon's theorem. Moment of resultant (sum) of concurrent individual forces equals sum of moments of individual forces.
Moment of a force about a given directed axis. Let u be a unit vector in the positive direction of a directed axis OL that passes through point O. Then, M_L = M_O * u = (r x F) * u is called the moment of F about axis OL. Result doesn't change if O replaced by O', another point on axis OL. Example: Moment of a force about a shaft axis, with force applied to a gear mounted on shaft.
Scalar (mixed, triple) product. (A x B) * C = det[A_x, A_y, A_z; B_x, B_y, B_z; C_x, C_y, C_z].
The x, y, z components of M_O are the moments of F about the x, y, z axes, respectively. In twodimensional problems in the xy plane, M_Oz = x F_y  y F_z, where x and y are the coordinates of a point on the line of action of F and M_Oz is positive counterclockwise. Subscript z sometimes dropped.
Resultant of coplanar forces, not concurrent at a point. Example: Body subject to several forces. Slide a pair of forces to point of concurrency of their lines of action. Find resultant. Repeat sliding and combining with resultant and another force. Continue until overall resultant F_R found. Note: force and moment balance on body is unchanged by this sliding and combining of forces. Analytically, can find overall resultant and location of its line of action from: F_Rx = F_x, F_Ry = F_y, (F_R d_R) = (Fd), where the d's are moment arms measured from a common moment center, and each Fd product is assigned a plus or minus sign according to the sense of the moment. The previous process may terminate with a pair of forces equal in magnitude, oppositely directed and not collinear, F and F. Such a pair of forces is called a couple. A couple has the property that its moment about any point is the same, or Fd, where F is the magnitude of either force, d is the perpendicular distance between the two forces, and account has been taken of the sense of the moment. This motivates representing the couple by a couple moment vector whose magnitude is Fd and whose direction is given by the righthand rule. Since the force and moment balance on a body is unchanged when a couple is moved parallel to itself, a couple or the couple moment can be applied anywhere on the body.
Couples in threedimensions. Couple moment is M = r_AB x F, where F is one of the couple forces and points A and B are on the lines of action of F and F. Couple moments add as vectors.
Resolution of a force into a force and a couple. The force and moment balance on a body is unchanged if a force is "moved" transversely and parallel to itself and a compensating couple moment is added to the body to maintain moment balance.
Resultant of any force system. The force and moment balance on a body is unchanged if all the forces are "moved" to a common arbitrary origin, say O, and compensating couple moments are introduced. The forces and couple moments can then be combined into a resultant force F_R = F, and a resultant couple moment M_R_O = (r x F) = M_O, where r is a relative position vector from O to any point on the line of action of F. Rectangular component forms of these equations: F_Rx = F_x, F_Ry = F_y, F_Rz = F_z, M_Rx = M_Ox, M_Ry = M_Oy, M_Rz = M_Oz.
Lecture 5
Equilibrium of a particle. A particle is in equilibrium when the resultant of all forces acting on it is zero. F = 0 is necessary and sufficient for equilibrium. Example: Tensions in knotted cords supporting mass in gravitational field.
Equilibrium of a rigid body. A body is in equilibrium when the resultant force and the resultant couple moment acting on the body are both zero. F = 0 and M_O = 0 are necessary and sufficient for equilibrium.
Freebody diagram. Before force and moment equations of equilibrium can be applied to a body, the body must be identified together with all the forces and couple moments acting on it. This is done by isolating the body from all other bodies in the form of a freebody diagram and identifying all forces and couple moments, known and unknown. Known forces and couple moments are drawn in their known directions, unknown forces and couple moments are drawn in assumed directions. All forces and couple moments are labelled with their magnitudes, symbolic or numerical. If in the solution of the force and moment equations of equilibrium, the answer for an unknown force or couple moment magnitude is negative, the actual direction of the unknown force or couple moment is opposite to that assumed.
Forces acting on a body may be support forces and support couple moments (called reactions), gravitational forces and contact forces. Types of supports: cable tension, pin, ball joint, roller, fixed.
Lecture 6
Equations of equilibrium. In three dimensions: F_x = 0, F_y = 0, F_z = 0, M_Ox = 0, M_Oy = 0, M_Oz = 0.
In two dimensions: F_x = 0, F_y = 0, M_O = 0.
Equilibrium in two dimensions. Example: A loaded beam with roller and pin supports.
Special cases: 1) Body with collinear forces. 2) Body with concurrent forces. 3) Body with parallel forces. 4) Body with general forces.
Alternative force and moment equilibrium equations. 1) Two force equations; one moment equation. 2) One force equation; two moment equations. 3) Three moment equations.
Equilibrium in three dimensions. Example: Ballandsocket and cable supported rod.
Some simple equilibrium cases: 1) Forces concurrent at a point. 2) Forces concurrent with a line. 3) Parallel forces. 4) General loading.
Two common equilibrium situations: 1) Body with two forces (must be collinear). 2) Body with three forces (must be coplanar and concurrent).
If there are more unknowns than equations of equilibrium available, then problem is statically indeterminate.
Lecture 7
Structures. An engineering structure is a system of interconnected members built to support or transfer forces and to withstand safely the loads applied to it. Will examine how forces internal to a structure can be determined. Will analyze trusses, frames, and machines. Will examine only statically determinate cases, that is, cases which can be analyzed entirely with equations of equilibrium only.
Plane truss. Basic unit of truss is triangle formed by three straight bars pinned together at their ends. This is a simplest noncollapsible structural unit. The triangle can be extended to create a larger noncollapsible structure by adding pairs of bars which are pinned to the already existing fixed joints. The resulting framework is called a simple truss. When all the members of a truss lie in one plane, truss called plane truss. Examples: Warren bridge truss; Howe roof truss. Trusses are designed to have all loads applied at the pins. Thus, truss members are twoforce members. Truss members are either in tension, compression or unloaded. Weight of members usually neglected.
Two methods of analysis are used to solve for forces in truss members: method of joints and method of sections.
Method of joints. Dismember entire truss into freebody diagrams of joint pins and members. Apply force equations of equilibrium in x and y directions to pins. Analyze joints in such a sequence that there are never more than two unknown member forces at any joint. Method of sections. Make exploratory separation cut through truss, cutting through no more than three members with unknown internal forces. Apply force and moment equations of equilibrium to one part of cut truss. Example: Howe truss.
Lecture 8
Frames and machines. Method of members. Pinconnected structures with members subjected to more than two forces are called frames (if structure supports loads) and machines (if structure is a mechanism that transforms an input force into an output force). Often, such structures can be analyzed by dismembering and applying equilibrium equations to parts. Example: Threebar frame; pliers.
Centroids and centers of gravity. Example: Thin plate hung in gravitational field edgewise from string. Imagine plate as broken up into small elements. At equilibrium, string line passes through point with coordinates xbar = x_i W_i / W and ybar = y_i W_i / W, where W_i is the weight of the ith small element of the plate, x_i and y_i are the corresponding xy coordinates in the plane of the plate, and W is the total weight of the plate. Point with coordinates xbar, ybar called center of gravity. If plate has uniform thickness and specific weight, xbar, ybar equations reduce to xbar = x_i A_i / A and ybar = y_i A_i / A, where A is the total area of the plate face. Now, xbar and ybar are the coordinates of the centroid of area A.
Lecture 9
Summations can be transformed into integrals. Integral of x dA over A called first moment of area A with respect to (or about) yaxis. Likewise for integral of y dA. Centroid lies on axis of symmetry of an area. To evaluate first moment integral, it is best to use lowest order element of area. Examples: Centroid of area in shape of triangle. Centroid of area in shape of circular sector.
Centers of gravity of threedimensional bodies. Centroids of volumes. Example: Centroid of a right circular cone.
Centroids of composite figures. Example: Area resulting from the contiguous addition of a triangle and a rectangle. Area resulting from the overlapping subtraction of an exparabolic area from a triangle.
Distributed forces. Line distribution. Distributed loading diagram w(x) versus x over an interval L, where w(x) is load intensity (force per unit length). Traditionally, positive w is assumed to act vertically downward. Resultant force equals the area under the w(x) curve; it's line of action passes through centroid of the area under the w(x) curve.
Area distribution. Distributed loading diagram p(x,y) versus x and y over an area A, where p(x,y) is load intensity (force per unit area). Traditionally, positive p is assumed to act vertically downward. Resultant force equals the volume under the p(x,y) surface; it's line of action passes through centroid of the volume under the p(x,y) surface.
A specific case when p(x,y) = p_o y, where p_o is a constant. Formula for resultant moment of load about xaxis leads to the integral of y^2 dA over A. This integral is called the second moment of area A about the xaxis, or the (area) moment of inertia of area A about the xaxis. It is usually denoted by I_x. Similarly, integral of x^2 dA is I_y. The integral of (x^2 + y^2) dA over A is called the polar moment of inertia J_O.
Parallel axis theorem. Allows "transfer" of area moment of inertia from centroidal axis to parallel noncentroidal axis: I = Ibar + A d^2, where Ibar is about centroidal axis, d is perpendicular distance between centroidal and noncentroidal axes, and I is about noncentroidal axis. Example: Area in form of triangle. Moments of inertia of composite areas. Example: Area in form of a "T".
Lecture 10
MECHANICS OF DEFORMABLE BODIES
Stress resultants (resultant internal loadings). Example: Cantilevered "Z" shaped bent bar, with rectangular cross section and with forces and couple moments applied at free end. Cut bar near wall transversely to its geometric axis. Isolate portion away from wall as a free body. This reveals that continuously distributed internal forces act over the cross section at the cut. The intensity of these distributed forces at a point on the cross section is called the stress vector. We can determine the resultant force F_R, and resultant couple moment M_R_O, acting at the centroid O of the cross section, that are statically equivalent to the distributed forces. Equilibrium implies that F_R and M_R_O are opposite to the resultant force and resultant couple moment system acting at the centroid that is equivalent to the applied end forces and couple moments. Define xyz axes with origin O at centroid of cross section, and with the xy axes aligned with the rectangular sides of the cross section. Then, the xcomponent of F_R is called shear force V_x, the ycomponent of F_R is called shear force V_y, and the zcomponent of F_R is called normal force N. The xcomponent of M_R_O is called bending moment M_x, the ycomponent of M_R_O is called bending moment M_y, and the zcomponent of M_R_O is called torque T. These rectangular scalar components are known collectively as stress resultants. In two dimensional problems, there are only three stress resultants: N, V and M. Traditionally, these don't carry any subscripts. Example: A frame with a transverse exploratory cut through one member. Find N, V and M.
Stress. To characterize the strength of a material, we introduce the concept of stress. Example: a uniform prismatic bar in tension due to axial tensile forces P acting at ends. Cut bar with an exploratory transverse planar cut. Isolate right part of bar as a free body. At left end, over surface at cut, continuously distributed internal forces act normal and away from cross section. Intensity of these distributed forces (force per unit area) is called normal stress, denoted as . Sufficiently distant from ends of complete bar, stress is practically uniform or constant. Hence, in between ends, normal stress equals axial load divided by crosssectional area, or = P/A. Units of stress are: N/m^2 or Pa; lb/in^2 or psi. Normal stress is positive when tensile; negative when compressive.
Lecture 11
One can show that if normal stress is uniform, then the axial loads applied at ends of bar must act at the centroid of cross section.
Bearing stress. Shear stress. Example: a clevis and bar connected by a single bolt inserted in a transverse hole passing through both parts. When clevis and bar are subjected to axial forces, both clevis and bar press against bolt over semicylindrical contact surfaces on opposite sides of bolt. Continuously distributed contact forces develop over contact surfaces with bolt. For design purposes, the intensity of these forces is represented by bearing stress which is the ratio of bearing load (resultant force over contact surface) divided by the bearing area defined as the projected area of the contact surface, projected onto a plane perpendicular to the line of action of the resultant force. This projected area is usually the product of the bolt diameter and a thickness such as the bar's thickness.
Freebody diagram of bolt reveals forces due to bar and clevis tend to shear bolt at two cross sections of bolt. Isolate central part of bolt bounded by these two cross sections. Continuously distributed forces act tangent to the two crosssectional surfaces. Intensity of distributed forces called shear stress, denoted by . Stress resultant associated with distributed forces is shear force V. Average shear stress is shear force divided by bolt crosssectional area (used in bolt or rivet design) or _ave = V/A_bolt.
Allowable stress. An allowable stress is a maximum stress used for sizing structural members to guard against failure of the member when in service; _all, _all.
Stresses on an oblique plane. Pass oblique exploratory cut through bar in tension. Isolate left part. Normal force and shear force act on oblique surface. Division of these forces by area of oblique surface gives normal stress and shear stress. Express these stresses as functions of angle of inclination of oblique surface, measured from axis of bar. Variation of these stresses with angle of inclination is sinusoidal. Shear stress is maximum on 45 degree oblique plane. Normal stress is maximum on transverse plane. Sign convention for stresses, particularly . Stresses on a rotated rectangular element isolated from bar. Shear stresses on opposite sides of element form couples. The resulting pair of couples oppose and cancel each other. Shear stresses on all four sides of element have equal magnitude.
Lecture 12
Generalize to an infinitesimal element in a state of plane stress. Consider a thin plate subject to inplane loads only. Isolate infinitesimal rectangular element from plate. Define rectangular stress components according to the notational and sign conventions of the theory of elasticity: _x, _y, _xy, _yx.
Strain. Needed for quantifying deformation of a body. Example: prismatic bar in tension. Divide elongation of bar by original length. This equals normal strain . If bar is in tension, corresponding strain is called tensile strain; as normal strain, it is positive. If bar is in compression, corresponding strain is called compressive strain; as normal strain, it is negative. The ratio of the elongation of any isolated partial length of the bar over the initial partial length is constant over the entire bar. So have condition of uniform strain. If bar has variable cross section, strain is no longer uniform; it is nonuniform. In this case, the definition of strain must account for variability, and we define normal strain at a point as the ratio of elongation of a piece of the bar of differential length over the differential length.
Shear strain. Small cubical element subject to shear stresses _xy only. Square shape parallel to stresses deformed into rhombus. Change in 90 degree angle called shear strain (measured in radians), _xy. Example: Square element with one diagonal that elongates and the other that contracts. Calculate normal strains in direction of edges and shear strain between adjacent edges when edges are aligned with xy axes.
Lecture 13
The tensile test, stressstrain diagram. Relationship between normal stress and normal strain established from tensile test. Bar with a test section is loaded in tension. Load over original crosssectional area (normal stress) is plotted against elongation of test section over original length of test section (normal strain). This is the stressstrain diagram. Example: structural (lowcarbon) steel. Diagram has an initial linear portion. Where linear portion ends, stress point called proportional limit. Thereafter, curve begins to bend in horizontal direction. Stress point at which curve becomes flat called yield point _Y. Elongation with little increase in stress called yielding. After some yielding, curve begins to rise for a while but then flattens again at a maximum point. The maximum point is called ultimate (or tensile) strength _u. Curve decreases thereafter until specimen fractures at stress point called rupture or breaking strength. The hump in curve at ultimate strength is misleading because at this point in the test, specimen narrows down locally or forms locally a neck. Phenomenon called necking. Crosssectional area at smallest part of neck is significantly smaller than original crosssectional area. So stress at neck is much higher than as calculated for stressstrain diagram. Actual stress at neck continues to rise after passage of ultimate strength. If specimen is unloaded from any stress level, unloading stressstrain path follows a curve that parallels linear portion of curve. There is a stress point between the proportional limit and the yield point, called the elastic limit, on the "knee" of the curve, up to which the specimen can be loaded and unloaded without permanent set or plastic deformation.
For materials that do no exhibit a yield point, yield strength _YS is defined which is a point at the intersection of the stressstrain curve with a line drawn parallel to the linear portion but offset to the right by 0.002 or 0.2% strain. Materials that can sustain large strains before fracture are called ductile; those that can sustain only small strains are called brittle. Ductile material examples: steel, aluminum. Brittle material examples: ceramics, cast iron, concrete, glass, some metallic alloys. Stressstrain diagrams in compression are also obtainable. For steel, the compression curve is nearly the same as the tension curve mirrored to negative stress and strain.
Shear stressshear strain diagrams can also be developed.
Allowable stress (Continued). To guard against permanent deformation or set in design, maximum allowable stress chosen for sizing structural members is well below proportional limit. Allowable stress (or working stress) is computed by dividing yield strength or ultimate strength by factor of safety greater than one.
Lecture 14
Hooke's law. Normal stress proportional to normal strain up to proportional limit. Material said to be linearly elastic because loading and unloading stressstrain curves are linear and coincident. Proportionality constant is modulus of elasticity or Young's modulus E. Hooke's law: = E.
Formula for elongation of linearly elastic prismatic bar under axial loading P, = PL / AE. Example: Deflections (displacements) at specific locations along two axially loaded solid cylinders joined in series.
Poisson's ratio. Axial elongation accompanied by lateral contraction. Poisson's ratio equals minus lateral strain divided by longitudinal strain. In linear elastic range, is constant (about 0.3 for metals).
Shear stressshear strain relation. Small cubical element subject to shear stresses _xy only. For linear elastic material: _xy = G _xy. G called shear modulus of elasticity or modulus of rigidity. G = E / [2 (1 + )].
Assumed all along materials homogeneous and isotropic, that is, mechanical properties do not vary with position and direction in material.
Lecture 15
Torsion. Bar with circular crosssection subject to torques (couples, twisting moments) at ends. Bar in pure torsion. Cross sections remain circular, crosssections remain plane, radial lines remain straight, that is, crosssections rotate as rigid bodies. Angle of rotation of one end of bar relative to other called angle of twist . Examination of deformation of a rectangular element at surface of bar aligned with bar axis shows element deformed into rhomboid. Deformation involves shear strain which is caused by shear stresses acting on sides of element. Relation between shear stress and bar radius c, bar length L, angle of twist and shear modulus: = (G / L) c.
Lecture 16
Reinterpret shear strain and shear stress analysis at surface of bar as occurring in interior at radial distance . This shows shear stress varies linearly with radial distance from center of shaft, = (G / L). Moment of shear stress distribution over cross section about center equals torque T. Angle of twist is proportional to torque and shaft length, and inversely proportional to polar moment of inertia of cross section J and G, = TL / (GJ). Shear stress is proportional to torque, radial distance from center, and inversely proportional to J, = T / J. Example: shaft and gear system transmitting power.
Shear force and bending moment in beams. Example: simply supported or simple beam with concentrated load or force. Pin (or hinge) support, roller support. Support forces (reactions). Example: cantilever beam with distributed load. Builtin (or fixed) support. Free end. Support forces and couple (reactions).
Will consider only statically determinate beams. Usually neglect weight of beam in computing reactions.
We will assume that beams have a vertical plane of symmetry in which all loads are applied. Thus, beam bending occurs in this plane.
Stress resultants in beams: axial force, shear force, and bending moment. Example: cantilever beam with end force inclined to beam centroidal axis. Isolate portion of beam between free end and a transverse exploratory cut. Continuously distributed forces act over cross section at cut. Stress resultants of these distributed forces consist of axial force N, shear force V, and bending moment M. Forces act through centroid of cross section; bending moment acts, as curved vector, in beam symmetry plane and is computed about centroidal axis of cross section. Sign convention for V and M: on rightfacing cross section at a cut, V positive down, M positive counterclockwise. Example: simple beam with concentrated load and couple. Shear diagram. Bendingmoment diagram. Example: cantilever beam with uniformly varying load.
Relationship between load, shear force, and bending moment. Distributed load acting downward is positive load intensity function w. Slope of shear diagram is minus the distributed load intensity, dV/dx = w. Slope of bending moment diagram is shear force, dM/dx = V. Down concentrated load P causes downward jump in shear diagram, V = P . A clockwise couple M_o causes a jump up in bending moment diagram, M = M_o. Example: simple beam with uniform and concentrated load.
Lecture 17
Stresses in beams. Normal stresses. Beam subjected to equal and opposite bending moments M at opposite ends is in pure bending. Assume beam lies along xaxis and is bent in xy plane of symmetry with concavity on top. Top longitudinal fibers compressed; bottom longitudinal fibers stretched. Somewhere in between, some fibers do not change length. These fibers form the neutral surface. Since the bending moment is constant along beam, every element of length dx must deform in the same way. Continuity of fibers from one element to the next demands that cross sections remain plane. So, beam bends into the shape of a circular arc with a center of curvature. Distance from center of curvature to neutral surface is radius of curvature, . All cross sections remain plane and normal to longitudinal fibers of beam. Adjacent cross sections rotate relative to each other. Intersection of neutral surface with cross section called neutral axis, N. A. Define xyz axes at cross section so that zaxis coincides with N. A. and yaxis points towards center of curvature. Then, normal strain in longitudinal fibers, a directed distance y from N. A., is _x = y / . If material is linearly elastic (Hookean), normal stress over cross section given by _x = Ey / . Stress resultants at cross section must equal zero longitudinal force and bending moment M. This implies (1) zaxis is centroidal axis, that is, N. A. passes through centroid of cross section, and (2) beam curvature = 1 / = M / EI, where I is area moment of inertia about zaxis or N. A. of cross section. Thus, _x = My / I, called flexure formula. _x called bending or flexural stress. Maximum tensile or compressive stress occurs in fibers farthest from N. A. of cross section. When shear forces present in beam, cross sections warp, and flexure formula is only an approximation. In most cases, approximation is excellent. Example: a simply supported beam with a uniform load and an "I" cross section. Example: a cantilever beam with an oblique transverse endforce and an inverse "U" cross section.
If we disregard whether a normal stress is tensile or compressive, the maximum normal stress is given by _m = Mc / I where c is distance from N.A. to outermost fiber.
THE END
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