ME 205
Call No. 22204
S. F. Felszeghy

Winter 2005

Text: Mechanics of Materials, 3rd Ed., F. P. Beer, E. R. Johnston, Jr., and J. T. De Wolf, McGraw-Hill, 2001.




Jan. 3 1.1 - 1.13 1.2, 1.18, 1.26, 1.31
Jan. 5 1.13, 2.1 - 2.3, 2.5, 2.6, 2.8 1.46, 2.16, 2.19, 2.30
2 Jan. 10 2.9 - 2.12 2.40, 2.52, 2.63, 2.72
Jan. 12 2.14, 2.15, 2.17, 3.1 - 3.5 2.80, 3.5, 3.17, 3.35
3 Jan. 19 3.5 - 3.7 3.39, 3.56, 3.70, 3.81
4 Jan. 24 3.13, 5.1 - 5.3 3.139, 5.13, 5.17, 5.44
  Jan. 26 5.3, 4.1 - 4.5 5.49, 4.1, 4.12, 4.26
5 Jan. 31 MIDTERM
  Feb. 2 4.5, 5.1, 5.4, 4.12 4.35, 5.61, 5.96, 4.115
6 Feb. 7 4.12, 4.14, 6.1, 6.2 4.133, 4.143, 6.4, 6.10
  Feb. 9 6.2 - 6.4, 7.1 - 7.3 5.76, 6.15(b), 5.82, 6.17(b)
7 Feb. 14 7.4, 7.9 6.21, 7.2, 7.4, 7.7
  Feb. 16 7.10, 7.11, 8.1, 8.2 7.11, 7.22, 7.32, 7.40
8 Feb. 21 MIDTERM
  Feb. 23 8.4 7.55, 7.101, 7.110, 7.132
9 Feb. 28 9.1 - 9.3, 5.5 8.35, 8.48, 9.4, 9.5
  Mar. 2 9.6, 9.7 9.35, 9.48, 9.67, 9.78
10 Mar. 7 10.1 - 10.3 10.10, 10.15
Mar. 9 10.4 10.22

FINAL EXAM:  Wednesday, March 16, 10:45 - 1:15 p.m.

Notebook containing solutions to problems is available in CSULA Library. Call No. is 559.

Concepts and Key Words

Lecture 1


Strength of materials (also known as mechanics of materials) is that branch of classical mechanics which deals with the static response of solid bodies under the action of applied forces. By response what is meant is the stresses (or the stress resultants, also known as internal forces) and the deformation of a body.

Example: a bar supported on a fulcrum used to balance a given weight at one end, with a force at the other end. The required balancing force can be found from the equations of static equilibrium. Usual assumptions: bar massless; bar has sufficient strength so that it will not break; bar has sufficient rigidity so that it will not bend excessively. Other questions one may ask: What material should be used for bar? Wood, metal, plastic? How large must the bar's cross section be so that it will neither break nor bend excessively? What is the least expensive material that will do the job? Answers to these questions are the concerns of engineering design. If the end of the bar, where the balancing force is applied, is instead built into a rigid wall, then there are, in number, more reactions than the number of static equilibrium equations available, and the problem of solving for the reactions is a "statically indeterminate" problem. In such a case, one needs to consider the deformation of the bar.

One could follow a "cut-and-try" approach to answering the above posed questions. A better approach is to develop some fundamental principles that govern the strength and rigidity of all solid bodies and use these principles in combination with a few experimentally determined mechanical properties of materials to predict how a body responds under the action of forces.

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 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 bar ends, stress is practically uniform or constant. Hence, there, normal stress equals axial load divided by cross-sectional area. Units of stress are: N/m^2 or Pa; lb/in^2 or psi. Normal stress is positive when tensile; negative when compressive. Product of constant normal stress and area is stress resultant or internal normal force. 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. Shearing 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 semi-cylindrical 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, _b, 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.

Free-body 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 cross sections. Continuously distributed forces act tangent to the cross-sectional surfaces. Intensity of distributed forces called shear stress, denoted by . Stress resultant associated with distributed forces called shear force. Average shear stress is shear force divided by bolt cross-sectional area (used in bolt or rivet design).

Lecture 2

Example: pin supported and connected two-bar, V shaped, assembly, with a clevis connection, and loaded pin. Find maximum value of average normal stresses in bars, bearing stresses on bars, and shearing stress in connecting pin.

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.

Generalize to an infinitesimal element in a state of plane stress. Thin plate subject to in-plane loads only. Isolate infinitesimal rectangular element from plate. Define rectangular stress components according to the notational and sign conventions of the theory of elasticity.

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.

Lecture 3

The tensile test, stress-strain diagram. Relationship between normal stress and normal strain established from tensile test. Bar with a test section is loaded in tension. Load over original cross-sectional area (normal stress) is plotted against elongation of test section over original length of test section (normal strain). This is the stress-strain diagram. Example: structural (low-carbon) 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. Cross-sectional area at smallest part of neck is significantly smaller than original cross-sectional area. So stress at neck is much higher than as calculated for stress-strain diagram. Actual stress continues to rise after passage of ultimate strength.

If specimen is unloaded from any stress level, unloading stress-strain 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 _Y is defined which is a point at the intersection of the stress-strain 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.

Stress-strain 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.

Allowable stress. To guard against permanent deformation or set in design, a maximum allowable stress is chosen for sizing structural members well below proportional limit. Allowable stress (or working stress) computed by dividing yield strength or ultimate strength by factor of safety greater than one.

Hooke's law. Normal stress proportional to normal strain up to proportional limit. Material said to be linearly elastic because loading and unloading stress-strain 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.

Lecture 4

Example: Deflections (displacements) at specific locations along two axially loaded solid cylinders joined in series.

Statically indeterminate structures. Cases arise where reactions and internal forces acting on and in axially loaded structural members cannot be obtained from the governing equations of equilibrium alone. Such structures called statically indeterminate. Such structures can be analyzed by taking into account elastic deformation. Two approaches possible. Illustration with steel reinforced concrete post. Method I: stiffness method or displacement method. A suitable displacement (deflection) is defined allowing introduction of deflection-internal force equations. Method II (superposition method): flexibility method or force method. An unknown reaction is removed (a fixity is released), and then reapplied to re-establish the original fixity.

Thermal stresses. Materials expand or contract if temperature raised or lowered. Elongation of free prismatic bar equals product of coefficient of thermal expansion, bar length, and temperature rise, _T = L T. If structure supported in statically indeterminate manner, thermal stresses develop when temperature of structure changed. Example solved by Method II: Normal stresses in series of cylinders, held between rigid walls, whose temperature is raised. If structure is statically determinate, no thermal stresses develop.

Lecture 5

Poisson's ratio; biaxial stress. Axial elongation is accompanied by lateral contraction. Poisson's ratio equals minus lateral strain divided by axial strain. In linear elastic range, is constant (about 0.3 for metals). Biaxial stress condition: normal stresses acting in two orthogonal directions, x and y. Strains associated with biaxial stresses obtained by superposition of strain effects from x and y stress components acting alone, assuming Hooke's law and Poisson effect.

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. For linear elastic material: _xy = G _xy. G called shear modulus of elasticity or modulus of rigidity. G = E / (2 (1 + )).

Stress-strain relations for plane stress. Assumed all along material homogeneous and isotropic, that is, mechanical properties do not vary with position and direction in material.

Torsion. Bar with circular cross-section subject to torques (couples, twisting moments) at ends. Bar in pure torsion. Cross sections remain circular, cross-sections remain plane, radial lines remain straight, that is, cross-sections rotate as rigid bodies. Angle of rotation of one end 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.

Reinterpret shear strain and shear stress analysis at surface 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. Example: shaft and gear system with statically indeterminate reactions.

Lecture 6

Torsion of thin-walled tubes. Tube with arbitrary cross-sectional shape. Product of shear stress in wall and wall thickness called shear flow. Shear flow is constant around tube wall centerline. Shear stress is proportional to torque and inversely proportional to wall thickness and area enclosed by centerline of tube wall, = T / (2t scriptA). Example: tube with "L" shaped cross-section.

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. Built-in (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.

Lecture 7

Sign convention for V and M: on right-facing cross section at a cut, V positive down, M positive counterclockwise.
Example: simple beam with concentrated load and couple. Shear diagram. Bending-moment 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 8

Stresses in beams. Normal stresses. Beam subjected to equal and opposite bending couples at opposite ends is in pure bending. Assume beam lies along x-axis and is bent in x-y 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 x-y-z axes at cross section so that z-axis coincides with N. A. and y-axis points towards center of curvature. Then, normal strain in longitudinal fibers 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 a bending moment, say M. This implies (1) z-axis 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 z-axis or N. A. Thus, _x = -My / I, called flexure formula. Maximum tensile or compressive stress occurs in fibers farthest from centroid of cross section. When shear forces present in beam, cross sections warp and flexure formula is only an approximation. In most cases, approximation is very good. Example: a simply supported beam with an overhang and an inverted "T" cross section (beam with one plane of symmetry).
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. Ratio I / c called (elastic) section modulus S. Examples: S for circular and rectangular cross sections.

Lecture 9

Design of beams (based upon normal bending stresses only). If overall maximum bending moment, regardless of sign, is M_max, and allowable bending stress is _all, then beam must have S greater than or equal to S_min = M_max / _all. Comparison of S values for rectangular and circular cross sections. Compare areas of square and circular cross sections with same S value. Square area is smaller. Biggest S value given by wide-flange section, or "W" shape beam. Example: selection, in design, of a lightest "W" beam section.

Eccentrically applied axial load on a cantilever beam. Neutral axis moves towards cross-section centroid as point of application of load moves away from centroid. For some point of application of load, normal stress distribution over cross section transitions from being all of one sign to being plus and minus.

Shear stresses in beams. Example: compare maximum bending stresses in beam made of six stacked unglued planks, with six stacked glued planks. Without glue, sliding and no shear stresses develop between planks. With glue, no sliding and shear stresses develop between planks. Presence of shear stresses reduces significantly the normal bending stresses in glued beam. Consider solid beam of narrow rectangular cross section. Assume shear stresses parallel to shear force V and uniform across beam width. Associated with shear stress on cross section, there are shear stresses between longitudinal layers of beam as well. Since top and bottom longitudinal surfaces of beam are stress free, shear stresses on cross section at top and bottom of cross section vanish. Isolate a beam element of length dx. Cut this element longitudinally a distance y_1 from N.A. Force equilibrium applied to brick element between cutting plane and nearest top, or bottom, beam surface, gives = VQ / Ib, where is shear stress at y_1, Q is first moment of cross section between y_1 and top, or bottom, beam surface, b is beam width. Extension of formula to non-rectangular symmetric cross sections for which b becomes local width t.

Lecture 10

Example: beam with rectangular cross section; shear stress distribution is parabolic about y_1 = 0 point, vanishing at outer surfaces.
Shear strain over a rectangular cross section is distributed parabolically; hence, cross sections, initially plane, warp.

Wide-flange beams. I shape; horizontal parts called flanges; vertical part called web. Shear stress distribution in web is shallow parabola; is about V / (web area).

Stresses under combined loading. Structural members often subjected simultaneously to axial, torsional, and flexural loading (bending).
Combined bending and axial loading. Example: cantilever beam subject to axial and transverse loads at end. If beam is long and slender, transverse bending may move line of action of axial load. Hence, bending moment in beam is dependent on transverse deflection. If beam is short and tall (in bending plane), lateral deflections have negligible effect on displacement of line of action of axial force. In this case, the stresses due to the axial force and the stresses due the transverse force can be computed independently and superimposed. This is what we will assume occurs hereafter. Under combined axial and transverse loading, neutral axis is shifted from centroidal position. Example: simple beam with rectangular cross section and an eccentric axial load applied parallel to beam. Normal force, shear force, and bending moment diagrams. Computation of maximum tensile and compressive stresses.

Example: cantilevered beam, with solid circular cross section, subject to eccentrically applied axial and transverse loads. Pass exploratory transverse cut through beam. Determination of force-couple system, referenced to centroid of outward facing cut end, that is equivalent to applied loads. Rectangular components of this force-couple system represent stress resultants that act at cross section. Stress resultants are normal force, shear forces, torque and bending moments. Calculation of normal and shear stresses on rectangular element on perimeter of cross section. Stresses due to each stress resultant are computed separately, and then summed keeping in mind conventions for positive directions.

Lecture 11

Stresses on an oblique plane (continued). Case of plane stress. Pass inclined exploratory plane through rectangular element in plane stress. Isolate triangular wedge. Apply equations of equilibrium. Express normal stress and shear stress on inclined plane in terms of rectangular stress components. Positive makes clockwise moment with respect to wedge. Numerical example: and vary sinusoidally with angle of inclination of exploratory plane; period of both functions is 180 degrees. The values _max and _min are principal stresses. Corresponding 's are zero. Planes on which principal stresses act called principal planes; their angles of inclination differ by 90 degrees. Maximum and minimum shear stresses. Planes on which _max and _min act are rotated 45 degrees with respect to principal planes. Example: calculation of principal stresses and planes; calculation of maximum and minimum shear stresses and the planes on which they act.

Lecture 12

Mohr's circle for plane stress. plotted against in - plane is a circle (called Mohr's circle) with radius R, centered at (C, O) on -axis. Construction of Mohr's circle when rectangular stress-component data are given. C = (_x + _y) / 2. R = (((_x - _y)/2)^2 + (_xy)^2)^0.5.
Proof of claim that horizontal and vertical coordinates of a point, say A, on Mohr's circle, located so that arc starting at point representing stress state on x-face and ending at point A subtends an angle 2 measured counterclockwise at the circle center, represent the and stresses on an oblique plane whose outward normal makes an angle with respect to the x-axis measured counterclockwise. Example: determining by Mohr's circle - principal stresses and planes, maximum stress and planes, and stresses on a rotated rectangular element. Mohr's circles for various stress conditions. Simple tension; simple compression; equal tension and compression; pure shear; equal biaxial tension (three dimensional interpretation).

Stresses in thin-walled pressure vessels. Cylindrical pressure vessel with inside gage pressure p and ratio of wall thickness t over inside radius r less than 0.1. Hoop (circumferential, tangential) stress is _1 = pr / t. Axial (longitudinal) stress is _2 = pr / 2t.

Lecture 13

Strains corresponding to plane stress. Generally, stresses cannot be measured directly. Instead stresses are computed from normal strain measurements. Suppose the strains _x, _y and _xy are given. Define orthogonal fibers a and b so that the a fiber makes a counterclockwise angle with respect to the x-fiber. Strains _a, _b and _ab can be expressed in terms of _x, _y and _xy. As in theory of elasticity, assume for positive _xy, right angle between x fiber and y fiber decreases. For purposes of Mohr's circle representation, assume for positive _ab, right angle between a fiber and b fiber increases. Then, plot of _ab/2 versus _a, with 2 (ccw) as parameter, gives Mohr's circle for strain analogous to Mohr's circle for stress. Connection between Mohr's circle for plane stress and Mohr's circle for corresponding strains. R_ = R_ E / (1+). (OC)_ = (OC)_ E / (1-).

Lecture 14

Design of beams (continued). Assume beams under transverse loading only. Recall maximum normal bending stress _m occurs in fiber farthest from N.A. Maximum transverse shear stress _m occurs at N.A. Beam design criteria: (1) _m <= _all; (2) _m <= _all. (1) usually governs long beams; (2) usually governs short, stubby beams. Exception: case of a long beam, but low transverse _all.
Exception: case of an I beam designed on the basis of principal stress _max <= _all. Principal stress _max is usually largest at flange to web junction.

Deflection of beams. Edge view of neutral surface called elastic curve (deflection curve). Let y be vertical deflection of elastic curve (positive up). From analytic geometry, for a plane shallow curve in x-y plane, y' is small, and curvature 1/ equals approximately y''. Thus, EIy'' = M, assuming beam deformation is due to bending moment only. Example: simple beam with uniform load, pinned at x = 0 and x = L. Indefinite integration (twice) of EIy'' = M gives y(x) containing two undetermined constants. Constants can be determined from boundary conditions: y(0) = 0 and y(L) = 0. Example: simple beam with concentrated load at x = a. Indefinite integration (twice) of EIy'' = M over intervals 0 <= x <= a and a <= x <= L produces four undetermined constants. These can be determined from the boundary conditions, and the continuity of y and y' at x = a.

Lecture 15

Singularity functions. Define function <x-a>^n as <x-a>^n = 0, if x <a, and <x-a>^n = (x-a)^n, if x >= a, where n = 0, 1, 2, 3, ... . With this function, can write a bending moment expression, for the left part of a transversely cut beam, that applies for all values of x. Integration of EIy'' = M introduces as many constants as there are boundary conditions.
Method of superposition. Because the equation EIy'' = M is a linear differential equation, the total deflection of a beam due to several loads is the sum of the deflections caused by the individual loads. Example 1: cantilever beam with uniform and concentrated end load. Application of tabulated beam deflection results. Example 2: cantilever beam with one intermediate concentrated load. Example 3: cantilever beam with uniform load over part of the beam. Application of result of Example 2 to solve this problem by integration. Example 4: simply supported beam with an overhang. Decomposition into simpler subproblems that have tabulated results.

Bar with eccentrically applied axial load. A vertical compression member with an axially sliding pin support at top end and fixed pin support at bottom end. Member is subjected to eccentrically applied axial compression loads at ends. This is a combined bending and axial load problem. What are the stresses in the member? We distinguish between two possibilities.

Lecture 16

(1) Bar is so short that lateral deflections produce negligible change in bending moment in bar. Stresses due to axial compressive force and bending moment can be superimposed. (2) Bar is so long that lateral deflections produce a significant change in bending moment in bar. Compression members that fall in category (2) called columns.

Columns with eccentric axial loads. Example: column with pinned-pinned ends. Equation governing lateral deflection: EIy'' = -P(e + y), where P is axial compressive load, and e is eccentricity of load. Solution. Plot P as function of y_max for different values of e > 0. Plot shows that for any e value, curve starts at (0, 0), then rises and becomes horizontally asymptotic at P = P_cr = ^2 EI/L^2 where L is length of column. P_cr called critical load. Knee of P versus y_max curve moves towards (0, P_cr) point as e approaches 0. Limitations of analysis.

Critical load calculated for column with centric applied load. Example: pinned-pinned ideal column with centric applied compressive load. Solutions: trivial and nontrivial. Nontrivial solutions y = C_1 sin n x/L exist for P = n^2^2 EI/L^2, where n = 1, 2, 3, ... , and C_1 is an arbitrary constant. The smallest of these is P_cr, and it is called Euler load. It is the smallest axial force that maintains the column in a bent shape. Physical interpretation. There are three possibilities when an axial force P is applied, and column is displaced from straight position. (1) P <P_cr, (2) P = P_cr , and (3) P > P_cr.

Lecture 17

More on the three possibilities. (1) When P <P_cr, we have the condition that column remains straight and undergoes axial shortening only. If we displace column laterally, it will return to its straight position when released. Column is said to be in stable equilibrium. Analogy: a bead at the bottom of a spherical bowl. (2) When P = P_cr, column will not return to its straight position if displaced laterally and released. Column is said to be in neutral equilibrium. Analogy: bead on a horizontal surface. (3) If P > P_cr, column will collapse if disturbed from its straight position. Column said to be in unstable equilibrium. This phenomenon of instability called buckling. Analogy: bead at top dead center of inverted spherical bowl.

Stresses in columns. The average compressive stress P_cr/A called critical stress _cr. Plot of _cr versus slenderness ratio L/r, where r is least radius of gyration of cross section, r = sqrt(I/A), is called Euler's curve. _cr is inversely proportional to the square of the slenderness ratio L/r. This curve governs ideal column behavior as long as _cr <= _Y (yield strength). Cross over from yielding to buckling occurs, with increasing L/r, at about L/r = 100 for structural steel.


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