교과목명 ( 국문) 고체역학 ( 영문) Solid Mechanics 담당교수( 소속) 박주혁( 기계공학과) 학수번호/ 구분/ 학점 004510/ 전필/3(3) 1반 전화( 연구실/HP) 3408-3771/010-6214-3771 강의시간/ 강의실화목9:00-10:15/ 충무관106 선수과목 공업역학/ 일반물리1 수강대상기계공학과 2학년 jhpark@sejong.ac.kr 연구실충무관1019/ 월13:00-15:00 교과목표교재참고도서과제도서수업방법과제물독서물학업성취평가방법 Fundamental principles of solid and structural mechanics will be presented. Application of the principles to mechanical & civil engineering structures will be emphasized. The course will further develop the concepts introduced in Statics and Dynamics. In particular, the notion of internal forces, and the concepts of stress and strain will be introduced. Mechanical properties of solids will be discussed, and the notion of statical indeterminacy will be introduced. The methods of structural analysis will be presented, including those based on energy concepts. 교재 : Mechanics of Materials, R. C. Hibbeler (Prentice Hall) 6th Edition 7판의수입차질로 6 판으로수업을합니다. 5 판번역서는시중에서구입가능합니다. 5,6,7 판의내용은큰차이가없으므로편한교재를선택하시기바랍니다. 참고문헌 : 1. Mechanics of Materials, Roy R. Craig, JR.. John Wiley &Sons, 2000 2. Mechanics of Materials, R.C. Hieler, Prentice Hall International, Inc., 1997 3. Mechanics of Materials, 6th Ed., James M. Gere 원서및번역서, 2004. 인터비젼 1. 재료역학, Gere&Timoshenko, 번역서, 1994. 반도출판사. 2. 재료역학, William F. Riley and Loren W. Zachary, 1993, 희중당. 3. 재료역학연습, 차경옥, 원화. 4. 고체역학, 3rd Ed., Beer & Johnston Jr., 번역서, 2004, 인터비젼 공학인증을위한기초자료를위하여본강좌는주 3시간수업으로이론강의와관련실험을수행하도록한다. 본강좌를위하여출석을엄하게관리하며매주과제를제출하며, 기술적인문답을위한교수와의면담을충실히수행하기를바랍니다. 1. 재료역학에관련된 3 가지실험 ( 인장실험, 스트레인게이지를이용한압력용기변형실 험, 보의처짐실험) 을실시하고실험보고서작성제출 2. 교재의관련단원연습문제를풀어서제출 3. 매단원에들어가기전예습하여 A4용지 2-3 매에요약하여제출할것( 선행학습) 전공에두번째로이수하는역학과목이므로학생들에게상당한부담이되는것으로알고있다. 현기계항공우주공학부교과과정상본주어진교재의전부를본과목에서다루는만큼예습과복습이병행되지않으면기대한학습성과를거두기가어렵다. 또한교과과정상진도가빨리진행되기에이에대한대비를스스로하기바람. 중간고사 30%, 기말고사 40%, 수시고사및과제 20%, 출석 10% ( 부정행위자는이유를불문하고 F 처리함, 또한과제물에대한부정행위도철저히가려낼예정임 )
주별이론강의교과내용 ( 교과목명: 고체역학) 2008-1 학기 주순번월일 / 교수내용수업형태및활용기자재 1 3/4 Principles of Statics (Ch 1) 1 3/6 Stress on Solids, Shear Stress and Allowable Stress Design (Ch 1) 3 3/11 Strain in Solids (Ch 2) 2 4 3/13 Introduction to Material Properties, Stress-Strain of Different Materials (Ch 3) 5 3/18 Material Strain Energy and Poisson's Ratio (Ch 3) 3 6 3/20 Introduction to Axial Members, Indeterminate Axial Members (Ch 4) 7 3/25 Stress Concentrations (Ch 4) 4 8 3/27 Introduction to Torsion Members, Behavior of Rod and Shafts in Torsion (Ch 5) 9 4/1 Plastic Response of Torsion Members (Ch 5) 5 10 4/3 Introduction to Bending, Drawing Shear and Moment Diagrams Quickly (Ch 6) 11 4/8 Flexural Properties of Beams, Unsymmetric Beams (Ch 6) 6 12 4/10 Composite Beams Including R/C Flexural Elements (Ch 6) 13 4/15 Stress Behavior in Beams - Elastic and Plastic (Ch 6) 7 14 4/17 Introduction to Transverse Shear, Shear Stress in Bending Beams (Ch 7) 8 15 4/22 중간고사성적입력및수강생열람기간 16 4/24
주별이론강의교과내용 ( 교과목명: 고체역학) 2008-1 학기 주순번월일 / 교수내용수업형태및활용기자재 9 17 4/29 Concept of Shear Flow in Built-up Sections, Shear Flow in Thin Walled Elements and Shear Center (Ch 7) 19 5/1 Combined Loadings and Resulting Stress (Ch 8) 21 5/6 Plane Stress Transformation, Principal Stresses (Ch 9) 10 23 5/8 Introduction to Mohr's Circle (Ch 9) 11 25 5/13 Stress Variations in Beam & Maximum Shear (Ch 9) Introduction to Bending Beams (Ch 12) 27 5/15 Beam Displacement - Integration and Moment Area Method (Ch 12) 29 5/20 Statically Indeterminate Beams (Ch 12) 12 31 5/22 Introduction to Columns, Euler's Formula for Slender Columns (Ch 13) 9 5/27 The Secant Formula (Ch 13) 13 10 5/29 Introduction to Energy Methods, Strain Energy of Loaded Elements (Ch 14) 11 6/3 Impact Loading by Energy Methods (Ch 14) 14 12 6/5 Make-up class 13 6/10 Principle of Virtual Work - I (Ch 14) 15 14 6/12 Principle of Virtual Work - II (Ch 14) 16 15 6/14 16 6/19 기말고사성적입력및수강생열람기간
강의목적및학습내용정리 CHAPTER 1 important principles of statics Use the principles to determine internal resultant loadings in a body Introduce concepts of normal and shear stress Discuss applications of analysis and design of members subjected to an axial load or direct shear Internal loadings consist of 1. Normal force, N 2. Shear force, V 3. Bending moments, M 4. Torsional moments, T Get the resultants using 1. method of sections 2. Equations of equilibrium Assumptions for a uniform normal stress distribution over x section of member ( σ = P/A) 1. Member made from homogeneous isotropic material 2. Subjected to a series of external axial loads that, 3. The loads must pass through centroid of cross section Determine average shear stress by using τ = V/Aequation 1. V is the resultant shear force on cross sectional area A 2. Formula is used mostly to find average shear stress in fasteners or in parts for connections Design of any simple connection requires that 1. Average stress along any cross section not exceed a factor of safety (F.S.) or 2. Allowable value of σallow or τallow 3. These values are reported in codes or standards and are deemed safe on basis of experiments or through experience CHAPTER 2 Define concept of normal strain Define concept of shear strain Determine normal and shear strain in engineering applications Loads cause bodies to deform, thus points in the body will undergo displacements or changes in position Normal strain is a measure of elongation or contraction of small line segment in the body Shear strain is a measure of the change in angle that occurs between two small line segments that are originally perpendicular to each other State of strain at a point is described by six strain components: 1. Three normal strains: e x,e y,e z 2. Three shear strains: γ xy,γ xz,γ yz
3. These components depend upon the orientation of the line segments and their location in the body Strain is a geometrical quantity measured by experimental techniques. Stress in body is then determined from material property relations Most engineering materials undergo small deformations, so normal strain e << 1. This assumption of small strain analysis allows us to simplify calculations for normal strain, since first order approximations can be made about their size CHAPTER 3 Show relationship of stress and strain using experimental methods to determine stress strain diagram of a specific material Discuss the behavior described in the diagram for commonly used engineering materials Discuss the mechanical properties and other test related to the development of mechanics of materials Tension test is the most important test for determining material strengths. Results of normal stress and normal strain can then be plotted. Many engineering materials behave in a linear elastic manner, where stress is proportional to strain, defined by Hooke s law, σ = Ee. E is the modulus of elasticity, and is measured from slope of a stress strain diagram When material stressed beyond yield point, permanent deformation will occur. Strain hardening causes further yielding of material with increasing stress At ultimate stress, localized region on specimen begin to constrict, and starts necking. Fracture occurs. Ductile materials exhibit both plastic and elastic behavior. Ductility specified by permanent elongation to failure or by the permanent reduction in cross sectional area Brittle materials exhibit little or no yielding before failure Yield point for material can be increased by strain hardening, by applying load great enough to cause increase in stress causing yielding, then releasing the load. The larger stress produced becomes the new yield point for the material Deformations of material under load causes strain energy to be stored. Strain energy per unit volume/strain energy density is equivalent to area under stress strain curve. The area up to the yield point of stress strain diagram is referred to as the modulus of resilience The entire area under the stress strain diagram is referred to as the modulus of toughness Poisson s ratio ( ν), a dimensionless property that measures the lateral strain to the longitudinal strain [0 ν 0.5] For shear stress vs. strain diagram: within elastic region, τ = Gγ, where G is the shearing modulus, found from the slope of the line within elastic region G canalsobeobtainedfromtherelationshipofg = E/[2(1+ ν)] Whenmaterialsareinserviceforlongperiodsoftime,creep and fatigue are important. Creep is the time rate of deformation, which occurs at high stress and/or high temperature. Design the material not to exceed a predetermined stress called the creep strength
Fatigue occur when material undergoes a large number of cycles of loading. Will cause micro cracks to occur and lead to brittle failure. Stress in material must not exceed specified endurance or fatigue limit CHAPTER 4 Analyze the effects of thermal stress, stress concentrations, inelastic deformations, and residual stress Determine deformation of axially loaded members Develop a method to find support reactions when it cannot be determined from equilibrium equations When load applied on a body, a stress distribution is created within the body that becomes more uniformly distributed at regions farther from point of application. This is the Saint Venant s principle. Relative displacement at end of axially loaded member relative to other end is determined from IfseriesofconstantexternalforcesareappliedandAE is constant, then Make sure to use sign convention for internal load P and that material does not yield, but remains linear elastic Superposition of load & displacement is possible provided material remains linear elastic and no changes in geometry occur Reactions on statically indeterminate bar determined using equilibrium and compatibility conditions that specify displacement at the supports. Use the load displacement relationship, d=pl/ae Change in temperature can cause member made from homogenous isotropic material to change its length by d = adtl. If member is confined, expansion will produce thermal stress in the member Holes and sharp transitions at x section create stress concentrations. For design, obtain stress concentration factor K from graph, which is determined empirically. The K value is multiplied by average stress to obtain maximum stress at x section, smax = Ks avg If loading in bar causes material to yield, then stress distribution that s produced can be determined from the strain distribution and stress strain diagram For perfectly plastic material, yielding causes stress distribution at x section of hole or transition to even out and become uniform If member is constrained and external loading causes yielding, then when load is released, it will cause residual stress in the material
CHAPTER 5 Discuss effects of applying torsional loading to a long straight member Determine stress distribution within the member under torsional load Determine angle of twist when material behaves in a linear elastic and inelastic manner Discuss statically indeterminate analysis of shafts and tubes Discuss stress distributions and residual stress caused by torsional loadings Torque causes a shaft with circular x section to twist, such that shear strain in shaft is proportional to its radial distance from its centre Provided that material is homogeneous and Hooke s law applies, shear stress determined from torsion formula, t =(Tc)/J Design of shaft requires finding the geometric parameter, (J/C) =(T/t allow ) Power generated by rotating shaft is reported, from which torque is derived; P = Tw Angle of twist of circular shaft determined from If torque and JG are constant, then For application, use a sign convention for internal torque and be sure material does not yield, but remains linear elastic If shaft is statically indeterminate, reactive torques determined from equilibrium, compatibility of twist, and torque twist relationships, such as f = TL/JG Solid noncircular shafts tend to warp out of plane when subjected to torque. Formulas are available to determine elastic shear stress and twist for these cases Shear stress in tubes determined by considering shear flow. Assumes that shear stress across each thickness of tube is constant Shear stress in tubes determined from t = T/2tA m Stress concentrations occur in shafts when x section suddenly changes. Maximum shear stress determined using stress concentration factor, K (found by experiment and represented in graphical form). t max =K(Tc/J) If applied torque causes material to exceed elastic limit, then stress distribution is not proportional to radial distance from centerline of shaft Instead, such applied torque is related to stress distribution using the shear stress shear strain diagram and equilibrium If a shaft is subjected to plastic torque, and then released, it will cause material to respond elastically, causing residual shear stress to be developed in the shaft
CHAPTER 6 Determine stress in members caused by bending Discuss how to establish shear and moment diagrams for a beam or shaft Determine largest shear and moment in a member, and specify where they occur Consider members that are straight, symmetric x section and homogeneous linear elastic material Consider special cases of unsymmetrical bending and members made of composite materials Consider curved members, stress concentrations, inelastic bending, and residual stresses Shear and moment diagrams are graphical representations of internal shear and moment within a beam. They can be constructed by sectioning the beam an arbitrary distance x from the left end, finding V and M as functions of x, then plotting the results Another method to plot the diagrams is to realize that at each pt, the slope of the shear diagram is a negative of the distributed loading, w = dv/dx; and slope of moment diagram is the shear, V = dm/dx. Also, the ( ve) area under the loading diagram represents the change in shear, DV = w dx. The area under the shear diagram represents the change in moment, DM = V dx. Note that values of shear and moment at any pt can be obtained using the method of sections A bending moment tends to produce a linear variation of normal strain within a beam. Provided that material is homogeneous, Hooke s law applies, and moment applied does not cause yielding, then equilibrium is used to relate the internal moment in the beam to its stress distribution That results in the flexure formula, s = Mc/I, where I and c are determined from the neutral axis that passes through the centroid of the x section If x section of beam is not symmetric about an axis that is perpendicular to neutral axis, then unsymmetrical bending occurs Maximum stress can be determined from formulas, or problem can be solved by considering the superposition of bending about two separate axes Beams made from composite materials can be transformed so their x section is assumed to be made from a single material A transformation factor, n = E 1/E 2 is used to do this. Once the transformation is done, the stress in the beam can be determined in the usual manner of the flexure formula Curved beams deform such that normal strain does not vary linearly from the neutral axis Provided that material of the curved beam is homogeneous, linear elastic and x section has axis of symmetry, then curved beam formula can be used to determine the bending stress, s = My/[Ae(R y)] Stress concentrations occur in members having a sudden change in their x section, such as holes and notches. The maximum bending stresses in such locations is determined using a stress concentration factor K, thatisfoundempirically,s=ks avg. If bending moment causes material to exceed its elastic limit, then normal strain remains linear, however stress distribution varies in accordance with shear strain diagram and balance
of force and moment equilibrium. Thus plastic and ultimate moments can be determined. If an applied plastic or ultimate moment is released, it will cause the material to respond elastically, thereby inducing residual stresses in the beam CHAPTER 7 Develop a method for finding the shear stress in a beam having a prismatic x section and made from homogeneous material that behaves in a linear elastic manner This method of analysis is limited to special cases of x sectional geometry Discuss the concept of shear flow, with shear stress for beams and thin walled members Discuss the shear center Transverse shear stress in beams is determined indirectly by using the flexure formula and the relationship between moment and shear (V = dm/dx). This result in the shear formula t =VQ/It. In particular, the value for Q is the moment of the area A about the neutral axis. This area is the portion of the x sectional area that is held on to the beam above the thickness t where t is to be determined If the beam has a rectangular x section, then the shear stress distribution will be parabolic, obtaining a maximum value at the neutral axis Fasteners, glues, or welds are used to connect the composite parts of a built up section. The strength of these fasteners is determined from the shear flow, or force per unit length, that must be carried by the beam; q = VQ/I If the beam has a thin walled x section then the shear flow throughout the x section can be determined by using q = VQ/I The shear flow varies linearly along horizontal segments and parabolically along inclined or vertical segments Provided the shear stress distribution in each element of a thin walled section is known, then, using a balance of moments, the location of the shear center for the x section can be determined. When a load is applied to the member through this pt, the member will bend, and not twist
CHAPTER 8 Analyze the stress developed in thin walled pressure vessels the stress analysis developed in previous chapters regarding axial load, torsion, bending and shear Discuss the solution of problems where several of these internal loads occur simultaneously on a member s x section A pressure vessel is considered to have a thin wall provided r/t 10. For a thin walled cylindrical vessel, the circumferential or hoop stress is s 1 = pr/t. This stress is twice as great as the longitudinal stress, s 2 = pr/2t. Thin walled spherical vessels have the same stress within their walls in all directions so that s 1 =s 2 = pr/2t Superposition of stress components can be used to determine the normal and shear stress at a pt in a member subjected to a combined loading. To solve, it is first necessary to determine the resultant axial and shear force and the resultant torsional and bending moment at the section where the pt is located. Then the stress components are determined due to each of these loadings. The normal and shear stress resultants are then determined by algebraically adding the normal and shear stress components. CHAPTER 9 Derive equations for transforming stress components between coordinate systems of different orientation Use derived equations to obtain the maximum normal and maximum shear stress at a pt Determine the orientation of elements upon which the maximum normal and maximum shear stress acts Discuss a method for determining the absolute maximum shear stress at a point when material is subjected to plane and 3 dimensional states of stress Plane stress occurs when the material at a pt is subjected to two normal stress components s x and s y and a shear stress t xy. Provided these components are known, then the stress components acting on an element having a different orientation can be determined using the two force equations of equilibrium or the equations of stress transformation. For design, it is important to determine the orientations of the element that produces the maximum principal normal stresses and the maximum in plane shear stress. Using the stress transformation equations, we find that no shear stress acts on the planes of principal stress. The planes of maximum in plane shear stress are oriented 45 from this orientation, and on these shear planes there is an associated average normal stress (s x + sy)/2. Mohr s circle provides a semi graphical aid for finding the stress on any plane, the principal normal stresses, and the maximum in plane shear stress.
To draw the circle, the s and t axes are established, the center of the circle [(s x + s y)/2, 0], and the controlling pt (s x,t xy )areplotted. The radius of the circle extends between these two points and is determined from trigonometry. CHAPTER 10 Apply the stress transformation methods derived in Chapter 9 to similarly transform strain Discuss various ways of measuring strain Develop important material property relationships; including generalized form of Hooke s law Discuss and use theories to predict the failure of a material When element of material is subjected to deformations that only occur in a single plane, then it undergoes plain strain. If the strain components e x, e y, and g xy are known for a specified orientation of the element, then the strains acting for some other orientation of the element can be determined using the plane strain transformation equations. Likewise, principal normal strains and maximum in plane shear strain can be determined using transformation equations. Straintransformationproblemscanbe solved in a semi graphical manner using Mohr s circle. Establish the e and g/2 axes, then compute center of circle [(e x + e y )/2, 0] and controlling pt [e, g/2], before plotting the circle. Radiusofcircleextendsbetweenthesetwoptsandisdeterminedfromtrigonometry. Absolute maximum shear strain equals the maximum in plane shear strain provided the in plane principal strains are of opposite signs. If the in plane principal strains are of same signs, then absolute maximum shear strain will occur out of plane and is determined from g max = e max /2. Hooke s law can be expressed in 3 dimensions, where each strain is related to the 3 normal stress components using the material properties E, andu, as seen in Eqns 10 18. If E and u are known, then G can be determined using G = E/[2(1 + u]. Dilatation is a measure of volumetric strain, and the bulk modulus is used to measure the stiffness of a volume of material. Provided the principal stresses for a material are known, then a theory of failure can be used as a basis for design. Ductile materials fail in shear, and here the maximum shear stress theory or the maximum distortion energy theory can be used to predict failure. Both theories make comparison to the yield stress of a specimen subjected to uniaxial stress. Brittle materials fail in tension, and so the maximum normal stress theory or Mohr s failure criterion can be used to predict failure. Comparisons are made with the ultimate tensile stress developed in a specimen.
CHAPTER 11 Design a beam to resist both bending and shear loads. Develop methods used for designing prismatic beams. Determine shape of fully stressed beams. Design of shafts based on both bending and torsional moments. Failure of a beam occurs when the internal shear or moment in the beam is a maximum. To resist these loadings, it is important that the associated maximum shear and bending stress not exceed allowable values as stated in codes. Normally, the x section of a beam is first designed to resist the allowable bending stress, then the allowable shear stress is checked. For rectangular sections, allow t 1.5 ( V max /A). For wide flange sections, we use allow t V max /A web For built up beams, the spacing of fasteners or the strength of glue or weld is determined using an allowable shear flow, q flow = VQ/I. Fully stressed beams are nonprismatic and designed such that each x section along the beam will resist the allowable bending stress. This will define the shape of the beam. A mechanical shaft generally is designed to resist both torsion and bending stresses. Normally, bending can be resolved into two planes, and so it is necessary to draw the moment diagrams for each bending moment component and then select the maximum moment based on vector addition. Once the maximum bending and shear stresses are determined, then depending upon the type of material, an appropriate theory of failure is used to compare the allowable stress to what is required. CHAPTER 12 Use various methods to determine the deflection and slope at specific pts on beams and shafts: 1. Integration method 2. Discontinuity functions 3. Method of superposition 4. Moment area method Use the various methods to solve for the support reactions on a beam or shaft that is statically indeterminate The elastic curve represents the centerline deflection of a beam or shaft. Its shape can be determined using the moment diagram. Positive moments cause the elastic curve to concave upwards and negative moments cause it to concave downwards. The radius of curvature at any pt is determined from 1/r = M/EI. Eqn of elastic curve and its slope can be obtained by first finding the internal moment in the member as a function of x.
If several loadings act on the member, then separate moment functions must be determined between each of the loadings. Integrating these functions once using EI(d 2 n/dx 2 ) = M(x) gives the eqn for the slope of the elastic curve, and integrating again gives the eqn for the deflection. The constants of integration are determined from the boundary conditions at the supports, or in cases where several moment functions are involved, continuity of slope and deflection at pts where these functions join must be satisfied. Discontinuity functions allow us to express the eqn of elastic curve as a continuous function, regardless of the no. of loadings on the member. This method eliminates the need to use continuity conditions, since the two constants of integration can be determined solely from the two boundary conditions. The moment area method if a semi graphical technique for finding the slope of tangents or the vertical deviation of tangents at specific pts on the elastic curve. The moment area method requires finding area segments under the M/EI diagram, or the moment of these segments about pts on the elastic curve. The method works well for M/EI diagrams composed of simple shapes, such as those produced by concentrated forces and couple moments. The deflection or slope at a pt on a member subjected to various types of loadings can be determined by using the principle of superposition. The table in the back of the book can be used for this purpose. Statically indeterminate beams and shafts have more unknown support reactions than available eqns of equilibrium. To solve them such problems, we first identify the redundant reactions, and the other unknown reactions are written in terms of these redundants. The method of integration or moment area theorems can be used to solve for the unknown redundants. We can also determine the redundants by using the method of superposition, where we consider the continuity of displacement at the redundant. The displacement due to the external loading is determined with the redundant removed, and again with the redundant applied and external loading removed. The tables in Appendix C of this book can be used to determine these necessary displacements.
CHAPTER 13 Discuss the behavior of columns. Discuss the buckling of columns. Determine the axial load needed to buckle an ideal column. Analyze the buckling with bending of a column. Discuss inelastic buckling of a column. Discuss methods used to design concentric and eccentric columns. Buckling is the sudden instability that occurs in columns or members that support an axial load. The maximum axial load that a member can support just before buckling occurs is called the critical load P cr. The critical load for an ideal column is determined from the Euler eqn, P cr = p 2 EI/(KL) 2, where K = 1 for pin supports, K = 0.5 for fixed supports, K = 0.7 for a pin and a fixed support, and K =2forafixedsupportandafreeend. If axial loading is applied eccentrically to the column, then the secant formula must be used to determine the maximum stress in the column. When the axial load tends to cause yielding of the column, then the tangent modulus should be used with Euler s eqn to determine the buckling load. This is referred to as Engesser s eqn. Empirical formulae based upon experimental data have been developed for use in the design of steel, aluminum, and timber columns. CHAPTER 14 Apply energy methods to solve problems involving deflection Discuss work and strain energy, and development of the principle of conservation of energy Use principle of conservation of energy to determine stress and deflection of a member subjected to impact Develop the method of virtual work and Castigliano s theorem Use method of virtual and Castigliano s theorem to determine displacement and slope at pts on structural members and mechanical elements When a force (or couple moment) acts on a deformable body it will do external work when it displaces (or rotates). The internal stresses produced in the body also undergo displacement, thereby creating elastic strain energy that is stored in the material. The conservation of energy states that the external work done by the loading is equal to the internal strain energy produced in the body. This principal can be used to solve problems involving elastic impact, which assumes the moving body is rigid and all strain energy is stored in the stationary body. The principal of virtual work can be used to determine the displacement of a joint on a truss or the slope and the displacement of pts on a beam or frame.
It requires placing an entire virtual unit force (or virtual unit couple moment) at the pt where the displacement (or rotation) is to be determined. The external virtual work developed is then equated to the internal virtual strain energy in thememberorstructure. Castigliano s theorem can also be used to determine the displacement of a joint on a truss or slope or the displacement of a pt on a beam or truss. Here a variable force P (or couple moment M) is placed at the pt where the displacement (or slope) is to be determined. The internal loading is then determined as a function of P (or M) and its partial derivative w.r.t. P (or M) is determined. Castigliano stheoremisthenappliedtoobtainthedesireddisplacement (or rotation).