CHAPTER OBJECTIVES Apply energy methods to solve problems involving deflection Discuss work and strain energy, and development of the principle of conservation.

CHAPTER OBJECTIVES 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

CHAPTER OBJECTIVES Use method of virtual and Castigliano’s theorem to determine displacement and slope at pts on structural members and mechanical elements

CHAPTER OUTLINE External Work and Strain Energy Elastic Strain Energy for Various Types of Loading Conservation of Energy Impact Loading *Principle of Virtual Work *Method of Virtual Forces Applied to Trusses *Method of Virtual Forces Applied to Beams *Castigliano’s Theorem *Castigliano’s Theorem Applied to Trusses *Castigliano’s Theorem Applied to Beams

14.1 EXTERNAL WORK AND STRAIN ENERGY
Work of a force: A force does work when it undergoes a displacement dx in same direction as the force. Work done is a scalar, defined as dUe = F dx. If total displacement is x, work becomes As magnitude of F is gradually increased from zero to limiting value F = P, final displacement of end of bar becomes .

Work of a force: For linear-elastic behavior of material, F = (P/)x. Substitute into Eqn 14-1 Suppose that P is already applied to the bar and another force P’ is now applied, so end of bar is further displaced by an amount ’. Work done by P (not P’) is then

Work of a force: When a force P is applied to the bar, followed by the force P’, total work done by both forces is represented by the area of the entire triangle in graph shown.

Work of a couple moment: A couple moment M does work when it undergoes a rotational displacement d along its line of action. Work done is defined as dUe = Md. If total angle of rotational displacement is  radians, then work If the body has linear-elastic behavior, and its magnitude increases gradually from zero at  = 0 to M at , then work is

Work of a couple moment: However, if couple moment already applied to the body and other loadings further rotate the body by an amount ’, then work done is

When loads are applied to a body and causes deformation, the external work done by the loads will be converted into internal work called strain energy. This is provided no energy is converted into other forms. Normal stress A volume element subjected to normal stress z. Force created on top and bottom faces is dFz = z dA = z dx dy.

Normal stress This force is increased gradually from zero to dFz while element undergoes displacement dz = z dz. Work done is dUi = 0.5dFz dz = 0.5[z dx dy]z dz. Since volume of element is dV = dx dy dz, we have Note that dUi is always positive.

Normal stress In general, for a body subjected to a uniaxial normal stress , acting in a specified direction, strain energy in the body is then If material behaves linear-elastically, then Hooke’s law applies and we express it as

Shear stress Shear stress cause element to deform such that shear force dF = (dx dy) acts on top face of element. Resultant displacement if  dz relative to bottom face. Vertical faces only rotate, thus shear forces on these faces do no work.

Shear stress Hence, strain energy stored in the element is Integrating over body’s entire volume to obtain strain energy stored in it Shear strain energy is always positive.

Shear stress Apply Hooke’s law  = /G,

Mutilaxial stress Total strain energy in the body is therefore

Mutilaxial stress Eliminate the strains using generalized form of Hooke’s law given by Eqns and After substituting and combining terms, we have

Mutilaxial stress If only principal stresses 1, 2, 3 act on the element, this eqn reduces to a simpler form,

14.2 ELASTIC STRAIN ENERGY FOR VARIOUS TYPES OF LOADING
Axial load: Consider a bar of variable and slightly tapered x-section, subjected to axial load coincident with bar’s centroidal axis. Internal axial force at section located from one end is N. If x-sectional area at this section is A, then normal stress  = N/A. Apply Eqn 14-8, we have

Axial load: Choose element or differential slice having volume dV = Adx, general formula for strain energy in bar is For a prismatic bar of constant x-sectional area A, length L and constant axial load N, integrating Eqn gives

EXAMPLE 14.1 Choose one of the 2 high-strength steel bolts to support a tensile loading. Determine the greatest amount of elastic strain energy that each bolt can absorb. Bolt A has a diameter of 20 mm for 50 mm of its length and root diameter of 18 mm within 6 mm threaded region. Bolt B has the same diameter throughout its 56 mm length and can be taken as 18 mm. For both cases, neglect extra material that makes up the thread. Take Est = 210(103) MPa, Y = 310 MPa.

EXAMPLE 14.1 (SOLN) Bolt A: For bolt subjected to maximum tension, Y will occur within the 6-mm region. This tension is

EXAMPLE 14.1 (SOLN) Bolt A: Apply Eqn to each region of the bolt,

EXAMPLE 14.1 (SOLN) Bolt B: From calculation above, it can also support a maximum tension force of Pmax = kN. Thus, By comparison, bolt B can absorb 20% more elastic energy than bolt A, even though it has a smaller x-section along its shank.

Bending moment: For the axisymmetric beam shown. Internal moment M, normal stress acting on element a distance y from neutral axis is  = My/I. If volume of element is dV = dA dx, where dA is area of exposed face and dx its length, elastic strain energy in beam is

Bending moment: Realize that area integral represents the moment of inertia of beam about neutral axis, thus

EXAMPLE 14.2 Determine the elastic strain energy due to bending of the cantilevered beam if beam is subjected to uniform distributed load w. EI is constant.

EXAMPLE 14.2 (SOLN) Establish the x coordinate with origin at the left side. Thus, internal moment is Applying Eqn yields

EXAMPLE 14.2 (SOLN) For x coordinate with origin on the the right side and extending +ve to the left. Thus, in this case Applying Eqn 14-17, we obtain the same result.

Transverse shear: Consider prismatic beam with axis of symmetry about the y axis. Internal shear V at section x results in shear stress acting on the volume element, having length dx and area dA, is  = VQ/It. Substitute into Eqn 14-11,

Transverse shear: Realize that integral in parentheses is evaluated over beam’s x-sectional area. To simplify, we define the form factor for shear as Form factor is dimensionless and unique for each specific x-sectional area. Substitute Eqn into above eqn,

EXAMPLE 14.4 Determine the strain energy in cantilevered beam due to shear if beam has a square x-section and is subjected to a uniform distributed load w. EI and G is constant.

EXAMPLE 14.4 (SOLN) From free-body diagram of arbitrary section, we have Since x-section is square, form factor fs = 6/5 and therefore Eqn becomes

EXAMPLE 14.4 (SOLN) Using results of Example 14.2, with A = a2, I = 1/12a4, ratio of shear to bending strain energy is Since G = E/2(1 + ) and   0.5 (sec 10.6), then as an upper bound, E = 3G, so that For L = 5a, contributions due to shear strain energy is only 8% of bending strain energy. Thus, shear strain energy is usually neglected in engineering analysis.

Torsional moment: Consider slightly tapered shaft. Section of shaft taken distance x from one end subjected to internal torque T. On arbitrary element of length dx and area dA, stress is  = T/J. Strain energy stored in shaft is

Torsional moment: Since area integral represents the polar moment of inertia J for shaft at section, Most common case occurs when shaft has constant x-sectional area and applied torque is constant, integrating Eqn gives

Torsional moment: If x-section is of other shapes than circular or tubular, Eqn is modified. For example, for a rectangular shaft with dimensions h > b,

EXAMPLE 14.5 Tubular shaft fixed at the wall and subjected to two torques as shown. Determine the strain energy stored in shaft due to this loading. G = 75 GPa.

EXAMPLE 14.5 (SOLN) Using method of sections, internal torque first determined within the two regions of shaft where it is constant. Although torques are in opposite directions, this will not affect the value of strain energy, since torque is squared in Eqn

EXAMPLE 14.5 (SOLN) Polar moment of inertia for shaft is Applying Eqn 14-22, we have

IMPORTANT A force does work when it moves through a displacement. If force is increased gradually in magnitude from zero to F, the work is U = (F/2), whereas if force is constant when the displacement occurs then U = F. A couple moment does work when it moves through a rotation. Strain energy is caused by the internal work of the normal and shear stresses. It is always a positive quantity.

IMPORTANT The strain energy can be related to the resultant internal loadings N, V, M, and T. As the beam becomes longer, the strain energy due to bending becomes much larger than strain energy due to shear. For this reason, shear strain energy in beams can generally be neglected.

14.3 CONSERVATION OF ENERGY
A loading is applied slowly to a body, so that kinetic energy can be neglected. Physically, the external loads tend to deform the body as they do external work Ue as they are displaced. This external work is transformed into internal work or strain energy Ui, which is stored in the body. Thus, assuming material’s elastic limit not exceeded, conservation of energy for body is stated as

Consider a truss subjected to load P. P applied gradually, thus Ue = 0.5P, where  is vertical displacement of truss at pt where P is applied. Assume that P develops an axial force N in a particular member, and strain energy stored is Ui = N2L/2AE. Summing strain energies for all members of the truss, we write Eqn as

Consider a beam subjected to load P. External work is Ue = 0.5P. Strain energy in beam can be neglected. Beam’s strain energy determined only by the moment M, thus with Eqn 14-17, Eqn written as

Consider a beam loaded by a couple moment M0. A rotational displacement  is caused. Using Eqn 14-5, external work done is Ue = 0.5M0. Thus Eqn becomes Note that Eqn is only applicable for a single external force or external couple moment acting on structure or member.

EXAMPLE 14.6 The three-bar truss is subjected to a horizontal force of 20 kN. If x-sectional area of each member is 100 mm2, determine the horizontal displacement at pt B. E = 200 GPa.

EXAMPLE 14.6 (SOLN) Since only a single external force acts on the truss and required displacement is in same direction as the force, we use conservation of energy. Also, the reactive force on truss do no work since they are not displaced. Using method of joints, force in each member is determined as shown on free-body diagrams of pins at B and C.

EXAMPLE 14.6 (SOLN) Applying Eqn 14-26,

EXAMPLE 14.6 (SOLN) Substituting in numerical data for A and E and solving, we get

EXAMPLE 14.7 Cantilevered beam has a rectangular x-section and subjected to a load P at its end. Determine the displacement of the load. EI is a constant.

EXAMPLE 14.7 (SOLN) Internal moment and moment in beam as a function of x are determined using the method of sections. When applying Eqn we will consider the strain energy due to shear and bending.

EXAMPLE 14.7 (SOLN) Using Eqns and 14-17, we have First term on the right side represents strain energy due to shear, while the second is due to bending. As stated in Example 14.4, the shear strain energy in most beams is much smaller than the bending strain energy.

EXAMPLE 14.7 (SOLN) To show this is the case, we require Since E  3G (see Example 14.4) then

EXAMPLE 14.7 (SOLN) Hence, if h is small and L relatively long, beam becomes slender and shear strain energy can be neglected. Shear strain energy is only important for short, deep beams. Beams for which L = 5h have more than 28 times more bending energy than shear strain energy, so neglecting only incurs an error of about 3.6%. Eqn (1) can be simplified to

14.4 IMPACT LOADING An impact occurs when one object strikes another, such that large forces are developed between the objects during a very short period of time.

14.4 IMPACT LOADING Solving and simplifying (st = W/k), Once max is computed, maximum force applied to the spring is

14.4 IMPACT LOADING For a case where the block is sliding on a smooth horizontal surface with known velocity  just before it collides with the spring. The block’s kinetic energy, 0.5(W/g)2 is transformed into stored energy in the spring.

14.4 IMPACT LOADING Ratio of equivalent static load Pmax to the load W is called the impact factor, n. Since Pmax = kmax and W = kst, then from Eqn , we express it as This factor represents the magnification of a statically applied load so that it can be treated dynamically. Using Eqn 13-34, n can be computed for any member that has a linear relationship between load and deflection.

14.4 IMPACT LOADING IMPORTANT Impact occurs when a large force is developed between two objects which strike one another during a short period of time. We can analyze the effects of impact by assuming the moving body is rigid, the material of the stationary body is linearly elastic, no energy is lost in the collision, the bodies remain in contact during collision, and inertia of elastic body is neglected. The dynamic load on a body can be treated as a statically applied load by multiplying the static load by a magnification factor.

EXAMPLE 14.8 Aluminum pipe is used to support a load of 600 kN. Determine the maximum displacement at the top of the pipe if load is (a) applied gradually, and (b) applied suddenly by releasing it from the top of the pipe at h = 0. Take Eal = 70(103) N/mm2 and assume that the aluminum behaves elastically.

EXAMPLE 14.8 (SOLN) (a) When load applied gradually, work done by weight is transformed into elastic strain energy in pipe. Applying conservation of energy,

EXAMPLE 14.8 (SOLN) (b) With h = 0, apply Eqn Hence The displacement of the weight is twice as great as when the load is applied statically. In other words, the impact factor is n = 2, Eqn

EXAMPLE 14.10 A railroad car assumed to be rigid and has a mass of 80 Mg is moving forward at a speed of  = 0.2 m/s when it strikes a steel 200-mm by 200-mm post at A. If the post is fixed to the ground at C, determine the maximum horizontal displacement of its top B due to the impact. Take Est = 200 GPa.

EXAMPLE (SOLN) Kinetic energy of the car is transformed into internal bending strain energy only for region AC of the post.. Assume that pt A is displaced (A)max, then force Pmax that causes this displacement can be determined from table in Appendix C.

EXAMPLE (SOLN) Substitute in numerical data yields Using Eqn (1), force Pmax becomes

EXAMPLE (SOLN) Refer to figure, segment AB of post remains straight. To determine displacement at B, we must first determine slope at A. Using formula from table in Appendix C to determine A, we have

EXAMPLE (SOLN) The maximum displacement at B is thus

*14.5 PRINCIPLE OF VIRTUAL WORK
Principle of Virtual Work was developed by John Bernoulli in 1717. It is a energy method of analysis and based on conservation of energy. Equilibrium conditions require the external loads to be uniquely related to the internal loads. Compatibility conditions require the external displacements to be uniquely related to the internal deformations.

When we apply a series of external loads P to a deformable body, these loadings will cause internal loadings u within the body. The external loads will be displaced Δ, and internal loadings will undergo displacements . Conservation of energy states that Based on this concept, we now develop the principle of virtual work to be used to determine the displacement and slope at any pt on a body.

Consider a body or arbitrary shape acted upon by “real loads” P1, P2 and P3.

There is no force acting on A, so unknown displacement Δ will not be included as an external “work term” in the eqn. We then place and imaginary or “virtual” force P’ on body at A, such that it acts in the same direction as Δ. For convenience, we choose P’ = 1. This external virtual load cause an internal virtual load u in a representative element of fiber of body. P’ and u is related by the eqns of equilibrium. Real loads at pt A displaced by , which causes element to be displaced dL.

Thus, external virtual force P’ and internal virtual load u “ride along” by Δ and dL respectively; these loads perform external virtual work of 1·Δ on the body and internal virtual work of u·dL on the element. Consider only the conservation of virtual energy, we write the virtual-work eqn as Virtual loadings Real displacements

P’ = 1 = external virtual unit load acting in direction of Δ. u = internal virtual load acting on the element. Δ = external displacement caused by real loads. dL = internal displacement of element in direction of u, caused by real loads. Virtual loadings Real displacements

The choice of P’ = 1 will give us a direct solution for Δ, Δ = ∑u dL. Similarly, for rotational displacement or slope of tangent at a pt on the body, virtual couple moment M’ having unit magnitude, is applied at a pt. Thus, a virtual load u is caused in one of the elements. Assume that real loads deform element by dL, rotation  can be found from virtual-work eqn: Virtual loadings Real displacements

M’ = 1 = external virtual unit couple moment acting in direction of . u = internal virtual load acting on an element.  = external rotational displacement in radians caused by the real loads. dL = internal displacement of element in direction of u, caused by real loads. Virtual loadings Real displacements

Internal virtual work: Terms on right-hand side of Eqns and represent the internal virtual work in the body. If we assume material behavior is linear-elastic and stress does not exceed proportional limit, we can formulate expressions for internal virtual work caused by stress. We then use eqns of elastic strain energy developed in chapter 14.2. They are listed in a table on next slide.

Internal virtual work:

Internal virtual work: Thus we can write the virtual-work eqn for a body subjected to a general loading as

*14.6 METHOD OF VIRTUAL FORCES APPLIED TO TRUSSES
Displacement at joint A caused by “real loads” P1 and P2, and since these loads only cause axial force in members, we need only consider internal virtual work due to axial load. Assume each member has a constant x-sectional area A, virtual load n and real load N are constant throughout member’s length. As a result, virtual work for entire truss is

1 = external virtual unit load acting on the truss joint in the stated direction of Δ. Δ = joint displacement caused by the real loads on the truss. n = internal virtual force in a truss member caused by the external virtual unit load. N = internal force in a truss member caused by the real loads. L = length of a member. A = x-sectional area of a member. E = modulus of elasticity of a member.

Temperature change: Truss members can change their length due to a change in temperature. Thus, we determine the displacement of a selected truss joint due to temperature change from Eqn 14-36, 1 = external virtual unit load acting on the truss joint in the stated direction of Δ. n = internal virtual force in a truss member caused by the external virtual unit load.

Temperature change: Δ = external joint displacement caused by the temperature change  = coefficient of thermal expansion of member. ΔT = change in temperature of member. L = length of member

Fabrication errors. Displacement in a particular direction of a truss joint from its expected position can be determined from direct application of Eqn 14-36, 1 = external virtual unit load acting on the truss joint in stated direction of Δ. n = internal virtual force in a truss member caused by the external virtual unit load.

Fabrication errors. Δ = external joint displacement caused by the fabrication errors. ΔL = difference in length of the member from its intended length caused by fabrication error.

Procedure for analysis Virtual forces n Place the virtual unit load on the truss at the joint where the desired displacement is to be determined. The load should be directed along line of action of the displacement. With unit load so placed and all real loads removed from truss, calculate the internal n force in each truss member. Assume that tensile forces are +ve and compressive forces are –ve.

Procedure for analysis Real forces N Determine the N forces in each member. These forces are caused only by the real loads acting on the truss. Again, assume that tensile forces are +ve and compressive forces are –ve. Virtual-work eqn Apply eqn of virtual work to determine the desired displacement.

Procedure for analysis Virtual-work eqn It is important to retain the algebraic sign for each of the corresponding n and N forces while substituting these terms into the eqn. If resultant sum ∑nNL/AE is +ve, displacement Δ is in the same direction as the virtual unit load. If a –ve value results, Δ is opposite to the virtual unit load.

Procedure for analysis Virtual-work eqn When applying 1·Δ = ∑n ΔTL, realize that if any members undergo an increase in temperature, ΔT will be +ve; whereas a decrease in temperature will result in a –ve value for ΔT. For 1·Δ = ∑n ΔL, when a fabrication error increases the length of a member, ΔL is +ve, whereas a decrease in length is –ve. When applying this method, attention should be paid to the units of each numerical qty.

Procedure for analysis Virtual-work eqn Notice however, that the virtual unit load can be assigned any arbitrary unit: pounds, kips, newtons, etc., since the n forces will have these same units, and as a result, the units for both the virtual unit load and the n forces will cancel from both sides of the eqn.

EXAMPLE 14.11 Determine the vertical displacement of joint C of steel truss. X-sectional area of each member is A = 400 mm2 and Est = 200 GPa.

EXAMPLE (SOLN) Virtual forces n We only place a vertical 1-kN virtual load at C; and the force in each member is calculated using the method of joints. Results are shown below. Using sign convention of +ve numbers for tensile forces and –ve numbers indicate compressive forces.

EXAMPLE (SOLN) Real forces N Applied load of 100 kN causes forces in members that can be calculated using method of joints. Results are shown below.

Virtual-work Equation Arranging data in the table below:
EXAMPLE (SOLN) Virtual-work Equation Arranging data in the table below: Member n N L nNL AB 100 4 BC 141.4 2.828 AC 1.414 141.4 565.7 CD 1 200 2 400 ∑ kN2·m

EXAMPLE (SOLN) Virtual-work Equation Thus Substituting the numerical values for A and E, we have

EXAMPLE 14.12 X-sectional area of each member of the steel truss is A = 300 mm2, and the modulus of elasticity for the steel members is Est = 210(103) MPa. (a) Determine the horizontal displacement of joint C if a force of 60 kN is applied to the truss at B. (b) If no external loads act on the truss, what is the horizontal displacement of joint C if member AC is fabricated 6 mm too short?

EXAMPLE (SOLN) a) Virtual forces n. A horizontal force of 1 kN is applied at C. The n force in each member is determined by method of joints. As usual, +ve represents tensile force and –ve represents compressive force.

EXAMPLE (SOLN) a) Real forces N. Force in each member as caused by externally applied 60 kN force is shown.

a) Virtual-work Equation
EXAMPLE (SOLN) a) Virtual-work Equation Since AE is constant, data is arranged in the table: Member n N L nNL AB 1.5 AC 1.25 75 2.5 CB 60 2 CD 0.75 45 50.625 ∑ 285 (kN)2·m

EXAMPLE (SOLN) a) Virtual-work Equation Substituting the numerical values for A and E, we have

EXAMPLE (SOLN) b) Here, we must apply Eqn Realize that member AC is shortened by ΔL = 6 mm, we have The –ve sign indicates that joint C is displaced to the left, opposite to the 1-kN load.

EXAMPLE 14.13 Determine the horizontal displacement of joint B of truss. Due to radiant heating, member AB is subjected to an increase in temperature ΔT = +60C. The members are made of steel, for which st = 12(10-6)/C and Est = 200 GPa. The x-sectional area of each member is 250 mm2.

EXAMPLE (SOLN) Virtual forces n. A horizontal 1-kN virtual load is applied to the truss at joint B, and forces in each member are computed.

EXAMPLE (SOLN) Real forces n. Since n forces in members AC and BC are zero, N forces in these members do not have to be determined. Why? For completeness, though, the entire “real” force analysis is shown.

EXAMPLE (SOLN) Virtual-work Equation. Both loads and temperature affect the deformation, thus Eqns and are combined, Negative sign indicates that roller B moves to the right, opposite to direction of virtual load.

*14.7 METHOD OF VIRTUAL WORK APPLIED TO BEAMS
Applying Eqn 14-36, virtual-work eqn for a beam is 1 = external virtual unit load acting on the beam in direction of Δ. Δ = displacement caused by the real loads acting on the beam. m = internal virtual moment in the beam, expressed as a function of x and caused by the external virtual unit load.

M = internal moment in the beam, expressed as a function of x and caused by the real loads. E = modulus of elasticity of material. I = moment of inertia of x-sectional area, computed about the neutral axis.

Similarly, for virtual couple moment and to determine corresponding virtual moment m, we apply Eqn for this case, Note that the integrals in Eqns and represent the amount of virtual bending strain energy stored in the beam. If concentrated forces or couple moments act on beam or distributed load is discontinuous, we’ll need to choose separate x coordinates within regions without discontinuities.

Procedure for analysis Virtual moments m or m. Place a virtual unit load on the beam at the pt and directed along the line of action of the desired displacement. If slope is to be determined, place a virtual unit couple moment at the pt. Establish appropriate x coordinates that are valid within regions of the beam where there is no discontinuity of real or virtual load.

Procedure for analysis Virtual moments m or m. With virtual load in place, and all the real loads removed from the beam, calculate the internal moment m or m as a function of each x coordinate. Assume that m or m acts in the +ve direction according to the established beam sign convention. Real moments. Using the same x coordinates as those established for m or m , determine the internal moments M caused by the real loads.

Procedure for analysis Real moments. Since +ve m or m was assumed to act in the conventional “positive direction,” it is important that +ve M acts in this same direction. This is necessary since +ve or –ve internal virtual work depends on the directional sense of both the virtual load, defined by  m or m , and displacement caused by M .

Procedure for analysis Virtual-work equation. Apply eqn of virtual work to determine the desired displacement Δ or slope . It is important to retain the algebraic sign of each integral calculated within its specified region. If algebraic sum of all the integrals for entire beam is +ve, Δ or  is in the same direction as the virtual unit load or virtual unit couple moment, respectively. If a –ve value results, Δ or  is opposite to virtual unit load or couple moment.

EXAMPLE 14.15 Determine the slope at pt B of the beam shown. EI is a constant.

EXAMPLE (SOLN) Virtual moments m. Slope at B is determined by placing a virtual unit couple moment at B. Two x coordinates must be selected to determine total virtual strain energy in the beam. Coordinate x1 accounts for strain energy within segment AB, and coordinate x2 accounts for the strain energy in segment BC. Internal moment m within each of these segments are computed using the method of sections.

EXAMPLE (SOLN) Real moments M. Using same coordinates x1 and x2 (Why?), the internal moments M are computed as shown.

EXAMPLE (SOLN) Virtual-work equation. Slope at B is thus Negative sign indicates that B is opposite to direction of the virtual couple moment.

EXAMPLE 14.16 Determine the displacement of pt A of the steel beam shown. I = 175.8(10-6) m4, Est = 200 GPa.

EXAMPLE (SOLN) Virtual moments m. Beam subjected to virtual unit load at A and reactions are computed. By inspection, two coordinates x1 and x2 must be chosen to cover all regions of the beam. For integration, it is simplest to use origins at A and C. Using method of sections, the internal moments m are shown.

EXAMPLE (SOLN) Real moments M. Reactions on beam are found first. Then, using same x coordinates as those found for m, internal moments M are determined.

EXAMPLE (SOLN) Virtual-work equation.

EXAMPLE (SOLN) Virtual-work equation. Substitute in data for E and I, we get The negative sign indicates that pt A is displaced upward.

*14.8 CASTIGLIANO’S THEOREM
This method was discovered in 1879 by Alberto Castigliano to determine the displacement and slope at a pt in a body. It applies only to bodies that have constant temperature and material with linear-elastic behavior. His second theorem states that displacement is equal to the first partial derivative of strain energy in body w.r.t. a force acting at the pt and in direction of the displacement.

Consider a body of arbitrary shape subjected to a series of n forces P1, P2, … Pn. Since external work done by forces is equal to internal strain energy stored in body, by conservation of energy, Ue = Ui. However, external work is a function of external loads Ue = ∑ ∫ P dx.

So, internal work is also a function of the external loads. Thus Now, if any one of the external forces say Pj is increased by a differential amount dPj. Internal work increases, so strain energy becomes Further application of the loads cause dPj to move through displacement Δj, so strain energy becomes

dUj = dPjΔi is the additional strain energy caused by dPj. In summary, Eqn represents the strain energy in the body determined by first applying the loads P1, P2, …, Pn, then dPj. Eqn represents the strain energy determined by first applying dPj, then the loads P1, P2, …, Pn. Since theses two eqns are equal, we require

Note that Eqn is a statement regarding the body’s compatibility requirements, since it’s related to displacement. The derivation requires that only conservative forces be considered for analysis.

*14.9 CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
Since a truss member is subjected to an axial load, strain energy is given by Eqn 14-16, Ui = N2L/2AE. Substitute this eqn into Eqn and omitting the subscript i, we have It is easier to perform differentiation prior to summation. Also, L, A and E are constant for a given member, thus

Δ = joint displacement of the truss. P = external force of variable magnitude applied to the truss joint in direction of Δ. N = internal axial force in member caused by both force P and loads on the truss. L = length of a member. A = x-sectional area of a member. E = modulus of elasticity of the material.

In order to determine the partial derivative N/P, we need to treat P as a variable, not numeric qty. Thus, each internal axial force N must be expressed as a function of P. By comparison, Eqn is similar to that used for method of virtual work, Eqn 14-39, except that n is replaced by N/P. These terms; n and N/P, are the same, since they represent the rate of change of internal axial force w.r.t. the load P.

Procedure for analysis External force P. Place a force P on truss at the joint where the desired displacement is to be determined. This force is assumed to have a variable magnitude and should be directed along the line of action of the displacement. Internal forces N. Determine the force N in each member caused by both the real (numerical) loads and the variable force P. Assume that tensile forces are +ve and compressive forces are –ve.

Procedure for analysis Internal forces N. Find the respective partial derivative N/P for each member. After N and N/P have been determined, assign P its numerical value if it has actually replaced a real force on the truss. Otherwise, set P equal to zero. Castigliano’s Second Theorem. Apply Castigliano’s second theorem to determine the desired displacement Δ.

Procedure for analysis Castigliano’s Second Theorem. It is important to retain the algebraic signs for corresponding values of N and N/P when substituting these terms into the eqn. If the resultant sum ∑ N (N/P) L/AE is +ve, Δ is in the same direction as P. If a –ve value results, Δ is opposite to P.

EXAMPLE 14.17 Determine the horizontal displacement of joint C of steel truss shown. The x-sectional area of each member is also indicated. Take Est = 210(103) N/mm2.

EXAMPLE (SOLN) External force P. Since horizontal displacement of C is to be determined, a horizontal variable force P is applied to joint C. Later this force will be set equal to the fixed value of 40 kN.

EXAMPLE 14.17 (SOLN) Internal forces N.
Using method of joints, force N in each member is found. Results are shown in table: Member N N/P N (P = 40 kN) L N(N/P)L AB 4000 BC 3000 AC 1.67P 1.67 66.67(103) 5000 556.7(106) CD 1.33P 1.33 -53.33(103) 283.7(106)

EXAMPLE (SOLN) Castigliano’s Second Theorem Applying Eqn 14-8 gives

EXAMPLE (SOLN) b) Here, we must apply Eqn Realize that member AC is shortened by ΔL = 6 mm, we have The –ve sign indicates that joint C is displaced to the left, opposite to the 1-kN load.

*14.10 CASTIGLIANO’S THEOREM APPLIED TO BEAMS
Internal strain energy for a beam is caused by both bending and shear. As pointed out in Example 14.7, if beam is long and slender, strain energy due to shear can be neglected. Thus, internal strain energy for a beam is given by Eqn 14-17; Ui = ∫M2 dx/2EI. We then substitute into Δi = Ui/Pi, Eqn and omitting subscript i, we have

It is easier to differentiate prior to integration, thus provided E and I are constant, we have Δ = displacement of the pt caused by the real loads acting on the beam. P = external force of variable magnitude applied to the beam in the direction of Δ.

M = internal moment in the beam, expressed as a function of x and caused by both the force P and the loads on the beam. E = modulus of elasticity of the material. I = moment of inertia of x-sectional area computed about the neutral axis.

If slope of tangent  at a pt on elastic curve is to be determined, the partial derivative of internal moment M w.r.t. an external couple moment M’ acting at the pt must be found. For this case The eqns above are similar to those used for the method of virtual work, Eqns and 14-43, except m and m0 replace M/P and M/M’, respectively.

If the loading on a member causes significant strain energy within the member due to axial load, shear, bending moment, and torsional moment, then the effects of all these loadings should be included when applying Castigliano’s theorem.

Procedure for analysis External force P or couple moment M’. Place force P on the beam at the pt and directed along the line of action of the desired displacement. If the slope of the tangent is to be determined, place a couple moment M’ at the pt. Assume that both P and M’ have a variable magnitude.

Procedure for analysis Internal moment M. Establish appropriate x coordinates that are valid within regions of the beam where there is no discontinuity of force, distributed load, or couple moment. Calculate the internal moments M as a function of P or M’ and the partial derivatives M/P or M/M’ for each coordinate of x.

Procedure for analysis Internal moment M. After M and M/P or M/M’ have been determined, assign P or M’ its numerical value if it has actually replaced a real force or couple moment. Otherwise, set P or M’ equal to zero. Castigliano’s second theorem. Apply Eqn or to determine the desired displacement Δ or . It is important to retain the algebraic signs for corresponding values of M and M/P or M/M’.

Procedure for analysis Castigliano’s second theorem. If the resultant sum of all the definite integrals is +ve, Δ or  is in the same direction as P or M’. If a –ve value results, Δ or  is opposite to P or M’.

EXAMPLE 14.20 Determine the slope at pt B of the beam shown. EI is a constant.

EXAMPLE (SOLN) External couple moment M’. Since slope at pt B is to be determined, an external couple moment M’ is placed on the beam at this pt. Internal moments M. Two coordinates x1 and x2 is used to determine the internal moments within beam since there is a discontinuity, M’ at B. x1 ranges from A to B, and x2 ranges from B to C.

EXAMPLE (SOLN) Internal moments M. Using method of sections, internal moments and partial derivatives are determined. For x1,

EXAMPLE (SOLN) Internal moments M. For x2,

EXAMPLE (SOLN) Castigliano’s second theorem. Setting M’ = 0 and applying Eqn 14-50, we have, Negative sign indicates that B is opposite to direction of couple moment M’.

EXAMPLE 14.21 Determine the vertical displacement of pt C of the steel beam shown. Take Est = 200 GPa, I = 125(10-6) m4.

EXAMPLE (SOLN) External force P. A vertical force P is applied at pt C. Later this force will be set equal to the fixed value of 5 kN.

EXAMPLE (SOLN) Internal moments M. Two x coordinates are needed for the integration since the load is discontinuous at C. Using method of sections, the internal moments and partial derivatives are determined as follows.

EXAMPLE (SOLN) Catigliano’s second theorem. Setting P = 5 kN and applying Eqn 14-49, we have

CHAPTER REVIEW 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.

CHAPTER REVIEW 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.

CHAPTER REVIEW The external virtual work developed is then equated to the internal virtual strain energy in the member or structure. 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.

CHAPTER REVIEW 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’s theorem is then applied to obtain the desired displacement (or rotation).

CHAPTER OBJECTIVES Apply energy methods to solve problems involving deflection Discuss work and strain energy, and development of the principle of conservation.

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CHAPTER OBJECTIVES Apply energy methods to solve problems involving deflection Discuss work and strain energy, and development of the principle of conservation.

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Presentation on theme: "CHAPTER OBJECTIVES Apply energy methods to solve problems involving deflection Discuss work and strain energy, and development of the principle of conservation."— Presentation transcript:

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