Copyright ©2014 Pearson Education, All Rights Reserved

Copyright ©2014 Pearson Education, All Rights Reserved
Chapter Objectives Formulate the strain energy stored in a member, when it is subjected to an axial load, bending moment, transverse shear or torsional moment. Apply the principle of conservation of energy to determine the stress and deflection of a member when it is subjected to impact. Copyright ©2014 Pearson Education, All Rights Reserved

In-class Activities Reading Quiz Applications Work of a force and strain energy Strain energy derived from bending moment Strain energy derived from transverse shear Strain energy derived from torsional moment Conservation of energy Impact loading Copyright ©2014 Pearson Education, All Rights Reserved

WORK OF A FORCE AND ENERGY
Work of a force: A force does work when it undergoes a displacement dx that is in the same direction as the force. If the total displacement is x, the work becomes For a linear-elastic material, if the force F increases gradually from zero to a value P=F, then work done is represented by the light-color-shaded triangle, i.e. Copyright ©2014 Pearson Education, All Rights Reserved

WORK OF A FORCE AND ENERGY (cont)
If an additional force P’ is subsequently added and it causes a further displacement ∆’, then the work done by P (not by P’ ) is represented by the dark color-shaped rectangle, i.e. Copyright ©2014 Pearson Education, All Rights Reserved

WORK OF A FORCE AND ENERGY (cont)
Strain Energy: Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM AKIALI LOAD
Consider a bar of variable yet slightly tapered cross section, which is subjected to an axial load coincident with the bar’s centroidal axis, For a prismatic bar of constant cross-sectional area A, length L, and constant axial load N, Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 1 One of the two high-strength steel bolts A and B shown in Fig. 14–8 is to be chosen to support a sudden tensile loading. For the choice it is necessary to 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 a root (or smallest) diameter of 18 mm within the 6 mm threaded region. Bolt B has “upset” threads, such that the diameter throughout its 56-mm length can be taken as 18 mm. In both cases, neglect the extra material that makes up the threads. Take Est = 210 MPa and σY = 310 MPa. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 1 (cont) Solutions Bolt A If the bolt is subjected to its maximum tension, the maximum stress of σY is 310 MPa will occur within the 6 mm region. This tension force is Applying Eq. 14–16 to each region of the bolt, we have Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 1 (cont) Solutions Bolt B Here the bolt is assumed to have a uniform diameter of 18 mm throughout its 56-mm length. Also, from the calculation above, it can support a maximum tension force of Pmax = N. Thus, By comparison, bolt B can absorb 36% more elastic energy than bolt A, because it has a smaller cross section along its shank. Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM BENDING MOMENT
Consider an axi-symmetric beam. The normal stress in the beam is Then, Or Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM TRANSVERSE STRAIN
We have Hence, Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM TRANSVERSE STRAIN (cont)
The shear stress is Then, Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM TRANSVERSE STRAIN (cont)
Writing leads to Form factor for a rectangular cross section Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 3 Determine the strain energy in the cantilevered beam due to shear if the beam has a square cross section and is subjected to a uniform distributed load w, Fig. 14–14a. EI and G are constant. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 3 (cont) Solutions From the free-body diagram of an arbitrary section, we have Since the cross section is square, form factor fs = 6/5, thus Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM TORSIONAL MOMENT
Consider a circular shaft or due. The shear stress is Then Copyright ©2014 Pearson Education, All Rights Reserved

STRAIN ENERGY DERIVED FROM TORSIONAL MOMENT (cont)
For the most common case of a prismatic shaft (or tube) of constant cross sectional and constant applied torque, Note: For rectangular-section having dimensions h > b, the torsional constant J = chb3, where Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 The tubular shaft in Fig. 14–17a is fixed at the wall and subjected to two torques as shown. Determine the strain energy stored in the shaft due to this loading. G = 75 GPa. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 (cont) Solutions Using the method of sections, internal torque is determined within the two regions of the shaft where it is constant. The polar moment of inertia for the shaft is Thus the strain energy is Copyright ©2014 Pearson Education, All Rights Reserved

CONSERVATION OF ENERGY
Physically, the external loads tends to deform the body so that the loads do external work Ue as they are displaced. The external work caused by the loads is transformed into internal work or strain energy Ui, which is stored within the body. When the loads are removed, the strain energy restores the body back to is original un-derformed position, provided that the material’s elastic limit is not exceeded. The conservation of energy is stated as Ue = Ui Copyright ©2014 Pearson Education, All Rights Reserved

CONSERVATION OF ENERGY (cont)
A truss A beam (subjected to a load P) A beam (subjected to a moment Mo) Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 5 The three-bar truss in Fig. 14–21a is subjected to a horizontal force of 20 kN. If the cross-sectional area of each member is 100 mm2, determine the horizontal displacement at point B. E 200 GPa. Copyright ©2014 Pearson Education, All Rights Reserved

IMPACT LOADINGS (cont)
If the weight W is applied statically (or gradually to the spring, the end displacement of the spring is ∆st = W/k. Using this simplification, the above equation becomes Once ∆max is computed, the maximum force applied to the spring can be determined from Note: if h = 0; then ∆max = 2 ∆st Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 6 The aluminum pipe shown in Fig. 14–26 is used to support a load of 600 kN. Determine the maximum displacement at the top of the pipe if the load is (a) applied gradually, and (b) applied by suddenly releasing it from the top of the pipe when h = 0. Take Eal = 70 GPa and assume that the aluminum behaves elastically. Copyright ©2014 Pearson Education, All Rights Reserved

Chapter Objectives Use the principle of virtual work to determine the displacement and slope at points on structural members and mechanical elements Use the Castigliano’s theorem to determine the displacement and slope at points on structural members and mechanical elements. Copyright ©2014 Pearson Education, All Rights Reserved

In-class Activities Reading Quiz Principle of virtual work Method of virtual work applied to beams Castigliano’s theorem Castigliano’s theorem applied to trusses Castigliano’s theorem applied to beams Copyright ©2014 Pearson Education, All Rights Reserved

READING QUIZ 1) Which of the following statements is incorrect for the principle of virtual work: It is based on the conservation of energy The material must behave elastically The external loadings, be it real or virtual, can be axial load, shear, bending moment or torsional loadings The virtual displacement can be translational or rotational Copyright ©2014 Pearson Education, All Rights Reserved

READING QUIZ (cont) 2) Which of the following statements is incorrect for the Castigliano’s theorem: The material must behave elastically The stress may exceed the proportional limit It applies only to bodies that have constant temperature It is based on conservation of energy Copyright ©2014 Pearson Education, All Rights Reserved

PRINCIPLE OF VIRTUAL WORK
It is based on the conservation of energy In general, the material does not have to behave elastically. Consider a deformable body of any shape or size and apply a series of external loads P to it. These loading will cause internal loadings u within the body. Since the body is deformable, the external loads will be displaced ∆, and the internal loadings will undergo displacements δ. Copyright ©2014 Pearson Education, All Rights Reserved

PRINCIPLE OF VIRTUAL WORK (cont)
Hence, by the principle of conservation of energy Ue = Ui, we have Now, consider an imaginary force P’ having a unit magnitude, acts together with P1, P2 and P3 on the body. Thus Hence, Copyright ©2014 Pearson Education, All Rights Reserved

For P1’ = 1, Similarly, if a virtual couple M’ , having a ‘unit’ magnitude, is applied at the point, the rotation θ at that point can be found from the virtual-work equation; i.e. Actually, the deformations can be caused by the axial load, shear, bending moment or torsional moment. And, the internal work can be developed in the body in different ways Copyright ©2014 Pearson Education, All Rights Reserved

Note: The expressions in the above table is valid only when the material behavior is linear-elastic and the stress does not exceed the proportional limit. In the expressions for the strain energy, the applied force/moment is assumed to apply gradually from 0 to its full value. In the expressions for the virtual work, the ‘full’ virtual loading is applied before the real loads cause displacement. Copyright ©2014 Pearson Education, All Rights Reserved

METHOD OF VIRTUAL FORCE APPLIED TO TRUSSES
The internal virtual work for a member is Virtual-work equation for the entire truss is 1 = external virtual unit it load Δ = joint displacement caused by the real loads n = internal virtual force in a truss member N = internal force in a truss member L = length of a member A = cross sectional area E = modular elasticity of member Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 1 Determine the vertical displacement of joint C of the steel truss shown in Fig. 14–32a.The cross-sectional area of each member is and Est = 200 GPa. Copyright ©2014 Pearson Education, All Rights Reserved

METHOD OF VIRTUAL FORCE APPLIED TO BEAMS
The virtual-work equation for the beam is If the slope of the tangent at a point on the beam’s elastic curve is to be determined, we have 1 = external virtual unit it load Δ = external joint displacement caused by the real loads m = internal virtual moment in beam M = internal moment in beam E = modulus of elasticity I = moment of inertia Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 2 (cont) Solutions The vertical displacement of point B is obtained by placing a virtual unit load at B, Fig. 14–36b. The internal moment M is shown in Fig. 14–36c. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 3 (cont) Solutions The slope at B is determined by placing a virtual unit couple moment at B, Fig. 14–37b. Using the method of sections the internal moments within each of these segments are shown in Fig. 14–37b. Using the same coordinates x1 and x2, the internal moments M are shown in Fig. 14–37c. Copyright ©2014 Pearson Education, All Rights Reserved

CASTIGLIANO’S THEOREM
It is based on conservation of energy The material must behave in linear-elastic manner It applies only to bodies that have constant temperature Copyright ©2014 Pearson Education, All Rights Reserved

CASTIGLIANO’S THEOREM (cont)
Derivation: For linear-elastic material, the principle of superposition applies and the resultant displacement does not depend on the sequence in which the forces are applied to the body. So, a small variation of the internal energy caused by a differential amount of dPj can be expressed as Copyright ©2014 Pearson Education, All Rights Reserved

CASTIGLIANO’S THEOREM (cont)
Derivation (cont): On the other hand, if the dPj is applied first and then followed by others P1, P2…Pn, then the internal energy can be written as Since the sequence of applying the forces does not matter, we have Copyright ©2014 Pearson Education, All Rights Reserved

CASTIGLIANO’S THEOREM APPLIED TO TRUSSES
The strain energy for a truss member is Then Procedures: External Force P Place a force P on the truss at the joint where the desired displacement is to be determined. This force is assumed to have a variable magnitude ad should be directed along the line of action of the displacement. Copyright ©2014 Pearson Education, All Rights Reserved

CASTIGLIANO’S THEOREM APPLIED TO TRUSSES (cont)
Procedures (cont): 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 positive and compressive forces are negative 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. Copyright ©2014 Pearson Education, All Rights Reserved

CASTIGLIANO’S THEOREM APPLIED TO TRUSSES (cont)
Procedures (cont): Castigliano’s Second Theorem Apply Castigliano’s Theorem to determine the desired displacement ∆. It is important to retain the algebraic sighs for corresponding values of N and ∂N/∂P when substituting these terms into the equation. If the resultant sum ΣN (∂N/∂P) L/AE is positive, ∆ is in the same direction as P. If a negative value results, ∆ is opposite to P. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 Determine the vertical displacement of joint C of the steel truss shown in Fig. 14–39a. The cross-sectional area of each member is A = 400 mm2 and Est = 200 GPa. Copyright ©2014 Pearson Education, All Rights Reserved

EXAMPLE 4 (cont) Solutions A vertical force P is applied to the truss at joint C, since this is where the vertical displacement is to be determined, Fig. 14–39b The reactions at the truss supports A and D are calculated and the results are shown in Fig. 14–39b Copyright ©2014 Pearson Education, All Rights Reserved

Copyright ©2014 Pearson Education, All Rights Reserved

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