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ENG5312 – Mechanics of Solids II 1 6. Energy Methods 6.1 External Work 6.1.1 Work of a Force The work done by a force is equivalent to the product of the component of the force acting in the direction of motion and the distance travelled. Ue If the force acts in the x -direction: Ue F ds x Fdx 0 If a force is applied to a prismatic beam in a gradual manner, i.e. the magnitude of the force increases from 0 to P , and the bar stretches by , when the material P x then: P 1 P2 1 xdx P 2 2 behaves in a linear-elastic manner F Ue x 0 (6-1) 6.1.2 Work of a Couple A couple moment does work as it goes through a rotation: Ue Md 0 If a moment is applied to a body with linear-elastic material behaviour such that the magnitude of the couple increases from 0 @ 0 to M @ then: Ue 0 M 1 2 d M (6-2) ENG5312 – Mechanics of Solids II 2 6.2 Strain Energy External work done by loads applied to a body will be converted into strain energy. This strain energy is cause by normal and shear stresses that deform the body. 6.2.1 Normal Stress Consider a body deformed by a normal stress EQS: The force on the top face is dFz z dxdy and if it is applied gradually as the element undergoes deformation d z z dz the work done by the force is (using Eq. (6-1)): Or 1 1 dUi dFz d z z dxdyz dz 2 2 1 dUi zz dV 2 So if a body is subjected to uni-axial normal stress, the strain energy is: Ui V dV Ui 2 2E dV V Note: Ui is always positive. 2 (6-3) For linear-elastic material behaviour, Hooke’s Law ( / E ) applies, and: (6-4) ENG5312 – Mechanics of Solids II 3 6.2.2 Shear Stress Consider an element subjected to shear stress, : The force dF dxdy on the top face will move dz . Assuming dF is applied gradually, and using Eq. (6-1): 1 1 dUi dxdydz dV 2 2 Or Ui V 2 dV (6-5) For linear-elastic behaviour, Hooke’s Law ( /G) applies, and: 2 Ui dV V 2G (6-6) ENG5312 – Mechanics of Solids II 4 6.2.3 Multi-axial Stress Consider an element subjected to a general state of stress. Assuming linear-elastic behaviour and all loads are applied gradually, the strain energy associated with each normal and shear stress can be added to give: Ui 1 2 x x V 1 1 1 1 1 yy zz xy xy yz yz xz xz dV (6-7) 2 2 2 2 2 Using the generalized Hooke’s Law: 1 x y z E y 1 zy x z E z 1 z x y E xy x xy G ; xy xy G ; xy xy G The strains can be eliminated from Eq. (6-7): Ui 1 2 2E V x 1 2 2 y2 z2 xy yz xz2 x y y z x z dV (6-8) 2G E And if only the principal stresses act on the element (i.e. 1, 2 and 3 ) Ui 1 2E V 2 1 22 32 E 1 2 2 3 1 3 dV (6-9) ENG5312 – Mechanics of Solids II 5 6.3 Elastic Strain Energy for Various Types of Loading 6.3.1 Axial Load Consider a bar with a slowly changing cross-section that is loaded centroidally. The internal load at x from one end is N , and the normal stress is N / A . using Eq. (6-4) the strain energy is: 2 N2 U i x dV dV 2 2E 2EA V V The volume dV can be expressed as Adx and: Ui L 0 N2 dx 2AE (6-10) If the cross-sectional area is constant: Ui N 2L 2AE Note: o L , Ui o A, Ui o E , Ui o i.e. something that is easy to distort will store more strain energy. (6-11) ENG5312 – Mechanics of Solids II 6 6.3.2 Bending Moment Application of a bending moment to a straight prismatic member results in a normal stress. Consider the element of area dA, y from the neutral axis, then ( My /I) , and using Eq. (6-4): Ui 2E dV V V 2 1 My dV 2E I The volume dV can be written as dV dAdx , so: 2 Ui L 0 M2 2EI 2 y dAdx 2 A Will give the strain energy in the member, and since 2 A Ui L 0 y dA I : M2 dx 2EI (6-12) Note: The bending moment needs to be expressed as a function of x , then Eq. (6-12) can be integrated. ENG5312 – Mechanics of Solids II 7 6.3.3 Transverse Shear Consider a prismatic beam with an axis of symmetry y . The internal shear force at x is V , and the shear stress on the element of area dA is VQ . Using Eq. (6-6) the strain energy is: It 2 2 1 VQ Ui dV dAdx It 2G 2G V V Or Ui L 0 fs A (6-13) A I2 A Q2 dA t2 (6-14) The strain energy can be written as; Ui L 0 Q2 dAdx t2 Defining the form factor, f s , which is a function of geometry: V2 2GI 2 f sV 2 dx 2GA (6-15) An example of the form factor calculation is given in the text. For a rectangular cross-section f s 6/5 . Note: Ui due to shear is usually much less than Ui for bending (se e.g.14.4, Hibbeler, 6e) and the shear strain energy stored in beams is usually neglected. ENG5312 – Mechanics of Solids II 8 6.3.4 Torsional Moment Consider a shaft with a gradually changing cross-section: If the shaft is subjected to an internal torque T at x from one end, the shear stress on the element dA at from the centroid is T /J , and using Eq. (6-6) the strain energy is; Ui V 2 1 T dV dAdx J 2G V 2G Or L Ui 0 T 2 2 dAdx 2GJ 2 A (6-16) But the polar moment of inertia, J, is defined as: J dA 2 (6-17) A Using Eq. (6-17) the strain energy can be written: Ui L 0 T2 dx 2GJ (6-18) If the shaft (or tube) has constant cross-sectional area: Ui T 2L 2GJ (6-19) ENG5312 – Mechanics of Solids II 9 6.4 Conservation of Energy The principal of conservation of energy states: Energy is a conserved property. It can neither be created nor destroyed; only its form can be altered from one form of energy to another. Only mechanical energy will be considered, but kinetic energy will be neglected since all loadings will be gradual. Conservation of energy would require that the external work done by applied loads (i.e. applied loads that cause deflections) must be equivalent to the strain energy developed in a body as it deforms. Ue Ui (6-20) If the loads are removed the stored strain energy will restore the body to its undeformed state (if the elastic limit has not been exceeded). 6.4.1 Trusses Consider a truss subjected to the load P . If the point of application of the load P deflects in the direction of P , and the load is increased gradually from 0 to P , then from Eq. (6-1): 1 Ue P 2 (6-21) This external work done on the body is stored as strain energy. If, due to P , the axial force N develops in a member, the strain energy stored in that member is N 2L from Eq. (6-11). To determine the total strain energy stored in the truss: 2AE N 2L Ui (6-22) 2AE ENG5312 – Mechanics of Solids II 10 Where the summation is over all the members in the truss. Conservation of energy requires Ue Ui , therefore: 1 N 2L P 2 2AE (6-23) The deflection caused by P can be evaluated after the axial forces in each member of the truss has been determined using statics. 6.4.2 Vertically Loaded Beams Consider a beam loaded with the vertical force P. The deflection at the point of application of P can be determined from the conservation of energy, Eq. (6-20), using Eqs. (6-1) and (6-12), for U e and Ui , respectively: 1 P 2 L 0 M2 dx 2EI (6-24) The bending moment would be written as a function of x . due to bending moment and shear, however, the strain Note: the beam deflects energy due to shear is usually neglected, thus the deflection can be written as a function of bending moment only. ENG5312 – Mechanics of Solids II 11 6.4.3 Beams Loaded with a Couple Consider a cantilever beam subjected to an applied moment Mo . The couple moment will cause the rotation at the point of application, and it does work due to this rotation: Ue Mo /2 from Eq. (6-2). The strain energy would be caused by the bending moment M , and Ui L 0 M2 dx from Eq. (6-12). 2EI Conservation of energy, Eq. (6-20), would require: 1 M 0 2 L 0 M2 dx 2EI (6-25) Where M is a function of x . Note: Application ofthe conservation of energy is limited to situations where only one applied load exists. For multiple applied loads, each load would have an associated external work and deflection, but there is only one conservation equation, so only one unknown deflection can be solved. ENG5312 – Mechanics of Solids II 12 6.5 Impact Loading Remember Mechanics II? Remember work-energy and conservation of energy methods? E.g. A weight is dropped from rest from a height h on to a linear spring, with spring constant k . What is the maximum deflection of the spring? o Conservation of energy: T1 V1 T2 V2 1 W h max k2max 2 o Or Work-Energy: Ue Ui strain energy in the spring. W h max o 1 kxdx k2max 2 The result can be rearranged to give: 2max 2W 2W max h 0 k k The quadratic equation can be solved to give the maximum root: o 0 o max max W 2 W W 2 h k k k (6-26) If the weight is applied statically (i.e. gradually) W k st or st W /k , and Eq. (6-26) can be written as: max st 2st 2 st h o Or h max st 1 1 2 st (6-27) ENG5312 – Mechanics of Solids II 13 o Where the term in the square root is the extra displacement due to dynamic loading. o Note: if h 0 , i.e. the weight W is released while it just touches the spring, max 2 st . E.g. A weight W travelling with velocity v on a frictionless horizontal surface impacts a linear spring, with spring constant k . What is the maximum deflection of the spring? o Conservation of Energy; T1 V1 T2 V2 1 W 2 1 2 v k max 2 g 2 o Or 2max o Wv 2 gk (6-28) A statically loaded spring would deflect st W /k , so Eq. (6-28) can be written as: max stv 2 g (6-29) How to convert this information into deflections of dynamically loaded members? i.e. How is impact loading simulated? Assume: i. The moving body is rigid. ii. The stationary body deforms in a linear-elastic manner (i.e. it behaves as a linear spring). iii. No energy is lost during the collision. iv. The bodies remain in contact during the collision. ENG5312 – Mechanics of Solids II 14 These are conservative assumptions, which lead to overestimates of forces (i.e. good for design purposes). With these assumptions, the deformable body behaves like a linear spring. i.e. an effective spring constant can be defined and Eqs. (6-27) or (6-29) can be used to determine max . An equivalent spring constant is not required. All that is needed is the static deflection, st , for use in Eq. (6-27). st can be obtained from the equation of the Hooke’s Law, Appendix C, or conservation of energy and strain elastic curve, energy. An impact factor, n , can bedefined from Eq. (6-27): h n 1 1 2 st So: max n st n W k Pmax nW k max And And the maximum stress is then: max n st . (6-30)