东南大学：《建筑力学 Architectural Mechanics》课程教学课件（英文讲稿）A20 Energy Method

Strain Energy due to Shearing StressesFor a beam subjected to a bending load.?Cd-dv1.bSetting dV= dA dx,QATdAdxF.S*2G1.bI.bS4dAdx62G1DefinetheformfactorforshearSdAIntermsoftheformfactorJAl2dxDimensionless2GAUniqueforeachspecificcross-sectional area11

Strain Energy due to Shearing Stresses • For a beam subjected to a bending load, • Setting dV = dA dx, • In terms of the form factor 2 2 * 1 S d 2 2 xy z z F S U dV V G G I b            2 * S 0 2 2 * S 2 0 1 d d 2 d d 2 L z A z L z A z F S U A x G I b F S A x GI b                   2 S S 0 d 2 L F U f x GA   * S z xy z F S I b   • Define the form factor for shear • Dimensionless • Unique for each specific cross￾sectional area 2 * 2 z S A z A S f dA I b         11

Form Factors for Shear. The form factor for rectangular cross-sectiondAhydy(bh)hh/21 bndy1.2062(bh /12)b. The form factor for circular cross-section121d/2~1.08dz011? For thin-walled circular tubes: fs = 1.95t dy? For thin-walled square tubes: fs = 2.3512

Form Factors for Shear • The form factor for rectangular cross-section     2 * S 2 2 2 2 2 2 3 2 d 1 d 1.20 2 4 12 z A z h h A S f A I b bh b h b y y b bh                                          h z b y dy • The form factor for circular cross-section         2 2 2 2 2 3/2 2 2 2 2 ' 4 S 2 1/2 4 2 ' 2 4 2 2 4 3 4 d d 1.08 64 2 4 d d y d d y d y d f z y d d y                                     z y d y • For thin-walled circular tubes: f dy S = 1.95. • For thin-walled square tubes: fS = 2.35. 12

Strain Energy due to Bending & Transverse Shear: For an end-loaded cantilever beamB. The total strain energy due to bothbending and transverse shear P2x?p2P?L33P2LL6U=U, +U.dxdx2EI5 2GA5GAC6EICp23P?L33Eh?18EI1-10GL?5GAL?6EI6EI. For steel. take E/G ～ 2.63Eh23Eh2oForh/L=1/5=0.0312o For h/L = 1/60.021710GL210GL23Eh23Eh2oForh/L=1/8:0.0122o For h/L= 1/10:0.007810GL210GL2. The strain energy due to transverse shear is of importance onlyin the case of very short deep beams, i.e., for large h/L ratios.13

• For an end-loaded cantilever beam Strain Energy due to Bending & Transverse Shear • The total strain energy due to both bending and transverse shear 2 2 2 2 3 2 0 0 2 3 2 3 2 2 2 6 3 d 2 5 2 6 5 18 3 1 1 6 5 6 10 L L b s P x P P L P L U U U dx x EI GA EI GA P L EI P L Eh EI GAL EI GL                         • For steel, take E/G ≈ 2.6 • The strain energy due to transverse shear is of importance only in the case of very short deep beams, i.e., for large h/L ratios. 13

Strain Energy due to a General State of Stress: Previously found strain energy due to uniaxial stress and planeshearing stress. For a general state of stress.U=(oxex +0yey, +0zez +Txyxy +Tyzyz +Tzxzx. With respect to the principal axes for an elastic, isotropic body,12ba+0 +02 -24(o.0 +00 +a.0)]u=u, +ud1-2v(ca + Ob + o.) = due to volume changeuy6E -b) + (ob -c) +(c-a)= due to distortionua120. Basis for the maximum distortion energy failure criteria,2ofor a tensile test specimenud<((ud-6G14

Strain Energy due to a General State of Stress • Previously found strain energy due to uniaxial stress and plane shearing stress. For a general state of stress,   u x x y y z z xy xy yz yz zx zx              2 1 • With respect to the principal axes for an elastic, isotropic body,             due to distortion 12 1 due to volume change 6 1 2 2 2 1 2 2 2 2 2 2 2                     d a b b c c a v a b c v d a b c a b b c c a G u E v u u u E u                    • Basis for the maximum distortion energy failure criteria,   for a tensile test specimen 6 2 G u u Y d d Y    14

Work and Energy under a Single Load:Strain energy may also be found fromthe work of the single load PxW=/Pdx0: For an elastic deformation.C. Previously,wefound the strainW=[Pdx=[kxdx=kx?=Px]energy by integrating the energydensity overthe volumeFor a uniform rod,. Knowing the equivalence betweenstrain energy and work,dv2EWPL(P/A)PLU-W=xAdxP/2AE2E2EA015

Work and Energy under a Single Load • Previously, we found the strain energy by integrating the energy density over the volume. For a uniform rod,   2 2 2 1 1 0 2 2 2 L U u dV dV E P A P L Adx E EA         • Strain energy may also be found from the work of the single load P1 ,   1 0 x W Pdx • For an elastic deformation, 2 1 1 2 1 2 1 1 0 0 1 1 W Pdx k xdx k x Px x x       • Knowing the equivalence between strain energy and work, 15 1 1 1 2 W P L U W x P AE    