16.21 Techniques of Structural Analysis and D Spring 2003 Unit #7-Concepts of work and energy Strain energy and potential energy of a beam brec sedans hoMe o the neutra xxis remain So Figure 1: Kinematic assumptions for a beam Kinematic assumptions for a beam: From the figure: AA'=u3(a1) Assume small deflections: B B",BB"=3+ duy. CC/= us()+ ul(1, I3). Assume planar sections normal to the neutral azis remain planar
16.21 Techniques of Structural Analysis and Design Spring 2003 Unit #7 - Concepts of work and energy Strain energy and potential energy of a beam Figure 1: Kinematic assumptions for a beam Kinemati ¯ c assumptions for a beam: From the figure: AA� = u3(x1). Assume small deflections: B� ∼ B��, B¯ B�� = u3 + du3. C¯ C� = u3(x) + u1(x1, x3). Assume planar sections normal to the neutral axis remain planar 1
after deformation. Then 3(x1) d u1(x1,x3) us(ai)is the only primary unknown of the problem From these kinematic assumptions we can derive a theory for beams Strains d 2 ∈11-dx1 r ∈1, plane stress E132dr3'dc1 1(-2+2)=0 Constitutive EIs dr Equilibrium: Apply equilibrium(in the undeformed configuration) to in tegral quantities (moment M and shear force V). Definitions of integral quantities as forces"equivalent"to the internal stresses V(x1)+ 13dA=0 M(r1)+011-3dA=0 replacing 11 A(x1) (-d2)14 A (10) A(x1) M1=EIO
� � � �� � after deformation. Then: u3 = u3(x1) (1) du3 u1(x1, x3) = −x3 (2) dx1 u3(x1) is the only primary unknown of the problem (3) From these kinematic assumptions we can derive a theory for beams. Strains: du1 d2u3 �11 = dx1 = −x3 dx2 (4) 1 �22 = �33 = −ν�11, plane stress (5) 1�du1 du3 � 1� du3 du3 � �13 = + = − + = 0 (6) 2 dx3 dx1 2 dx1 dx1 Constitutive: d2u3 σ11 = E�11 = −Ex3 dx2 (7) 1 Equilibrium: Apply equilibrium (in the undeformed configuration) to integral quantities (moment M and shear force V ). Definitions of integral quantities as forces “equivalent” to the internal stresses: V (x1) + σ13dA = 0 (8) � A(x1) M(x1) + σ11x3dA = 0 (9) A(x1) replacing σ11: � d2u3 2 � d2 M u3 2 (x1) = −E dx2 x3 dA = E dx2 x3dA (10) A(x1) 1 1 �A(x1) d2u3 M(x1) = EI(x1) dx2 (11) 1 2
Also note M 11 With these definitions we can apply equilibrium as shown in the figure q M V w+dv M+dM dv Fra=0: v-gdaI-v-dv=0- (13) dm 0: -M+M+dM-Vdc,+ 2 da V|(14) d/dM d 2m Replacing equation(11)in the last expression dx2 da2)+q(x1)=0 Fourth order differential equation governing the deflections of beams. Needs 4 boundary conditions. Examples L case a u43(0)=0,a3(0)=0,3(L)=0,t3(L)=0 case a u3(0)=0,u3(0)=0.,a3(L)=0,u3(L) 3
Also note: Mx3 σ11 = − (12) I With these definitions we can apply equilibrium as shown in the figure: � dV Fx3 = 0 : V − qdx1 − V − dV = 0 → = −q (13) dx1 � dx2 MB 1 = 0 : −M + M + dM − V dx1 + q = 0 → dM dx1 = V (14) 2 d �dM � d2M dx1 dx1 = −q → dx2 = −q (15) 1 Replacing equation (11) in the last expression: d2 � d2u3 � dx2 EI dx2 + q(x1) = 0 (16) 1 1 Fourth order differential equation governing the deflections of beams. Needs 4 boundary conditions. Examples: a) b) L L x1 x1 • case a u3(0) = 0, u� 3(L) = 0, u��� 3(0) = 0, u�� 3 (L) = 0. • case a u3(0) = 0, u�� 3(L) = 0. 3(0) = 0, u3(L) = 0, u�� 3
Strain energy of a beam Start from the general definition of strain g ade for a linear elastic material we concluded 2 Classical beam theory: 011+0, all other stress components are zero U==0 E22 d 2 ∈11 0=1Er2 (20) E: dr2 d x2 EIO dx2 Iso note U M(x1) Complementary strain energy of a beam 1a2 M.3 2E 2E( I The complementary strain energy is then U 1/M2 2 E12 dAde M2 2/ EI Potential of the external forces:
� � � � � Strain energy of a beam Start from the general definition of strain energy density: � �ij U = σijd�ij (17) 0 for a linear elastic material we concluded: � = 1 U σij �ij (18) 2 Classical beam theory: σ11 �= 0, all other stress components are zero. U� = 1 σ11�11 = 1 E� 2 (19) 2 2 11 d2u3 � 1 2 �d2u3 �2 �11 = −x3 dx2 → U = Ex3 dx2 (20) 1 2 1 (21) � 1 2 �d2u3 �2 U = UdV = V 2 Ex3 dx2 dV (22) V 1 � L �d2u3 �2 � 1 2 = 2 0 E dx2 x3dAdx1 (23) 1 A(x) U = 1 2 � L 0 EI(x) �d2u3 dx2 1 �2 dx1 (24) also note: (25) U = (26) 1 2 L 0 M(x1) d2u3 dx2 1 dx1 Complementary strain energy of a beam Complementary strain energy density: � � �11 1 σ2 11 1 Uc = �11dσ11 = = 0 2 E 2E The complementary strain energy is then � 1 � L M2 � Uc = UcdV = V 2 0 EI2 A(x1) �−Mx3 �2 (27) I 2 x3dAdx1 (28) Uc = 1 2 � L 0 M2 EI dx1 (29) Potential of the external forces: 4
q()u3(1)d 1- Pus(L)-Mu3(L The total potential energy of the beam is d 2 El 2-(dx2 dx1+qu3+Pu3(L)+Mu3(L)(32) This is the expression we gave the first day of class
� � � � � S tiuidS − V V = − fiuidV (30) � L 0 V = − q(x1)u3(x1)dx1 − Pu3(L) − Mu3(L) (31) The total potential energy of the beam is: Π(u3) = � L 0 �1 2 �d2u3 EI dx2 �2 dx1 + qu3 + Pu3(L) + Mu3(L) (32) This is the expression we gave the first day of class!. 5