Bending of Beams Chapter Objectives Develop formulas to find the deflection and stresses in a beam made of composite materials. Develop formulas for symmetric beams that are narrow or wide. Develop formulas for nonsymmetric beams that are narrow or wide. 6.1 Introduction To study mechanics of beams made of laminated composite materials,we need to review the beam analysis of isotropic materials.Several concepts applied to beams made of isotropic materials will help in understanding beams made of composite materials.We are limiting our study to beams with transverse loading or applied moments. The bending stress in an isotropic beam(Figure 6.1 and Figure 6.2)under an applied bending moment,M,is given by 2 (6.1) where z=distance from the centroid I=second moment of area The bending deflections,w,are given by solving the differential equation 431 2006 by Taylor Francis Group,LLC
431 6 Bending of Beams Chapter Objectives • Develop formulas to find the deflection and stresses in a beam made of composite materials. • Develop formulas for symmetric beams that are narrow or wide. • Develop formulas for nonsymmetric beams that are narrow or wide. 6.1 Introduction To study mechanics of beams made of laminated composite materials, we need to review the beam analysis of isotropic materials. Several concepts applied to beams made of isotropic materials will help in understanding beams made of composite materials. We are limiting our study to beams with transverse loading or applied moments. The bending stress in an isotropic beam (Figure 6.1 and Figure 6.2) under an applied bending moment, M, is given by1,2 , (6.1) where z = distance from the centroid I = second moment of area The bending deflections, w, are given by solving the differential equation σ = Mz I 1343_book.fm Page 431 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
432 Mechanics of Composite Materials,Second Edition M M FIGURE 6.1 Bending of a beam. Neutral axis FIGURE 6.2 Curvature of a bended beam. EI dw dr? =-M, (6.2) where E=Young's modulus of the beam material. d-w The term of dx2 is defined as the curvature w Kx=- a2 (6.3) 2006 by Taylor Francis Group,LLC
432 Mechanics of Composite Materials, Second Edition , (6.2) where E = Young’s modulus of the beam material. The term of is defined as the curvature , (6.3) FIGURE 6.1 Bending of a beam. FIGURE 6.2 Curvature of a bended beam. M M z x O z Neutral axis ρ EI d w dx M 2 2 = − d w dx 2 2 κx w x = − ∂ ∂ 2 2 1343_book.fm Page 432 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Bending of Beams 433 Mid-plane Neutral axis FIGURE 6.3 Laminated beam showing the midplane and the neutral axis. giving EIK:=M. (6.4) The formula for the bending stress is only valid for an isotropic material because it assumes that the elastic moduli is uniform in the beam.In the case of laminated materials,elastic moduli vary from layer to layer. 6.2 Symmetric Beams To keep the introduction simple,we will discuss beams that are symmetric and have a rectangular cross-section3(Figure 6.3).Because the beam is sym- metric,the loads and moments are decoupled in Equation(4.29)to give Kx D (6.5) or Kx M Ky =[D My (6.6) Now,if bending is only taking place in the x-direction,then 2006 by Taylor Francis Group,LLC
Bending of Beams 433 giving . (6.4) The formula for the bending stress is only valid for an isotropic material because it assumes that the elastic moduli is uniform in the beam. In the case of laminated materials, elastic moduli vary from layer to layer. 6.2 Symmetric Beams To keep the introduction simple, we will discuss beams that are symmetric and have a rectangular cross-section3 (Figure 6.3). Because the beam is symmetric, the loads and moments are decoupled in Equation (4.29) to give (6.5) or . (6.6) Now, if bending is only taking place in the x-direction, then FIGURE 6.3 Laminated beam showing the midplane and the neutral axis. zc Mid-plane z Neutral axis EIκ =x M M M M D x y xy x y xy ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎤ ⎦ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ κ κ κ κ κ κ x y xy x y xy D M M M ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎤ ⎦ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ −1 1343_book.fm Page 433 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
434 Mechanics of Composite Materials,Second Edition My =0,Mxy=0 M: = (6.7) 0 that is, K=DiM (6.8a) Ky=Di2M (6.8b) Kw DieM:, (6.8c) where D are the elements of the [D]-1 matrix as given in Equation(4.28c) Because defining midplane curvatures(Equation 4.15), K=-02 dx2 wo Ky=- (6.9) Ky=-2 子wa dxdy the midplane deflection wo is not independent of y.However,if we have a narrow beam-that is,the length to width ratio (L/b)is sufficiently high, we can assume that wo=wo(x)only. K=-=DiM,. (6.10) dx2 Writing in the form similar to Equation(6.2)for isotropic beams, 2006 by Taylor Francis Group,LLC
434 Mechanics of Composite Materials, Second Edition , , (6.7) that is, (6.8a) (6.8b) , (6.8c) where are the elements of the [D]–1 matrix as given in Equation (4.28c). Because defining midplane curvatures (Equation 4.15), , , (6.9) , the midplane deflection w0 is not independent of y. However, if we have a narrow beam — that is, the length to width ratio (L/b) is sufficiently high, we can assume that w0 = w0(x) only. . (6.10) Writing in the form similar to Equation (6.2) for isotropic beams, My = 0 Mxy = 0 κ κ κ x y xy x D ⎡ M ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎤ ⎦ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ −1 0 0 κx x = D M∗ 11 κy x = D M∗ 12 κxy = D Mx ∗ 16 Dij ∗ κx w x = − ∂ ∂ 2 0 2 κy w y = − ∂ ∂ 2 0 2 κxy w x y = − ∂ ∂ ∂ 2 2 0 κx x d w dx = − = D M∗ 2 0 2 11 1343_book.fm Page 434 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Bending of Beams 435 Mb dx2 E,I' (6.11) where b=width of beam Ex=effective bending modulus of beam I=second moment of area with respect to the x-y-plane From Equation(6.8a)and (6.11),we get 12 Ex=HDi (6.12) Also, 1奶3 (6.13) 12 M=M,b. (6.14) To find the strains,we have,from Equation (4.16), ∈x=ZKx (6.15a) ∈g=zKy (6.15b) Yy=zKw· (6.15c) These global strains can be transformed to the local strains in each ply using Equation(2.95): r (6.16) The local stresses in each ply are obtained using Equation(2.73)as 2006 by Taylor Francis Group,LLC
Bending of Beams 435 , (6.11) where b = width of beam Ex = effective bending modulus of beam I = second moment of area with respect to the x–y-plane From Equation (6.8a) and (6.11), we get . (6.12) Also, (6.13) . (6.14) To find the strains, we have, from Equation (4.16), (6.15a) (6.15b) . (6.15c) These global strains can be transformed to the local strains in each ply using Equation (2.95): . (6.16) The local stresses in each ply are obtained using Equation (2.73) as d w dx M b E I x x 2 0 2 = − E h D x = ∗ 12 3 11 I bh = 3 12 M M= xb ∈ =x x zκ ∈ =y y zκ γ κ xy xy = z ∈ ∈ ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ = ⎡ ⎣ ⎤ ⎦⎡ ⎣ ⎤ ⎦⎡ ⎣ ⎤ ⎦ ∈ ∈ − 1 2 12 1 γ γ k x y xy R T R ⎡ ⎣ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ k 1343_book.fm Page 435 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC