66 Mechanics of Composite Materials,Second Edition Arbitrary plane AP FIGURE 2.5 Stresses on an infinitesimal area on an arbitrary plane. section,a force AP is acting on an area of AA.This force vector has a com- ponent normal to the surface,AP,and one parallel to the surface,AP.The definition of stress then gives △P =lim aM-0△A' t;lim △P (2.3a,b) AM4-→0△A The component of the stress normal to the surface,o is called the normal stress and the stress parallel to the surface,t,is called the shear stress.If one takes a different cross-section through the same point,the stress remains unchanged but the two components of stress,normal stress,o,and shear stress,t,will change.However,it has been proved that a complete definition of stress at a point only needs use of any three mutually orthogonal coordi- nate systems,such as a Cartesian coordinate system. Take the right-hand coordinate system x-y-z.Take a cross-section parallel to the yz-plane in the body as shown in Figure 2.6.The force vector AP acts 2006 by Taylor Francis Group,LLC
66 Mechanics of Composite Materials, Second Edition section, a force ΔP is acting on an area of ΔA. This force vector has a component normal to the surface, ΔPn, and one parallel to the surface, ΔPs. The definition of stress then gives , . (2.3a,b) The component of the stress normal to the surface, σn, is called the normal stress and the stress parallel to the surface, τs, is called the shear stress. If one takes a different cross-section through the same point, the stress remains unchanged but the two components of stress, normal stress, σn, and shear stress, τs, will change. However, it has been proved that a complete definition of stress at a point only needs use of any three mutually orthogonal coordinate systems, such as a Cartesian coordinate system. Take the right-hand coordinate system x–y–z. Take a cross-section parallel to the yz-plane in the body as shown in Figure 2.6. The force vector ΔP acts FIGURE 2.5 Stresses on an infinitesimal area on an arbitrary plane. Arbitrary plane ΔPs ΔP ΔA ΔPn σn A Pn A = →limΔ Δ 0 Δ τs A Ps A = → limΔ Δ 0 Δ 1343_book.fm Page 66 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 67 △P △P AA Cross-section FIGURE 2.6 Forces on an infinitesimal area on the y-z plane. on an area AA.The component AP:is normal to the surface.The force vector AP,is parallel to the surface and can be further resolved into components along the y and z axes:AP and AP.The definition of the various stresses then is x=lim △2 △M-0△A △P. Twy=lim 44-0△A' Ux=lim △2 (2.4a-c） 44-→0△A Similarly,stresses can be defined for cross-sections parallel to the xy and xz planes.For defining all these stresses,the stress at a point is defined generally by taking an infinitesimal cuboid in a right-hand coordinate system 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 67 on an area ΔA. The component ΔPx is normal to the surface. The force vector ΔPs is parallel to the surface and can be further resolved into components along the y and z axes: ΔPy and ΔPz. The definition of the various stresses then is , . (2.4a–c) Similarly, stresses can be defined for cross-sections parallel to the xy and xz planes. For defining all these stresses, the stress at a point is defined generally by taking an infinitesimal cuboid in a right-hand coordinate system FIGURE 2.6 Forces on an infinitesimal area on the y–z plane. z ΔPz ΔA Cross-section ΔPy ΔP ΔPx y x σx A Px A = →limΔ Δ 0 Δ τxy A Py A = → limΔ Δ 0 Δ τxz A Pz A = → limΔ Δ 0 Δ 1343_book.fm Page 67 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
68 Mechanics of Composite Materials,Second Edition G H FIGURE 2.7 Stresses on an infinitesimal cuboid. and finding the stresses on each of its faces.Nine different stresses act at a point in the body as shown in Figure 2.7.The six shear stresses are related as Txy=Tyx Tye=Tay! Tzx=Tz· (2.5a-c） The preceding three relations are found by equilibrium of moments of the infinitesimal cube.There are thus six independent stresses.The stresses o,, and o are normal to the surfaces of the cuboid and the stressest and tw are along the surfaces of the cuboid. A tensile normal stress is positive,and a compressive normal stress is negative.A shear stress is positive,if its direction and the direction of the normal to the face on which it is acting are both in positive or negative direction;otherwise,the shear stress is negative. 2.2.2 Strain Similar to the need for knowledge of forces inside a body,knowing the deformations because of the external forces is also important.For example, a piston in an internal combustion engine may not develop larger stresses than the failure strengths,but its excessive deformation may seize the engine. Also,finding stresses in a body generally requires finding deformations.This is because a stress state at a point has six components,but there are only three force-equilibrium equations (one in each direction). 2006 by Taylor Francis Group,LLC
68 Mechanics of Composite Materials, Second Edition and finding the stresses on each of its faces. Nine different stresses act at a point in the body as shown in Figure 2.7. The six shear stresses are related as , , . (2.5a–c) The preceding three relations are found by equilibrium of moments of the infinitesimal cube. There are thus six independent stresses. The stresses σx, σy, and σz are normal to the surfaces of the cuboid and the stresses τyz, τzx, and τxy are along the surfaces of the cuboid. A tensile normal stress is positive, and a compressive normal stress is negative. A shear stress is positive, if its direction and the direction of the normal to the face on which it is acting are both in positive or negative direction; otherwise, the shear stress is negative. 2.2.2 Strain Similar to the need for knowledge of forces inside a body, knowing the deformations because of the external forces is also important. For example, a piston in an internal combustion engine may not develop larger stresses than the failure strengths, but its excessive deformation may seize the engine. Also, finding stresses in a body generally requires finding deformations. This is because a stress state at a point has six components, but there are only three force-equilibrium equations (one in each direction). FIGURE 2.7 Stresses on an infinitesimal cuboid. G F σzz σxx σyy τzy τzx τyx τyz τxz τxy H E A B z y x C τ τ xy = yx τ τ yz = zy τ τ zx = xz 1343_book.fm Page 68 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
Macromechanical Analysis of a Lamina 69 Q D Ax C Ay (x,y) A 444444= FIGURE 2.8 Normal and shearing strains on an infinitesimal area in the x-y plane. The knowledge of deformations is specified in terms of strains-that is, the relative change in the size and shape of the body.The strain at a point is also defined generally on an infinitesimal cuboid in a right-hand coordi- nate system.Under loads,the lengths of the sides of the infinitesimal cuboid change.The faces of the cube also get distorted.The change in length cor- responds to a normal strain and the distortion corresponds to the shearing strain.Figure 2.8 shows the strains on one of the faces,ABCD,of the cuboid. The strains and displacements are related to each other.Take the two perpendicular lines AB and AD.When the body is loaded,the two lines become A'B'and A'D'.Define the displacements of a point (x,y,z)as u=u(x,y,z)=displacement in x-direction at point (x,y,z) v=v(x,y,z)=displacement in y-direction at point (x,y,z) ww(x,y,z)=displacement in z-direction at point (x,y,z) The normal strain in the x-direction,E,is defined as the change of length of line AB per unit length of AB as A'B'-AB Ex lim AB-0 AB (2.6) where 2006 by Taylor Francis Group,LLC
Macromechanical Analysis of a Lamina 69 The knowledge of deformations is specified in terms of strains — that is, the relative change in the size and shape of the body. The strain at a point is also defined generally on an infinitesimal cuboid in a right-hand coordinate system. Under loads, the lengths of the sides of the infinitesimal cuboid change. The faces of the cube also get distorted. The change in length corresponds to a normal strain and the distortion corresponds to the shearing strain. Figure 2.8 shows the strains on one of the faces, ABCD, of the cuboid. The strains and displacements are related to each other. Take the two perpendicular lines AB and AD. When the body is loaded, the two lines become A′B′ and A′D′. Define the displacements of a point (x,y,z) as u = u(x,y,z) = displacement in x-direction at point (x,y,z) v = v(x,y,z) = displacement in y-direction at point (x,y,z) w = w(x,y,z) = displacement in z-direction at point (x,y,z) The normal strain in the x-direction, εx, is defined as the change of length of line AB per unit length of AB as , (2.6) where FIGURE 2.8 Normal and shearing strains on an infinitesimal area in the x–y plane. Q′ D′ C′ B′ B x C A D y Δy Δx (x,y) θ1 θ2 P′ A′ εx AB A B AB AB = ′ ′ − → lim0 1343_book.fm Page 69 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC
70 Mechanics of Composite Materials,Second Edition AB=(AP)+(BP)2, =V△x+(x+△x,y)）-x,y+[o(x+△x,)-o(x,y, AB=△x. (2.7a,b) Substituting the preceding expressions of Equation(2.7)in Equation(2.6), Ex=lim △-→0 Ar Using definitions of partial derivatives 割食 ou Ex=ax' (2.8) because au1, Ox k1, x for small displacements. The normal strain in the y-direction,e is defined as the change in the length of line AD per unit length of AD as Ey=lim. A'D-AD ΓAD0AD (2.9) where 2006 by Taylor Francis Group,LLC
70 Mechanics of Composite Materials, Second Edition (2.7a,b) Substituting the preceding expressions of Equation (2.7) in Equation (2.6), . Using definitions of partial derivatives (2.8) because , , for small displacements. The normal strain in the y-direction, εy is defined as the change in the length of line AD per unit length of AD as , (2.9) where A B′ ′ = (A P′ ′) + (B′ ′ P ) , 2 2 = + [ ( Δ Δ x u x + x y, ) − u(x y, )] +[ ( v x + Δx y, ) − 2 v x( , y)] , 2 AB = Δx. u x x y u x y x x ε = + + − → lim ( , ) ( , / Δ Δ 0 1 2 2 1 ) ( , ) ( , ) Δ Δ x Δ v x x y v x y x ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ + ⎡ + − ⎣ ⎢ ⎤ ⎦ ⎥ ⎧ ⎨ ⎪ 2 ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ − 1 x u x v x ε = + ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ + ∂ ∂ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ 1 2 2 2 1 / ⎦ ⎥ ⎥ − 1 x = u x ε , ∂ ∂ ∂ ∂ << u x 1 ∂ ∂ << v x 1 y AD A D AD AD ε = ′ ′ − → lim0 1343_book.fm Page 70 Tuesday, September 27, 2005 11:53 AM © 2006 by Taylor & Francis Group, LLC