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Macromechanics of a Laminate Textbook: Mechanics of Composite MaterialsMechanics of Composite Materials Author: Autar KawAutar Kaw

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Figure 4.1

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CHAPTER OBJECTIVES CUnderstand the code for laminate stacking sequence CDevelop relationships of mechanical and hygrothermal loads applied to a laminate to strains and stresses in each lamina CFind the elastic stiffnesses of laminate based on the elastic moduli of individual laminas and the stacking sequence CFind the coefficients of thermal and moisture expansion of a laminate based on elastic moduli, coefficients of thermal and moisture expansion of individual laminas, and stacking sequence

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Laminate Behavior elastic moduli the stacking position thickness angles of orientation coefficients of thermal expansion coefficients of moisture expansion

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Figure 4.2

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Figure 4.3

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Classical Lamination Theory

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Figure 4.4

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Global Strains in a Laminate

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Figure 4.5

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Figure 4.6

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Stresses in a Lamina in a Laminate

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Forces and Stresses

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Forces and Strains

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Integrating terms

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Forces and Strains

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Moments and Strains

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Forces, Moments, Strains, Curvatures

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Steps

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6. Solve the six simultaneous Equations (4.29) to find the midplane strains and curvatures. 7. Knowing the location of each ply, find the global strains in each ply using Equation (4.16). 8. For finding the global stresses, use the stress-strain Equation (2.103). 9. For finding the local strains, use the transformation Equation (2.99). 10. For finding the local stresses, use the transformation Equation (2.94).

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Figure 4.7

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Problem A [0/30/-45] Graphite/Epoxy laminate is subjected to a load of N x = N y = 1000 N/m. Use the unidirectional properties from Table 2.1 of Graphite/Epoxy. Assume each lamina has a thickness of 5 mm. Find a)the three stiffness matrices [A], [B] and [D] for a three ply [0/30/-45] Graphite/Epoxy laminate. b)mid-plane strains and curvatures. c)global and local stresses on top surface of 30 0 ply. d)percentage of load N x taken by each ply.

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Solution A) From Example 2.4, the reduced stiffness matrix for the 0 0 Graphite/Epoxy ply is

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From Equation (2.99), the transformed reduced stiffness matrix for each of the three plies are

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The total thickness of the laminate is h = (0.005)(3) = m. The mid plane is m from the top and bottom of the laminate. Hence using Equation (4.20), the location of the ply surfaces are h 0 = m h 1 = m h 2 = m h 3 = m

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From Equation (4.28a), the extensional stiffness matrix [A] is

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The [A] matrix

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From Equation (4.28b), the coupling stiffness matrix [B] is

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The B Matrix

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From Equation (4.28c), the bending stiffness matrix [D] is

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The [D] matrix

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B) Since the applied load is N x = N y = 1000N/m, the mid- plane strains and curvatures can be found by solving the following set of simultaneous linear equations (Equation 4.29).

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Mid-plane strains and curvatures

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C) The strains and stresses at the top surface of the 30 0 ply are found as follows. First, the top surface of the 30 0 ply is located at z = h 1 = m. From Equation (4.16),

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Table 4.1 Global strains (m/m) in Example 4.3 Ply #Position εxεx εyεy 1(0 0 )Top Middle Bottom (10 -8 ) (10 -7 ) (10 -7 ) (10 -6 ) (10 -6 ) (10 -6 ) (10 -6 ) (10 -6 ) (10 -6 ) 2(30 0 )Top Middle Bottom (10 -7 ) (10 -7 ) (10 -7 ) (10 -6 ) (10 -6 ) (10 -6 ) (10 -6 ) (10 -7 ) (10 -7 ) 3(-45 0 )Top Middle Bottom (10 -7 ) (10 -7 ) (10 -7 ) (10 -6 ) (10 -6 ) (10 -6 ) (10 -7 ) (10 -6 ) (10 -6 )

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Using the stress-strain Equations (2.98) for an angle ply,

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Table 4.2 Global stresses (Pa) in Example 4.3 Ply #Positionσxσx σyσy τ xy 1(0 0 )Top Middle Bottom (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) 2(30 0 )Top Middle Bottom (10 4 ) (10 5 ) (10 5 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) 3(-45 0 )Top Middle Bottom (10 5 ) (10 4 ) (10 4 ) (10 5 ) (10 4 ) (10 4 ) (10 5 ) (10 4 ) (10 4 )

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The local strains and local stress as in the 30 0 ply at the top surface are found using transformation Equation (2.94) as

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Table 4.3 Local strains (m/m) in Example 4.3 Ply #Positionε1ε1 ε2ε2 γ 12 1(0 0 )Top Middle Bottom (10 -8 ) (10 -7 ) (10 -7 ) 5.955(10 -6 ) 5.134(10 -6 ) 4.313(10 -6 ) (10 -6 ) (10 -6 ) (10 -6 ) 2(30 0 )Top Middle Bottom 4.837(10 -7 ) 7.781(10 -7 ) 1.073(10 -6 ) 4.067(10 -6 ) 3.026(10 -6 ) 1.985(10 -6 ) 2.636(10 -6 ) 2.374(10 -6 ) 2.111(10 -6 ) 3(-45 0 )Top Middle Bottom 1.396(10 -6 ) 5.096(10 -7 ) (10 -7 ) 1.661(10 -6 ) 1.800(10 -6 ) 1.940(10 -6 ) (10 -6 ) (10 -6 ) (10 -7 )

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Table 4.4 Local stresses (Pa) in Example 4.3 Ply #Positionσ1σ1 σ2σ2 τ 12 1(0 0 )Top Middle Bottom (10 4 ) (10 4 ) (10 4 ) (10 4 ) 5.359(10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) 2(30 0 )Top Middle Bottom (10 4 ) (10 5 ) (10 5 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) 3(-45 0 )Top Middle Bottom (10 5 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 4 ) (10 3 ) (10 3 )

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D) The portion of the load N x taken by each ply can be calculated by integrating the stress through the thickness of each ply. However, since the stress varies linearly through each ply, the portion of the load N x taken is simply the product of the stress at the middle of each ply (See Table 4.2) and the thickness of the ply. Portion of load N x taken by 0 0 ply = 4.464(10 4 )(5)(10 -3 ) = N/m Portion of load N x taken by 30 0 ply = 1.063(10 5) (5)(10 -3 ) = N/m Portion of load N x taken by ply = 4.903(10 4 )(5)(10 -3 ) = N/m The sum total of the loads shared by each ply is 1000 N/m, ( ) which is the applied load in the x-direction, N x.

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Percentage of load N x taken by 0 0 ply Percentage of load N x taken by 30 0 ply Percentage of load N x taken by ply

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Figure 4.8

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