8.1: Nomenclature of composite materials

8.1.4 Nomenclature of composite materials

Composite materials are identified by the name of the fiber followed by the name of the matrix. For example, AS4/3501-6 denotes the carbon fiber AS4 and the epoxy matrix 3501-6. The data in table 8.2 is taken from Herakovich (1998, p. 245), and it lists typical properties for AS4/3501-6 and T300/5208 carbon fiber-reinforced epoxy composites.

Table 8.2   Material properties of selected CFRP lamina

Property	Units	AS4/3501-6	T300/5208
Axial modulus E1	GPa	148	132
	Msi	21.5	19.2
Transverse modulus E2	GPa	10.5	10.8
	Msi	1.46	1.56
Major Poisson’s ratio ν21	dimensionless	0.30	0.24
Major Poisson’s ratio ν23	dimensionless	0.59	0.59
Shear modulus G12	GPa	5.61	5.65
	Msi	0.81	0.82
Shear modulus G23	GPa	3.17	3.38
	Msi	0.46	0.49
Density	g/cm^3	1.52	1.54
	lb./in.^3	0.055	0.056
Ply thickness tply	mm	0.127	0.127
	in.	0.005	0.005
Fiber volume fraction Vf	dimensionless	0.62	0.62

Example 8.1 Transformed reduced stiffness matrix for an off-axis ply

Determine the transformed reduced stiffness matrix of T300/5208 carbon/epoxy for a 30-degree off-axis lamina in U.S. customary units.

Solution.   From table 8.2 , , $v_{21}=0.24$, and . The minor Poisson’s ratio is . The reduced stiffness matrix is computed from eq. (8.30) and eq. (8.31); i.e.,

The transformed reduced stiffness matrix is given by $\left[\bar{Q}\right]=\left[T_{\sigma 1}\right]\left[Q\right]{\left[T_{\sigma 1}\right]}^T$, the reduced stiffness by eq. (8.27), and the transform matrix [T_σ1] by eq. (8.28). The matrix product is

and the result is

■

8.1.5 Laminated wall

Laminates are made by stacking the unidirectional lamina, also called plies, at different fiber orientations. The plies are usually bound together by the same matrix material that is used within the lamina. Laminates are designated by the ply angle stacking sequence. A $\left[\begin{array}{ccc}45-45\hfill & 0\hfill & 90\hfill \end{array}\right]$ stacking sequence denotes a 4-ply laminate with plies at 45, −45, 0, and 90 degrees with respect to the longitudinal z-axis. A ${\left[\begin{array}{cccc}45\hfill & -45\hfill & 0\hfill & 90\hfill \end{array}\right]}_2$ stacking sequence denotes an 8-ply laminate with plies at 45, −45, 0, 90, 45, −45, 0, and 90 degrees. A stacking sequence denotes an 8-ply symmetric laminate with plies at 45, −45, 0, 90, 90, 0, −45, and 45 degrees. The assumptions of lamination theory are

• The laminate consists of perfectly bonded layers or lamina.
• Each layer is a homogeneous material with known effective properties.
• Each layer is in a state of plane stress.
• Individual layers can be isotropic or orthotropic.

Consistent with thin-walled bar theory in chapter 3, we assume that the strains ${\epsilon }_{\mathbf{\text{zz}}}$, ${\epsilon }_{\mathbf{\text{ss}}}$, and ${\gamma }_{\mathbf{\text{zs}}}$ are spatially uniform through the thickness of the wall. That is, there is no local bending of the laminated wall. The laminate can stretch and shear in-plane as membrane. For a laminate of Np-plies, the material law for the k-th ply, where $k=1,2,\dots ,Np$, is obtained from eq. (8.30) as

Even though the strains are uniform through the thickness of the wall, note that the stresses are piecewise constant through the thickness of the wall since the transformed reduced stiffness matrix changes from ply to ply. Let the origin of the thickness coordinate ζ be at the midplane of the laminate, such that $-t/2\le \zeta \le t/2$, where t denotes the total thickness of the laminated wall. The stress resultant n_s, the axial stress resultant n_z, and the shear flow q are defined by integrals through the thickness of the wall of the corresponding stresses; i.e.,

where ζ = ζ_k at the bottom of the k-th ply, and $\zeta ={\zeta }_{k+1}$ at the top of the k-th ply. Denote the thickness of the k-th ply by t_k such that ${\zeta }_{k+1}-{\zeta }_k=t_k$. Substitute for the stresses from Hooke’s law (8.32) into eq. (8.33) to get

The last result is written as

where [A] is the in-plane stiffness matrix. Elements of the in-plane stiffness matrix are computed by the sum

Stiffness elements A₁₁ and A₂₂ correspond to in-plane extensional stiffnesses in the s- and z-directions, respectively. Element A₆₆ corresponds to a shear stiffness in the s- z plane, stiffnesses A₂₁ = A₁₂ are Poisson’s type terms, and stiffnesses A₆₁ = A₁₆ and A₆₂ = A₂₆ couple in-plane shear and extension. The in-plane stiffness matrix depends on the content of the layers in the laminate, and is independent of the stacking sequence of the layers through the thickness of the laminate.

Example 8.2 In-plane stiffness matrix for a laminate with two plies

Consider a two-ply $\left[\begin{array}{cc}\phi \hfill & -\phi \hfill \end{array}\right]$ laminate with plies of equal thickness t∕2.

(a)  Determine the [A] matrix

(b)  Evaluate the [A] matrix for T300/5208 with φ = 30^° and $t/2=0.005$ in.

Solution to part (a).   The transformed reduced stiffnesses are given by eq. (8.31) in which $m=cos\phi$ and $n=sin\phi$. Note that stiffnesses ${\bar{Q}}_{11}$, ${\bar{Q}}_{22}$, ${\bar{Q}}_{66}$, and ${\bar{Q}}_{21}$ are even functions of the ply angle φ, and stiffnesses ${\bar{Q}}_{61}$ and ${\bar{Q}}_{62}$ are odd functions of φ. Thus,

$$\begin{array}{cc}\left[A\right]=\left[\begin{array}{ccc}\hfill {\bar{Q}}_{11}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{12}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{16}\left(\phi \right)\hfill \\ \hfill {\bar{Q}}_{21}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{22}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{26}\left(\phi \right)\hfill \\ \hfill {\bar{Q}}_{61}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{62}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{66}\left(\phi \right)\hfill \end{array}\right]\frac{t}{2}+\left[\begin{array}{ccc}\hfill {\bar{Q}}_{11}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{12}\left(\phi \right)\hfill & \hfill -\left({\bar{Q}}_{16}\left(\phi \right)\right)\hfill \\ \hfill {\bar{Q}}_{21}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{22}\left(\phi \right)\hfill & \hfill -\left({\bar{Q}}_{26}\left(\phi \right)\right)\hfill \\ \hfill -\left({\bar{Q}}_{61}\left(\phi \right)\right)\hfill & \hfill -\left({\bar{Q}}_{62}\left(\phi \right)\right)\hfill & \hfill {\bar{Q}}_{66}\left(\phi \right)\hfill \end{array}\right]\frac{t}{2}& \\ \left[A\right]=\left[\begin{array}{ccc}\hfill {\bar{Q}}_{11}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{12}\left(\phi \right)\hfill & \hfill 0\hfill \\ \hfill {\bar{Q}}_{21}\left(\phi \right)\hfill & \hfill {\bar{Q}}_{22}\left(\phi \right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill {\bar{Q}}_{66}\left(\phi \right)\hfill \end{array}\right]t.& \end{array}$$

Solution to part (b).   From the T300/5208 example on page 213 , , , and . Thus,

■

8.1.6 Balanced and specially orthotropic laminates

A laminate consisting of off-axis plies with positive fiber angles φ_i and off-axis plies with negative fiber angles  − φ_i, $=1,2,3,\dots ,i_{max}$, with each φ_i-ply and  − φ_i-ply having the same thickness and material properties, is called a balanced laminate. For example, a stacking sequence ${\left[30/-30\right]}_{2S}$ is a balanced laminate consisting of eight plies if each 30^°-ply and  − 30-ply have the same thickness and material properties. For a balanced laminate the in-plane stiffness coefficients $A_{16}=A_{61}=0$, and $A_{26}=A_{62}=0$ as example 8.2 illustrates. The in-plane material law for a balanced laminate reduces to the form

In eq. (8.37) resultants n_s and n_z are independent of the shear strain γ_zs, and the shear flow q is independent of the normal strains ε_ss and ε_zz. That is, there is no coupling between in-plane extension and shear. Laminates whose material law is given by (8.37) are also said to be specially orthotropic. Laminates consisting of only 0^° and 90^° plies are specially orthotropic laminates, since the product $mn=cos\phi sin\phi =0$ in the last two rows of (8.31) results in ${\bar{Q}}_{16}={\bar{Q}}_{26}=0$ for these laminates. Hence, a [0∕90] laminate has coupling stiffnesses $A_{16}=A_{26}=0$ as can be recognized from eq. (8.36). Another example of a specially orthotropic laminate is a stacking sequence ${\left[\pm 45/0/90\right]}_S$.

8.2 Composite thin-walled bar with a closed cross-sectional contour

The analysis in this section was published by Johnson, et al., (2001), and it is also reviewed by Vasiliev and Morozov (2013). We consider free bending and torsion of a thin-walled bar with a closed cross-sectional contour as depicted in figure 8.4. The laminated wall consists of unidirectional FRP layers. The external traction components acting on the lateral surface of the bar p_n(s, z), p_s(s, z), and p_z(s, z) appearing in eq. (3.42) on page 38 are prescribed to be zero for all values of s and z. Thus, distributed force intensities $f_x=f_y=f_z=0$ in eq. (3.42), and distributed moment intensities $m_x=m_y=m_z=0$ in eq. (3.45) all vanish. The differential equilibrium equations (3.53), (3.56), (3.54), and (3.61) on page 40 are satisfied for

Fig. 8.4   Closed cross-sectional bar subject to free bending and torsion.

Hence, the axial force N, shear forces V_x and V_y, and the torque M_z are uniform along the length L of the bar. Bending moment equilibrium equations (3.55) and (3.57) on page 40 are satisfied by

where $M_x^0$ and $M_y^0$ are the bending moments acting on the cross section at z = 0.

Consider a free body diagram of the stress resultants acting on a segment of the wall with dimensions Δs-by-Δz is shown in figure 8.5.

Fig. 8.5   Stress resultants acting on an element of the wall.

Force equilibrium leads to

Expand the functions n(s, z), q(s, z), and n_s(s, z) in a Taylor series about s and z to get

Division of eq. (8.41) by the product Δs·Δz, followed by taking the limit as Δs→0 and Δz→0 leads to the differential equations

From eq. (3.6) on page 32 the derivative of the unit tangent vector is $\frac{\partial \widehat{t}}{\partial s}=\frac{-\widehat{n}}{R_s}$, where R_s is the radius of curvature of the contour. The differential equations of equilibrium at coordinates s and z in the wall are

From the last two equations in (8.42) we get

That is, the circumferential stress resultant vanishes and the shear flow is independent of the longitudinal coordinate z.

8.2.1 Anisotropic Hooke’s law for the cross section

Set n_s = 0 in eq. (8.35), and solve for the normal strain ε_ss to eliminate it in the material law. We write the resulting material law in several forms to be used in subsequent developments:

$$\begin{array}{cc}{n}_z=B{\epsilon }_{zz}+bq\hspace{1em}q=B_s{\gamma }_{sz}+B_z{\epsilon }_{zz}\mathrm{\text{, and}}& \text{(8.44)}\\ {\epsilon }_{zz}=\frac{1}{B}\left(n_z-bq\right)\hspace{1em}{\gamma }_{sz}=\frac{1}{B}\left(aq-b{n}_z\right).& \text{(8.45)}\end{array}$$

The coefficients in the previous equations are

$$\begin{array}{cc}B=A_{22}-A_{12}{A}_{21}/A_{11}-b{B}_z,& \text{(8.46)}\\ B_s=A_{66}-A_{61}{A}_{16}/A_{11}\hspace{1em}{B}_z=A_{62}-A_{12}{A}_{61}/A_{11}\mathrm{\text{, and}}& \text{(8.47)}\\ a=\frac{1}{B_s}\left(A_{22}-A_{12}{A}_{21}/A_{11}\right)\hspace{1em}b=B_z/B_s.& \text{(8.48)}\end{array}$$

The stiffness parameters b and B_z represent the shear-extension coupling of the laminated wall, since they are directly related to stiffnesses A₆₁ and A₆₂ by eqs. (8.47) and (8.48). In a specially orthotropic laminate $A_{61}=A_{62}=0$, so $b=B_z=0$. There is no material coupling between shear and extension in a specially orthotropic laminate.

The second assumption is traditional for the beam theory and states that the axial strain is a linear function of coordinates x and y. From eq. (3.30) on page 35 the axial normal strain along the contour (ζ = 0) is

where w(z) is the axial displacement of the cross section, ϕ_x(z) is the rotation of the cross section about the x-axis, and ϕ_y(z) is the rotation of the cross section about the negative y-axis. Refer to figure 8.4. Substitute eq. (8.49) for the strain in the first equation of (8.44) to get the normal stress resultant as

Substitute the previous expression for the normal stress resultant into the definition of the bar resultant N in eq. (3.39) on page 37 to get

where

In eq. (8.52) the modulus-weighted extensional stiffness of the cross section of the beam is denoted by S, the modulus-weighted first moment of the cross-sectional area about the x-axis by S_x, and the modulus-weighted first moment of the cross-sectional area about the y-axis by S_y. We now locate the origin of the x-y coordinates at the modulus-weighted centroid of the cross section. Let $x\left(s\right)=X\left(s\right)-X_c$ and $y\left(s\right)=Y\left(s\right)-Y_c$, where X(s) and Y(s) are the Cartesian coordinates of the contour with respect to an arbitrary origin at point O (see Fig. 3.1 on page 29). The coordinates (X_c, Y_c) of the modulus-weighted centroid are determined from

Since $S_x=S_y=0$, eq. (8.51) is written as

The bending moments M_x and M_y acting in the cross section are determined from the normal stress resultant n_z by

Substitute eq. (8.50) for the normal stress resultant into these expressions for the bending moments to get

The modulus-weighted second moments of the cross section appearing in eq. (8.56) are defined by

Solve for the gradients of the bending rotations eq. (8.56) and write the result as

where

and

Substitute eq. (8.54) for the axial displacement gradient, and eq. (8.58) for the bending rotation gradients, into eq. (8.50) to express the normal stress resultant as

where