Question 6.2: Consider the carbon/epoxy simply supported beam discussed in...
Consider the carbon/epoxy simply supported beam discussed in Example 6.1. For ready reference, we repeat the details here.
Dimensions: l = 500 mm, b = 20 mm, and h = 6 mm
Ply sequence: [0°/90°/0°], each ply being 2 mm in thickness
Determine the maximum displacement and longitudinal in-plane stress \sigma_{xx} at the center of the beam. The beam is under a central point load of 100 N applied parallel to the plies. The point load is applied over an area of 6 mm × 20 mm.
Material properties are as follows:
E_1=125 \ GPa,E_2=10 \ GPa,\nu _{12}=0.25,and \ G_{12}=8 \ GPaFor the given material properties and ply sequence, the transformed reduced stiffness matrix and the laminate compliance matrices are as follows:
[\bar{Q} ]^{(1)}=[\bar{Q} ]^{(3)}=\begin{bmatrix} 125.6281&2.5126&0\\2.5126&10.0503&0\\0&0&8 \end{bmatrix} \times 10^3 \ MPa \\ [\bar{Q} ]^{(2)}=\begin{bmatrix} 10.0503&2.5126&0\\2.5126&125.6281&0\\0&0&8 \end{bmatrix} \times 10^3 \ MPa \\ [A^\ast ]=\begin{bmatrix} 1.9163&-0.0991&0\\-0.0991&3.4362&0\\0&0&20.8333 \end{bmatrix} \times 10^{-6}(MPa.mm)^{-1} \\ [D^\ast ]=\begin{bmatrix} 0.4595&-0.0806&0\\-0.0806&3.8907&0\\0&0&6.9444 \end{bmatrix} \times 10^{-6}(MPa.mm^3)^{-1} \\ [B^\ast ]=[C^\ast ]=0Effective extensional stiffness of the beam is given by Equation 6.142 as follows:
E_{xx}^{ex}I_{zz}=\frac{b^3}{12A_{11}^\ast }=\frac{20^3}{12\times 1.9163\times 10^{-6}}=347.8926\times 10^6 \ N.mm^2z-coordinates of different plies are as follows:
z_0=-3 \ mm,z_1=-1 \ mm,z_2=1 \ mm, \ and \ z_3=3 \ mmDisplacement under the point load is the maximum displacement, and it is obtained by using effective extensional stiffness in Equation 6.73 as
(w_0)_{max}=-\frac{Pl^3} {48E_{xx}^bI_{yy}}\quad\quad\quad\quad\quad (6.73) \\ (w_0)_{max}=-\frac{100\times 500^3}{48\times 347.8926\times 10^6} =-0.75 \ mmMaximum bending moment is M = 12,500 N ⋅ mm
Maximum effective bending stress occurs at the bottom and top of the beam at the center. They are given by Equation 6.150:
At the bottom of the beam under the central point load,
\sigma _{xx}^{eff}=\frac{yM}{I_{zz}} \quad\quad\quad\quad\quad (6.150) \\ \left(\sigma _{xx}^{eff}\right) _{max}=\frac{10\times 12500}{6\times 20^3/12}=31.25 MPa
which implies N_{xx} = 31.25 \times 6 = 187.5 N/mm. \ N_{xx} is positive (tensile) at the bottom of the beam and negative (compressive) at the top.
To determine the in-plane stresses in the plies at the outermost face of the beam, we proceed as follows. The midplane strains and curvatures are given by
\begin{Bmatrix} \varepsilon _{xx}^0\\\varepsilon _{yy}^0\\\gamma _{xy}^0 \end{Bmatrix} =\begin{bmatrix} 1.9163&-0.0991&0\\-0.0991&3.4362&0\\0&0&20.8333 \end{bmatrix} \times \begin{Bmatrix} 187.5\\0\\0 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 3.593\\-0.186\\0 \end{Bmatrix} \times 10^{-4} \\ \begin{Bmatrix} \kappa _{xx} \\ \kappa _{yy} \\\kappa _{xy} \end{Bmatrix} =\begin{Bmatrix} 0\\0\\0 \end{Bmatrix}Then, the global strains
\begin{Bmatrix} \varepsilon _{xx}\\\varepsilon_{yy}\\ \varepsilon _{xy}\end{Bmatrix} =\begin{Bmatrix} 3.593\\-0.186\\0 \end{Bmatrix} \times 10^{-4}Note that the strains are the same in all the plies. Global stresses in the outermost plies (0^\circ ) are obtained as
\begin{Bmatrix} \sigma _{xx}\\\sigma _{yy}\\\tau_{xy} \end{Bmatrix} ^{(1)}=\begin{bmatrix} 125.6281&2.5126&0\\2.5126&10.0503&0\\0&0&8 \end{bmatrix} \times \begin{Bmatrix} 3.593\\-0.186\\0 \end{Bmatrix} \times 10^{-1}=\begin{Bmatrix} 45.1\\0.7\\0\end{Bmatrix} MPa
The local stresses are obtained by transformation as follows:
\begin{Bmatrix} \sigma _{11}\\\sigma _{22}\\ \tau_{12} \end{Bmatrix} ^{(1)}=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} \begin{Bmatrix} 45.1\\0.7\\0 \end{Bmatrix} =\begin{Bmatrix} 45.1\\0.7\\0 \end{Bmatrix} MPa
Similarly, the global and local stresses in the middle ply (90^\circ ) are obtained as
\begin{Bmatrix} \sigma _{xx}\\\sigma _{yy}\\ \tau_{xy} \end{Bmatrix} ^{(2)}=\begin{bmatrix} 10.0503&2.5126&0\\2.5126&125.6281&0\\0&0&8 \end{bmatrix} \times \begin{Bmatrix} 3.593\\-0.186\\0 \end{Bmatrix} \times 10^{-1}=\begin{Bmatrix} 3.6\\-1.4\\0 \end{Bmatrix} MPa
\begin{Bmatrix} \sigma _{11}\\\sigma _{22}\\ \tau_{12} \end{Bmatrix}^{(2)} =\begin{bmatrix} 0&1&0\\1&0&0\\0&0&-1 \end{bmatrix} \begin{Bmatrix} 3.6\\-1.4\\0 \end{Bmatrix} =\begin{Bmatrix} -1.4\\3.6\\0 \end{Bmatrix} MPav