Question 6.3: Consider a carbon/epoxy cantilevered T-section beam with the...
Consider a carbon/epoxy cantilevered T-section beam with the following dimensions: length l = 500 mm, flange width b = 20 mm, web height h = 30 mm, and flange thickness t = 4 mm. Thickness of the web is the same as that of the flange. The flange and the web are composed of 0^\circ plies, each ply being 0.5 mm in thickness. Determine the maximum displacement if the beam is under a tip point load of 100 N. 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 matrix are obtained as follows (detailed calculations are not shown):
[\bar{Q} ]=\begin{bmatrix} 125.6281&2.5126&0\\2.5126&10.0503&0\\0&0&8 \end{bmatrix} \times 10^3 \ MPa \ (samme \ for \ all \ the \ plies) \\ [A^\ast ]=\begin{bmatrix} 2&-0.5&0\\-0.5&25&0 \\ 0&0&31.25 \end{bmatrix} \times 10^{-6} \ (MPa.mm)^{-1} \\ [D^\ast ]=\begin{bmatrix} 1.5000&-0.3750&0\\-0.370&18.7500&0\\0&0 &23.4375 \end{bmatrix} \times 10^{-6} \ (MPa.mm^3)^{-1}Distances to the flange and web centroids from the neutral axis are obtained as
z_{c1}=\frac{h+t_1}{2\left[1+\left(b/h\right) \left(\left(A^\ast _{11})^{(2)}/(A_{11}^\ast\right)^{(1)} \right) \right] }=\frac{30+4}{2\left[1+\left(20/30\right) \right] } =10.2 \ mm \\ z_{c2}=\frac{h+t_1}{2\left[1+\left(h/b\right) \left(\left(A^\ast _{11})^{(1)}/(A_{11}^\ast\right)^{(2)} \right) \right] }=\frac{30+4}{2\left[1+\left(30/20\right) \right] } =6.8 \ mmEffective bending stiffness of the beam is
\begin{matrix} E_{xx}^bI_{yy}&=&\frac{bz_{c_1}^2}{(A^\ast _{11})^{(1)}}+\frac{b}{(D^\ast _{11})^{(1)} }+\frac{h}{12(A^\ast _{11})^{(2)}} (h^2+12z_{c_2}^2)\\ &=&\frac{20\times 10.2^2}{2.0\times 10^{-6}}+\frac{20}{1.5\times 10^{-6}} +\frac{30}{12\times 2.0\times 10^{-6}}[30^2+12\times 6.8^2] \\ &=& 2.872\times 10^9 \ MPa.mm^4 \end{matrix}Displacement under the point load is the maximum displacement and it is given by (Equation 6.129)
(w_0)_{max}=-\frac{Pl^3}{3E_{xx}^bI_{yy}}\quad\quad\quad\quad\quad (6.129) \\ (w_0)_{max}=-\frac{100\times 500^3}{3\times 2.872\times 10^9} =-1.45 \ mmThe stresses and strains vary along the length. Bending moment is maximum at the fixed end of the cantilevered beam and we shall find the stresses and strains at the top and bottom faces of the beam at the fixed end.
In order to find the bending stresses in the flange, we determine the nonzero stress resultants as follows (refer Equations 6.175 and 6.176):
Midplane strains and curvatures in the flange are
\begin{Bmatrix} \varepsilon _{xx}^0 \\ \varepsilon _{yy}^0 \\ \gamma _{xy}^0 \end{Bmatrix} ^{(1)}=\begin{bmatrix} 2&-0.5&0\\-0.5&25&0\\0&0&31.25 \end{bmatrix} \times \begin{Bmatrix} 88.788\\0\\0 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 177.576\\-44.394\\0 \end{Bmatrix} \times 10^{-6}and
\begin{Bmatrix} \kappa _{xx}^0 \\ \kappa _{yy}^0\\\kappa _{xy}^0 \end{Bmatrix} ^{(1)}=\begin{bmatrix} 1.5000&-0.3750&0\\-0.3750&18.7500&0\\0&0&23.4375 \end{bmatrix} \times \begin{Bmatrix} 11.606\\0\\0 \end{Bmatrix} \times 10^{-6}=\begin{Bmatrix} 17.409\\-4.325\\0 \end{Bmatrix} \times 10^{-6}Global strains at the top face of the flange are
\begin{Bmatrix} \varepsilon _{xx}\\\varepsilon _{yy}\\\gamma _{xy} \end{Bmatrix}^{(1)}=\begin{Bmatrix} 177.576\\-44.394\\0 \end{Bmatrix}\times 10^{-6}+2\times\begin{Bmatrix} 17.409\\-4.352\\0 \end{Bmatrix}\times 10^{-6}=\begin{Bmatrix} 212.394\\-53.098\\0 \end{Bmatrix}\times 10^{-6}Global stresses at the top face of the flange are
\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} 212.394\\-53.098\\0 \end{Bmatrix}\times10^{-3}=\begin{Bmatrix} 26.5\\0\\0 \end{Bmatrix} \ MPaIt can be seen that the local stresses are the same as the global stresses.
Then, we turn our attention to the web stresses, for which we find the only nonzero stress resultant in the web as follows (refer Equation 6.178):
(at the web-to-flange interface)
(N_{xx})^{(2)}=\frac{-(30/2+6.8)\times (100\times 500)}{2\times 10^{-6}\times 2.872\times 10^9}=-189.763 \ N/mm(at the bottom of the beam)
Midplane strains and curvatures in the web (bottom face of the beam) are
and
\begin{Bmatrix} \kappa _{xx}^0\\\kappa _{yy}^0\\\kappa _{xy}^0 \end{Bmatrix} =\begin{Bmatrix} 0\\0\\0 \end{Bmatrix}Global strains in the web (bottom face of the beam) are
\begin{Bmatrix} \varepsilon _{xx}\\\varepsilon _{yy}\\\gamma _{xy} \end{Bmatrix}^{(2)} =\begin{Bmatrix} -379.526\\94.882\\0 \end{Bmatrix} \times 10^{-6}Global stresses in the web (bottom face of the beam) are
\begin{Bmatrix} \sigma _{xx}\\\sigma _{yy}\\\tau _{xy} \end{Bmatrix}^{(2)} =\begin{bmatrix} 125.6281&2.5126&0\\2.5126&10.0503&0\\0&0&8 \end{bmatrix} \times\begin{Bmatrix} -379.526\\94.882\\0 \end{Bmatrix} \times10^{-3}=\begin{Bmatrix} -47.44\\0\\0 \end{Bmatrix} \ MPaNote that the stresses in the beam are predominantly uniaxial—tensile at the top face of the beam and compressive at the bottom.