Question 6.1: Consider a carbon/epoxy simply supported beam with the follo...

For the given material properties and ply sequence, the transformed reduced stiffness matrix and the laminate bending compliance matrix are obtained as follows (detailed calculations are not shown):

[\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

Effective bending stiffness of the beam is

E^b_{xx}I_{yy}=\frac{b}{D^\ast _{11}} =\frac{20}{0.4595\times 10^{-6}} =43.5256\times 10^6 N.mm²

z-coordinates of different plies are as follows:

z_0=-3 mm, z_1=-1 mm, z_2=1 mm, and z_3=3 mm

Displacement under the point load is the maximum displacement and it is given by (Equation 6.73)

(w_0)_{max}=-\frac{Pl^3}{48E^b_{xx}I_{yy}} \quad\quad\quad\quad\quad (6.73) \\ (w_0)_{max}=-\frac{100\times 500^3}{48\times 43.5256\times 10^6} =-5.98  mm

Maximum bending stresses occur at the bottom and top of the beam at the center. They are given by (Equation 6.72):
At the bottom of the beam under the central point load,

\left(\sigma ^{(k)}_{xx}\right)_{max}=\pm \frac{Plh}{8b}\left(\bar{Q}^{(k)}_{11}D^\ast _{11}+\bar{Q}^{(k)}_{12}D^\ast _{12}+\bar{Q}^{(k)}_{16}D^\ast _{16}\right) \quad\quad\quad\quad\quad (6.72) \\ \left(\sigma _{xx}^{(1)}\right) _{max}=\frac{100\times 500\times 6}{8\times 20\times 10^3}\times (125.6281\times 0.4595-2.5126\times 0.0806) =107.86 MPa  (tensile)

At the top of the beam under the central point load,

\left(\sigma _{xx}^{(3)}\right) _{max}=-\frac{100\times 500\times 6}{8\times 20\times 10^3}\times (125.6281\times 0.4595-2.5126\times 0.0806) =-107.86 MPa (compressive)

For determining the interlaminar normal stresses in the beam under the point load, we need to consider the local applied load distribution. (Note that under a strictly pointed force, the local normal stress would be infinite!) Then, we see that for the given “pointed load” of 100 N over a local beam length 20 mm, the “uniformly distributed load” is q = 5 N/mm.
Now, we utilize Equation 6.47 and determine the interlaminar normal stresses as follows: At the bottom face of the beam, there is no applied load and as a result, the interlaminar normal stress at the bottom face in the first ply is zero, that is,

Comparing this with Equation 6.47 and substituting z = −3 = z_0 for the first ply, we readily find

Then, at the top of the first ply,

\sigma _{zz}^{(1)}(250,-1)=-\frac{5}{20}\times \left(125.6281\times 0.4595-2.5126\times 0.0806\right) \times \left(\frac{-1+27}{6} \right) \times 10^{-3}=-0.0623 MPa  (compressive)

Equating stresses at the interface between first and second plies, the interlaminar normal stress at the bottom of the second ply is

\sigma _{zz}^{(2)}(250,-1)=-0.0623 MPa  (compressive)

Comparing this with Equation 6.47 and substituting z = −1 = z_1 for the second ply, we readily find

C_3^{(2)}=-0.0623  MPa

Then, at the top of the second ply,

\sigma _{zz}^{(2)}(250,1)=-\frac{5}{20}\times \left(10.0503\times 0.4595-2.5126\times 0.0806\right) \times \left(\frac{1+1}{6} \right)\times 10^{-3}-0.0623 =-0.0628 MPa  (compressive)

Comparing this with Equation 6.47 and substituting z = 1 = z_1 for the third ply, we readily find

C_3^{(3)}=-0.0628 MPa

Then, at the top of the third ply,

\sigma _{zz}^{(3)}(250,3)=-\frac{5}{20}\times \left(125.6281\times 0.4595-2.5126\times 0.0806\right) \times \left(\frac{27-1}{6} \right) \times 10^{-3}-0.0628=-0.125  MPa (compressive)

Note that the interlaminar normal stress at the top of the third ply under the pointed load, as expected, is equal in magnitude to the local applied stress.