Question 14.6: Design a cantilever beam of length 400 mm and width 25 mm to...

Option-1
Solid rectangular cross section (plies normal to loading direction)
Let us consider a ply sequence consisting only 0° plies. For this ply sequence and the given material properties, the reduced stiffness matrix and transformed reduced stiffness matrix are given by

[Q]=[\bar{Q}]=\begin{bmatrix} 130,407.524&1630.049&0\\1630.094&6520.376&0\\0&0&5015.674 \end{bmatrix}

Let the height of the beam be denoted by h. Then, noting that the \bar{Q} matrix is the same for all the plies, the bending stiffness matrix is given by

[D]=\frac{h^3}{12}\begin{bmatrix} 130,407.524&1630.094&0\\1630.094&6520.376&0\\0&0&5015.674 \end{bmatrix}

The bending compliance [D^\ast ] matrix can be obtained by using the equations described earlier (refer Equations 5.60 and 5.67, Chapter 5) . The element in [D^\ast ] that we need for beam design is D^\ast _{11} . Note that for symmetric laminates, [D^\ast ] is simply the inverse of [D]. Then, D^\ast _{11} can be conveniently determined as follows (refer Equation 5.163, Chapter 5):

[\acute{D} ]=[D]-[B][A]^{-1}[B]\quad\quad\quad\quad\quad (5.60) \\ [D^\ast ]=[\acute{D} ]^{-1} \quad\quad\quad\quad\quad (5.67) \\ D^\ast _{11}=\frac{D_{22}D_{66}-D^2_{26}}{Det[D]}\quad\quad\quad\quad\quad (5.163)

With little arithmetic, we can show that

D^\ast _{11}=\frac{3}{32,500h^3}

and the effective bending modulus (refer Equation 6.15, Chapter 6)

E^b_{xx}=\frac{12}{h^3D^\ast _{11}} =130,000 \ MPa

The area moment of inertia about the bending axis is

I_{yy}=\frac{bh^3}{12}

Then, the tip deflection (δ) of the beam is given by (refer Equation 6.129, Chapter 6)

\delta =\frac{Pl^3}{3E_{xx}^bI_{yy}}=\frac{5\times 400^3}{3\times 130,000\times (25\times h^3/12)} =\frac{5120}{13h^3}

Now, we can readily find that h ≥ 7.3 so that δ ≤ 1.0. Let us provide a beam thickness of 8.0 mm, that is, h = 8.0. Then,

[D]=\begin{bmatrix} 5,564,054.357&69,550.677&0\\69,550.677&278,202.709&0\\0&0&214,002.091 \end{bmatrix}

and the bending compliance matrix is

[D^\ast ]=\begin{bmatrix} 180.288&-45.072&0\\-45.072&3605.769&0\\0&0&4672.852 \end{bmatrix} \times 10^{-9}

Axial bending stresses are given by (refer Equation 6.126, Chapter 6)

\sigma_{xx}^{(k)}(x,z)=\frac{P(l-x)z}{b}\begin{pmatrix} \bar{Q}_{11}^{(k)}D^\ast _{11}+\bar{Q}_{12}^{(k)}D^\ast _{12}+\bar{Q}_{16}^{(k)}D^\ast _{16} \end{pmatrix}

Then, the maximum axial bending stresses are obtained as follows:

The axial bending stresses are much within the respective tensile and compressive strengths.
The laminate for the beam is to be made by laying up 32 prepreg plies, all aligned in the same direction, in a matched-die-mold.
Note that we have verified the beam design only for longitudinal bending stresses. In a beam, interlaminar stresses may be high. The process of interlaminar stress determination is rather laborious as it involves sequential operations going through each ply in the laminate. In the present case, there are 32 plies and a manual method is not practical. For further details and demonstration, the reader may refer to Example 6.1.

Option-2
Solid rectangular cross section (plies parallel to loading direction)
With all 0° plies, the orientation of the plies w.r.t. the loading direction is insignificant and Option-2 is equivalent to Option-1. However, if plies at nonzero directions are used, plies parallel to the loading direction would result in higher bending stiffness.
Option-3: Box-Section
Let us consider a box-section with overall height of 20 mm, width of 25 mm, and flange/web thickness of 2 mm. Let us consider 0° plies for the flanges and the webs. The beam cross section is symmetric and the vertical distances from the centroid of the beam to the flange and web centroids are readily obtained as

z_{c1}=z_{c4}=9 \ mm \ (flanges)

and

z_{c2}=z_{c3}=0 \ (webs)

The extensional and bending stiffness matrices are determined as

[A]=\begin{bmatrix} 260,815.048&3260.188&0\\3260.188&13,040.752&0\\0&0&10,031.348 \end{bmatrix} \\ [D]=\begin{bmatrix} 86,938.349&1086.729&0\\1086.729&4346.917&0\\0&0&3343.783 \end{bmatrix}

The compliance matrices are

[A^\ast ]=\begin{bmatrix} 3.846&-0.962&0\\-0.962&76.923&0\\0&0&99.687 \end{bmatrix} \times 10^{-6} \\ [D^\ast ]=\begin{bmatrix} 11.538&-2.885&0\\-2.588&230.769&0\\0&0&299.062 \end{bmatrix} \times 10^{-6}

The effective flexural rigidity of the beam is given by (refer Equation 6.208, Chapter 6)

E_{xx}^bI_{yy}=\frac{bz^2_{c1}}{(A^\ast _{11})^{(1)} }+\frac{b}{(D^\ast _{11})^{(1)}} +\frac{h}{6(A^\ast _{11})^{(2)}}(h^2+12z^2_{c2}) -\frac{bz^3_{c3}}{(A^\ast _{11})^{(4)}} +\frac{b}{(D^\ast _{11})^{(4)}} \quad\quad\quad\quad\quad (6.208) \\ E_{xx}^{fl}I_{yy}=\left\lgroup\frac{25\times 9^2}{3.846}+\frac{25}{11.538}+\frac{16^2}{3\times 3.846}\times \frac{16}{2}+\frac{25}{11.538} \right\rgroup \times 10^6 =708.355\times 10^6

The tip deflection is then given by

\delta=\frac{Pl^3}{3E_{xx}^bI_{yy}}=\frac{5\times 400^3}{3\times 708.355\times 10^6} =0.15 \ mm

It can be seen that the box-section offers a far more efficient solution in terms of lower tip deflection at a lower mass.