Composite Plate Force Constants Solution

Force Constants of a Composite Plate

Problem Analysis: We have a composite plate formed by welding a steel plate and an aluminium plate together. Both plates have length $l$ and width $b$. The thickness of the steel plate is $t_s$ and the aluminium plate is $t_a$. Their Young’s moduli are $Y_s$ and $Y_a$ respectively. We need to determine the equivalent force constant ($k$) along the $x$, $y$, and $z$ axes.

Fundamental Concept: The force constant (stiffness) $k$ of a solid block is given by: $$ k = \frac{YA}{L} $$ Where $Y$ is Young’s Modulus, $A$ is the cross-sectional area perpendicular to the force, and $L$ is the length of the material parallel to the direction of the force.

1. Along the X-axis (Length)

When a force is applied along the x-axis, both plates undergo the same extension ($\Delta x$). This indicates that the materials act as springs connected in Parallel.

$$ k_x = k_{steel} + k_{aluminium} $$

For both plates, the length along the force is $l$.

Steel: Area $A_s = b \cdot t_s$ $\Rightarrow k_s = \frac{Y_s (b t_s)}{l}$
Aluminium: Area $A_a = b \cdot t_a$ $\Rightarrow k_a = \frac{Y_a (b t_a)}{l}$

Substituting these into the parallel formula:

$$ k_x = \frac{Y_s b t_s}{l} + \frac{Y_a b t_a}{l} $$

$$ k_x = \frac{b}{l} (Y_s t_s + Y_a t_a) $$

2. Along the Y-axis (Thickness)

When a force is applied along the y-axis, the stress passes vertically through one plate and then the other. The total compression/extension is the sum of the individual deformations. This indicates a Series combination.

$$ \frac{1}{k_y} = \frac{1}{k_{steel}} + \frac{1}{k_{aluminium}} \quad \Rightarrow \quad k_y = \frac{k_s k_a}{k_s + k_a} $$

Here, the cross-sectional area for both is $A = l \cdot b$. The lengths along the force are their respective thicknesses.

Steel: Length $= t_s$ $\Rightarrow k_s = \frac{Y_s (lb)}{t_s}$
Aluminium: Length $= t_a$ $\Rightarrow k_a = \frac{Y_a (lb)}{t_a}$

Substituting into the series formula:

$$ k_y = \frac{ \left( \frac{Y_s lb}{t_s} \right) \left( \frac{Y_a lb}{t_a} \right) }{ \frac{Y_s lb}{t_s} + \frac{Y_a lb}{t_a} } $$

Factoring out $lb$ from the numerator and denominator:

$$ k_y = lb \left[ \frac{ \frac{Y_s}{t_s} \cdot \frac{Y_a}{t_a} }{ \frac{Y_s}{t_s} + \frac{Y_a}{t_a} } \right] = lb \left[ \frac{ \frac{Y_s Y_a}{t_s t_a} }{ \frac{Y_s t_a + Y_a t_s}{t_s t_a} } \right] $$

$$ k_y = \frac{lb Y_s Y_a}{Y_s t_a + Y_a t_s} $$

3. Along the Z-axis (Width)

Similar to the x-axis, when force is applied along the z-axis (width), both plates deform by the same amount. This is again a Parallel combination.

$$ k_z = k_{steel} + k_{aluminium} $$

For both plates, the length along the force is $b$.

Steel: Area $A_s = l \cdot t_s$ $\Rightarrow k_s = \frac{Y_s (l t_s)}{b}$
Aluminium: Area $A_a = l \cdot t_a$ $\Rightarrow k_a = \frac{Y_a (l t_a)}{b}$

Substituting these values:

$$ k_z = \frac{Y_s l t_s}{b} + \frac{Y_a l t_a}{b} $$

$$ k_z = \frac{l}{b} (Y_s t_s + Y_a t_a) $$