# Problem 4: Rigid Polymer Foam Density Under Load --- ## 1. **Introduction** Rigid polymer foams, such as Divinycell H200, are widely used in composite materials for their lightweight and structural properties. Understanding how their density changes under mechanical loading is crucial, especially in applications where the foam acts as a core in sandwich structures. When subjected to compressive stress, the foam deforms, altering its volume and, consequently, its density, assuming mass conservation. The change in density depends on the volumetric strain, which arises from the normal and lateral strains under applied stress. **Explanation:** The introduction establishes the context of the problem by describing the material, its application, and the physical principles involved—specifically, the relation between applied stress, induced strain, and resulting density change. This background is essential because it frames the relevance of using mechanical properties (such as Young's modulus and Poisson's ratio) and the concept of volumetric strain to analyze how the foam's density evolves under load. --- ## 2. **Relevant Formulas Required To Solve The Question** ### **List of Key Formulas:** 1. **Volumetric Strain:** \[ \frac{\Delta V}{V_i} = (1+\varepsilon_x)(1+\varepsilon_y)(1+\varepsilon_z) - 1 \] 2. **Density Change (Conservation of Mass):** \[ \rho_f = \frac{m}{V_f},\qquad \rho_i = \frac{m}{V_i} \] \[ \frac{\rho_f}{\rho_i} = \frac{V_i}{V_f} \] So, \[ \rho_f = \frac{\rho_i}{1 + \frac{\Delta V}{V_i}} \] 3. **Strain Relationships under Uniaxial Compression:** - Axial strain: \[ \varepsilon_x = \frac{\sigma}{E} \] - Lateral strains (using Poisson's ratio, \( \nu \)): \[ \varepsilon_y = \varepsilon_z = -\nu \varepsilon_x \] **Explanation:** Each formula is directly applicable to this problem. The volumetric strain formula measures the total fractional change in volume due to deformation along all three axes. The density change formula, derived from mass conservation, connects initial and final densities via volume change. The strain relationships under uniaxial compression relate the imposed stress to the resulting strains, employing Young's modulus and Poisson's ratio. These formulas together allow for direct calculation of the foam’s density under mechanical load. --- ## 3. **Step-by-Step Solution** --- ### **Step 1: Calculate Axial and Lateral Strains for \( \sigma = 3\,\text{MPa} \)** - **Axial strain:** \[ \varepsilon_x = \frac{\sigma}{E} = \frac{3\,\text{MPa}}{310\,\text{MPa}} = 0.009677 \] - **Lateral strains:** \[ \varepsilon_y = \varepsilon_z = -\nu \varepsilon_x = -0.4 \times 0.009677 = -0.003871 \] **Explanation:** The stress is applied in one direction (uniaxial compression), producing an axial strain and two equal lateral strains in the perpendicular directions due to Poisson's effect. These strains quantify the actual deformation experienced by the foam under the given stress. --- ### **Step 2: Compute Volumetric Strain** \[ \frac{\Delta V}{V_i} = (1+\varepsilon_x)(1+\varepsilon_y)(1+\varepsilon_z) - 1 \] \[ = (1+0.009677)(1-0.003871)(1-0.003871) - 1 \] \[ = (1.009677)(0.996129)(0.996129) - 1 \] First, calculate \( (0.996129)^2 \): \[ 0.996129^2 = 0.992273 \] Now multiply by \( 1.009677 \): \[ 1.009677 \times 0.992273 = 1.001797 \] Subtract 1: \[ 1.001797 - 1 = 0.001797 \] So, \[ \frac{\Delta V}{V_i} = 0.001797 \] **Explanation:** The volumetric strain formula aggregates the effects of axial compression and lateral expansion/shrinkage, yielding the net fractional change in volume. The calculation above shows that the foam experiences a very slight increase in volume under the given compressive loading. --- ### **Step 3: Find the Change in Density, \( \Delta \rho \) (for \( \sigma = 3\,\text{MPa} \))** \[ \rho_f = \frac{\rho_i}{1 + \frac{\Delta V}{V_i}} \] \[ = \frac{200\,\text{kg/m}^3}{1 + 0.001797} \] \[ = \frac{200}{1.001797} = 199.64\,\text{kg/m}^3 \] \[ \Delta \rho = \rho_f - \rho_i = 199.64 - 200 = -0.36\,\text{kg/m}^3 \] **Explanation:** The density decreases slightly because the foam’s volume increases under compressive stress, while the mass remains constant. This formula directly links volumetric strain to the density change, illustrating the inverse relationship between volume and density for a constant mass. --- ### **Step 4: Find the Density at Failure (\( \sigma = 5.4\,\text{MPa} \))** - **Axial strain at failure:** \[ \varepsilon_x = \frac{\sigma_{\text{fail}}}{E} = \frac{5.4}{310} = 0.017419 \] - **Lateral strains at failure:** \[ \varepsilon_y = \varepsilon_z = -0.4 \times 0.017419 = -0.006968 \] - **Volumetric strain at failure:** \[ \frac{\Delta V}{V_i} = (1 + 0.017419)(1 - 0.006968)^2 - 1 \] \[ (1 - 0.006968)^2 = 0.993056 \] \[ 1.017419 \times 0.993056 = 1.010720 \] \[ 1.010720 - 1 = 0.010720 \] - **Density at failure:** \[ \rho_f = \frac{200}{1.010720} = 197.88\,\text{kg/m}^3 \] **Explanation:** Applying the same logic as before, but with the failure stress, the calculation reveals the final density at the threshold of structural failure. The increased volumetric strain causes a further (but still small) decrease in density, consistent with the material’s expected behavior under larger deformation. --- ## **Conclusion** - **Change in density for \( \sigma = 3\,\text{MPa} \):** \( \boxed{\Delta \rho = -0.36\,\text{kg/m}^3} \) - **Density at failure (\( \sigma = 5.4\,\text{MPa} \)):** \( \boxed{197.88\,\text{kg/m}^3} \) **Summary:** Under a uniaxial compressive stress of \( 3\,\text{MPa} \), the foam density decreases slightly by \( 0.36\,\text{kg/m}^3 \). At the failure stress of \( 5.4\,\text{MPa} \), the foam’s density is \( 197.88\,\text{kg/m}^3 \). These results illustrate how mechanical loading affects the foam’s density due to volume change, with mass remaining constant throughout the process.