\(16.14\) A continuous and aligned fiber-reinforced composite having a cross- sectional area of \(1130 \mathrm{~mm}^{2}\left(1.75 \mathrm{in} .^{2}\right)\) is subjected to an \(\mathrm{ex}\) ternal tensile load. If the stresses sustained by the fiber and matrix phases are \(156 \mathrm{MPa}\) \((22,600 \mathrm{psi})\) and \(2.75 \mathrm{MPa}(400 \mathrm{psi})\), respectively; the force sustained by the fiber phase is \(74,000 \mathrm{~N}\left(16,600 \mathrm{lb}_{6}\right)\); and the total longitudinal strain is \(1.25 \times 10^{-3}\), determine (a) the force sustained by the matrix phase, (b) the modulus of elasticity of the composite material in the longitudinal direction, and (c) the moduli of elasticity for fiber and matrix phases.

To find the force sustained by the matrix phase, we will first determine the individual forces for fiber and matrix phases, and then subtract the force of fiber phase from the total force to find the force of matrix phase. The total force acting on the composite material can be found using the stress values and the cross-sectional area. Total force = Fiber Stress × Area + Matrix Stress × Area Let's use the given values to compute for the total force: Fiber Stress = \(156 \mathrm{MPa}\) Matrix Stress = \(2.75 \mathrm{MPa}\) Cross-sectional Area = \(1130 \mathrm{mm}^{2}\) Total Force = \((156 × 10^6 N/m^2)(1130 × 10^{-6} m^2) + (2.75 × 10^6 N/m^2)(1130 × 10^{-6} m^2)\) Calculating this expression, we have: Total Force = \(\approx 178,900 \mathrm{N}\) Now, let's subtract the given force sustained by the fiber phase to get the force sustained by the matrix phase: Force sustained by matrix phase = Total Force - Force sustained by fiber phase Given, Force sustained by fiber phase = \(74,000 \mathrm{N}\) Force sustained by matrix phase = \(178,900 \mathrm{N} - 74,000 \mathrm{N} \approx 104,900 \mathrm{N}\)

To find the modulus of elasticity of the composite material in the longitudinal direction, we can use the formula: Modulus of Elasticity (Composite) = \(\frac{\text{Total Stress}}{\text{Total Strain}}\) Total Stress = \(\frac{\text{Total Force}}{\text{Cross-sectional Area}}\) Given: Total Strain = \(1.25 \times 10^{-3} \mathrm{~(unitless)}\) We have already calculated the Total Force as \(178,900\,\text{N}\). Now we can use this value to find Total Stress: Total Stress = \(\frac{178,900\,\text{N}}{1130\,\text{mm}^2}\) Calculating this expression, we have: Total Stress = \(\frac{178,900\,\text{N}}{1130 \times 10^{-6}\,\text{m}^2} = 158.225\,\text{MPa}\) Now we can proceed to calculate the Modulus of Elasticity (Composite): Modulus of Elasticity (Composite) = \(\frac{158.225\,\text{MPa}}{1.25 \times 10^{-3}}\) Calculating this expression, we have: Modulus of Elasticity (Composite) \(\approx 126,580\,\text{MPa}\)

To find the moduli of elasticity for fiber and matrix phases, we will use the following formula: Modulus of Elasticity (Phase) = \(\frac{\text{Stress (Phase)}}{\text{Strain (Phase)}}\) We were given both stress values for Fiber and Matrix Phases, but we still need to find the strain values for both phases. Given: Total Strain = Fiber strain + Matrix Strain = \(1.25\times 10^{-3}\) We can now find the individual strains using their individual Stress and Force values: Fiber Strain = \(\frac{\text{Fiber Force}}{\text{Fiber Stress} \times \text{Cross-sectional Area}}\) Matrix Strain = \(\frac{\text{Matrix Force}}{\text{Matrix Stress} \times \text{Cross-sectional Area}}\) Using the given values: Fiber Strain = \(\frac{74,000\,\text{N}}{(156 \times 10^6\,\text{N}/\text{m}^2)(1130 \times 10^{-6}\,\text{m}^2)} \approx 0.721\,\text{(unitless)}\) Matrix Strain = \(\frac{104,900\,\text{N}}{(2.75 \times 10^6\,\text{N}/\text{m}^2)(1130 \times 10^{-6}\,\text{m}^2)} \approx 37.651\,\text{(unitless)}\) Now that we have the Strain values for both Fiber and Matrix Phases, we can find their Moduli of Elasticity: Modulus of Elasticity (Fiber) = \(\frac{156\,\text{MPa}}{0.721} \approx 216.368\,\text{GPa}\) Modulus of Elasticity (Matrix) = \(\frac{2.75\,\text{MPa}}{37.651} \approx 73.037\,\text{MPa}\) So, the moduli of elasticity for the fiber and matrix phases are approximately \(216.368\,\text{GPa}\) and \(73.037\,\text{MPa}\), respectively. In conclusion: (a) The force sustained by the matrix phase is approximately \(104,900\,\text{N}\). (b) The modulus of elasticity of the composite material in the longitudinal direction is approximately \(126,580\,\text{MPa}\). (c) The moduli of elasticity for the fiber and matrix phases are approximately \(216.368\,\text{GPa}\) and \(73.037\,\text{MPa}\), respectively.