composite modulus in long and perp
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Catégorie :Category: nCreator TI-Nspire
Auteur Author: oONOLTZOo
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 2.83 Ko KB
Mis en ligne Uploaded: 10/10/2024 - 11:01:46
Uploadeur Uploader: oONOLTZOo (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a4245325
Type : Classeur 3.0.1
Page(s) : 1
Taille Size: 2.83 Ko KB
Mis en ligne Uploaded: 10/10/2024 - 11:01:46
Uploadeur Uploader: oONOLTZOo (Profil)
Téléchargements Downloads: 1
Visibilité Visibility: Archive publique
Shortlink : http://ti-pla.net/a4245325
Description
Fichier Nspire généré sur TI-Planet.org.
Compatible OS 3.0 et ultérieurs.
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A continuous and aligned glass fibrereinforced compositeconsists of 40 vol% of glass fibres having a modulus of elasticity of 69 GPaand 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4GPa a. Compute the modulus of elasticity of this composite in the longitudinaldirection. b. If the cross-sectional area is 250 mm2 and a stress of 50 MPa isapplied in this longitudinal direction, compute the magnitude of the loadcarried by each of the fibre and matrix phases. c. Determine the strain that issustained by each phase when the stress in part is applied. Given Data Volume fraction of glass fibers, Vf=0.40V Volume fraction of polyester resin, Vr=0.60V Modulus of elasticity of glass fibers, Ef=69 GPa Modulus of elasticity of polyester resin, Er=3.4 GPa Cross-sectional area, A=250 mm^2 =250 ×^ 10 6 m^2 Applied stress, Ã=50 MPa=50 × 10^6 Pa Part (a): Modulus of Elasticity of the Composite The modulus of elasticity of a continuous and alignedcomposite in the longitudinal direction can be calculated using the rule ofmixtures: Ecomposite = Vf Å Ef+Vr Å ErE Ecomposite = (0.40 * 69 GPa)+(0.60 * 3.4 GPa) Ecomposite = (27.6 GPa)+(2.04 GPa)=29.64 GPa Part (b): Load Carried by Each Phase The load (F) carried by the composite can be calculatedusing the applied stress: F=Ã * A= F=(50*10^6 Pa) * (250 * 10^ 6 m^2)=12.5 N Now, to find the load carried by each phase, we will use thefollowing relationships: The load carried by the fibers, Ff Ff = Ãf * Af The load carried by the matrix, Fr_Fr : Fr= Ãr * Ar Using the volume fractions, the loads can be expressed as: Ff= Vf * F and Fr = Vr * F Substituting the values: Ff= 0.40 * 12.5 N=5.0 Fr= 0.60 * 12.5 N=7.5 Part (c): Strain Sustained by Each Phase Applied Stress Calculation : The applied stress is given as Ã= 50 MPa = 50 * 10 * ^6 Pa Using the Composite Modulus of Elasticity : Theoverall strain in the composite is related to the applied stress and themodulus of elasticity of the composite: õc=Ã / Ecomposite We previously calculated the modulus of elasticity of thecomposite: Ecomposite = 29.64 GPa = 29.64 × 10 * ^9 Pa Calculating the Strain in the Composite : Substituting the values: õc=50×10*^6 Pa/ 29.64 * 10*^ 9 Pa = 0.00169 Calculation of the Transverse Modulus When stress is applied perpendicular to the direction offiber alignment, the modulus of elasticity of the composite in the transversedirection can be calculated using the inverse rule of mixtures : 1 / Ec,transverse = Vf / Ef + Vr / Er Calculating Each Term : For glass fibers: 0.4069= 0.0058 GPa*^ 1 = 5.8 × 10^ 3 GPa * ^ 1 For polyester resin: 0.603.4 = 0.1765 GPa*^ 1 = 1.765 * 10^ 1 GPa*^ 1 Combining the Terms : 1Ec,transverse= 0.0058 + 0.1765 = 0.1823 GPa *^ 1 Calculating the Transverse Modulus : Ec,transverse=1 / 0.1823 = 5.49 GPa Made with nCreator - tiplanet.org
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Compatible OS 3.0 et ultérieurs.
<<
A continuous and aligned glass fibrereinforced compositeconsists of 40 vol% of glass fibres having a modulus of elasticity of 69 GPaand 60 vol% of a polyester resin that, when hardened, displays a modulus of 3.4GPa a. Compute the modulus of elasticity of this composite in the longitudinaldirection. b. If the cross-sectional area is 250 mm2 and a stress of 50 MPa isapplied in this longitudinal direction, compute the magnitude of the loadcarried by each of the fibre and matrix phases. c. Determine the strain that issustained by each phase when the stress in part is applied. Given Data Volume fraction of glass fibers, Vf=0.40V Volume fraction of polyester resin, Vr=0.60V Modulus of elasticity of glass fibers, Ef=69 GPa Modulus of elasticity of polyester resin, Er=3.4 GPa Cross-sectional area, A=250 mm^2 =250 ×^ 10 6 m^2 Applied stress, Ã=50 MPa=50 × 10^6 Pa Part (a): Modulus of Elasticity of the Composite The modulus of elasticity of a continuous and alignedcomposite in the longitudinal direction can be calculated using the rule ofmixtures: Ecomposite = Vf Å Ef+Vr Å ErE Ecomposite = (0.40 * 69 GPa)+(0.60 * 3.4 GPa) Ecomposite = (27.6 GPa)+(2.04 GPa)=29.64 GPa Part (b): Load Carried by Each Phase The load (F) carried by the composite can be calculatedusing the applied stress: F=Ã * A= F=(50*10^6 Pa) * (250 * 10^ 6 m^2)=12.5 N Now, to find the load carried by each phase, we will use thefollowing relationships: The load carried by the fibers, Ff Ff = Ãf * Af The load carried by the matrix, Fr_Fr : Fr= Ãr * Ar Using the volume fractions, the loads can be expressed as: Ff= Vf * F and Fr = Vr * F Substituting the values: Ff= 0.40 * 12.5 N=5.0 Fr= 0.60 * 12.5 N=7.5 Part (c): Strain Sustained by Each Phase Applied Stress Calculation : The applied stress is given as Ã= 50 MPa = 50 * 10 * ^6 Pa Using the Composite Modulus of Elasticity : Theoverall strain in the composite is related to the applied stress and themodulus of elasticity of the composite: õc=Ã / Ecomposite We previously calculated the modulus of elasticity of thecomposite: Ecomposite = 29.64 GPa = 29.64 × 10 * ^9 Pa Calculating the Strain in the Composite : Substituting the values: õc=50×10*^6 Pa/ 29.64 * 10*^ 9 Pa = 0.00169 Calculation of the Transverse Modulus When stress is applied perpendicular to the direction offiber alignment, the modulus of elasticity of the composite in the transversedirection can be calculated using the inverse rule of mixtures : 1 / Ec,transverse = Vf / Ef + Vr / Er Calculating Each Term : For glass fibers: 0.4069= 0.0058 GPa*^ 1 = 5.8 × 10^ 3 GPa * ^ 1 For polyester resin: 0.603.4 = 0.1765 GPa*^ 1 = 1.765 * 10^ 1 GPa*^ 1 Combining the Terms : 1Ec,transverse= 0.0058 + 0.1765 = 0.1823 GPa *^ 1 Calculating the Transverse Modulus : Ec,transverse=1 / 0.1823 = 5.49 GPa Made with nCreator - tiplanet.org
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