Question

Q1: A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 31%. The remainder is vinyl ester. The density of the vinyl ester is 0.651 g/cm³, and its modulus of elasticity is 4.89 GPa. The density of E-glass is 2.685 g/cm³, and its modulus of elasticity is 71 GPa. A section of composite 1.00 cm by 25.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the modulus of elasticity of the composite in GPa.

Q2:A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 37%. The remainder is vinyl ester. The density of the vinyl ester is 0.726 g/cm³, and its modulus of elasticity is 4.6 GPa. The density of E-glass is 2.798 g/cm³, and its modulus of elasticity is 71 GPa. A section of composite 1.00 cm by 50.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the density of the composite in g/cm^3.

^Q3:A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 26%. The remainder is vinyl ester. The density of the vinyl ester is 0.657 g/cm³, and its modulus of elasticity is 4.05 GPa. The density of E-glass is 2.857 g/cm³, and its modulus of elasticity is 81 GPa. A section of composite 1.00 cm by 40.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the mass of vinyl ester in the section in kilograms.

Q4:A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 36%. The remainder is vinyl ester. The density of the vinyl ester is 0.886 g/cm³, and its modulus of elasticity is 3.25 GPa. The density of E-glass is 2.889 g/cm³, and its modulus of elasticity is 71 GPa. A section of composite 1.00 cm by 50.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the mass of E-glass fibers in the section in grams.

Q5: A fiberglass composite is composed of a matrix of vinyl ester and reinforcing fibers of E-glass. The volume fraction of E-glass is 41%. The remainder is vinyl ester. The density of the vinyl ester is 0.786 g/cm³, and its modulus of elasticity is 4.98 GPa. The density of E-glass is 3.016 g/cm³, and its modulus of elasticity is 72 GPa. A section of composite 1.00 cm by 50.00 cm by 200.00 cm is fabricated with the E-glass fibers running longitudinal along the 200 cm direction. Assume there are no voids in the composite. Determine the mass of vinyl ester in the section in grams.

Q6: A tensile test uses a test specimen that has a gage length of 50 mm and an area = 218 mm². During the test, the specimen yields under a load of 98,101 N. The corresponding gage length = 50.32 mm - this is at the 0.2 percent yield point. The maximum load of 169,143 N is reached at a gage length = 61 mm. If fracture occurs at a gage length of 69 mm, determine the percent elongation in % - enter your answer as a whole number, not as a fraction.

Q7: A tensile test uses a test specimen that has a gage length of 50 mm and an area = 210 mm². During the test, the specimen yields under a load of 95,602 N. The corresponding gage length = 50.396 mm - this is at the 0.2 percent yield point. The maximum load of 136,786 N is reached at a gage length = 60.028 mm. Fracture occurs at a gage length of 71 mm. Determine the engineering strain at the maximum load.

Q8: A tensile test uses a test specimen that has a gage length of 50 mm and an area = 223 mm². During the test, the specimen yields under a load of 95,039 N. The corresponding gage length = 50.24 mm - this is at the 0.2 percent yield point. The maximum load of 166,212 N is reached at a gage length = 64 mm. Fracture occurs at a gage length of 67 mm. If the speciman necked to an area of 109.7 mm² determine the percent reduction in area - enter your answer as a whole number, not as a fraction.

Q9: A tensile test uses a test specimen that has a gage length of 50 mm and an area = 215 mm². During the test, the specimen yields under a load of 97,371 N. The corresponding gage length = 50.25 mm - this is at the 0.2 percent yield point. The maximum load of 162,418 N is reached at a gage length = 64 mm. Fracture occurs at a gage length of 71 mm. Determine the tensile strength in MPa.

Q10: A tensile test uses a test specimen that has a gage length of 50 mm and an area = 223 mm². During the test, the specimen yields under a load of 96,612 N. The corresponding gage length = 50.26 mm - this is at the 0.2% yield point. The maximum load of 167,406 N is reached at a gage length = 61 mm and the specimen fractures at a gage length of 71 mm. Determine the modulus of elasticity in MPa.

Answer 1

1.

By Young's Modulus of Elasticity formulae we know,

Ec= EmVm+EfVf

Ec= 4.89*0.69+71*.31

Ec= 25.384 GPa

(Ec= Modulus of Elasticity of Composite, Em= Modulus of Elasticity of Matrix, Ef= Modulus of Elasticity of Fibre)

(Vm= Volume of Matrix, Vf=Volume of Fibre)

2.

Density distribution

Dc= DmVm+DfVf

Dc= 0.726*0.63+2.798*0.37

Dc= 1.4926 g/cm3

(Dc= Density of composite, Dm= Density of Matrix,Df=Density of Fibre)

(Vm= Volume of Matrix, Vf=Volume of Fibre)

3.

Density distribution

Dc= DmVm+DfVf

Dc= 0.657*0.74+2.857*0.26

Dc= 1.229 g/cm3

Mass of the composite = Density of the composite* Volume of the composite

Mass= Dc*Vc

= 1.229*1*40*200

= 9832 g or 9.832 kg

(Dc= Density of composite, Dm= Density of Matrix,Df=Density of Fibre)

(Vm= Volume of Matrix, Vf=Volume of Fibre)

4.

Volume of E glass = 0.36*Volume of Composite

= 0.36*1*50*200

= 3600 cm3

Mass of E-glass in the composite= Density of E-glass* Volume

= 2.889*3600

Mass of E-glass in the section = 10400.4 g

5.

Volume of Vinyl Ester =0.59*Volume of Compsite

= 0.59*1*50*200

= 5900 cm3

Mass of Vinyl Ester in the composite=Density of Vinyl Ester*Volume of Vinyl Ester in the Composite

= 0.786*5900

Mass of Vinyl Ester in the composite= 4637.4 g

6.

Gauge length initial= 50 mm

Gauge length finsl= 69 mm

Elongation =(Final - Initial)/ Initial

= (69-50)/50

= 0.38

% Elongation = 0.38*100 = 38 %

7.

Engineering Strain =(Final Dimension - Initial Dimension)/ Initial Dimension

= (60.028-50)/ 50 (* At maximum load gauge length = 60.0258)

= 0.20056

8.

Percentage reduction in area=(Initial Area-Final Area)/ Initial Area

= (223-109.7)/223

= 51 %

9.

Ultimate Tensile Strength = Ultimate Load/ Area of C/S

= 162418/215

= 755.43 MPa

10.

According to Hooke's Law,

Modulus of Elasticity = Stress / Strain (In the elastic region)

Stress= Load / Area (At Yield Point)

Strain= (Final Dimension- Initial Dimension)/ Initial Dimension (At the Yield Point)

Therefore, Stress = 96612 / 223

= 433.237 MPa

Strain = (50.26-50)/50

= 0.0052

Modulus of Elasticity = Stress/Strain

= 433.237/ 0.0052

= 83314.8 MPa