Fibre Volume Fraction and Stiffness Properties in Composite Materials

CIVIL ENGINEERING

Problem 1
Given data,
Thickness of pultruded material = 0.25
Top layer CFM 1 oz CFM, t= 0.02
Rovings 100 -113 yield Eglass Rovings, t= 0.085 in
Mid-layer CFM 1 oz CFM, t = 0.04 in
Rovings 100 -113 yield Eglass Rovings, t= 0.085 in
Bottom layer CFM 1 oz CFM, t = 0.04 in
Area of one 113 yield roving = 2.615 * 10-3 in2
Weight of Eglass continuous filament mats (CFM) = 1.0 oz/ft2
Eglass = 10.5 *106 psi = 10.5*106 * 0.7303 = 7.66*106 kg/cm2
vglass = 0.2
Epolyster = 4.7 * 106 psi = 4700000*0.7303 = 3.4*106 kg/cm2
Vpolyster = 0.38
ECFM = 1.26 * 106 psi = 1.26*106*0.7303 = 0.92 * 106 kg/cm2
vCFM = 0.35
Tensile strength CFM = 12.45 Ksi
V CFM = 0.35
Tensile strength CFM = 12.45 Ksi
a) Fibre volume fraction in the roving layer
Eglass = 10.5 * 106 psi = 10.5*106 *0.7303 = 7.66* 106 kg/cm2
vglass = 0.2
ECFM = 1.26 * 106 psi *0.7303 = 9.2 * 105
vCFM = 0.35

According to elastic young modus (E), it can be stated that stress is directly proportional to
strain that can be represented as
Ϭ α ε
Ϭ = Eε
E = Ϭ/ε
On another hand, Poisson ratio can be elucidated as a proportion of decreased in a lateral
measurement to the increase in the length in a sample of material. Therefore, mathematically,
it can be represented as
V = εlateral/ εaxial
By applying above given data, it can be said that
VCFM = 0.35
Vglass = 0.2
Tensile strength CFM = 12.45 Ksi = 875.32 Kg/cm2
Tensile strength CFM = F/A = 875.32
A = F/875.32 ------ (1)
Weight of the E-glass continuous filament mats (CFM) = 1.0 oz/ft2 = 2.63 gm
In order to determine fibre volume fraction, it can be determined by using rule of mixture,
which can be represented as follow:
Vf = JAf / dPf ---- (2)
Where,
J = number of layers
Af = Areal weight of the fabric
d = Thickness of product