CEE 595: Composites for Construction Midterm Exam Solution

This document presents a comprehensive solution to a Civil Engineering midterm exam focusing on composite materials, specifically addressing problems related to pultruded materials used in structural applications. The solution meticulously calculates the fiber volume fraction, engineering stiffness properties (E1, E2, G12, Poisson's ratios) for both the roving and CFM layers, and determines the [A] and [D] matrices of the layup, and also calculates the effective engineering in-plane extensional properties. The assignment explores the properties of E-glass/Polyester composites, including the analysis of laminate weight and physical properties of different manufacturing processes such as resin-infused and open mold methods. Additionally, the solution analyzes a composite material for construction, calculating density and other relevant parameters. The solutions are derived using the rule of mixtures and inverse rule of matrix. This resource is designed to assist students in understanding and solving complex problems related to composite materials in civil engineering.

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

Pf = Fibre density
Vf = Fibre volume fraction
Vf = 113*2.63 / 0.04* Pf
Vf = 297.19/ 0.04* Pf -------- (3)
Fibre density (Pf) = m/v = 2.63/ A. h----- (4)
= 2.63/ 2.615*10-3*0.04
= 2.63 / 0.1046*10-3
= 25.143 *103 ---------- (5)
Putting the value of equation (5) into (3), we get
Vf = 297.19/ 0.04* 25.143*103
Vf = 297.19 / 100.57
Vf = 2.95
b)
The engineering stiffness properties of the roving layer has been described below:
E1 = Axial stiffness of composite
E2 = Transverse stiffness of composite
G12 = Shear stiffness of composite roving acting in 1 direction on a plane with normal in 2
direction
According to rule of mixture,
E1 = EfVf + EmVm ------ (6)
Ef = the fibre modulus of elasticity along the fibre axis
Em = – the modulus of elasticity of the matrix
Vf = the fibre volume fraction
Vm = the matrix volume fraction
By putting the value in Equation (6), we get

E1 = 7.66*106 *2.95 + 3.4*106* 3.5
E1 = 7.66*2.95*106 / 3.4*3.5*106
E1 = 22.597/11.9
E1 = 1.9 --- (7)
The Transverse stiffness of composite can be determined by using inverse rule of matrix
(IRM) rule. Mathematically, it can be represented as
E2 = Ef*Em / EmVf + EfVm
E2 = 3.4*106*7.66*106 / 3.4*106*2.95 + 7.66*106*3.5
E2 = 26.044*1012/ (10.03+26.81)106
E2 = 26.044/36.84 * 106
E2 = 7.07 * 105
f = volume fraction of fibre
In order to determine the value of Shear stiffness of composite roving, it is necessary to
determine the value of volume fraction of fibre. It can be presented, mathematically below:
E1 = (1- f) Em + f*Ef ----- (8)
The equation (8) can also be represented as
E1 = Em + f (Ef + Em)
By putting the value we get
E1 = 3.4*106 + f (7.66*106 + 3.4*106)
E1 = 3.4 *106 +f (11.06 * 106)
1.9 - 3.4*106 = f *(11.06 * 106)
- 1.5 * 106 = f (11.06 * 106)
f = -1.5 *106/ 11.06*106
f = -0.14 ------ (9)
With the help of rule of mixture for poisonous ratio V12
V12 = fVf + (1-f)Vm
V12 = 0.14*2.95 + (1- (-0.14))Vm

V12 = 0.413+ 114*3.5
V12 = 0.413 +3.99
V12 = 4.403
c)
d)
e)
f)
Problem 2
a)
b)
Laminate weight
Laminate weight Resign infused Open mold
Fiber 0.18 0.18
Resin 0.09 0.21
Total 0.27 0.39
Physical properties
Resign infused Open mold
Density (g/cc) 1.87 1.59
Fibre content (% by wt.) 68% 46%
Thickness (in) 0.028 0.047
Laminate Moduli
Resign infused Open mold
Ex 2.03 1.86
Ey 2.83 1.86
Gxy 0.59 0.41
Ex flex 2.69 1.77
Ey flex 2.69 1.77

c)
Composite for construction
Width – 6 inch
Volume fraction (Vf) = Pmat – Pmatrix / Pmat - Pmatrix
Density (Pmat) = mmat /Vmat
Massmat = weight / g ----- [g = 9.8 m/s2]
Vmat = l*b*h
= 6*2.54
= 15.24 cm3
Weight =743.89* 10-2
Mass = 743.89-10-2/9.8
= 0.76 kg
L = 0.02 *2.54 = 0.0508 cm
B = 0.04*2.54 = 0.1016 cm
H = 0.02*2.54 = 0.0508 cm
V= 2.6*105 cm3
Density of matrix (Pmatrix) = 0.76/ 2.6*105
= 2.9 *105 kg/cm3
Density of matrix (Pmatrix) = 0.029 kg/m3
d)