This document contains the solutions to homework problems 4.5, 4.6, and 4.11. Problem 4.5 involves calculating the effective Young's modulus, shear modulus, and Poisson's ratio of a composite with known fiber and matrix properties. Problem 4.6 generalizes these calculations for different fiber/matrix materials. Problem 4.11 constrains the fiber and matrix moduli based on requirements for the effective Young's moduli. The problem is solved by setting up equations relating the fiber volume fraction and moduli and solving them simultaneously.

1. Homework # 4
Problems : 4.5 , 4.6 , 4.11
4.5
𝐸𝑓 = 230 𝐺𝑝𝑎 , 𝐸 𝑚 = 4.6 𝐺𝑝𝑎 , 𝐺𝑓 = 1.84 𝐺𝑝𝑎 , 𝐺 𝑚 = 1.77 𝐺𝑝𝑎 , 𝜗𝑓 = 0.25 ,
𝜗 𝑚 = 0.3 , 𝑉𝑓 = 0.4 , 𝑉𝑚 = 0.6
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅 ∶ 𝑬 𝟏 , 𝑬 𝟐 , 𝑮 𝟏𝟐, 𝝑 𝟏𝟐
𝐸1 = 𝐸 𝑚 𝑉𝑚 + 𝐸𝑓 𝑉𝑓 & 𝐸2 =
𝐸 𝑚 . 𝐸𝑓
𝑉𝑓 𝐸 𝑚 + 𝑉𝑚 𝐸𝑓
𝐺12 =
𝐺 𝑚 . 𝐺𝑓
𝐺𝑓 𝑉𝑚 + 𝐺 𝑚 𝑉𝑓
& 𝜗12 = 𝜗𝑓 𝑉𝑓 + 𝜗 𝑚 𝑉𝑚
𝐸1 = 4.6 ∗ 0.6 + 230 ∗ 0.4 = 94.76 𝐺𝑝𝑎
𝐸2 =
4.6 ∗ 230
0.4 ∗ 4.6 + 0.6 ∗ 230
= 7.5658 𝐺𝑝𝑎
𝐺12 =
1.84 ∗ 1.77
0.6 ∗ 1.84 + 0.4 ∗ 0.4
= 2.57658 𝐺𝑝𝑎
𝜗12 = 0.25 ∗ 0.4 + 0.3 ∗ 0.6 = 0.28

2. 4.6
𝑹𝒆𝒒𝒖𝒊𝒓𝒆𝒅 ∶ 𝑬 𝟏 , 𝑬 𝟐 , 𝑮 𝟏𝟐, 𝝑 𝟏𝟐 For E-glass , S-glass , Carbon T-300 :
𝐸1 = 𝐸 𝑚 𝑉𝑚 + 𝐸𝑓 𝑉𝑓 & 𝐸2 =
𝐸 𝑚 . 𝐸𝑓
𝑉𝑓 𝐸 𝑚 + 𝑉𝑚 𝐸𝑓
𝐺12 =
𝐺 𝑚 . 𝐺𝑓
𝐺𝑓 𝑉𝑚 + 𝐺 𝑚 𝑉𝑓
& 𝜗12 = 𝜗𝑓 𝑉𝑓 + 𝜗 𝑚 𝑉𝑚
𝐺23 = 𝐺 𝑚
𝑉𝑓 + Ƞ23(1 − 𝑉𝑓)
Ƞ23 1 − 𝑉𝑓 + 𝑉𝑓 𝐺 𝑚 /𝐺𝑓
& Ƞ23 =
3 − 𝜗 𝑚 + 𝐺 𝑚 /𝐺𝑓
4 1 − 𝜗 𝑚

3. 4.11
𝐸1 = 𝐸 𝑚 𝑉𝑚 + 𝐸𝑓 𝑉𝑓 & 𝐸2 =
𝐸 𝑚 . 𝐸𝑓
𝑉𝑓 𝐸 𝑚 + 𝑉𝑚 𝐸𝑓
𝑉𝑚 = 1 − 𝑉𝑓
𝐸1 > 30 𝐺𝑝𝑎 𝑎𝑛𝑑 3.5𝐸2 > 𝐸1
∴ 𝐸1 > 30𝐺𝑝𝑎 𝑎𝑛𝑑 𝐸2 < 8.5714
∴ 𝐸1 = 𝐸 𝑚 1 − 𝑉𝑓 + 𝐸𝑓 𝑉𝑓
𝐸2 =
𝐸 𝑚 𝐸 𝑓
𝑉 𝑓 𝐸 𝑚 + 1−𝑉 𝑓 𝐸 𝑓
𝒂𝒔𝒔𝒖𝒎𝒆 𝑽 𝒇 = 𝟎. 𝟒
∴ 0.6𝐸 𝑚 + 0.4𝐸𝑓 > 30
∴
−𝐸 𝑚 𝐸𝑓
0.4𝐸 𝑚 + 0.6𝐸𝑓
> −8.5714

4. 𝐸 𝑚 >
10
6
30 − 0.4𝐸𝑓 ……………………… (1
−10
6
30−0.4𝐸 𝑓 𝐸 𝑓
0.4∗
10
6
30−0.4𝐸 𝑓 +0.6𝐸 𝑓
> −8.5714 …………………….(2
From (1) and (2) we can get
𝐸𝑓 & 𝐸 𝑚 𝑎𝑛𝑑 𝑓𝑟𝑜𝑚 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝐸1 𝑤𝑒 𝑐𝑎𝑛 𝑔𝑒𝑡 𝑉𝑓