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Homework # 3
1
Given ∝𝑓 𝑎𝑛𝑑 ∝ 𝑚 . .
as α is a thermal expansion coeff.
And volume fractions of fibers and matrix are given .
Determine :
∝1 and ∝2
given the assumptions of stress and strain :
1/ ∈1= ∈𝑓= ∈ 𝑚
2/ 𝛿2 = 𝛿𝑓 = 𝛿 𝑚
_________________________________________________
Solution :
1_st direction ..
∆𝑥 𝛿𝑥
Equilibrium of forces :
𝛿1 𝐴1 = 𝛿𝑓 𝐴𝑓 + 𝛿 𝑚 𝐴 𝑚
∆𝑥. 𝛿𝑓 . 𝐴𝑓
∆𝑥. 𝐴
+
∆𝑥. 𝛿 𝑚 . 𝐴 𝑚
∆𝑥. 𝐴
= 𝛿1
𝑉𝑓 𝛿𝑓 + 𝑉𝑚 𝛿 𝑚 = 𝛿1
Using generalized Hook’s law :
∈ =
𝛿
𝐸
+ 𝛼𝑇
∈. 𝐸 = 𝛿 + 𝛼𝑇𝐸
𝛿 = 𝐸 ∈ −∝ 𝑇
So ..
𝑉𝑓 𝐸𝑓 ∈𝑓−∝𝑓 𝑇 + 𝑉𝑚 𝐸 𝑚 ∈ 𝑚 −∝ 𝑚 𝑇 = 𝐸1 ∈1−∝1 𝑇
From assumption 1 :
𝑉𝑓 𝐸𝑓 ∈1−∝𝑓 𝑇 + 𝑉𝑚 𝐸 𝑚 ∈1−∝ 𝑚 𝑇 = 𝐸1 ∈1−∝1 𝑇
𝑉𝑓 𝐸𝑓 ∈1−∝𝑓 𝑇 + 𝑉𝑚 𝐸 𝑚 ∈1−∝ 𝑚 𝑇 = 𝐸1. ∈1− 𝐸1. 𝑇. ∝1
∈1 𝑉𝑓 𝐸𝑓 + 𝑉𝑚 𝐸 𝑚 − 𝑇 𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚 = 𝐸1. ∈1− 𝐸1. 𝑇. ∝1
𝐸1 = 𝑉 𝑓 𝐸 𝑓 + 𝑉 𝑚 𝐸 𝑚
𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚 = 𝐸1 ∝1
∝1=
𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚
𝐸1
∝1=
𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚
𝑉𝑓 𝐸𝑓 + 𝑉𝑚 𝐸 𝑚
2_st direction ..
lm+lf
lf
∆𝑙 = ∈𝑓 𝑙𝑓 + ∈ 𝑚 𝑙 𝑚 = ∈2 𝑙𝑓 + 𝑙 𝑚
∈2= ∈𝑓 𝑉𝑓 + ∈ 𝑚 𝑉𝑚
𝛿2
𝐸2
+ ∝2 𝑇 =
𝛿𝑓
𝐸𝑓
+∝𝑓 𝑇 𝑉𝑓 +
𝛿 𝑚
𝐸 𝑚
+∝ 𝑚 𝑇 𝑉𝑚
Using assumption .. 2
𝛿2
𝐸2
+ ∝2 𝑇 =
𝛿2
𝐸𝑓
+∝𝑓 𝑇 𝑉𝑓 +
𝛿2
𝐸 𝑚
+∝ 𝑚 𝑇 𝑉𝑚
𝛿2
𝐸2
+ ∝2 𝑇 =
𝛿2
𝐸𝑓
+∝𝑓 𝑇 𝑉𝑓 +
𝛿2
𝐸 𝑚
+∝ 𝑚 𝑇 𝑉𝑚
𝛿2
1
𝐸2
−
𝑉𝑓
𝐸𝑓
−
𝑉𝑚
𝐸 𝑚
+ 𝛼2 𝑇 = 𝑇 𝛼𝑓 𝑉𝑓 + 𝛼 𝑚 𝑉𝑚
𝐸2 =
𝐸 𝑚 . 𝐸𝑓
𝑉𝑓 𝐸 𝑚 + 𝑉𝑚 𝐸𝑓
∴ 𝛼2 = 𝛼 𝑓 𝑉 𝑓 + 𝛼 𝑚 𝑉 𝑚

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Homework three

  • 1. Homework # 3 1 Given ∝𝑓 𝑎𝑛𝑑 ∝ 𝑚 . . as α is a thermal expansion coeff. And volume fractions of fibers and matrix are given . Determine : ∝1 and ∝2 given the assumptions of stress and strain : 1/ ∈1= ∈𝑓= ∈ 𝑚 2/ 𝛿2 = 𝛿𝑓 = 𝛿 𝑚 _________________________________________________ Solution : 1_st direction ..
  • 2. ∆𝑥 𝛿𝑥 Equilibrium of forces : 𝛿1 𝐴1 = 𝛿𝑓 𝐴𝑓 + 𝛿 𝑚 𝐴 𝑚 ∆𝑥. 𝛿𝑓 . 𝐴𝑓 ∆𝑥. 𝐴 + ∆𝑥. 𝛿 𝑚 . 𝐴 𝑚 ∆𝑥. 𝐴 = 𝛿1 𝑉𝑓 𝛿𝑓 + 𝑉𝑚 𝛿 𝑚 = 𝛿1 Using generalized Hook’s law : ∈ = 𝛿 𝐸 + 𝛼𝑇 ∈. 𝐸 = 𝛿 + 𝛼𝑇𝐸 𝛿 = 𝐸 ∈ −∝ 𝑇 So .. 𝑉𝑓 𝐸𝑓 ∈𝑓−∝𝑓 𝑇 + 𝑉𝑚 𝐸 𝑚 ∈ 𝑚 −∝ 𝑚 𝑇 = 𝐸1 ∈1−∝1 𝑇
  • 3. From assumption 1 : 𝑉𝑓 𝐸𝑓 ∈1−∝𝑓 𝑇 + 𝑉𝑚 𝐸 𝑚 ∈1−∝ 𝑚 𝑇 = 𝐸1 ∈1−∝1 𝑇 𝑉𝑓 𝐸𝑓 ∈1−∝𝑓 𝑇 + 𝑉𝑚 𝐸 𝑚 ∈1−∝ 𝑚 𝑇 = 𝐸1. ∈1− 𝐸1. 𝑇. ∝1 ∈1 𝑉𝑓 𝐸𝑓 + 𝑉𝑚 𝐸 𝑚 − 𝑇 𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚 = 𝐸1. ∈1− 𝐸1. 𝑇. ∝1 𝐸1 = 𝑉 𝑓 𝐸 𝑓 + 𝑉 𝑚 𝐸 𝑚 𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚 = 𝐸1 ∝1 ∝1= 𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚 𝐸1 ∝1= 𝑉𝑓 𝐸𝑓 ∝𝑓 + 𝑉𝑚 𝐸 𝑚 ∝ 𝑚 𝑉𝑓 𝐸𝑓 + 𝑉𝑚 𝐸 𝑚
  • 4. 2_st direction .. lm+lf lf ∆𝑙 = ∈𝑓 𝑙𝑓 + ∈ 𝑚 𝑙 𝑚 = ∈2 𝑙𝑓 + 𝑙 𝑚 ∈2= ∈𝑓 𝑉𝑓 + ∈ 𝑚 𝑉𝑚 𝛿2 𝐸2 + ∝2 𝑇 = 𝛿𝑓 𝐸𝑓 +∝𝑓 𝑇 𝑉𝑓 + 𝛿 𝑚 𝐸 𝑚 +∝ 𝑚 𝑇 𝑉𝑚 Using assumption .. 2 𝛿2 𝐸2 + ∝2 𝑇 = 𝛿2 𝐸𝑓 +∝𝑓 𝑇 𝑉𝑓 + 𝛿2 𝐸 𝑚 +∝ 𝑚 𝑇 𝑉𝑚 𝛿2 𝐸2 + ∝2 𝑇 = 𝛿2 𝐸𝑓 +∝𝑓 𝑇 𝑉𝑓 + 𝛿2 𝐸 𝑚 +∝ 𝑚 𝑇 𝑉𝑚
  • 5. 𝛿2 1 𝐸2 − 𝑉𝑓 𝐸𝑓 − 𝑉𝑚 𝐸 𝑚 + 𝛼2 𝑇 = 𝑇 𝛼𝑓 𝑉𝑓 + 𝛼 𝑚 𝑉𝑚 𝐸2 = 𝐸 𝑚 . 𝐸𝑓 𝑉𝑓 𝐸 𝑚 + 𝑉𝑚 𝐸𝑓 ∴ 𝛼2 = 𝛼 𝑓 𝑉 𝑓 + 𝛼 𝑚 𝑉 𝑚