MatSci 201 HW7 2008 Solutions - MatSci 201 Introduction to...

MatSci 201: Introduction to Materials
Homework #7: Due Monday, December 1, 2008 by 5 PM
Submit in class or in the hanging folder outside 3035 Cook Hall
No late homework accepted.
Mechanical Behavior and Ceramics
1. (a) For an iron bar of length 10 cm and cross-section area of 1 cm
2
, calculate the
force required in newtons (N) to produce an elongation of 0.1 mm, assuming
elastic deformation and a Young’s modulus of 200 GPa. Be careful about
units.
(b) For the same iron bar, we first apply stress of 400 MPa. This produces a total
strain of 0.3%. The stress is then reduced to zero. The material sustains a
permanent strain because the elastic limit is exceeded. Calculate the value of
this permanent strain.
(c) The same iron bar discussed above has surface cracks 2 mm long. The
relevant fracture toughness is 20 MPa-m
1/2
(2 × 10
7
Pa-m
1/2
). Assume the
geometric factor Y to be 1.0. What is the maximum stress this iron bar can
withstand before fracture?
Solution
(a) First, write down the expression relating force to the extension as follows:
F = σ ⋅ A = E ⋅ ε ⋅ A = E
Δ L
L






A
where symbols have their usual meanings.
Substituting E = 200 × 10
9
Pa, A = 10
-4
m
2
, L = 0.1 m, and Δ L = 10
-4
m, we have:
F = 200 × 10
9
⋅
10
− 4
10 − 1






⋅ 10
− 4
⋅ = 2 × 10
4
N
(b) Given that
ε
total
= ε
elastic
+ ε
plastic
=
σ
E
+ ε
plastic
, the plastic or permanent strain is equal to:
ε
plastic
= ε
total
−
σ
E
= 0.003 −
400
200
= 0.003 − 0.002 = 0.001 = 0.1%
(c) The relevant formula is
K
c
= Y σ
c
π a
, so that the critical stress is given by:
σ
c
=
K
c
Y π a
=
20 × 10 6
1 ⋅ π ⋅ 2 × 10 − 3 m
= 252 × 10 6 = 252 MPa

2. Callister, Question 6.19
6.19 We are asked to ascertain whether or not it is possible to compute, for brass, the
magnitude of the load necessary to produce an elongation of 1.9 mm (0.075 in.). It is first