Page 207 - Mechanical Behavior of Materials

208 Chapter 5 Stress–Strain Relationships and Behavior

Determine the following in terms of σ z and the elastic constants of the material:

(a) The stress that develops in the y-direction.
(b) The strain in the z-direction.
(c) The strain in the x-direction.

(d) The stiffness E = σ z /ε z in the z-direction. Is this apparent modulus equal to the elastic
modulus E from a uniaxial test on the material? Why or why not?
(e) Assume that the compressive stress in the z-direction has a magnitude of 75 MPa and

that the block is made of a copper alloy, and then calculate σ y , ε z , ε x , and E .
Solution Hooke’s law for the three-dimensional case, Eq. 5.26, is needed. The situation posed
requires substituting ε y = 0 and σ x = 0, and also treating σ z as a known quantity.
(a) The stress in the y-direction is obtained from Eq. 5.26(b):

1
0 = σ y − ν (0 + σ z ) , σ y = νσ z Ans.
E
(b) The strain in the z-direction is given by substituting this σ y into Eq. 5.26(c):

1 − ν 2
1
ε z = σ z − ν (0 + νσ z ) , ε z = σ z Ans.
E E

(c) Thestraininthe x-direction is given by Eq. 5.26(a) with σ y from part (a) substituted:
ν (1 + ν)
1
ε x = 0 − ν (νσ z + σ z ) , ε x =− σ z Ans.
E E

(d) The apparent stiffness in the z-direction is obtained immediately from the equation
for ε z :

σ z E

E = = 2 Ans.
ε z 1 − ν
(e) For the copper alloy, Table 5.2 provides constants, E = 130 GPa = 130,000 MPa,
and ν = 0.343. The compressive stress requires that a negative sign be applied, so that σ z =
−75 MPa. Substituting these quantities into the equations previously derived gives

σ y = νσ z = (0.343)(−75 MPa) =−25.7MPa Ans.

1 − ν 2 1 − 0.343 2 −6
ε z = σ z = (−75 MPa) =−509 × 10 Ans.
E 130,000 MPa