Deformation - Orthopaedic Biomaterials in Research and Practice - page 30

Table 2.2 units of stress a

System

British CGS SI

Pound/inch 2 (psi) Dyne/cm 2 Newton/m 2 (Pa)

1 = 6.895 × 10 4 = 6.895 × 10 3

1.45 × 10 −5 = 1 = 0.1

1.45 × 10 −4 = 10 = 1

1 kgf/m 2 = 1.422 × 10 −3 psi = 98.07 dynes/cm 2 = 9.807 Pa.

a

Since strain is the ratio of two lengths, it has no dimensions. It is
either expressed as the ratio or multiplied by 100 and expressed as a
percent.
Similarly, if the initial width is

) (Figure 2.8c), then the elon-

′ ww
0

( 2

gation ( L 0 is one-half as much (= 0.5*( L − L 0 )) for a given force.

Thus, the effect of the force is to produce an elongation inversely propor-
tional to the initial width in the two-dimensional case. In three dimen-
sions, the force produces an elongation that is inversely proportional to
the cross-sectional area. We can remove this proportionality, to be able
to express a constant effect of the force on the rod, irrespective of cross-
sectional area, by defining a new quantity, stress :

′ − ′ )

σ = F / A 0

where A 0 is the initial cross-sectional area. Stress has the dimensions
of force per unit area and may be expressed in several ways as seen in
Table 2.2.
Stress is found by multiplying a scalar (1/area) times a vector (force),
so it is a vector quantity and is often shown as such, especially to denote
stress distributions.

PROBLEM 2.2
Which part experiences the greater internal compressive stress?

A. A stainless steel bar with a rectangular cross section measuring
1.2 cm × 3.0 cm under an axial load of 1.25 kN, or

B. A titanium alloy rod 3.5 cm in diameter under an axial load of
2 kN.

ANSWER:

For A, σ = 1250/(0.012 × 0.03) = 3.47 MPa
For B, σ = 2000/(3.14 × (0.035/2) 2 ) = 2.08 MPa

Therefore, the correct answer is A , even though the load in A is less
than that in B . Note that stress depends only on dimensions, not on the
material from which the object is made.