Page 106 - Mechanical Behavior of Materials

Section 3.8 Materials Selection for Engineering Components 107

P
d = 2r

L

v
max

Figure 3.29 Cantilever beam.

the order of 2 or 3. Weight is critical, so the mass m of the beam must be minimized. Finally, the
diameter, d = 2r, of the cross section may be varied to allow the material chosen to meet the various
requirements just noted.
A systematic procedure can be followed that allows the optimum material to be chosen in this
and other analogous cases. To start, classify the variables that enter the problem into categories as
follows: (1) requirements, (2) geometry that may vary, (3) materials properties, and (4) quantity to
be minimized or maximized. For the beam example, with ρ being the mass density, these are

1. Requirements: L, P, X
2. Geometry variable: r
3. Material properties: ρ, σ c
4. Quantity to minimize: m

Next, express the quantity Q to be minimized or maximized as a mathematical function of the
requirements and the material properties, in which the geometry variable does not appear:

Q = f 1 (Requirements) f 2 (Material) (3.1)

For the beam example, Q is the mass m, so that the functional dependencies needed are

m = f 1 (L, P, X) f 2 (ρ, σ c ) (3.2)

with the beam radius r not appearing. Note that all the quantities in f 1 are constants for a given
design, whereas those in f 2 vary with material.
For the procedure to work, the equation for Q must be expressed as the product of two separate
functions f 1 and f 2 , as indicated. Fortunately, this is usually possible. The geometry variable cannot
appear, as its different values for each material are not known at this stage of the procedure.
However, it can be calculated later for any desired values of the requirements. Once the desired
Q = f 1 f 2 is obtained, it may be applied to each candidate material, and the one with smallest or
largest value of Q chosen, depending on the situation.