Timothy A. Philpot - Mechanics of materials _ an integrated learning system-John Wiley (2017)

Equivalent Beams
Before considering a beam made of two materials, let us first examine what is required so that
two beams of different materials can be considered equivalent. Suppose that a small rectangular
aluminum bar having an elastic modulus E alum = 70 GPa is used as a beam in pure bending
(Figure 8.10a). The bar is subjected to an internal bending moment M = 140,000 N · mm,
which causes the bar to bend about the z axis. The width of the bar is 15 mm, and the height of
the bar is 40 mm (Figure 8.10b); therefore, its moment of inertia about the z axis is I alum =
80,000 mm 4 . The radius of curvature ρ of this beam can be computed from Equation (8.6):
1 M
140,000 N⋅mm
= =
2 4
ρ EI (70,000 N/mm )(80,000 mm )
alum
∴ ρ = 40,000 mm
The maximum bending strain caused by the bending moment can be determined from
Equation (8.1):
1 1
ε x =− y =− ( ± 20mm) = ± 0.0005 mm/mm
ρ 40,000 mm
Next, suppose that we want to replace the aluminum bar with wood, which has an elastic
modulus E wood = 10 GPa. In addition, we require that the wooden beam be equivalent to the
aluminum bar. The question, then, becomes, “What dimensions are required in order for
the wooden beam to be equivalent to the aluminum bar?”
But what is meant by “equivalent” in this context? To be equivalent, the wooden beam
must have the same radius of curvature ρ and the same distribution of bending strains ε x as
the aluminum bar for the given internal bending moment M. To produce the same ρ for the
140 N · m bending moment, the moment of inertia of the wooden beam must be increased to
I
M 140,000 Nmm ⋅
= ρ =
(40,000 mm) = 560,000 mm
E 10,000 N/mm
wood 2
Thus, the wooden beam must be larger than the aluminum bar in order to have the same
radius of curvature. However, equivalence also requires that the wood beam exhibit the
same distribution of strains. So, since strains are directly proportional to y, the wooden
beam must have the same y coordinates as the aluminum bar; in other words, the height of
the wooden beam must also be 40 mm.
The moment of inertia of the wooden beam must be larger than that of the aluminum
bar, but the height of both must be the same. Therefore, the wooden beam must be wider
than the aluminum bar if the two beams are to be equivalent:
I
wood
3
3
bh bwood
(40mm)
= = = 560,000 mm
12 12
∴ b = 105 mm
wood
4
4
ρ
M
y
O
(a) Bar subjected to pure
bending
z
15 mm
y
x
40 mm
M
(b) Cross-sectional dimensions
of aluminum bar
In this example, a wooden beam that is 105 mm wide and 40 mm tall is equivalent to an
aluminum bar that is 15 mm wide and 40 mm tall (Figure 8.10c). Since the elastic moduli
of the two materials are different (by a factor of 7 in this case), the wooden beam (which z
40 mm
has the lesser E) must be wider than the aluminum bar (which has the greater E)—wider
in this case by a factor of 7.
105 mm
Now, if the two beams are equivalent, are the bending stresses the same? The bending
stress produced in the aluminum bar can be calculated from the flexure formula:
(c) Cross-sectional dimensions
of equivalent wooden beam
(140,000 Nmm)(20mm)
⋅
σ alum =
= 35 MPa
FIGURE 8.10 Equivalent
4
80,000 mm
beams of aluminum and wood.
271
FLExuRAL STRESSES IN bEAMS
OF TwO MATERIALS
y