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

Bending Stresses at the intersection of the two Materials
The joint between the timber (1) and the aluminum plate (2) occurs at y = 3.3487 in. At
this location, the bending strain in both materials is identical: ε x = 614 mε. Since the
elastic modulus of the aluminum is six times greater than the elastic modulus of the wood,
the bending stress in the aluminum, calculated as
σ =− n My (
=− − 60,000 lb
6 ⋅ in.)(3.3487 in.)
x2 = 6,263 psi = 6,263 psi (T)
4
I
192.5 in.
t
is six times greater than the bending stress in the wood:
My ( 60,000 lb in.)(3.3487 in.)
σ x1 =− =− − ⋅ = 1, 044 psi = 1,044 psi (T)
4
I
192.5 in.
t
This result can also be seen by applying Hooke’s law to each material. For a normal strain
ε x = 614 mε, the normal stress in wood timber (1), from Hooke’s law, is
σ
−6
= E ε = (1,700,000 psi)(614 × 10 in./in.) = 1, 044 psi = 1,044 psi (T)
x1 1 x
and the normal stress in aluminum plate (2) is
σ
−6
= E ε = (10,200,000 psi)(614 × 10 in./in.) = 6,263 psi = 6,263 psi (T)
x2 2 x
mecmovies
ExAmpLES
m8.16 Determine the bending stresses in a composite beam,
using the transformed-section method.
m8.18 Given allowable stresses
for two materials, determine the
largest allowable moment that
can be applied about the horizontal
axis of the beam cross
section shown.
.
.
m8.17 Given allowable stresses for the aluminum and
brass materials, determine the largest allowable moment that
can be applied about the z axis to the beam cross section.
m8.19 Given allowable
stresses for
wood and steel materials,
determine the
largest allowable moment
and, in turn, the
maximum distributed
load that can be applied
to a simply supported
beam.
.
. .
.
.
.
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