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Architecture 324
Structures II
Combined Materials
• Strain Compatibility
• Transformed Sections
• Flitched Beams
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Structures II
Slide 2/25
Strain Compatibility
Axial
With the two materials bonded together,
both will act as one and the deformation
in each is the same.
Therefore, the strains will be the same in
each under axial load, and in flexure
stains are the same as in a solid section,
i.e. linear.
Source: University of Michigan, Department of Architecture
Flexure
In flexure, if the two materials are at the
same distance from the N.A., they will
have the same strain at that point.
Therefore, the strains are “compatible”.
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Structures II
Slide 3/25
Strain Compatibility (cont.)
The stress in each material is determined
by using Young’s Modulus
Source: University of Michigan, Department of Architecture
  E
Care must be taken that the elastic limit of
each material is not exceeded, in either
stress or strain.
Source: University of Michigan, Department of Architecture
University of Michigan, TCAUP
Structures II
Slide 4/25
Transformed Sections
Because the material stiffness E can vary
for the combined materials, the Moment of
Inertia, I , needs to be calculated using a
“transformed section”.
In a transformed section, one material is
transformed into an equivalent amount of
the other material. The equivalence is
based on the modular ratio, n.
EA
n
EB
Source: University of Michigan, Department of Architecture
Based on the transformed section, Itr can
be calculated, and used to find flexural
stress and deflection.
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Structures II
Slide 5/25
Flitched Beams & Scab Plates
• Compatible with wood structure, i.e. can be
nailed
• Lighter weight than steel section
• Less deep than wood alone
• Stronger than wood alone
• Allow longer spans
• Can be designed over partial span to
optimize the section (scab plates)
Source: University of Michigan, Department of Architecture
Source: University of Michigan, Department of Architecture
University of Michigan, TCAUP
Structures II
Slide 6/25
Analysis Procedure:
1.
Determine the modular ratio(s).
Usually the weaker (lower E)
material is used as a base.
2.
Construct a transformed
section by scaling the width of
the material by its modular, n.
3.
Determine the Moment of Inertia
of the transformed section.
4.
Source: University of Michigan, Department of Architecture
EA
n
EB
Calculate the flexural stress in
each material using:
fb 
Mc n
Transformed material
Base material
I tr   I   Ad
I tr
2
Transformation equation or solid-void
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Structures II
Slide 7/25
Analysis Example:
For the composite section, find the
maximum flexural stress level in each
laminate material.
fb 
Mc n
I tr
Source: University of Michigan, Department of Architecture
Determine the modular ratios for each
material.
Use wood (the lowest E) as base
material.
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Structures II
Slide 8/25
Analysis Example cont.:
Determine the transformed width of
each material.
Construct a transformed section.
Source: University of Michigan,
Department of Architecture
Transformed Section
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Structures II
Slide 9/25
Analysis Example cont.:
Construct a transformed section.
Calculate the Moment of Inertia for
the transformed section.
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Structures II
Slide 10/25
Analysis Example cont.:
Find the maximum moment.
Source: University of Michigan,
Department of Architecture
By diagrams
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or
Structures II
by summing moments
Slide 11/25
Analysis Example cont.:
Calculate the stress for each material
using the modular ratio to convert I.
Itr /n = I
Compare the stress in each material
to limits of yield stress or the safe
allowable stress.
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Structures II
Slide 12/25
Pop Quiz
Material A is A-36 steel (Data Sheet D-4,
Gr S-1) E = 29,000 ksi
Material B is aluminum (Data Sheet D-10,
Gr A-2) E = 10,000 ksi
Source: University of Michigan, Department of Architecture
  E
If strain, 1 = 0.001
What is the stress in each material at that
point?
A steel _______ ksi
B aluminum ______ ksi
Care must be taken that the elastic limit of
each material is not exceeded, in either
stress or strain.
Source: University of Michigan,
Department of Architecture
University of Michigan, TCAUP
Structures II
Slide 13/25