Dissimilar material such as a composite
structure:
what if E and I are not constant??
y
assuming bending only; Mz applied; determine Iz
In this cross section, the upper region has a
modulus = E2 where the remainder has modulus
E1
E2
as with Euler bending, plane sections remain
plane etc....
εx =
−y
R
z
NA
NA
ε is axial strain
E1
y the distance from the neutral (z) axis
R the radius of curvature
σ x = E⋅ε x =
−E⋅y
Hooke applies (although E is now dependent on y)
signs consistent with Shames 11.2
R
pure (only) bending =>
⌠
Fx =  σ x dA = 0 =

⌡
⌠
−

⌡
E( y )⋅y
R
dA =
E( y )⋅y
R
E
suppose we define a parameter Ti such that
then:
⌠
−

⌡
T =
i
E
dA
i
net axial force = 0
that is, a fraction of a reference modulus E1
1
−E ⌠
1
Ti( y )⋅y dA
R 
⌡
"transfer" Ti to the area, in such a way that y is not affected =>
Ti( y )⋅y⋅dA = y ⋅ ( Ti( y )⋅dA) = y ⋅ dy⋅ ( Ti( y )⋅dz)
which means "transfer" to dz, and in rectangular shape, equivalent to applying to z dimension, for thin walled
vertical sections that's the thickness. that is over the different moduli:
⌠
⌠
 T ( y )⋅y dA = T ⋅b⋅ y dy
i
i

⌡
⌡
1
in a horizontal thin walled section, y ~ constant => can still apply to thickness:
y
y
→
E2
z
=>
NA
NA
←
E1
Ti*b
z
E1
E1
←
←
→
b
b
→
E1
E2
→
←
→
t
←
T2*t
=>
E1
E1
→
⌠
 T ( y )⋅y dA = 0
i

⌡
←
→
t
←
t
=> NA is at the cg of the transformed section
now continue looking at the bending moment:
−E ⌠
⌠
−1 ⌠
1 

2
2
M z = − σ x⋅y dA =
⋅  E( y )⋅y dA =
⋅
T ⋅y dA
i


R
R
⌡
⌡
⌡
define
εx =
−y
R
E
T =
i
⌠
2
Iz_tr =  T ⋅y dA
i

⌡
E
σ x = E⋅ε x =
i
1
σ x = −E⋅( y )⋅
1
=>
R
−E( y )⋅y
R
Mz
E ⋅Iz_tr
=
using the relation defined above
and again we can apply Ti to the z dimension in
vertical sections and to the y in horizontal
Mz
E ⋅Iz_tr
1
= −E( y )⋅
Mz
E ⋅Iz_tr
⋅y
where we have written E(y) as it varies
1
Mz
Mz
⋅y = −Ti⋅( y )⋅
⋅y
1 E ⋅I
Iz_tr
1 z_tr
⋅y = −Ti⋅( y )⋅E ⋅
1
2
assume for plot like text
R := −1
−y
ε x( y ) :=
R
10
y
0
10
(plane sections remain plane)
10
0
10
ε x( y)
to evaluate stress
at yi (where modulus changes, substitute E(y) = Ti*E1 =>
Iz_tr := 1
i := 0 .. 1
Ti( y ) :=
 0
yy :=  0.7

 1.0 
M z := −1
y := 0, 0.01 .. 1
∑ TTi⋅(yyi < y ≤ yyi+1)
Mz
Iz_tr
 1

 0.5 
summation is to superpose the values appropriate to the (y)
dependence
i
σ x( y ) := −Ti( y )⋅
TT :=
Mz
⋅y
1 I
z_tr
⋅y
σ_ref ( y ) := −TT ⋅
hence the stress variation is
1
1
y
y
Ti( y) 0.5
0.5
0
0
0.2
0.4
0.6
0
0.8
σ x( y) , σ_ref ( y)
0
0.5
y
3
1
x := 0 .. 10
In := 1
v( x) := 1
n := 1 .. 2
E := 1
dg := 1
I := 1
A := 1
n
d := 1
constant := 0
z := 0 .. 10
i0 := 1
n
n
n := 1 .. 2
V( x) := 1
σ z := 6
t := 1
N0 := 0
τ xz := 1
τ yz := 4
σ 1 := 1
τ xy := 1 σ 2 := 2
σ 3 := 3
4
σ y := 5
σ Y := 4
σ x := 2