SANDWICH CONSTRUCTIONS
Zdeněk Padovec
1
Sandwiches – what it is?
 constructions for bending – weight savings
 skins for bending load, core for shear load
 analogy to I profile
2
Sandwiches – what it is?
 skins – metal, composite (unidirectional, fabrics,
random reinforcement, …)
 core – foam, honeycomb, balsa wood, cork, …
3
Sandwiches – what it is?
 classic lamination theory
 no shear deformations
 classic lamination
deformation
theory with transverse shear
 core transmits both shear and bending load and we deal with
it as one of the layers, consider as isotropic
𝐍
𝐀
𝐌 = 𝐁
𝐐𝒔
𝟎
𝟎
𝐁
𝐃
𝟎
𝟎 𝛆𝐦
𝟎
𝐤
𝐅 𝛄𝒔
 sandwich theory
 core transmits just shear load, consider as isotropic
𝐀
𝐂
𝟎
𝐁
𝐃
𝟎
𝟎
𝐍
𝐌 =
𝐐𝒄
𝟎 𝛆𝐦
𝟎
𝐤
𝐅 𝛄𝒄
4
Classic lamination theory
 layered laminates




each lamina is orthotropic and quasi homogeneous
thickness << width, length
plane stress
displacement of particular points are very small
ideal, infinitely thin joint between laminas – continuous
displacements, linear in through thickness directions
 Kirchhoff ´s hypothesis, εz  0
 linear relationship between stress and strain
5
Classic lamination theory
 forces and moments working on lamina
ℎ𝑘
𝑁𝑥 =
ℎ𝑘
𝜎𝑥𝑥 𝑑𝑧 , 𝑁𝑦 =
ℎ𝑘−1
ℎ𝑘
𝑀𝑥 =
𝜎𝑦𝑦 𝑑𝑧, 𝑁𝑥𝑦 =
ℎ𝑘−1
ℎ𝑘
𝜎𝑥𝑥 𝑧𝑑𝑧 , 𝑀𝑦 =
ℎ𝑘−1
ℎ𝑘
𝜎𝑥𝑦 𝑑𝑧
ℎ𝑘−1
ℎ𝑘
𝜎𝑦𝑦 𝑧𝑑𝑧, 𝑀𝑥𝑦 =
ℎ𝑘−1
𝜎𝑥𝑦 𝑧𝑑𝑧
ℎ𝑘−1
6
Classic lamination theory
 σ is substituted according to Hookes´s law
𝑛
𝑁𝑥
𝑁𝑦 =
𝑁𝑥𝑦
𝑘=1
𝑛
𝑀𝑥
𝑀𝑦 =
𝑀𝑥𝑦
𝑘=1
ℎ𝑘
ℎ𝑘−1
ℎ𝑘
ℎ𝑘−1
𝑄11
𝑄21
𝑄61
𝑄11
𝑄21
𝑄61
𝑄12
𝑄22
𝑄62
𝑄12
𝑄22
𝑄62
𝜀𝑥𝑥
0
ℎ𝑘 𝑄
𝑄16
11
0
𝑄26 𝜀𝑦𝑦 𝑑𝑧 +
𝑄21
𝑄66
0
ℎ𝑘−1 𝑄61
𝛾𝑥𝑦
𝜀𝑥𝑥
𝜀
𝑁𝑥
𝐴11 𝐴12 𝐴16 𝐵11 𝐵12 𝐵16 𝑥𝑥 0
𝑁𝑦
𝐴21 𝐴22 𝐴21 𝐵21 𝐵22 𝐵26 𝜀𝑦𝑦
𝑁𝑥𝑦
0
𝐴61 𝐴62 𝐴66 𝐵61 𝐵62 𝐵66
𝛾
=
.
𝑀𝑥
𝐵11 𝐵12 𝐵16 𝐷11 𝐷12 𝐷16 𝑥𝑦
𝑀𝑦
𝐵21 𝐵22 𝐵26 𝐷21 𝐷22 𝐷26 𝑘𝑥
𝐵61 𝐵62 𝐵66 𝐷61 𝐷62 𝐷66 𝑘𝑦
𝑀𝑥𝑦
𝑘𝑥𝑦
𝑄16
𝑄26
𝑄66
𝑘𝑥
𝑘𝑦 𝑧𝑑𝑧
𝑘𝑥𝑦
𝑄12
𝑄22
𝑄62
𝑄16
𝑄26
𝑄66
𝑘𝑥
𝑘𝑦 𝑧 2 𝑑𝑧
𝑘𝑥𝑦
0
ℎ𝑘 𝑄
𝑄16
11
0
𝑄26 𝜀𝑦𝑦 𝑧𝑑𝑧 +
𝑄21
𝑄66
0
ℎ𝑘−1 𝑄61
𝛾𝑥𝑦
0
𝑄12
𝑄22
𝑄62
𝑛
𝐴𝑖𝑗 =
1
𝐵𝑖𝑗 =
2
𝐷𝑖𝑗 =
1
3
𝑄𝑖𝑗
𝑘
ℎ𝑘 − ℎ𝑘−1
𝑘
ℎ𝑘 2 − ℎ𝑘−1 2
𝑘
ℎ𝑘 3 − ℎ𝑘−1 3
𝑘=1
𝑛
𝑄𝑖𝑗
𝑘=1
𝑛
𝑄𝑖𝑗
𝑘=1
7
Classic lamination theory
 A…membrane stiffness of a laminate
 B...bending – extension coupling stiffness of a laminate
 D…bending stiffness of a laminate
𝐍
𝐀 𝐁 𝛆𝐦 𝟎
=
𝐌
𝐁 𝐃 𝐤
𝛆𝐦
𝐤
𝟎
𝐍
𝐀
𝐁
=
𝐁 𝐃 𝐌
8
Lamination theory with transverse shear deformation
 normal to the middle plane WILL NOT be normal after
the deformation –Kirchhoff hypothesis DOES NOT work
𝟎
𝐍
𝐀 𝐁 𝟎 𝛆𝐦
𝐌 = 𝐁 𝐃 𝟎
𝐤
𝐐𝒔
𝟎 𝟎 𝐅 𝛄𝒔
 𝐐𝒔 is resultant of outer forces which acting in normal
direction to the plate and which are causing shear
deformations (index s – whole sandwich), F is shear
stiffness matrix of the laminate
 𝛄𝒔 are shear strains
9
Lamination theory with transverse shear deformation
𝑄𝑦
𝐹44
=
𝐹54
𝑄𝑥
0
𝐹45 γ𝑦𝑧
0
𝐹55
γ𝑧𝑥
ℎ
𝐹𝑖𝑗 =
ℎ𝑘 − ℎ𝑘−1 𝐶𝑖𝑗
´
𝑘
, 𝑖, 𝑗 = 4, 5.
𝑘=1
 core transmits both shear and bending load and
we deal with it as one of the layers, consider as
isotropic
10
Sandwich theory
 linear relationship between stress and strain
 core thickness is much more bigger then skin thickness
 core displacements u and v in x and y direction is linear
in through thickness
 skin displacements u and v are constant in their whole
thickness
 through thickness displacement w is independent on
coordinate z - 𝜀𝑧𝑧 = 0
 core transmits JUST shear stresses
 low skin thickness – through thickness shear stress and
normal stress in z direction = 0
𝐍
𝐀
𝐌 = 𝐂
𝐐𝒄
𝟎
𝐁
𝐃
𝟎
𝟎
𝟎 𝛆𝐦
𝟎
𝐤
𝐅 𝛄𝒄
11
Sandwich theory
 index 1 – upper skin, index 2 – lower skin
𝐴𝑖𝑗 = 𝐴𝑖𝑗 1 + 𝐴𝑖𝑗 2 , 𝐵𝑖𝑗 =
ℎ
2
𝐴𝑖𝑗 2 − 𝐴𝑖𝑗 1 , 𝐶𝑖𝑗 = 𝐶𝑖𝑗 1 + 𝐶𝑖𝑗 2 , 𝐷𝑖𝑗 =
𝑛1
𝐴𝑖𝑗 1 =
𝑄𝑖𝑗
𝑘
ℎ𝑘 − ℎ𝑘−1
𝑄𝑖𝑗
𝑘
ℎ𝑘 − ℎ𝑘−1
𝑘
ℎ𝑘 2 − ℎ𝑘−1 2
𝑘
ℎ𝑘 2 − ℎ𝑘−1 2
ℎ
2
𝐶𝑖𝑗 2 − 𝐶𝑖𝑗 1
𝑘=1
𝑛2
𝐴𝑖𝑗 2 =
𝐶𝑖𝑗
1
1
=
2
𝐶𝑖𝑗 2 =
𝐹𝑖𝑗 =
1
2
𝑘=1
𝑛1
𝑄𝑖𝑗
𝑘=1
𝑛2
𝑄𝑖𝑗
𝑘=1
ℎ𝐶𝑖𝑗 ´𝑐
 dissymmetric sandwich plate, anisotropic core
12
Sandwich theory
 index 1 – upper skin, index 2 – lower skin
𝐴𝑖𝑗 = 2𝐴𝑖𝑗 1, 𝐵𝑖𝑗 = 0, 𝐶𝑖𝑗 = 0, 𝐷𝑖𝑗 = ℎ𝐶𝑖𝑗 2
𝑛1
𝑨𝒊𝒋 𝟏 =
𝑪𝒊𝒋 𝟏
1
=
2
𝑛1
𝑄𝑖𝑗
𝑘
ℎ𝑘 − ℎ𝑘−1 = 𝑨𝒊𝒋 𝟐 =
𝑘=1
𝑄𝑖𝑗
𝑛2
𝑄𝑖𝑗
𝑘=1
𝑘
ℎ𝑘 2 − ℎ𝑘−1 2 = − 𝑪𝒊𝒋 𝟐
𝑘=1
ℎ𝐺𝑖𝑗 𝑐 ,
1
=
2
𝑘
ℎ𝑘 − ℎ𝑘−1
𝑛2
𝑄𝑖𝑗
𝑘
ℎ𝑘 2 − ℎ𝑘−1 2
𝑘=1
𝐹𝑖𝑗 =
𝐹44 = 𝐹55 = ℎ𝐺𝑐 , 𝐹45 = 0
 symmetric sandwich plate, isotropic core
 no coupling between resultants N and modified
curvatures k and also between moments M and
middle plain strain 𝜀𝑚
0
13
Sandwich theory
 symmetrical sandwich plate, isotropic core
 no coupling between resultants N and modified
curvatures k and also between moments M and
0
middle plain strain 𝜀𝑚
0
N
A 0 0 εm
M = 0 D 0
k
Q𝑐
0 0 F γ𝑐
14
Exercise
 F = 100 N, l = 400 mm, b= 40 mm, h1 = h2 = 3 mm, skins
are symmetric to the middle plane, 7 layers, each skin
[0|±30|90|-+30|0], hk0,428 mm, C fibers, epoxy
matrix, EL=1,299·105 MPa, ET=0,1399·105 MPa,
GLT=0,045·105 MPa, GTT´=0,03·105 MPa, νLT=0,28.
 Core thickness h = 10 mm, isotropic material EC=75
MPa, νC=0,34. Displacement under the force = ?
 a) sandwich theory
UPPER SKIN
 b) lamination theory
h1
F
CORE
l/2
l/2
h
h2
LOWER SKIN
15
Exercise
 a) SANDWICH THEORY
 core just shear stress, skins just bending stress
 we can derive 𝑤𝐹 =





𝐹𝑙 3
𝐷11 ∗
48𝑏
𝐹55 ∗ 1
1 + 12 ∗ 2
𝐷11 𝑙
𝐷11 ∗ and 𝐹55 ∗
we need to identify elements
symmetric sandwich – B = 0
1 𝑛2
2
2
2
2
𝐷𝑖𝑗 = ℎ𝐶𝑖𝑗 , 𝐶𝑖𝑗 =
𝑄
ℎ
−
ℎ
𝑖𝑗 𝑘 𝑘
𝑘−1
2 𝑘=1
can be proved D*= D-1
𝑄𝑖𝑗 …reduced stiffness matrix for each layer of
𝑘
laminated skin
 𝐷11 ∗ = 6,67984056 ∙ 10−5 N −1 m−1
16
Exercise
 a) SANDWICH THEORY
 core just shear stress, skins just bending stress
 we can derive 𝑤𝐹 =
𝐹𝑙 3
𝐷11 ∗
48𝑏
𝐹55 ∗ 1
1 + 12 ∗ 2
𝐷11 𝑙
𝐷11 ∗ and 𝐹55 ∗
 we need to identify elements
 symmetric sandwich – B = 0
 shear stiffness matrix F depends on core parameters
𝐸𝑐
= 2 1 + 𝜐𝑐 → 𝐺𝑐 = 27,985 Mpa
𝐺𝑐
 𝐹44 = 𝐹55 = ℎ𝐺𝑐 = 2,7985N −1 m, 𝐹45 = 0
𝐹44
∗
 𝐹55 =
= 3,573 ∙ 10−6 m−1 N, 𝛥𝐹…determinant F
𝛥𝐹
𝑤𝐹 = 1,12 mm
17
Exercise
 a) SANDWICH THEORY
 core has ZERO SHEAR STIFFNESS, skins just bending
stiffness


𝐹𝑙 3
we can derive 𝑤𝐹 =
𝐷11 ∗
48𝑏
∗
𝐷11 = 6,67984056 ∙ 10−5 N −1 m−1
𝑤𝐹 = 0,22 mm
18
Exercise
 b) LAMINATION THEORY
 core transmits both shear and bending load and we
deal with it as one of the layers, consider as
isotropic
 symmetric laminate – B = 0
𝟎
𝐍
𝐀 𝟎 𝟎 𝛆𝐦
 𝐌 = 𝟎 𝐃 𝟎
𝐤
𝐐𝒔
𝟎 𝟎 𝐅 𝛄𝒔

𝐹𝑙 3
𝐹55 ∗ 1
∗
we can derive 𝑤𝐹 =
𝐷
1 + 12 ∗ 2
48𝑏 11
𝐷11 𝑙
1 𝑛
3
3
𝐷𝑖𝑗 =
𝑄
ℎ
−
ℎ
𝑖𝑗 𝑘 𝑘
𝑘−1
3 𝑘=1

 can be proved D*= D-1
 𝐷11 ∗ = 5,015552603 ∙ 10−5 N −1 m−1
19
Exercise
 b) LAMINATION THEORY
𝟎
𝐍
𝐀 𝟎 𝟎 𝛆𝐦
 𝐌 = 𝟎 𝐃 𝟎
𝐤
𝐐𝒔
𝟎 𝟎 𝐅 𝛄𝒔



𝐹𝑙 3
𝐹55 ∗ 1
∗
we can derive 𝑤𝐹 =
𝐷
1 + 12 ∗ 2
48𝑏 11
𝐷11 𝑙
𝐹𝑖𝑗 = ℎ𝑘=1 ℎ𝑘 − ℎ𝑘−1 𝐶𝑖𝑗 ´ , 𝑖, 𝑗 = 4, 5
𝑘
𝐹44 = 2,085127931 ∙ 107 N −1 m, 𝐹55 = 2,470842217
107 N −1 m, 𝐹45 = 0
𝐹44
∗
𝐹55 =
= 4,047203 ∙ 10−8 m−1 N,
𝛥𝐹

 𝛥𝐹…determinant F
𝑤𝐹 = 0,18 mm
20
∙
Exercise
 b) LAMINATION THEORY
 core is one of the layers and it has bending stiffness,
no through thickness shear stresses
 symmetric laminate – B = 0
𝟎
𝐀
𝟎
𝟎
𝐍
𝛆𝐦
 𝐌 = 𝟎 𝐃 𝟎
𝐤
𝟎
𝟎 𝟎 𝟎
𝟎
𝐹𝑙 3
 we can derived 𝑤𝐹 =
𝐷11 ∗
1
3
48𝑏
3
𝑛
𝑄
ℎ
𝑖𝑗 𝑘 𝑘 −
𝑘=1
 𝐷𝑖𝑗 =
ℎ𝑘−1 3
 can be proved D*= D-1
 𝐷11 ∗ = 5,015552603 ∙ 10−5 N −1 m−1
21
Exercise
 b) LAMINATION THEORY
 core is one of the layers and it has bending stiffness,
no through thickness shear stresses
 symmetric laminate – B = 0
𝑤𝐹 = 0,17 mm
22
Conclusion
 deflection of sandwich beam
with shear [mm]
without shear
[mm]
Sandwich theory
1,12
0,22
Lamination theory
0,18
0,17
Experiment
0,51
FEM (solid)
0,71
23