Laminated Composite Beams and Columns

CHAPTER

LAMINATED COMPOSITE BEAMS AND COLUMNS

6

High modulus carbon fiber reinforced composites are successfully used to fabricate laminated composite beams loaded with axial and transverse forces and to reinforce traditional metal and concrete beam elements to increase their stiffness. Various approaches to the analysis and design of composite beams have been discussed by Vasiliev (1993), Gay, Hoa, and Tsai (2003), Kollar and Springer (2003), Dekker (2004), and Vinson and Sierakowski (2004). This chapter is concerned with the analysis of static, stability, and dynamic problems using elastic beam theory, allowing for transverse shear deformation.

6.1 BASIC EQUATIONS Consider a beam as shown in Fig. 6.1 and loaded with a transverse distributed load (p, q) and/or concentrated forces (R, F), and end forces and moment (Vl , Nl , Ml ). The analysis of such a beam is based on a specific feature of the structure, namely, the height h is much smaller than its length l. This feature allows us to introduce some assumptions concerning the distribution of beam displacements through the beam height which, in turn, enables us to reduce the problem of the analysis of beams to a solution of ordinary differential equations. We also assume that the beam is loaded only in the xz-plane (see Fig. 6.1), so that the structure is in a state of plane stress. Single out the shaded element of the beam as shown in Fig. 6.2 and assume that the stresses acting on this element are uniformly distributed over the element width b. The equilibrium equations for this element have the following form: b

@σx @ 1 ðbτ xz Þ 5 0; @x @z

@ @τ xz 50 ðbσz Þ 1 b @x @z

(6.1)

in which bðzÞ is the width of the beam cross section (see Fig. 6.1). The stresses σz and τ xz must satisfy the boundary conditions on the beam surfaces. Note that in accordance with Fig. 3.1, we measure the z-coordinate from some reference plane which is located at distances e and s from the beam bottom and top surfaces (see Fig. 6.1). Then, the boundary conditions can be presented as σz ðx; z 5 2 eÞ 5 2 p; σz ðx; z 5 sÞ 5 2 q;

τ xz ðx; z 5 2 eÞ 5 0 τ xz ðx; z 5 sÞ 5 0

(6.2)

Since the beam height h is relatively small, we assume that the beam deflection, uz , does not depend on coordinate z (see Fig. 6.1), whereas the axial displacement, ux , is a linear function of z.

Advanced Mechanics of Composite Materials and Structures. DOI: https://doi.org/10.1016/B978-0-08-102209-2.00006-2 © 2018 Elsevier Ltd. All rights reserved.

377

378

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

uz z q

Fm

q

bk s

V

dz

z

dx

N

t

e

M

l p

x ux

Vl

Ml Nl

dz

b(z) b1

Rm

xm

h

t

p

FIGURE 6.1 A beam loaded with surface and end forces and moment.

σ zb + b

∂ (σ z b) ∂z

σ xb dz

τ xz b

b(z )

dx

τ xzb

τ xzb + b

∂τ xz ∂x

σ xb + b

∂σ x ∂x

σ xb

FIGURE 6.2 Stress state of a beam element.

These assumptions are the same as those introduced in Section 3.1 for thin laminates. Thus, in accordance with Eqs. (3.1) and (3.2), we have uz 5 wðxÞ; ux 5 uðxÞ 1 zθðxÞ

(6.3)

Here, w is the beam deflection, whereas u and θ are the axial displacement and the angle of rotation of the beam cross section, respectively. The axial strain is specified by the first equation of Eq. (3.3), that is, εx 5 u0 ðxÞ 1 zθ0 0

in which ð. . .Þ 5 dð. . .Þ=dx. Using Hooke’s law, we get


0 0 σx 5 Eεx 5 E u 1 zθ

(6.4)

for the axial stress, where E is the axial modulus of the beam material. It follows from this equation 0 that σx depends on two functions, u0 ðxÞ and θ ðxÞ, that can be replaced by the corresponding stress resultants, that is, with the internal axial force N acting in the reference plane and directed along

6.1 BASIC EQUATIONS

379

the beam axis, and the internal bending moment M exerted on the reference plane and acting in the xz-plane, that is, ðs N5

ðs σx bdz;

M5

2e

σx bzdz

(6.5)

2e

(see Fig. 6.1). Substitution of Eq. (6.4) for σx into Eq. (6.5) yields 0

0

0

N 5 B11 u 1 C11 θ0 ;

M 5 C11 u 1 D11 θ

(6.6)

where the coefficients ðs B11 5

ðs Ebdz;

C11 5

2e

ðs D11 5

Ebzdz; 2e

Ebz2 dz

(6.7)

2e

are similar to the stiffness coefficients specified by Eqs. (3.6)(3.8). The constitutive equations, Eq. (6.6) can be simplified by a suitable choice of reference plane. Introduce a new coordinate for the shaded beam element t5z1e

(6.8)

(see Fig. 6.1) which changes from t 5 0 (z 5 2 eÞ to t 5 h (z 5 sÞ. Then, the stiffness coefficients in Eq. (6.7) become B11 5 I0 ;

C11 5 I1 2 eI0 ;

D11 5 I2 2 2eI1 1 e2 I0

(6.9)

in which ðh In 5

Ebtn dt; n 5 0; 1; 2

(6.10)

0

Now assume that the coordinate of the reference plane e (see Fig. 6.1) determines the location of the beam neutral axis, so that the internal axial force N acting along this axis induces only an axial displacement u, whereas the bending moment M causes only a rotation of the cross section. Then, it follows from Eq. (6.6) that C11 5 0, and the second equation of Eq. (6.9) yields e5

I1 I0

(6.11)

Under this condition, Eq. (6.6) reduce to 0

0

B 5 I0 ; D 5 I2 2

I12 I0

N 5 Bu ; M 5 Dθ

(6.12)

where (6.13)

are the axial and bending stiffnesses of the beam. Substituting Eq. (6.12) into Eq. (6.4) and using Eq. (6.8), we arrive at the following final expression for the axial stress: 

σx 5 E

N M 1 ðt 2 eÞ B D



(6.14)

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CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Consider now the shear stress τ xz . According to the first assumption in Eq. (6.3), the beam height does not change with deformation of the beam. So, the distribution of the shear stress over the beam height does not affect the beam behavior which is governed by the resultant transverse shear force V only. In accordance with Eq. (3.12), ðs

ðh

V5

τ xz bdz 5

τ xz bdt

2e

(6.15)

0

The second part of this equation is written with the aid of Eq. (6.8). The transverse shear force V (see Fig. 6.1) is linked with the average shear strain γ (see Eq. 3.14) by the constitutive equation V 5 Sγ

(6.16)

in which 0

γ5θ1w

(6.17)

The shear stiffness coefficient S is defined and discussed in Section 3.5. To determine this coefficient, we average the shear strain γ xz as ðh ðh 1 1 τ xz dt γ5 γ dt 5 h xz h G 0

(6.18)

0

where G 5 Gxz is the shear modulus of the beam material in the xz-plane. Since the actual distribution of shear stress over the beam height, as mentioned earlier, does not affect the beam deformation, we can introduce the averaged shear stress τ a 5 V=bh which has the same resultant force V as the actual stress τ xz . Taking τ xz 5 τ a in Eq. (6.18), we arrive at Eq. (6.16) in which h2 S5 Ð h dt 0 bG

(6.19)

For a homogeneous rectangular cross section with b 5 constant and G 5 constant, we get S 5 Gbh

(6.20)

The internal axial force N, bending moment M and transverse shear force V must satisfy the equilibrium equations following from Eq. (6.1) and boundary conditions in Eq. (6.2). To derive these equations, consider the first equation of Eq. (6.1) and using the transformation defined by Eq. (6.8) present it in the form @ @σx ðbτ xz Þ 5 2 b @t @x

Integrating with respect to t from t 5 0 and taking into account the boundary conditions, Eq. (6.2), according to which τ xz ðt 5 0Þ 5 0, we get τ xz 5 2

ðt 1 @σx bdt b @x 0

6.1 BASIC EQUATIONS

381

Substituting the axial stress specified by Eq. (6.14) into this equation, we have 2 t 3 ð ðt 1 4N 0 M0 τ xz 5 2 Ebdt 1 Eðt 2 eÞbdt5 b B D 0

(6.21)

0

Now, take t 5 h and again use the boundary conditions, Eq. (6.2) (according to which τ xz ðt 5 hÞ 5 0), to get N0 M0 I0 1 ðI1 2 eI0 Þ 5 0 B D

where the integrals In ðn 5 0; 1; 2Þ are specified by Eq. (6.10). Applying Eq. (6.11), we can prove that the second term is zero, so that 0

N 50

(6.22)

This is the first equilibrium equation for the beam under study showing that if the beam is not loaded with any distributed axial forces, which is the case, the internal axial force does not change along the beam axis. Thus, Eq. (6.21) reduces to ðt M0 τ xz 5 2 Eðt 2 eÞbdt bD

(6.23)

0

Substitute this result into Eq. (6.15) for the transverse shear force to get V52

ðh ðt M0 dt Eðt 2 eÞbdt D 0

(6.24)

0

Calculating the integral by parts and using Eqs. (6.10), (6.11), and (6.23), we get in several steps ðh

ðt

ðh

ðh

dt Eðt 2 eÞbdt 5 h Ebtdt 2 0

0

0

0

ðh

ðh

0

0

Ebt2 dt 2 e@h Ebdt 2

0

1 EbtdtA

(6.25)

5 hI1 2 I2 2 eðhI0 2 I1 Þ 5 2 I2 1 eI1 1 hðI1 2 eI0 Þ 5 2 D

Thus, Eq. (6.22) yields V 5M

0

(6.26)

This is the moment equilibrium equation for the beam element. Using Eq. (6.26), we can present the following final form of Eq. (6.23) for the shear stress: τ xz 5 2

ðt V Eðt 2 eÞbdt bD 0

in which e is specified by Eq. (6.11).

(6.27)

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CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Consider the so-called shear correction factor discussed, for example, by Birman and Bert (2002). In its application to homogeneous beams with rectangular cross section, Eq. (6.20) for the beam shear stiffness is modified as S 5 kGbh

(6.28)

in which k is the shear correction factor. The concept of the shear correction factor (k 5 2/3 and k 5 0.889) was introduced by Timoshenko (1921, 1922). Reissner (1945) actually used k 5 5/6 as proposed by Goens (1931), whereas Mindlin (1951) proposed k 5 π2 =12. This problem is further discussed in Section 6.7.4. Recall that Eq. (6.19) for S is derived under the condition that the shear stress τ xz is averaged through the beam height. Now we have Eq. (6.27), which specifies the actual distribution for the shear stress. Substituting Eq. (6.27) into the second part of Eq. (6.18), we arrive at Eq. (6.16) in which S52

Ð h dt 0 bG

ðt

Dh

(6.29)

Eðt 2 eÞbdt

0

For a rectangular homogeneous beam, D 5 Ebh3 =12, e 5 h/2 and Eq. (8.29) yields S 5 Gbh. Hence, k 5 1 in Eq. (6.28). Thus, we have two different equations for the beam shear stiffness, that is, Eqs. (6.19) and (6.29). This contradiction is natural because the analysis is based on approximate equations, Eq. (6.3). For a homogeneous beam, both equations give the same result S 5 Gbh For laminated beams a question arises as to what equation should be used for practical analysis. The difference between Eqs. (6.19) and Eq. (6.29) is most pronounced for sandwich beams with a lightweight core. Consider the cross section shown in Fig. 6.3 for which b 5 h0 5 25 mm and δ 5 2:5 mm. The relevant parameters for aluminum facings are E 5 70 GPa, ν 5 0:3, and G 5 26.9 GPa, whereas for the core we take E0 5 140 MPa, ν 0 5 0, and G0 5 70 MPa. Thus, E=E0 5 500, G=G0 5 384, and h0 =δ 5 10. For the beam shown in Fig. 6.3, Eq. (6.19) gives S 5 63 kN, whereas Eq. (6.29) yields S 5 58:2 kN, that is, the difference is 8.2%. However, Eq. (6.19) is much simpler than Eq. (6.29) and for this reason it is recommended for practical analysis.

b δ

h0 δ FIGURE 6.3 Cross section of a sandwich beam.

6.1 BASIC EQUATIONS

383

Return to the equilibrium equations and consider the second equation of Eq. (6.1) which can be presented as @ @τ xz ðbσz Þ 5 2 b @x @t

Substituting τ xz from Eq. (6.24), integrating with respect to t from t 5 0, and taking into account the boundary conditions in Eq. (6.2) according to which σz ðt 5 0Þ 5 2 p, we arrive at 2 t t 3 ð ð 1 4V 0 σz 5 dt Eðt 2 eÞbdt 2 pb1 5 b D 0

(6.30)

0

in which b1 5 bðt 5 0Þ (see Fig. 6.1). Taking t 5 h and using the boundary conditions in Eq. (6.2) according to which σz ðt 5 hÞ 5 2 q, we get ðh ðt V0 dt Eðt 2 eÞbdt 5 pb1 2 qbk b 0

0

where bk 5 bðt 5 hÞ (see Fig. 6.1). Using Eq. (6.25) for the integral in the left-hand side of this equation, we arrive at the last equilibrium equation for the beam element, that is, 0

V 1 p 5 0;

p 5 pb1 2 qbk

(6.31)

This is the equilibrium equation for the beam element, providing the equilibrium in the z-direction (see Fig. 6.1). Now, Eq. (6.30) allows us to determine the transverse normal stress as 2 t t 3 ð ð 1 4p σz 5 2 dt Eðt 2 eÞbdt 1 pb1 5 b D 0

(6.32)

0

Thus, we have constructed the theory of composite beams which includes the equilibrium equations, Eqs. (6.22), (6.26), and (6.31), that allow us to determine the stress resultants N, V, and M, and the constitutive equations, Eqs. (6.12) and (6.16), (6.17) that can be used to find the displacements u, w, and the angle of rotation θ. These equations can be divided into two independent groups. The first group describes the axial loading and includes Eqs. (6.22) and (6.12), that is, 0

N 5 0;

N 5 Bu

0

(6.33)

This set of equations is of the second order and requires two boundary conditions (one for each end). At the fixed end u 5 0, whereas at the free end N 5 0. For the end x 5 l loaded with an axial force Nl , N 5 Nl . Note that the force must be applied to the neutral axis of the beam. Otherwise, the force must be moved to the neutral axis and the additional bending moment must be applied in the end cross section. The second group of equations, Eqs. (6.26) and (6.31) and Eqs. (6.12), (6.16), and (6.17), that is, 0

M 5 V; 0 M 5 Dθ ;

0

V 1p 50 0 V 5 Sðθ 1 w Þ

(6.34)

384

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

describes the beam bending. This set is of the fourth order and requires four boundary conditions (two for each end of the beam). At a clamped end w 5 0 and θ 5 0, at a hinged end w 5 0 and M 5 0, whereas for a free end V 5 0 and M 5 0. For the end x 5 l loaded with the transverse force Vl and the bending moment Ml , we have V 5 Vl and M 5 Ml .

6.2 STIFFNESS COEFFICIENTS The stiffness coefficients B, D, and S depend on the beam structure. For homogeneous or quasihomogeneous (consisting of identical layers) beams with rectangular cross sections (see Fig. 6.4), E 5 constant, G 5 constant, and b 5 constant and Eqs. (6.10), (6.11), and (6.20) yield I0 5 Ebh; h e5 ; 2

1 I1 5 Ebh2 ; 2

B 5 Ebh;

D5

1 I2 5 Ebh3 3

1 Ebh3 ; 12

S 5 Gbh

The axial and shear stresses specified by Eqs. (6.14) and (6.27) are σx 5

   1 12M h N1 2 t2 ; bh h 2

τ xz 5

6V tðh 2 tÞ bh3

For laminated beams (see Fig. 6.5), it follows from Eqs. (3.42) and (3.46) that I0 5

k X Ei bi hi ;

I1 5

i51

k 1X Ei bi hi ðti21 1 ti Þ 2 i51

k
2  1X Ei bi hi ti21 1 ti21 ti 1 ti2 3 i51 I2 h2 B 5 I0 ; D 5 I2 2 1 ; S5 P hi I0 k i51 Gi bi

I2 5 e5

I1 ; I0

t

h

b

FIGURE 6.4 Rectangular cross section.

(6.35)

6.2 STIFFNESS COEFFICIENTS

385

bk k

i tk = h t i−1

ti

2

t1

1 b1 FIGURE 6.5 Laminated cross section.

where hi 5 ti 2 ti21 is the thickness of the i-th layer. The axial stress in the i-th ply is σðiÞ x 5 Ei

  N M 1 ðzi 2 eÞ B D

(6.36)

where ti21 2 e # zi # ti 2 e. The shear stress acting between the i-th and the (i 1 1)-th layers is ði;i11Þ τ xz 52

i X
 V Ej bj hj tj21 1 tj 2 2e 2Dbi;i11 j51

(6.37)

where bi;i11 is the lower value of bi and bi11 . Composite tows can be used to reinforce metal profiles as shown in Fig. 6.6. In this case, Eq. (6.10) for coefficients In is generalized as ðh In 5

Ebtn dt 1

m X

Ej Aj tjn ðn 5 0; 1; 2Þ

j51

0

Here m is the number of composite tows, Ej and Aj are the modulus and the cross-sectional area of the j-th tow and tj is the tow coordinate (see Fig. 6.6). The axial stress in the composite tow is σðjÞ x 5 Ej

   N M
1 tj 2 e B D

Finally, consider a laminated beam whose layers are parallel to the plane of bending as shown in Fig. 6.7. The structure is symmetric with respect to the plane of bending (xz-plane). For such a beam, h e5 ; 2

B5h

k X

Ei bi ;

D5

i51

The axial stress is σðiÞ x 5 Ei



k h3 X Ei bi ; 12 i51

N M 1 z B D



S5h

k X i51

Gi bi

386

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

tj

FIGURE 6.6 Metal profiles reinforced with composite tows.

z

h h/ 2 b1 b 2

bi

bk

FIGURE 6.7 Laminated beam with layers parallel to the plane of bending.

Repeating the derivation of Eq. (6.27) for the shear stress, we arrive at τ ðiÞ xz

  V h2 2 2z 5 Ei 2D 4

6.3 BENDING OF LAMINATED BEAMS Consider the problem of bending described by Eq. (6.34). Integration of these equations with respect to x from x 5 0 yields

6.3 BENDING OF LAMINATED BEAMS

V 5 V0 2 Vp 2 VR M 5 M0 1 V0 x 2 Mp 2 MR M0 V0 2 x1 x 2 θp 2 θR θ 5 θ0 1 D 2D  1
M0 2 V0 3 x 2 x 1 wp 1 wR w 5 w 0 1 V 0 x 2 Mp 2 MR 2 θ 0 x 2 S 2D 6D

387

(6.38)

where V0 , M0 , θ0 , and w0 are the initial values of V, M, θ, and w corresponding to the cross section x 5 0. These values can be found from the boundary conditions at the beam ends x 5 0 and x 5 l (see Fig. 6.1). The load terms with subscript “p” correspond to the distributed loads and have the following form: ðx Vp 5

ðx Mp 5

pdx; 0

θp 5

Vp dx;

ðx 1 Mp dx; D

0

ðx wp 5

0

θp dx

(6.39)

0

For uniform pressure p 5 p 0 , Vp 5 p 0 x;

Mp 5

1 p x2 ; 2 0

θp 5

1 p x3 ; 6D 0

wp 5

1 p x4 24D 0

(6.40)

The load terms with subscript “R” in Eq. (6.38) correspond to the concentrated forces Rm and Fm shown in Fig. 6.1. These terms can be written with the aid of Eq. (6.39) if we present them in the form p 5 R m δðx 2 xm Þ where R m 5 Rm 2 Fm and δ is the delta function. Using the rules of integration of this function, we get from Eq. (6.39) VR 5

n X

VRðmÞ ;

MR 5

m51

n X

MRðmÞ ;

θR 5

m51

n X

θðmÞ R ;

wR 5

m51

n X

wðmÞ R

(6.41)

m51

where n is the number of the beam cross sections in which the forces act and for xm , x VRðmÞ 5 0;

MRðmÞ 5 0;

θRðmÞ 5 0;

wðmÞ R 50

(6.42)

and for x $ xm VRðmÞ 5 R m ;

MRðmÞ 5 R m ðx 2 xm Þ; wRðmÞ 5

θðRmÞ 5

Rm ðx2xm Þ3 6D

Rm ðx2xm Þ2 ; 2D

(6.43)

The solution given by Eq. (6.38) is universal and allows us to study both statically determinate and redundant beams using one and the same procedure. This solution can be applied also to multisupported beams. Introducing forces Fm as the support reactions at the support cross sections x 5 xm , we can find Fm using the conditions wðx 5 xm Þ 5 0. The second term with S in Eq. (6.38) for w accounts for the transverse shear deformation. As can be seen, the allowance for this deformation practically does not hinder the analysis of the beam. If the shear deformation is neglected, we must take S-N in Eq. (6.38) for w. As a result, we arrive at the solution corresponding to classical beam theory. To demonstrate the application of the general solution provided by Eq. (6.38), consider a beam similar to that supporting the passenger floor of an airplane fuselage shown in Fig. 6.8. Since the cross section x 5 0 is clamped, we must take w0 5 0 and θ0 5 0 in Eq. (6.38). The beam consists of

388

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

R

R

Q

x

(A)

x=c 0.5

x=l

V

(B)

M 0.05

(C)

w 0.001

(D)

FIGURE 6.8 Distributions of the normalized shear force V 5 V=Ql (B), bending moment M 5 M=Ql2 (C), and deflection w 5 Dw=Ql4 (D) along the x-axis, of the clamped beam (A).

two parts corresponding to 0 # x , c and c # x # l (see Fig. 6.8). For the first part, p 5 2 qbk (see Fig. 6.1), so we take p 5 2 Q, where Q 5 qbk . Then, Eqs. (6.39) and (6.42) yield Vpð1Þ 5 2 Qx;

Mpð1Þ 5 2

1 2 Qx ; 2

θpð1Þ 5 2

Q 3 x ; 6D

wð1Þ p 52

Q 4 x 24D

VR 5 0; MR 5 0; θR 5 0; wR 5 0

and the solution in Eq. (6.38) can be written as V1 5 V0 1 Qx 1 M1 5 M0 1 V0 x 1 Qx2 2 M0 V0 2 Q 3 x x1 x 1 θ1 5 6D D 2D   1 1 M0 2 V0 3 Q 4 x 2 x 2 w1 5 V0 x 1 Qx2 2 x 2D 6D S 2 24D

Consider the second part for which p 5 0. Then, the load terms in Eq. (6.39) become Vpð2Þ 5

ðx

ðc pdx 5 2

0

Qdx 5 2 Qc 0

(6.44)

6.3 BENDING OF LAMINATED BEAMS

Mpð2Þ 5

ðx

ðc Vp dx 5

0

Vpð1Þ dx 1

ðx

389

1 Vpð2Þ dx 5 2 Qcð2x 2 cÞ 2

c

0

0c 1 ð ðx 1 1@ Qc 2 ð2Þ ð1Þ ð2Þ A ðc 1 3x2 2 3cxÞ θp 5 Mp dx 5 Mp dx 1 Mp dx 5 2 D D 6D ðx 0

wð2Þ p 5

0

ðx

ðc θp dx 5

0

θð1Þ p dx 1

0

0

ðx

θpð2Þ dx 5 2

 Qc
3 4x 2 c3 1 4xc2 2 6x2 c 24D

c

The support reaction R (see Fig. 6.8) is treated as the unknown concentrated force. Then, Eq. (6.43) yield R R ðx2cÞ2 ; wR 5 ðx2cÞ3 VR 5 R; MR 5 Rðx 2 cÞ; θR 5 2D 6D Finally, we get for the second part of the beam V2 5 V0 1 Qc 2 R 1 M2 5 M0 1 V0 x 1 Qcð2x 2 cÞ 2 Rðx 2 cÞ 2  M0 V0 2 Qc
2 R x1 x 1 c 1 3x2 2 3cx 2 ðx2cÞ2 θ2 5 D 2D 6D 2D   1 1 M0 2 V0 3 x 2 x w2 5 V0 x 1 Qcð2x 2 cÞ 2 Rðx 2 cÞ 2 2D 6D S 2  Qc
3 R 4x 2 c3 1 4xc2 2 6x2 c 1 ðx2cÞ3 2 24D 6D

(6.45)

The solution obtained, Eqs. (6.44) and (6.45), includes three unknown parameters: V0 , M0 , and R, which can be found from two symmetry conditions, that is, V2 ðx 5 lÞ 5 0, θ2 ðx 5 lÞ 5 0, and the condition for the reaction force w1 ðx 5 cÞ 5 w2 ðx 5 cÞ 5 0. The result is as follows V0 5 R 2 Qc;

1 1 M0 5 Qc2 ð1 2 4ks Þ 2 Rcð1 2 6ks Þ 4 3 1 3lð1 1 4ks Þ 2 2c R 5 Qc 2 4lð1 1 3ks 2 3c

(6.46)

where ks 5 D=Sc2 . For numerical analysis, neglect the shear deformation by taking ks 5 0. Then, the solution in Eq. (6.46) reduces to V0 5 2

Qcð5l 2 4cÞ ; 2ð4l 2 3cÞ

M0 5

Qc2 ð6l 2 5cÞ ; 12ð4l 2 3cÞ

R5

Qcð3l 2 2cÞ 2ð4l 2 3cÞ

The dependencies of the normalized shear force, the bending moment, and the beam deflection on the axial coordinate are presented in Fig. 6.8. Beam deflections are often used as approximation functions in the solutions of plate bending problems (see Chapter 7: Laminated Composite Plates). Solutions of typical beam problems are presented in Table 6.1. As an example, consider a simply supported I-beam loaded with uniform pressure q (see Fig. 6.9). The beam is made of an aluminum alloy for which E 5 70 GPa and G 5 26:9 GPa. At

390

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Table 6.1 Solutions for the Beams Loaded With Uniform Pressure for Typical Boundary Conditions Case

Beam Type

Solution q

1

bk

w x l

q

2

bk w x l

q

3

bk

w x l

V 5 2 qbk ðl 2 xÞ 1 M 5 qbk ðl2xÞ2 2 qbk 2 ð3l 2 3lx 1 x2 Þx θ5 6D qbk  3 x 2 4lx2 1 6l2 x w52 24D D 1 12 ð2l 2 xÞx S 1 V 5 qbk ð2x 2 lÞ 2 1 M 5 2 qbk ðl 2 xÞx 2 qbk 3 ðl 2 6lx2 1 4x3 Þ θ5 24D qbk  3 l 2 2lx2 1 x3 w52 24D D 1 12 ðl 2 xÞx S V 5 qbk ðx 2 2l Þ 1 qbk ð6x3 2 6lx 1 l2 Þx M5 12 qbk ð2x2 2 3lx 1 l2 Þx θ5 12D   qbk D xðl 2 xÞ 1 12 ðl 2 xÞx w52 S 24D

q

q

δ δ

h0

w l

x FIGURE 6.9 A simply supported I-beam.

b

δ δ

6.3 BENDING OF LAMINATED BEAMS

391

b4 = b 4

b3 = δ 3

t3 e

2 1

t0 = 0

t1

t4 = h

t2

b1 = b2 = b

FIGURE 6.10 Coordinates of the layers.

the bottom, where the maximum tensile stress acts, the beam is reinforced with a unidirectional carbon-epoxy layer with modulus of elasticity Ec 5 140 GPa and shear modulus Gc 5 3:5 GPa. The beam dimensions are l 5 1250 mm; δ 5 2:5 mm; h0 5 100 mm; b 5 50 mm

(6.47)

The layer coordinates are shown in Fig. 6.10. The beam is composed of four layers with the following parameters: b1 5 50 mm; t0 5 0; t1 5 2:5 mm; h1 5 2:5 mm; E1 5 140 GPa; G1 5 3:5 GPa ðlayer1Þ b2 5 50 mm; t2 5 5 mm; h2 5 2:5 mm; E2 5 70 GPa; G2 5 26:9 GPa ðlayer2Þ b3 5 2:5 mm; t3 5 105 mm; h3 5 100 mm; E3 5 70 GPa; G3 5 26:9 GPa ðlayer3Þ b4 5 50 mm; t4 5 h 5 107:5 mm; h4 5 2:5 mm; E4 5 70 GPa; G4 5 26:9 GPa ðlayer4Þ

(6.48)

The beam strength is analyzed in accordance with the following procedure. 1. Determine the maximum shear force, bending moment, and deflection. The beam under study corresponds to Case 2 in Table 6.1 from which it follows that 1 Vm 5 V ðx 5 0Þ 5 2 qbl 2   l 1 5 2 qbl2 Mm 5 M x 5 2 8   l 5qbl4 48D 52 ð1 1 αÞ; α 5 Wm 5 w x 5 2 5Sl2 384D

(6.49)

2. Determine the stiffness coefficients. First, calculate the I-coefficients specified by Eq. (6.35), that is, I0 5

4 X

Ei bi hi 5 5:25  104 GPa  mm2

i51

I1 5

4 1X

2

i51

Ei bi hi ðti21 1 ti Þ 5 1:95  106 GPa  mm3

392

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

I2 5

4 1X 2 Ei bi hi ðti21 1 ti21 ti 1 ti2 Þ 5 1:66  108 GPa  mm4 3 i51

The coordinate of the neutral axis can be found from Eq. (6.11), which yields e5

I1 5 37:14 mm I0

The bending stiffness of the beam is calculated according to Eq. (6.13) as D 5 I2 2

I12 5 0:936  108 GPa  mm4 I0

For a beam without the composite layer D 5 0:573  108 GPa  mm4 , that is, the composite layer increases the beam bending stiffness by 63%. The transverse shear stiffness of the beam is specified by Eq. (6.35), which give S5 P 4

h2

hi i51 Gi bi

5 7:68  103 GPa  mm2

3. Calculate the axial stress using Eqs. (6.36) and (6.49), according to which σðiÞ x 5 Ei

Mm qbl2 h ðt 2 eÞ ðt 2 eÞ 5 2 Ei 8D D

e where t 5 t=h, e 5 ; and ti21 # t # ti . For the beam with dimensions as per Eqs. (6.47) and h (6.48), the maximum tensile stress in the composite layer corresponds to t 5 0 and is equal to ð2Þ σð1Þ x 5 542q. The maximum tensile stress in the metal part is σ x ðt 5 t1 Þ 5 253q. The maximum compressive stress in the metal part is σðx4Þ ðt 5 hÞ 5 2 514q. 4. Calculate the shear stress. The most dangerous is the shear stress which acts between composite layer 1 and metal layer 2 and can cause delamination. This stress is specified by Eq. (6.37) which yields ð1;2Þ τ xz 52

Vm ql E1 bh1 ðh1 2 2eÞ E1 b1 h1 ðt1 2 2eÞ 5 4D 2Db

ð1;2Þ For the beam under study, τ xz 5 2 4:2q. 5. Calculate the maximum deflection. The deflection is specified by the third equation of Eq. (6.49) in which α 5 0:075. Thus, the allowance for transverse shear deformation increases the maximum deflection by 7.5%.

6.4 NONLINEAR BENDING To demonstrate the concept of nonlinear bending, consider Fig. 6.11. A conventional simply supported beam is shown in Fig. 6.11A. As can be seen, the beam deflection (dashed line) causes the displacements of the beam supports toward each other. The deflection is small and the curve slope is also so small that it can be neglected. In this case, the beam equations can be written for the

6.4 NONLINEAR BENDING

393

q

(A)

z x=−

q

l 2

x=

l 2 x

(B)

FIGURE 6.11 Bending (A) and bending-stretching (B) of the beam.

N0 dx dw N0

ds

α2

α1

FIGURE 6.12 A curved element of the beam axis.

initial rectilinear shape of the beam resulting in Eqs. (6.33) and (6.34). For the beam in Fig. 6.11B, in contrast to the beam in Fig. 6.11A, the ends do not move under bending. The deflection of this beam is also small (it is even smaller than the deflection of the beam shown in Fig. 6.11A), however, the obvious difference between the length of the dashed curve and the initial length of the beam allows us to conclude that the beam axis experiences tension under bending. The slope of the beam axis shown in Fig. 6.11B is small, so that the first equation of Eq. (6.33) is still valid and shows that the internal axial force N which appears as a result of the beam bending does not depend on the axial coordinate, so that N 5 N0 5 constant. To determine this force, consider a curved element of the beam axis shown in Fig. 6.12 from which it follows that ds 5

  pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi2 1 dx2 1 dw2 5 1 1 ðw0 Þ dxD 1 1 ðw0 Þ2 dx 2

0

The last part of this equation can be obtained if we take into account that w 5 dw=dx is much smaller than unity. So, the elongation of the beam axis due to bending is ε5

ds 2 dx 1 0 2 5 ðw Þ dx 2

394

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

pn dx FIGURE 6.13 Imaginary pressure simulating the action of forces N0 .

This elongation should be added to u0 in Eq. (6.33), and the final expression for the axial force becomes   1 0 N0 5 B u 1 ðw0 Þ2 2

(6.50)

0

Solving this equation for u and integrating the resulting equation with respect to x from x 5 2 l=2 (see Fig. 6.11B), the following expression for uðxÞ is derived: 1 1 u 5 N0 x 2 B 2

ðx

ðw0 Þ2 dx 1 c

2l=2

where c is the constant of integration. Since uðx 5 2 l=2Þ 5 0 (see Fig. 6.11B), c 5 0. However, uðx 5 l=2Þ is also zero, and we arrive at the following equation for N0 : B N0 5 2

l=2 ð

ðw0 Þ2 dx

(6.51)

2l=2

As can be seen, the dependence of N0 on w is not linear. The axial force N0 also affects the equilibrium equation, particularly the second equation of Eq. (6.34) presenting the projection of the forces on the axis normal to the beam axis. The effect of N0 can be simulated if we introduce some imaginary pressure shown in Fig. 6.13. Static equivalence of the forces in Figs. 6.12 and 6.13 gives Pn dx 5 2 N0 α1 1 N0 α2 0

Here, α1 and α2 are small angles, such that α1  w0 and α2  w 1 w00 dx. Then, Pn 5 N0 w00 , and the second equation of Eq. (6.34) can be generalized as 0

00

V 1 N0 w 1 p 5 0

(6.52)

where p is the acting transverse load. The rest of Eq. (6.34) remain the same, that is,
0 0 0 M 5 V; M 5 Dθ ; V 5 S θ 1 w

(6.53)

Reduce the set of equations, Eqs. (6.52) and (6.53), to one equation for the beam deflection. For this purpose, first, substitute V from the first equation of Eq. (6.53) into Eq. (6.52) to get 00

00

M 1 N0 w 1 p 5 0

Second, eliminate θ from the last two equations of Eq. (6.53), that is, M5D

 0  V 2 w00 S

(6.54)

6.4 NONLINEAR BENDING

395

Third, substitute V 0 in this equation with its expression from Eq. (6.52) and determine 00

M 5 2 Dð1 1 λÞw 2

D p S

(6.55)

where λ 5 N0 =S. Substituting Eq. (6.55) into Eq. (6.54), we finally arrive at 00

wIV 2 k2 w 5

  1 D N0 p 2 p 00 ; k2 5 Dð1 1 λÞ S Dð1 1 λÞ

(6.56)

Consider the solution of this equation for the beam shown in Fig. 6.11B for which, as earlier, p 5 2 qbk 5 2 Q (see Fig. 6.1). The solution of Eq. (6.56) is symmetric with respect to x 5 0 and has the following form: w 5 C1 1 C2 coshkx 1

Q 2 x 2N0

The bending moment is specified by Eq. (6.55), which yields

  Q M 5 2 D k2 C1 ð1 1 λÞcoshkx 1 N0

The constants of integration C1 and C2 can be found from the boundary conditions wðx 5 6 l=2Þ 5 0 and Mðx 5 6 l=2Þ 5 0. The resulting solution is w5

      Q D coshkx x2 l2 QD coshkx 1 21 2 ; M5 12 N0 N0 coshη N0 coshη 2 8

(6.57)

in which η 5 kl=2. To find N0 , apply Eq. (6.51). Substitution of the first equation of Eq. (6.57) into Eq. (6.51) gives the following transcendental equation: "    # N0 l Q2 l3 2D l 1 D2 k 2 1 l 2 cosh η 2 sinhη 1 sinh η cosh η 2 50 2 2B k 2 2N02 24 N0 cosh η 2 2N02 ðcoshηÞ2 k

(6.58)

Recall that in accordance with the foregoing derivation

vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi lu N0 u  η5 u  N0 t 2 D 11 S

To simplify the solution, neglect the shear deformation taking S-N. Then, Eq. (6.58) can be transformed into the following dimensionless form:      q2 1 1 1 1 1 2 tanh η 1 sinhη cosh η 2 1 51 1 2 3 η 4N ðcosh ηÞ2 η N 24 N Ql4 in which q 5 D

rffiffiffiffi B ; D

N5

N0 l2 ; D

η5

1 pffiffiffiffi N 2

The function N ðq) is presented in graphical form by Reismann (1988).

396

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

For the beam shown in Fig. 6.11B and having square ðb 5 hÞ uniform cross section, the dependencies of the normalized maximum deflection wm 5 wðx 5 0Þ on the loading parameter q are shown in Fig. 6.14. As can be seen, for relatively low loads (q # 60Þ the nonlinear solution (curve 2) practically coincides with the linear one. The normalized maximum stress σm 5

σm l2 ; Eh2

σm 5 E

  N0 Mm h 6 B D 2

is presented in Fig. 6.15. As can be seen, the loading parameter q, for which the linear solution is valid, is much smaller for the maximum compressive stress than for the maximum tensile stress. It should be taken into account that the ultimate value of the loading parameter is limited by material

wm /h 1.0

1

0.8

2

0.6 0.4 0.2

q 0

200

400

600

800

1000

FIGURE 6.14 Dependencies of the normalized maximum deflection on the loading parameter corresponding to the linear (1) and nonlinear (2) solutions.

σm 6

1 4

2

2

q 0

200

–2 –4

1

400

600

800

1000

2

FIGURE 6.15 Dependencies of the normalized maximum stress on the loading parameter corresponding to the linear (1) and nonlinear (2) solutions.

6.4 NONLINEAR BENDING

397

z,w l /2

l/ 2

P

h

A

c

P A

x

e

FIGURE 6.16 Eccentric tension of a laminated beam.

strength. For example, for an aluminum beam with l/h 5 10, the material yields at q 5 10 for which the linear solution is valid. However, for composite beams, the situation can be different. For example, for a unidirectional glass-epoxy composite that has relatively low modulus and high strength (see Table 1.5), the ultimate value of q can reach 250, which is far beyond the limit for the linear solution. In conclusion, consider another nonlinear problem. Suppose that the laminated beam in Fig. 6.16 is loaded with two tensile forces P applied at the beam ends x 5 6 l=2. If the forces are applied to the beam neutral axis, that is, at points A with coordinate e, then no bending is observed. A natural question arises as to what happens if the forces are applied at some points with coordinate c which is not equal to e. The problem of a beam subjected to eccentric tension (for a homogeneous rectangular beam e 5 h=2 and a force which does not induce any beam bending passes through the center of the cross section) can be studied using Eq. (6.56) in which p 5 0 and N0 5 P, that is, 00

wIV 2 k2 w 5 0;

k2 5

P ; Dð1 1 λÞ

λ5

P S

(5.59)

The solution of this equation which satisfies the boundary conditions wðx 5 6 l=2Þ 5 0 is w 5 2 Cðcoshη 2 coshkxÞ

(6.60)

where η 5 kl=2. The rotation angle and the bending moment can be found with the aid of Eqs. (6.53) and (6.55), that is, θ 5 2 kC ð1 1 λÞsinh kx; M 5 2 Dk2 Cð1 1 λÞcoshkx

(6.61)

Assume that the beam in Fig. 6.16 is clamped, so that θðx 5 6 l=2Þ 5 0. Then, the first equation of Eq. (6.61) yields C 5 0 and w 5 0. Thus, bending of a clamped beam does not occur irrespective of the location of the force. Now assume that the beam ends are hinged and can rotate around points A (see Fig. 6.16) under a bending moment M0 5 Pðc 2 eÞ. Taking Mðx 5 6 l=2Þ 5 M0 in the second equation of Eq. (6.61), we can present Eq. (6.60) as   coshkx w 5 ðC 2 eÞ 1 2 coshη

(6.62)

As an example, consider a two-layer beam composed of a unidirectional glass-epoxy and carbon-epoxy layers with the following parameters: h1 5 1 mm, E1 5 60 GPa, h2 5 1 mm, and E2 5 140 GPa. The beam length and width are l 5 250 mm; b 5 25 mm. Calculate the I-coefficients in Eq. (6.35), that is, I0 5 5  103 GPa  mm2 ;

I1 5 6  103 GPa  mm3 ;

I2 5 8:67  103 GPa  mm4

398

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

− w/h 0.3

α =1 0.2

0 .1

0.1

0 .01

x/l

0 0

0.1

0.2

0.3

0.4

0.5

FIGURE 6.17 Deflection of the beam under eccentric tension.

Then, the beam stiffness in Eq. (6.13), and the coordinate of the neutral axis in Eq. (6.11) become B 5 5  103 GPa  mm2 ;

D 5 4:5  103 GPa  mm4 ;

e 5 1:6 mm

The axial strain of the beam under tension is specified by Eq. (6.12), that is, 0

ε5u 5

P B

(6.63)

The maximum strains of the carbon-epoxy and glass-epoxy composites are (see Table 1.5) εc 5 1:43% and εg 5 3%. To find the ultimate force P, we should substitute ε5εc into Eq. (6.62) to get P 5 71:5 kN. Introduce the ratio α 5 P=P and dimensionless coordinate x 5 x=l (see Fig. 6.16). pffiffiffi ffi Then, η 5 15:71 α and in accordance with Eq. (6.62) w5

 pffiffiffiffi  w cosh31:42 αx 5 ðc 2 e Þ 1 2 h cosh15:71

Let the force P be applied at the layers’ interface, so that c 5 h1 , c 5 0:5, and e 5 e=h 5 0:8. The functions wðxÞ are presented in Fig. 6.17 for various values of the loading parameter. As can be seen, under a force equal to the ultimate value ðα 5 1Þ, bending is concentrated in the vicinity of the beam ends.

6.5 BUCKLING OF COMPOSITE COLUMNS Consider a column compressed by an axial force T as shown in Fig. 6.18. For a particular value of the force T, which is called the critical value, the initial rectilinear shape becomes unstable, and the column axis becomes curved as shown in Fig. 6.18 with dashed lines. According to the static buckling criterion, the critical force Tc is the minimum force under which two equilibrium states of the column are possible, that is, the initial rectilinear shape and the slightly curved shape. To apply this criterion, we need to induce some deflection (see the dashed lines in Fig. 6.18) and to determine

6.5 BUCKLING OF COMPOSITE COLUMNS

T

399

T

T

l

w

x

(A)

(B)

(C)

FIGURE 6.18 Axial compression and buckling modes of a column with simply supported ends (A), with clamped end and free end (B), and with clamped ends (C).

whether this deflection can be supported by the acting force T. Note that this small deflection is artificially induced because the column is analyzed assuming that it is “perfect” (i.e., it is perfectly straight, no transverse forces acting, etc.). In real structures, the tendency to bending always exists. For example, a small eccentricity of the load application (see Fig. 6.16) causes bending from the very beginning of the loading process (see Fig. 6.17). The equilibrium of a compressed column with some small deflection w can be described by Eq. (6.59) in which P 5 2 T, that is, 00

wIV 1 k2 w 5 0;

k2 5

T ; Dð1 2 λÞ

λ5

T S

(6.64)

Note that this equation has a trivial solution w 5 0 which corresponds to the initial rectilinear shape of the column. We need to find out whether the critical force Tc exists, for which Eq. (6.64) has a nonzero solution. The general solution of Eq. (6.64) is w 5 C1 x 1 C2 1 C3 sin kx 1 C4 cos kx

(6.65)

Applying Eqs. (6.53) and (6.55), we can also find θ 5 2 kð1 2 λÞðC3 cos kx 2 C4 sin kxÞ 2 C1 M 5 Dk2 ð1 2 λÞðC3 sin kx 1 C4 cos kxÞ

(6.66)

V 5 Dk3 ð1 2 λÞðC3 cos kx 2 C4 sin kxÞ

Consider the simply supported column shown in Fig. 6.18A for which wðx 5 0Þ 5 wðx 5 lÞ 5 0;

M ðx 5 0Þ 5 M ðx 5 lÞ 5 0

(6.67)

For these boundary conditions, Eqs. (6.65) and (6.66) give the following algebraic equations written in terms of the constants of integration Ci ði 5 1; . . . ; 4Þ: C2 1 C4 5 0; C4 5 0;

C1 l 1 C2 1 C3 sinkl 1 C4 coskl 5 0 C3 sinkl 1 C4 coskl 5 0

400

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

The last two equations yield C3 sinkl 5 0

Obviously, if C3 5 0, then all the constants are zero and w 5 0, which corresponds to the initial shape of the column. So, C3 6¼ 0 and sinkl 5 0

(6.68)

As can be readily seen, C3 is the only one nonzero constant and Eq. (6.65) reduces to w 5 C3 sinkx

(6.69)

Note that within the approach under discussion the constant C3 cannot be determined. Recall that Eq. (6.64) is valid for small deflections only. So, for the unknown C3 , we can formally assume that it is small enough so that the deflection, Eq. (6.69), is also small. The nonzero deflection given by Eq. (6.69) exists if the parameter k satisfies Eq. (6.68) from which kl 5 nπ ðn 5 1; 2; 3; . . .Þ. Using the corresponding Eq. (6.64) for k, we get the values of forces Tn 5

π2 n2 D   ðn 5 1; 2; 3; . . .Þ π2 n2 D l2 1 1 Sl2

(6.70)

for which the nonzero deflection of the column can take place. Only one of these values, that is, the smallest one, has a physical meaning and represents the critical force. If we neglect the shear deformation taking S-N, the result is evident: the minimum value of Tn takes place when n 5 1, that is, TE 5

π2 D l2

(6.71)

This is a well known Euler formula for the critical load. In the general case, the situation looks more complicated. However, it can be proved that n 5 1 in Eq. (6.70) as well. Indeed, using Eq. (6.71), that equation can be transformed as follows Tn 5

n2 TE TE 1 1 n2 S

Consider the inequality Tn . T1 for n . 1. This inequality reduces to n . 1 which is true. So, in the general case, n 5 1 and the critical force is determined by Tc 5

TE TE 11 S

(6.72)

For n 5 1, Eq. (6.69) specifies the mode of buckling w 5 C3 sin

πx l

shown in Fig. 6.18A with the dashed line. In conclusion, consider the case λ 5 1, or T 5 S which is sometimes associated with the so00 called shear mode of buckling. In this case, Eq. (6.64) reduces to w 5 0 whose solution is

6.6 FREE VIBRATIONS OF COMPOSITE BEAMS

401

w 5 C1 x 1 C2 . Applying the boundary conditions in Eq. (6.67), we get C1 5 C2 5 0 and w 5 0. Thus, the case λ 5 1 corresponds to the straight column and T 5 S is not a critical load. Anyway, the force T 5 S is always higher than Tc given by Eq. (6.72). Indeed, the inequality T 5S . Tc is equivalent to the inequality S . 0. The result obtained for the critical force, that is, Eq. (6.72), can be used for columns with various boundary conditions substituting the corresponding expression for TE instead of that given by Eq. (6.71). Particularly for the cantilever (see Fig. 6.18B) and clamped (see Fig. 6.18C) columns, we have, respectively, TE 5

π2 D ; 4l2

TE 5

4π2 D l2

It should be noted that a buckling analysis must be accompanied by a study of stresses because composite beams can fail before buckling if the stresses in the layers reach the material ultimate compressive stresses. The stresses in the layers are specified by Eq. (6.36) in which N 5 2 T and M 5 0, that is, σðiÞ x 5 2 Ei

T B

Buckling can take place if these stresses calculated for T 5Tc are less than material strength.

6.6 FREE VIBRATIONS OF COMPOSITE BEAMS Composite beams are good models allowing us to discuss some specific features of the dynamic behavior of laminated composite structures. The equations of motion of the beam element shown in Fig. 6.2 can be written adding the inertia terms to the equilibrium equations, Eq. (6.1), that is, b

@σx @ @2 ux 1 ðbτ xz Þ 2 ρb 2 5 0 @z @x @t

b

@τ xz @ @2 uz 1 ðbσz Þ 2 ρb 2 5 0 @x @t @z

in which t is the time and ρ is the material density. Applying the assumptions introduced in Section 6.1, that is, using Eq. (6.3) for the displacements and Eqs. (6.5) and (6.15) for forces and moment, we can arrive at the following equations for composite beams [these equations are generalizations of the equilibrium equations, Eqs. (6.22), (6.26), and (6.31)]: @N @2 u @2 θ 2 Bρ 2 2 Cρ 2 5 0 @x @t @t @M @2 u @2 θ 2 V 2 Cρ 2 2 Dρ 2 5 0 @x @t @t @V @2 w 2 Bρ 2 2 pðx; tÞ 5 0 @x @t

(6.73)

402

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

where Bρ 5 J0 ;


 Cρ 5 J0 eρ 2 e ;

Dρ 5 J2 2 2eJ1 1 e2 J0 ðh Jn 5 bρtn dt

ðn 5 0; 1; 2Þ

(6.74)

0

are the inertia coefficients which are analogous to the stiffness coefficients in Eq. (6.9). For laminated beams, Jn 5

k
 1 X n11 bi ρi tin11 2 ti21 n 1 1 i51

(6.75)

(see Fig. 6.5). The parameter eρ in Eq. (6.74) eρ 5

J1 J0

(6.76)

is similar to the coordinate of the neutral axis e defined by Eq. (6.11). In particular, Bρ is the inertia term corresponding to translational displacement of the beam element in Fig. 6.2, Dρ corresponds to the element rotation, and Cρ is the coupling term linking transverse and rotational motions of the beam element. In the general case, eρ does not coincide with e because the integrals In [see Eq. (6.10)] depend on the stiffness distribution, whereas the coefficients Jn determined by Eq. (6.75) are governed by the density distribution through the height of the beam cross section. However, there are some particular cases for which eρ 5 e. For homogeneous beams (E 5 constant and ρ 5 constant), we have Ðh eρ 5 e 5

0 Ðh

btdt ;

C50

bdt

0

For beams with a symmetric laminated structure, h eρ 5 e 5 ; 2

C50

(6.77)

The forces and moments in Eq. (6.73) are linked with the corresponding deformations by the constitutive equations, Eqs. (6.12), (6.16), and (6.17), that is, 0

N 5 Bu ;

0

M 5 Dθ ;


0 V 5S θ1w

(6.78)

Substituting Eq. (6.78) into Eq. (6.73), we can derive the motion equations in terms of displacements. Taking p 5 0 for the case of free vibrations, we get @2 u @2 u @2 θ 2 Bρ 2 2 Cρ 2 5 0 2 @x @t @t   @2 θ @w @2 u @2 θ D 2 2S θ1 2 Cρ 2 2 Dρ 2 5 0 @x @x @t @t   2 2 @θ @ w @ w S 1 2 2 Bρ 2 5 0 @x @x @t B

(6.79) (6.80) (6.81)

6.6 FREE VIBRATIONS OF COMPOSITE BEAMS

403

In the general case,eρ 6¼ e and Cρ 6¼ 0, so that Eq. (6.79) which describes the axial vibrations includes the rotation angle θ and is coupled through this term with Eqs. (6.80) and (6.81) which describe the flexural vibrations of the beam. Consider here the flexural vibrations. For this purpose, assume that the beam has a symmetric structure, so that Eq. (6.77) hold. Then, Eqs. (6.80) and (6.81) can be transformed to the following equation written in terms of the beam deflection: D

  Bρ D @4 w Bρ Dρ @4 w @4 w @2 w 2 D 1 1 1 Bρ 2 5 0 ρ 4 2 2 4 @x @x @t @t S @t S

(6.82)

In this equation, the terms including S allow us to take into account the shear deformation, whereas the terms with coefficient Dρ allow for the rotation inertia of the beam cross section. Following Uflyand (1948), Eq. (6.82) can be presented in the form   4 2 @4 w 1 1 @ w 1 @4 w 2@ w 2 1 1 1 C 50 @x4 @t2 C12 C22 @x2 @t2 C12 C22 @t4

or



@2 1 @2 2 2 2 2 @x C1 @t

where



 @2 w 1 @2 w @2 w 1 C2 2 5 0 2 2 2 2 @x @t C2 @t sffiffiffiffiffi S C2 5 ; Bρ

sffiffiffiffiffiffi D ; C1 5 Dρ

rffiffiffiffiffi Bρ C5 B

To clarify the physical meaning of the coefficients C1 and C2 , consider a beam with rectangular cross section (see Fig. 6.19) for which D5

1 1 Ebh3 ; S 5 Gbh; Bρ 5 bhρ; Dρ 5 ρbh3 12 12

Then,

sffiffiffiffi E ; C1 5 ρ

(6.83)

sffiffiffiffi G C2 5 ρ

are the velocities of normal stress (C1 ) and shear stress (C2 ) wave propagation along the beam axis. Neglecting rotatory inertia ðDρ 5 0Þ or shear deformation ðS-NÞ in Eq. (6.82), we actually assume that C1 or C2 become infinitely high. w x h l b FIGURE 6.19 Free vibrations of a simply supported beam.

404

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

For qualitative analysis, consider flexural vibrations of a simply supported beam shown in Fig. 6.19 in which case the beam deflection can be presented as wðx; tÞ 5 Wm sin

πmx sinωm t l

(6.84)

where m is the number of the vibration mode (the mode corresponding to m 5 1 is shown in Fig. 6.19 with a dashed line), Wm is the amplitude, and ωm is the frequency of the m-th vibration mode. Recall that the number of the beam vibrations per 1 second (Hz) can be found as νm 5

ωm 2π

Substituting the deflection, Eq. (6.84), into Eq. (6.82) results in the following equation for the frequency:  

πm2 

πm2 Bρ Dρ 4 DBρ Dρ 1 50 ωm 2 Bρ 1 ω2m 1 D l l S S

(6.85)

Neglecting both transverse deformation ðS-NÞ and rotatory inertia ðDρ 5 0Þ, we arrive at a result corresponding to classical beam theory, that is, πm l

ω0m 5

sffiffiffiffiffi D Bρ

(6.86)

If we neglect the rotatory inertia ðDρ 5 0Þ and take into account the shear deformation, Eq. (6.85) yields ω0m ωm 5 pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 αS

where ω0m is specified by Eq. (6.86) and αS 5

π2 m2 D l2 S

allows for shear deformation. For a homogeneous beam with a rectangular cross section, using Eq. (6.83), we get αS 5

π2 m2 Eh2 12Gl2

Thus, the shear deformation reduces the frequency and the effect depends on the ratios E/G, h/l, and the mode number m. Consider now the effect of rotatory inertia. Taking S-N in Eq. (6.85), we have ω0m ωm 5 pffiffiffiffiffiffiffiffiffiffiffiffiffi 1 1 αr

where αr 5

π2 m2 Dρ l2 Bρ

6.6 FREE VIBRATIONS OF COMPOSITE BEAMS

405

Again, for a homogeneous beam with a rectangular cross section αr 5

π2 m2 h2 12l2

(see Fig. 6.19). As can be seen, the effect depends on the beam relative thickness and mode number. Consider coupled longitudinal-transverse vibrations of the beams (for which Cρ 6¼ 0Þ described by Eqs. (6.79)(6.81). Taking, for the sake of brevity, Dρ 5 0 and S-N, we can transform these equations into the following form: BD

@6 w @6 w @6 w @4 w @4 w 2 Bρ D 4 2 2 Cρ2 2 4 1 BBρ 2 2 2 B2ρ 4 5 0 6 @x @x @t @x @t @x @t @t

Substituting the deflection from Eq. (6.84), we arrive at the following equation for the frequency:



 Bλ2m 2 Bρ ω2m Dλ4m 2 Bρ ω2m 2 Cρ2 λ2m ω4m 5 0

in which λm 5 πm=l. Taking into account Eq. (6.74) for Cρ , we can present this equation in the form h
2 i
 ω4m B2ρ 1 2 λ2m eρ 2e 2 Bρ λ2m ω2m Dλ2m 1 B 1 BDλ6m 5 0

It follows from this equation that the effect of vibration coupling can be evaluated by comparing the parameter αc 5

2 π2 m2
e2eρ 2 l

with unity. If αc {1, the coupling effect can be ignored. Finally, consider flexural vibrations of beams loaded with axial compressive force T (see Fig. 6.18). For this purpose, we need to modify Eq. (6.81), adding to it the term with the axial force as in Eq. (6.52), that is, present it as   @θ @2 w @2 w @2 w 1 2 2 T 2 2 Bρ 2 5 0 S @x @x @x @t

(6.87)

Taking, for the sake of brevity, Dρ 5 0 and Cρ 5 0 in Eq. (6.80), we get the second equation D

  @2 θ @w 50 2 S θ 1 @x2 @x

(6.88)

These two equations, Eqs. (6.87) and (6.88) can be reduced to the following equation:   T @4 w DBρ @4 w @2 w @2 w D 12 2 1 T 2 1 Bρ 2 5 0 S @x4 @x @t S @x2 @t2

For a simply supported beam (see Fig. 6.18A), substituting the deflection in accordance with Eq. (6.84), we finally get vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

 u u1 2 T 1 1 αm u αm S u ωm 5 ω0m t αm 11 S

(6.89)

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CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

where αm 5 π2 m2 D=l2 . As can be seen, a compressive force reduces the frequency of flexural vibrations. For the first mode (m 5 1), we have α1 5 TE , where TE is the Euler critical force specified by Eq. (6.71), and Eq. (6.89) becomes vffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   u T u u1 2 TT 1 1 E E u S ω1 5 ω01 u t TE 11 S

For the critical force T 5 Tc in Eq. (6.72), ω1 5 0, that is, compressive loading up to the critical load results in zero frequency of flexural vibrations. A tensile force, naturally, increases the frequency.

6.7 REFINED THEORIES OF BEAMS AND PLATES The composite beam, being a simple but representative model, allows us to consider the so-called “higher-order” theories of beams, plates, and shells intensively discussed in the literature. Note that although the equations presented in the text following are valid only for beams, the main references are made to plates, for which such theories have been originally constructed. Consider two typical problems of beam theory, that is, the cantilever beam shown in Fig. 6.20 and the simply supported beam shown in Fig. 6.21. Both beams are homogeneous and have rectangular cross sections. z,w

h/2

F

M

h

x, u

h/2

V

l

dx

b

FIGURE 6.20 A cantilever beam.

z, w q x, u l FIGURE 6.21 A simply supported beam.

6.7 REFINED THEORIES OF BEAMS AND PLATES

407

6.7.1 FIRST-ORDER THEORY The theory described in the foregoing sections is sometimes referred to as the “first-order” theory and is based on the approximations given by Eq. (6.3) for the displacements which, for the case of bending, yield ux 5 zθðxÞ;

uz 5 wðxÞ

(6.90)

The corresponding static variables are the internal bending moment and the transverse shear force specified by Eqs. (6.5) and (6.15), that is, ð2

ð2

h

h

M5

σx bzdz;

V5

2h2

τ xz bdz

(6.91)

2h2

This approximation reduces the theory to Eq. (6.34), that is, 0

0

M 5 V;

V 1p 50
0 V 5 Sγ; γ5 θ1w

0

M 5 Dθ ;

(6.92) (6.93)

For a beam with a rectangular cross section D5

1 Ebh3 ; 12

S 5 Gbh

(6.94)

(see Fig. 6.20). The normal and shear stresses are specified by Eqs. (6.12) and (6.27) which, for the stiffnesses in Eq. (6.94), give σx 5

12M z; bh3

τ xz 5

  6V h2 2 2 z bh3 4

(6.95)

For the cantilever beam (see Fig. 6.20), the solution of Eqs. (6.92) and (6.93) is F 5 V;

6F M 5 2 F ðl 2 xÞ; θ52 ð2l 2 xÞx 3  Ebh 2 2F Eh x w5 3lx 2 x2 1 Ebh3 2G

(6.96)

For a simply supported beam (see Fig. 6.21), the solutions are given in Table 6.1 (Case 2) and have the following form: 1 1 M 5 2 qbðl 2 xÞx V 5 2 qbðl 2 2xÞ; 2  2   q
3 q Eh2 2 3 3 2 3 θ5 ; w 5 2 ðl 2 xÞ x l 2 6lx 1 4x l 2 2lx 1 x 1 2Eh3 2Eh3 G

(6.97)

Traditionally, the main shortcoming of the “first-order” theory is associated with the shear stress distribution following from Eq. (6.90). Indeed, the exact constitutive equation for the shear stress is   @ux @uz 1 τ xz 5 Gγ xz 5 G @z @x

(6.98)

408

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Substituting Eq. (6.90), we get


0 τ xz 5 G θ 1 w

(6.99)

which does not depend on z. Thus, the stress τ xz in Eq. (6.99) does not satisfy the boundary conditions (see Fig. 6.20) according to which   h 50 τ xz z 5 6 2

(6.100)

However, as discussed in Section 3.1, Eq. (6.99) is not correct. The displacements in Eq. (6.90) approximate the distribution of the actual displacements through the beam height. It is known that approximate functions must not be differentiated, as has been done to derive Eq. (6.99). The second equation in Eq. (6.90) states that uz does not depend on z, which means that the beam element shaded in Fig. 6.20 is absolutely rigid in the z direction. As is known, the distribution of forces over the surface of an absolutely rigid body does not affect the body motion. Only the forces resultant, that is, the transverse shear force V for the beam, affects the beam deflection. Integration in the second equation of Eq. (6.91) allows us to avoid differentiation of ux in Eq. (6.98) with respect to z. Indeed, using Eqs. (6.90), (6.91), and (6.98), we get in several steps ð2 h

V5 2h2

 ð2  @ux @uz 1 dz τ xz bdz 5 bG @z @x h

2h2

      h h 0 0 2 ux 2 1 hw 5 bGhðθ 1 w Þ 5 bG ux 2 2

Thus in the theory under consideration, the constitutive equation exists for the shear force V only and does not exist for the shear stress τ xz . The shear stress following from the equilibrium equation is specified by the second equation of Eq. (6.95) and, naturally, satisfies the boundary conditions as per Eqs. (6.100). The same situation occurs in classical beam theory in which Eq. (6.98) simply does not exist, so there is no temptation to use it and the shear stress is found from the equilibrium equation.

6.7.2 HIGHER-ORDER THEORIES Consider the “higher-order” theories which are widely discussed in the literature. These theories are based on the displacement approximations which generalize Eq. (6.90) as ux 5 zu1 ðxÞ 1 z2 u2 ðxÞ 1 z3 u3 ðxÞ 1 ? 1 zk uk ðxÞ

(6.101)

uz 5 wðxÞ

(6.102)

To be specific, we shall discuss the so-called “third-order” theory. To derive the equations of this theory, take k 5 3 in Eq. (6.101), that is, ux 5 zu1 ðxÞ 1 z2 u2 ðxÞ 1 z3 u3 ðxÞ; uz 5 wðxÞ

(6.103)

6.7 REFINED THEORIES OF BEAMS AND PLATES

409

Following the conventional procedure, substitute Eqs. (6.103) into Eq. (6.98) to get 0

τ xz 5 Gðu1 1 w 1 2u2 z 1 3u3 z2 Þ

Apply the boundary conditions in Eq. (6.100) and determine the functions u2 and u3 . The resulting displacement field becomes ux 5 zu1 2

4z3
0 u1 1 w ; 2 3h

uz 5 wðxÞ

(6.104)

The displacements in Eq. (6.104) have been originally proposed by Vlasov (1957) and Ambartsumyan (1958) and are used in numerous papers and books to construct refined plate and shell theories (e.g., Reddy, 2004). It seems, at first glance, that the displacements in Eq. (6.103) are more accurate than the linear approximation given by Eq. (6.90). However, this is not actually the case. In general, introduction of force constraints into the kinematic field is not the best way to improve it. To show this, apply the constitutive equations and find the stresses corresponding to the displacements presented by Eq. (6.104), that is,   @ux 4z3 0 0 00 5 E zu1 2 2 ðu1 1 w Þ σx 5 Eεx 5 E @x 3h     @ux @uz 4z2 0 τ xz 5 Gγ xz 5 G 1 5 Gðu1 1 w Þ 1 2 2 @z @x h

(6.105)

Following Ambartsumyan (1987), introduce the bending moment and the shear force in Eq. (6.91). Substituting the stresses given by Eq. (6.105), we get M5E

  bh3 0 1 00 u1 2 w ; 4 15


2 0 V 5 Gbh u1 1 w 3

(6.106)

Consider first the cantilever beam (see Fig. 6.20) for which the equilibrium equations, Eq. (6.92) for p 5 0, have the solution given by Eq. (6.96). Integration of Eq. (6.106) yields   3F 12F x2 2 3 lx 2 2 C1 10Ghb Eh b 2   6Fx 6F x3 1 3 lx2 2 1 C1 x 1 C2 w5 5Ghb Eh b 3 u1 5

(6.107)

where C1 and C2 are the constants of integration which should be determined from the boundary conditions at x 5 0 (see Fig. 6.20). According to these conditions, we should put wðx 5 0Þ 5 0 and, taking into account that u1 is actually the angle of the cross section rotation, put u1 ðx 5 0Þ 5 0: Then, Eq. (6.107) reduce to u1 5 2

6F ð2l 2 xÞx; Eh3 b

w5

  2F 3Eh2 2 3lx 2 x x 1 Eh3 b 4G

(6.108)

Comparing this result with Eq. (6.96), we can conclude that u1 coincides with θ, whereas the deflection has a different coefficient in the term corresponding to the shear deformation (0.75 instead of 0.5). However, the main difference between the solutions is more significant. Indeed, for the clamped edge of the beam, Eq. (6.96) of the “first-order” theory give θ 5 0 and w 5 0, and Eq. (6.90) for the displacements yield ux ðx 5 0Þ 5 0 and uz ðx 5 0Þ 5 0. Thus, the displacements are

410

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

identically zero at the clamped edge of the beam. But for the “third-order” theory, Eq. (6.104) and (6.108) give ux ðx 5 0Þ 5 2

2Fz3 ; Gh3 b

uz ðx 5 0Þ 5 0

Thus the boundary conditions for the clamped end ðux 5 0Þ cannot be satisfied within the framework of the “third-order” theory. A similar conclusion can be derived for a simply supported beam shown in Fig. 6.21. Substitution of the bending moment M and the shear force V (neither of which depend on the shear deformation) from Eq. (6.97) into Eq. (6.106) and integration yield the following solution:   q 3Eh2 3 2 3 ðl 2 2xÞ l 2 6lx 1 4x 2 2Eh3 10G   q 6Eh2 3 2 3 w52 ðl 2 xÞ x l 2 2lx 1 x 2 2Eh3 5G u1 5

(6.109)

in which the constants of integration are found from the boundary conditions wðx 5 0Þ 5 wðx 5 lÞ 5 0 (see Fig. 6.21). Comparing this solution with the corresponding equation in Eq. (6.97), we can conclude that the results coincide within the accuracy of the terms including the shear modulus G. The bending moment M specified by the first equation of Eq. (6.106) and Eq. (6.109) satisfies the boundary conditions (see Fig. 6.21), according to which Mðx 5 0Þ 5 M ðx 5 lÞ 5 0. However, the axial stress is not zero at the beam ends. Substituting Eq. (6.109) into the first equation of Eq. (6.105), we arrive at σx 5 2

  6q qE 3h2 2 2 2z z ð l 2 x Þxz 1 h3 Gh3 10

(6.110)

At the beam ends (x 5 0 and x 5 l), the first term vanishes, but the second term gives nonzero stress, which cannot exist at the ends of a simply supported beam. Usually, simply supported beams and plates are studied applying a trigonometric series, each term of which satisfies the boundary conditions wð0Þ 5 wðlÞ 5 0 and Mð0Þ 5 M ðlÞ 5 0, that is, u1 5

X

Um cosλm x;

w5

m

X

Wm sinλm x

(6.111)

m

in which λm 5 πm=l (see Fig. 6.21). Substituting expansions (6.111) into Eq. (6.106), we get   bh3 X 1 λm Wm 2 Um sinλm x 15 m 4 X 2 ðλm Wm 1 Um Þcosλm x V 5 Gbh 3 m

M5E

(6.112)

Decomposing the load p 5 2 qb into a similar series p 5 2 qb 5 2 b

X m

qm sinλm x

(6.113)

6.7 REFINED THEORIES OF BEAMS AND PLATES

411

and substituting Eqs. (6.112) and (6.113) into the equilibrium equations, Eq. (6.92), we arrive at the following set of algebraic equations derived in terms of Wm and Um : E

  h2 2 1 λm λm Wm 2 Um 5 2Gðλm Wm 1 Um Þ 4 5 2 Ghλm ðλm Wm 1 Um Þ 5 2 qm 3

Solving these equations for Wm and Um , we have Wm 5 2

  12qm Eh2 2 λ ; 1 1 10G m Eh3 λ4m

Um 5

  12qm Eh2 2 λ 1 2 40G m Eh3 λ3m

(6.114)

Substitution of Eqs. (6.111) and (6.114) into Eq. (6.105) yields the following equation for σx : σx 5 2

 X 12z X qm Ez 3h2 2 2 2z sinλ x 1 qm sinλm x m h3 m λ2m Gh3 10 m

Retaining a finite number of terms in the series of this equation, we can arrive at the conclusion that σx ðx 5 0Þ 5 σx ðx 5 lÞ 5 0. However, for an infinite number of terms, the second series converges, and in accordance with Eq. (6.113) becomes equal to the second term in Eq. (6.110), which means that σx is not zero at the beam ends. Thus, the “third-order” theory does not allow one to satisfy the boundary conditions which are satisfied within the framework of the “first-order” theory. It is important that, following Ambartsumyan (1987), we used the displacements in Eq. (6.104) in conjunction with the equilibrium equations, Eq. (6.92), which follow from the free-body diagrams for the shaded element of the beam shown in Fig. 6.20. In more recent versions of the “third-order” theory (e.g., Reddy, 2004), the governing equations are derived with the aid of the variational principle. To discuss this approach, apply the minimum condition for the total potential energy of the beam, according to which h=2 ðl ð ðl
 δT 5 b dx σx δεx 1 τ xz δγ xz dz 2 b pδwdx 5 0 2h=2

0

(6.115)

0

Consider the “first-order” theory. Using Hooke’s law and displacements as per Eq. (8.90), we get the following equation for the axial stress: σx 5 Eεx 5 E

@ux 0 5 Eθ z @x

(6.116)

As has been noted earlier, the constitutive equation for the shear stress in the “first-order” theory exists only in integral form, Eq. (6.93). To apply this form, we need to use Eq. (6.91) for V and replace γ xz with the average shear strain γ 5 θ 1 w0 . Taking into account Eq. (6.116) and Eq. (6.91) for the bending moment, we can present Eq. (6.115) in the following form: ðl 0



0 Mδθ 1 V ðδθ 1 δw0 Þ 2 pδw dx 5 0

(6.117)

412

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Integration by parts yields the following variational equations and natural boundary conditions: 0

M 2 V 5 0; x5l ½Mδθx50

5 0;

0

V 1p 50

(6.118)

½Vδwx5l x50

(6.119)

50

As can be seen, Eq. (6.118) coincide with equilibrium equations, Eq. (6.92), whereas the boundary conditions, Eq. (6.119) give M 5 0 and V 5 0 which are equivalent to σx 5 0 and τ xz 5 0 for the free end [see Eq. (6.95)], and for the clamped end give θ 5 0 and w 5 0 which are equivalent to ux 5 0 and uz 5 0 [see Eqs. (6.90)]. Thus, the variational formulation of the “first-order” theory is equivalent to the direct formulation based on the equilibrium equations and the corresponding boundary conditions. The situation is totally different for the “third-order” theory for which we have constitutive equations, Eq. (6.105), for both normal and shear stresses. Substituting Eq. (6.105) into Eq. (6.115), and integrating with respect to z, we get ðl




0  0 Mδu1 2 H δu1 1 δw00 1 V ðδu1 1 δw0 Þ 2 Qðδu1 1 δw0 Þ 2 pδw dx 5 0

0

where, M and V are specified by Eq. (6.91) and H5

4b 3h2

h=2 ð

σx z3 dz;

Q5

2h=2

4b h2

h=2 ð

τ xz z2 dz

(6.120)

2h=2

The corresponding variational equations and the natural boundary conditions are 0

0

M 2 H 2 V 1 Q 5 0; ½ðM2HÞδu1 x5l x50 5 0;

0

0

V 2 Q 1 H 00 1 p 5 0

½Hδw0 x5l x50 5 0;

½ðV 2QÞδwx5l x50 5 0

(6.121) (6.122)

in which, in accordance with Eqs. (6.91), (6.105) and (6.120) Ebh3
0 00  4u1 2 w ; 60
2 0 V 5 Gbh u1 1 w ; 3

M5

Ebh3
0 00  16u1 2 5w 1260 2 Q 5 Gbhðu1 1 wÞ 15

H5

(6.123)

The equations obtained, Eq. (6.121) in conjunction with Eq. (6.123), are of the sixth order which corresponds to three natural boundary conditions in Eq. (6.122). For a cantilever beam shown in Fig. 6.20, at the clamped end u1 5 0, w0 5 0; and w 5 0. Correspondingly, it follows from Eq. (6.104) that ux 5 0 and uz 5 0 at x 5 0. Thus, the boundary conditions at the clamped end can be satisfied. However, it follows from Eq. (6.105) that, under these conditions, τ xz ðx 5 0Þ 5 0 and the force F acting at x 5 l (see Fig. 6.20) is not balanced. This result is expected because the variational equations, Eq. (6.121), do not coincide with equilibrium equations, Eq. (6.92), and hence, the equilibrium equations of a beam element as a solid are violated. The reason for this is associated with the formulation of the displacement field. In contrast to Eq. (6.90) of the “first-order” theory which allow for two mutually independent degrees of freedom for the shaded beam element shown in Fig. 6.20 (i.e., displacement w and rotation angle θ), in Eq. (6.104) of the “third-order” theory

6.7 REFINED THEORIES OF BEAMS AND PLATES

413

the displacement and the rotation are not mutually independent, and the variational equations, Eq. (6.121), cannot provide the equilibrium of the beam element as a solid.

6.7.3 CONSISTENCY CONDITIONS Energy consistency conditions for the refined beam, plate, and shell theories have been proposed by Vasiliev and Lurie (1990a, b, and 1992). According to these conditions, equilibrium equations following from the free body diagram for the structure element and the boundary conditions corresponding to the assumed displacement field must coincide with the variational equations and the natural boundary conditions providing the minimum of the structure’s total potential energy. As follows from Section 6.7.2, the “first-order” theory is energy consistent. In case the energy consistency conditions are not satisfied, there can exist, in principle, two versions of the refined theories. The theories of the first type can be referred to as energy inconsistent, but statically consistent. As an example of such a theory, consider classical beam theory. Because transverse shear deformation is ignored in classical beam theory, we must put the shear stiffness S-N in Eq. (6.93). Then, as follows from these equations, γ 5 0, θ 5 w0 , and M 5 2 Dwv

Consider now Eq. (6.117) for the variation of the total potential energy of the beam. Since θ 1 w0 5 0 the second term in this equation is zero and it reduces to ðl

ðl ðM δθ0 2 pδwÞdx 5 2 ðDδwv 1 pδwÞdx 5 0

0

(6.124)

0

It is clear that classical beam theory cannot be energy consistent because the beam element must have two mutually independent degrees of freedom, that is, displacement w and rotation θ whereas the functional in Eq. (6.124) includes only deflection w. Integrating by parts, we arrive at the following variational equation: Mv 1 P 5 0

(6.125)

Compare this equation with the equilibrium equations, Eq. (6.34), which follow from the free body diagram for the beam element. As can be seen, the direct equilibrium approach yields two equilibrium equations, whereas the energy approach gives only one equation. Thus the theory is not energy consistent. However, it is statically consistent because Eq. (6.125) follows from Eq. (6.34). The theories of the second type can be referred to as energy inconsistent and statically inconsistent. The example of such a theory is the “third order” theory described by Eq. (6.121) derived by the energy method. These equations are not equilibrium equations because they do not provide equilibrium of a beam element as a solid.

6.7.4 THEORY OF ELASTICITY SOLUTION Return to the “third-order” theory. Irrespective of the energy inconsistency of Eq. (6.121) and (6.123), the corresponding governing equations have the specific structure which shows us a way to

414

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

improve the “third-order” theory. Integrating Eq. (6.121), substituting Eq. (6.123), and performing some transformations, we arrive at the following two equations with regard to u1 and w:  G
30C1 x2 1 C2 x 1 C3 2 1260C1 2 Eh 0 w 5 4u1 2 30C1 x2 2 C2 x 2 C3

00

u1 2 k2 u1 5 2 672

(6.126)

in which k2 5 840G=Eh2 and C1 , C2 and C3 are the constants of integration. As can be seen, Eq. (6.126), in addition to the traditional polynomial solution, contain a solution which includes the exponential function Expð2 kxÞ which rapidly vanishes at a distance from the beam end and corresponds to the so-called boundary-layer solution. Goldenveizer (1958) was the first who stated that the nonlinear approximation of the displacement distribution through the plate thickness [as in Eq. (6.104)] must be supplemented with the boundary-layer solutions. To do this, consider substitution of Eq. (6.108) for u1 and w corresponding to the cantilever beam shown in Fig. 6.20 into Eq. (6.104) for ux and uz to get the displacements   6F E 3 z ð 2b 2 x Þxz 1 Eh3 b 3G   2F 3Eh2 x u0z 5 3 3lx 2 x2 1 Eh b 4G

u0x 5 2

(6.127)

which are treated now as the basic approximation. Following Vasiliev and Lurie (1972), and Vasiliev (2010), approximate the total displacements ux and uz supplementing the displacements in Eq. (6.127) with a set of boundary-layer solutions which vanish at a distance from the clamped end of the beam (see Fig. 6.20), that is, ux 5 u0x 1 u x ; uz 5 u0z 1 u z ðxÞ

where u x 5 C0 f0 ðzÞ 1

N X

e2sk x fk ðzÞ

(6.128)

(6.129)

k51

Here, C0 is a constant coefficient, f0 ðzÞ 5 z, and fk ðzÞ (k 5 0; 1; 2 . . .Þ is a complete system of functions which should be found (note that without f0 the functions fk do not form a complete system). Substituting Eqs. (6.128) and (6.129) into the equations of Hooke’s law, find the stresses X @ux 12F 5 2 3 ðl 2 xÞz 2 E sk e2sk x fk ðzÞ bh @x k " #     X @ux @uz 6F h2 du z 2 2sk x dfk 1 5 G C0 1 2z 1 1 e τ xz 5 G @z @x dz dx Gbh3 4 k σx 5 E

(6.130)

Apply the equilibrium equation, that is, the first equation of Eq. (6.1) in which b 5 constant @σx @τ xz 1 50 @x @z

(6.131)

Substituting for the stresses their expressions, Eq. (6.130), we arrive at the following equation for fk ðzÞ ðk 5 1; 2; 3; . . .Þ: d 2 fk 1 α2k fk 5 0 dz2

6.7 REFINED THEORIES OF BEAMS AND PLATES

415

The antisymmetric solution with respect to z of this equation is fk ðzÞ 5 Ck sinαk z;

α2k 5 s2k

E G

(6.132)

Now, Eq. (6.130) for τ xz becomes "

τ xz 5 G C0 1

#   X 6F h2 du z 2 2sk x 2 z 1 C α e cosα z 1 k k k Gbh3 4 dx k

(6.133)

According to the boundary conditions, Eq. (6.102), τ xy ðz 5 6 h=2Þ 5 0. Thus X du z 5 2 C0 2 Ck αk e2sk x cosλk ; dx k

1 λk 5 αk h 2

(6.134)

Integrating this equation and taking into account that for the clamped end (see Fig. 6.20) uz ðx 5 0Þ 5 0, we get uz 5 2 C0 x 1

X

Ck

k

αk 2sk x ðe 2 1Þ cosλk sk

(6.135)

Substituting Eq. (6.134) into Eq. (6.133), we arrive at the following expression for the shear stress "

#   X 6F h2 2 2sk x 2z 2 τ xz 5 G Ck αk e ϕk ðzÞ Gbh3 4 k

(6.136)

ϕk ðzÞ 5 cosλk 2 cosαk z

(6.137)

in which Apply the equilibrium equations, Eqs. (6.91) and (6.92) according to which h=2 ð

V 5b

τ xz dz 5 F

(6.138)

2h=2

for the cantilever beam (see Fig. 6.20) under consideration. Substituting Eq. (6.136) into Eq. (6.138), we derive the following equation for the parameter λk : tanλk 5 λk

(6.139)

As can be readily checked, the functions ϕk(z) in Eq. (6.137) and their derivatives are orthogonal on the interval [2h/2, h/2], that is, 8 > : 0; n ¼ 6 k 2h=2 h=2 ð

(

h=2 ð

0

0

ϕk ϕn dz 5 2h=2

2 2 2 λ sin λk ; n 5 k h k 0; n 6¼ k

(6.140)

Moreover, the following integral conditions are also valid: h=2 ð

h=2 ð

ϕk dz 5 0; 2h=2

0

h=2 ð

ϕk dz 5 0; 2h=2

0

zϕk dz 5 0 2h=2

(6.141)

416

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS


 0 It follows from Eq. (6.141) that the system of functions fk 5 Ck =αk ϕk is complete if it is supplemented with function f0 5 z [see Eq. (6.129)] in which this property of functions fk is used. To determine the constants C0 and Ck , apply the boundary condition for the clamped end of the beam, that is, ux ðx 5 0Þ 5 0. Using Eqs. (6.127) and (6.128), and (6.132) and (6.137), we have ux ðx 5 0Þ 5 2

X Ck 0 2F 3 z 1 C0 z 1 ϕ ð zÞ 5 0 3 Gbh αk k k

(6.142)

Multiplying this equation by z, integrating from h/2 to h/2 and using Eq. (6.141), we get 0 C0 5 3F=ð10GbhÞ.
Multiplying by ϕk ðzÞ, integrating, and using Eq. (6.140), we find 2 Ck 5 F= Gbλk sinλk . Thus, we have constructed the mathematically exact solution of the problem. To study the solution, present the explicit expressions for the displacements and the stresses. The final expression for the axial displacement follows from Eqs. (6.127) and (6.142), that is, ux 5 2

X sin αk z 6F F 3z 2z3 2 3 1 ð2l 2 xÞxz 1 e2sk x 2 3 Eh b Gb 10h h k λk sin λk

!

(6.143)

This displacement satisfies exactly the boundary conditions. We have two ways to simplify it. First, we can neglect the second term and arrive at the “first-order” theory, Eqs. (6.90) and (6.96). This approach looks attractive because all the boundary conditions are satisfied. Second, we can neglect the boundary-layer part of the solution, that is, omit the sum in Eq. (6.143). In this case, we arrive at the displacement that does not satisfy the boundary condition as per Eq. (6.142) and does not coincide with Eq. (6.127) for u0x which corresponds to the “third-order” theory. Thus, the “third-order” theory displacement must be supplemented with the boundary-layer solution. The beam deflection is specified by Eqs. (6.127), (6.128) and (6.135) which give rffiffiffiffi   2F 3Eh2 F E X cosλk 2 ð1 2 e2sk x Þ x2 uz 5 3 3lx 2 x 1 Eh b Gb G k λ2k sin λk 5G

(6.144)

Direct calculation shows that the second term in this equation corresponding to the boundarylayer solution is negligible. Returning to the shear correction factor in Eq. (6.28), we can conclude that for the exact solution (without the boundary-layer part) k 5 5=6, for the “first-order” theory [see Eq. (6.96)], k 5 1, and for the “third-order” theory [see Eq. (6.108)], k 5 2=3: The stresses corresponding to the foregoing exact solution can be found from Eqs. (6.130), (6.136) and presented in the following final form: rffiffiffiffi 12F 2F E X sinαk z 2sk x ð l 2 x Þz 2 e bh3 bh G k λk sinλk   6F h2 2F X cosλk 2 cosαk z 2sk x 2 z2 2 e τ xz 5 3 bh bh k λk sinλk 4 σx 5 2

(6.145)

(6.146)

It follows from these equations that the solution contains a polynomial part, which is referred to as a penetrating solution because it does not vanish at a distance from the beam clamped end, and a boundary-layer part. It is important that as follows from Eq. (6.141), the boundary-layer solution is

6.7 REFINED THEORIES OF BEAMS AND PLATES

417

self-balanced, that is, it does not contribute to the bending moment and the shear force. Introduce the normalized normal stress and dimensionless coordinates as follows σ 5 σx

bh2 x z ; x5 ; z5 l h 6F

and present Eq. (6.145) as h σ 5 2 2ð1 2 x Þz 2 3

rffiffiffiffi E X sin 2λk z 2s k x e  G k λk sin λk

where 2l sk 5 h

rffiffiffiffi E λk G

and λk are the roots of Eq. (6.139) listed in Table 6.2. The higher roots corresponding to kc10 can be approximately found as λk  ð2k 1 1Þ=ð2πÞ. For a beam with parameters h=l 5 0:1, E=G 5 10, we have the following stress distribution:

  σ 5 2 ð1 2 x Þσ ð0Þ ðz Þ 1 0:1054 σ ð1Þ ðz Þe2s 1 x 1 σ ð2Þ ðz Þe2s 2 x 1 σ ð3Þ ðz Þe2s 3 x 1 . . .

(6.147)

ðiÞ

The functions σ ðz Þ for i 5 0; 1; 2; 3 and the corresponding values of the parameter s i are presented in Fig. 6.22. As can be seen, the higher the gradient of the function σðiÞ ðz Þ, the higher the rate with which the corresponding boundary-layer solution vanishes at a distance from the clamped end x 5 0. For example, the contribution of the first boundary-layer solution σð1Þ ðz Þ to the maximum stress σðz 5 0:5Þ at a distance from the clamped end x 5 0:01 (which is 0.1 of the beam thickness) does not exceed 0.2 %. Thus, the improvement of the penetrating solution provided by the boundary-layer solutions is more of a formal nature. Moreover, for practical analysis, the penetrating solution looks more realistic than the derived exact solution. Indeed, at the corner of the clamped edge (x 5 0; z 5 h=2 in Fig. 6.20) the series in Eq. (6.145) does not converge, giving rise to infinitely high normal stresses at these corners. This problem is discussed further in Section 6.7.5. Consider the shear stress acting in the clamped cross section of the beam. Taking x 5 0 in Eq. (6.146), we get τ xz ðx 5 0Þ 5

  6F h2 2F X ϕk ðzÞ 2 2 z 2 bh3 4 bh k λk sinλk

(6.148)

Table 6.2 Nonzero Roots of the Equation tanλk 5 λk ðk 5 1; 2; 3. . .10Þ k

1

2

3

4

5

λk

4.4934

7.7253

10.9041

14.0661

17.2208

k

6

7

8

9

10

λk

20.3713

23.5194

26.6661

29.8116

32.9563

418

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

z

z

0.5

0.5

0.4

0.4

0.3

0.3

0.2

0.2

0.1

0.1

σ 0

s1 = 28 .42

–0.2

–0.4

–0.6

–0.8 –1.0

10 ⋅ σ

(0)

0.3

0.2

0.1

0

z 0.5

0.4

0.4

0.3

0.3

s 2 = 48 .86

s 3 = 68 .96

0.2

0.1 0.1

–0.3

z

0.5

0.2

0.2

–0.1 –0.2

(1)

10 ⋅ σ 0

0.1

(2)

–0.1 –0.2

0.2

0.1

10 ⋅ σ 0

–0.1

(3)

–0.2

FIGURE 6.22 Functions σðiÞ ðzÞði 5 0; 1; 2; 3 Þand the corresponding values of parameter si .



 where ϕk ðzÞ is given by Eq. (6.137). Since ϕk z 5 6 h=2 5 0, we have τ xz z 5 6 h=2 5 0, so the boundary conditions are satisfied on the beam top and bottom surfaces. Now, use the orthogonality conditions in Eq. (6.141) and decompose the quadratic function in the first term in Eq. (6.148) into a series with respect to the functions ϕk ðzÞ. Taking into account the fact that in accordance with Eq. (6.141) the system of functions ϕk ðzÞ is complete only in conjunction with unity, we have X h2 2 z2 5 C 1 Bk ϕk ðzÞ 4 k

where C and Bk are some constant coefficients. Multiplying this equation first by unity and second by ϕn ðzÞ, integrating from h/2 to h/2, and using Eqs. (6.140) and (6.141), we can determine C and Bk and finally arrive at h2 h2 2 z2 5 4 6

X ϕ ðzÞ k 112 sin λk λ k k

!

6.7 REFINED THEORIES OF BEAMS AND PLATES

419

Substituting this result into Eq. (6.148), we get τ xz ðx 5 0Þ 5

F bh

(6.149)

As can be seen, the shear stress at the clamped end of the beam does not depend on z. This result is expected, because it follows directly from the constitutive equation, Eq. (6.98) for the shear stress. Indeed, at the clamped end of the beam [see Fig. (6.20)] ux 5 0 and, hence, ð@ux Þ=@z 5 0, whereas uz does not depend on z. So, at x 5 0 the shear stress does not change through the beam height and is specified by Eq. (6.149). However, this equation is valid only for the open interval 2 h=2 , z , h=2. At the interval ends z 5 6 h=2, the shear stress must be zero due to the symmetry of the stress tensor (τ xz 5 0 at z 5 6 h=2 according to the boundary conditions). Thus, the distribution of the shear stress over the beam height for a clamped edge of the beam is not continuous, that is, τ xz is constant for 2 h=2 , z , h=2 and zero for z 5 6 h=2. Naturally, the derivative @τ xz =@z is infinitely high at the corner points (x 5 0, z 5 6 h=2), and as follows from the equilibrium equation, Eq. (6.131), the normal stress σx is singular at the corner points.

6.7.5 NONCLASSICAL ELASTICITY SOLUTION It follows from the previous section that the solution corresponding to classical theory of elasticity is singular, that is, the shear stress distribution over the clamped edge of the beam shown in Fig. 6.20 is not continuous, whereas the normal stresses in the corners of this edge are infinitely high. This result cannot be treated as correct both mathematically and physically. First, it is obvious that the derivatives of the stresses which enter the equilibrium equation, Eq. (6.131), do not exist at the corner points with coordinates x 5 0, z 5 6 h=2 (see Fig. 6.20) and hence, this equation is not valid at these points. Second, cantilever beams similar to those shown in Fig. 6.20 are widely used in engineering and are analyzed with the aid of the equations of strength of materials (mechanics of materials) which do not provide singular solutions. The foregoing results according to which the maximum stress in the beam is infinitely high actually mean that the cantilever beams cannot work which is obviously not the case. Singularities, being quite natural for mathematics, cannot exist in reality. Singular solutions of physical problems mean that the mathematical model of the problem is not consistent and does not adequately correspond to the physical problem. Vasiliev and Lurie (2014) have shown experimentally by testing glass beams that the stress singularity does not reduce the beam strength and, hence, does not have physical nature. The classical theory of elasticity is based on the so-called phenomenological model of a solid according to which an infinitesimal element of a solid has the same properties as the whole solid. This means that the equilibrium equation, Eq. (6.131), and the rest of equations of the theory are derived for an infinitely small element of a solid. Apply a more general model of a solid (Vasiliev & Lurie, 2015a, b) assuming that it consists of the elements with small but finite dimensions, a as shown in Fig. 6.23. Consider an arbitrary point A with coordinates x and z and introduce local coordinates α and β in the vicinity of this point such that 2 a=2 # α # a=2 and 2 a=2 # β # a=2. Let some continuous field function f (stress, strain, or displacement) be specified in the domain

420

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

z Tz

4 a 2

Txz

3

β A ( x, z )

Txz

α

a 2

1

a 2

x

Tx

a 2

2

FIGURE 6.23 Element of a solid with finite dimensions.

½2a=2; a=2 with the derivatives existent up to the third order. Decompose this function into a Taylor series as   @f @f 1 @2 f 2 @2 f @2 f 2 αβ 1 f ðx; z;α; β Þ 5 f ðx; zÞ 1 α 1 β 1 α 1 2 β @x @z 2! @x2 @x@z @z2  3  3 3 3 1 @ f 3 @ f 2 @ f @ f 3 2 1 α α 1 3 β 1 3 αβ 1 β 3! @x3 @x2 @z @x@z2 @z3

(6.150)

Assume that the function f represents the stress σx , σz , or τ xz and construct the equilibrium equation for the element shown in Fig. 6.23. The resultants of the stresses σx and τ xz acting on the sides of this element in the x direction are a ð2 R223;124 ðσx Þ 5

  a σx x; z; α 5 6 ; β dβ 5 2

a 2      a @σx a2 @2 σx 1 @2 σx a3 @3 σx @3 σ x 1 6 ðx; yÞ 1 1 5 a σx 6 2 @x 3 @z2 8 @x2 48 @x3 @x@z2 2

a ð2 R324;122 ðτ xz Þ 5

  a dα 5 τ xz x; z; α; β 5 6 2

a 2      a @τ xz a2 1 @2 τ xz @2 τ xz a3 @3 τ xz @3 τ xz 6 ðx; yÞ 1 1 1 5 a τ xz 6 8 3 @x2 @z2 48 @x2 @z @z3 2 @z 2

(6.151)

(6.152)

Here, symbol (x, y) means that changing x to y and y to x we can write similar equations for the y direction. The equilibrium equations can be presented as R223 ðσx Þ 2 R124 ðσx Þ 1 R324 ðτ xz Þ 2 R124 ðτ xz Þ 5 0 ðx; yÞ

6.7 REFINED THEORIES OF BEAMS AND PLATES

421

Substituting the resultants with their expressions given by Eqs. (6.151) and (6.152), we get Lx ðT Þ 5

@Tx @Txz 1 5 0 ðx; yÞ @x @z

in which T ðTx ; Tz ; Txz Þ 5 σðσx ; σz ; τ xz Þ 1

(6.153)

  a2 @2 @2 σðσx ; σz ; τ xz Þ 1 24 @x2 @z2

(6.154)

As can be seen, Eq. (6.153) has the same form as the traditional equilibrium equation, Eq. (6.131), but it includes the generalized stresses T instead of traditional stresses σ. Using the traditional definition of strain as the ratio of the absolute deformation of the element in Fig. 6.23 to the corresponding initial dimensions, we get for the normal and shear strains  1
Ex 5 u223 2 u124 ðx; zÞ x x a  1
2 u122 1 u223 2 u124 Exz 5 u324 x z z a x

(6.155)

Here

u223;124 5 x

1 a

a ð2

  a ux x; z; α 5 6 ; β dβ 2

a 2 2 a  ð2  1 a 5 x; z; α; β 5 6 dα u324;122 u x x a 2 a 2 2

ðx; zÞ (6.156) ðx; zÞ

are the average displacements of the edges of the element in Fig. 6.23 expressed in terms of the actual displacements ux and uz . Decomposing these displacements with the aid of Eq. (6.150) (i.e., taking f 5 ux and f 5 uz , and substituting the resulting expressions into Eqs. (6.154) and (6.155), we finally get Ex 5

@Ux ; @x

Ez 5

@Uz ; @z

@Ux @Uz 1 @z @x

(6.157)

  a2 @2 @2 uðux ; uz Þ 1 24 @x2 @z2

(6.158)

Exz 5

where U ðUx ; Uz Þ 5 uðux ; uz Þ 1

are the generalized displacements similar to the generalized stresses in Eq. (6.154). The generalized displacements have a simple physical meaning: they correspond to the displacements of the element in Fig. 6.23 averaged over its area. Using Eq. (6.150), we have 1 ux 5 2 a

ZZa=2 ux ðx; z; α; β Þdxdβ 5 Ux 2a=2

ðx; zÞ

(6.159)

422

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Now assume that the generalized stresses and strains are linked by Hooke’s law, that is, Tx 5

E ðEx 1 νEz Þ 1 2 ν2

ðx; zÞ;

Txz 5

E Exz 2ð1 1 νÞ

(6.160)

For a homogeneous stress-strain state (i.e., not depending on coordinates), T 5 σ, U 5 u and Eq. (6.160) reduce to the traditional Hooke’s law. To complete the theory, apply the principle of virtual displacements to the element in Fig. 6.23 to get ZZ

½Lx ðT ÞδUx 1 Lz ðTÞδUz dxdz 5 0

(6.161)

Here, L are specified by Eq. (6.153) and δU are the variations of the average displacements of the element in Eq. (6.159). Performing variation in Eq. (6.161), we arrive at the following natural boundary conditions: Tx δUx 5 0;

Txz δUz 5 0

(6.162)

Thus, the proposed generalized theory reduces to equilibrium equations, Eq. (6.153), straindisplacement relationships, Eq. (6.157), constitutive equations, Eq. (6.160), and the boundary conditions in Eq. (6.162). As can be seen, these equations completely coincide with the corresponding equations of the classical theory of elasticity, but include generalized stresses, strains and displacements given by Eqs. (6.154), (6.157), and (6.158). Consequently, the generalized solution of the problem coincides with the traditional solution of the classical theory of elasticity. Actual stresses and displacements can be found from Eqs. (6.154) and (6.158). In conclusion, a main principal point needs to be made. As can be seen, Eqs. (6.154) and (6.158) include some structural parameter a which is not known and must be determined experimentally. However, the constructed theory is of a phenomenological nature, that is, it does not allow for the actual microstructure of a solid. Consequently, the structural parameter a, in principle, cannot be determined using this theory and possible macrostructural experiments. To avoid this dispffiffiffi crepancy, we put (see Vasiliev & Lurie, 2015a, b) a 5 2i 6r, where i2 5 2 1 and r is the real parameter that can be found experimentally. As a result, Eqs. (6.154) and (6.158) reduce to  2  @ σ @2 σ 1 @x2 @z2  2  @ u @2 u 1 U 5 u 2 r2 @x2 @z2

T 5 σ 2 r2

(6.163) (6.164)

Now consider bending of the cantilever beam shown in Fig. 6.20. To find the axial displacement ux , we should use Eq. (6.164) in which U is specified by the corresponding solution of the classical theory of elasticity, that is, by Eq. (6.143). The resulting equation is presented as @2 ux @2 ux 1 6F 1 2 2 2 ux 5 2 ð2l 2 xÞxz 2 @x @z r Ebh3 r2 F 2 Gbr 2

X sinαk z 3z 2z3 e2sk x 2 3 1 2 h 10h sinλ λ k k k

!

(6.165)

6.7 REFINED THEORIES OF BEAMS AND PLATES

423

The general solution of this equation can be given by

  6F  2 4F 3h2 z2 2 2 2 3r z x 1 2ðr 2 lxÞ z 1 2 Ebh3 Gbh3 40 2 F X sinα z 
k  e2sk x 2 Gb k λ2k sinλk r 2 s2k 2 α2k 2 1 X X
 1 A z3 1 3rxz e2x=r 1 Ak1 e2ηk x sinαk z 1 Ak2 e2ξk x sinβ k z

ux 5

k

(6.166)

k

where η2k 5 α2k 1

1 ; r2

ξ 2k 5 β 2k 1

1 ; r2

βk 5

πð2k 2 1Þ h

The terms including the force F in Eq. (6.166) are the particular solutions of Eq. (6.165), whereas the other terms represent the solutions of the homogeneous equation corresponding to Eq. (6.165). These last terms allow us to satisfy the boundary condition ux ðx 5 0; zÞ 5 0 at the clamped end of the beam (see Fig. 6.20). Using this condition and Eqs. (6.139) and (6.140), we get F 
 Gbλ2k sinλk r 2 s2k 2 α2k 2 1    48F E h2 2k 2 1 2 2 2 r sin r 1 Ak2 5 G 40 2 Ebh2 π2 ð2k21Þ2 A5

2F ; Gbh

Ak1 5

(6.167)

Thus, the axial displacement is specified by Eqs. (6.166) and (6.167). The final expression is rather cumbersome and is omitted. The beam transverse displacement uz satisfies the equation following from Eqs. (6.164) and (6.144), that is,   d 2 uz 1 2F 3Eh2 2 2 u 5 2 3lx 2 x 1 z r2 Ebh3 r 2 dx2 5G rffiffiffiffi X F E cosλk ð1 2 e2sk x Þ 1 Gbr 2 G k λ2k sinλk

The solution of this equation satisfying the boundary condition uz ðx 5 0Þ 5 0 at the clamped end of the beam (see Fig. 6.20) is     6F x 6F 2 2 ð Þ 1 2r x x l 2 l 2 x 1 Ebh3 3 5Gbh r ffiffiffi ffi   12Flr2 2x=r F EX 1 e2sk x 2 r 2 s2k e2x=r 2 e 2 11 Ebh3 Gb G k λ3k r 2 s2k 2 1

uz 5

(6.168)

Determine the stresses. The shear stress τ xz can be found from Eq. (6.163) in which Txz is specified by Eq. (6.146), that is,   @2 τ xz @2 τ xz 1 6F h2 2 2 z 1 2 τ 5 2 xz @x2 @z2 r2 bh3 r2 4 2F X cosλk 2 cosαk z 2sk z 1 e 2bhr 2 k λk sinλk

(6.169)

424

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

The general solution of this equation is

( )   6F h2 2F X 1 cosαk z 2 2 
 e2sk x
2 2 z 2 2r 1 τ xz 5 2 3 bh bh k 4 λk sinλk r 2 s2k 2 α2k 2 1 λ2k r 2 s2k 2 1 X z 1Bcosh 1 Bk e2sk x cosγ k z r k

(6.170)

where γ 2k 5 s2k 2 r12 . As earlier, the terms with F in Eq. (6.170) are the particular solutions of Eq. (6.169), whereas the other terms are the solutions of the homogeneous to Eq. (6.169)
equation corresponding  which allow us to satisfy the boundary conditions τ xz x; z 5 6 h=2 5 0 on the longitudinal top and bottom surfaces of the beam (see Fig. 6.22). Using these conditions and Eqs. (6.139) and (6.140), we get B5

12Fr 2
; h=2r

bh3 cosh

Bk 5

bhλ2k

2Fr 2 α2k 

2 
2  2 2 r sk 2 1 r sk 2 α2k 2 1 cos γ k h=2

The final expression for the shear stress is τ xz 5

"
 #) cosh z=r h2
 2 z2 2 2r2 1 2 4 cosh h=2r ( ) 2F X 1 cosαk z r 2 α2k cosγ k z 



 
 1 2 e2sk x 1 bh k λk sinλk r2 s2k 2 α2k 2 1 λ2k r 2 s2k 2 1 λ2k r2 s2k 2 1 r2 s2k 2 α2k 2 1 cos γ k h=2

6F bh3

(

(6.171)

The series entering this solution converge and, in contrast to Eq. (6.146), it specifies the continuous distribution of τ xz over the beam height. Finally, determine the normal stress σx which can be found from Eq. (6.163) in which Tx is specified by Eq. (6.145), that is, @2 σx @2 σx 1 12F 1 2 2 2 σx 5 3 2 ðl 2 xÞz 2 r bh r @x @z rffiffiffiffi 4F E X sinαk z 2sk x 1 e bhr 2 G k λk sinλk

(6.172)

Since there are no boundary conditions imposed on σx , we can take only the particular solution of Eq. (6.172), which has the form σx 5 2

12F 2F ðl 2 xÞz 2 bh3 bh

rffiffiffiffi EX e2sk x sinαk y 
 G k λk sinλk 1 2 r 2 s2k 2 α2k

(6.173)

The series in this solution converges and, as opposed to the classical case given by Eq. (6.145), this solution is not singular. The solutions obtained in this study include three experimental constants: E, G, and r. To determine these constants, the experimental beam shown in Fig. 6.24 has been studied. The beam is made of hybrid carbon-glass-epoxy plies of composite material made by angle-ply circumferential winding (see Section 2.6). The beam has the following geometric parameters

6.7 REFINED THEORIES OF BEAMS AND PLATES

425

FIGURE 6.24 Experimental composite beam.

(Fig. 6.20): l 5 300 mm, b 5 89.7 mm, h 5 53.8 mm. The axial modulus Ex 5 105:84 GPa has been measured under axial compression of the beam. To find the shear modulus G, a bending test has been used. For a beam with the length l, which is much larger than the beam height h, we can neglect the boundary-layer parts of the solution and present the beam deflection in Eq. (6.168) as uz ðxÞ 5

i 6F h 2 x 6F x l 2 1 2r 2 ðl 2 xÞ 1 x 3 Ebh 3 5Gbh

(6.174)

The deflection of the beam end is uz ðlÞ 5

4Fl3 6Fl 1 5Gbh Ebh3

Measuring uz ðlÞ corresponding to the applied force F, we get G 5 5.926 GPa. To determine the parameter r, calculate the shear strain εxz . Using Eqs. (6.166)(6.168) and neglecting the boundary-layer parts of the solution, we have for a relatively long beam εxz 5

  @ux @uz 12Fr 2 6F h2 2 2 1 5 2 z 1 2 2r @z @x Ebh3 Gbh3 4

(6.175)

The maximum value of this strain corresponds to z 5 0, that is, εm xz 5

  3Fr 2 6F h2 2 2 2r 1 Gbh3 4 Ebh3

Solving this equation for r 2 , we get

  h2 2Gbh m  12 εxz r2 5
3F 8 1 2 E=G

(6.176)

426

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

FIGURE 6.25 Bending of the experimental beam.

Thus, to determine r, we need to measure εm xz . The idea of the experiment is demonstrated in Fig. 6.25. A laser 1 whose beam is focused on the screen at point A is placed on the top surface of a composite beam. A laser 2 whose beam is focused at point B of the screen is fixed at the level of the middle plane of the beam cross section. The lasers fixed to the beam are shown in Fig. 6.24. Under the action of the load F applied to the beam end, the points A and B move to their new locations at A1 and B1 , respectively. The angles of the laser beam rotations are α1 5 ðAA1 Þ=L and α2 5 ðBB1 Þ=L where L is the distance between the end cross section of the composite beam and the screen (see Fig. 6.25). Then, the shear strain can be calculated as εm xz 5 α1 2 α2 . For a load F 5 24.5 kN, the experiment yields εm 5 0:001271. Then, Eq. (6.176) gives r 5 0.0381h 5 xz 2.05 mm. Note that the experimental beam has been assembled of the layers having a thickness of about 1.5 mm. To demonstrate a specific feature of the theory, consider the shear stresses at a distance from the clamped end of the beam. Neglecting the boundary-layer part of the solution in Eq. (6.171), we get 6F τ xz 5 3 bh

(

"
 #) cosh z=r h2 2 2
 2 z 2 2r 1 2 4 cosh h=2r

Comparing this stress with the shear strain in Eq. (6.175), we can conclude that Hooke’s law for shear, that is, τ xz 5 Gεxz , is satisfied only if r 5 0. For r 6¼ 0, Hooke’s law for shear is violated. The reason is associated with the nonzero gradients of the functions τ xz ðzÞ and εxz ðzÞ, because of which τ xz and εxz do not coincide with the corresponding generalized stress and strain Txz and Exz . The fact that Hooke’s law is not valid for the stress and strain fields with gradients has been demonstrated experimentally by Andreev (1976). Consider the results of the numerical analysis for the beam with the foregoing parameters. The dependence of the normalized beam deflection uz 5 uz =um , where um 5 4Fl3 =Ebh3 is the maximum deflection corresponding to the conventional solution of strength of materials, on coordinate x 5 x=l is shown in Fig. 6.26. The solid line corresponds to the strength of materials solution (i.e., the classical beam theory), whereas the dashed line denotes the solution given by Eq. (6.168). Note that

6.7 REFINED THEORIES OF BEAMS AND PLATES

427

FIGURE 6.26 Dependencies of the normalized deflection on the axial coordinate corresponding to strength of materials (ss) and the solution in Eq. (6.168) (s s s).

FIGURE 6.27 Distributions of the normalized normal stress acting in the fixed cross section of the beam over the beam thickness corresponding to strength of materials (ss), Eqs. (6.145) (s s s) and (6.173) (      ).

the results of the calculation for the first-order theory, Eq. (6.96), practically coincide with the dashed line. The distribution of the normalized stress σx 5 σx =σm , where σm 5 2 6Fl=bh2 is the maximum stress corresponding to strength of materials solution, over the height of the fixed cross section of the beam (x 5 0; y 5 y=hÞ is presented in Fig. 6.27. The solid line demonstrates the strength of materials (and the first-order theory) solution, the dashed line corresponds to the singular solution of the classical theory of elasticity given by Eq. (6.145), whereas the dotted line shows a refined solution obtained as per Eq. (6.173). As can be seen, the refined solution provides a maximum stress which is 12% higher than that calculated using the strength of materials formula. The distribution of the normalized shear stress τ xz 5 τ xz =τ m , where τ m 5 3F=2bh is the maximum stress corresponding to the strength of materials theory, over the height of the beam clamped cross section (x 5 0; y 5 y=hÞ is shown in Fig. 6.28. The solid line corresponds to the strength of materials

428

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

FIGURE 6.28 Distributions of the normalized shear stress acting in the fixed cross section of the beam over the beam thickness corresponding to strength of materials (ss), Eqs. (6.146) (s s s), and (6.171) (       ).

solution, the dashed line shows the discontinuous solution of the classical theory of elasticity as per Eq. (6.146), whereas the dotted line presents the solution given by Eq. (6.171). As can be seen, the shear stress, specified by the new solution is continuous.

6.8 LAMINATED BEAMS Consider laminated beams. The first solution for a laminated cantilever beam was obtained by S.G. Lekhnitskii in 1935 (Lekhnitskii, 1935) and reviewed by Carrera (2003). The biharmonic equation of elasticity theory for a laminated beam was solved with the aid of a polynomial representation of the stress function for each layer. Thus, S.G. Lekhnitskii has constructed a penetrating solution ignoring boundary-layer solutions. Equations describing the combination of a penetrating and boundary-layer solutions have been derived by Bolotin (1963). The laminate consists of a system of load-carrying layers alternating with core layers which take only transverse stresses (see Fig. 6.29). The axial displacements of the load-carrying layers correspond to classical plate theory, according to which 0

0

uðiÞ x 5 ui 2 zi wi ;

uði11Þ 5 ui11 2 zi11 wi11 x

(6.177)

0

(Bolotin & Novichkov, 1980), where ð. . .Þ 5 dð. . .Þ=dx, zi is counted from the middle surface of the i-th layer (2hi =2 # zi # hi =2Þ, and wi is the deflection of the i-th layer. The axial stress in the i-th layer is 00  ðiÞ ðiÞ ðiÞ
0 σðiÞ x 5 A11 εx 5 A11 ui 2 zi wi

For the core layers, the axial stress is zero and the shear strain does not depend on z, that is, ði;i11Þ γ xz 5

1 hi;i11



ui11 2 ui 1

  1
0 1
0 0 0 wi hi 1 wi11 hi11 1 wi11 1 wi 2 2

6.8 LAMINATED BEAMS

429

z i +1 hi +1

u i +1

h i , i +1 zi hi

ui

FIGURE 6.29 Distribution of the axial displacement over the height of a laminated beam.

The transverse normal strain of the core layers is εði;i11Þ 5 z

1 ðwi11 2 wi Þ hi;i11

The governing equations for the functions ui and wi and the corresponding natural boundary conditions are derived with the aid of the variational principle. The total order of these equations depends on the number of layers. The solutions of numerous problems have been obtained by Bolotin and Novichkov (1980). They include the penetrating solutions and the system of boundarylayer solutions which describe local bending of load-carrying layers, as well as shear and normal strains of the core layers rapidly vanishing from the beam edge. A theory that is more realistic for composite laminates (that do not have core layers) has been proposed by Grigolyuk and Chulkov (1964). In this theory, the axial stiffness of the core layers is not ignored. A similar theory with membrane load-carrying layers have been developed by Elpatievskii and Vasiliev (1965). The main shortcoming of the theories under consideration is the high order of the governing equations (which depends on the number of layers) in which the penetrating solution is mixed with boundary-layer solutions. In modern versions of the theory (Reddy, 2004), the penetrating solution (the corresponding displacement is shown in Fig. 6.29 with a dashed line) is singled out and the shear deformation of the layers is taken into account. The resulting expressions for the displacements become P ux 5 uðxÞ 1 zθðxÞ 1 i ui ðxÞϕi ðzÞ P uz 5 w 1 i wi ψi ðzÞ

(6.178)

where z is the global normal coordinate of the laminate and ϕi ðzÞ, ψi ðzÞ are some functions that provide continuity of the displacements, the derivatives of which with respect to z are not continuous at the layer interfaces. Different forms of such functions have been presented by Carrera (2003). Since three functions for the whole laminate (u; θ; and w) are introduced in Eq. (6.178), the number of functions ui and wi can be respectively reduced. For example, we can take for the first and the last (k-th) layers u1 5 0, w1 5 0, uk 5 0 (Reddy, 2004).

430

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

z2 h

2

x

F

z1

1

h0 h

l

b

FIGURE 6.30 A cantilever sandwich beam.

In the simplest version of the theory, the deflection is assumed to be the same for all the layers, that is, uz 5 wðxÞ, whereas the functions ϕi ðzÞ in Eq. (6.178) are some linear functions providing the continuity of the axial displacement ux with respect to z (Carrera, 2003). Applying the continuity conditions for the shear stresses at the layer interface surfaces, we can express ui ðxÞ in terms of two functions θðxÞ and wðxÞ (Carrera, 2003). As a result, the order of the governing equations for this theory does not depend on the number of layers. The theory constructed in this way is similar to the “higher-order” theory for homogeneous beams discussed earlier. Return to the general theory based on Eqs. (6.177) or (6.178). The most pronounced difference between this theory and the theory based on a linear distribution of the axial displacement through the beam height is manifested in sandwich beams with stiff load-carrying layers and a shear deformable lightweight core. Consider as an example the cantilever sandwich beam shown in Fig. 6.30. The axial displacements of the first and the second layers (see Fig. 6.30) following from Eqs. (6.177) and (6.178) are   H 0 2 z1 θðxÞ 2 z1 w ðxÞ u1 5 uðxÞ 2 2   H 0 2 z2 θðxÞ 2 z2 w ðxÞ u2 5 uðxÞ 2 2

(6.179)

Here, uðxÞ, θðxÞ, and wðxÞ are the axial displacement, rotational angle, and the deflection (the same for both layers), H 5 h0 1 h, 2h=2 # zi # h=2, ði 5 1; 2Þ. The shear strain in the core is γ5

     1 h h H 0 0 2 u1 z1 5 1 w 5 ðθ 1 w Þ u2 z2 5 2 h0 2 2 h0

Then, the normal stresses in the layers and the shear stress in the core become     H 0 0 σð1Þ 5 E u 2 2 z1 θ 2 z1 w00 2     H 0 0 ð2Þ 00 σ 5E u 2 2 z2 θ 2 z2 w 2 H 0 τ 5 G ðθ 1 w Þ h0

(6.180)

6.8 LAMINATED BEAMS

431

where E is the modulus of the layer and G is the shear modulus of the core material. The forces and the moments in the layers are h=2 ð

N1 5 b 2h=2 h=2 ð

N2 5 b 2h=2 h=2 ð

M1 5 b

  H 0 σð1Þ dz1 5 Bl u 2 θ0 2

  H 0 σð2Þ dz2 5 Bl u 1 θ0 2 (6.181)


0

σð1Þ z1 dz1 5 2 Dl θ 1 w

 00

2h=2 h=2 ð

M2 5 b


0  σð2Þ z2 dz2 5 Dl θ 2 w00

2h=2

where Bl 5 Ebh; Dl 5

1 Ebh3 12

(6.182)

are the membrane and bending stiffnesses of the layers. The equilibrium equations for the layers (see Fig. 6.31A) can be written as 1 0 M1 2 V1 1 τbh 5 0; 2

0

M2 2 V2 1

1 τbh 5 0 2

(6.183)

From Eq. (6.180) for τ, (6.181) for M, and Eq. (6.183) we can find the transverse forces in the layers, that is,
00  1 1 τbh 5 2 Dl θ 1 w000 1 GbhH ðθ 1 w0 Þ 2 2
00  1 1 0 000 V2 5 M2 1 τbh 5 Dl θ 2 w 1 GbhH ðθ 1 w0 Þ 2 2 0

V1 5 M1 1

Then, the total shear force V 5 V1 1 V2 1 bh0 τ

becomes


0  000 V 5 2 2Dl w 1 S w 1 θ

(6.184)

S 5 GbH

(6.185)

where is the shear stiffness of the beam. Finally, the total axial force and bending moment can be presented as 0

N 5 N1 1 N2 5 2Bl u ;

00

M 5 M1 1 M2 5 2 2Dl w 1

1 0 Bl H 2 θ 2

(6.186)

432

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

V1+ V1′dx

(A)

N1 + N1′dx

N1 M 1 V1

N2 M2

V2

τ

M2 + M2′ dx

τ

V 2 + V2′dx N2 + N2′ dx M1+ M1′dx

dx

(B)

V + V ′dx N

N + N ′dx M + M ′dx

M V

dx FIGURE 6.31 Free-body diagrams for a sandwich beam: equilibrium of the layers (A), and equilibrium of the beam element (B).

The equilibrium equations of the beam element follow from Fig. 6.31B and have the form 0

N 5 0;

0

M 5 V;

0

V 50

(6.187)

The boundary conditions for the clamped end of the beam (see Fig. 6.30), that is, uðx 5 0Þ 5 0;

θðx 5 0Þ 5 0;

0

w ðx 5 0Þ 5 0

(6.188)

provide with regard to Eq. (6.179) u1 ðx 5 0Þ 5 0 and u2 ðx 5 0Þ 5 0. For the end x 5 l (see Fig. 6.30), we have N 5 0;

V 5 F;

M50

(6.189)

The first equations in Eqs. (6.186)(6.189) yield N 5 0 and u 5 0. Eq. (6.187) for V and the corresponding boundary condition in Eq. (6.189) give V 5 F. Applying Eq. (6.184), we get
0  22Dl ww 1 S w 1 θ 5 F

(6.190) 0

Now, according to the second equilibrium equation in Eq. (6.187), M 5 F and M 5 Fx 1 C1 . Since Mðx 5 lÞ 5 0, we arrive at M 5 2 Fðl 2 xÞ. Then, the second equation of Eq. (6.186) yields 1 0 22Dl wv 1 Bl H 2 θ 5 2 Fðl 2 xÞ 2

6.8 LAMINATED BEAMS

Integration gives 0

2Dl w 1

433

  1 x2 Bl H 2 θ 5 F lx 2 1 C2 2 2

Using the boundary conditions in Eq. (6.188), we have C2 5 0 and 0

2Dl w 1

 1 F
Bl H 2 θ 5 2lx 2 x2 2 2

(6.191)

Thus, the problem of a beam in bending is reduced to Eqs. (6.190) and (6.191). Eliminating θ from these equations and integrating the resulting equation, we arrive at 00

w 2 k2 w 5 2

  FS x3 Bl H 2 2 1 x 1 C3 lx 2 2Dl Bl H 2 3 S

(6.192)

where C3 is the constant of integration and k2 5

Dt S 1 ; Dt 5 2Dl 1 Bl H 2 Dl Bl H 2 2

(6.193)

Here, Dt is the total bending stiffness of the beam. The solution of Eq. (6.192) consists of two parts. The first part is the polynomial (i.e., penetrating solution) and the second part is the boundary-layer solution. Retaining the boundary-layer solution in the vicinity of the clamped end only, we get w 5 C4 e2kx 1

  FBl H 2 2Dl Fl 2 F 3 FSl Dl Bl H 2 12 x1 x 2 x 1 2 C 3 2Dt 6Dt 2Dl Bl H 2 2Dt S Dt Dt S

The constants of integration C3 and C4 are found from the boundary conditions wð0Þ 5 0 and θð0Þ 5 0. Then, the final expressions for w and θ become w5

  rffiffiffiffiffiffiffiffiffi   FBl H 2 2Dl Bl Dl
2kx Fl 2 F 3 12 e 21 1x 1 H x 2 x 2Dt S Dt Dt S 2Dt 6Dt     F 2Dl 2Dl Bl H 2
2kx 2 1 2 e θ5 1 2 2 2lx 1 x Bl H 2 Dt Dt S

(6.194)

These expressions include the parameter d 5 Dl =Dt which is the ratio of the bending stiffness of the layer to the total bending stiffness of the beam. Using Eqs. (6.182) and (6.193) for the stiffness coefficients, we get d5

h2

h2 ; 1 6H 2

H 5 h0 1 h

For practical sandwich beams, this parameter is normally rather small. For example, for h0 5 15mm, h 5 1mm (see Fig. 6.30), d 5 0:00065. Thus, we can simplify Eq. (6.194) taking d 5 0. Then, w5

  F 6Dt x; 3lx 2 x2 1 6Dt S

θ52

F ð2l 2 xÞx 2Dt

where Dt 5 Bl H 2 =2. These equations coincide exactly with Eq. (6.96) which correspond to a linear distribution of the axial displacement through the beam height.

434

CHAPTER 6 LAMINATED COMPOSITE BEAMS AND COLUMNS

Table 6.3 Normalized Maximum Deflections of Sandwich Beams E, GPa

G, MPa

l, mm

h, mm

h0 , mm

wð1Þ m

ð2Þ wm

wð3Þ m

70 70 70 70 70

55.7 915 915 38.5 38.5

280 480 280 280 280

0.7 0.7 0.7 1.0 2.4

3.7 4.8 5.1 17.0 18.8

0.990 0.969 0.856 0.329 0.171

0.992 0.983 0.986 0.923 0.921

1.027 1.020 1.033 0.985 1.053

Taking into account the transverse normal strain of the core εz , we arrive at an additional boundary-layer solution rapidly vanishing at a distance from the clamped end.
 For simply supported sandwich beams (see Fig. 6.21) loaded with pressure q 5 q0 sin πx=l , the comparison of the maximum deflection wm corresponding to various beam models with the experimental results wem obtained by Alexandrov, Bryukker, Kurshin, and Prusakov (1960) is presented in Table 6.3. In this table, wm 5 wm =wem , wð1Þ m corresponds to the classical beam theory for which S-N, wð2Þ specifies the results corresponding to the displacements given by m Eq. (6.179), whereas deflections wð3Þ m show the solution based on the linear approximation of the ð3Þ axial displacement. It follows from Table 6.3 that wð2Þ m and wm are characterized by a comparable accuracy for all the beams. Thus, the theory constructed in Section 6.1 allows us to describe the beam behavior with reasonable accuracy and is generalized in Chapter 7, Laminated Composite Plates for composite plates.

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