(17.22) Ⅲ The case of a multilayer plate such that for any two plies k and m one has in the plane of the plate 11 : Then Equations 17.19 and 17.20 reduce to (17.23) The preceding particular cases constitute a severe restriction among the variety of practical laminations. Nevertheless we will conserve in the following the simplified forms of Equations 17.21, 17.22, and 17.23 because they well show the direct connection between the warpings h x and h y and the transverse shear forces Q x and Q y , respectively. 17.6.3.2 Consequences Setting h x and h y in the forms: (17.24) The constitutive Equations 17.10 and 17.11 are written as: Ⅲ 11 Such a limiting case is rare in practice, because it imposes in particular: . 1. cylindrical bending about y axis G xz k ∂ 2 h x ∂ z 2 z E 11 k EI 11 E 12 k EI 12 + ˯ ʈ Q x –= 2. cylindrical bending about x axis G yz k ∂ 2 h y ∂ z 2 z E 22 k EI 22 E 12 k EI 12 + ˯ ʈ Q y –= E 33 k E 33 m G xy k G xy m a km == E ij k E ij m a km i and j" 1, 2, 3== G xz k ∂ 2 h x ∂ z 2 z E 11 k EI 11 E 12 k EI 12 + ˯ ʈ Q x –= G yz k ∂ 2 h y ∂ z 2 z E 22 k EI 22 E 12 k EI 12 + ˯ ʈ Q y –= h x x, y, z() Q x hG xz ·Ò gz()¥= h y x, y, z() Q y hG yz ·Ò pz()¥= ˛ Ô Ô ˝ Ô Ô ¸ Q x hG xz ·Ò ∂ w 0 ∂ x q y + ˯ ʈ Q x hG xz ·Ò G xz gd zd zd h/2– h/2 Ú += TX846_Frame_C17 Page 331 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC then by setting : (17.25) Ⅲ then by setting : (17.26) There appear two transverse shear coefficients k x and k y which require the knowledge of the functions g(z) and p(z) for their calculations. 17.6.4 Warping Functions Ⅲ Boundary conditions: We have assumed that the upper and lower faces of the plate were free of any shear. Then the transverse shear in Equations 17.17 and 17.18 leads to Ⅲ then with Equation 17.25: Ⅲ then with [17.26]: Ⅲ Continuity at the interfaces: The continuity of the transverse shear at the interfaces between layers results from the assumed perfect bonding between two plies (see Paragraph 15.1.2). One then has at the interface between two consecutive plies k and k + 1: k x 1 1 hG xz ·Ò G xz gd zd zd h/2– h/2 Ú – ˯ ʈ = Q x hG xz ·Ò k x ∂ w 0 ∂ x q y + ˯ ʈ = Q y hG yz ·Ò ∂ w 0 ∂ y q x – ˯ ʈ Q y hG yz ·Ò G yz pd zd zd h/2– h/2 Ú += k y 1 1 hG yz ·Ò G yz pd zd zd h/2– h/2 Ú – ˯ ʈ = Q y hG yz ·Ò k y ∂ w 0 ∂ y q x – ˯ ʈ = ∂ w 0 ∂ y q y + ˯ ʈ Q x hG xz ·Ò dg dz + 0 for zh/2±== k x gd zd + 0 for zh/2±== ∂ w 0 ∂ y q y – ˯ ʈ Q y hG yz ·Ò dp dz – 0 for zh/2±== k y pd zd + 0 for zh/2±== t xz k t xz k+1 ; t yz k t yz k+1 == TX846_Frame_C17 Page 332 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC then with Equations 17.17, 17.18 and Equations 17.25, 17.26: Ⅲ Formulation of the warping functions: Let us substitute to g(z) and p(z) the functions g 0 (z) and p 0 (z) such that: g 0 (z) and p 0 (z) are called the warping functions. Then the boundary conditions and the interface conditions simplify, and Equations 17.23 allow one to formulate the problems that permit a simple calculation of warping functions g 0 (z) and p 0 (z). One obtains (17.27) (17.28) The antisymmetric functions g 0 and p 0 are then defined in a unique manner. 17.6.5 Consequences Ⅲ Form of the transverse shear stresses: Equations 17.17 and 17.18 then take the simple forms: (17.29) G xz k k x dg k dz + ˯ ʈ G xz k+1 k x dg k+1 dz + ˯ ʈ = G yz k k y dp k dz + ˯ ʈ G yz k+1 k y dp k+1 dz + ˯ ʈ = g 0 z() gz() zk x ; p 0 ¥ z()+ pz() zk y ¥+== d 2 g 0 dz 2 z hG xz ·Ò G xz k E 11 k EI 11 E 12 k EI 12 + ˯ ʈ ¥–= dg 0 dz 0 for zh/2±== G xz k dg 0k dz G xz k+1 dg 0k+1 dz for zz k == Ó Ô Ô Ô Ì Ô Ô Ô Ï d 2 p 0 dz 2 z hG yz ·Ò G yz k E 22 k EI 22 E 12 k EI 12 + ˯ ʈ ¥–= dp 0 dz 0 for zh/2±== G yz k dp 0k dz G yz k+1 dp 0k+1 dz for zz k == Ó Ô Ô Ô Ì Ô Ô Ô Ï t xz Q x G xz k hG xz ·Ò dg 0 dz ; t yz ¥ Q y G yz k hG yz ·Ò dp 0 dz ¥== TX846_Frame_C17 Page 333 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC Ⅲ Transverse shear coefficients: One obtains these coefficients from the Equation 17.5: Ⅲ here is : noting that : One obtains : (17.30) Ⅲ here is : leading to: (17.31) In summary, in the absence of body forces (inertia forces, example), the bending behavior uncoupled from the membrane behavior of a thick laminated plate can be simplified in a few particular cases noted below. The characteristic relations are summarized in the following table. Bending Behavior (no in-plane stress resultants) homogeneous orthotropic plate/orthotropic axes : x, y, z or Laminated plate/midplane symmetry/orthotropic axes of plies: x,y,z/same Poisson ratios v xy and v yx for all plies/cylindrical bending about x or y axis. or Laminated plate/midplane symmetry/orthotropic axes of plies: x,y,z/elastic constants are proportional from one ply to another ᭹ Equilibrium relation: Ú h/2– h/2 E 11 EI 11 E 12 EI 12 + ˯ ʈ h x zzd 0= E 11 EI 11 E 12 EI 12 + ˯ ʈ Q x hG xz ·Ò g 0 kz x –()¥ zzd h/2– h/2 Ú 0= E 11 EI 11 E 12 EI 12 + ˯ ʈ z 2 zd h/2– h/2 Ú C 11 EI 11 C 12 EI 12 + C 11 C 12 C 12 2 – C 11 C 12 C 12 2 – 1.== = k x E 11 EI 11 E 12 EI 12 + ˯ ʈ g 0 zzd h/2– h/2 Ú = Ú h/2– h/2 E 22 EI 22 E 12 EI 12 +() h y zzd 0= E 22 EI 22 E 12 EI 12 + ˯ ʈ Q y hG yz ·Ò p 0 kz y –()¥ zzd h/2– h/2 Ú 0= k y E 22 EI 22 E 12 EI 12 + ˯ ʈ p 0 zzd h/2– h/2 Ú = ∂ Q x ∂ x ∂ Q y ∂ y p z ++ 0; ∂ M y ∂ x ∂ M xy ∂ y Q x –– 0; ∂ M xy ∂ x ∂ M x ∂ y Q y ++ 0=== TX846_Frame_C17 Page 334 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC ᭹ Constitutive relations: with ᭹ Stresses Ⅲ stresses within the ply : s x : cf. [17.14]; s y : cf. [17.15]; t xy : cf. [17.16] Ⅲ transverse shear stresses (17.32) ᭹ Warping functions Ⅲ g 0 (z) is the solution of the problem: Ⅲ p 0 (z) is the solution of the problem: ᭹ Transverse shear coefficients k x and k y : Ⅲ They are given by the formula: M y M x – M xy – Q x Q y Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ C 11 C 12 00 0 C 21 C 22 00 0 00C 33 00 000 hG xz ·Ò k x 0 000 0 hG yz ·Ò k y ∂q y ∂ x ∂q x ∂ y – ∂q y ∂ y ∂q x ∂ x – ∂ w 0 ∂ x q y + ∂ w 0 ∂ y q x – Ó ˛ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ ÔÔ ÔÔ ÔÔ Ï¸ = C[] 1– 1 EI = t xz Q x G xz k hG xz ·Ò dg 0 dz = ; t yz Q y G yz k hG yz ·Ò dh 0 dz = d 2 g 0 dz 2 z hG xz ·Ò G xz k E 11 k EI 11 E 12 k EI 12 + ˯ Á˜ ʈ –= dg 0 dz 0= for zh/2±= G xz k dg 0k dz G xz k+1 dg 0k+1 dz = for zz k = Ó Ô Ô Ô Ô Ì Ô Ô Ô Ô Ï d 2 p 0 dz 2 z hG yz ·Ò G yz k E 22 k EI 22 E 12 k EI 12 + ˯ Á˜ ʈ –= dp 0 dz 0= for z h/2±= G yz k dp 0k dz G yz k+1 dp 0k+1 dz = for zz k = Ó Ô Ô Ô Ô Ì Ô Ô Ô Ô Ï k x E 11 EI 11 E 12 EI 12 + ˯ ʈ g 0 zz; k y E 22 EI 22 E 12 EI 12 + ˯ ʈ p 0 zzd h/2– h/2 Ú =d h/2– h/2 Ú = TX846_Frame_C17 Page 335 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC 17.6.6 Interpretation in Terms of Energy We will limit ourselves to the surface energy density due to transverse shear stresses as: Substituting Equation 17.29, one obtains The first integral can be rewritten as: or, taking into account Equation 17.27: where one recognizes Equation 17.30 of the transverse shear coefficient k x , The first integral is reduced to Following a similar approach for the second integral and taking into account Equations 17.28 and 17.31 for the transverse shear coefficient k y , the surface energy due to transverse shear takes the form: 17.7 EXAMPLES Examples for plates in bending are shown in details in Part Four of this book, in Chapter 18, “Applications.” We give here a few useful elements to treat these examples. W t 1 2 t xz g xz t yz g yz +()zd h/2– h/2 Ú 1 2 t xz 2 G xz t yz 2 G yz + ˯ ʈ zd h/2– h/2 Ú == W t 1 2 Q x 2 G xz hG xz ·Ò 2 g 0 d zd ˯ ʈ 2 z 1 2 +d h/2– h/2 Ú Q y 2 G yz hG yz ·Ò 2 p 0 d zd ˯ ʈ 2 zd h/2– h/2 Ú = 1 2 Q x 2 hG xz ·Ò 2 G xz d zd g 0 g 0 d zd ˯ ʈ g 0 d 2 g 0 z 2 d – zd h/2– h/2 Ú 1 2 Q x 2 hG xz ·Ò 2 G xz g 0 dg 0 dz h/2– h/2 hG xz ·Ò E 11 EI 11 E 12 EI 12 + ˯ ʈ g 0 zzd h/2– h/2 Ú + Ó ˛ Ì ˝ ϸ 1 2 k x Q x 2 hG xz ·Ò W t 1 2 k x Q x 2 hG xz ·Ò 1 2 k y Q y 2 hG yz ·Ò += TX846_Frame_C17 Page 336 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC 17.7.1 Homogeneous Orthotropic Plate Ⅲ Warping functions: Equation 17.27 becomes 12 Ⅲ Transverse shear stresses and shear coefficients: One deduces from Equation 17.32: (17.33) (17.34) In an analogous manner starting from Equation 17.28: then: (17.35) (17.36) Remark: In Application 18.3.7 (Chapter 18), one treats the case of a thick homogeneous orthotropic plate in cylindrical bending about the y axis. The plate supports a uniformly distributed load. One can consider there the strong influence of transverse shear in bending. Two characteristics of the plate then apply directly on the deflection: 12 g 0 is, as g, antisymmetric in z (see Equation 17.4). E 11 k E 11 = ; E 12 k E 12 ; E 22 k E 22 = ; G xz k = G xz = d 2 g 0 dz 2 zh E 11 E 22 E 11 E 22 E 12 2 – 12 h 3 E 12 2 E 11 E 22 E 12 2 – 12 h 3 – ˯ Á˜ ʈ – z 12 h 2 ¥–== dg 0 dz 0 for zh/2±== then 12 dg 0 dz 3 2 14 z 2 h 2 – ˯ ʈ ; g 0 3 2 z 1 4 3 z 2 h 2 – ˯ ʈ == t xz Q x h 3 2 14 z 2 h 2 – ˯ ʈ ¥= k x 12 h 3 3 2 1 4 3 z 2 h 2 – ˯ ʈ z 2 zd h/2– h/2 Ú = k x 6 5 = p 0 z() g 0 z()= t yz Q y h 3 2 14 z 2 h 2 – ˯ ʈ ¥= k y 6 5 = TX846_Frame_C17 Page 337 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC Ⅲ The relative thickness h/a, where a is the length of the bent side of the plate Ⅲ The ratio E x /G xz (for certain combinations of fiber/matrix, this ratio becomes large compared with unity; for example, for unidirectional) 17.7.2 Sandwich Plate The plate consists of two orthotropic materials: Material (1) for the skins Material (2) for the core (see Figure 17.3) Assuming the proportionality of elastic coefficients for the two materials leads to (see Section 17.6.3): one deduces from there: Figure 17.3 Sandwich Plate then C ij E ij z 2 zd h/2– h/2 Ú E ij ➀ z 2 zd H 1 /2– H 2 /2– Ú E ij ➁ z 2 zdE ij ➀ z 2 zd H 2 /2 H 1 /2 Ú + H 2 /2– H 2 /2 Ú +== C ij E ij ➀ H 1 3 H 2 3 – 12 ˯ ʈ E ij ➁ H 2 3 12 += C ij E ij ➀ a H 1 3 12 with a H 1 3 12 ¥ H 1 3 H 2 3 – 12 a 12 + H 2 3 12 == 1 E I 11 - C 11 C 11 C 22 C 12 2 – E 11 ➀ E 11 ➀ E 22 ➀ E 12 ➀ () 2 – 12 a H 1 3 ¥== 1 E I 12 C 12 – C 11 C 22 C 12 2 – E 12 ➀ – E 11 ➀ E 22 ➀ E 12 ➀ () 2 – 12 a H 1 3 ¥== TX846_Frame_C17 Page 338 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC 17.7.2.1 Warping Functions Ⅲ From the above one can write in Equation 17.27 13 : In addition : Equation 17.27 then can be written as: Ⅲ In Equation 17.28, one obtains an analogous formulation. In effect, one can write The problem [17.28] is then written as: Ⅲ Remark: These problems are identical to that which allows the calculation of the warping function for the bending of a sandwich beam, and one can consider it in Chapter 18, application 18.3.5. One can then carry out the same steps of calculation. The results obtained are shown below. 13 See Equations 17.2. E 11 k EI 11 E 12 k EI 12 + ˯ ʈ E x k E x ➀ 12 a H 1 3 ¥ E x k E x ➀ H 1 3 H 2 3 –() 12 E x ➁ H 2 3 12 + == hG xz ·ÒG xz ➀ H 1 H 2 –()G xz ➀ H 2 += d 2 g 0 dz 2 z– E x k G xz k 12 G xz ➀ H 1 H 2 –()G xz ➁ H 2 + E x ➀ H 1 3 H 2 3 –()E x ➁ + H 2 3 ¥¥= dg 0 dz 0 for zH 1 ±/2== G xz dg 0 dz continuous for zH 2 ±/2= Ó Ô Ô Ô Ì Ô Ô Ô Ï E 22 k EI 22 E 12 k EI 12 + ˯ ʈ E y k E y ➀ 12 a h 3 ¥ E y k E y ➀ H 1 3 H 2 3 –() 12 E y ➁ H 2 3 12 + == d 2 p 0 dz 2 z E y k G yz k 12 G yz ➀ H 1 H 2 –()G yz ➁ + H 2 E y ➀ H 1 3 H 2 3 –()E y ➁ + H 2 3 ¥¥–= dp 0 dz 0 zH 1 /2±== G yz dp 0 dz continuous for zH 2 /2±= Ó Ô Ô Ô Ì Ô Ô Ô Ï TX846_Frame_C17 Page 339 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC 17.7.2.2 Transverse Shear Stresses Ⅲ Stress t xz : (17.37) Ⅲ Stress t yz : (17.38) 17.7.2.3 Transverse Shear Coefficients (17.39) k y is given by an expression formally identical to that in which the index x is replaced by y. In Application 18.3.8 we treat the case of a rectangular sandwich plate in cylindrical bending, clamped on one side and subjected to a uniform linear force on another. The plate is free on the other sides. One shows the influence of transverse shear on the deflection. This influence increases when: Ⅲ The mechanical characteristics (moduli) of the core are weaker than those of the skins. Ⅲ The relative thickness of the core is important (thin skins). Ⅲ The relative thickness of the plate is large (thick plate). H 2 2 z H 2 2 ££– ; t xz Q x 6 E x ➁ H 2 2 4 z 2 – ˯ ʈ E x ➀ H 1 2 4 H 2 2 4 – ˯ ʈ + E x ➀ H 1 3 H 2 3 –()E x ➁ H 2 3 + ¥¥= H 2 2 z H 1 2 ££ : t xz Q x 6 E x ➀ H 1 2 4 z 2 –() E x ➀ H 1 3 H 2 3 –()E x ➁ H 2 3 + ¥¥= H 2 2 z H 2 2 ££– ; t yz Q y 6 E y ➁ H 2 2 4 z 2 –()E y ➀ H 1 2 4 H 2 2 4 –()+ E y ➀ H 1 3 H 2 3 –()E y ➁ H 2 3 + ¥¥= H 2 2 z H 1 2 ££ ; t yz Q y 6 E y ➀ H 1 2 4 z 2 –() E y ➀ H 1 3 H 2 3 –()E y ➁ H 2 3 + ¥¥= k x a x 8 E x ➀ H 1 3 H 2 3 –()E x ➁ H 2 3 +[] E x ➁ G xz ➁ H 2 3 E x ➀ H 1 2 4 5 E x ➁ E x ➀ – ˯ ʈ H 2 2 + Ó Ì Ï º= … E x ➀ () 2 G xz ➀ 4 5 H 1 5 H 2 5 5 H 1 2 H 2 3 –+ ˯ ʈ + ˛ ˝ ¸ 3b x E x ➀ H 1 2 H 2 2 –() E x ➀ H 1 3 H 2 3 –()E x ➁ H 2 3 + + with a x 12 G xz ➀ H 1 H 2 –()G xz ➁ H 2 + E x ➀ H 1 3 H 2 3 –()E x ➁ H 2 3 + ¥= b x a x 16 H 2 E x ➀ G xz ➀ H 2 2 3 H 1 2 G xz ➀ G xz ➁ 1– ˯ Á˜ ʈ + Ó Ì Ï = H 2 2 G xz ➀ G xz ➁ 1 2 3 E x ➁ E x ➀ – ˯ Á˜ ʈ – ˛ ˝ ¸ TX846_Frame_C17 Page 340 Monday, November 18, 2002 12:33 PM © 2003 by CRC Press LLC [...]... 2002 12: 40 PM PART IV APPLICATIONS We have grouped in this last part of the book exercises and examples for applications These have various objectives and different degrees of difculties Leaving aside (except for special cases) the cases that are too academic, we will concern ourselves with applications of concrete nature, with an emphasis on the numerical aspect of the results A few of these applications. .. Monday, November 18, 2002 12: 40 PM 18 APPLICATIONS 18.1 LEVEL 1 18.1.1 Simply Supported Sandwich Beam Problem Statement: 1 The following gure represents a beam made of duralumin that is supported at two points It is subjected to a transverse load of F = 50 daN Calculate the deectiondenoted as D of the beam under the action of the force F 2 We separate the beam of duralumin into two parts with equal thickness... parallel pipe made of polyurethane foam, making the skins of a sandwich beam having essentially the same mass as the initial beam (in neglecting the mass of the foam and the glue) The beam is resting on the same supports and is subjected to the same load F Calculate the deection caused by F, denoted by D Compare with the value of D found in Part 1 (Take the shear modulus of the foam to be: Gc = 20 MPa.)... bending moment shear D = 1.22 mm Comparing with the deection D found in Part 1 above: D 14 = D 1 Remarks: The sandwich conguration has allowed us to divide the deection by 14 without signicant augmentation of the mass: with adhesive lm thickness 3 0.2 mm and a specic mass of 40 kg/m for the foam, one obtains a total mass of the sandwich: m = 700 g (duralumin) + 50 g (foam) + 48 g (adhesive) This... of the beam, and I is its exure moment of inertia It corresponds to a critical speed for a beam in rotation, which should not be reached during the operation The carbon/epoxy unidirectional has Vf = 60% ber volume fraction The thickness of a cured ply is 0 .125 mm 1 Give the characteristics of a suitable shaft of carbon/epoxy composite One will make use of the tables in Section 5.4.2 and will use a... 3, and a minimum thickness of 2.5 mm, the resistance condition: Mt 300 - Ê MPa 2 3 2pr e leads to a radius of the tube of r 43 mm From this we nd a mass of: (rsteel = 7,800 kg/m ): 3 msteel = 10.5 kg The saving in mass of the composite solution over the steel solution is 73% The real saving is higher because it takes into account the disappearance of the intermediate bearing and of one part. .. carbon/epoxy composite 18.1.6 Wing Tip Made of Carbon/Epoxy Problem Statement: Wing tip refers to a part of airplane wing as shown in Figure 18.1 It is made of a sandwich structure with carbon/epoxy skins (Figure 18.2) xed to the rest of the wing by titanium borders as shown Under the action of the aerodynamic forces (Figure 18.3), the wing tip is subjected to bending moments, torsional moments, and shear... the core of the sandwich structure transmits only shear forces, and the skins support the exural moments This is represented in Figure 18.4(b); the skins resist in their respective planes the in-plane stress resultants: Nx, Ny, and Txy Figure 18.5 shows the values of these stress resultants 7 Recall the expression for the rotational kinetic energy of a mass m placed at a radius r and 2 2 2 2 1 1 1... Problem Statement: Consider a helicopter blade mounted on the rotor mast as shown schematically in the following gure 3 Or the orthotropic axes: see Chapter 12, Equation 12. 9 â 2003 by CRC Press LLC TX846_Frame_C18a Page 348 Monday, November 18, 2002 12: 40 PM The characteristics of the rotor are as follows: Rotor with three blades; rotational speed: 500 revolutions per minute The mass per unit length... with the drag and its consequences One examines the helicopter as immobile with respect to the ground (stationary ight in immobile air) In neglecting the weight of the blade compared with the load application and in assuming innite rigidity, the relative equilibrium conguration in uniform rotation is as follows: â 2003 by CRC Press LLC TX846_Frame_C18a Page 349 Monday, November 18, 2002 12: 40 PM 1 Justify . + C 11 C 12 C 12 2 – C 11 C 12 C 12 2 – 1.== = k x E 11 EI 11 E 12 EI 12 + ˯ ʈ g 0 zzd h/2– h/2 Ú = Ú h/2– h/2 E 22 EI 22 E 12 EI 12 +() h y zzd 0= E 22 EI 22 E 12 EI 12 + ˯ ʈ Q y hG yz ·Ò . Plate then C ij E ij z 2 zd h/2– h/2 Ú E ij ➀ z 2 zd H 1 /2– H 2 /2– Ú E ij ➁ z 2 zdE ij ➀ z 2 zd H 2 /2 H 1 /2 Ú + H 2 /2– H 2 /2 Ú +== C ij E ij ➀ H 1 3 H 2 3 – 12 ˯ ʈ E ij ➁ H 2 3 12 += C ij E ij ➀ a H 1 3 12 with a H 1 3 12 ¥ H 1 3 H 2 3 – 12 a 12 + H 2 3 12 == 1 E I 11 - C 11 C 11 C 22 C 12 2 – E 11 ➀ E 11 ➀ E 22 ➀ E 12 ➀ () 2 – . E 12 EI 12 + ˯ ʈ h x zzd 0= E 11 EI 11 E 12 EI 12 + ˯ ʈ Q x hG xz ·Ò g 0 kz x –()¥ zzd h/2– h/2 Ú 0= E 11 EI 11 E 12 EI 12 + ˯ ʈ z 2 zd h/2– h/2 Ú C 11 EI 11 C 12 EI 12