where: 18.1.15 Manipulator Arm of Space Shuttle Problem Statement: A manipulator arm is made of two identical tubular columns in carbon/epoxy (V f = 60%; thin cylindrical tubes of revolution) with pins as shown in Figure 18.6. Among the different geometric configurations found when the arm is deployed, one can consider the geometries noted as (a), (b), and (c) in Figure 18.7. F represents the concentrated inertial force. Note the following: E x = Longitudinal modulus of elasticity of the tube in x direction (Figure 18.6) G xy = Shear modulus in the tangent plane x, y (Figure 18.6) I = Quadratic moment of flexure of a cross section (annular) of the tube with respect to its diameter Figure 18.6 V f E 2024 E m – E f E m – = V f 31%= TX846_Frame_C18a Page 392 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC Solution: 1. Starting from the relations of flexure and torsion of composite tubes (see Section 5.4.5, Figure 5.31): one obtains for the components of displacement at the end: Ⅲ Configuration (a): Ⅲ Configuration (b): Ⅲ Configuration (c): Remark: For configurations (a) and (b), one obtains a displacement that is as small as the modulus E x is large. Then (see Section 5.4.2, Tables 5.4 and 5.5), G xy is relatively small, which means that E x /G xy >> 1. The displacement of config- uration (c) is much larger than the others. This will create problems when operating the arm. 2. The deflections are identical for configurations (a) and (c) if then: E x I d 2 v dX 2 M f G xy I o d q X dX ; M t == D Y Fᐉ 3 3E x I = D x F ᐉ/2() E x I ᐉ 2 ᐉ 2 ¥¥ F ᐉ/2() 3 3E x I – Fᐉ 3 6E x I –== D y F ᐉ/2() 2E x I ᐉ 2 ˯ ʈ 2 ¥ Fᐉ 3 16E x I == D y F ᐉ/2() 3 3E x I 2 F ᐉ/2() G xy I o ᐉ 2 ᐉ 2 ¥¥+¥ º == º Fᐉ 3 8E x I 2 3 E x 2G xy + ˯ ʈ 1 3 1 8 2 3 E x 2G xy + ˯ ʈ = E x G xy 4= TX846_Frame_C18a Page 394 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC 3. In looking for the modulus G xy to be as high as possible, one reads on Tables 5.4 and 5.5 (Section 5.4.2.) a ratio E x /G xy = 3.9 (Ӎ4) for the composition: 4. The maximum value of the longitudinal modulus of elasticity observed on Table 5.4 is This corresponds to a shear modulus (Table 5.5): The same deflection as the previous one for the configuration (c) can be obtained by then: The tube with thickness e¢ and modulus E x ¢ will be more stiff for configuration (a) but will have a mass multiplied by 3.5 to keep the stiffness of configuration (c). 5. Configurations (a) and (c) are the more deformable. One then has to write with ᐉ = 15 m; I = p r 3 e; r = 0.15 m; (F/D) min = 10 4 N/m; E x = 75,407 MPa e ≥ 14 mm E¢ x 134,000 MPa= G¢ xy 4,200 MPa= Fᐉ 3 8E¢ x I¢ 2 3 E¢ x 2G¢ xy + ˯ ʈ Fᐉ 3 3E x I = I¢ I p r 3 e¢ p r 3 e 3E x 8E¢ x 2 3 E¢ x 2G¢ xy + ˯ ʈ 3.5== = e¢ e 3.5= F D y 3E x I ᐉ 3 F D ˯ ʈ min ≥= TX846_Frame_C18a Page 395 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC The ply thickness being 0.13 mm, one obtains 108 layers oriented as follows: 6. The specific mass of the laminate is indicated in Section 3.3.3, as r = 1530 kg/m. 3 The distributed mass of the arm is then: with the angular acceleration indicated in Figure 18.8, one obtains the following inertial load: We then deduce from there: Ⅲ The deflection at the end due to the concentrated mass: Ⅲ The deflection at the end due to distributed load 19 : from which we can obtain a total deflection: The rigidity (F/D total ) appears to be well related essentially to the concentrated inertial load at the extremity of the arm. 19 Result obtained from the differential equation: m ᐉ 2 p re r ¥ 20.2 kg/m== D concent. 100ᐉ 3 3E x I = EI x d 2 v/dX 2 0.81 6 ᐉ 2 23X/ᐉ()– X/ᐉ() 3 +[]–= D distributed 11 120 0.81ᐉ 4 E x I ¥= D total 100ᐉ 3 3E x I 10.033+() # 100ᐉ 3 3E x I = TX846_Frame_C18a Page 396 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC 18.2 LEVEL 2 18.2.1 Sandwich Beam: Simplified Calculation of the Shear Coefficient Problem Statement: Represented below is the cross section of a sandwich beam. The thickness of the skins is small compared with that of the core. Under the action of a shear load T, the shear stresses in the section are assumed to vary in a piecewise-linear fashion 20 along the y direction. The constitutive materials, denoted as 1 and 2, are assumed to be isotropic, or transversely isotropic. The shear moduli are denoted as G 1 for material 1 (skin) and G 2 for material 2 (core). The beam has a width of unity. 1. Calculate the shear coefficient k for flexure in the plane x,y. 2. Give a simplified expression for the case—current in the applications— where G 1 >> G 2 with the notations for the thicknesses: Solution: 1. Let W be the strain energy due to shear stresses. One has (Equation 15.17): In the upper skin, one has 20 This representation of the shear stresses is only approximate. One will find in Application 18.3.5 the results concerning a more precise distribution of these stresses. In fact, the approximate representation of the shear proposed here is better approximated than the skins of the sandwich structure will have a small thickness as compared to that of the core. e p H 1 H 2 – 2 ; e c H 2 == dW dx 1 2 kT 2 GS·Ò 1 2 t xy 2 G i yd section Ú == t xy H 1 2y– H 1 H 2 – t o ¥= TX846_Frame_C18b Page 397 Monday, November 18, 2002 12:43 PM © 2003 by CRC Press LLC On the other hand in the core: t xy = t 0 Then with: One deduces from there the maximum shear stress t o : Strain energy: After calculation: one then has Then: with (Equation 15.16): ͗GS͘ = G 1 (H 1 – H 2 ) + G 2 H 2 : 2. Case where G 2 << G 1 : One can rewrite then: T t xy y 1¥d() section Ú = t o T 2 H 1 H 2 + ¥= dW dx 1 2 t xy 2 Gi yd Ú t o 2 G 2 y t o 2 G 1 H 1 2y–() 2 H 1 H 2 –() 2 yd H 2 /2 H 1 /2 Ú +d o H 2 /2 Ú == 1 2 t xy 2 G i yd Ú t o 2 2 H 2 G 2 H 1 H 2 – 3G 1 + ˯ ʈ 2T 2 H 1 H 2 +() 2 H 2 G 2 H 1 H 2 – 3G 1 + ˯ ʈ == 1 2 kT 2 GS·Ò 2T 2 H 1 H 2 +() 2 H 2 G 2 H 1 H 2 – 3G 1 + ˯ ʈ = k 4 GS·Ò H 1 H 2 +() 2 H 2 G 2 H 1 H 2 – 3G 1 + ˯ ʈ = k 4 G 1 H 1 H 2 –()G 2 H 2 +[] H 1 H 2 +() 2 H 2 G 2 H 1 H 2 – 3G 1 + ˯ ʈ = k 4 GS·Ò e c 2e p e c ++() 2 e c G c ¥ 1 2 3 e p G c e c G p += << 1 k # GS·Ò e c 2 1 e p e c + ˯ ʈ 2 e c G c ¥ TX846_Frame_C18b Page 398 Monday, November 18, 2002 12:43 PM © 2003 by CRC Press LLC One obtains the following simplified form, valid if e p << e c and G c << G p : 18.2.2 Procedure for Calculation of a Laminate Problem Statement: Consider a balanced carbon/epoxy laminate with respect to the 0∞ direction (or x), having midplane symmetry. The plies have the orientations 0∞, 90∞, +45∞, –45∞ with a certain proportions (recall that there are as many plies of +45∞ as there are –45∞). This laminate is subjected to a unit uniaxial stress s ox = 1 MPa (see following figure). Propose a procedure to establish a simple program allowing one to obtain the following: 1. The modulus of elasticity E x of the laminate and the Poisson coefficient n xy 21 2. The stresses in each ply and in the orthotropic axes of this ply 22 3. The Hill–Tsai 23 expression for each ply 4. The largest stress s ox max admissible without failure of any ply One gives in the following the characteristics of the unidirectional plies (identical) making up the laminate: Carbon/epoxy ply with V f = 60% fiber volume fraction E ᐉ = 134,000 MPa 24 ; E t = 7000 MPa; G lt = 4200 MPa; n lt = 0.25 Fracture strengths: s ᐉ tension = 1,270 MPa; s ᐉ compression = 1130 MPa s t tension = 42 MPa; s t compression = 141 MPa t ᐉt = 63 MPa 21 See Equation 12.8. 22 These are the stresses s ᐉ , s t , t ᐉt (see, for example, Equation 11.1). 23 See Chapter 14. 24 See Section 3.3.3, Table 3.4. k GS·Ò 1 G c e c 2e p +() = TX846_Frame_C18b Page 399 Monday, November 18, 2002 12:43 PM © 2003 by CRC Press LLC Solution: Recall first the procedure for calculation (see also Section 12.1.3.): 1. Modulus E x and Poisson coefficient n xy : The behavior of a laminate having midplane symmetry and working in its plane can be written as (Equation 12.7): [a] with: where e k is thickness of ply k, and h is the total thickness of the laminate. is the stiffness matrix for the ply k in the x,y axes (see Equation 11.8), as: [b] Note that (%), (%), (%), (%) are the respective proportions of the plies in the directions 0∞, 90∞, +45∞, –45∞. The previous terms can be written as: [c] Here the terms , , and their symmetrical counterparts are zero because the laminate is balanced (see Equation 11.8). The relation denoted as [a] above is then inverted and can be written as: [d] s ox s oy t oxy Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ 1 h A 11 A 12 A 13 A 21 A 22 A 23 A 31 A 32 A 33 e ox e oy g oxy Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ = 1 h A ij E ij k e k h k 1 st ply= n th ply  = E ij [] k s x s y t xy Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ E 11 E 12 E 13 E 21 E 22 E 23 E 31 E 32 E 33 e ox e oy g oxy Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ = ply k ply k ply k p 0∞ p 90∞ p 45∞ p 45∞– 1/h()A ij 1 h A ij E ij 0∞ p 0∞ E ij 90∞ p 90∞ E ij 45∞ p 45∞ E ij 45– ∞ p 45∞– +++= 1 h A 13 1 h A 23 e ox e oy g oxy Ó ˛ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ Ï¸ 1 E x - n yx E y 0 - n xy E x 1 E y 0 00 1 G xy s ox s oy t oxy Ó ˛ ÔÔ ÔÔ Ì ˝ ÔÔ ÔÔ Ï¸ = TX846_Frame_C18b Page 400 Monday, November 18, 2002 12:43 PM © 2003 by CRC Press LLC where , , , , are the global moduli and Poisson coefficients of the laminate. Here this laminate is subjected to a uniaxial stress s ox = 1 MPa, then: One obtains as well the modulus and the Poisson coefficient required: 2. Stresses in the ply: The previous result gives us the global strains of the laminate, strains that each ply should follow as: For a ply k, the relation mentioned above in [b] is then written as: [e] This gives the stresses in ply k, expressed in the coordinates x, y. One can express them in the orthotropic axes of the ply (axes ᐉ, t of the following figure, and Equation 11.4 recalled below): [f] E x E y G xy n xy n yx e ox s ox E x 1 MPa E x MPa() ; e oy n xy E x s ox – n xy E x MPa() 1 MPa¥–== = = E x MPa() 1 e ox = n xy e – oy E x 1 MPa() MPa ¥= e ox 1 E x s ox ; e oy n xy E x s ox ; g oxy 0== = s x s y t xy Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ E 11 E 12 E 13 E 21 E 22 E 23 E 31 E 32 E 33 e ox e oy 0 Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ = ply k ply k s ᐉ s t t ᐉt Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ c 2 s 2 2cs– s 2 c 2 2cs sc sc– c 2 s 2 –() s x s y t xy Ó ˛ ÔÔ Ì ˝ ÔÔ Ï¸ c q cos= s q sin= = ply k ply k TX846_Frame_C18b Page 401 Monday, November 18, 2002 12:43 PM © 2003 by CRC Press LLC 3. Hill-Tsai expression: It is written as (Equation 14.7): One can then calculate the values ( a 2 ) k required for each ply k. 4. The largest stress s ox max without fracture: The stresses s ᐉ , s t , and t ᐉt are calculated for a uniaxial stress: s ox = 1 MPa. Now apply the maximum stress found s ox max (MPa). The stresses s ᐉ , s t , and t ᐉt in the ply k are multiplied by the ratio: and the critical value of the Hill-Tsai expression is obtained as: With the values ( a 2 ) k found in the previous question for the Hill-Tsai expression between brackets, one obtains Then: Examination of each ply will lead to a different value for s ox max . One has to keep the minimum value as the critical stress that should initialize damage (failure of a ply) as: 18.2.3 Kevlar/Epoxy Laminates: Evolution of Stiffness Depending on the Direction of the Load Problem Statement: Consider the balanced laminates of Kevlar/epoxy with V f = 60% fiber volume fraction, working in their planes, with the following compositions: a 2 s ᐉ 2 s ᐉ 2 s t 2 s t 2 s ᐉ s t s ᐉ 2 t ᐉt 2 t ᐉt 2 +–+= rupture rupture rupture rupture s ox max 1 MPa s ox max 2 1 MPa() 2 s ᐉ 2 s ᐉ 2 s t 2 s t 2 s ᐉ s t s ᐉ 2 t ᐉt 2 t ᐉt 2 +–+ Ó ˛ Ì ˝ ϸ k 1= rupture rupture rupture rupture s ox max 2 a k 2 1MPa() 2 = s ox max 1 MPa a k = s ox max min 1 a k = TX846_Frame_C18b Page 402 Monday, November 18, 2002 12:43 PM © 2003 by CRC Press LLC [...]... (aEh)y, and h (aEh)xy from the Equation 12.18 This calculation requires knowledge of the terms a E 1 , a E 2 , and a E 3 of each ply (Equations 12.17 and 11.10 and numerical values in Tables 1.3 and 1.4 of Section 1.6) For that, one has to know the coefficients of expansion aᐉ and at of a ply in its principal axes (ᐉ,t) It can be written (Equations 10.7 and 10.8 and numerical values in Tables 1.3 and 1.4... in its principal axes (ᐉ,t); one has (Equation 10.2 and those that follow and numerical values in Tables 1.3 and 1.4 in Section 1.6.) Eᐉ = 74,000 ¥ 0.25 + 4,000 ¥ 0.75 = 21,500 MPa nᐉt = 0.25 ¥ 0.25 + 0.4 ¥ 0.75 = 0.36 1 E t = 4000 - = 5240 MPa 4000 0.75 + ¥ 0.25 74,000 1 G ᐉt = 140 0 = 1840 MPa 140 0 0.75 + - ¥ 0.25 30,000 n tᐉ = ( 5240/21,500... E ᐉ and The moduli of elasticity and coefficients of expansion are given in Section 27 3.3.3, Table 3.4 Then: 90∞ a E1 = 0.237 With the known values eox and eoy, one has sx = 7021 ¥ 115 ¥ 10 -6 -6 + 1717 ¥ (–704 ¥ 10 ) – (–160)(0.237) = 37.5 MPa In an analogous manner: 90∞ 90∞ 90∞ s y = E 21 e ox + E 22 e oy – DT a E 2 27 Recall also the property n tᐉ / E t = n ᐉᒑ / E ᐉ (see Sections 3.1 and 3.2 and. .. 12:43 PM Problem Statement: The thicknesses e1 and e2 are small relative to the average radius of the tube, denoted as r 1 Give the numerical values of E2 and n2 (noting that the moduli of elasticity of epoxy resins and polyester resins are equivalent) 2 When taking into account the temperature variation, denoted as DT, the mechanical behavior of the polymer and of the reinforcement, respectively, can... 56,600¯ 56,600 6 8030 Ⅲ Laminate (c): The proportions of 25% along the directions 0∞ and 90∞ can be obtained from Table 5.9 In this view one has to evaluate by extrapolation, 26 starting from the values corresponding to the percentages of 20% and 30%, as : Ex = (1/2) (28,260 + 35,400) = 31,830 MPa 26 See also Application 18.2 .14 © 2003 by CRC Press LLC TX846_Frame_C18b Page 405 Monday, November 18, 2002... –45∞ 4 For the other materials, one obtains immediately (Section 1.6): 6 2 E/r (steel) = 26.3 ¥ 10 (m/s) 6 2 E/r (Duralumin-2024) = 26.8 ¥ 10 (m/s) 6 2 E/r (Titanium-TA6V) = 23.9 ¥ 10 (m/s) Remark: The notion of specific modulus is particularly important for aeronautical construction When one compares on the above diagram the performances of Kevlar/epoxy with those of steel, Duralumin, and titanium, one... specific modulus with q for each of the laminates 4 Compare with the specific moduli of conventional materials, steel, aluminum alloys Duralumin-2024, and titanium alloy TA6V Solution: Each of the balanced laminates constitutes a thin plate of orthotropic material, with orthotropic axes x, y, z (see figures above and below) The constitutive relation corresponds with Equation 12.9 For a balanced laminate, this... Ï ¸ Ô Ô 1 Ô Ô Ô Ô ˝ + a 2 DT Ì 1 ˝ Ô Ô Ô Ô Ô 0 Ô ˛ Ó ˛ 1 n – 2 E 2 1 -E2 0 2 where one can recognize the strains and stresses in each of the materials Starting with an assembly (polymer + reinforcement) not stressed nor strained at ambient temperature (20∞C), which is heated up to 140 ∞C (a) Write the equations for the external equilibrium of the assemblage (b) Write the equality of the strains Deduce... components (polymer and glass/polyester reinforcement) as well as their strains 3 Being subjected to high temperature, the internal tube in polymer obeys creep law The stresses calculated previously do not remain constant in time They evolve and stabilize at a certain final state When this state is achieved, if one separates the internal polymer envelope (by imagination) from its reinforcement and cools it... at 140 ∞C already crept, one cools the whole reinforced tube quickly, from 140 ∞C to 20∞C Calculate the final stresses in the assembly, denoted as s ¢¢ , s ¢¢ , s ¢¢ , s ¢¢ at the end of the cooling 1x 1y 2x 2y Remark Solution: 1 We will use for the elastic characteristics of a unidirectional ply of glass/ polyester at Vf = 0.6 those of a glass/epoxy ply from Table 3.4 For a laminate at ±45∞, Table 5.14 . h -1 ( a Eh) y , and h -1 ( a Eh) xy from the Equation 12.18. This calculation requires knowledge of the terms , , and of each ply (Equations 12.17 and 11.10 and numerical values in Tables 1.3 and 1.4. the coefficients of expansion a ᐉ and a t of a ply in its principal axes (ᐉ,t). It can be written (Equations 10.7 and 10.8 and numerical values in Tables 1.3 and 1.4 of Section 1.6): From which. as: [c] Here the terms , , and their symmetrical counterparts are zero because the laminate is balanced (see Equation 11.8). The relation denoted as [a] above is then inverted and can be written as: [d] s ox s oy t oxy Ó