Plastics Engineered Product Design Episode 5 pps

140 Plastics Engineered Product Design in orthotropic materials, with one direction parallel, and one at right angles to the fibers. Multilayer plates, in which layers of fabric or of roving are laid up parallel or perpendicular to each other, are also orthotropic. If the same number of strands or yarns is found in each principal direction (balanced construction), the strength and elastic properties are the same in those directions but not at intermediate angles; if the number of strands or yarns is different in the two principal directions (unbalanced construction), the strength and elastic properties are different in those directions as well as at all intermediate angles. In the foregoing discussion the direction perpendicular to the plane of the plate has been neglected because the plate is assumed to be thin and the stresses are assumed to be applied in the plane of the plate rather than perpendicular to it. This assumption, which considerably simplifies the theory, carries through all of the following discussion. It is true, of course, that properties perpendicular to the plane of the plate are undoubtedly different than in the plane of the plate, and in thick plates this difference has to be taken into account, particularly when stresses are not planar. For isotropic materials, such as mat-reinforced construction, if E is the modulus of elasticity in any reference direction, the modulus El at any angle to this direction is the same, and the ratio El/E is therefore unity. Poisson’s ratio v is similarly a constant in all directions, and the shearing modulus G = €/2 (1 + v). If v, for example, is 0.3, G/E = 0.385 at all angles. These relationships are shown in Fig. 2.33. The following familiar relationships between direct stress (r and strain, E, and shearing stress z and strain y hold: E = o/f (2-88) y= z/G (2-89) A transverse strain (contraction or dilation) is caused by CT equal to &J = -v& (2-90) For orthotropic materials, such as fabric and roving-reinforced construction, EL and ET are the elastic moduli in the longitudinal (L) and transverse (T) directions, GLT is the shearing modulus associated with these directions, vLT~s the Poisson’s ratio giving the transverse strain caused by a strett in the longitudinal direction, and vLT is Poisson’s ratio giving the longitudinal strain caused by a stress in the transverse direction. The modulus at any intermediate angle is El, and if 01 is a stress applied in the 1 -direction at an angle a with a longitudinal direction (Fig. 2.34, top), the stress o1 causes a strain 2 - Design Optimization 141 1.00 0.90 0.80- 0.70 cs 0.60 E 6 0.50 0.40 0.30 020 0.10 W 0- Modulus of elasticity, shear modulus, and Poisson's ratio for isotropic material such as mat-reinforced plastics - - b 3 - 3 - - - - q" - - 0.70 0.60 V i 0.30 0.10 o,zol Degrees Elastic constants of unbalanced orthotropic material E,, GIB and vIz are all functions of the angle between direction of stress and the longitudinal axis (warp direction) of the material Factors ml and m2 account for direct and shear strains caused by shear and direct stresses, respectively Angle 00 is longitudinal direction and angle 900 is transverse direction 0.50 0.40 0.30 0.10 0 -0.10 0.20 3 142 Plastics Engineered Product Design El = dEl (2-91) in which El may be found from (2-92) This relationship is plotted as E1/EL in Fig. 2.34, in which 0" corres- ponds to the longitudinal direction and 90" to the transverse direction. A transverse strain E2 = V12E1 is caused by o1 In which (Fig. 2.34) (2-93) (2-94) Unlike isotropic materials, stress q, when applied at any angle except 0" and 90", causes shear distortion and the shear strain y12 is found from ri2 = mldEL (2-95) in which (Fig. 2.34) EL EL )} (2-96) ( ET GLT - - - - -cos2 a 1 + 2vLT+ - - - EL 1 EL m, =sin 2 a vLT+ I ET 2 GLT A shearing stress n12 applied in the 1-2 directions causes a shear strain Yl2 ri2 = T121G12 (2-97) in which (Fig. 2.34) This relationship is plotted as G12/GLTin Fig. 2.34. Unlike isotropic materials, stress r12 causes a strain g1 in the 1-direction = - mlzl 2L /E (2-99) and a strain in the 2-direction E~ = - m2T1 2L /E in which (Fig. 2.34) (2- 100) E 1 EL -sin2 a 1 +2vLT+ EL - )} (2-101) ET GLT 2 - Design Optimization r I I 11) I I I I I I - - - IIlfllll~l' 0 10 20 30 40 50 60 70 80 90 100 143 +2.00 . 8 2 c) +1.w Figure 2.35 Elastic constants of balanced orthotropic material. Constants and angles have same meaning as previous figure -1.00 t- (2- 102) The two values of Poisson's ratio are related : VLT/ VLT = WET In plotting Fig. 2.34 the following values were used: EL = 5,000,000 psi ET = 500,000 psi GLT = 550,000 psi vLT = v,. = 0.450 VTL = v9p = 0.045 These values, for example, might correspond to a parallel glass filament reinforced panel employing an intermediate polyester resin. When the orthotropic material is balanced, the longitudinal and transverse properties are the same, that is, EL = ET and VLT = VLT= VTL. The properties are symmetrical about the 45" angle, as shown in Fig. 2.35, in which the following values were used: EL = ET = 3,000,000 psi GLT = 500,000 psi VLT= VTL = 0.20 144 Plastics Engineered Product Design These values might correspond, for example, to a square-weave or sym- metrical satin-weave fabric-reinforced construction. As an example of the application of the foregoing equations, the tensile stress ol acting on the small plate at the top of Fig. 2.34 is 10,000 psi, the shear stress z12 is 4000 psi, and the angle a is 30". Then from Fig. 2.34, El& = 0.367 or f1 = 0.367 x 5,000,000 = 1,830,000 psi G12/GLT = 0.81 or G12 = 0.81 x 550,000 = 445,000 psi v12 = -0.0286 m, = 4.66 m2 = 4.98 Then, strains caused by q are E1 &2 = - (-0.0286) 5.45 x = 0.1 6 x x2 = -4.66 X 10,000/5,000,000 = -9.32 x = 10,000/1,830,000 = 5.45 x and strains caused by z12 are x2 = 4,000/550,000 = 7.28 x &1 = -4.66 x 4,000/5,000,000 = -3.73 x &2 = -4.98 X 4,000/5,000,000 = -3.98 x Total strains, therefore, are x2 = -2.04~ 10-3 = 1.72 x 10-3 = -3.82 x 10-3 (2-103) (2-104) (2- 1 05) (2- 1 06) (2- 107) (2-108) Problems involving Fig. 2.35 can be solved in an analogous manner. It must be kept in mind that Eqs. 2-92,2-94,2-96,2-98, and 2-101 are valid and useful if the fibers and the resin behave together in accordance with the assumptions upon which their derivation is based. If only the values of Eo En G,, and vLT are available, the intermediate values of E,, G12, v12, and the values of m1 and m2 can be estimated by means of these equations. Composite Plates Fibrous reinforced plates in practice are ofien made up of several layers, and the individual layers may be of different construction, such as mat, fabric, or roving. Furthermore, the various layers may be oriented at different angles with respect to each other in order to provide the best combination to resist some particular loading condition. Outside loads or stresses applied to a composite plate of this typc result in internal stresses which are different in the individual layers. External direct 2 - Design Optimization 145 Composite panel with layers a and b of different orthotropic materials oriented at arbitrary angles ci and p with respect to applied stresses ol, o2 and y,* stresses may result not only in internal direct stresses but in internal shear stresses, and external shear stresses may result in internal direct stresses as well as internal shear stresses. Fig. 2.36 depicts a small composite plate made up of materials a and b having principal longitudinal and transverse directions L, and T,, and Lb and Tb, respectively. Several layers of each are present but their total thicknesses are t, and tb, respectively, and the overall thickness is t. Outside stresses q, 02, and r12 are applied in the 1 and 2 directions, as shown. The 1-direction makes an angle a with La3 and a reverse angle p with Lb. The angle a is considered to be positive and the angle p negative. The internal stresses o,,, 02, r12a, and 016, 026, r12b in the individual layers can be found by observing that the sums of the internal stresses in the 1 and 2 directions must equal the external stresses in these directions, and that the strains must be the same in all layers. These relationships may be written in the following forms: (2-109) (2-110) (2-111) 146 Plastics Engineered Product Design &2,3 = &2b = &2 (2- 1 1 3) fi2o=fi2b= fi2 (2-114) Strains and stresses are induced in each layer. Because the layers are firmly bonded together the strains are the same in the LZ and b layers, and are equal to the strains in the whole plate: (2-115) (2-11 6) (2-117) Solution of the foregoing Eqs. 2-109 to 2-117 leads to the following simultaneous equations: Allola + A12°2a A13Z12a = - taib (z - V21b 9 - rnlbE '12) (2-118) 26 Lb A31010 f A3202a f A3357120 = - -mIb 01 -m2b 2 + - Tl2) (2-120) '( tb Lb G12b in which: All=-+- 1 1 A %Z-V2lb A - rnla rnlb €Lata €Lbtb 13 - €2ata EZbtb 12 - €lata €lbtb A %-V12b A,,=-+- 1 1 A !?k-m26 €Lata €Lbtb 23 - €lata €lbtb €20~0 E2btb 21 - where AZl = A12, numerically. 1 1 A33 = - + - G120ta G12btb 2 - Design Optimization 147 2.37 Fibrous glass-reinforced plastic thin-wall cylinder. (a) internal pressure alone and (b) internal pressure plus twisting moment '2 c1= 19,200 psi B (0) p = 800 psi ri = 5.00 in. rg = 4.80 in. Example: The application of the foregoing expressions may be illustrated by a cylindrical pressure vessel as shown in Fig. 2.37a. The wall of this vessel, having an external radius of 5 in., and wall thickness of 0.20in., may be considered to be a thin plate. It is subjected to an internal pressure of 800 psi. the circumferential stress crl and the longitudinal stress o2 in the wall are calculated pro = 19,200 psi (32 = 2t pro = 9,600 psi O' = -i- The stresses acting on a small part of the wall are therefore as shown in Fig. 2.37a. 148 Plastics Engineered Product Design Three types of construction will be investigated as shown in Fig. 2.37 (l), (2), (3). All three employ the balanced fabric having the characteristics shown in Fig. 2.35. In (1) the fabric is simply wrapped in layers a and b with the L and Tdirections laid in the circumferential and axial directions. In (2) the layers are laid at 45" to the axis of the cylinder, and in (3) they are laid at alternate 30" angles in left-hand and right-hand spirals as shown. In each instance ta = tb = 0.10 in. Referring to Fig. 2.35, it is seen that for Case 1 €1, = flb = €20 = €26 = 3 x lo6 psi = V2la = = 0.20 m,,= mlb= m2,= m2b = 0 All = A22, A12 = A21 A18 = A31 = A32 = A23 = 0 Eqs. 2-118 to 2-120 therefore become A33Z120 = Solution of these equations and reference to Eqs. 2-109 to 2-1 11 show that 01, = 01b = 01 = 19,200 psi 020 = 02b = 02 = 9,600 psi 2120 = 212b = 0 This proves what might have been expected intuitively; because of symmetry with respect to the 1-2 directions chosen; the internal direct stresses o1, olb, Oza, and 02b are equal to the imposed stresses ol and 02, and there is no internal shear stress. The same result is found for Case 2. In this balanced fabric ml = m2 = 0 at 45", there is no shear distortion caused by direct stress, and shear therefore is zero. In Case 3: G12, = G12b = 1.82 X 0.5 X lo6 = 0.91 X lo6 VI20 = V12b = vsp = v6p = v210 = v126=0.523 m,, = m80" = 0.775, = mlb = -mla = -0.775 m2, = m60' = 0.775, = m2b = -m2, = -0.775 The values of mlb and m2b are negative because the 30" angles of orientation of the longitudinal direction Lb of layers b is measured in the negative direction whereas it is positive for the a-layers. 2 - Design Optimization 149 Equations 22 become The first two of these equations are exactly like the first two equations for Cases 1 and 2 and show that the internal direct stresses are equal to the imposed, that is 01 02, = qb = 02 = = 01 b = 01 = 19,200 psi 9,600 psi The third equation, however, is not equal to zero, and its solution, together with equation 19c, shows that T~~~ = 6750 psi 2126 = 6750 psi Appreciable shear stresses are set up within the body of the cylinder wall when layers are oriented as in Case 3, even though no shear forces are applied to the cylinder itself. The shear stresses in layers b are oriented in the direction opposite to the shear stresses in layers a. The difference in the shear stresses between the two layers must be taken up by shear in the adhesive bond between them, that is, in the layer of resin that holds the fiber-reinforced layers together. The difference is 6750 - ( - 6750) = 13,500 psi This shear stress in the resin bonding the layers together is therefore seen to be high. In Cases 1 and 2 the orientation of the fibers with respect to the 1-2 directions chosen resulted in zero shear stresses associated with those directions, whereas in Case 3 the shear stresses were not zero. In all three cases, symmetry of the fiber orientations with respect to the stress directions resulted in internal stresses equal to the external stresses. These are special cases. In the more general case the internal direct stresses in the individual layers are not necessarily equal to the external direct stresses, nor are they the same in the various layers. Furthermore, even symmetrical Case 3 leads to internal shear stresses when external shear stresses are absent. In the more general case it is still more true [...]... 2.94 G12a = 0.93 x 0 .5 x I O 6 = 0.4 65 x I O 6 psi 0.07 in 45" = € 6 = 0 .52 6 x 3 x lo6 2 = 1 .57 8 x I O 6 psi = V2lb = 0 .57 9 mlb = m2b = 0 G 2 = 2 .5 ~ 0 x I O 6 1b 5 =1. 250 x I O 6 psi tb = a Elb = Solving for the various constants and substituting in Eqs 2-118 to 2-120 11.241201, -5. 6641 01, -4.04610ib - + - 5. 664102, - 4.046162, = +190,180 23.203002, - 4 .51 56 212, = -21 ,I 50 4 .51 5602, + 26.4668212,... x 5 x 106 0.2 85 0.2 85 x IO6 1 150 psi 0.2 x 5 x 106 0.2 85 + 0.1 x 3 x 106 0.326 x IO6 13 15 psi 0.1 35 0.2 x 5 x 106 Plan e 0.2 15 + 0.1 x 3 x 106 0.324 x IO6 1310 psi 0.3 65 EjAj These would be the critical planes because they represent planes between layers of different materials, and consequently the resin alone would largely carry the stress The shear stress at the neutral axis would 154 Plastics Engineered. .. M = 3,600 in.-lb 7,800 in.-lb 10,200 in.-lb 7,600 in.-lb 4,400 in.-lb 3 ,50 0 in.-lb o,,/in.lb 0-0 b-b c-c d-d e-e f- f 0.3 85 in 0.1 85 in 0.0 85 in 0.1 15 in 0.3 15 in 0.41 5 in 5 x 106 3x 1x 1x 5x 3x 106 106 106 106 106 11.1 psi 3.19 psi 0.49 psi 0.66 psi 9.07 psi 7.1 6 psi 40,000/11.1 25, 000/3.19 5, 000/0.49 5, 000/0.66 40,000/9.07 25, 000/7.16 = = = = = If, for example, the beam were a simple beam carrying... in thick, and are oriented at 15" to the circumferential direction as shown Layers 6, of balanced material having the properties of Fig 2. 35, are a total of 0.07 in thick and are oriented at 45" as shown Referring to Fig 2. 35, the properties are found to be b- layers a- layers t, = 0.13 in a = 15' El, = 0.703 X 5~ I O 6 = 3 .51 5 x lo6 psi € , = 0.109 x 5 x 106 2 = 0 .54 5 x I O 6 psi ~ 7 2 , = 0.193 mi,... proposed design; (b) comparison of critical stress and/or deformation values with design criteria to ensure that the proposed design will satisfy product requirements and materials limitations; and (c) modification of the proposed design to obtain optimum satisfaction of product requirement For metallic materials, component design is usually strength limited so 166 Plastics Engineered Product Design. .. Plastic ABS Acetal homopolymer Acetal copolymer Nylon (0.2 wto/o) Polycarbonate Polymethyl methacrylate Poisson‘s ra tio Shear modulus MPa Shear stress 0. 35 0.36 0. 35 0. 35 0.34-0.43 0.37 0. 35 9 65 660 1340 1000 51 .2 30.0 65. 5 53 .0 66.4 41 .5 44.6 7 85 MPa Brittleness Brittleness identifies material easily broken, damaged, disrupted, cracked, and/or snapped Brittleness can result from different conditions... 3 Design Parameter 1 75 Figure 3.6 Tensile stress-strain behavior of ductile plastics Plastic ABS Aceta I homo po Iy m e r Acetal copolymer Acrylic Nylon Phenolic Polyethylene PoIy p ropy1e ne PoIy s ty re ne Polysulfone Modulus M Pa Yield stress MPa 2,700 3,100 55 69 61 72 82 62 30 35 25 70 2,800 3,000 2,400 19,300 1,200 1,400 3,100 2 ,50 0 Elongation h at yield, O Elongation a t break, Yo 2 .5 75 75. .. moment at the center 2 Design Optimization 153 Cross section of: (a) isotropic beam (b) composite beam made in layers o f different materials, and (c) composite beam having properties [where all E are x lo6 psi] E, = 5, E, = 3, E, = 1, E = 5, and E5 = 3; also (where ‘T are IO3 psi) ‘T, = 40, 0, 25, , ‘T, = 5, ( , = 40, and o5= 25 3 of the span would be WL/4 Setting this equal to 3 ,50 0 in.-lb gives the... 170 Plastics Engineered Product Design Stress-strai n behavior The information presented throughout this book is used in different loading equations As an example stress-strain data may guide the designer in the initial selection of a material Such data also permit a designer to specify either design stresses or strains safely within the proportional/elastic limit of the material However for certain products... o loads (Courtesy of Plastics FALLO) TENslLEMODULUS - FLEXURALCOMPRESSIVE - SHEARTORSION OTHERS - - -CREEP -FATIGUE -TORSION -RAPID -MOTION -COMBINED OTHERS SIRESSES 162 Plastics Engineered Product Design The behavior of materials (plastics, steels, etc.) under dynamic loads is important in certain mechanical analyses of design problems Unfortunately, sometimes the engineering design is based on the . b-b 1 0.2 x 5 x 106 0.2 85 0.2 85 x IO6 1 150 psi 131 5 psi 0.2 x 5 x 106 + 0.1 x 3 x 106 0.2 x 5 x 106 + 0.1 x 3 x 106 0.2 85 0.1 35 0.2 15 0.3 65 0.326 x IO6. 3 x lo6 = 3 .51 5 x lo6 psi = 0 .54 5 x IO6 psi mlb = m2b = 0 = 1 .57 8 x IO6 psi €2, = 0.109 x 5x 106 = V2lb = 0 .57 9 ~72, = 0.193 G12b = 2 .5 ~0 .5 x IO6 mi, =. the 45& quot; angle, as shown in Fig. 2. 35, in which the following values were used: EL = ET = 3,000,000 psi GLT = 50 0,000 psi VLT= VTL = 0.20 144 Plastics Engineered Product Design