PART IV APPLICATIONS We have grouped in this last part of the book exercises and examples for applica- tions.These have various objectives and different degrees of difficulties.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 should be used as validation tests for numerical models. 2003 by CRC Press LLC
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 difficulties. 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 should be used as validation tests for numerical models. TX846_Frame_C18a Page 341 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
18 APPLICATIONS 18.1 LEVEL 1 18.1.1 Simply Supported Sandwich Beam Problem Statement: 1.The following figure 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 deflection-denoted as A-of the beam under the action of the force F 0=10cm F=50 daN h=5mm 1) e=500 mm F=50 daN ep=2.5 mm 2) 瓶 ep ec=25 mm 安 2.We separate the beam of duralumin into two parts with equal thickness ep=2.5 mm,by imaginarily cutting the beam at its midplane.Each half is bonded to a 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 deflection caused by F denoted by A'.Compare with the value of A found in Part 1.(Take the shear modulus of the foam to be:G=20 MPa.) Solution: 1.We will use the classical formula that gives the deflection at the center of the beam on two supports: F △= 48E1 with I= 12 2003 by CRC Press LLC
18 APPLICATIONS 18.1 LEVEL 1 18.1.1 Simply Supported Sandwich Beam Problem Statement: 1. The following figure 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 deflection—denoted 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 ep = 2.5 mm, by imaginarily cutting the beam at its midplane. Each half is bonded to a 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 deflection 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.) Solution: 1. We will use the classical formula that gives the deflection at the center of the beam on two supports: D F 48EI ----------- with I bh3 12 = = -------- TX846_Frame_C18a Page 343 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
For duralumin (see Section 1.6):E=75,000 MPa.One finds △=16.7mm 2.Denoting by W the elastic energy due to flexure,one has' 高+高r heam with 1 (G两*G.(e.+2e,)×b Using Castigliano theorem,one has aF then: ∫尚=+∫高票 0≤x≤E/2:M=Fx/2;T=-F12 2≤x≤e:M=t-T=F2 r"高受×+nc- …+总-x-+× △'= Fe3 Fe k 8(ED+4(G两 Approximate calculation: (En=B,xe,xbxe+e2+E eb 2 12 then: (EI)=7090 MKS+7.8 MKS with E=60 MPa (cf.1.6) negligible TTo establish this relation,see Chapter 15,Equation 15.17. 2 See calculation of this coefficient in 18.2.1,and more precise calculation in 18.3.5. 2003 by CRC Press LLC
For duralumin (see Section 1.6): E = 75,000 MPa. One finds 2. Denoting by W the elastic energy due to flexure, one has1 with2 : Using Castigliano theorem, one has then: Approximate calculation: then: 1 To establish this relation, see Chapter 15, Equation 15.17. 2 See calculation of this coefficient in 18.2.1, and more precise calculation in 18.3.5. D = 16.7 mm W 1 2 -- M2 · Ò EI ----------- x 1 2 -- k · Ò GS ------------T 2 dx beam Ú d + beam Ú = k · Ò GS ------------ # 1 Gc ec + 2ep ( ) ¥ b --------------------------------------- D¢ ∂W ∂F = -------- D¢ M · Ò EI ----------- dM dF ------- x k · Ò GS ------------T dT dF ------ dx beam Ú d + beam Ú = 0 £ £ x /2: M = = Fx/2; T F– /2 /2 £ £ x : M F 2 = = --( ) – x ; T F /2 D¢ 1 · Ò EI ----------- Fx 2 ------ x 2 ¥ -- x F 2 --( ) – x ( ) – x 2 ----------------dxº /2 Ú d + 0 /2 ÚÓ ˛ Ì ˝ Ï ¸ = º k · Ò GS ------------- F 2 -- dx 2 – ¥ –------ F 2 -- dx 2 ¥ ------ /2 Ú + 0 /2 ÚÓ ˛ Ì ˝ Ï ¸ + D¢ F3 48 · Ò EI ----------------- F 4 ------ k · Ò GS = + ------------ · Ò EI # Ep ep b ec + ep ( )2 2 ---------------------- Ec ec 3 b 12 ¥ ¥ ¥ + ¥ ------- · Ò EI 7090 MKS 7.8 MKS negligible = = + with Ec 60 MPa ( ) cf. 1.6 TX846_Frame_C18a Page 344 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
one obtains for△'： △'=0.18mm+1.04mm bending shear moment △'=1.22mm Comparing with the deflection A found in Part 1 above: 14 Remarks: The sandwich configuration has allowed us to divide the deflection by 14 without significant augmentation of the mass:with adhesive film thickness 0.2 mm and a specific 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 corresponds to an increase of 14%with respect to the case of the full beam in Question 1. The deflection due to the shear energy term is close to 6 times more important than that due to the bending moment only.In the case of the full beam in question 1,this term is negligible.In effect one has: k=1.2 for a homogeneous beam of rectangular section,then: GS =8.27×108 (with G=29,000 MPa,Section 1.6).The contribution to the deflection A of the shear force is then: ∫会r乐=002mmxA 18.1.2 Poisson Coefficient of a Unidirectional Layer Problem Statement: Consider a unidirectional layer with thickness e as shown schematically in the figure below.The moduli of elasticity are denoted as E(longitudinal direction) and E,(transverse direction). 2003 by CRC Press LLC
one obtains for D¢: Comparing with the deflection D found in Part 1 above: Remarks: The sandwich configuration has allowed us to divide the deflection by 14 without significant augmentation of the mass: with adhesive film thickness 0.2 mm and a specific mass of 40 kg/m3 for the foam, one obtains a total mass of the sandwich: m = 700 g (duralumin) + 50 g (foam) + 48 g (adhesive) This corresponds to an increase of 14% with respect to the case of the full beam in Question 1. The deflection due to the shear energy term is close to 6 times more important than that due to the bending moment only. In the case of the full beam in question 1, this term is negligible. In effect one has: k = 1.2 for a homogeneous beam of rectangular section, then: (with G = 29,000 MPa, Section 1.6). The contribution to the deflection D of the shear force is then: 18.1.2 Poisson Coefficient of a Unidirectional Layer Problem Statement: Consider a unidirectional layer with thickness e as shown schematically in the figure below. The moduli of elasticity are denoted as E (longitudinal direction) and Et (transverse direction). D¢ = 0.18 mm 1.04 mm + bending shear moment D¢ = 1.22 mm D D¢ ----- 14 1 = ----- k GS ------ 8.27 10–8 = ¥ k GS ------ T dT dF ------dx Ú = 0.02 mm << D TX846_Frame_C18a Page 345 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC
Show that two distinct Poisson coefficients ve,and vie are necessary to charac- terize the elastic behavior of this unidirectional layer.Numerical application:a layer of glass/epoxy.V=60%fiber volume fraction. Solution: Let the plate be subjected to two steps of loading as follows: b 1.A uniform stress oe along the e direction:the changes in lengths of the sides can be written as: b 2.A uniform stress o,along the t direction:for a relatively important elongation of the resin,one can only observe a weak shortening of the fibers along e.Using then another notation for the Poisson coefficient,the change in length can be written as: b a Now calculating the accumulated elastic energy under the two loadings above: When o is applied first,and then o,is applied, 1 W=soxaxexAb+-ic,×bxexha,.+axaxex△b, When o,is applied first,and then o is applied, W'=30×bxexha2+50×a×e×△b,+o,×bxe×△@ The final energy is the same: W=W' 2003 by CRC Press LLC
Show that two distinct Poisson coefficients nt and nt are necessary to characterize the elastic behavior of this unidirectional layer. Numerical application: a layer of glass/epoxy. Vf = 60% fiber volume fraction. Solution: Let the plate be subjected to two steps of loading as follows: 1. A uniform stress s along the direction: the changes in lengths of the sides can be written as: 2. A uniform stress st along the t direction: for a relatively important elongation of the resin, one can only observe a weak shortening of the fibers along . Using then another notation for the Poisson coefficient, the change in length can be written as: Now calculating the accumulated elastic energy under the two loadings above: When s is applied first, and then st is applied, When st is applied first, and then s is applied, The final energy is the same: Db1 b -------- s E ----- ; Da1 a --------- nt E = = –------s Db2 b -------- nt Et –------st ; Da2 a --------- st Et = = ---- W 1 2 --s a e Db1 1 2 = ¥ ¥ ¥ + --st ¥ b e ¥ ¥ Da2 + s ¥ a e ¥ ¥ Db2 W¢ 1 2 --st b e Da2 1 2 = ¥ ¥ ¥ + --s ¥ a e ¥ ¥ Db1 + st ¥ b e ¥ ¥ Da1 W W= ¢ TX846_Frame_C18a Page 346 Monday, November 18, 2002 12:40 PM © 2003 by CRC Press LLC