Solutions to Problems 215 Problem 3.2 Let W be the width of the cross-section in Fig.3.3(see book).Also,let Wf and Wm be the widths of the fiber and matrix,respectively. △W时/Wf △L/L vm = △Wm/Wm △L/L △w/=-6wAg △Wm=-vmwm△L △W=△W于+△Wm =-(6w1+wm） -(W+)兴 △W/W Γ△L/L =V时+vmvm where Wf/W=Vf and Wm/W=Vm.Then,we obtain: h2=vi2V时+vmVm Problem 3.3 Let W be the width of the cross-section in Fig.3.3 (see book).Also,let WI and Wm be the widths of the fiber and matrix,respectively.Also,from equilibrium,we have o2 =o2=a2. 时=贴=g "=02=E"e=Em△W" Wm Aws=wI o Wm △Wm=Em2 W=△W时+△W" W W
Solutions to Problems 215 Problem 3.2 Let W be the width of the cross-section in Fig. 3.3 (see book). Also, let Wf and W m be the widths of the fiber and matrix, respectively. νf 12 = −∆Wf /Wf ∆L/L νm = −∆W m/W m ∆L/L ∆Wf = −νf 12Wf ∆L L ∆W m = −νmW m ∆L L ∆W = ∆Wf + ∆W m = − νf 12Wf + νmW m ∆L L ∆W W = − νf 12 Wf W + νm W m W ∆L L −∆W/W ∆L/L = νf 12V f + νmV m where Wf /W = V f and W m/W = V m. Then, we obtain: ν12 = νf 12V f + νmV m Problem 3.3 Let W be the width of the cross-section in Fig. 3.3 (see book). Also, let Wf and W m be the widths of the fiber and matrix, respectively. Also, from equilibrium, we have σf 2 = σm 2 = σ2. σf 2 = σ2 = Ef 2 ε f 2 = Ef 2 ∆Wf Wf σm 2 = σ2 = Emεm 2 = Em ∆W m W m ∆Wf = Wf Ef 2 σ2 ∆W m = W m Em σ2 ε2 = ∆W W = ∆Wf + ∆W m W
216 Solutions to Problems (+) W W/W wm/W E2= E E'm 1 2二 1 vf Vm where wf/W=Vf and Wm/W=Vm. Problem 3.4 The following is a listing of the modified MATLAB function E2 called E2Modified.Note that this modified function is available with the M-files for the book on the CD-ROM that accompanies the book. function y E2Modified(Vf E2f Em,Eta,NU12f,NU21f,NUm,E1f,p) %E2Modified This function returns Young's modulus in the % transverse direction.Its input are nine values: Vf -fiber volume fraction 名 E2f- transverse Young's modulus of the fiber % Em Young's modulus of the matrix Eta stress-partitioning factor % NU12f -Poisson's ratio NU12 of the fiber NU21f Poisson's ratio NU21 of the fiber % NUm Poisson's ratio of the matrix % E1f longitudinal Young's modulus of the fiber % p parameter used to determine which equation to use: % p =1-use equation (3.4) % p=2-use equation (3.9) % p=3-use equation (3.10) % p=4 -use the modified formula using (3.23) Use the value zero for any argument not needed in the calculations. Vm 1 Vf; if p =1 y 1/(Vf/E2f Vm/Em) elseif p ==2 y 1/((Vf/E2f Eta*Vm/Em)/(Vf Eta*Vm)) elseif p ==3 deno E1f*Vf Em*Vm; etaf =(E1f*Vf ((1-NU12f*NU21f)*Em NUm*NU21f *Eif)*Vm)/deno;
216 Solutions to Problems = Wf Ef 2 + W m Em σ2 W ε2 = Wf /W Ef 2 + W m/W Em σ2 ε2 = 1 E2 σ2 1 E2 = V f Ef 2 + V m Em where Wf /W = V f and W m/W = V m. Problem 3.4 The following is a listing of the modified MATLAB function E2 called E2Modified. Note that this modified function is available with the M-files for the book on the CD-ROM that accompanies the book. function y = E2Modified(Vf,E2f,Em,Eta,NU12f,NU21f,NUm,E1f,p) %E2Modified This function returns Young’s modulus in the % transverse direction. Its input are nine values: % Vf - fiber volume fraction % E2f - transverse Young’s modulus of the fiber % Em - Young’s modulus of the matrix % Eta - stress-partitioning factor % NU12f - Poisson’s ratio NU12 of the fiber % NU21f - Poisson’s ratio NU21 of the fiber % NUm - Poisson’s ratio of the matrix % E1f - longitudinal Young’s modulus of the fiber % p - parameter used to determine which equation to use: % p = 1 - use equation (3.4) % p = 2 - use equation (3.9) % p = 3 - use equation (3.10) % p = 4 - use the modified formula using (3.23) % Use the value zero for any argument not needed % in the calculations. Vm = 1 - Vf; if p == 1 y = 1/(Vf/E2f + Vm/Em); elseif p == 2 y = 1/((Vf/E2f + Eta*Vm/Em)/(Vf + Eta*Vm)); elseif p == 3 deno = E1f*Vf + Em*Vm; etaf = (E1f*Vf + ((1-NU12f*NU21f)*Em + NUm*NU21f *E1f)*Vm)/deno;
Solutions to Problems 217 etam =(((1-NUm*NUm)*E1f -(1-NUm*NU12f)*Em)*Vf +Em*Vm)/deno; y =1/(etaf*Vf/E2f etam*Vm/Em); elseif p ==4 EmPrime Em/(1 NUm*NUm); y 1/(Vf/E2f Vm/EmPrime); end Problem 3.5 The transverse modulus E2 is calculated in GPa using the three different formulas with the MATLAB function E2 as follows.Note that the three values obtained are comparable and very close to each other. >>E2(0.65,14.8,3.45,0,0,0,0,0,1) ans 6.8791 >E2(0.65,14.8,3.45,0.5,0,0,0,0,2) ans 8.7169 >>E2(0.65,14.8,3.45,0,0.3,0.3,0.36,85.6,3) ans 7.6135 Problem 3.6 >y(1)=E2(0,14.8,3.45,0,0,0,0,0,1) y= 3.4500 >y(2)=E2(0.1,14.8,3.45,0,0,0,0,0,1) y= 3.45003.7366
Solutions to Problems 217 etam = (((1-NUm*NUm)*E1f - (1-NUm*NU12f)*Em)*Vf + Em*Vm)/deno; y = 1/(etaf*Vf/E2f + etam*Vm/Em); elseif p == 4 EmPrime = Em/(1 - NUm*NUm); y = 1/(Vf/E2f + Vm/EmPrime); end Problem 3.5 The transverse modulus E2 is calculated in GPa using the three different formulas with the MATLAB function E2 as follows. Note that the three values obtained are comparable and very close to each other. >> E2(0.65, 14.8, 3.45, 0, 0, 0, 0, 0, 1) ans = 6.8791 >> E2(0.65, 14.8, 3.45, 0.5, 0, 0, 0, 0, 2) ans = 8.7169 >> E2(0.65, 14.8, 3.45, 0, 0.3, 0.3, 0.36, 85.6, 3) ans = 7.6135 Problem 3.6 >> y(1) = E2(0, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 >> y(2) = E2(0.1, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366
218 Solutions to Problems >y(3)=E2(0.2,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.0750 >y(4)=E2(0.3,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.4809 >y(5)=E2(0.4,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.9766 >>y(6)=E2(0.5,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.97665.5956 >y(7)=E2(0.6,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.97665.59566.3905 >y(8)=E2(0.7,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.97665.59566.3905 7.4486 >y(9)=E2(0.8,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.97665.59566.3905 7.44868.9266
218 Solutions to Problems >> y(3) = E2(0.2, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 >> y(4) = E2(0.3, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 >> y(5) = E2(0.4, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 >> y(6) = E2(0.5, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 5.5956 >> y(7) = E2(0.6, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 5.5956 6.3905 >> y(8) = E2(0.7, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 5.5956 6.3905 7.4486 >> y(9) = E2(0.8, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 5.5956 6.3905 7.4486 8.9266
Solutions to Problems 219 >>y(10)=E2(0.9,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.97665.59566.3905 7.44868.926611.1363 >>y(11)=E2(1,14.8,3.45,0,0,0,0,0,1) y= 3.45003.73664.07504.48094.97665.59566.3905 7.44868.926611.136314.8000 >>z(1)=E2(0,14.8,3.45,0.4,0,0,0,0,2) z= 3.4500 >>z(2)=E2(0.1,14.8,3.45,0.4,0,0,0,0,2) 2= 3.45004.1402 >>z(3)=E2(0.2,14.8,3.45,0.4,0,0,0,0,2) Z= 3.45004.14024.8933 >z(4)=E2(0.3,14.8,3.45,0.4,0,0,0,0,2) z= 3.45004.14024.89335.7182 >>z(5)=E2(0.4,14.8,3.45,0.4,0,0,0,0,2) 2= 3.45004.14024.89335.71826.6258 >>z(6)=E2(0.5,14.8,3.45,0.4,0,0,0,0,2) Z= 3.45004.14024.89335.71826.62587.6290
Solutions to Problems 219 >> y(10) = E2(0.9, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 5.5956 6.3905 7.4486 8.9266 11.1363 >> y(11) = E2(1, 14.8, 3.45, 0, 0, 0, 0, 0, 1) y = 3.4500 3.7366 4.0750 4.4809 4.9766 5.5956 6.3905 7.4486 8.9266 11.1363 14.8000 >> z(1) = E2(0, 14.8, 3.45, 0.4, 0, 0, 0, 0, 2) z = 3.4500 >> z(2) = E2(0.1, 14.8, 3.45, 0.4, 0, 0, 0, 0, 2) z = 3.4500 4.1402 >> z(3) = E2(0.2, 14.8, 3.45, 0.4, 0, 0, 0, 0, 2) z = 3.4500 4.1402 4.8933 >> z(4) = E2(0.3, 14.8, 3.45, 0.4, 0, 0, 0, 0, 2) z = 3.4500 4.1402 4.8933 5.7182 >> z(5) = E2(0.4, 14.8, 3.45, 0.4, 0, 0, 0, 0, 2) z = 3.4500 4.1402 4.8933 5.7182 6.6258 >> z(6) = E2(0.5, 14.8, 3.45, 0.4, 0, 0, 0, 0, 2) z = 3.4500 4.1402 4.8933 5.7182 6.6258 7.6290