hw1_solutions - AE 221 Aerospace Structures II Spring 2004...

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AE 221 – Aerospace Structures II – Spring 2004 Chapter 1 – Review – Solution To learn more about Matlab commands and their arguments, type help and then the command that you need help with. For example, if you want to find out what linspace does, enter help linspace. Go to http://www.cds.caltech.edu/~murray/courses/cds110/fa02/cds101/lectures/L1.3_matl ab-primer.pdf for a Matlab tutorial. 1. Since () 2 1 x stp is zero for 2 1 x < and unity for 2 1 x , + + + < + + = 2 1 x 4 x 6 x 3 x 2 1 x 4 x 6 x x f 2 3 3 for for ) ( . The commands below illustrate one of the ways to generate a plot in Matlab. >> x1=linspace(0,.5,51); >> y1=x1.^3+6*x1+4; >> x2=linspace(.5,1,51); >> y2=x2.^3+3*x2.^2+6*x2+4; >> plot(x1,y1,x2,y2) It can be seen that a discontinuity exists at 2 1 x = .
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2. >> x=linspace(0,1); >> y1=3*x.*sin(x); >> y2=3*x.*sin(x)+cos(x); >> y3=3*x.*sin(x)+2*cos(x); >> plot(x,y1,x,y2,x,y3) 3. First, we calculate symbolically the function g using the following commands. >> syms a L x >> f=3*x*sin(x)+a*cos(x); >> g=int(0.5*((diff(f,x))^2+f^2),x,0,L) g = 9/2*L-9/2*L*cos(L)^2+3/2*a*cos(L)*sin(L)-3/2*a*L+3/2*L^3+1/2*a^2*L Then we generate the corresponding plot in Matlab. << a=linspace(0,3); << L=ones(1,100); << g=9/2*L-9/2*L.*cos(L).^2+3/2*a.*cos(L).*sin(L)- 3/2*a.*L+3/2.*L.^3+1/2*a.^2.*L;
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<< plot(a,g) 4. We let = = = 2 3 2 1 4 2 2 2 L 4 L 2 0 b C C C x L 3 0 L 2 0 L 2 L 3 L 2 L 3 1 A , ,. >> syms C1 C2 C3 L >> A=[1 3*L 2*L^2;3*L 2*L^2 0;2*L^2 0 3*L^4]; >> x=[C1;C2;C3]; >> b=[0;2*L;4*L^2]; >> x=A\b x = [ 34/29] [ -22/29/L] [ 16/29/L^2] 5. >> w = dsolve('D4w=x+3','w(0)=0','Dw(0)=1','w(1)=0','D2w(1)=0','x') w = 1/120*x^5+1/8*x^4+3/20*x^3-77/60*x^2+x >> x=linspace(0,1); >> plot(x,1/120*x.^5+1/8*x.^4+3/20*x.^3-77/60*x.^2+x)
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6. >> [y,z]=meshgrid(-1:.2:3,0:pi/20:pi); >> q=2.*y.*z.*sin(z)+2.*exp(abs(y)).*cos(5.*z); >> surf(y,z,q)
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7a) 3 1 1 1 33 22 11 ii = + + = + + = δ b) 3 jj ii ij ij = = = OR in expanded form 3 1 0 0 0 1 0 0 0 1 2 2 2 2 2 2 2 2 2 2 33 2 32 2 31 2 23 2 22 2 21 2 13 2 12 2 11 ij ij = + + + + + + + + = + + + + + + + + = c) In expanded form: ik jk ij 33 2 33 33 33 23 32 13 31 3 j j 3 12 32 13 22 12 12 11 2 j j 1 11 2 11 31 13 21 12 11 11 1 j j 1 0 = = = + + = = = + + = = = + + = M 8a) () ( ) jr kq kr jq jr kq kr jq jr kq kr jq jr kq kr jq jq rk kq rj jr qk kr qj jr kq kr jq jq ki ir kq ji ir jr ki iq kr ji iq jr kq kr jq jq ki kq ji ir jr ki kr ji iq jr kq kr jq ii kr kq ki jr jq ji ir iq ii iqr ijk 3 3 3 ε = = + + = + + = + = = det b) kr kr kr kr kr kr jj rk kj rj jr jk kr ii kr kr jj ki kj ji ir jr ki kr ji ij jr kj kr jj ii kr kj ki jr jj ji ir ij ii ijr ijk 2 3 3 6 3 3 = + + = + + = + = = det
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c) Using the result from part c, substituting r for k, we get 6 2 kk ijk ijk = = δ ε 9a) 6 3 3 mm ii = + = + b) ik kj ij a a = c) 0 jki ij = 10. We can use Cauchy’s equation j ij i n T σ = , where j n is the unit vector normal to the surface.
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This note was uploaded on 01/18/2010 for the course AE 322 taught by Professor Lambros during the Spring '04 term at University of Illinois at Urbana–Champaign.

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hw1_solutions - AE 221 Aerospace Structures II Spring 2004...

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