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Hw solution 5
1.
Hw solution #5
: Problem 5.2: Given: Θ1= 45 Θ2= 45 Solution : From solved example in 5.1 Q= 20.874 3.260 0 3.260 11.898 0 0 0 3.789 At Θ=45 (Q11)’=Q11*cosΘ^4+2*(Q11+2Q66)sin^2Θ*cos^2(Θ)+Q22*sin^ 4(Θ) (Q11)’=8.193+5.419=13.612 (Q12)’=(Q11+Q22- 4Q66)sin^2(Θ)*cos^2(Θ)+Q12*(sin^4(Θ)+cos^4(Θ)) =6.034 (Q22)’=Q11*sin^4(Θ)+2(Q12+2Q66)sin^2(Θ)*cos^2(Θ)+Q22*sin ^4(Θ)
2.
(Q22)’=13.612 (Q16)’=2.509+(-.265)=2.244 (Q26)’=2.509-.265=2.244 (Q66)’=4.6685+1.8945=6.563 m^2 n^2 2mn T=
n^2 m^2 -2mn -mnmn m^2-n^2 T= 0.5 0.5 1 0.5 0.5 -1 -0.5 0.5 0 ____________________________________________ probem :5-5 : Given : W=10 mm t=2mm L=150 mm F=200 N Θ=10 E1=19.981 υ12=0.274 E2=11.389
3.
t=2 mm w=10 L=150 mm ϵ11
1/E1 -υ12/E1 0 ϵ22 = υ12/E1 1/E2 0 ɣ12 0 0 1/G12 σ11=F11/A=200/10*2*10^-3 = 10000 -υ12/E1=0.274/19.98*10^9=.0137*10^-9 1/E1=.05*10^-9 1/E2=.0878*10^-9 1/G12=.2639*10^-9
4.
ϵ11 .05 -.0137
0 ϵ22 = -.0137 .08787 0 *10^-9* ɣ12 0 0 0.2639 10000 0 0 ϵ11=10000*0.05*10^-9= 5*10^-6 ϵ22=-1.37*10^-7 ɣ12 =0 m=cosΘ n=sin Θ ϵxx m^2 n^2 -2mn ϵ11 ϵyy = n^2 m^2 2mn ϵ22 ϵxymn -mn m^2-n^2 ɣ12
5.
T^-1= 0.969 .0301
-.342 .0301 .969 .342 .171 -.171 .939 ϵxx=(5*10^-6*0.969)+.0301*10^-7*-1.37 ϵxx=4.84*10^-6 ϵyy=4.84*10^-6 ϵxy=0
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