1) The problem calculates the transformed stiffness matrix Q' for a material at an angle of 45 degrees and determines that (Q11)' = 13.612, (Q12)' = 6.034, and (Q22)' = 13.612. 2) It then calculates the stresses and strains in a beam with given dimensions, materials properties, and a applied load. It determines the normal strains are ε11 = 5x10^-6 and ε22 = -1.37x10^-7 and the shear strain is γ12 = 0. 3) It then transforms the strains to the global coordinate system using a transformation matrix and determines the transformed normal strains are εxx = εyy

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