The engineering constants for an orthotropic material are found to be Ej= 27.579 GPa E2=20.684 GPa E3=21.374 GPa V12 = 0.2 G12=41.368 GPa V23 = 0.4 G23=48.263 GPa V13 = 0.6 G31=13.790 GPa Find the stiffness matrix [C] and the compliance matrix [S] for the preceding orthotropic material.

Q11 Find the vertical displacement at B. AE = 2000 for all members. [Ans: 0.052] 3 A 4 B 5

Problem E8.3 Figure E8.4 shows a material chart with the two indices of Exercise E8.2 as axes. Identify and plot coupling lines for selecting materials for a column with F= 10° N and H=3m (the same conditions as above), and for a second column with F= 10³ N and H= 20m. M2 = Density* Price / (Modulus^0.5), ($/m³)/(MPa^0.5) 10- 0.1- Technical ceramics Zirconia Aluminium Silicon carbide Soda-lime glass Cast iron Carbon steel 1 Concrete Ti alloys CFRP Cement Polymers Composites GFRP Brick Granite Al alloys Bamboo Softwood Nontechnical ceramics Natural materials.... Hardwood TTTT PTFE PP Phenolics 103 Cu alloys PE Metals and alloys TTTT 10 10² M₁ = Density * Price/ Compressive strength, ($/m³)/(MPa) Pb alloys 104 MFA 16 105

What is the magnitude of only the x component of BG?