Table Q1. Typical Values of Lamina Strengths for Several Composites at Room Temperature. Question 2 (25 marks) The filament-wound pressure vessel described in the Example on Page 46, Lecture Note 3 is fabricated from E-glass/470-36 E-glass/vinyl ester having the lamina strengths listed in Table Q1. Determine the internal pressure p, which would cause failure of the vessel according to (a) the maximum stress criterion and (b) the Tsai-Hill criterion. A filament-wound cylindrical pressure vessel of mean diameter d=1m and wall thickness t=20mm is subjected to an internal pressure, p. The filament-winding angle =53.1 from the longitudinal axis of the pressure vessel, and the glass/epoxy material has the following properties: E1=40 GPa,E2=10GPa,G12=3.5GPa, and v12=0.25. By the use of a strain gage, the normal strain along the fiber direction is determined to be 1= 0.001. Determine the internal pressure in the vessel. d=1m,t=20mm,=53.1E1=40GPa,E2=10GPa,G12=3.5GPa,andv12=0.251=0.001 Solution: From mechanics of materials, the stresses in a thin-walled cylindrical pressure vessel are given by x=2tpr=2(0.02)0.5p=12.5p,xy=0y=tpr=0.020.5p=25p (Note that r=d/2=0.5m. Since the given strain is along the fiber direction, we must transform the above stresses to the 12 axes by using the below equation: 1212=[T]xyxy[T]=c2s2css2c2cs2cs2csc2s2c=cosands=sin d=1m,t=20mm,=53.1E1=40GPa,E2=10GPa,G12=3.5GPa,andv12=0.251=0.001 Solution (cont'd): From the transformation, the stresses along the 12 axes are 1=xcos2+ysin2+2xysincos=(12.5p)(0.6)2+(25p)(0.8)2+0=20.5p(MPa)1212=[T]xyxy2=xsin2+ycos22xysincos=(12.5p)(0.8)2+(25p)(0.6)20=17.0p(MPa)[T]=c2s2css2c2cs2cs2csc2s212=xsincos+ysincos+xy(cos2sin2)c=cosands=sin=(12.5p)(0.8)(0.6)+(25p)(0.6)(0.8)+0=6.0p(MPa) d=1m,t=20mm,=53.1E1=40GPa,E2=10GPa,G12=3.5GPa,andv12=0.251=0.001 Solution (cont'd): From the stress-strain relationship in the 12 coordinates (principal coordinates) 1=E11E1v122=40(103)20.5p40(103)0.25(17.0p)=0.001 1212=E11E1v120E2v21E21000G1211212 Therefore, the resulting pressure is p=2.46MPa