Question 6.S.P.2: A [(0/±30/90)2/45/20]s laminate is fabricated using a graphi...
A [(0/±30/90)_{2}/45/\overline{20}]_{s} laminate is fabricated using a graphite–epoxy material system. Each ply has a thickness of 0.125 mm. Determine the number of plies in the laminate, the total laminate thickness, and the z-coordinate of each ply interface.
An ordered listing of all fiber angles that appear in the laminate is as follows:
[\underset{\overset{\uparrow }{ply 1} }{0° } , 30° , -30° , 90° , 0° , 30° , -30° , 90° , 45° , \underset{\overset{\uparrow }{ply 10 \\(midplane)} }{20° }, 45° , 90° , -30° , 30° , 0° , 90° , -30° , 30° , \underset{\overset{\uparrow }{ply 19} }{0° } ]The laminate contains a total of 19 plies, the fiber angles appear symmetrically about the midplane of the laminate, and the midplane passes through the center of the 20° ply. Because all plies are made of the same composite material system, they all have the same thickness. The total laminate thickness is therefore t = 19 (0.125 mm)=2.375 mm.
Ply interface positions are:
z_{0}=-t/2=-1.1875 mm z_{1}=z_{0}+t_{1}=-1.0625 mm z_{2}=z_{1}+t_{2}=-0.9375 mm
z_{3}=z_{2}+t_{3}=-0.8125 mm z_{4}=z_{3}+t_{4}=-0.6875 mm z_{5}=z_{4}+t_{5}=-0.5625 mm
z_{6}=z_{5}+t_{6}=-0.4375 mm z_{7}=z_{6}+t_{7}=-0.3125 mm z_{8} = z_{7}+t_{8}=-0.1875 mm
z_{9}=z_{8}+t_{9}=-0.0625 mm z_{10}=z_{9}+t_{10}=0.0625 mm z_{11}=z_{10}+t_{11}=0.1875 mm
z_{12}=z_{11}+t_{12}=0.3125 mm z_{13}=z_{12}+t_{13}=0.4375 mm z_{14} = z_{13}+t_{14}=0.5625 mm
z_{15}=z_{14}+t_{15}=0.6875 mm z_{16}=z_{15}+t_{16}=0.8125 mm z_{17} = z_{16}+t_{17}=0.9375 mm
z_{18}=z_{17}+t_{18}=1.0625 mm z_{19} =z_{18}+t_{19}=1.1875 mm
Note that the total laminate thickness equals the difference between z_{19}and z_{0}, as expected: t=z_{19}-z_{0}=1.1875 mm-(-1.1875 mm)=2.375 mm.