Question 10.1: Determine the programmed bandwidth for four rovings of 12 k ...
Determine the programmed bandwidth for four rovings of 12 k carbon fibers so as to obtain an average ply thickness of 0.6 mm. Assume the carbon filament diameter and fiber volume fraction as 7 µm and 0.6, respectively.
The cross-sectional area of filaments in one roving is estimated as
CSA_{fib}=\frac{12,000\times \pi\times 0.007^2}{4} =0.4618 \ mm^2The cross-sectional area of composite ply for four rovings is then given by
CSA_{com}=\frac{4\times 0.4618}{0.6}=3.0788 \ mm^2The programmed bandwidth required for an average ply thickness of 0.6 mm is obtained as
b=\frac{3.0788}{0.6}=5.13 \ mmRelated Answered Questions
Question: 10.2
Verified Answer:
The required programmed bandwidth is arrived at as...