Answer:
The value of critical length = 3.46 mm
The value of volume of fraction of fibers = 0.43
Step-by-step explanation:
Given data
= 800 M pa
D = 0.017 mm
L = 2.3 mm
= 5500 M pa
= 18 M pa
= 13.5 M pa
(a) Critical fiber length is given by
![L_(c) = \sigma_(f) ((D)/(2 \sigma_(c) ) )](https://img.qammunity.org/2021/formulas/engineering/high-school/207uzxdy1jp45uv0sd7egjk4oaqgde403j.png)
Put all the values in above equation we get
![L_(c) =5500 ((0.017)/((2) (13.5)) )](https://img.qammunity.org/2021/formulas/engineering/high-school/40k99cbqwhu8iywl9ioq7s6ohsqrk5o8ow.png)
mm
This is the value of critical length.
(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by
![V_(f) = (\sigma_T - \sigma_m)/((L\sigma_c)/(D) - \sigma_m )](https://img.qammunity.org/2021/formulas/engineering/high-school/j7s8r6jzkrk2n9ysxo9lnpa0tmglaiux5v.png)
Put all the values in above formula we get
![V_(f) = (800-18)/(((2.3)(13.5))/(0.017) - 18 )](https://img.qammunity.org/2021/formulas/engineering/high-school/vroiso5czeq2mnz8iyxc0y1coi432lxzhf.png)
= 0.43
This is the value of volume of fraction of fibers.