Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer:
The value of critical length = 3.46 mm
The value of volume of fraction of fibers = 0.43
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
Put all the values in above equation we get
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
Put all the values in above formula we get
= 0.43
This is the value of volume of fraction of fibers.