Question

An experiment of fiber pull out test is performed to measure the interfacial shear strength between the fiber and the matrix (as shown in figure). The embedded length of fiber in the matrix is 65 mm and the fiber radius is 5 mm. If the maximum pull-out load is 750 N, what is the interfacial shear strength from the test? 354 kPa 734.6 kPa 183.65 kPa 367.3 kPa

Answer 1

Answer:

Ans will be 734.6 kPa

Explanation:

There is mistake in nptel content.

the actual formula is given in the photo.

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Answer 2

The correct answer is 367.3kPa

Given Data :

Fiber radius (r) = 5 mm = 0.005 m

Fiber diameter (d) = 10 mm = 0.01 m

Embedded Length (l) = 65 mm = 0.065 m

Load ( f) = 750 N

Formula to be used :

T is the interfacial Shear Strength

$T = \frac{f}{\pi \: d \: l}$

Putting the values in formula :

$\: \: \: \: T \: = \frac{ 750}{3.14 \times0.01 \times 0.065 } \\ T = 367466.921 \ \\ T = 367.3kpa$

Hence, the correct answer is 367.3 kPa

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