Hi All..

Just wanted to know in the attached picture how would you know to use sin60 in the calculation when finding the force is there a method? why is in not cos60?

(..the square is meant to be a material with a 15mm crack)

Thank you

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- Nov 16th 2008, 09:58 AM #1
## using sin or cos in equ.

Hi All..

Just wanted to know in the attached picture how would you know to use sin60 in the calculation when finding the force is there a method? why is in not cos60?

(..the square is meant to be a material with a 15mm crack)

Thank you

- Nov 16th 2008, 10:08 PM #2

- Joined
- Dec 2007
- Posts
- 11

- Nov 17th 2008, 03:24 PM #3

- Nov 18th 2008, 03:28 AM #4
A large steel plate under uniform applied tensile stress contains a crack, which is remote from the edges of the plate. The crack is 15mm (2a) in length and is inclined at an angle 60degrees to the applied stress (I have labeled this F in the attached picture). The fracture toughness (K) of the steel is 23MPam^1/2. Calculate the fracture stress ($\displaystyle \sigma_f$) of the plate.

Y, the geometric factor is 1 as the crack is not influenced by anything else. a, the crack length is 7.5mm since the crack has two tips.

$\displaystyle K = Y\sigma_f\sqrt{\pi{a}}$

and the solution is

$\displaystyle \sigma_f = \frac{23\times10^6}{\sin60\sqrt{\pi\times7.5\times 10^{-3}}}$

Yeah sorry I didn't post the whole question before.. here it is.. so I wanted to know why we use sin60 in the answer?

Thank you