Cosine Rule

[Jkls]

Hello,

I'm having problems understanding the following

A = x sin²o + y cos²o

so

A = 1/2 (x +y) - 1/2 (x-y) cos(2o)

Help would be greatly appreciated.

 

[Deleted member 4993]

Hello,

I'm having problems understanding the following

A = x sin²o + y cos²o

so

A = 1/2 (x +y) - 1/2 (x-y) cos(2o)

Help would be greatly appreciated.

Does your problem look like:

\(\displaystyle A \ = \ X * cos(\theta) \ + \ Y * sin(\theta)\)

 

[Jkls]

I've attached a screen shot from the text showing exactly what it looks like.

 

[Deleted member 4993]

I've attached a screen shot from the text showing exactly what it looks like.

Okay, that is a stress-transformation problem. Simplify the right-hand-side of the equation - and you'll get the Left-hand-side.

start with:

cos(2Φ) = cos²(Φ) - sin²(Φ)

then

½(σ₁ + σ₃) - ½(σ₁ - σ₃)*cos(2Φ)

= ½(σ₁ + σ₃) - ½(σ₁ - σ₃)*[cos²(Φ) - sin²(Φ)]

and continue.....

 

[soroban]

Hello, Jkls!

Another approach . . . a bit longer than Subhotosh's.

\(\displaystyle \sigma_n \;=\;\sigma_1\sin^2\!\phi + \sigma_3\cos^2\!\phi\)

. . . . \(\displaystyle =\;\frac{1}{2}(\sigma_1 + \sigma_3) - \frac{1}{2}(\sigma_1-\sigma_3)\cos(2\phi) \quad\hdots\quad(5.3a)\)

I assume that \(\displaystyle (5.3a)\) is the equation number.


We have:

\(\displaystyle \sigma_n \;=\;\sigma_1\sin^2\!\phi + \sigma_3\cos^2\!\phi\)

. . ..\(\displaystyle =\;\sigma_1(1-\cos^2\!\phi) + \sigma_3\cos^2\!\phi\)

. . . \(\displaystyle =\;\sigma_1 - \sigma_1\cos^2\phi + \sigma_3\cos^2\!\phi\)

. . ..\(\displaystyle =\;\sigma_1 - (\sigma_1 - \sigma_3)\cos^2\!\phi \)

. . . \(\displaystyle =\;\sigma_1 - (\sigma_1-\sigma_3)\dfrac{1 + \cos2\phi}{2}\)

. . ..\(\displaystyle =\; \sigma_1 - \frac{1}{2}(\sigma_1 -\sigma_3) - \frac{1}{2}(\sigma_1-\sigma_3)\cos2\phi \)

. . ..\(\displaystyle =\;\sigma_1 - \frac{1}{2}\sigma_1 + \frac{1}{2}\sigma_3 - \frac{1}{2}(\sigma_1-\sigma_3)\cos2\phi\)

. . . \(\displaystyle =\;\frac{1}{2}\sigma_1 + \frac{1}{2}\sigma_3 - \frac{1}{2}(\sigma_1-\sigma_3)\cos2\phi\)

. . . \(\displaystyle =\;\frac{1}{2}(\sigma_1+\sigma_3) - \frac{1}{2}(\sigma_1-\sigma_3)\cos2\phi\)

 

[Jkls]

Thanks very much, those were very helpful. Much appreciated.

 

[Jkls]

If I can, could I please continue here with another question?

I'm having difficulty understanding how, in the denominator, AC = AB/cos(Ѳ)