Is the Complex Direct Stress Equation Correctly Simplified?

[Mechaman]

Attached image with problem.

2 Questions;

1.
δn.|AB| = δx.|BC|.cosΘ
becomes
δn = δx.|BC|.cosΘ / |AB|

This is dividing the entire "δx.|BC|.cosΘ" equation by |AB| or just the |BC| part? Is there a difference?

2.
Half way down page, boxed equation. How does δx.|BC|.cosΘ / |AB| = δxcos^2Θ and then 1/2(δx+δy)
Is this calculus? Should I find this in a log table I've been searching but haven't found where to look this up.

Thanks for any help.

 

[SteamKing]

Attached image with problem.
View attachment 89360
2 Questions;

1.
δn.|AB| = δx.|BC|.cosΘ
becomes
δn = δx.|BC|.cosΘ / |AB|

This is dividing the entire "δx.|BC|.cosΘ" equation by |AB| or just the |BC| part? Is there a difference?

It's simple trig. |BC| / |AB| = cos (θ)

2.
Half way down page, boxed equation. How does δx.|BC|.cosΘ / |AB| = δxcos^2Θ and then 1/2(δx+δy)
Is this calculus? Should I find this in a log table I've been searching but haven't found where to look this up.

A log table? No, it's not calculus, either. It has more to do with transforming coordinates and using trig identities than with anything exotic like log tables or calculus.

You can find a derivation for these formulas along with discussions on deriving Mohr's circle for plane stress.

https://www.uwgb.edu/dutchs/structge/mohrcirc.htm

The basic idea is to assume the wedge is in equilibrium and to calculate the forces which keep it in equilibrium from the stresses on the sides of the wedge.

 

[Mechaman]

Thanks I understand the first part now I have the double angle forumula.

I partly understand the second part until the last factoring:

δx(1+cos2θ)1/2 + δy(1-cos2θ)1/2
1/2(δx+δy) + 1/2(δx-δy)cos2θ

I know this type of factoring looks familiar its been a while since I've done this type of stuff.