The last part of your first equation should probably be
s cos^2q = s[(1+ cos 2q)/ 2]
That is just a trigonometric identity. The same applies to
sinq cosq = [sin 2q / 2]
I don't know what s' or t' mean, nor nor where you found that in The Theory of Elasticity. Are you referring tto the book by Timoshenko?
Using mechanics of materials principles, (i.e. equations of mechanical equilibrium applied to a free-body diagram), derive the following equations:
s’ = s cos2q = s[(1+ cos 2q) / 2]
t' = s sinq cosq = s [sin 2q / 2]
s=sigma, q=theta, t=tau
I have looked in "Theory of Elasticity" at their proof, but still cannot get the answer to be the above equations, please help!
2 answers
The above set of equation is a degenerated form of equation (13') in the Theory of Elasticity, namely:
σ=σx cos²α + σy sin²α
τ=(1/2)sin(2α)(σy-σx)
where
σx=principal stress in the x-direction
σy=principal stress in the y-direction
At this position, τxy=0 (no shear stress).
Setting σy=0 will yield the above equations.
Check if you have omitted a negative sign in the second equation.
Equation (13') applies to the situation of stresses measured at an angle α (≡q) from the principal directions, namely σ and τ.
The complete derivation is given on the page preceding equation (13'), chapter 2, section 9 "stress at a point".
σ=σx cos²α + σy sin²α
τ=(1/2)sin(2α)(σy-σx)
where
σx=principal stress in the x-direction
σy=principal stress in the y-direction
At this position, τxy=0 (no shear stress).
Setting σy=0 will yield the above equations.
Check if you have omitted a negative sign in the second equation.
Equation (13') applies to the situation of stresses measured at an angle α (≡q) from the principal directions, namely σ and τ.
The complete derivation is given on the page preceding equation (13'), chapter 2, section 9 "stress at a point".
1 answer