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!
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?
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".
To derive the equations s' = s cos²(2q) and t' = s sin(q) cos(q), we need to apply the principles of mechanics of materials, particularly the equations of mechanical equilibrium applied to a free-body diagram.
Let's start by considering a small element of a material, such as a small cube or rectangle with sides dx, dy, and dz. We will analyze the stresses and strains in this element.
1. First, consider the stress component s'. To derive this equation, we will focus on the normal stress s in the x-direction acting on one face of the element.
- Draw a free-body diagram of the element and apply mechanical equilibrium equations. The forces acting on the element in the x-direction include the normal stress acting on the face, s*dx*dy, and the shearing stress t' acting on the same face, t'*dx*dy.
- Apply the equation of mechanical equilibrium in the x-direction: ΣFx = 0.
(s*dx*dy) - (s*dx*dy)cos(2q) - (t'*dx*dy)sin(2q) = 0
- Simplify the equation: s - s*cos(2q) - t'*sin(2q) = 0.
- Rearrange the equation to solve for s': s' = s*cos²(2q) + t'*sin(2q)*cos(2q).
- To simplify further, use the double-angle identity sin(2q) = 2*sin(q)*cos(q).
- Substitute the value of sin(2q) in the equation: s' = s*cos²(2q) + 2*t'*sin(q)*cos(q)*cos(2q).
- Simplify the equation using the identity cos(2q) = 2*cos²(q) - 1: s' = s*cos²(2q) + 2*t'*sin(q)*cos(q)*(2*cos²(q) - 1).
- Further simplify: s' = (2*s*cos²(q) - s)*cos²(q) + 4*t'*sin(q)*cos(q)*cos²(q).
- Finally, simplify using the identity cos²(q) = (1 + cos(2q))/2: s' = s*(1 + cos(2q))/2 + 2*t'*sin(q)*cos(q)*(1 + cos(2q))/2.
- This yields the derived equation: s' = s*(1 + cos(2q))/2.
2. Now, let's consider the shear stress component t'.
- Draw a free-body diagram of the small element again, focusing on the shearing stress t' acting on the x-face.
- Apply the equation of mechanical equilibrium in the x-direction: ΣFx = 0.
(t'*dx*dy)cos(2q) - (s*dx*dy)sin(2q) = 0
- Simplify the equation: t'*cos(2q) - s*sin(2q) = 0.
- Rearrange the equation to solve for t': t' = s*sin(2q)/cos(2q).
- Simplify further using the trigonometric identity sin(2q)/cos(2q) = tan(2q).
- Substitute the value of tan(2q): t' = s*tan(2q)/1.
- Simplify to the final equation: t' = s*sin(q)*cos(q)/2.
So, using the principles of mechanics of materials and the equations of mechanical equilibrium, we have derived the given equations s' = s*cos²(2q) and t' = s*sin(q)*cos(q)/2.