Using mechanics of materials principles (ie. equations of mechanical equilibrium applied to a free-body diagram), derive the following equations.
sigma' = sigma cos^2 θ = sigma(1+cos 2θ / 2)
and tau = sigma sin θ cos θ = sigma( sin2θ / 2)
The equations given would apply to a particular case of plane stress condition where σx and σy are the principal stresses, thus the shear stress τxy is zero. Furthermore, σx equals zero, i.e. it is a case of uniaxial tensile or compressive stress.
The derivation of the general formula for the general case is presented in the classic work "Theory of Elasticity" by Timoshenko and Goodier (chapter 2 sec. 9) and is the basis of the Mohr's circle. Therefore, the same derivation will be found in many sources including web-pages.
Here's one place you may look:
http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/plane_stress.cfm#transform
The Mohr's circle is a related subject that may eventually interest you, if not for the time being.
To derive the equations, we will start with the equations of mechanical equilibrium in two dimensions:
ΣFx = 0 (sum of forces in the x-direction is zero)
ΣFy = 0 (sum of forces in the y-direction is zero)
ΣM = 0 (sum of moments about any point is zero)
We will consider a small rectangular element with sides dx and dy within a larger beam or material. Let's assume that the stress acting on the small element is given by σ.
1. Derivation of sigma' = σ cos^2 θ:
Step 1: Create a free-body diagram of the small rectangular element. Assume that the normal stress σ is acting on the element in the x-direction and the angle it makes with the x-axis is θ.
Step 2: Resolve the normal stress into its x and y components:
σx = σ cos θ (stress component in the x-direction)
σy = σ sin θ (stress component in the y-direction)
Step 3: Apply the equation of mechanical equilibrium in the x-direction:
ΣFx = σx × dy - σ' × dy = 0
σx = σ' (since the element is small, the stress acting on both faces in the x-direction is approximately the same)
Step 4: Substitute the expression of σx from step 2 into the equation:
σ' = σ cos θ
Step 5: Rewrite cos^2 θ as (1 + cos 2θ) / 2 using the double-angle identity:
σ' = σ(1 + cos 2θ) / 2
Therefore, we have derived sigma' = σ cos^2 θ = σ(1 + cos 2θ) / 2.
2. Derivation of tau = σ sin θ cos θ:
Step 1: Create a free-body diagram of the small rectangular element. Assume that the shear stress tau is acting on the element at an angle θ with respect to the x-axis.
Step 2: Resolve the shear stress into its x and y components:
τx = τ cos θ (shear stress component in the x-direction)
τy = τ sin θ (shear stress component in the y-direction)
Step 3: Apply the equation of mechanical equilibrium in the y-direction:
ΣFy = τy × dx - τ × dx = 0
τy = τ (since the element is small, the shear stress acting on both faces in the y-direction is approximately the same)
Step 4: Substitute the expression of τy from step 2 into the equation:
τ = τ sin θ
Step 5: Rewrite sin θ cos θ as sin2θ / 2:
τ = σ sin θ cos θ = σ( sin2θ / 2)
Therefore, we have derived tau = σ sin θ cos θ = σ( sin2θ / 2).
To derive the equations sigma' = sigma cos^2 θ and tau = sigma sin θ cos θ, we can start by considering a small element of a material under stress. This small element can be represented by a three-dimensional free-body diagram.
Let's assume that the small element is subjected to a normal stress sigma and a shear stress tau. We need to decompose these stresses into their x and y components, using the angle θ as shown in the figure below.
|\
| \
| \
sigma'| \
| \ <------ y-axis
| \
| \
|_______\
shear stress (tau)
^
| <------ x-axis
The normal stress sigma' acting on a plane inclined at an angle θ with respect to the x-axis can be expressed as sigma' = sigma cos^2 θ.
To derive this equation, we use the principle of mechanical equilibrium along the x-axis. In the x-direction, the forces acting on the small element should add up to zero. Therefore, we have:
sum of x-components of stress = 0
sigma' * cos θ + tau * sin θ = 0 ----(1)
However, we know that tau = sigma sin θ cos θ. So, we substitute this relation into equation (1):
sigma' * cos θ + sigma * sin θ * cos θ = 0
Dividing through by sigma cos θ, we get:
sigma' = -sigma sin θ
As -sin θ = sin (-θ), we can represent this as:
sigma' = sigma sin (-θ)
Using the trigonometric identity sin (-θ) = -sin θ, we have:
sigma' = -sigma sin θ
Now, we define the positive normal stress σ' as:
σ' = -sigma sin θ
Finally, recalling that sin (-θ) = -sin θ, we have:
σ' = sigma sin θ
To derive the equation tau = sigma sin θ cos θ, we use the principle of mechanical equilibrium along the y-axis. In the y-direction, the forces acting on the small element should add up to zero. Therefore, we have:
sum of y-components of stress = 0
tau + sigma' * sin θ = 0 ----(2)
Substituting for sigma' = sigma sin θ, we get:
tau + sigma * sin θ * sin θ = 0
Simplifying:
tau = -sigma sin^2 θ
Now, using the trigonometric identity sin^2 θ = 1 - cos^2 θ, we have:
tau = -sigma (1 - cos^2 θ)
Multiplying through by -1, we get:
tau = sigma (cos^2 θ - 1)
Finally, rearranging the equation:
tau = sigma (1 - cos^2 θ)
Since cos^2 θ = 1 - sin^2 θ, we can substitute this relation into the above equation:
tau = sigma (sin^2 θ)
This equation can also be represented as:
tau = sigma sin θ cos θ
Therefore, we have derived the equations sigma' = sigma cos^2 θ and tau = sigma sin θ cos θ by applying the equations of mechanical equilibrium to a free-body diagram of a small element under stress.