Principal Stresses and Finding Shear forces

[mm391]

Homework Statement

I am trying to find the Normal Stress of face AB in the attached picture, the shear stress in BC and AC and the maxium shear stress in side the rod.

σ_x = 40N/mm^2
σ_y = 62.5N/mm^"
Area =314mm^2

Homework Equations

I have attached the relavent equation underneath the problem digram

The Attempt at a Solution

Solving σ₂ i get ζ_xy to be 50N/mm^2

And then use that to find σ₁. Is σ₁ the normal stress in face AB. How do I go n to work out the rest.

To find P σ₁=P/A you are given area so you can rearrange to find P.
Does ζ_xy=ζ_yx

 

[Chestermiller]

Homework Statement

I am trying to find the Normal Stress of face AB in the attached picture, the shear stress in BC and AC and the maxium shear stress in side the rod.

σ_x = 40N/mm^2
σ_y = 62.5N/mm^"
Area =314mm^2

Homework Equations

I have attached the relavent equation underneath the problem digram

The Attempt at a Solution

Solving σ₂ i get ζ_xy to be 50N/mm^2

And then use that to find σ₁. Is σ₁ the normal stress in face AB. How do I go n to work out the rest.

To find P σ₁=P/A you are given area so you can rearrange to find P.
Does ζ_xy=ζ_yx

The simplest way to solve this problem is to make use of dyadic tensor notation. The first step is to express the stress tensor in terms of its principal components:

\mathbf{σ}=\frac{P}{A}\mathbf{ii}

where i is a unit vector directed to the right and parallel to P. Note that there is only one non-zero principal component of the stress tensor in this problem, namely P/A.

The unit vector i can be expressed in terms of the unit vectors in the x- and y-directions (on the figure) by the equation:

\mathbf{i}=-\mathbf{i_x}sin(\alpha)-\mathbf{i_y}cos(\alpha)

If we substitute this relationship into the above equation for the stress, we obtain:

\mathbf{σ}=\frac{P}{A}\mathbf{i_xi_x}sin^2(\alpha)+\frac{P}{A}\mathbf{i_yi_y}cos^2(\alpha)+\frac{P}{A}(\mathbf{i_xi_y}+\mathbf{i_yi_x}) sin(\alpha)cos(\alpha)

From this equation, it follows that

σ_x=\frac{P}{A}sin^2(\alpha)
σ_y=\frac{P}{A}cos^2(\alpha)
τ_{xy}=\frac{P}{A}sin(\alpha)cos(\alpha)

You can use the first two of these equations to solve for P from your input data. You can also use the first two of these equations to solve for α. Once you know α, you can use the third equation to solve for τ_xy