Statics

3D Analysis

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Contents:

  1. 3D Components
  2. 3D Equilibrium Equations
  3. Structure Analysis

3D Force Components

The addition of a third dimension alters the process of vector decomposition of an arbitrary force slightly:

Fx=Fcos⁡θxF_x=F\cos\theta_x

Fy=Fcos⁡θyF_y=F\cos\theta_y

Fz=Fcos⁡θzF_z=F\cos\theta_z

F=Fx2+Fy2+Fz2F=\sqrt{F^2_x+F^2_y+F^2_z}

F→=Fxi^+Fyj^+Fzk^\displaystyle\bm{\overrightarrow{F}}=F_x\bm{\hat{i}}+F_y\bm{\hat{j}}+F_z\bm{\hat{k}}

F→=F(cos⁡θxi^+cos⁡θyj^+cos⁡θzk^)\displaystyle\bm{\overrightarrow{F}}=F(\cos\theta_x\bm{\hat{i}}+\cos\theta_y\bm{\hat{j}}+\cos\theta_z\bm{\hat{k}})

The subscript for the angles denotes the axis that the angle is measured from.

We can describe the vector component of the last equation with respect to the forces unit vector:

F(cos⁡θxi^+cos⁡θyj^+cos⁡θzk^)=Fn^F(\cos\theta_x\bm{\hat{i}}+\cos\theta_y\bm{\hat{j}}+\cos\theta_z\bm{\hat{k}})=F\bm{\hat{n}}

A unit vector has a magnitude of 1, and can be derived with the following definition:

n^=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^(x2−x1)2+(y2−y1)2+(z2−z1)2=n→∣n→∣\displaystyle\bm{\hat{n}}=\dfrac{(x_2-x_1)\bm{\hat{i}}+(y_2-y_1)\bm{\hat{j}}+(z_2-z_1)\bm{\hat{k}}}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}}=\dfrac{\bm{\overrightarrow{n}}}{\left|\bm{\overrightarrow{n}}\right|}

3D Equilibrium Equations

For a structure to be static, every point in the structure must satisfy the following:

Fx=0F_x=0

Fy=0F_y=0

Fz=0F_z=0

Mx=0M_x=0

My=0M_y=0

Mz=0M_z=0

Structure Analysis

The method of joints and frame analysis is completed using the exact same process outlined in the 2D structure analysis sections. The only difference between 2D and 3D analysis is the application of the three dimensional equilibrium equations. Because of more complex geometries, 3D analysis is typically more difficult than 2D, however taking a problem one step at a time will make the process much more manageable.