3D shearing w. r. t. $x$-axis

Question

Shearing in $X$ axis is achieved by using the following shearing equations-

$X_{new} = X_{old}, Y_{new} = Y_{old} + Sh_y × X_{old}, Z_{new} = Z_{old} + Sh_z × X_{old}.$

My question is during $Y_{new}$ and $Z_{new}$ calculations why $Sh_y,Sh_z$ is multiplying with $X_{old}$ respectively, why not any other constant? Every books, websites mention just formula not the reason.

Can anyone give pictorial representations how shearing is happening with explanation.

Answer 1

Let's start in 2-D.

Then, in the acception provided in Wikipedia, A "shear keeping fixed the $x$ coordinate" , or otherwise told a " shear along the $y$ axis" will be a deformation like the one depicted.

Each point will move parallel to the $y$ axis by a quantity which is proportional to its distance from it, i.e.proportional to the $x$ coordinate.

In fact , $A=(x,y)$ will move to $A'(x',y')$ in this way $$ A = \left( {x,y} \right)\; \to \quad A' = \left( {x,y'} \right) = \left( {x,y + \Delta y} \right) = \left( {x,y + x\tan \theta } \right) $$ that is $$ \left\{ \begin{array}{l} x' = x \\ y' = y + kx \\ \end{array} \right. $$

Passing to 3-D, imagine the picture above reproduced on every section at constant $z$.
Clearly we will have $$ \left\{ \begin{array}{l} x' = x \\ y' = y + kx \\ z' = z \\ \end{array} \right. $$ i.e. a movement along $y$, proportional to $x$, at constant $x,z$.
It is also clear, by symmetry that keeping the slide to be in the $y$ direction, thus at constant $x,z$, the magnitude could be proportional to $z$ instead than $x$. That means that we are tilting the planes parallel to $x,z$ (normal to $y$) around the $z$ axis in the first case and around the $x$ axis in the second.

Legend:

• take a parallelepiped (grey-yellow)
• apply a shear in the $y$ direction, proportional to $x$ (grey -> lavender, yellow -> blue): the planes are tilted around their intersection with the plane $y,z$, parallel to $z$;
• if the shear is still in the $y$ direction, but proportional to $z$, then you have the second picture, with a tilt around the intesections with the $x,y$ plane, parallel to the $x$ axis.

Finally we can compose the shear above with a shear along $z$, proportional to $x$, at constant$x,y$ to give $$ \begin{array}{l} (I)\left\{ \begin{array}{l} x' = x \\ y' = y + kx \\ z' = z \\ \end{array} \right.\quad (II)\left\{ \begin{array}{l} x'' = x' \\ y'' = y' \\ z'' = z' + mx \\ \end{array} \right.\quad \Rightarrow \\ \Rightarrow \quad (II) \circ (I) = (II) \circ (I) = \left\{ \begin{array}{l} x'' = x \\ y'' = y + kx \\ z'' = z + mx \\ \end{array} \right. \\ \end{array} $$ which is the expression you are considering