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