This article shows the example calculation for Warping Angle and Warping factor for a Quadrilateral element.
Warping Angle Calculation Example of a Quadrilateral Element:

Let "ABCD" is the quadrilateral element with the nodal coordinates A(10,15,0), B(25,25,0), C(35,15,5), and D(25,5,5) as shown in the figure below.

With ABCD quadrilateral element two sets of triangular planes are possible ABD &BCD and ABC & ACD.
Step2: Angle Between ABD & BCD:

First Construct two Triangular planes ABD (Plane1) & BCD (Plane2) as shown in the figure above.

Now let us write equations of the first set of planes shown below.
Equation of the plane1, ABD is:
Equation of the plane2, BCD is:
Normal Form of Equations of Plane (1) & (2):
The Direction Cosines of Plane1 & Plane2 are:
So, The angle between two planes,
Angle between Plane1 &2 is (θ1 )=42.998 degree
Step2: Angle Between ABC & ACD:

Construct two Triangular planes ABC (Plane1) & ACD (Plane2) as shown in the figure below.

Now let us write equations of the first set of planes shown below.
Equation of the plane3, ABC is:
Equation of the plane4, ACD is:
Normal Form of Equations of Plane (3) & (4):
The Direction Cosines of Plane3 & Plane4 are:
So, The angle between two planes,
Angle between Plane3 &4 is (θ2 )=50.860 degree
From Step1 & Step2: The Warping Angle is maximum of the two angle calculated
Warping Angle = Max(θ1, θ2)=50.86°
2. Warping Factor Calculation Example of a Quadrilateral element

Let "ABCD" is the quadrilateral element with the nodal coordinates A(10,15,0), B(25,25,0), C(35,15,5), and D(25,5,5) as shown in the figure below.
Direction Cosines of AC :
Direction Cosines of BD :

Now, We have to find the common normal to AC & BD.

Let, The direction cosines of Common normal is l,m &n.
Since, AC ⊥ Common Normal
Similarly, Since, BD⊥Common Normal
So, the direction cosines of common Normal are:
Equation of a plane Passing Through A&C, whose normal is perpendicular to AC is:
Since it is passing through A(10,15,0)
So, Equation of Plane Passing through AC is:
Similarly, Equation of a plane Passing Through B&D, whose normal is perpendicular to BD is:
Since it is passing through B(25,25,0) ⟹ p=50
So, Equation of Plane Passing through BD is:
Now, let's find out the perpendicular distance between nodal points to the Average Plane.
The Perpendicular distance of Plane (b) from pointA(10,15,0) is:
Perpendicular distance of Average plane from PointA is, AA'=4.811/2=2.4055 (=h)
Similarly, BB′′=CC′′=DD′′= 4.811 and BB′=CC′=DD′=2.4055
A’B’C’D’ is the Average normal plane of the Quad Element ABCD, which is equidistance from all Nodes of Quad Element
Find the Projection of Nodes on the Average Normal Plane:
Projection of Node “A” on average normal plane is A’.
Since, Plane A’B’C’D’ is parallel to Planea & Planeb.
So, the direction cosines of Plane A’B’C’D’ are as follows:
Let us assume, the projection of NodeA on Average Normal plane is A′ (x,y,z).
So, the direction cosines of AA′ are,
So, the direction cosines of line AA′ are ⇒
By equating these values with l, m, n, we will get x, y and z values.
So, the projection of NodeA is, A’(10.463,15.463,2.315)
Similarly, the Projections of other nodes B, C and D are,
B′(24.537,24.537,2.315), C′(35.463,15.463,2.685), D’(19.537,4.537,2.685)
Area of the projected Quadrilateral on Average Normal Plane A'B'C'D':
Area of the Quad Element,∆A′B′C′ D′=∆A′B′C′+∆A′C′D′
A’B’=17.374, B’C’=14.207, C’D’=20.046, D’A’=14.207, A’C’=25.495
S1=(Perimeter of A′B′C′)/2=28.538,
S2=(Perimeter of A′C′D′)/2=29.874
Area, A=√(S(Sa)(Sb)(Sc))
Area of the projected quadrilateral on average normal plane, A=A1+A2=117.8757+141.9319=259.8
So, the Warping factor is the ratio between "distance between two parallel planes and Square root of the area of the projected quadrilateral on average normal plane.
Warping Factor= 2h/√A= (4.811)/√(259.8) = 0.298