1. Flexural loading of a beam
Compared to metals, plastics have low stiffness. However, the plastics allow far greater design freedom, so that using a ribbed design can compensate the lack of stiffness. The deflection of a beam under flexural load is inversely proportional with the product of the E-modulus and the moment of inertia of the cross section. In order to maintain equal stiffness in bending, the following should apply:
 
                        E[metal].I[metal] = E[plastic].I[plastic]  
 
Since a wall thickness in excess of 4 mm should be avoided, just increasing the thickness of a part usually not feasible, and from a cost perspective not desirable. Figure 1 compares a EI of a 1.5 by 10 mm steel cross section to a rectangular cross section of equivalent stiffness in Polyamide 66, 30% GF, conditioned and to a U-section of equivalent stiffness in Polyamide 66, conditioned.
 
Steel:               E = 210.000 MPa
                        I = 1/12 * Width * Height³ = (10* 1.5³)/12 = 33.75 mm^4 
                        EI[Steel] = 210.000 * 33.75 = 7,087,500 N.mm² 

 
Equivalent stiffness on a rectangular cross section in PA66, conditioned:
 
PA66, 30% GF: E = 6000 N/mm² (Conditioned)
                        7,087,500 = 6000*I[PA66]
                        I[PA66]=7,087,500/6000 = 1181 mm^4 = 1/12*Width* Height³ 
                        Width = 10 mm, so
                        Height³ = (12 * 1181)/10 = 1417 mm³
                        Height = 11.25 mm
 
Now the same for a U-section with a uniform 3 mm wall section:
 
PA66, 30% GF:  E = 6000 N/mm²
                         7,087,500 = 6000*I[PA66]
                         I[PA66]=7,087,500/6000 = 1181 mm^4
 
Using the appropriate formula, it follows that the height is 12.5 mm
 
This means the height is only 1.5 mm higher than a full cross section. At the same time, surface area of the cross section, directly proportional to the weight of the part, is 87 mm² instead of 112.5 mm²! Conclusion: use ribs whenever you can, both for cycle time and raw material usage.
 




2. Deflection of beams
Designing for stiffness for a beam under flexural load is than quite simple: to get maximum stiffness, ensure you have a cross section with the highest possible moment of inertia around the axis of bending, within the following design rules:
·         Maximum wall section should not exceed 4 mm
·         Ribs should be thinner than the base surface to avoid sink marks
·         Ensure ribs are rounded of at the base to avoid stress concentration
·         Take measures for de-moulding, i.e a draft angle on the ribs of at least 1° and avoid long, thin cores
 
It also makes sense to ensure the cross section has a high moment of inertia in the areas where the bending load is highest. For instance, when a cantilever beam is loaded, the tip of the beam does need to have a particular high moment of inertia, however the end where it is fixed does need a high resistance to bending. Taken this into account saves cost, both in terms of material and cycle time, and saves weight.
 
The figure at the top give an overview of beams of a constant cross section behave under load.





3. Moments of inertia
The figure above shows how to calculate the moment of inertia for various cross sections.





4. Torsional loading of a beam
Torsional loads are more difficult to deal with in thermoplastic design. The basic formula for the angular twist of a beam is:
 
                       q=TL/KG
 
where            q=angular twist in radians
                     L = Length of the beam [mm]
                     T=Torsion Moment [N.mm]
                     K=Cross section dependend factor
                     G=Shear modulus of a material [MPa]
 
For isotropic materials, the shear modulus G is given by the equation:
 
                        G=E/2(1+u)
 
Where             G= Shear modulus
                      E=Tensile Modulus
                      u=Poisson ratio = 0.25 to 0.35 for PA66
 
For unreinforced Polyamide 66, this formula gives a value close to reality. For reinforced thermoplastics, the shear modulus is lower than this value, as these materials are non-isotropic. Depending on how the orientation is compared to the fibre direction, G for Polyamide 66 GF 30 (conditioned) varies from1200 to 2000 Mpa, whereas the value based on the formula of an isotropic material would lead you to expect a value of 2500 Mpa.
 
Compensating for the low shear modulus by using a cross section with a high K-value is also not easy. Closed cross sections give the highest K-value, but it is usually not possible to make closed cross sections with thermoplastics (only by using gas assist injection moulding in relatively simple geometries).
 
To maximize torsional stiffness, the following guidelines can be given:
·         To maximize the K-value, concentrate as much material as possible in the centre of a cross section
·         Use cross ribbing
·         Use Diagonal ribbing






5. Tensile loading of a beam
Deformation under tensile loading is given quite simply by the formula:
 
Dl=F/A
 
Where Dl=change of length
F=applied force
A=surface area of the bar
 
To make the parts as stiff as possible, the surface area needs to be maximized, within the limits of good design. (Wall section not to exceed 4 mm)


6. Loading of flat plates
Formulas that approximate the deflection of flat plates are quite complex and involved. They can be used only for simple geometries:
·         Uniform thickness which is thin compared to the dimensions of the plate
·         Deflection less than one half of the plate thickness
·         Isotropic, homogeneous material
·         No ribbing, bosses etc. on the surface
For most situations, calculations with these formulas are only of limited use, and we recommend the use of finite element calculations for more accurate results. However, clearly putting diagonal ribbing on a large flat area will greatly enhance stiffness.