## Simcenter 3D: Analysis of a Pinned Lug

Not all loading conditions are created equal. In this blog you will learn when and how to use Siemens Simcenter 3D to analyze complex loading conditions that are outside of the range for hand calculations. Simcenter 3D can be used to solve these types of problems to calculate stress, reduce weight and improve design as your project requires.

### Challenges:

• Stresses must be within acceptable range
• Save weight by reducing stress
• Properly analyze complex loading conditions outside the range of hand calculations

### Values:

• Use Simcenter 3D with contact to predict lug stresses
• Hand calculations to validate lug stress analysis results
• When to use FEA vs hand calculations

The following pinned lug geometry will be analyzed using Simcenter Nastran.  The lug is made of Aluminum 7050-T7452 per AMS 4108.  A nominal diameter 3/8” pin (0.375 inches, 9.525 mm), made of AISI 4130 HT90, which is a low alloy steel, bears on the face of the lug.  A surface-to-surface contact between the pin and lug face is used.  A chamfer is also applied on the inside edge where the pin contacts the lug.

Lug Dimensions

The pin finite element model is shown below.  A constraint at the wall restrains the lug in translation and rotation (123456 degrees of freedom).  For simplicity, a 1000 lb (4448 N) load is applied. As a stepping stone in the analysis a simple no gap (e = 0%) model will be analyzed.

Of key importance is that the pin and lug have dissimilar meshes.  It’s common for contact methodologies to require coincident meshes, which puts additional burden on the user.  Simcenter 3D and Simcenter Nastran (formerly known as NX Nastran) support dissimilar meshes.

FEM of the Pin-Lug in Simcenter 3D

A small amount of static friction is applied (μ=0.1) to improve convergence.  Contact is simply done by selecting the two CAD surfaces (the face of the pin and the lug face).  Initial penetration is automatically calculated from the geometry (and is 0.0).

Contact in Simcenter 3D

The stress at the midplane of the lug for the Unit Load condition is taken to be 11,288 psi (11.288 ksi, 77.828 MPa).  Note that there is artificial peaking at the edge of the lug (not shown).  The hand calculation shows the actual FEM stress is lower than the hand calculation.

Maximum Principal Stress for Lug with No Gap

The hand calculation was performed with a  P = 1000 lb unit case for validation of the stress results.  First, we’ll calculate the nominal stress using the area represented by the dashed region.

Area for Nominal Stress Calculation

$\sigma_{nominal} = \frac{P}{A} = \frac{1000}{(H - d)h} = \frac{1000}{(1.036 - 0.375) * 0.45} = 3361 \text{psi}$

The nominal stress does not consider the effects of the hole, so we need to adjust the nominal stress using:

$\sigma_{max} = K_t \sigma_{nominal}$

Where Kt is composed of a geometric correction and a correction due to a material mismatch between the pin and the lug (steel vs. aluminum).

Typically, Chart 5.12 from Peterson would be used to calculate the Kt for the lug.  However, for large thickness to diameter ratios (h/d = 0.45/0.375 = 1.2) it does not apply.

Chart 5.11 from Peterson (typically used for round end lugs with h/d < 0.5) cannot be used to calculate the effects of a rounded lug since  h/d = 0.45/0.375 = 1.2 (thickness to diameter ratio). The key reason for this mismatch is the high thickness of the lug, which increases pin bending and thus pin bearing on the lug face.

Chart 5.12:  Stress concentration factors  for round-ended lugs, c / H > 1.5, h / d < 0.5  .

The alternative is to use Chart 5.11, which is meant for square end lugs with an h/d > 0.5.  Thankfully, a well round end lug with adequate radius (typical is H/2, which is the case) is similar to a square end lug.  The FEA will show the results match very well.

The parameter  c/H = 0.518/1.036 = 0.5.  Then d/H = 0.375/1.036 = 0.36 .  Then, using Chart 5.12 from Peterson as well the limiting Kte being Kt,100 (see the figure below), we can simply use Kte = Kt,100 = 3.2.

Chart 5.11: Stress concentration factors  for square-ended lugs, c  / H > 1.5, h /d < 0.5 .

An additional correction for the material mismatch between the pin and lug is also necessary.  Chart 5.12 from Peterson can be used to calculate this correction factor.  Note that this correction is only necessary because a thick lug is used.  If h/d < 0.5, then it’s not necessary.  For an h/d  =1.2 and Epin/Elug = Esteel/Ealuminum = 3, we can find that Kte‘ / Kte = 1.16.

Stress concentration factors Kte‘ for thick lugs.  Square or round ended lugs with h/d >0.5 and 0.3≤ d/H ≤ 0.6.

The combined stress concentration factor is found as:

$K_{t} = K_{te} \frac{K_{te}^{'}}{K_{te}} = 3.2*1.16 = 3.71$

Finally, the peak stress can be found as:

$\sigma_{max} = K_t \sigma_{nominal} = 3.71 * 3362 = 12,479\text{psi}$

The error between the FEA and the hand calculation is 10% despite the assumptions.  This error is driven by the small gap.

$Error = \frac{\sigma_{FEA}}{\sigma_{handcalc}} - 1 = \frac{11,288}{12,479} - 1 = -0.095 = -9.5%$

In a second case, the same lug geometry will be used, but with a 5% gap instead of no gap. As before, the Simcenter Nastran analysis will be shown first.

5% Pin-Lug Gap Distance

The peak stress in the lug is simply taken as 12,652 psi (12.652 ksi, 87.232 MPa).

In the case of a 5% gap, we must consider the more complicated version of the following formula:

$K_{te} = K_{t,0.2} + f(K_{t,100} - K_{t,0.2})$

f can be taken from the sub-chart of Chart 5.11 in Peterson.

Gap Tolerance Factor

The following formula must then be applied to correct for the nominal gap value.

$K_{te} = K_{t,0.2} + f(K_{t,100} - K_{t,0.2}) = 3.45 = 0.2(3.2 - 3.45) = 3.4$

The combined stress concentration factor is found as:

$K_{t} = K_{te} \frac{K_{te}^{'}}{K_{te}} = 3.4*1.16 = 3.94$

Finally, the peak stress can be found as:

$\sigma_{max} = K_t \sigma_{nominal} = 3.94 * 3362 = 13,259\text{psi}$

Again, the hand calculation is very accurate (within 5% of the FEA stress).  Note that as the pin diameter becomes smaller, it’s stiffness will decrease.  This decrease in stiffness combined with the thick lug will cause the pin to bend and the predicted stress and the FEM stress to diverge.

$Error = \frac{12,652}{13,259} - 1 = -0.045 = -4.5%$

In conclusion, a hand calculation can get results that are within 5-10% of a finite element analysis for simple geometry.  For lug designs that match the source data well (e.g., square and round lugs within the limits of the thickness to diameter ranges), hand calculations are appropriate.  For designs outside the typical design space, Simcenter 3D can be used to analyze the configurations with no additional complexity.  These geometric features include lugs with cutouts, which will dramatically change the stress field.  Updating the model to include nonlinear materials and eccentric loading is a simple process as well.

In a future post, it’s planned to revisit this lug analysis and discuss interference and how that can be beneficial to fatigue performance.

## Written by Steven Doyle

Steve Doyle is a Senior Aerospace Engineer with a background in structural analysis, CFD, and programming.  He’s also the author of pyNastran.  Outside of dorky projects, he likes to hike and rock climb!