Hileman's Blog

Let’s start with two pieces of sheet metal that are riveted together and determine how the rivets will contribute to position variation.    

Sheet Metal Riveted Assembly

In this example I am using a #4 rivet that has the following specifications     

• Rivet Diameter = 3.18mm +/- 0.08 mm
• Hole Diameter = 3.3mm min, 3.4 mm max (3.35 +/- 0.5)

#4 Rivet Assembly

Let’s develop the tolerance equation.  There are at least two different ways to develop the tolerance equation.  One is to use the radius of the components.  The other is to use the the diameter of the components.  In the end, we will end up with the same answer.  The radial method takes advantage of symmetry; we are analyzing only half of the feature, or rivet assembly.  This means that the standard deviation that we calculate applies to only half of the riveted assembly.  In other words, we are able to calculate the variation in the plus or minus direction only.  In order to estimate the variation in both plus and minus direction, we would need to double the standard deviation that we calculate.  I prefer the radial method just because it’s seems to suite my mind better, but both the radial and diametral methods are valid.     

Rivet Tolerance Loop

• A = Radius of Hole in Plate A, 1.675 mm
• B = Radius of Rivet, 1.59 mm
• C = Gap Between Rivet and Hole Plate A 0.085mm
• D = Gap Between Hole in Plate B and Rivet, 0.085mm
• E = Radius of Rivet, 1.59 mm
• F = Radius of Hole in Plate B, 1.675 mm

The tolerance equation is       

A - B – C + D + E – F = 0       

1.675 - 1.59 - 0.085 + 0.085 + 1.59 - 1.675 = 0       

0 = 0       

The sum equals 0, which is good because we we started at the surface of the hole in Plate A and ended on the coincident surface of the hole in Plate B.       

Let’s apply the variation.  The tolerance of the diameter of the rivet is 0.08 mm. The radial tolerance is half of the diametrical tolerance, or 0.04 mm.  I will model this as a triangular distribution because the supplier has not provided any quality limits with respect to the tolerance.  The standard deviation is       

Rivet SD = 2*T/5 = 2(0.04)/5 = 0.016 mm       

The radial clearance is 0.085mm.  When the rivet is installed, it could end up any where in the hole.  The gap could be 0.0 mm, or as high as 0.085 mm.  The gap cannot be higher than 0.085mm because we are using the radial, or symmetry method, and nothing “exists” below the center of the hole or rivet.  The standard deviation is       

Clearance SD = (0.85 – 0.0)/3.5 = (0.085)/3.5 = 0.024 mm       

According to the rivet specifications, the minimum recommended hole diameter is 3.3 mm, and the maximum hole diameter is 3.4 mm.  From this, we can conclude that the tolerance for the diameter of the hole is +/- 0.05 mm.  This is a tight tolerance for a punched hole.  The generally accepted recommended tolerance by sheet metal fabricators is +/- 0.08 mm, so that is what I will use.  The radial tolerance will be half of 0.08mm, or 0.04 mm.  I will also assume a Cp = 1.00 for the hole diameter.  The standard deviation for the hole diameter is       

Hole SD = 2T/6 = 2(0.04)/6 = 0.0133 mm       

Let’s sum the standard deviations as we did in previous posts.  Note that calculation if for only the plus or minus direction.       

 σ_Total = (σ_HoleA² + σ_Rivet² + σ_Gap² + σ_Gap² +  σ_Rivet² + σ_HoleB²)^1/2       

σ_Total = ((0.0133)² + (0.016)² + (0.024)² + (0.024)² +  (0.016)² + (0.0133)²)^1/2       

σ_Total = 0.045 mm       

σ_{+/- Total} = 0.09mm       

This is a sizable amount of variation and is on par with the standard deviation of a punched feature to a bend, 0.08mm.  This analysis assumes that rivet is inserted perfectly perpendicular to the sheet metal surface.  Rivets can be installed crooked, but I think this is still a pretty good estimate for the variation we should expect to see.       

Next let’s look at a typical tab and slot.      

Tab and Slot Assembly

Determine the appropriate tab and slot clearance.  In order to determine the variance this feature, we need to determine the appropriate (or smallest) clearance for the slot about the tab.  A tab and slot can be used as a positioning mechanism in two different directions.  Perpendicular to the face of the tab, and parallel to the face of the tab.  Let’s look at perpendicular to the tab first.  The sheet metal is 1 mm thich, or about 19 gauge.  The tolerance range on sheet metal thickness is about 10%.  For 1 mm thick sheet metal, the tolerance is 0.4 mm.  It’s difficult to imagine that is that much, but based on my findings from various sources, it is indeed 10%.  If we assume the quality of this tolerance limit has a Cp = 1.00, the standard deviation can be calculated as follows.       

Tab Thickness SD =2 T/6 = 2(0.1)/6 = 0.033 mm       

A stamped feature will usually have a tolerance of 0.08 mm.  If we assume a Cp = 1.00, the standard deivation is       

Slot SD = 2T/6 = 2(0.08)/6 = 0.027 mm       

We should apply the rigor for the tab and slot as we would any other tolerance analysis.        

• A = Width of Slot, ???
• B = Gap Between Slot and Tab, ???
• C = Tab Thickness, 1.0 mm
• D = Gap Between Tab and Slot, ???

The tolerance equation is       

A – B – C – D = 0       

Solve for the gap.  If we assume the tab is centered in the slot, then B and E are equal.  The tolerance equation reduces to       

Gap = (A – B)/2       

This is the gap between the tab and slot on each side of the tab.  Notice that this tolerance equation is also the equation used for the Gap’s standard deviation.  If I had not through the rigorous development of the tolerance equation, then I could have estimated the gap standard deviation to be the standard deviation of the slot plus the standard deviation of the tab width, rather than this sum divided by two!  I could have made the clearance between the tab and the slot too much.       

The specification limits on the Gap will be 0.0 mm.  The standard deviation for the gap will be       

σ_Gap = ((σ_Slot² + σ_Tab²)^1/2)/2       

σ_Gap = (((0.027)² + (0.033)²)^1/2)/2       

 σ_Gap = 0.021 mm       

If we want a 4 sigma fit between the tab and the slot, then the clearance on each side of the tab should be       

 Clearance = 4(0.021) = 0.084       

In most designs I have seen for tab and slots, 0.2mm on each side is used, but we don’t need that much clearance.  0.084 mm is close enough to 0.1mm, and 0.1mm is easier to remember, so that’s what I will use.  Let’s examine how the tab and slot effect position.  The slot width we will use is 1.2mm       

In this analysis, let’s use the “radial” method.       

• A = Half Slot Width, 0.06 mm
• B = Half Sheet Metal Thickness, 0.5 mm
• C = Gap, 0.1mm

Tab and Slot Loop

The tolerance equation is       

A – B – C = 0       

0.6 – 0.5 -0.1 = 0       

0 = 0       

Let’s apply the variation       

σ_Slot² = 0.027/2 = 0.0135 mm       

σ_Tab² = 0.033/2 = 0.0165 mm       

The nominal gap is 0.1 mm.  It could be as small a 0.0 mm, or as big as 0.1 mm.  Since this is a clearance, we should use a uniform distribution.       

 σ_Gapt² = (Max – Min)/3.5 = (0.1)/3.5 = 0.0285       

The total standard deviation is       

σ_Total = (σ_Slot² + σ_Tab² + σ_Gap²)^1/2       

σ_Total = ((0.0135)² + (0.0165)² + (0.0285)²)^1/2        

σ_Total =  0.036 mm       

σ_{+/- Totall} = 0.072 mm       

Notice that the tab and slot provide less positional variation than the rivet assembly.  0.18 mm may not seem like much, but is a 20% improvement.  This improvement could be the difference in a high quality design and an average design.       

Let’s examine one more alignment scheme, and that is a tab and slot where the positional variation is parallel to the tab.       

We will follow the same method as before.       

Tab Width SD =2 T/6 = 2(0.08)/3 = 0.027 mm       

A stamped feature will usually have a tolerance of 0.08 mm.  If we assume a Cp = 1.00, the standard deviation is       

Slot SD =2 T/6 = 2(0.08)/3 = 0.027 mm       

We should apply the rigor for the tab and slot as we would any other tolerance analysis.        

• A = Width of Slot, ???
• B = Gap Between Slot and Tab, ???
• C = Tab Thickness, 12.0 mm
• D = Gap Between Tab and Slot, ???

Tab Width

The tolerance equation is       

A – B – C – D = 0       

Solve for the gap.  If we assume the tab is centered in the slot, then B and E are equal.  The tolerance equation reduces to       

Gap = (A – B)/2       

The specification limits on the Gap will be 0.0 mm.  The standard deviation for the gap will be       

σ_Gap = ((σ_Slot² + σ_Tab²)^1/2)/2       

σ_Gap = (((0.027)² + (0.027)²)^1/2)/2       

 σ_Gap = 0.019 mm       

If we want a 4 sigma fit between the tab and the slot, then the clearance on each side of the tab should be       

 Clearace = 4(0.019) = 0.08       

In this analysis, let’s use the “radial” method.       

• A = Half Slot Width, 6.08 mm
• B = Half Tab Width, 6.0 mm
• C = Gap, 0.08mm

Tab Width Loop

The tolerance equation is       

A – B – C = 0       

6.08 - 6.0 -0.08 = 0       

0 = 0       

Let’s apply the variation       

σ_Slot² = 0.027/2 = 0.0135 mm       

σ_Tab² = 0.027/2 = 0.0135 mm       

The nominal gap is 0.08 mm.  It could be as small a 0.0 mm, or as big as 0.08 mm.  Since this is a clearance, we should use a uniform distribution.       

 σ_Gapt² = (Max – Min)/3.5 = 2(0.08)/3.5 = 0.023       

The total standard deviation is       

σ_Total = (σ_Slot² + σ_Tab² + σ_Gap²)^1/2       

σ_Total = ((0.0135)² + (0.0135)² + (0.023)²)^1/2       

σ_Total =  0.03 mm       

σ_{+/- Totall} = 0.06 mm       

This is a substantial improvement over a rivet, and almost a 16% improvement over the tab and slot when using the tab thickness rather the tab width.  A half shear with a clearance hole would yield very similar results to the tab and slot just analyzed.  The down side of the half shear is that if it used for long sheet metal parts, the parts can bow, and the half shear will not engage before the adjoining sheet metal prior to being assembled.       

Now that we have gone through this analysis, we don’t need to go through it again.  In a large assembly, if a #4 rivet is in the tolerance loop, I can apply a standard deviation of 0.9 mm.  If a tab and slot is in the loop, a stadard deviation of 0.072mm or 0.06mm.  For future analysis, I have just saved myself a ton of work, and improved the quality at the same time.  Who says you need to pick two between Speed, Quality and Cost?       

So what’s this post all about?  It’s about two things really.  One is that we should not assume rivets, tab and slots, or half shears  locate anything precisely.  Two is that a detaileded tolerance analysis means paying attention to the details.

Tags: Cp, half shear, Quality, rivets, tab and slot, tolerance, tolerance analysis

This entry was posted on Thursday, June 24th, 2010 at 1:09 am and is filed under People Management, Quality, Tolerance Analysis. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.