## A Look at Bolt Lug Strength

### By: Daniel Lilja

All action designers are concerned primarily with producing a safe and functional product. All other considerations such as weight, finish, eye appeal and other cosmetics are secondary. From a safety standpoint, the strength of the bolt lugs is of prime concern. Bolt lug shear strength depends on several inputs. Some are fairly obvious, but others may not be at first glance. The number of lugs on the bolt is in the obvious category, as are the axial length of the lugs and the width of the lug. Or more precisely, not the actual width but the length of the radial segment of the lug that’s connected with the bolt body. Also, the strength of the material is important.

The shear strength is calculated from these factors, as I will explain in detail a little later. This shear strength must be greater than the amount of backwards thrust on the bolt generated by the cartridge being fired. So another number that must be calculated is bolt thrust. Bolt thrust is fairly simple to calculate and depends on the inside diameter of the cartridge case being fired and the chamber pressure.

Also of interest, but not necessarily crucial to safety, is the amount of flex or spring in the lugs while they’re under the thrust load.

So then, we will look at three different calculations related to bolt lug strength: lug shear strength, bolt thrust as generated by the cartridge case, and lug flex.

## Shear Strength

Unless a cartridge case undergoes a complete head separation upon firing, the side walls of the brass case will stick against the chamber walls. Under some circumstances they may absorb as much as half the thrust. Case walls or a chamber that are oily will reduce this friction. So the action designer will not take this aspect into consideration when designing the lugs to more closely simulate a complete case head failure.

Also, in the formula that will follow the calculated strength value will be reduced by half, adding an additional safety margin of two. This is in keeping with generally accepted engineering practice for suddenly applied loads.

The formula for calculating lug shear strength is:

**LS=(L*LL*NL*YS)/2 **

Where:

LS is the calculated lug shear strength

L is the length of the arc segment

LL is the axial length of the lugs

NL is the number of lugs on the bolt

YS is the yield strength of the material the lugs are made from

As mentioned, this number is then divided in half for a safety margin of two.

The only thorn in this formula is the length of the arc segment. We need to determine the total area in shear. Because the root of the lugs are joined to the bolt body on a radius, the length of the radial segment needs to be calculated. The formula for this calculation looks like:

**L=.01745*R*ANG **

where:

R=the radius of the bolt body

ANG=arccosine of the angle of the segment or

ANG=arccosine(x/SQR(1-x^2))+1.5708

where:

x=1-(R-.5*SQR(4*R^2-W^2))

where:

R=the radius of the bolt body

W=the width of a bolt lug

If this all seems a little imposing, we’ll list a short computer program in GWBASIC later that will hopefully make more sense.

The type and hardness of the steel the bolt is made from is important too. This part of the equation falls under the yield strength input. The yield strength is the maximum amount of pressure the steel can take without becoming permanently deformed. Up to this point it will return to its original condition. Since 4140 type chrome-moly is probably the most common type of steel used in bolts, I’m including a list of the yield strength for this steel at various Rockwell `C’ hardness values.

ROCKWELL ‘C’ HARDNESS | 4140 YIELD STRENGTH PSI |

20 | 83,500 |

22 | 87,000 |

30 | 135,000 |

34 | 148,750 |

37 | 159,000 |

42 | 178,000 |

46 | 195,000 |

49 | 211,000 |

## Bolt Thrust

Bolt thrust is easy to calculate. Only two inputs are required. They are peak chamber pressure in PSI and as mentioned, the inside area of the case head that the gas pressure can work on. The formula then is:

THRUST=AREA*CPSI Where:

AREA=3.1416*(HS/2)^2

HS=the diameter of the inside of the case head.

I sectioned some cases and measured the inside diameters and found that they were as follows:

CARTRIDGE CASE | INSIDE DIAMETER (HS) |

222 | .300″ |

PPC | .370″ |

308 | .385″ |

MAGNUM | .420″ |

378 WBY MAG | .500″ |

50 BMG | .680″ |

The thrust is measured in pounds per square inch.

## Bolt Flex

Bolt flex or spring is an interesting measure that is easily calculated using the numbers we have already determined for area in shear and bolt thrust. Another number that we will introduce into the equation is the shear modulus of elasticity. For steel this number is 11,500,000 pounds and is a constant. It does not change regardless of the type of steel being used or the heat treatment of the steel.

The equation looks like:

FLEX=THRUST/(SA*11500000)

where:

SA=the shear area of the bolt lugs or:

SA=L*LL*NL

Or more simply put, the area in shear for all lugs is multiplied times the shear modulus, and this number is divided into the thrust. Often the resulting number will be in the .001″-.002″ range. This explains why it becomes necessary to bump the shoulders of brass cases back after a few reloadings. New brass is elastic enough that it will return to its original shape, but with progressive loadings the brass becomes more plastic until it does not return to its original form at all. Cases become sticky, bolt lift more difficult and eventually the cases have to be replaced. With very high pressure loads this can happen on the first firing.

Also, the lug abutments in the receiver are set back a small amount too, compounding the problem. In fact it is possible for machining operations on the action (such as scope base screw or guard screw holes, magazine cutouts, and feed ramps) to weaken it to the point that the lugs have more strength than the action. In single shot bolt actions like we will be reviewing, this would not be the case, but with light weight repeaters this is entirely possible and quite probable, at least with the bottom side abutment.

So from a bolt flex standpoint we can see that the more lug area in shear, the less case stretch we will have. And this explains why higher pressure loads cause case stretching and sticky bolt lift. The obvious ways to increase shear area are: increasing the number of lugs, lengthening the lugs, or making them wider. Increasing the minor diameter of the lugs would help too, but to a lesser extent.

**September 2003:**

I received the following in an e-mail from Al Harral aka: Varmint Al.

*Hi Dan. I have used the LS-DYNA Finite Element code to calculate the stress level and stretching in a Stolle Panda rifle action. It was in response to a post I made about reducing the friction between the brass and chamber in a rifle. One of the responses was about your custom action stretching calculations.*

*In the results, I have linked to your page as a reference and a comparison. I am in no way trying to belittle your calculation but do point out the limitations of such a calculation when compared to a more detailed Finite Element calculation.*

*Therefore, I thought I would give you a heads up on the analysis I have done. It is posted on my web site here: http://www.varmintal.net/abolt.htm*

Al used some very powerful software and graphics to make some more detailed calculations on a similar action. This is worth looking at if you’re interested in this subject.

The following is computer code in GWBASIC that lists a short program that will calculate the above formulas:

10 CLS:PRINT””:PRINT”* * * * CODE THRUST * * * * ”

20 PRINT” WRITTEN AND COPYRIGHT BY DANIEL LILJA, APRIL 1990″

30 PRINT””:PRINT”ENTER THE HEAD SIZE OF THE CARTRIDGE THE

BARREL WILL BE CHAMBERED FOR:”

40 INPUT”1=.222; 2=PPC; 3=.308; 4=MAGNUM; 5=378 WBY MAGNUM; 6=.50

BMG”;QHS

50 INPUT”ENTER THE PEAK CHAMBER GAS PRESSURE (PSI):”;CPSI

60 INPUT”ENTER THE NUMBER OF BOLT LUGS ON THE BOLT:”;NL

70 INPUT”ENTER THE MINOR DIAMETER OF THE BOLT LUG (INCHES):”;BMD

80 INPUT”ENTER THE AXIAL LENGTH OF ONE BOLT LUG (INCHES):”;LL

90 INPUT”ENTER THE WIDTH OF ONE BOLT LUG (INCHES):”;W

100 INPUT”ENTER THE YIELD STRENGTH OF THE BOLT STEEL (PSI):”;YS

110 REM: BEGIN CALCULATIONS

120 IF QHS=1 THEN HS=.3

130 IF QHS=2 THEN HS=.37

140 IF QHS=3 THEN HS=.385

150 IF QHS=4 THEN HS=.42

160 IF QHS=5 THEN HS=.5

170 IF QHS=6 THEN HS=.68

180 AREA=3.1416*(HS/2)^2

190 THRUST=AREA*CPSI

200 REM CALCULATE LENGTH OF ARC SEGMENT OF BOLT LUG

210 R=BMD/2:REM RADIUS OF BOLT BODY

220 H=R-.5*SQR(4*R^2-W^2):REM HEIGHT OF ARC SEGMENT

230 X=1-H/R

240 ANG=(-1*ATN(X/SQR(1-X^2))+1.5708)*114.6:REM ARCCOSINE OF ANGLE

OF SEGMENT

250 L=.01745*R*ANG:REM LENGTH OF ARC SEGMENT

260 LS=(L*LL*NL*YS)/2:REM LUG STRENGTH IN POUNDS DIVIDED BY TWO

FOR SAFETY FACTOR

270 SA=L*LL*NL:REM AREA IN SHEAR FOR ALL LUGS

280 FLEX=THRUST/(SA*11500000#):REM SHEAR MODULUS OF ELASTICITY

290 PRINT””:PRINT'”THE APPROXIMATE BOLT THRUST IN POUNDS

IS:”;INT(THRUST/10+.5)*l0

300 PRINT”THE APPROXIMATE BOLT LUG SHEAR STRENGTH IN POUNDS

IS:”;INT(LS/10+.5)*l0

310 PRINT”THE LUG SHEAR AREA (SQ. INCHES):”;:PRINT USING”.###”;SA

320 PRINT”THE AMOUNT OF BOLT LUG FLEX FOR THESE CONDITIONS

(INCHES):”;:PRINT USING”.####”;FLEX

330 END