Something is wrong with my figuring but it would seem that the 5/8-11 Gr.2 screw wins on strength overall. If possible someone check my math so I also know the formula and how it works.
wanna convert the measurements lol - Cruiser
a 1/4"
Bolt is about a M6
a 5/16"
Bolt is about a M8
a 3/8"
Bolt is about a M9.5 or M10
a ½"
Bolt is about a M12 or M13
As for threads per inch, consider them to be course grade with the exception of the 5/16-24 and 1/4-28 which would be National Fine grade.
“A note from McMaster is that metric sizes are 4.8 carbon steel, comparable to Grade 2.” This would mean that the normal metric
bolt is not going to be very strong.
I wouldn't begin to know the differences in the
rating specs--like DIN933, DIN 912 and so forth and how they apply to use...or where to read up on that. But FWIW, I'm torn between the titanium and the Gr8 Zinc...and thats only a guess on my percieved strength of the metals and
thickness of the bolts... - Wolf
Wolf you get the
point, there is no ready or easy table for this and it is very unfortunate. This is my
point in posting the survey. We assume we know what the heck we are doing, might even specify a Titanium
bolt and think it’s strength is extreme, but it’s not with study. We also have to balance bulk in general verses
bolt grade by instinct.
Din numbers along with Rockwell Hardness are hardness of steel which roughly equates to grades of bolts except it only accounts for the surface hardness and not the flex. Glass has decent surface hardness for instance. The rough equation would be the minimum tensile strength in combination with surface hardness. That is if you can find a chart for the tensile strength and hardness of the various grades of
bolt.
Than it’s also size and mass verses sheer and tensile strength. Is a 5/8" Grade 2
bolt really going to be stronger than a alloy 1/4-20
bolt of alloy steel, and what
effect will fine threading of a
bolt have on the strength?
There are some things that you can cross off the list as un-important information for the moment where strength is involved and that’s coating of the metal - or at least normally with those excepting the black oxide coating which normally dictates a alloy steel if not otherwise mentioned. Surface coatings will be a detail for engineer stress in being able to hold the
nut but not the overall strength of the
bolt itself. Alloy steel or Aircraft Grade as it’s trade name might be is about a grade 9 or 10 in strength and while black oxide coatings can be applied to normal steel bolts of grade 2 which is normal when the
bolt is not listed for grading or in this case the grade 5
bolt, it’s usually only applied to alloy steel. Ultra Coated bolts while resistant to corrosion does not otherwise add to strength. The coating might have a slight difference in overall sheer due to the surface hardness but after that it’s not much difference between a Ultra Coated
bolt and a Zinc
bolt.
Titanium is seemingly a extreme strength material but it’s going to be more a alloy of it mixed with steel as otherwise it would be probably too high in carbon content for tensile strength in general, plus expensive due to the difficulty in working. It’s primary use is in resistance to corrosion, salt water and chemicals. Rockwell hardness is B75 and Minimum Tensile Strength is 62,000psi. From the hardness and psi, it would appear that Titanium is not very strong, only slightly stronger than Grade 2 steel.
Stainless Steel in general is not graded the same. It’s certainly harder than standard steel but overall it’s not normally stronger than say a Grade 5
bolt. Grade 18-8 is standard, type 316 is slightly stronger in tensile strength. Type 316 is B85-95, and has a tensile strength of 85,000psi. Grade 18-8 is B85-B95 with 80,000psi.
Military Specification bolts means the
bolt has tighter tolerances than normal grades of
bolt. In the case of a grade 3
bolt of ½", is it going to be stronger than a grade 5
bolt of 3/8". Good question in mass verses strength.
Grade 8 is a “medium-carbon alloy steel that has been quenched and tempered for greater strength”. “Rockwell hardness is C33-39. Min. Tensile strength is 150,000 psi.” Grade 5 is C25-34 with a tensile strength of 120,000psi. for these sizes. Grade 2 is B70 and 60,000psi.
Alloy steel has a better thread to
nut fitting which adds to strength, are traceable in materials meaning that out of liability the materials are up to specification even Military Spec. “These screws are heat treated, have a black oxide finish, and exceed the tensile strength of grade 8.” - McMaster Carr Hardness is C39-45 and minimum Tensile Strength is 180,000psi.
In general we can probably cross out the 5/16-18x2" Titanium Hex Head Screw #94081a591 having checked the description since it would be much less strong than a Grade 8
bolt of the same size. We can also cross out the 1/4-28x3" Gr.8 Hex Head Screw of Ultra-coating #91286a151 since a alloy steel 1/4"
bolt will be stronger.
After that, it’s a question of what is the formula for bulk verses tensile strength than also bulk verses sheer strength which would be Rockwell Hardness. I don’t know but will present the formula for each, but it’s probably going to be safe to assume that any 1/4"
bolt is not going to have a better strength in either dimension than a 5/8"
bolt. We can safely cross off a 1/4"
bolt from the list.
This leaves us with a choice of:
5/16-24x2" Gr.8 Hex Head Screw of Zinc plating #91257a611
3/8-16x2" Gr.5 Hex Head Screw of Black Oxide Coating #92965a632
1/2-13x2" Gr.3 Military Spec Hex Head Screw of type 18-8 Stainless Steel #92245a722
5/8-11x3" Gr.2 Hex Head Screw of Zinc Plating. #91309a806
If you are not up to speed at this
point than going back and studying before you specify would be useful.
Before we get into the formula, is it safe or not to assume that something 5/16" is going to have half the tensile strength as something double the size at 5/8"? Thus unless the 5/16" Gr.8
bolt has a tensile strength of at least 120,000 psi it’s out. In this
bolt it would thus it is not crossed out yet. That is given for bulk you are only doubling up the PSI as a factor. If not than bulk equals less than tensile strength and the 5/16"
bolt would be crossed out. On the other
hand since we are also considering sheer strength, we should probably cross out the 5/16"
bolt given the Time to get out actual engineering books as it would seem that’s about missing from McMaster.
From Machinery’s Handbook #26 by Erik Oberg, by Industrial Press, NY. 2000 ISBN: 0-8311-2625-6, we learn that a Rockwell hardness scale B is for medium hardness steel verses scale C is for hardness greater than B-100. In other words for anything on the C-chart in comparison to the B-Chart, add 100 to the numbers listed on the C-Chart as they compare to the B-Chart.
Tension is O = F/A
O = Simple normal tensile or compressive strength in pounds per square inch.
F = External force in pounds
A = Cross-Sectional area in square inches.
Compression is O = - F/A
Sheer is t = F/A
t = Simple sheer stress in pounds per square inch.
(Note the actual symbols do not translate)
This is simple force without other factors into it.
P1476
Preload for Bolts In Shear. - In shear-loaded joints with members that slide, the joint members transmit shear loads to the fasteners in the joint and the preload must be sufficient to hold the joint members in contact. In joints that do not slide (i.e., there is no relative motion between joint members), shear loads are transmitted within the joint by fractional forces that mainly result from the preload. Therefore, preload must be great enough for the resulting friction forces to be greater than the applied shear force. With high applied shear loads, the shear stress induced in the fastener during application of the preload must also be considered in the bolted-joint design. Joints with combined
axial and shear loads must be analyzed to ensure that the bolts will not fail in either tension or shear.
General Application of Preload. - Preload values should be based on joint requirements, as outlined before. Fastener applications are generally designed for maximum utilization of the fastener material; that is to say, the fastener size is the maximum required to preform its function and a maximum safe preload is generally applied to it. However, if a low-strength fastener is replaced by one of higher strength, for the sake of convenience or standardization, the preload in the replacement should not be increased beyond that required in the original fastener.
To utilize the maximum amount of
bolt strength, bolts are sometimes tightened to or beyond the yield
point of the material. This practice is generally limited to ductile materials, where there is considerable difference between the yield strength and the ultimate (breaking) strength, because low-ductility materials are more likely to fail due to unexpected overloads when preloaded to yield. Joints designed for primary static load conditions that use ductile bolts, with a yield strain that is relatively far from the strain at fracture, are often preloaded abouve the yield
point of the
bolt material. Methods for tightening up t and beyond the yield
point include tightening by feel without special tools, and the use of electronic equipment designed to compare the applied torque with the angular rotation of the fastener and detect changes that occur in the elastic properties of fasteners at yield.
Bolt loads are maintained below the yield
point in joints subjected to cyclic loading and in joints using bolts of high-strength material where the yield strain is close to the strain at fracture. For these conditions, the maximum preloads generally fall within the following ranges: 50 to 80 per cent of the minimum tensile ultimate strength; 75 to 90 per cent of the minimum tensile yield strength or proof load; or 100 per cent of the observed proportional limit or onset of yield.
P1478
Preload Adjustments. - Preloads may be applied directly by
axial loading or indirectly by turing of the
nut or
bolt. When preload is applied by turning of nuts or bolts, a torsion load component is added to the desired
axial bolt load. The combined loading increases the tensile stress on the
bolt. It is frequently assumed that the additional torsion load component dissipates quickly after the driving force is remove and , therefore, can be largely ignored. This assumption may be reasonable for fasteners loaded near to or beyond yield strength, but for critical applications where
bolt tension must be maintained below yield, it is important to adjust the
axial tension requirements to include the effects of the preload tension. For this adjustment, the combined tensile stress (vol Mises stress)
Ftc is psi (Mpa) can be calculated from the following:
Ftc = square root of F(2/t) + 3 (2/s)
Where Ft is the
axial applied tensile stress in pse (Mpa) and Fs is the sheer stress in psi (Mpa) caused by the torsion load application.
P1489
Working Strength of Bolts.
... the following empirical formula was established for the working strength of bolts used for packed joints or joints where the elasticity of a gasket is greater than the elasticity of the studs or bolts.
W = St (0.55d² - 0.25d)
In this formula, W=working strength of
bolt or permissible load, in pounds, after allowance is made for initial load due to tightening; St = allowable working stress in tension, pounds per square inch; and d = normal outside diameter of stud or
bolt, inches. A somewhat more convenient formula, and one that gives approximately the same results, is W = St (A-0.25d)
In this formula, W,St,and d are previously given and A = area at the root of the thread, square inches.
Example: What is the working strength of a 1-inch
bolt that is screwed tightly in a packed joint when the allowable working stress is 10,000psi? W=10,000 (0.55 x 1 - 0.25 x 1) = 3,000 pounds approx.
Given all of this of the bolts left, we can assume the following with the simple formula:
5/16-24x2" Gr.8 Hex Head Screw (0.2603" id) 150,000psi (0.408862² - 0.078125) = 150,000 (0.1671715 - 0.078125) = 150,000 (0.0890465) = 13,356.975#
3/8-16x2" Gr.5 Hex Head Screw (0.2970"id) 120,000psi (0.4665127². - 0.09375) = 120,000 (0.217634 - 0.09375) = 120,000 (0.123884) = 14,866.08#
1/2-13x2" Gr.3 Military Spec Hex Head Screw of type 18-8 Stainless Seel (0.4056"id) 80,000psi (0.6370962² - 0.125) = 80,000 (0.4058915 - 0.125) = 80,000 (0.2808915) = 22,471.32#
5/8-11x3" Gr.2 Hex Head Screw (0.5119"id) 60,000psi (0.8040669² - 0.15625) = 60,000 (0.6465235 - 0.15625) = 60,000 (0.4902735) = 29,416.41#
Going back to the 1/4-20x3" Alloy Steel
Socket Head Cap Screw, it has a tensile strength of 180,000psi. and a minor dia. of 0.1876" for a class 2A fit. This is 180,000 (0.2946727² - 0.0625) = 180,000 (0.086832 - 0.0625) = 180,000 (0.024332) = 4,379.76#.
There is one X factor I have not figured out yet: A = area at the root of the thread, square inches.
Example: What is the working strength of a 1-inch
bolt that is screwed tightly in a packed joint when the allowable working stress is 10,000psi? W=10,000 (0.55 x 1 - 0.25 x 1) = 3,000 pounds approx.
You will note that it's (.55 x 1 - 0.25 x 1) above. Where 0.55 x 1 comes from I do not know because it should be 0.622" at the root of a standard coarse 1"-8tpi class 1A
bolt. This would give us 10,000 (0.9770065² - 0.25 x 1) = 10,000 (0.9545717 - 0.25) = 10,000 (0.7045717) = 7,045.717#
Something is wrong with my figuring but it would seem that the 5/8-11 Gr.2 screw wins on strength overall. This also means that I got it wrong with I think I guessed the 5/16-24 Gr.8 screw.