All PARCO structural extrusions have a 2° taper that spring locks fasteners as they are tightened. Fasteners will not loosen, even under heavy vibration.
Contact us to discuss your requirements of Aluminum Extrusion Prototype Manufacturing. Our experienced sales team can help you identify the options that best suit your needs.
The table (below) indicated the amount of torque needed in foot lbs. to activate the 2°drop lock feature of the extrusion. The nut and bolt combination is pre-loaded when tightened to the minimum torque rating. The pre-loaded state makes a vibration-proof connection.
An aluminum extrusion alloy is a predetermined mixture of one or more elements together with aluminum to be heated and hydraulically pressed through an extrusion die. Some common elements alloyed with aluminum include copper, magnesium, manganese, chromium, silicon, iron, nickel and zinc. PARCO extrusions use the 6063 alloy. In addition to aluminum, the major alloying elements for this alloy include Mg .45-.9% and Si .20-.6%.
Being a full-service manufacturer, we can send you raw materials (Parts and Extrusion):
We also have our own line of t-slotted aluminum extrusions that allows for the modularity, flexibility, and strength that industry needs today.
PARCO EXTRUSION BENEFITS:
Linear adjustment in X, Y, and Z-axis
All fractional extrusions are on 1/2”, 3/4”, 1”, or 1 1/2” center lines. All Metric extrusions are on 30mm, 40mm, and 45mm centerlines
All Metric extrusions are on 10mm, 15mm, 2.5mm, and 30mm centerlines
Use fractional or metric sized fasteners
Lightweight and easy to use
Fractional extrusions are stocked in 92”, 120”, and 240” lengths
Metric extrusions are stocked in 2.33M, 6M, or specified in profile lengths
No welding required to assemble your design, which means no heat stress or warping
Easy to fabricate
Add to your design or change it at any time – flexibility is the key
*As a partner with an extruder, there are many other alloys that we can extrude. If your application calls for a different alloy, let us know and we can extrude it for you.
Related Articles: T-Slot Aluminum – Aluminum Yield Strength – Aluminum Alloy and Temper – Procurement & Certifications – ISO Standards for Aluminum
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS
FIRST NAME
*
LAST NAME
*
*
MESSAGE
*
ADDITIONAL DETAILS
Thanks. We have received your request and will respond promptly.
Log In
Are you an
Engineering professional?
Join Eng-Tips Forums!
*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.
Promoting, selling, recruiting, coursework and thesis posting is forbidden.
Eng-Tips Posting Policies
Structural Design with Aluminum Extrusions
thread507-378734 Forum Search FAQs Links MVPsForum
Search
FAQs
Links
Want more information on Rapid Tooling For Medical Devices? Feel free to contact us.
MVPs
(Civil/Environmental)
(OP)
22 Jan 15 18:21I’m designing some aluminum extrusions for use as the structural framework for large horizontal sliding doors. As most of you know, trying to find the correct allowable stresses to use for aluminum extrusions is no simple task because the allowable stresses are dependent upon the Alloy, the slenderness ratio and the member configuration.I’m revising a spreadsheet I developed a few years ago to include a more detailed analysis of the extruded member I’m using for the vertical edge to the door panel.So the first thing is to compute the section properties of the member, which in my case is made up of two extrusions, one a rectangular tube, the other an odd shaped, non-symmetrical “channel”. The rectangular tube is inserted into the “channel” to reinforce the section. Once I have the combined section properties, I have to compute the allowable stresses which are dependent upon the slenderness ratio, which is based on the unbraced length of the member. I have braces, horizontal girts at 21” o.c. (Lb = 21”), therefore S = Lb/ry = 21.0/.6092 = 34.5.Now there are a whole slew of choices as seen on the attached sheet for computing the allowable stresses. In working through the equations for S1, S2, and >S2 (see the circled area of my spreadsheet), the allowable stresses are really small, 3.8 ksi and even -2.9 ksi! The allowable stresses just seem unreasonably small considering my past design experiences and I don’t understand the negative number. I can space the girts closer which would reduce S, but it does not help much.This is where I’m stuck and would like some input from the group.
(Mechanical)
22 Jan 15 18:58That's not a spreadsheet; that's an image of a spreadsheet.We can't see the formulae, so we can't check them.I'd guess that the negative number you worry about derives its sign from the negative number used for wind suction near the top of the page, but we can't check your work from here.3-ish ksi is a reasonable design stress for soft aluminum.You didn't mention an alloy or a temper, so it's conservative, unless you're buying a strong alloy, heat treated, facts not in evidence.
Mike Halloran
Pembroke Pines, FL, USA
(Civil/Environmental)
(OP)
22 Jan 15 19:35Thanks Mike,
The alloy is noted on the first page, 6063-T5 and the equations are in the first column, but you are correct, you wouldn't can't see them or manipulate them. The S1, S2, and >S2 are strictly a function of the "base" allowable stress - a constant * b/t.
The negative numbers pop out thus: S2 = 14.7 - .370*34.5 = -2.9 in the second column. In the first column I used S2 = 16.6 - .70*34.5 = 3.8 just to see the difference the two equations make. The come from different specification sections, see 3.4.8 and 3.4.11 on the first page.
My problem really is deciding on which equations to use and if the chosen one results in a negative number, what are my options.
(Petroleum)
22 Jan 15 23:08You use the stress that is appropriate for the slenderness. S1, S2 and S3 are slenderness ratio thresholds for the allowable stress calculations. For instance, if your slenderness falls between S1 and S2, you use the allowable stress calculated for S2. For a given slenderness ratio, oddball results for calculations you don't need to do are simply superfluous and have no meaning.
(Petroleum)
22 Jan 15 23:14Your equation in line 65 is wrong. You don't use the minimum of the three allowable stresses, you use the one appropriate for your slenderness ratio. Also, this only checks the outstanding flange whose measurements are b and t. Have you checked the gross section for compression ( tension, bending, etc )? Also, since this is a wind case, have you used the correct allowable stress usually increased for transitory loading?
(Petroleum)
22 Jan 15 23:29On page 3 line 60 the yield strength of 6063-T5 is shown as 30ksi. Is this correct? 6063-T6 is 30ksi but I think 6063-T5 is 25ksi. Also note that the yield stress for 6063-T5 is different in compression and tension, you have used 30ksi which is the tensile yield strength for 6063-T6 but this calculation is for a compression load case.
On page 3 line 62 you show the design stress to be 8.7 for S1. Why is this not 13.5 ksi as shown in the red box on page 1? Why calculate it at all?
For 16.6 - 0.37 x (b/t) to come out to -2.9, (b/t) must equal 52.7. Using 52.7 to calculate the stress for S > S2 I get 3.19 ksi but your result is 4.9. Is there something I am doing wrong?
(Civil/Environmental)
(OP)
23 Jan 15 00:02IFRs:
Thank you, you have clarified a number of my problems, not the least of which was thinking in terms of single value properties like steel, when an aluminum alloy has multiple values depending on thickness, tension, compression, shear, or bearing. So, the Fy = 30,000 ksi has no place in my calcs. The 3063-T5 Alloy has the following minimum properties:
Ftu = 22 ksi Tensile Ultimate
Fty = 16 ksi Tensile Yield
Fcy = 16 ksi Compressive Yield
Fsu = 13 ksi Shear Ultimate
Secondly, I was using out of date references, I finally found and downloaded the "Aluminum Design Manual - Specifications & Guidelines for Aluminum Structures", the June 2005 edition, by the Aluminum Association, Inc. Much clearer than the references I was using.
And, yes you are correct, that the section has to be as a whole and then each element of the section has to be reviewed and a weighted average of the allowable stress used.
Thank you for your comments!
(Structural)
23 Jan 15 00:42You seem to have some confusion as to the alloy you are using. Please clarify. Is it 6063T5 or other? Makes a big difference in aluminum. You also reference a 3000 series. These alloys are not typically used for structural extrusions.
Is your structure welded?
(Civil/Environmental)
(OP)
23 Jan 15 04:11The alloy is 6063-T5, and the member it is used in is a extruded, not welded. I think Mike has sent me on the right track, thanks for all the input! Once I get it worked out, I'll do a follow up post here.
(Petroleum)
23 Jan 15 04:34If you are designing in aluminum, it would be best to pay for the 2015 version of the Aluminum Design Manual. Worth it in my opinion, for any professional.
Why are you extruding in 6063-T5 when -T6 gets you much higher strengths, or even 6061-T6 for more?
I don't understand why you are using a "weighted average of the allowable stress"
Are you looking at combined bending and compression?
Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.
Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.
Click Here to join Eng-Tips and talk with other members! Already a Member? Login
The company is the world’s best 5-Axis CNC Machining For Automotive Parts supplier. We are your one-stop shop for all needs. Our staff are highly-specialized and will help you find the product you need.