Tubing tensile strength
Can anybody tell me the tensile strength of 1 3/4" x .049" 4130 steel tubing?
Also 1 1/2" x .049"?
Yield strength of what Wicks is currently selling is 75,000 psi. (You should use yield strength for most structural calculations, not the ultimate tensile strength)
For a member loaded in pure tension simply use the cross sectional area of the tubing to calculate the strength of the member.
For members loaded in shear, compression, or bending it gets more complicated.
OK, I'm a bit math challenged, but area of a circle is pi times radius squared, so 1 3/4" solid would be 2.4052 sq. in., the "hole" in a .049" wall tube would be 2.1435 sq. in., subtracting that would give 0.2617 sq. in. Multiplied by 75,000 would give 19,627.5 lbs yield strength, correct?
Ugg - it is too early for math.
For a square tube Area = H^2 - h^2, so in this case H=1.75, h=1.652, so Area = 0.333396 in^2
Strength of the member (in tension) would then be the yield strength x the area, so 75,000psi x 0.333396 square inches = 25004.7 pounds. Divide that by a factor of safety (usually 2 but aircraft designers sometimes use 1.5).
Remember this is only good for pure, evenly loaded tension - if the member is loaded in any other way or if the load is concentrated somehow the loads at a point in the member may be higher & you can get a point failure which will start a crack or a tear in the member.
Last edited by Mike Switzer; 10-15-2012 at 07:31 AM.
Reason: cant type without coffee
Thanks Mike, I'm trying to figure out a suitable size for a wing strut for a 1928 high wing monoplane, gross weight around 2,000 lbs., so 1 3/4" x .049" (large fairing on the back) should be plenty.
The compressive load from negative g's in flight and ground wind loads needs to be considered for struts.
usually the strut column load capapability is much less than tension and depends on many factors.
Now that I have woken up I realized you are using round tube (I am using square in mine) but you get the idea.
Will the strut have a single bolt at each end, so that it is free to rotate at the joint? If so you can figure that it is only loaded in tension (in flight, compression on the ground). If the ends are fixed (welded or multiple bolts) the loading won't be as simple - there will probably be a moment acting at the ends unless it is designed like a truss forcing all the members into either tension or compression.
Wicks & Spruce both sell aerodynamic tubing, you might want to check it out, it will save you making a fairing.
For compression loads there are several equations depending on length - Euler & Johnson are most common.
Originally Posted by Bill Berson
Thanks Mike (been away from the forum for a bit...), the struts are built the way they are to copy how they were in 1928, tube structure covered in fabric
Upper end has a horizontal bolt, and lower end has an adjustment fork with vertical bolt onto tab on fuselage.
Last edited by Andrew King; 10-29-2012 at 11:27 AM.
In 1928 I feel sure they were using 1018, so if you just duplicate the dimensions with 4130 it should probably be good - but you should try to do a little research to verify that.
The yield strength of current 1018 is 32 kpsi for hot rolled & 54 kpsi for cold rolled. 1018 used to be as high as 70 kpsi for cold rolled, and I would have to do some research to be sure, but I don't think it was available until the mid to late 30s at the earliest - it was used in the war effort in some designs because 4130 was not available or cost more.