There are really two questions there combined into one question. Schedule
80 is a pipe wall thickness designation....It does not specify alloy - there
are approximately 30+ alloy designations for pipe:
http://www.key-to-steel.com/Articles/Art24.htm
But making it simple, it it is assumed that the material is standard black
steel pipe to ASTM 53, typical strength has a 52,000 psi yield. That would
be about 70,000 psi ultimate strength, and a hardness of about 79 Rockwell
B.
Normalized 4130 pipe, has a hardness of about Rockwell C 30, with an
ultimate tensile strength of about 130,000, and a yield strength
approximately of 110,000 psi.
Just based on strength, the 4130 is better. But our failure criterion is
bending, or plastic deformation. Now we have to compare the wall
thicknesses and the strength, so see which works better. This is where the
metallurgist, being the jack-of-all-trades, instantly transforms into a
stress engineer.
The maximum stress in a beam is given by s=Mc/I, where s is the stress, M is
the moment, c is the distance of the outer fiber from the neutral axis, and
I is the moment of inertia.
I for a thin walled tube (good enough in this illustration) is: I =pi*t8r^3
Assuming a 2" diameter, the I for the two cases are:
4130 (2" dia. x 0.18" wall): 0.565
A53 (2" dia x 0.25" wall): 0.785
c is the fiber distance from the neutral axis, and is c = r + t/2differs for
each one because of the wall thickness:
4130 (2" dia. x 0.18" wall): 1.09
A53 (2" dia x 0.25" wall): 1.13
M is unknown, and is a function of the length of the mast, and the
windloading. For the purposes of this, the masts are the same length, and
the windloading is the same.
Failure is yielding of the material, or when the stress equals the yield
strength (less any safety factor - we will ignore that for the time being).
For the 4130 pipe, yield is 110,000 psi = s = Mc/I = 1.93M
For the A53 pipe, Yield is 50,000 psi = s = Mc/I = 1.44M
The max moment at failure can be calculated by rearranging:
for 4130 M = 110,000/1.93 = 56,994 ft-lbs
for A53 M = 50,000/1.44 = 34,722 ft-lbs
In other words, the 4130 can take about 1.64 times the moment than the A53
pipe, even with a thicker wall. Since the lengths of the mast are assumed
to be the same, then the 4130 can take 1.64 times the load that A53 can
before bending can occur, for this specific example, and specific material
conditions.
The next question is, is this strength necessary for the application? It
depends on the windload of the antennas, and the length of the mast. From
that calculation, and a suitable safety factor, the maximum antenna loading
for a specific mast can be determined.
I hope that this clarifies some things - finnaly I have been able to answer
someone's question and contribute, instead of just asking questions..... :)
Scott MacKenzie, PhD
Metallurgist
KB0FHP
-----Original Message-----
From: towertalk-bounces@contesting.com
[mailto:towertalk-bounces@contesting.com]On Behalf Of Mike Bragassa
Sent: Thursday, July 13, 2006 11:39 PM
To: towertalk@contesting.com
Subject: [TowerTalk] Chrome-moly VS: steel
Importance: High
For you metallurgists in the group:
Re: 20 ft mast pipe
How does a 0.18 inch chromemoly pipe compare to a 0.25 inch schedule 80
pipe?
73, Mike, K5UO
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