"C" channel beam loads

[JohnTheWelder]

Background:
Hobby welder trying to design a utility trailer, no engineering background.

Problem:
I'm trying to determine if a 3" steel (A36) "C" channel (4.1 lbs/ft) can support a 1500 lb. point load at the center of a 78" span. Here are my calculations:

Shear force = 750 lbs.

Max Bending Moment (at center) = (750) (39 in.) = 29250 in./lbs.


Bending Stress (3"-4.1 channel) = (29250 in./lbs.) (1.5 in.) / 1.66 = 26431 psi

Bending Stress (4"-5.4 channel) = (29250 in./lbs.) (2.0 in.) / 3.83 = 15274 psi


Horz Shear Stress (center of 3" channel) = (750)(.605)(.75) / (1.66)(.170) = 1206 psi

A36 tensile yield stress = 36ksi.

For bending stress, I assume that the 3&quot; &quot;C&quot; channel is ok since 26431 < 36000. Or do I need to allow for a safety margin, say maximum design stress of only 60% of yield stress?

For horz shear stress, I don't know what the allowable value should be, so I'm clueless here. Where can I find the allowable horz shear stress for 3&quot;-4.1 A36 &quot;C&quot; channel?

Are my calculations correct?

Should I just give up and go back to playing with Legos?

Thanks

 

[pianomarty]

Hi Welder

I agree with the way you've done the moment calculation,
except you've been a little careless with units; the
moment should be inch-lbs, not inches PER lb. Since
the moment of inertia (1.66) is in units of in^4, and the distance, you up with lbs/in^2, as you should.

I'm not sure I understand your shear calculation. I would
just divide the 750 lbs. over the whole cross-section
of the beam (1.21 in.) to get a shear of 600 lbs/in^2.
However, when I look at your units, you've got (area)x(depth) on top divided by (inertia)x(thickness) on the bottom....so your units work out correctly to lbs/in^2.
Do you know something I don't know?

(Either way, the shear looks much too low to be the
governing factor.)

Marty

 

[JohnTheWelder]

Marty,

My horz shear stress formula came from the Univ of Wisconsin-Stout's online &quot;Statics and Strength of Materials&quot; course. It's defined as follows:

Horz Shear Stress = VAy'/Ib

V = Shear force at location along the beam where we wish to find from the horizontal shear stress

A = cross sectional area, from point where we wish to find the shear stress at, to an outer edge of the beam cross section (top or bottom)

y' = distance from neutral axis to the centroid of the area A.

I = moment of inertia for the beam cross section.

b = width of the beam at the point we wish to determine the shear stress.


Am I using the wrong formula? Also, is the 3&quot; beam sufficient?

Sorry about getting fast and loose with the units. College was 22 years ago, and I was a business major.

 

[pianomarty]

Yes, I looked up some references on the internet and what you've done is right. The only detail that you might question is your value of .75 for the distance to the centroid. That figure should probably be a little higher. But it's not available on the website where I found the rest of the beam data:

http://ourworld.compuserve.com/homepages/MJVanVoorhis/t404.htm

I can't say I understand these shear questions very well, but I'm pretty sure that in structural design, the allowable shear stresses are just a little less than the allowable tensile stresses. So in your case, it is quite a low stress.

Marty

 

[desertfox]

Hi JohnTheWelder

I was just looking at your problem and was wondering how the ends of the channel were fixed.
I assume they are welded at each end and if they are it makes a difference to the beam calculations.
I am working on the assumption that your calculations are based on simply supported beam theory which if the channel is welded each end is incorrect.
Also the channel in the attitude you are using it would create twisting in addition to bending because the load you are applying does not pass through the flexural centre of the beam which for the channel would outside the section a distance away from the vertical channel wall.So before I put pen to paper can you tell me the end fixings of the beam and if possible where the point load is applied horizontally along the 1.5&quot; flange

regards

desertfox

 

[JohnTheWelder]

Hi desertfox,

I just knew someone was going to ask about the ends of the channel! Ok, let me describe what I am trying to do...

The goal is to make a 82&quot; x 148&quot; frame that will be the support for the floor of a dump bed utility trailer. The outer perimeter of the frame is a rectangle constructed from 5&quot;-6.7 channel, 78&quot; x 144&quot; (inside dimension), with the flanges facing out. The 3&quot;-4.1 channel would be used as intermediate supports for the floor of the trailer, spanning the 78&quot; dimension on 18&quot; centers.

The ends of the 3&quot; channel would be welded to the flat 5&quot; face of the 5&quot; channel. The actual floor surface would be 12ga or 14ga steel, welded to the frame.

After thinking about your comment on twist, I'm hoping that the floor sheating will stabilize any twist in the 3&quot; channel.

As far as loading goes, I'm assuming 1,000 lbs/ft of floor, evenly distributed, for a total load of 12,000 lbs.

I can't address your question regarding the location of the point load on the flange. Why? Because my limited knowledge of engineering restricts me to using fuzzy phrases like &quot;evenly distributed&quot;. Bottom line, I don't know what to tell you...

 

[desertfox]

Thanks JohnTheWelder

I will get back to you shortly

 

[Tuck]

Your allowable shear stress is approximated by 50% of your tensile stress. If the value shear stress is well below,in yourcase 18ksi, you're fine.

 

[desertfox]

Hi JohnTheWelder

Based on my calculations your 3&quot; beam in bending would be okay. My calculations are based on the beam being fixed at both ends ie:- ends cannot rotate (welded,in practice the &quot;c&quot; section would probably be somewhere in between a fixed end beam and a simply supported beam.
So my figure for maximum stress is:-

13,295psi just under half the the uts
you specify.

Normally the maximum shear stress of a material is about 50 to 60% of the u.t.s so your calculation shows this to be okay.I have based this calculation on the &quot;c&quot; section being prevented from twisting from your earlier message.
I also worked out that if you used a 70mm x 70mm x 3.6mm wall box section this would give you equal strength in both directions and would be capable of handling the bending too.
The weight per foot of this box section would be 5lb per foot and it would also be a lot better at taking torsional loads whether generated from trailer loading or during trailer use.

hope this is of help

regards desertfox

 

[PMD123]

The allowable stress for shear is .4 x 36 ksi per AISC Code for A36. Only count the web area in a shear calculation. You do not need to calculated the horizontal rolling shear unless you are connection 2 sections etc. together.

The maximum bending stress for the channel would be .6 x 36 ksi and could be less.

Fb=allowable stress = 12x10^3/(l*d/Af)
l= length between compression flange braces say 78&quot;
d/Af = 7.78 for a c3x4.1 and 8.52 for a C4x5.4

Therefore the C3x4.1 allowable stress = 19775 psi for 78 ' unbraced length of compression flange and C4 x 5.4 is 18057 psi.

Don't forget impact forces that are possible as well as fatigue limits.