Angle Beam 1

[Plastickmat]

I have to calculate maximum shear for an angle beam of 44W steel.

The setup goes that its actually a sliding bender applying a "moment" onto my beam. The bender has a "Travel" distance I'm trying to find the worst case scenarios where the tension would create the largest stress on it.

Anyone have a few tips of how to start this, thanks.


Mat

 

[Plastickmat]

Please be more clear on the following : unit tension? Isn't that what I did with my "1KN" ? Here's the sketch I did.

Also : fcy? and how do I calculate the stress your speaking of from your unit tension. These are the steps I'm confused with.

 

[rb1957]

if your pic is right, then your problem is easier ... the maxiomum moment in the angle is 1/2 the applied moment.

yes, unit tension = 1kN, max moment in angle = 130kN.mm, reaction = 0.44kN.

ok, good, now wiki "bending stress in beam" or something like, or find a strength of materials text. you're looking for something like stress = My/I.

something to consider is that your applied moment isn't along a principal axis of the section, you should be able to find that too (wiki or texts).
you should be able to determine the principal axes of your angle cross-section (the quick answer is the minor (weak) axis is through the mid pts of the two legs of the angle, the major (strong) axis is normal to this, thru the elbow of the angle (this includes several simplifying assumptions). You applied moment will be broken into two components acting along the principal axes and resultant stress = M1*y1/I1 + M2*y2/I2.

good luck !

 

[Plastickmat]

Why is your max moment 130kN.mm?

I understand what you mean with the component moments and having the angle in this position was one of the challenges I realised.

Since we're using a unit tension, where does it come together that I have the official 'max value' if my calculations are based on arbitrary values.

Thanks

 

[BAretired]

If T = 1 kN, the applied moment is 0.261 kN-m. The reactions at each end are 0.437 kN as shown on the sketch, one up and one down. Force F is 0.261/b, one up and one down where b is the distance between them.

Beam moment varies linearly from 0 to -Ra to Rc to 0 at left support, first F, second F and right support respectively. In the above, a is distance from left support to first F and c is distance from second F to right support.

This is elementary statics which you should know if you are making these calculations. I think your employer will require the services of a structural engineer if the results are to be sent out to customers.



BA

 

[rb1957]

trying to find the maximum tension allowed, you're using a unit tension load to develop the critical internal stresses (= key design limitations).

and you know your maximum allowable stress (fty, fcy) and the SF you want in the design.

so you can factor the unit tension up untill the internal stress calculated equals the maximum allowable ... T = Tu*(allowable/stress)/SF

you can calc the maximum moment on the beam, draw a moment diagram. you'll see it starts at zero (pinned ends), increases to a maxmimum, drops where your external moment is applied, and increases again. 1/2ing the external value applies when the load is introduced at the mid-length of the beam; different positions will produce a higher beam moment.

 

[Plastickmat]

Many thanks guys,

Ya basic statics I am still learning, I see it becomes very elementary for guys like you which is nice to see. All of you have been very helpful I'll be crunching up some more with your help!

Thanks a lot
Mat

 

[rb1957]

get a strength of materials textbook, look up online reference's MIT's opencourseware is a good resource.

 

[Plastickmat]

Just realised something don't know if it makes a difference to your understanding,


The sketch including moment calculations only represents the "sliding part" on top of the main beam. Basic statics I guess again, but is it more logical to represent the reactions calculated on the top part in two forces on the main beam or is it "ok" to represent the same moment on the beam as another ponit moment.?

Basically what makes the setup easier to analyse, because on the slider it generates a moment on the "center point" for rotation, but on the main beam the slider's position creates a punctual moment in different points.

Thanks

 

[rb1957]

yes, the slider is probably not applied a moment to the angle beam (it would do if it was clamped over it, like a sleeve on a tube) ... a couple is more reasonable.

it should actually slightly reduce the peak moment, ie using the moment is conservative for the beam bending moment.

 

[BAretired]

The sketch including moment calculations only represents the "sliding part" on top of the main beam. Basic statics I guess again, but is it more logical to represent the reactions calculated on the top part in two forces on the main beam or is it "ok" to represent the same moment on the beam as another ponit moment.?

To calculate end reactions on the beam, it makes no difference whether you consider a "point" moment or a force couple. You must use a couple if you want to draw the beam moment diagram correctly.

Basically what makes the setup easier to analyse, because on the slider it generates a moment on the "center point" for rotation, but on the main beam the slider's position creates a punctual moment in different points.

What is a "punctual" moment? A moment which is on time? So far as end reactions are concerned, the position of the point moment (or couple) makes no difference to the end reactions of the beam but changes the shape of the bending moment diagram.

By the way, you should draw the shear force and bending moment diagrams of the beam for a clearer understanding of the problem.


BA

 

[Plastickmat]

A punctual moment is a moment that is on time, yes.

Think I've got it nailed down a little better now, thanks for the support guys!