basic design of steel beams

[visitor]

When we are designing a steel plate girder to support a floor slab, we normally check the stresses in the flanges as longitudinal axial stress due to longitudinal bending.
But there is considerable local bending of the flange transverse ( ie. as a cantilever from web plate). But normally this is not taken into account.

How to consider this bending and combine it with longitudinal bending stresses ?
Please advice.

 

[ishvaaag]

You may use a Von Mises Criteria with all the stresses present. In any case try to assign investigated values to the transversal flange bending stresses. If you have to think about the value yourself, estimate the compatible deflection at the tip of the flange when the slab is present and loaded, this maybe you an idea. Also, this effect for many cases should either be small or well covered by the safety factors of use, since composite rolled beams could also require such concept calculation and is rarely seen. So you would estimate the limit or allowable stress depending if LRFD or ASD, then use it as limit for the Von Mises criteria of failure for all stresses present.

A further comment proceeds. If most of the examples you see in the books for cases like yours omit this, it may be you are right, but more likely, the in the books described procedure has shown to be safe enough. If one thinks this way (and for some cases may be very well needed) we would need be continuously adding new stresses and safety factors for apparently not contemplated effects, such stress concentrations and so on. It is in the nature of the codes to precisely offer safe procedures of design in the omission of such technical delicacies difficult to implement in ordinary practices.

Steel, elastic buckling apart, also enjoys MAGNIFIC in-member strength safety factors, due to the use of yield value as reference, not ultimate (overstrength) stress. This is precisely put unto play at the detail level, the plasticity allowing precisely the misconsideration of such finesses as long as the to yield referenced code methods are used.

 

[vonlueke]

> How to consider this bending and combine it with longitudinal bending stresses? <

Below I'm answering your question using only fundamental concepts and neglecting in-plane shear stress (if it's significant, though I would assume it's probably not). Of course it might be wiser to also check a detailed reference on plate girder analysis, such as Salmon, Charles G., Steel Structures Design and Behavior, if you have one.

Let x axis be beam longitudinal axis, y axis be horizontal transverse axis, and z axis be vertical axis. Let origin be at beam centroid. Let w = beam uniform distributed load applied to top flange. Let external moments Mx = Mz = 0. Let P = beam external axial load, if any, where positive value means tension, negative means compression. Let b = top flange width, t = top flange thickness, z1 = distance from centroid to upper extreme fiber. Let Iy = beam moment of inertia about y axis. Let s1x = top flange top surface x-direction normal stress, s1y = top flange top surface y-direction normal stress, s2x = top flange bottom surface x-direction normal stress, s2y = top flange bottom surface y-direction normal stress. Notice, with above coordinate system, positive My value places top flange in x-direction tension (beam curvature concave downward). Input negative value for My in regions where beam curvature is concave upward. Let's assume your units are N and mm, though any will do.

All along top flange top surface, s1x = (My)(z1)/(Iy) + P/A. All along top flange bottom surface, s2x = (My)(z1-t)/(Iy) + P/A. Also, treating each side of top flange as cantilever, per your suggestion, Mx' = 0.5[(w/b)(1 mm)](0.5*b-y)^2. Maximum y-direction stress occurs at or near y=0 (the web). Hence, s1y = [Mx'(0)](0.5t)/[(1 mm)(t^3)/12]. s2y = -s1y.

Then, von Mises combined equivalent stress (for biaxial stress state) on top flange top surface is s1vm = (s1x^2 - s1x*s1y + s1y^2)^0.5. And von Mises stress on top flange bottom surface is s2vm = (s2x^2 - s2x*s2y + s2y^2)^0.5.

Then, if using ASD (allowable stress design), compute MS = yield margin of safety = {(Fy)/[FS*max(s1vm,s2vm)]} - 1, where Fy = beam material yield strength = 250 MPa for A36 steel, and FS = 1.67, a fundamental factor of safety against yield used in AISC. If MS is zero or positive, above analysis indicates flange under combined bending stress is not yielding. Good luck.

 

[visitor]

Thanks Ishvaaq & vonlueke

As you said, the von Mises equation is good for checking the above. In general , no code specifies the check as per
Mises equation. But BS5400 - Part 3 specifies this method for checking the longitudinally stiffened flanges of beams.

Can we apply the same check for designing the orthotropic deck slab of bridges. (ie. to check the deck plate for the
combined stresses due to the longitudinal bending stresses and the transverse stresses due to local bending etc)?
How does the deflection of the deck plate be calculated due to the two directional bending of flange plates ?

I wish to get your reply.


Thanks

 

[ishvaaag]

In spite of the fact of the Von Mises check being good to analyze the failure of the steel ductile material, the intricacies of orthotropic deck -girder, box or truss- design make me think better follow reliable procedures explicited elsewhere. Furthermore it is part of the orthotropic deck concept to accept some iterated local plastifications, then of course at such localizations the Von Mises criterium would show the structure to fail (locally) and hence would be contrary to the engineering criteria that have deemed the orthotropic decks fit for bridge use. In reality this may happen as well at some connections if you go Von Mises instead of using the code approved procedure: think for example in the limit squashing stresses in plates bearing shafts in shear connections.

 

It seems to me the topic is getting close to a stiffened panel design. A typical stiffened panel has longitudinal, transverse, shear and normal pressure acting all together. There are two checks (if neglect fatigue for now) for the design 1) yield strength using von Mises or principal stress criteria and 2) buckling strength. #1 is very straight forward but #2 is quite complicated. There are 3 codes available for steel structures, Dnv 30.1, ABS buckling check and API 2V. You can see sample program at

http://www.lanxun.com/pce.