Pressure Load

I suspect that you would want lbs/in of length on a beam because that makes the most sense for this type of problem - in which case, you would multiply psi by the width of the beam.

Suppose the beams are spaced ten feet apart and a concrete floor slab spanning from beam to beam is six inches thick (weighing approximately 75 pounds per square foot). Suppose the design live load is 150 pounds per square foot and the carpet plus ceiling plus partition load is 20 pounds per square foot. The total load on the floor system, neglecting the weight of beams is then 75 + 150 + 20 = 245 psf.

Now, in order to design the beam, you want to know the uniform service load per lineal foot. That would be 10 * 245 psf = 2450 plf. If you are using modern design methods, you would want to know the factored load. That would be 1.25(75 + 20) + 1.5(150) = 344 psf (the load factor for dead load and live load is 1.25 and 1.5 respectively). So the factored load on the beam would be 10 * 344 = 3440 plf.

You would calculate the bending moment and shear using factored loads. You would calculate deflection using service loads. Then you would select a beam having adequate strength to resist factored loads and adequate stiffness to hold deflection to whatever standard you decide is appropriate.

Hi Guru:

thanks for the reply.Well lets say I have a rectangler plate with pressure loading on it.If I have to analyze that as a beam then should I multiply the pressure by length of the plate or width to get uniform load of lb/in.?

thanks

If you are looking at stresses in plates -

Try Roark's Formulas for Stress and Strain

OR

http://www.efunda.com/formulae/solid_mechanics/plates/casestudy_list.cfm

You should not analyze a rectangular plate as a beam. You should consider it as a two way structure and use one of the classical solutions as suggested by the other Guest on Post #4. For simply supported rectangular plates, you could use the Navier solution. For other boundary conditions, you could attempt to analyze it using the theory of elasticity. For a satisfactory approximation, you might want to employ "Yield Line Theory".

When the length to width ratio is very large (say greater than 2:1), a rectangular plate is sometimes approximated as a one way slab. In that case, consider a strip one inch wide with a load of "q" lb/in. Then you can treat it as a beam, but remember, in the end regions, there will be two way action, so if you are designing a concrete slab, you will need to reinforce for "corner levers".

Hi,

Lets say I have a plate of 30 inch in length and 11 inch width and 0.125 inch thick with pressure load 25 psi on the`plate. Then as per your suggestion i can treat it as a beam with 25*11=275lbs/inch load on 30 inch length.(I did'nt consider unit width as I know the total width).is it okay?

where is thickness of plate ciming into the caln?

thanks

If the plate is simply supported on all four edges, you could treat the plate as a one way slab spanning 11". For a one inch width of plate, the load is 25 pounds per inch and the bending moment is 25*11²/8 = 378"#. The section modulus is 0.125²/6 = 0.0026 in³. The maximum stress in the plate is 378/0.0026 = 145,150 psi which is too high, so you need a thicker plate.

If the plate is supported only on two edges such that it spans 30" then your approach would be correct but the bending stresses would be even higher than above. Isn't it simpler to consider a one inch width with a load of 25#/" than an 11" width with a load of 275#/"?

Hi

thanks for the reply.

But Now I will come to my class problem. I have a 100 inch length and 40 inch wide & 6 in depth plate , it has honeycomb structure what I mean 0.125 inch face sheet of aluminum on either side and rest (6-.125*2)in honeycomb made of alumnium.It is supportted on 4 sides.Pressure load on this is 75 psi. I need to calculate the stiffness of this plate for this load.

So for beam of 100 in span I applied 75*40=3000 lbs/in load and calculated the deflection using equation of a beam simplly supported on both the sides.So do uthink its right.

I know to check the stresses its not simple because it is composite.But I am just checking the stiffness using hand calculation.

Once I got the deflection lets say it 3 inches at the center then stiffness would be total load on the plate which is (100*40*75=300000 lbs) devided by deflection, which is 300000/3=100000lbs/in.

Do u think I am right in this.

Hope I am not disturbing u.

Thanks

CR4 does not condone members working out homework problems for students, so I cannot do the problem for you, but you might want to think about a couple of things.

A rectangular plate spans in both directions. When the length is more than twice the width, it is reasonable to treat it as a one way span...in the short direction! That is not theoretically correct, but the study of engineering is full of approximations, so it seems to be a reasonable approach and it is conservative.

Your actual problem consists of a honeycomb structure with cover plates top and bottom. Assuming the honeycomb material is isotropic (has the same properties in each orthogonal direction) you are probably justified in assuming that it acts similarly to a solid plate. Why then would you consider the span to be 100"? Would it not tend to span in the shorter direction?

Hi,

Isn't considering a span of 100inch conservative over 40.As the length is more deflection is more.

Also in ur calculation I didn't understand the equation of bending moment 25*112/8 = 378"#. And The section modulus 0.1252/6 = 0.0026 in3 (how did u consider the inertia in to accaount).

Thnks

Hi,

Don't you think that the equation you used i.e.25*11²/8 should be actually 25*11³/8.

The pressure is 25psf, so the UDL on the width is 25*11, so the concentrated load on the width is (25*11*11), then by applying the equation of BM of simply supported with UDL i.e P*L²/8 it comes out to be (25*11)*(11²/8).

No, I don't think so. Think of it as a one inch wide strip of steel or aluminum plate. The load on the plate is 25 psi, so the load on a one inch strip is 25 pounds per lineal inch. The bending moment on this narrow beam is thus 25*11²/8 or 378"#.

If you wish to talk about the entire plate, then the load would be 25*30 = 750#/" and the bending moment would be 11,340"#. Of course, this would not be accurate as the end regions of the plate could not deflect due to simple supports on all four sides.

Considering the plate as a one way slab is an approximation and is roughly valid only in the central region of the plate. A finite element analysis would find that the central deflection is marginally less than predicted by this simple theory because the plate does carry some load in the long direction and because the Poisson ratio affects the way a plate deflects. Elsewhere, the plate would dish down from zero at all edges to the maximum deflection at the midpoint of the plate.

It is also interesting to note that the corners of the plate, if not held down would rise above the supports due to so called "corner levers". But the simple support implies resistance to vertical movement, so the plate is considered held down at all points. If, instead of metal plate, this was a concrete slab, you could expect diagonal cracks across the corners resulting from tension in the top fibers.

Hi,

Suppose the beams are spaced ten feet apart and a concrete floor slab spanning from beam to beam is six inches thick (weighing approximately 75 pounds per square foot). Suppose the design live load is 150 pounds per square foot and the carpet plus ceiling plus partition load is 20 pounds per square foot. The total load on the floor system, neglecting the weight of beams is then 75 + 150 + 20 = 245 psf.

Now, in order to design the beam, you want to know the uniform service load per lineal foot. That would be 10 * 245 psf = 2450 plf.

My qyestion for this answer is

Is this 2450 plf acting as a uniform load on slab in the length direction or width direction, why did'nt u consider the width to muplitply pressure to get uniform load.

thanks

Suppose that the dead plus live load of a floor system is 245 pounds per square foot. Suppose that the floor area is 40' x 60'. Let us say, that we decide upon a framing system with beams spanning 40' spaced at 10' centers and slabs spanning from beam to beam (probably not a particularly good choice of framing systems, but I am trying to illustrate a design method).

In days gone by, when engineers used elastic design, we would have said that the load on each beam is 10*245 = 2450 pounds per lineal foot of span or 2.45k/'. Since the beam spans 40', we would have calculated the bending moment as:

M = wL²/8 = 2.450*(40)²/8 = 490'k

We would then have checked the steel handbook for a beam which could provide a resisting moment of 490'k and would have made a selection. Naturally, we would confirm that the shear capacity was adequate and that the deflection met the requirements of the building code.

Today, the procedure would be a little different, but not much...we would calculate the factored bending moment, select a beam capable of resisting the factored moment, check factored shear, unfactored deflection and eventually we would select a beam section which satisfies all of these requirements. And that selection is what we would specify on the framing plan.

I am not sure that I have fully answered your question...if not please respond.

HI ,

I understand what u r tring to say but what is the width of the beam which u considered for 40 inch length beam. Because this width is not coming in to any of ur calculation.

Or is it 10 inch spacing between beams is the width of the beam?Am i right?

Thanks

I don't believe you and I are communicating very well. I was talking about an arbitrary framing system consisting of beams spaced at ten foot centers, spanning forty feet (not forty inches) and carrying a floor load of 245 psf.

The width of beam would likely be about 6" or 7" but the width of beam is of no practical interest at all and plays no role in the calculation. If the beams are spaced ten feet apart, each beam carries ten feet of floor load, in this case, 10*245 = 2450#/'.

This particular discussion does not relate to a rectangular plate. It relates to a system of slabs and beams in a more or less typical framing system for a large floor area.

Are you purposely being obtuse? Because if you are, I will withdraw from this discussion immediately. If, on the other hand, you are acting in good faith and really want answers to questions which are troubling you, I will try to help in any way I can. I simply don't understand your problem.

If anyone else sees something that I am missing, please join in.

How do I convert a pressure (lbs/in^2) into a uniform line load (lbs/in) on a beam? Do I multiply the pressure x length of the beam or do I multiply the pressure x width? thanks

Look at your units, first multiple your pressure by the area it is acting on, giving you lb's only then divide by the length for lb/in. this is your UDL, lb's only is you virtual point load for analysis.

Hi,

why should I multiply by length , why not width to get lb/in.

Thanks