Tributary Areas for Gravity Loads

8/2/2019 Tributary Areas for Gravity Loads

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Tributary Areas for Gravity Loads

Last Revised: 03/24/2009

If the beam is supporting a floor, roof, or wall that has a pressure loading normal to the surface,

the total force on the beam equals the area of surface supported (i.e. the tributary area) times thepressure on the surface.

Consider a series of floor joists (repetitive beam members) supporting a floor system as shown inthe framing plan in Figure TA.2.1.

Figure TA.2.1

Sample Floor Framing System

Taking a closer look at a single joist, as shown in Figure TA.2.2, you can see that the floorsystem spans as a continuous beam across evenly spaced supports. Note that the floor spans

from joist to joist instead of in the same direction as the joist since the floor is substantiallystiffer (try the deflection calcs if you want!) in the short direction. In this situation, the floor


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system will transfer half of a span's uniformly distributed load to the joist on either end of thefloor span. So, it can be said that the joist supports all the load on the area shown (the hatched

area). Each joist in the system will likewise support the floor system, so that all of the floor areais accounted for.

Figure TA.2.2Floor Joist Tributary Area

The hatched area is referred to as the tributary area for the joist. It's dimension transverse to thejoist is half the distance to the next joist on either side (also known as the tributary width) and it's

length is the length of the joist. The total load (in force units) on the joist equals the tributaryarea (area units) times the uniform pressure loading (force per unit area).

The two dimensional loading diagram is constructed by multiplying tributary width (length units)

by the uniform pressure loading (force per unit area) to get the distributed load magnitude (forceper unit length of joist). This can be expressed mathematically as:

w = q tw

where:


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y w = magnitude of the distributed load (force per unit length)y q = the magnitude of the uniform load (force per unit area)y tw = the tributary width (length).

Note that tw = s if the joist spacing is uniform.

Another way to look at this is to consider w to be a representative unit length of the joist. Thearea that it supports equals the tributary width times the unit length. The load w that that unit

length supports equals the tributary area (1*tw) times the uniform pressure load q. Hence theload per that unit length is w = 1*tw*q = q tw.

The idealized beam loading diagram is shown in Figure TA.2.3.

Figure TA.2.3

Idealized Beam Loading Diagram

Noticing that each joist transfers half of its load to each supporting member (i.e. the reactions

each equal wL/2), we can now draw the loading diagram for one of the supporting girders.

As the girder collects the joist reactions, we can draw the girder load diagram as having a seriesof point loads. Figure TA.2.4 shows such a case for a typical girder supporting evenly spaced

joist reactions of equal magnitude.

Figure TA.2.4

Girder Loading Diagram


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In order to do the analysis we need to have designed the joists so that we know where each joistis located.

To side track for a moment, consider the possibility that we could approximate the series of pointloads by an equivalent distributed load. The equivalent distributed load could be computed by

y adding up all the point loads and dividing by the girder length, ory dividing a point load, P, by the point load spacing, S.

You will get the same answer either way if the reactions are equal and the spacings are equal.

Since we are designing beams for shear, moment, and deflection, approximating the series of

point loads as a uniform load will only work if the values for shear, moment and deflection arenearly the same or greater than the values obtained from an analysis of a series of point loads.

Let's check this out.

Consider a beam of length L, that supports a series of point loads of magnitude P.

The next three figures compare the results for shear and moment from analysis that consider theloads as point loads and a an equivalent uniform load.

Figure TA.2.5a

Internal Force Comparison when S = L/2


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Figure TA.2.5b


Figure TA.2.5c



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Notice that, as the number of loads increases, the difference between the results for the series of

point loads begins to come closer to the uniform load results.

Generally, the approximate method is used whenever the joist spacing is less than or equal to L/4since the results are pretty close and the uniformly distributed load is easier to analyze than a

series of point loads.

So, with the above in mind, lets take a look at one of the girders in Figure TA.2.1. We'll startwith the girder on grid line 1 between grids A and B. Instead of computing the joist reactions,

we can see that each joist deposits half its load on each of the supporting girder. Therefore, sincethe floor pressure is uniform, we can say that the girder supports the sum of half the areas of each

of the joists. Graphically, we can draw a line down the center of each supported joist and saythat all the area between the line and the girder is tributary to the girder. You can see this in

Figure TA.2.6. The distance of the tributary area in the direction of the joists is the tributarywidth.

Figure TA.2.6

Area Tributary to Girder 1,AB


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The load diagram for the beam would be that of a simply supported, uniformly loaded beamhaving a load intensity:

w = q tw

Where tw, in this case is seven (7) feet. Notice that the other girder on grid 1 has the same load

intensity. You should be able to say way that is so at this point.

We can repeat this exercise for all the girder in the framing plan. Note that all the floor area

must be accounted for! See Figure TA.2.7 to the tributary area assignments for all the girders.

Figure TA.2.7

Girder Tributary AreasClick on image for Powerpoint animation


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Click on the Figure to get a powerpoint animation that dynamically illustrates the girder tributary

areas.

Next we look at the columns.

Each column supports either one or two, simply supported, uniformly loaded girders. Each

girder adds half it's supported load to each supporting column. Hence, each column supportshalf the area supported by each contributing girder.

For example, Figure TA.2.8 shows the area tributary to the column at the intersection of grids 1

& B. This area represents half the area supported by girder 1,AB and half the area supported bygirder 1,BC.

Figure TA.2.8

Column 1B Tributary Area


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As all the load on the floor system is supported by the nine columns, we can draw a diagramillustrating the areas that are tributary to each column. Again... all the area must be accounted

for and no part of the area is to be counted twice. Figure TA.2.9 shows the diagram for areatributary to the columns. You can click on the figure to see a powerpoint animation of the areas.

Figure TA.2.9

Column Tributary AreasClick on image for Powerpoint animation


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The load on each column can be determined by multiplying the Tributary Area for each column

by the uniform load intensity, q.

Hopefully, you are starting to see the usefulness of this method. You can determine the load on

any member of this floor framing plan in any order! Also the analysis of the girders issomewhat simplified.

Now, lets look at a few more challenging framing layouts.

Framing that is not perpendicular to the supported member

A rather common situation is the one illustrated in Figure TA.2.10. In this layout, some of theframing is perpendicular to it's supports and others are not.

Figure TA.2.10

Floor Framing PlanClick on image for Powerpoint animation


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Each joist has the same uniform load intensity, w = q s, but has a different length. The designerwill need to decide whether to design for the worst case and use the same for all joists or

decrease the size as the joists get shorter.

To find the loading on the two girders, we can readily identify their tributary areas as being half

that supported by each joist, so we can draw a line down the center of the joists to divide the twotributary areas as shown in Figure TA.2.11.

Figure TA.2.11

Areas Tributary to the Girders


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In this case, if you are observant, you will notice that each girder supports half of all the joists

which support all the floor, so it follows that each girder supports half the total floor load. Thequestion now is: How is applied to each girder?

Let's start with girder AB.

In this case the joists are perpendicular to the girder. Each joist reaction can be distributed over a

length of girder equal to the joist spacing, s. This means that the linear load intensity is greater atthe "A" end of the girder. The 2D load intensity, w, at the A end of the girder equals:

wA = q tw = q (L1/2)

The load intensity at the "B" end of the girder equals zero since tw is zero at this point. Theresulting beam load diagram (not including beam self weight) is shown in Figure TA.2.12. If

beam self weight is to be included then a uniform load equal to the beam weight per unit lengthshould be added to the loading.


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Figure TA.2.12

Girder AB Load Diagram

Another way to arrive at the value for wA is to recognize that the distribution is linearly varyingfrom zero then solve the following triangle equation for wA:

q (Trib. Area) = 0.5 L2 wA

The total load from the diagram equals the tributary area times the load intensity.

Another thing to note is that the load diagram follows the shape of the tributary area diagram in

this case. This is always true when the supported framing is perpendicular to the member. Thisis not precisely true for other situations, as we will now see.

Consider girder BC. In this case the supported framing is not perpendicular to the girder.

A common mistake here is to assume that peak load in the loading diagram occurs where a lineperpendicular to the girder passes through the center of the longest joist. This in not right! Note

that the longest joist (and the most heavily loaded) transfers all it's load to the "C" end of thegirder, making that the largest load intensity. Since joist length's vary linearly, the resulting

beam loading diagram is of the same shape as beam loading diagram for girder AB.

As seen in Figure TA.2.13, a joist that is coming into the girder at an angle U from perpendicular

spreads it's load over a length s/cos U of the girder. The load intensity per unit length of girderthen becomes:

wj = [q*(s (Lj/2))] / [s / cos U] = 0.5 q Lj cos U

where:

y (s (Lj/2)) = the tributary area of the joist that is supported by the girdery s / cos U = the length of girder over which the joist reaction is distributed.


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Figure TA.2.13

Load from a Joist

From this derivation, we can conclude that the load intensity at "C" end of the girder equals

wC = 0.5 q L1 cos U

Alternately, you can find wC by recognizing that the load on the girder has a triangular

distribution and then set up the expression that equates the tributary load to the shape of the loaddiagram:

q (Trib. Area) = 0.5 sqrt (L12

+ L22) wC

This results in the load diagram given in Figure TA.2.14.

Figure TA.2.14

Load Diagram for Girder BC


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Now let's consider the three columns.

Each column supports one or two ends of the girders. Unfortunately the girders are notuniformly loaded so we cannot say that the girders transfer half their load to each column. When

we add it the uniform weight of the beams we get load diagrams of the general shape shown inFigure TA.2.15.

Figure TA.2.15

General Loading Diagram for Girders AB & BC

Since we now have a member with a non-uniform load, we need to actually compute the

reactions for the girders then apply them to the columns. The tributary area method is not veryuseful for these columns in this case.

Note, however, that if the beam self weight is ignored and W2 = 0, then you can say that the

reaction at "A" is 2/3 of the total load and the reaction at "B" is 1/3 of the total load on the


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girder. With uniform pressure, the column at the "A" end can be said to support 2/3 of thebeam's tributary area and the "B" end supports 1/3 the beam's tributary area. In the case of the

floor system in Figure TA.2.10, this means that each column supports 1/3 of the total floor area.

To see a powerpoint animation that highlights different tributary areas for this problem, click

here.

Trial Problems

You can download a PDF file of the various floor configurations shown in Figure TA.2.16. Tryyour hand at identifying the tributary areas and drawing the loading diagrams for the various

girders. Where it is convient to use the tributary area method, identify the areas tributary to thecolumns and walls that support the joists and girders.

If you have difficulty, take the problems to your instructor for personalized assistance.

Figure TA.2.16Sample Framing Plans


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Tributary Areas for Gravity Loads