W Tributary Loads - static and strength of materials PDF

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Tributary Load or Load Tracing This analysis is how loads are transferred in a structural system, regardless of materials or sizes of members from the point of load down to the foundation in the ground. This analysis helps to size the structural members with a whole building in mind. Gravity loads travel along load paths that start from • the flooring system (deck) • to secondary supports (joists that carry the deck) • primary support members (beams or girders) This hierarchy is shown on exhibits 1 (figure 3-14 ), 2 (Platform frame) and 3a, 3b and 3c Note a couple of things in terms of relations: 1) Decking is a spanning element - a structure carrying often a uniform code load per square foot i.e. 100 lbs per square foot). 2) Decking spans between the joists (distance "a" in exhibit 1, 2'-0 in exhibit 2..); the deck distributes the uniform load across its surface to the joists EQUALLY. 3) The joists thus have a uniform load applied to them - calculated as load per linear foot w = (load per sf) x (tributary width), where this width is the deck span between joists (we will come back to this). Thus deck or floor load multiplied by the tributary width is called "w" page 1 of 9

4) These joists often span between beams. The joists EQUALLY distributing their loads to either beam (ie, half the joist load goes to one beam the other half to the other beam). Each joist carries Load W = w x joist span (called L) 5) The beams thus carry a series of joists at concentrated loads points; because the joist distribute their load equally to either supporting beam, the point loads are W/2 at every joist placement. 6) Sometimes beams are then carried by girders (larger beams). Like the beams with joists, where the beams are carried by girders, the girders have concentrated or point loads where the beam rests. 7) The whole assembly then sits on columns - which have their own load limitations. 8) At this point you will note that we DO NOT take into account the weight of the actual members unless it is specifically noted. One variation to this occurs at the end of the bay - where the last joist carries only half the load (because its tributary width is only 1/2 the size) see exhibits 1 and 2 (hand drawn figures and/or overheads) We will concentrate on floor and column tributary analysis, but all components of a building carry loads in this mannerLoad tracing is very well documented in the book on reserve Statics and Strengths of Materials for Architecture and Building Construction, Chapter 4 page 2 of 9

Methods of Analysis There are two methods we will use - either one works - to understand the loads that each member in this framing system has to carry. Method 1: mathematical chain Method 2: calculation by area The first is far more accurate (gives closer load readings on decks and joists), the second is quicker. Both give the same loads on the ultimate collection beams and columns and thus foundation. See exhibit 4 (hand drawn figures and/or overheads) Also, we can take a shortcut from deck to larger supporting beam (almost ignoring the actual spacing of joists) by realizing that when the joists are closely spaced, the beam almost carries a uniform load directly from the deck. See exhibit 4b (hand drawn figures and/or overheads)

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Structural Diagrams

20 ft C1

This is the kind of diagram to represent the loads that you will see - it is also how the structural engineer puts his documents together. Note the different loads and the diagrams showing the direction of joists 20 ft

20 ft

20 ft

20 ft

20 ft C3

B1 3 J1

40 ft

B1

B2

B3

B4

B5

B6

B7

C6 B1 6 C2

C5

C3

C7

B1 5

J2 20 ft

B8

B9

B1 0

C4

B1 2

B1 1 J3 B1 4 C8

B1 7

C9

For the Framing Plan shown, analyzethe load on each member by filling out the Tributary Load Worksheets, for the members indicated below; Given:

All joists are spaced 24" o.c. Dead Loads on deck: 100 PSF (in shaded area) 150 PSF (in all other areas)

All "Free Body Diagrams," or FBD, Should Include Spans (L), Magnitude of Uniform (W) and Concentrated Loads (P). Both Reactions Should be Included on the FBD. All columns should calculate total axial load. Required Calculations for FBD or column loading diagrams: Joists: J1, J2, J3 Beams: B1, B2, B3.....B15, B16 Columns: C1, C2, C3.... C6, C7 Note: All the FBD here have no concentrated loads. page 4 of 9

Some other rules of thumb: This a structural bay that is 20 ft by 20 ft Beams B11 and B12 in fact lie either under or at the same level as the joists J3 (see exhibit 5) and carry no load. Beams B 16 and B17 on the other hand equally carries the load of joists J3 (that is the full deck load). each of these beams carry half the deck load and thus half the joist loads. Beams B16 and B17 carry their loads to columns C6 and C7, and C8 and C9 respectively.

20 ft C6

B1 1

B1 6

B1 2

J3

C8

C7

B1 7

C9

If this bay were standing alone - each column would carry half the laod that the beam carries, or 1/4 of the deck load. Because this bay is on the south east corner of the floor diagram shown on the previous page, - Column C6 carries 1/2 beam loads of B6 and B15, as well as B16 and B11. - Column C7 carries 1/2 beam loads of B16 and B7 as well as B12 . - Column C8 carries 1/2 beam loads of B11 and B14 as well as B17 . and - Column C9 at the corner only carries 1/2 beam loads of B17 and B12 .

You can see too the importance of knowing which way the joists span - as the beams not carrying the joists have no deck loads transferred to them have no loads on them. When in doubt - joists span the least dimension of a bay page 5 of 9

Doing analysis problems Free Body Diagram

Member Design.

RL=

the pre-drawn Tributary Load work sheets helps keep this straight Tributary Width

Uniform Uniform Load Load Coefficient W w

Total Tributary Concentr. Area Load

Max. Shear Max P Moment

=RR

Step 1: either the problem or you will designate the joists, beams and columns to analyze - use J1 etc, B1 etc and C1 etc. Step 2: Ascertain uniform or point loading across the diagram watch for changes or variations (see diagram on page 4) Step 3: Draw free body diagram of the individual spanning member - start with joists, then move to beams and then to trusses - as each one rests on the next. Step 4: Try to group similar spanning members to solve together all joists and then separately all beams with same span and same load. This way you may have to solve for far fewer situations. Step 5: For each unique set of joists (ONLY) calculate tributary width - the spacing of joist indicating how much deck loads each joist carries. Remember that the end joists have half the tributary width. Step 6: Calculate for each group of joists, w = load/sf x tributary width Step 7: Calculate for each group of joists, W = w x length of joist page 6 of 9

Step 8: For each joist calculate the tributary area = tributary width x joist length Step 9: If there is a concentrated load note it on diagram Step 10: Now complete free body diagram, knowing the uniform load W and any concentrated loads, using the methods of beam equilibrium - to solve for reactions right and left; note these reactions become the concentrated loads on the supporting beams Step 11: Given the freebody diagram, calculate the maximum shear and moment in the beam (you can now use the information found on "page 77" if the laods match those diagrams or the diagrams on the next page). - If the total load is a simple uniform load use the equations; - If it is a compound of several uniform loads, treat each separately and do a free body diagram approach and add the max V and max M (i.e. different diagrams for two overlapping uniform loads and additional point loads) Step 12: Repeat for beams, then girders (if necessary), noting these have point or concentrated loads (if done mathematically); once to beams you can use the area method (that is one beam carries half the bay's deck load, again depending on joist direction) Remember too, beams on the outside of the floor have loads to only one side, while beams in the middle of the floor have loads often from both sides. When in doubt, do a full free body diagram to figure V and M.

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For uniform load Free Body Diagram (FBD)

Vmax = wL/2 = W/2

"w" in k/ft

Mmax = wL2/8 = WL/8 L/

L/

2

RL = w x L/2 K

2

RR= w x L/2 K

VL = +W/2 K ShearDiagram

+A 1

V= 0

-A 2

VR = - W/2 K Mmax

Bending Moment Diagram

M= 0 Member Design.

Diagram of Column & Tributary Area

Total Axial Load on Column (P)

Step 13: Using the Axial Load Worksheet (above) analyze the columns that support the beams; half the load of each beam goes to the column at the end. As you can see a fast method is the area method - it can be used alone or to check the mathematical method. Axial load P is simply the additive reaction at each end of the beam for all the beams loading onto one column. page 8 of 9

Sample analysis problems The three problems of varying complexity are attached: Problem 1 - wood frame has uniform sized bays and uniform loads with 1 bay carrying a heavier uniform load. Joists all run left to right with 2'-0 spacing. Problem 2 - Introduces a girder that carries some of the beams Problem 3 - Introduces point loads such as walls

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