INTRODUCTION - BEAMCHEK

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INTRODUCTIONtoSTRUCTURAL DESIGNA Beginner’s guide to Gravity Loadsand Residential Wood Structural DesignbyAlbert Hilton CohenCopyright AHC 1998-2002 All rights reserved.No part of this publication may be reproduced or transmitted inany form or by any means, electronic or mechanical, includingphotocopy, without permission in writing form the publisher, ACSoftware Inc. 330 Dayton St. #6 Edmonds, WA, USA 98020Please visit our website for a free demoof BeamChekTM Structural Design Softwarewww.BeamChek.com1

INTRODUCTIONThis guide is intended as introduction to residential gravity loads, loadpaths and structural wood design. Further study is recommended prior todesigning structures. Proper structural design engineering requires athorough understanding of construction materials, construction practices,engineering principles and local building codes. The followingpublications are suggested as important reference materials are a startingpoint for an education in structural engineering. There are many bookspublished on this subject by McGraw-Hill, Inc., Wiley & Sons, Inc.,Craftsman Book Company and others.Simplified Engineering for Architects and BuildersParker/Ambrose, author Wiley & Sons, publisherDesign of Wood StructuresDonald Breyer, author McGraw-Hill, publisherNational Design Specification, NDS & design values supplementAmerican Forest & Paper Association, AF&PA1111 - 19th St. N.W., Suite 800Washington, D.C. 20036Design manual published by your local Wood Products Association.Manual of Steel ConstructionAmerican Institute of Steel Construction, Inc., AITC1 East Wacker Drive, Suite 3100Chicago, IL 60601Your Building Code: The S.B.C.C.I., B.O.C.A., C.A.B.O., or the U.B.C.Uniform Building Code, International Conference of Building Officials5360 S. Workman Mill RoadWhittier, CA 90601Product literature from your local truss manufacturers association.Product literature and design data for structural composite lumber fromyour local suppliers. The national companies for these products are TrusJoist MacMillan, Boise Cascade, Louisiana-Pacific and Georgia- Pacific.2

1. FORCES AND LOAD TERMINOLOGYFORCESStructures are subjected to many kinds of natural forces. The mostbasic force is gravity which is always at work and usually acts uponbuildings in a simple, downward direction. Sideways or lateral forcescan be produced by wind and earthquakes. Wind passing over a roofcan also create suction which is an uplift force. Lateral forces vary inintensity based on the building’s location on our planet, whereas gravityacts similarly on all buildings. Other forces include impact loads,temporary loads such as construction materials stockpiled while thebuilding is being constructed, and moving loads caused by automobilesor construction equipment. The term force is used interchangeably withload and sometimes weight. This booklet deals with the vertical forcescreated by gravity. Lateral and moving loads require special analysisand are separate subjects.EQUILIBRIUMThe goal of the whole design process is to achieve an equilibrium ofthe forces acting upon a structure. Without equilibrium the building willmove and that is not good! Equilibrium must be accomplished for thebuilding as a whole and for all the parts or smaller assemblies within thebuilding as well. For all of the forces acting downward due to gravity,an equal, opposite force calleda reaction must be pushing up.FORCESFORCESIn other words, as the loadstravel down load pathsthrough the structure, eachelement such as beams andposts, must be capable ofsupporting or reacting to theloads above it. All of theloads acting on a structure willFORCESultimately accumulate in thefoundation and must be metwith an equivalent reactionfrom the earth below.REACTIONS (Earth)3

TYPES OF LOADSVertical loads fall into two categories called live loads and deadloads. When these two are combined they are referred to as the totalload. Dead loads are the actual weights of all the permanentcomponents of a structure such as wood framing, roofing, plywoodsheathing, and insulation. On occasion, permanent equipment such aslarge air conditioners can be considered dead loads. These are loads thatwill be acting upon the structure throughout its life. Live loads on theother hand are transient items such as furniture, people, and snow. Theanticipated weight of live loads to be used for building design arespecified in the building code that is in force where the building will beconstructed. Local building officials will also have site specificrequirements for certain live loads such as for anticipated snow fall. Thebuilding use or occupancy can also affect the design load requirements.Note: The loading examples included in this booklet may or may notrepresent the live load requirements of the building department havingjurisdiction where your building will be constructed.You should contact your building official to confirm the floor live loadbased on the type of occupancy and the roof live load based on the localhistory of snow fall. If the snow load is large, inquire whether areduction is allowed for steep pitched roofs.TERMINOLOGY OF LOADSThe structural design for gravity loads involves evaluating eachmember for performance under the anticipated live loads, dead loads,and a combined force of live load plus dead load often called the totalload or “TL”. The design process starts at the roof and continues downto the foundation. This is opposite the actual construction which startsat the bottom and works up. Loads are described in terms of pounds. Anoften used symbol for pounds is # or lbs. When designing largestructures with large loads, engineers will often use the term kipssymbolized by k. One kip is equal to 1000#. Kips simplify calculationsby dropping the last three zeros. In residential design we deal withlower weights and use pounds for greater accuracy.A.) PT LOADA Point Load is a concentrated load in pounds at a specific location.This may be the location of bearing of a beam or a post.4

B.) PSFPounds per square foot is used to describe loads on flat surfaces suchas floors and roofs. Each square foot of the surface has the same load.To total the load on an area, multiply the Area times the PSF.C.) PLFPounds per lineal foot is used to describe loads on walls or longmembers such as beams. The beam receives an equal load for each footof length.Example: Beam ‘A’ has 2 sq ft of contributing load on each side (atributary load). The load on each sq ft is 100 PSF. Therefore 2 ft 2 ft a tributary width of 4 ft x 100 PSF 400 PLF along the beam.Note: Rafters and floor joists have a tributary load equal to theirspacing, i.e., 12” on center, 16” on center, etc. Their PLF PSF xspacing in feet. To convert inches to feet, divide by 12. Example: 16inches / 12 1.333 ft.2 ft2 ftEach Sq. Ft. 100#BEAM AD.) UNIFORM LOADA uniform load is a continuous load along the entire length of amember and is expressed in PLF. A partial uniform load is alsoexpressed in PLF, but does not run the entire length of the member.Note: The ends of joists and rafters bearing on a wall or beam eachproduce a small point load and when spaced 24”oc or less (in a uniformmanner) they can be considered to produce uniform loading.5

E.) TRIBUTARY WIDTHTributary loading or tributary width is the accumulation of loads thatare directed toward a particular structural member.Example: Tributary width is 7 ft 5 ft 12 ft. If the load is 100 PSF,the load to the beam would be 12 ft x 100 PSF 1200 PLF. The leftwall has 7 ft of tributary width and would receive a load of 700 PLF.The right wall has 5 ft of tributary width and gets a load of 500 PLF.TRIBUTARY WIDTH7 ftLOADLOADWALL1/2 Span5 ft7 ftLOADBEAM5 ftLOADWALL1/2 SpanNote: No matter where the beam is located in relationship to the wallsit will still have a tributary width of 12 ft which is one half the distancebetween the walls. The tributary width to each outside wall will be onehalf the distance between the outside wall and beam.F.) UNIFORM INCREASING LOADSOccasionally you will need to deal with triangles. Triangular areasare sometimes designed into floor plans and are also sometimes presentin residential roofs. Triangular areas can contribute a uniformincreasing load to a structural member. Most often an increasing loadstarts at one end of the member as a zero load and increases to the otherend where it is at a maximum load. Complicated or unusual triangularshapes can be solved by trigonometry when encountered. Structuraltriangles are usually “right triangles” which have one angle equal to 90degrees. Here are some short cuts for working with simple triangles.All triangles have three angles. The sum of these angles always equal180 degrees. If you know two of the angles you can solve for the third.Right equilateral triangles have two equal sides and a 90 degree angle.If you know the length of the two equal sides you can find thehypotenuse (the long side) by multiplying the length of a short sidetimes 1.414. If you know the length of the longest side, divide it by1.414 to find the short sides.6

45 10 FT X 1.414 14.14 FT.10 FT90 AREA bxh/245 HT(h)BASE (b)10 FTThe area of a triangle with a 90 corner can be found by multiplyingthe two short sides and then dividing by 2. Other triangles can be solvedby multiplying the base times the height and dividing by 2.LOADING DIAGRAMSA load diagram is a working sketch of the loads present on a structuralmember and is recommended before tackling a complex loadingproblem. The diagramming convention is to select one end of themember as the left end and locate the loads and their distances towardthe right. Beams that overhang a support at one end are shown at theright. The reactions at the supports may be of different magnitudes andyou’ll need to keep them organized as they may be used or accumulatedfor a beam or post load later in your design analysis. The left reaction iscalled R1, the right reaction is R2. The reactions are the locations of thesupports such as a wall or post under the member. By identifying thedistance, or start and end locations, in relationship to the reactions theloads can be accurately placed on the structural member.Note: Overhangs are often referred to as cantilevers by the building0Distance0Point LdDistanceEndPoint LdEndStartStartPartial UniformUniform LoadPartial UniformBeam self-weight (a uniform load)R1OVERHANGSPAN“Backspan” of OverhangR2trade. In structural design, a cantilever is a whole different animal andshould not be confused with overhanging structural members.7

THE SPANThe span of a structural member is the horizontal distance from face toface supports, plus one half the required length of bearing at each end.The minimum bearing length for wood members is 1-1/2” bearing onwood and 3” bearing on masonry. If the member is continuous oversupports, the span is the distance between centers of the supports.SPAN is 10’- 1½”SPAN is 12’- 1½”10’-0”This distance is calledthe “clear span”12’-0”The design span is thehorizontal distance plus 1/2 therequired bearing lengthSIMPLE SPANSOne half the requiredbearing length of1-1/2”at each end ¾”This is the minimum.It might be larger due tothe load and the woodspecies value for Fc .SPAN is 14’- 3½”14’- 0”5½” post(5½” / 2 2¾”)bearing lengthMULTIPLE SPAN CONTINUOUS BEAMBACK SPAN is 16’- 4½”OH is 4’- 3¾”16’- 0” BACKSPANR14’- 0”R2OVERHANGING BEAM8OVERHANG7½”post (net)(7½” / 2 3¾”)bearing length

DEAD LOAD WEIGHT OF MATERIALSMaterialApproximate lbs/ft²Wood shinglesAsphalt or fiber glass shinglesClay tileConcrete tileSlate 1/4 in.2.02-39 - 128 - 1210.0Rigid Insulation(per in.)Batt or blown insulation (per in.) (varies!)0.20.31/2” plywood sheathing3/4” plywood sheathing2” deckingconcrete (light weight fill, per in.)1/2” gypsum wall board5/8” gypsum wallboard1” Plaster1.52.24.28.02.02.58.0Double glazed skylight or wood frame window 8 - 10Hardwood floor (3/4”)Ceramic tile (3/4” thinset)Carpet & padStucco (7/8” on wire & felt)4” Brick wall ( per sq ft)Wood siding3.85.03.010.040.02.5Framing members in PSFNominal12” oc16” oc24” .42.22.93.74.44x framing members in PLF4x65.04x86.84x108.64x1210.49

DEAD LOAD ASSEMBLY EXAMPLESROOFAsphalt shingles (1 layer)Asphalt shingles (future layer)1/2” plywoodInsulation2x8 rafters @ 16” oc2x6 ceiling joists @ 16” oc5/8” gypsum wall boardPSF2.52.51.52.52.21.72.5TOTAL15.4Roof dead load is typically between 10 and 20 PSF.As pitch increases the dead load increases.EXTERIOR WALLWood Siding1/2” Plywood sheathingInsulation2x6 Studs @ 16” oc1/2” gypsum wallboardPSF2.51.51.21.72.0TOTAL8.9Walls are typically assumed at 10 PSF, note that windows arealso approximately 10 PSF.An 8 ft high wall including double top plate and sill plateweighs about 80 PLF.An 8 ft Interior wall weighs 45 to 55 PLF.FLOORCarpet & pad3/4” Plywood sheathingInsulation2x10 Joists @ 16” ocPSF3.02.22.52.8TOTAL10.5Floor dead load is typically between 10 PSF to 15 PSF.Remember to add the ceiling material if floor is over livingspace, i.e., the second floor of a two story house.10

2. LOAD PATH EXAMPLESUNIFORM LOADS ON RAFTERS AND FLOOR JOISTSRemember that rafters and floor joists are just small beams with lighterload requirements. They are usually spaced at regular intervals and areoften connected to a communal diaphram of plywood or similarmaterial. Their loading condition is usually a simple uniform load.ONE SQUARE FOOT LOADS, ALL ALONG A RAFTEROR FLOOR JOIST (THIS IS A UNIFORM LOAD)Note: The PLF is the same as PSF in this case because thetributary width is only one foot wide. Loads on adjacentrafters are not shown for clarity.12”12”1/21/212 IN.RAFTERS ORFLOOR JOISTSSPACED AT 12” O.C.SUPPORTING WALL,BEAM OR HEADERAT EACH END11

OTHER RAFTER AND JOIST SPACINGSTHE PLF ALL ALONG THIS RAFTER OR FLOOR JOIST ISONE SQUARE FOOT x 1.33 BECAUSE THE SPACING IS 16” o.c.( 16” / 12” 1.33 )Similarly 24” o.c. would equal the load in PSF x 2.016”16”1/21/212 IN.16 IN.RAFTERS ORFLOOR JOISTSSPACED AT 16” O.C.SUPPORTING WALL,BEAM OR HEADERAT EACH END12

ROOF UNIFORM LOAD DISTRIBUTIONThese examples use an assumed load of 40# PSF. You will be designingwith two sets of loads: Live Loads and Total Loads which are dead load live load.The live load appropriate to your locale is specified by your buildingoffical. The dead load is the actual weight of the construction materials tobe used in the project.Note: The dead load will increase as the roof pitch increases because therafters and roof surface become longer. This is especially important whendesigning for tile and slate roofs which are 0”ROOF LOAD 40 PSF9.0 ft x 40 PSF 360 PLFON EACH WALLDon’t forget theeaves, snow fallson them ”2’-0”This would be typical of atrussed roof or a small spanroof with a 1x ridge board.ROOF LOAD 40 PSF9.0 ft x 40 PSF 360 PLFON EACH WALL132’-0”

5’-6”3.3’-6”3’-6”5’-6”(2’ 3½’)TypicalSTRUCTURAL RIDGEBEAMRIDGE BEAM LOAD IS3.5 ft x 2 x 40 PSF 280 PLF2’-0”14’-0”2’-0”ROOF LOAD 40 PSF5.5 ft x 40 PSF 220 PLFON EACH WALL4.NON-STRUCTURAL RIDGEA2’B5’-0”ROOF LOAD 40 PSFLOAD A (OUTSIDE WALL) 2.5’ 2’ x 40 PSF 180 PLFLOAD B (ATTIC WALLS) 5’ 2.5’ x 40 PSF 300 PLFLOWER SIDE WALLS ½ B A 330 PLFCENTER WALL ½ B ½ B 300 PLFNote: Ceiling joists would bedesigned with a 300# point loadat mid-span due to the attic wallssupporting the rafters, plus auniform live and dead load.B5’-0”5’-0”A5’-0”2’ATTICWALLS½ of Beach �5.3’-6”3’-6”5’-6”WALLWALL SUPPORTED RIDGEROOF LOAD 40 PSF5.5 ft x 40 PSF 220 PLFON EACH OUTSIDE WALL2’-0”CENTER WALL LOAD IS3.5 ft x 2 x 40 PSF 280 PLF147’-0”7’-0”2’-0”

FLOOR UNIFORM LOAD DISTRIBUTIONThese examples use an assumed load of 50 PSF.You will be designing with two sets of loads:Live Loads and Total Loads (dead load live load)The live load appropriate to your locale is specified by your buildingoffical. The dead load is the actual weight of the construction materialsto be used in the project. Common residential loads are 40 PSF liveload and 10 PSF dead load, for a total load of 50 PSF.Note that the formulas in Example 3 provide for more accuratereactions. Beginning designers often try to be overly exact in theircalculations only to find that a close approximation will often result inthe same beam size selection. Accuracy is important, but the range ofaccuracy should to be appropriate to the task.7’-0”1.7’-0”FULL SPAN FLOOR JOISTS14’-0”FLOOR LOAD 50 PSF(14 ft / 2) x 50 PSF 350 PLFON EACH WALL2.4’-0”4’-0”4’-0”4’-0”FLOOR WITHCENTER SUPPORTBEAMFLOOR LOAD 50 PSF4 ft x 50 PSF 200 PLFON EACH WALL16’-0”(16 ft / 2) x 50 PSF 400 PLFON CENTER BEAM15

7’-0”3.a 2’-0”7’-0”OVERHANGINGFLOOR JOISTSBACKSPANl 14’-0”FLOOR LOAD w 50 PSF(14 ft / 2) x 50 PSF 350 PLFAPPROX. ON LEFT WALLActually 342.85 PLF w/2l (l ²- a²)7 ft 2 ft x 50 PSF 450 PLF APPROX. ON RIGHT WALLActually 457.14 PLF w/2l (l a)²l 8’-0” w 50 PSF4.CONTINUOUS BEAMWITH 2 EQUAL SPANS3 x 8ft x 50 PSF / 8 150 PLFON END WALLS3wl/85wl/83wl/85wl/8BEAM(or wall)16’-0”5.4 x 8 ft x 50 PSF / 10 160 PLFON END SUPPORTS8’-0”Total at centersupport is 10wl/810 x 8 ft x 50 PSF / 8 500 PLFON CENTER SUPPORTCONTINUOUS BEAMWITH 3 EQUAL SPANS8’-0”l 8’-0”w 50 1 x 8 ft x 50 PSF / 10 440 PLFON INTERIOR11wl/1024’-0”Check: 24’ x 50# 1200# Total , 160# 440# 440# 160# 1200# Total, OK16

MULTI-STORY LOAD PATH EXAMPLELoads are the same on each side in this example because the house is symetrical.10’ x 40 PSF 400 PLF2’-0”10’-0”Tributary5’-0”7’ x 40 280 PLF5’-0”RIDGE SUPPORTUpper Wall Dead Load11.75’ht. x 8 PSF 94 PLFWall Dead Load8’ht. x 10 80 PLF10’-0” Tributary width5’-0”2nd Floor Load5’ x 50 PSF 250 PLF2nd Floor load10’ x 50 PSF 500 PLFLower Wall Dead Load8’ht. x 8 PSF 56 PLFSIZE HEADER ‘A’5’-0”5’-0”5’-0” Trib. width5’-0”First Floor load5’ x 50 PSF 250 PLFPosts and pierpads at 6 ft ocSIZE GIRDER ‘B’20’-0”172’-0”

MULTI-STORY EXAMPLE LOADS:Given: ROOF 40 PSF, FLOOR 50 PSFHEADER A:Roof and eave 280 PLFUpper wall 80 PLF2nd Floor 250 PLFTotal uniform load 600 PLF6x Header clear span is 10 ftplus bearing area of 3” / 2 1.5”Total design span is 10.125 ftTotal weight 10.125 x 600 PLF 6075#Reactions are equal: 6075# / 2 3038# eachGIRDER B:Roof 400 PLFUpper wall 94 PLF2nd Floor 500 PLFLower wall 56 PLFFirst Floor 250 PLFTotal uniform load 1300 PLFTo continue.Girder ‘B’ has a span is 6 feet.Total weight on Girder ‘B’ 6ft x 1300 PLF 7800#The reactions for the Girder would be 7800# / 2 3900# ateach end. Therefore, posts and pier pads in the crawl spacewould need to be capable of supporting 3900# from half thegirder at each side which is a total of3900# 3900# 7800#.18

COMPLEX LOADING EXAMPLEGiven PSF loads: Roof LL 30#, DL 10#; Floor LL 40#, DL 10#; Wall DL 10#3 ft20 ft9 ft2 ft10 ft400 PLF1800#HEADER A 9 ft1800#WALL UNDERHEADER 80 PLF8 ftWALL80 PLFWALL80 PLF500 PLF480 PLF480 PLFHEADER B 14 ftR1Joists and rafters runfront to back in thisR2014 ft12 ft3 ftPOINT LOADS1800# EACHPARTIAL UNIFORMLOADS AT 400 PLFFLOOR & WALL UNIFORM LOAD 580 PLFBEAM SELF-WEIGHT UNIFORM LOADR1R2LEFT REACTIONRIGHT REACTIONLOAD LOCATION DIAGRAM FOR HEADER B19

LOADS ON HEADER AUNIFORM LOADLOAD FROM ROOF 10 ft TRIB. LOAD x 40 PSF 400 PLF400 PLF x 9ft 3600# TOTALONE HALF TO EACH END 1800# POINT LOADSLOADS ON WALLUNIFORM LOADROOF LOAD TO WALLS BEYOND HEADER.10 ft TRIB. LOAD x 40 PSF 400 PLFLOADS ON HEADER BUNIFORM LOADLOAD FROM FLOOR10 ft TRIB. LOAD x 50 PSF 500 PLFplusWALL DEAD LOAD (DL)8 ft HEIGHT x 10 PSF 80 PLFPOINT LOADSLEFT POINT LOAD AT 3 ft 1800#RIGHT POINT LOAD AT 12 ft 1800#Distance is always from the left, thereforeRight Point Load is at 3 ft 9 ft 12 ftLEFT PARTIAL UNIFORM LOAD400 PLF FROM ROOFLOCATE FROM LEFT END:START AT 0 ft, END AT 3 ft (3 ft long)RIGHT PAR

Design of Wood Structures Donald Breyer, author McGraw-Hill, publisher National Design Specification, NDS & design values supplement American Forest & Paper Association, AF&PA 1111 - 19th St. N.W., Suite 800 Washington, D.C. 20036 Design manual published by your local Wood Products Association. Manual of Steel Construction

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