Floor Systems Framing of Floors Residential Architectural Drafting

Floor Systems & Framing of Floors Residential Architectural Drafting

Floor System Types — Conventional Dimensional Lumber Framing

Floor System Types — Open Web Floor Joist

Floor System Types — Truss Joist Floor Framing

Floor System Types — Post and Beam Floor Framing

Design Criteria for Structural Loading § Load Types l l l Dead loads Live loads Dynamic loads

Dead Loads § Definition: (DL) loads that make up the actual weight of the structure, such as walls, floors, roofs and any permanently fixed loads such as furnace, air conditioner or other service equipment. Materials that make up the walls, such as, studs, plywood, insulation, sheet rock, nails, glue, etc. are DL. § Building codes specify a minimum of 10#/sq ft for floors and ceilings, DL = 10#/sqft

Live Loads § Definition: (LL) loads that are fluctuating and changing through the use of the building. These loads include: people, furniture, and exterior weather related items such as, ice, snow, rain, etc. § Building codes specify the amount of live load upon type of use or occupancy. Codes differ, common residential LL = 40#/sq ft

Dynamic Loads § Definition: loads imposed on the structure by outside natural forces, such as wind and earthquake. § Wind loads l l Shear wall design used to resist wind pressure Uplift forces placed upon the roof § Earthquake loads l Seismic loads causing lateral forces on entire structure

Typical Loads for Residential Construction See Text

Framing Spacing Practices § Code based, acceptance varies § Spacing Options l l 12” OC 16” OC Common Spacing 19. 2” OC 24” OC

Load Consideration for one Joist § Considering 12” OC joist spacing 12” 15’-0” SPAN 1’ X 15’ = 15 SQFT X 50 = 750# § Considering 16” OC joist spacing 16” 12’-0” SPAN 1. 33’ X 12’ = 15. 96 SQFT X 50 = 798# § Considering 19. 2” OC joist spacing 19. 2” 18’-0” SPAN 1. 6’ X 18’ = 28. 8 SQFT X 50 = 1440# § Considering 24” OC joist spacing 24” 10’-0” SPAN 2’ X 10’ = 20 SQFT X 50 = 1000#

Sizing Joist Using Span Tables § Loading reactions of wood members l For every action there is an equal and opposite reaction, creates a “state of rest” § Deflection Allowances (Stiffness) § Floor = 1/360 Two types of actions or stresses l l § Roof = 1/240 Fiber Bending Stress (Fb)--a bending stress Modulus of Elasticity (E)--stiffness of structure • considered as the deflection or amount of sag when structural members are given a load.

Table Values § Look up values for lumber type & grade l Normal Duration for fiber stress (Fb) • Typical consideration for floor loads l Modulus of Elasticity (E) § See Text Fb = fiber stress in bending E = Modulus of Elasticity (Stiffness)

Construction Lumber Considerations — Wood Type & Quality § Wood Type Available in Area l l Douglas Fir-Larch (North) Hemlock-Fir (North) Spruce-Pine-Fir (North) Southern Pine § Wood Quality or Grade Value l l l Select Structural (Best) No. 1/No. 2 (Normally specified) No. 3 (Worst)

Required: Find the Fiber Stress in Bending and the Modulus of Elasticity of a 2 x 8 Douglas fir-Larch for grade No. 1/No. 2 (see text) DOUGLAS FIR-LARCH Fb E

Required: Find the Fiber Stress in Bending and the Modulus of Elasticity of a 2 x 10 Hemlock-Fir for grade No. 1/No. 2 Hem-Fir Fb E

See Text Joist Sizing & Spacin Problem #1: Span = 11’-8” 1 Hem-Fir 1 st Step--find E: E = 1. 6 2 nd Step--Use E and find size to fit span 3 rd Step--find Fb value (2 x 8): 2 * Fb = 1, 380 4 th Step--determine if Fb works with E Solution = 4 2 x 8 @19. 2 OC Va l is u ue 1, 3 nder 8 so i 0 wo t rks !

See Text Problem #2: Span = 14’-9” Douglas Fir 1 st Step--find E: E = 1. 6 2 nd Step--Use E and find sizes to fit span 3 rd Step--find Fb value (2 x 10): Joist Sizing & Spacing 1 2 e lu r 4 Va ove it is 45 ’t 10 esn do rk! Fb = 1, 045 4 th Step-determine if Fb works with E 5 th Step--using Fb 4 Solution: 2 x 10 @ 16 OC(Fb is wo find working column 5 5

§ Loads l l 40 LL 10 DL § 1 --Lumber type § 2 --Lumber grade § 3 -Spacing § 4 --Span Solution: DF #22 x 10 @ 16” OC will span 14’-11” Span Table (not in text)

Handout on Structural Analysis #1 § Use both charts in text § Remember that if all values in the “E” column apply and work then the modulus of Elasticity is the tendency of failure § If values are adjusted in the Fb row then the Fiber stress in bending is the tendency of failure

Anything wrong here?

Beam Types § § § § § 1 --Solid timber beam 2 --Built-up dimensional lumber beam 3 --Glued Laminated beam 4 --Parallel strand lumber beam (PSL) 5 --Laminated veneer lumber beam (LVL) 6 --Truss I-Joist beam 7 --Box or Plywood beam 8 --Flitch beam (wood and steel) 9 --Steel beams

Beam Type—Solid Lumber Beam

Beam Type—Built-up Dimensional Lumber Beam § Dimensional lumber (2 x 6, 2 x 8, 2 x 10, 2 x 12) nailed and/or glued together § Vertical placement

Beam and Joist Attached with joist hangers § Joist are attached to beams with metal joist hangers § What type of beam is shown?

Beam Type — Glued Laminated § Dimensional lumber placed horizontally and glued together

Beam Type — Parallel Strand Lumber Beam § See classroom example

Beam Type — Laminated Veneer Lumber Beam § Laminated Veneer Lumber (LVL) § Made of ultrasonically graded douglas fir veneers with exterior adhesives under heat and pressure § 1 3/4” wide x (5 1/2 to 18”) depth

Beam Type — Truss I-Joist Beam § Laminated or Solid wood (top and bottom chords) § OSB or Plywood web

Beam Type — Box or Plywood Beam § 2 x @ 12” or 16” structure with plywood skin § Designed by architect or engineer

Beam Type — Flitch Beam § A sandwich of wood and steel

Beam Type — Steel Beams § S shape (American Standard shape) l Often called an Ibeam § W & M shapes l Wide flange design § C shape § Channel shape S-I Shape W or M Shape C--Channel Shape

Beam Type — Steel Beams § Drawing Callouts: l l Shape, Nominal height x Weight/foot Example: W 10 x 25

Reaction § Reaction is the portion of the load that is transferred to the bearing points of the beam § A simple beam reaction to a load would be at the end supports. Each end would support or be required to carry half the total load

Calculating the Reactions of a Beam § Total load on beam should equal reaction loads: Reaction formula l 25 x 900 = 22500# R = wl § R 1 = 15/2 x 900# = 6750# 2 § R 2 = 10/2 x 900# = 4500# W = uniform load l = § R 3 = (15/2 + 10/2) x 900 =11250# length of span W = 900 #/ linear foot Span = 10’-0” Span = 15’-0” R 1 R 3 R 2

Simple Beam Design § Simple beam has a uniform load evenly distributed over the entire length of the beam and is supported at each end. § Uniform load = equal weight applied to each foot of beam

§ Terminology Tributary area l Joist/Rafter of beam l Beam/Girder l Post/Column l Span l Tributary area § Conditions of Design l Uniform load over length of beam l Beam supported at each end Beam span 15’-0” Simple Beam Design

Simple Beam Design l 16’ x 15’ = 240 sq ft § Total Load on Beam l 240 x 50#/sq ft = 12, 000# § Load at each supporting end l 12, 000/2 = 6000# Tributary area of beam Beam span 15’-0” § Tributary area

Determine the size of a Solid Wood Beam using Span Table § 1)Determine the tributary area and calculate the total load (W) for the beam LL = 50# 10 x 12 x 63 = 7560 TLD DL = 13# 20’-0” § Select beam size from table 12’-0” BEAM 10’-0”

7560 TLD w/ span of 12’ § Solution = 4 x 14 Beam

Reading the Steel Table § Table values of load are given in kips l 1 kip = 1000 lbs § Shape and nominal size across the top § Weight per foot is given below designation § Span is located along the left side of table

Example of Using Steel Table 30’-0” BEAM 18’-0” § Calculate load: 18 x 30 x 60 = 32400 TLD § Select Beam W 18 x 40

Steel I-Beam Table

Glued-Laminated Beam Table

Columns and Post

Steel Column Table

Wood Post Table

Load Considerations § § § § First floor loads (DL + LL) = 50#/sq ft First floor partitions (DL) = 10#/sq ft Second floor loads (DL + LL) = 50#/sq ft Second floor partitions (DL) = 10#/sq ft If Truss design no loads on interior structure(DL) If rafter/ceiling joist design (DL) = 20#/sq ft Roof load regionally varies (LL) = 20 -50#/sq ft

Beam Sizing and Post Spacing Trial & Error Method 1 --Locate tributary area 2 --Determine various conditions placing post to shorten the beam span 3 --Go to tables & choose beam 4 --Smaller beams are less expensive and usually better

Crawl Space Floor Joist, Beam/Post

Handout on Structural Analysis #2 § Before doing calculations sketch problem to visualize conditions § Calculate the tributary loads for beams and columns conditions § Use Handout charts and tables and select beams and columns for conditions