Structural Analysis by Hand - VBCOA

1 1 VBCOA Region 5 May 15, 2014 Structural Analysis by Hand2 PresenterBrian Foley, CountyDeputy Building 324 18423 LogisticsExitsRestroomsCell PhonesNo Smoking24 Questions?SURE Go ahead and ask your question!This class is interactive!There are no stupid questions!5 Attendees Inspector Plan reviewer Contractor Engineer Architect Designer Suppliers Attorney6 Objectives Determine loading requirements for joists and beams Apply loads using free body diagrams Calculate moment and deflection Analyze compliance for flexure, shear and deflection Analyze simple steel beam Analyze spread footings37 Assumptions No Structural theory Knowledge of arithmetic and basic algebra Simple loading Uniform loads Point loads at midspan Simple spans Wood, LVL, steel Residential construction8 Math 1019 Operation SequencePlease Excuse My Dear Aunt Sally Parenthesis Exponent Multiplication Division Addition + (3 1) / 2 4 x 3 Step 1.

2 Parenthesis 52+ 2/ 2 4 x 3 Step 2: Exponents25+ 2 / 2 4 x 3 Step 3: Multiply 25 + 2 / 2 12 Step 4: Divide 25 + 1 12 Step 5: Add26 12 Step 6: Subtract1411 Mathematical Formulas Volume of a box Length x Width x Height Variables L = Length W = Width H = Height V= Volume V = L x W x H V = LWHH eight12 Mathematical Formulas V = LWH Values L = 5 W = 2 H = 6 V = 5 x 2 x 6 V = (5)x(2)x(6) V = (5)(2)(6) V = 60 feet3 Height513 You Try It, Find V Values 2 w = 3 h = 6 14 You Try It, Find V Values: 2 , w 3 , h 6 Convert h value from inches to feet: 6 = Insert values into formula: 1 ft315A Word About Units Always include units with every calculation Ensure all calculations are completed in the same units (inches, feet)616 Vertical Load PathVertical load path transfers gravity load (snow): to roof sheathing to rafters to top plate to studs to bottom plate to foundation & footings to ground17 Design MethodologiesLoad & Resistance Factor Design (LRFD) Applied loads adjusted up Resistance capacity of Structural member adjusted down Compare values:capacity > loadsAllowable Stress Design (ASD) Actual stress calculated using applied loads Structural member s allowable stresses calculated Compare values.

3 Allowable > actual18 Joist/Beam Analysis719 Sample First Floor Framing Plan2x8 joists16"16"(2) 1 x9 MicrolamHem-Fir #24'-0"10'-0"12'-0"20 Step 1: Determine uniform dead load Units for uniform dead load pounds per square foot lbs/ft2 psf Dead Load: weight of structure Assume 10 PSF for floors Assume 15 PSF for roofs21 Step 2: Determine uniform live load (psf) Live Load: weight produced byuse and occupancy People Furniture Vehicles Units: pound per squarefoot (psf) IBC Table IRC Table 2: Determine uniform live load (psf) For floors: IRC Table 2: Determine uniform live load (psf) For roofs, greatest of live load or snow load Snow load: IRC Section Northern Virginia counties 20 30 psf24 Step 2: Determine uniform live load (psf) Roof live load: IRC Table 3: Calculate tributary width (feet) Load influence distance from each side of a framing member Joists: half the distance to next adjacent joists on each side Beams: half of the joists span that bear on each side of the beam26 Step 3: Calculate tributary width (feet)16"16"' "16216216 feettoConvertTWEXAMPLE: JOIST8"8"Tributary width27 Step 3: Calculate tributary width (feet)EXAMPLE: BEAM4'-0"10'-0"12'-0"Tributary width1028 Step 3: Calculate tributary width (feet)EXAMPLE: BEAM2'5''721024 TW29 Step 4: Calculate linear load (plf) w = uniform load xTW Units.

4 Pounds per linear foot = lbs/ft = plf EXAMPLE:JOIST:wLL= (40)( ) = plfwDL= (10)( ) = plfw= plfBEAM:wLL= (40)(7) = plfwDL= (10)(7) = plfw= plf30 Free Body Diagram A two dimensional graphic symbolization of a Structural member which models bearing locations and loading elements Linear load, w: Point load, P: Bearing locations (reaction), R:100 plf500 lbs1131 Free Body plf10 -0 Beam350 plf12 -0 wlRR32 Example: Free Body Diagram On plans provided, first floor joist adjacent fireplace hearth extension:15 -4 33 You Try It Draw the free body diagram of the sunroom beam Show load and its valueTotal uniform load = 50 psfTributary width = 5 Total linear load = (50)(5) = 250 plf Show span length250 plf10 -4 1234 Step 5: Bending Analysis Flexure, bending, moment, torque Highest at midspan for uniform loadPulling stress or tension on bottom face of member35 Step 5A: Determine F b(psi) Allowable bending stress, F b The maximum bending stress permissible for a specified Structural member Units for stress: pounds per square inch lbs/in2 psi36 Step 5A: Determine F b(psi) Raw value based on wood species: Fb Adjusted allowable bending stress,F b= FbCMCFCrCD, where.

5 CM= Service condition (wet or dry) CD= Load duration (normal or snow) Cr= Repetitive use (joists, 3+ply beams) CF= Member size (2x?)1337 Step 5A: Determine F b(psi) Use tables in chart based on: Species Service condition (wet or dry) Load duration (normal or snow) Single (2 ply beam) or repetitive (joists) Member size EXAMPLE: Joists: Repetitive, dry, 2x8, normal load duration, Hem Fir#2: F b= 1,173 psi Beam: Single, dry, (2)1 x9 , normal load duration, Microlam:F b= 2,684 psi38 Step 5B: Determine b (in), d (in), S (in3)EXAMPLE:261bdS ) )( (61)2(: SBEAMS ection ) )( (61: SJOIST39 EXAMPLE:JOIST: S= in3 BEAM:S=(2)( )= in3 Step 5B: Determine S (in3)1440 Step 5C: Determine Span Length, l(ft) Calculate Moment, M (lbs ft)lw82wlM where:l= span length, ftw= total linear load, plfM = moment, lbs-ft41 Step 5C.

6 Determine Span Length, l(ft) Calculate Moment, M (lbs ft)where:l= span length, ftw= total linear load, plfM = moment, lbs-ftJOIST:w = plfl = 10'BEAM:w = 350 plfl = 12' )10)( (2 MEXAMPLE:82wlM ftlbs300,68)12)(350(2 M42 Step 5D: Calculate fb(psi) Actual bending stress, fb The bending stress a specified Structural member is experiencing under maximum applied loadwhere:S= section modulus, in3M = moment, lbs-ftfb= actual bending stress, psiSMfb12 1543 Step 5D: Calculate fb(psi)where:S= section modulus, in3M = moment, lbs-ftfb= actual bending stress, psiSMfb12 JOIST:M = lbs-ftS = in3 BEAM:M = 6,300 lbs-ftS = ) )(12( , )300,6)(12( bf44 Step 5E: Compare F bwith fbIf fb F b, then member is goodfor bendingEXAMPLE:JOIST: fb= psi < F b= 1,173 psiOK!

7 BEAM: fb= 1, psi < F b= 2,684 psi OK!45 Step 6: Shear Analysis Shear is similar to a cutting stress Highest at ends = reaction Wood shear Analysis uses shear value at a distance from the end equal to member s depthMember experiencesa slicing action invertical plane1646 Step 6A: Determine F v(psi) Allowable shear stress, F v The maximum shear stress permissible for a specified Structural member Units for stress: pounds per square inch lbs/in2 psi47 Step 6A: Determine F v(psi) Allowable shear stress Raw stress based on wood species: Fv Adjusted allowable shear stress: F v ,F v=FvCMCD, where: CM= Service condition (wet or dry) CD= Load duration (normal or snow)48 Step 6A: Determine F v(psi) Use tables based on.

8 Species Service condition (wet or dry) Load duration (normal or snow)EXAMPLE:Joist: Dry, Hem Fir#2: F v= 150 psiBeam: Dry, Microlam:F v= 285 psi1749 Step 6B: Calculate Area, A (in2) ) )( (:JOIST ) )( )(2(:BEAM AA=bdArea:50 Step 6B: Calculate Area, A (in2)EXAMPLE:JOIST: A= in2 BEAM:S=(2)( )= in251 Step 6C: Calculate Shear, V (lbs)where:l= span length, ftw= total linear load, plfd = member depth, in 122dlwVJOIST:w = plfl = 10'd= "BEAM:w = 350 plfl = 12'd= " , V1852 Step 6D: Calculate fv(psi) Actual shear stress, fv The shear stress a specified Structural member is experiencing under maximum applied loadwhere:fv= actual shear stress, psiA= area, in2V = shear, lbsAVfv23 53where:A= area, in2V = shear, lbsfv= actual shear stress, psiAVfv23 JOIST:V = lbsA = in2 BEAM:V = 1, = ) )(2() ,1)(3( ) )(2() )(3( vfStep 6D: Calculate fv(psi)54 Step 6E: Compare F vwith fvIf fv F v, then member is goodfor shearEXAMPLE:JOIST:fv= psi < F v= 150 psi OK!

9 BEAM:fv= psi < F v= 285 psi OK!1955 Step 7: Deflection Analysis "Sag" a member experiences Analysis is based on live load only Highest at midspanDeflection, 56 Step 7A: Determine allowable deflection (in) Use Table l= span length of member, in.(span length must be converted from feet to inches)57 Step 7B: Calculate allowable deflection (in)EXAMPLE: Floor,JOIST:BEAM: " )10)(12(max " )12)(12(max 36012maxl 2058 Step 7C: Calculate I(in4)EXAMPLE:3121bdI bd43in250) )( (121)2(: IBEAMM oment of ) )( (121: IJOIST59 Step 7C: Calculate I(in4)EXAMPLE:JOIST: I= in4 BEAM:I=(2)(125)=250 in460 Step 7D: Determine E(psi) Modulus of elasticity, E The mathematical description of a Structural member s elastic characteristics Units for modulus of elasticity: pounds per square inch lbs/in2 psi2161 Step 7D: Determine E(psi) Modulus of elasticity Raw value based on wood species: E Adjusted modulus of elasticity: E E =E CM, where.

10 CM= Service condition (wet or dry)62 Step 7D: Determine E(psi) Use tables based on: Species Service condition (wet or dry)EXAMPLE:Joist: Hem Fir#2, dry: E = x 106psiBeam: Microlam, dry:E = x 106 psi63 Step 7E: Calculate actual deflection (in)where: act = actual deflectionl= span length, ftwLL= linear live load, plfE = modulus of elasticity, psiI= moment of inertia, 2264 Step 7E: Calculate actual deflection, act(in)JOIST:wLL= plfl = 10'E= in4 BEAM:wLL= 280 plfl = 12'E= 250 in4" ) )( ()10)( )( (64 actEXAMPLE:" )250)( ()12)(280)( (64 act65 Step 7F: Compare maxto actIf act max, then member is goodfor :Joist: act= " < max= " OK!Beam: act= " < max= " OK!