Problem 1. Lab #12 Solution

Transcription

1 Lab #1 Solution Problem 1 Given Truss Roof Dead Load of 16 psf, Snow Live load of 16 psf, Truss spacing of 4" o.c. Lumber is No. DF-L Bolt holes are 7/" diameter, MC < 19%, Temp < 100 degree F, and Ci = 1.0 Determine size of bottom tension chord and compression chords for truss wu := 64 plf x := 10.5 ft Joint Loads J1 := wu x J1 = 67 J := wu x J 336 = Reactions for truss wu R := R = 1344 lb Tension in bottom chord - method of joints Tension is equal to horizontal component of diagonal member at left end T:= ( R J) 10.5 T = lb Tension is parallel to grain F T := 575 psi from Table 4A, pg 31 from NDS supp manual ( No DF-L) Adjustment Factors C D := 1.15 section.3. - snow & dead load C T := 1.0 C M := 1.0 C F := 1.3 Assume x 6 initially C I := 1.0 Value for allowable tension F Tp := F T C D C M C T C F C I F Tp = 59.6 psi Find required area T A req := A F req = 3.51 Tp

2 Nominal size of x 6 b := 1.5 h := Calculate area of x 6 with 7/" diameter hole A1 := b h A1 = 6.93 A x 6 will satisfy 7 Check x 4 b = 1.5 h := 3.5 A := b h A = 3.93 A x 4 also will satisfy Determine size for compression chord Length of compression chord L1 := L1 = Compression in chord C := ( R J) L1 C = Start with x 6 F B := 900 psi F C := 1350 psi from Table 4A, pg 31 from NDS supp manual ( No DF-L) E := psi C Fc := 1.1 for compression C Fb := 1.3 for bending Section Properties pg. 14 of supp NDS A :=.5 in S x := in 3 Check Axial Compression C fc := fc = 36.4 psi A Factors for compression C D = 1.15 C M = 1 C T = 1 C Fc = 1.1 C i := 1.0 Modulus Value from Table C M := 1.0 C t := 1.0 C T := 1.0 CT factor - do not have compression chord that is smaller than x 4 E p := EC M C t E p =

3 Determine Cp value Section K ce := 0.3 visually graded lumber L1 = ft h = 5.5 depth of x 6 section c := 0. sawn lumber K ce E p F ce := F ce = 3.19 L1 1 h F star := F C C D C M C T C Fc C i F star = F ce 1 + F star C p := c 1 + F ce F star c F ce cf star C p = 0.41 F Cp := F star C p F Cp = 71.7 psi > fc = 36.4 psi Compression OK At connections, although buckling is prevented, compression is checked using reduced area but Cp=1.0 assuming a bolt diameter plus 1/" for the opening diameter 7 A net := A net = 6.9 in If Cp=1 then allowable stress is The acting axial stress is C = psi < F A star = OK net Bending Check Assuming a pinned end chord and load on horizontal plane wu x 1 M := M = 1054 in lb M f b := f S b = psi x

4 Factors for Bending C D = 1.15 C M = 1 C t = 1 C L := 1.0 C Fb = 1.3 C fu := 1.0 C i := 1.0 C r := 1.0 C f := 1.0 F Bp := F B C D C Fb F Bp = psi Actual bending moment is larger than allowable! f b = psi > F Bp = psi Increase section to a x A := 10. in S x := in 3 C Fb := 1. for bending M f b := S x f b = 05 psi F Bp := F B C D C Fb F Bp = 14 psi Interaction - no bending about weak axis and axial load is concentric - Section actual compression on x allowable compression on x C Fc := 1.05 table 4A NDS supp pg 30 C f c := f A c = 93 psi F Cp := F C C D C Fc C p F Cp = 66 psi K ce = 0.3 h3 := 7.5 depth of x K ce E p F ce1 := F ce1 = 1430 psi L1 1 h3 Interaction eq. f c F Cp f b + = 0.99 < 1.0 OK f c F Bp 1 F ce1 Solution - for tension chord ( x4 or larger ok) for compression chord ( x required)

5 Problem Given a simply supported roof beam Subjected to 000 lbs (D+L) applied at the middle of the span With a design span of ft, MC<19%, under normal temp conditions Determine least grade required for a 4 x DF-L Use deflection limit of L/360 for the total deflection assuming 50% of the concentrated load is permanent Compression zone of beam is prevented from buckling and wood is seasoned. Length of bearing is 6 inches Neglect self weight L:= ft P := 000 lbs Determine Adjustment factors for bending Properties for 4 x A := 5.3 in C F := 1.3 Table 4A - for 4 x beam C M := 1.0 S x := in 3 I x := in 4 C t := 1.0 C D := 1.0 for live load Moment on Beam M := PL M = 4000 lb ft 4 M1 Bending stress on beam f b := S x f b = psi For Bending Find grade of DF-L that has Fb > fb F B := f b C F F B = psi From tables on pg 31 of NDS supp. Select Structural Grade has F B := 1500 psi For Shear parallel to grain For select structural F v := 10 psi Factors C D = 1 C M = 1 C t = 1 C i = 1 actual shear stress on wood F v = 10 psi < fv = psi OK 3.50 P fv := fv = Largest shear is at support R=0.5P A

6 For Deflection - Section 3.5 E := psi for select structural grade all factors for E are = 1.0 Kcr := 1.5 for seasoned lumber Immediate deflection due to long term component of design load L =.5P ( L 1) 3 LT := Use dead load only ( P = 1000 lbs) 4 E I x LT = 0.07 in Short term deflection due to normal component of design load Use live load ( P = 1000 lbs) ST := LT T := Kcr LT + ST T = 0.1 Deflection limit is L/360 L1 360 = 0.67 in > T = 0.1 OK minimum E could be.1.67 E = psi This corresponds to a grade of No. or better Bearing Section 3.10 Bearing perpendicular to grain Factors - C M = 1 C t = 1 C i = 1 C b := 1.0 bearing at support locations Bearing at point of load R := 0.5 P bearing area A b := in R P f br := f A br = 47.6 psi f br := f b A br = 95. psi b Bearing perp to grain - OK for any grade of DF-L Solution: Choose select structural for grade