Problem 7.3
Given
Member size 2x4 lower chord Load D = 20 psf
S = 55 psfStress grade and species Select Structural S-P-F (South) Fasteners ignore Adjustment factors CM = 1
Ct = 1
Ci = 1Assume pin-connected truss at 4 ft OC.
Size (NDS Supplement table 1B)
Ag = 5.250 in.2
An = Ag - 1.5 in. x (XX in. + 1/16 in.) = XX in.2 (NDS section 11.1.2.2 and appendix L table L1)
Load Combinations
Load Symbol Given Value (psf)
(acting downward)dead load D 20 weight of ice Di 0 earthquake load E 0 load due to fluids F 0 flood loads Fa 0 load due to sub-surface pressure H 0 live load L 0 roof live load Lr 0 rain load R 0 snow load S 55 self-straining force T 0 wind load W 0 wind-on-ice Wi 0
ASCE 7 IBC ASD Load Combination Possible Values (psf) 1 16-8 D + F 20 2 16-9 D + H + F + L + T 20 3 16-10 D + H + F + (Lr or S or R) 75 4 16-11 D + H + F + 0.75(L + T) + 0.75(Lr or S or R) 61.25 5 16-12 D + H + F + (W or 0.7E) 20 6 16-13 D + H + F + 0.75(W or 0.7E) + 0.75L + 0.75(Lr or S or R) 61.25 7 16-14 0.6D + W + H --A 8 16-15 0.6D + 0.7E + H --A A Combinations 7 and 8 are not applicable because S is acting in the same direction (downward) as D. See highlighted paragraph in ASCE 7 section C2.4.
(D)max = 20 psf
(D+S)max = 75 psf (D+0.75S) involves the same load types and is smaller.
Adjusted Design Values
NDS Supplement table 4A Cr = 1 (more than 24 in. OC)
CM = 1 (given)
CF = 1.5 for Fb, 1.5 for Ft, 1.15 for Fc
Cfu = 1 (not a beam)NDS Supplement section 2.3
CD = 0.9 (D) or 1.15 (D+S) depending on load combination
Ct = 1 (given)NDS Supplement section 3.3 CL = 1 (no compression)
NDS Supplement section 3.7
CP = 1 (not a column)
NDS Supplement section 3.10
Cb = 1 (not enough info given)
NDS Supplement section 4.3 Ci = 1 (given)
NDS Supplement section 4.4
CT = 1 (not compression chord)
Property Reference Design
Values (psi)
(Table 4A)Adjustment Factors (Table 4.3.1) Adjusted Design
Values (psi)CD CM Ct CL CF Cfu Cr CP Ci CT Cb bending stress Fb 1,300 0.9 or 1.15 1 1 1 1.5 1 1 1 -- tension stress parallel to grain Ft 575 0.9 or 1.15 1 1 1.5 1 776 or 992 shear stress parallel to grain Fv 135 0.9 or 1.15 1 1 1 -- compression stress perpendicular to grain Fc⊥ 335 1 1 1 1 -- compression stress parallel to grain Fc 1,200 0.9 or 1.15 1 1 1.15 1 1 -- modulus of elasticity (or MOE) E 1,300,000 1 1 1 -- modulus of elasticity for stability calculations Emin 470,000 1 1 1 1 --
Actual Stress for (D) Load Combination (ASD)
Load at pins A and C = 20 psf x 2.5 ft x 4 ft = 200 lb (these pass directly into ground connections and do not influence truss)
Load at other pins in upper chord = 20 psf x 5 ft x 4 ft = 400 lbHeight of truss peak = 10 ft x 5/12 = 4.167 ft (needed for drawing truss)
Height of intermediate nodes = 5 ft x 5/12 = 2.083 ft (estimated from given diagram)Load in bottom chord = 1.44 kip (T) (Autodesk Robot, negative values represent tension)
ft = P/An = 1,440 lb / 5.250 in.2 = 274 psi ≤ 776 psi ✓
Actual Stress for (D+S) Load Combination (ASD)
Load at pins A and C = 75 psf x 2.5 ft x 4 ft = 750 lb (these pass directly into ground connections and do not influence truss)
Load at other pins in upper chord = 75 psf x 5 ft x 4 ft = 1,500 lbLoad in bottom chord = 5.4 kip (T) (Autodesk Robot, negative values represent tension)
ft = P/An = 5,400 lb / 5.250 in.2 = 1,029 psi ≤ 992 psi ✘