Problem 6.33
Given
beam-to-column connection
Ct = 1
4x12 No. 1 DF-L with MC ≤ 19%, A = 12 in., B = 5 in., D+S
lb = 5 in. (measured parallel to grain) < 6" ✓ (NDS section 3.10.4)
A = 12 in. > 3 in. ✓
Cb = (lb + 0.375)/lb = 1.075
NDS Supplement table 4A Cr = 1
CM = 1
Cfu = 1
CF = 1.1 for Fb, 1 for Ft, 1 for FcNDS Supplement section 2.3
CD = 1.15 (S)
Ct = 1NDS Supplement section 3.3 CL = 1 (not enough info)
NDS Supplement section 3.7
CP = 1 (not a column)
NDS Supplement section 4.3 Ci = 1
NDS Supplement section 4.4
CT = 1 (not a truss)
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 1,000 1.15 1 1 1 1.1 1 1 1 -- tension stress parallel to grain Ft 675 1.15 1 1 1 1 --
--shear stress parallel to grain Fv 180 1.15 1 1 1 -- compression stress perpendicular to grain Fc⊥ 625 1 1 1 1.075 672 compression stress parallel to grain Fc 1,500 1.15 1 1 1 1 1 -- modulus of elasticity (or MOE) E 1,700,000 1 1 1 -- modulus of elasticity for stability calculations Emin 620,000 1 1 1 1 --
fc = P/A < F'c⊥
P < F'c⊥A = 672 lb x (3.5 in. x 5 in.) = 11,760 lb
5-1/8x33 DF glulam, 24F-V4, MC = 18%, A = 0 in., B = 12 in., D+S
lb = 12 in. (measured parallel to grain) < 6" ✘ (NDS section 3.10.4)
A = 0 in. > 3 in. ✘
Cb = 1
NDS Supplement table 5A CV = 1 (not enough info)
Cfu = 1
CM = 0.8 for Fb, 0.8 for Ft, 0.875 for Fv, 0.53 for Fc⊥, 0.73 for Fc, 0.833 for E and Emin
NDS section 2.3
CD = 1.15 (S)
Ct = 1
NDS section 3.3 CL = 1 (not enough info)
NDS section 3.7
CP = 1 (not a column)
NDS section 5.3
Cc = 1 (not curved)
Property Reference Design
Values (psi)
(Table 5A)Adjustment Factors (Table 5.3.1) Adjusted Design
Values (psi)CD CM Ct CL CV Cfu Cc CP Cb bending stress F+bx
F-bx2,400
1,8501.15 0.8 1 1 1 1 1 --
--tension stress parallel to grain Ft 1,100 1.15 0.8 1 -- shear stress parallel to grain Fvx 265 1.15 0.875 1 -- compression stress perpendicular to grain Fc⊥x 650
6500.53 1 1 345
345compression stress parallel to grain Fc 1,650 1.15 0.73 1 1 -- radial tension Frt -- 1.15 1 1 -- modulus of elasticity (or MOE) Ex 1,800,000 0.833 1 -- modulus of elasticity for stability calculations Ex min 930,000 0.833 1 --
fc = P/A < F'c⊥
P < F'c⊥A = 345 lb x (5.125 in. x 12 in.) = 21,220 lb
6x16 No. 1 DF-L with MC = 20%, A = 8 in., B = 10 in., D+Lr
lb = 10 in. (measured parallel to grain) < 6" ✘ (NDS section 3.10.4)
A = 8 in. > 3 in. ✓
Cb = 1
NDS Supplement table 4A Cr = 1
CM = 0.85 for Fb, 1 for Ft, 0.97 for Fv, 0.67 for Fc⊥, 0.8 for Fc, 0.9 for E and Emin
Cfu = 1
CF =NDS Supplement section 2.3
CD = 1.25 (Lr)
Ct = 1NDS Supplement section 3.3 CL = 1 (not enough info)
NDS Supplement section 3.7
CP = 1 (not a column)
NDS Supplement section 4.3 Ci = 1
NDS Supplement section 4.4
CT = 1 (not a truss)
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 1,000 1.25 0.85 1 1 1 1 1 -- tension stress parallel to grain Ft 675 1.25 1 1 1 --
--shear stress parallel to grain Fv 180 1.25 0.97 1 1 -- compression stress perpendicular to grain Fc⊥ 625 0.67 1 1 1 419 compression stress parallel to grain Fc 1,500 1.25 0.8 1 1 1 -- modulus of elasticity (or MOE) E 1,700,000 0.9 1 1 -- modulus of elasticity for stability calculations Emin 620,000 0.9 1 1 1 --
fc = P/A < F'c⊥
P < F'c⊥A = 419 lb x (5.5 in. x 10 in.) = 23,030 lb
Deflection Limit
CD is not used to calculate F'c⊥ because Fc⊥ is based on a 0.04" deformation limit in ASTM D143. (textbook section 6.8)