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 Fc

NDS Supplement section 2.3

CD = 1.15 (S)
Ct = 1

 NDS 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-bx		 2,400
1,850		 1.15 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
650		   0.53 1           1 345
345
compression 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 = 1

 NDS 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)