Chapter 5 SolutionsPage 1 of 7 Design of Wood Structures-ASD/LRFD 7th Edition solutions manual by Donald Breyer, Kelly Cobeen, David Pollock Jr. Download more: https://solutionsmanualbank.com/download/design-of-wood-structures- asdlrfd-seventh-7th-edition-solutions-manual (6th edition also avaiable link) DESIGN OF WOOD STRUCTURES – ASD/LRFD (7th edition) 5.1 a. t = 1.5 in. b. t = 1.375 in. c. t ≤ ¾ in. for tightly curved members 1 1 3 3 3 5.2 a. b 3 in., 5 in., 6 in., 8 in., and 10 in. 8 8 4 4 4 3 1 1 b. b = 3.0 in., 5.0 in., 6 in., 8 in., and 10 in. 4 2 2 5.3 a. The combination symbol is made up of two parts. The first part is the reference bending design value in hundreds of psi followed by the letter “F”. The second part indicates the layup combination, and visually graded combinations have a “V”. For example, 24F-V4 b. The “V” is replaced by an “E”. For example, 24F-E4 c. Axial combinations are specified with numerical values. For example: 1, 2, 3, etc. d. Framing, Industrial, Architectural, Premium. They do not affect grading for strength. 5.4 The most common width (thickness) for resawn glulams is 2½ in. This glulam width is produced by “ripping” a glulam fabricated from 2 x 6 lamination stock. NDS Supplement Tables 1C and 1D give section properties for 2½ in. width glulams. Other widths may by fabricated by ripping wider glulams. See Fig. 5.3. 5.5 Horizontal or vertical finger joints. See Fig. 5.7. 5.6 Edge joints may be glued, but full contact is often difficult to maintain. As an alternative, edge joints (without glue) may be staggered in adjacent laminations. 5.7 Lamination layup is generally not specified; only the minimum grade is specified. Chapter 5 Solutions Page 2 of 7 5.8 Higher quality laminations are located at the outer surface, coinciding with the location of maximum bending stress. Lower quality laminations are used in the inner portion of the cross section. Special tension laminations may be required in the tension zone. Chapter 5 Solutions Page 3 of 7 5.9 a. bending combination: Fbx = 2000 psi, E = 1.5x106 psi Stress Class b. bending combination: Fbx = 2400 psi, E = 1.8x106 psi Stress Class c. bending combination: Fbx = 3000 psi, E = 2.1x106 psi Stress Class with Southern Pine laminations d. bending combination: Fbx = 2000 psi, visually graded Combination 7 lams with Douglas Fir-Larch outer and inner laminations e. bending combination: Fbx = 2400 psi, visually graded Combination 5 lams with Douglas Fir-Larch outer laminations and Hem-Fir inner laminations f. bending combination: Fbx = 2400 psi, E-rated Combination 11 lams with Hem-Fir outer and inner laminations g. bending combination: Fbx = 2600 psi, visually graded Combination 2 lams with Southern Pine outer and inner laminations h. axial combination: Visually graded Douglas Fir-Larch Combination 5 lams i. axial combination: Visually graded Hem-Fir Combination 16 lams j. axial combination: Visually graded Southern Pine Combination 48 lams 5.10 b = 5.125 in. (width); d = 12 in. (depth); L = 21 ft (length) CV = volume effect factor adjusts Fbx for other sizes bd 2 (5.125)(28.5) 2 5.11 a. S 693.8 in3 vs 693.8 in3 OK x 6 6 bd 3 (5.125)(28.5) 3 I 9887 in4 vs. 9887 in4 OK x 12 12 b. CD = 1.15, CM = 1.0 1 C (21)(12)(5.125) 10 0.879 (32)(28.5)(5.125) V For bending about the strong (x-x) axis: Fbx ' 2400(1.15)(1.0)(0.879) 2427 psi ' Fvx 265(1.15)(1.0) 305 psi E x' 1.8x10 6 (1.0) 1.8x10 6 psi ' c. Fbx 2427(0.8) 1942 psi with CM = 0.8 for Fb ' Fvx 305(0.875) 267 psi with CM = 0.875 for Fv E x' 1.8x10 6 (0.833) 1.5x10 6 psi with CM = 0.833 for E Chapter 5 Solutions Page 4 of 7 5.12 a. Same results as Problem 5.11, part a. b. CM = 1.0; λ = 0.8 (for L due to occupancy) 1 C (21)(12)(5.125) 10 0.879 (32)(28.5)(5.125) V For bending about the strong (x-x) axis: F+bnx = F+bx (KF) = 2400 psi (2.54) = 6096 psi = 6.10 ksi F+ + bnx’ = F bnx (Φb)(λ)(CM)(CV) = 6.10 ksi (0.85)(0.8)(1.0)(0.879) = 3.65 ksi Fvnx = Fvx (KF) = 265 psi (2.88) = 763 psi = 0.763 ksi Fvnx’ = Fvnx (Φv)(λ)(CM) = 0.763 ksi (0.75)(0.8)(1.0) = 0.458 ksi Ex’ = Ex (CM) = 1.8x106 psi (1.0) = 1800 ksi c. Wet service conditions: CM < 1.0 F+bnx’ = 3.65 ksi (0.8) = 2.92 ksi with CM = 0.8 for Fb Fvnx’ = 0.458 ksi (0.875) = 0.401 ksi with CM = 0.875 for Fv Ex’ = 1800 ksi (0.833) = 1500 ksi with CM = 0.833 for E bd 2 (5.125)(28.5) 2 5.13 a. S 693.8 in3 vs 693.8 in3 OK x 6 6 bd 3 (5.125)(28.5) 3 I 9887 in4 vs. 9887 in4 OK x 12 12 b. CD = 1.15, CM = 1.0 1 C (21)(12)(5.125) 10 0.879 (32)(28.5)(5.125) V For bending about the strong (x-x) axis: Fbx ' 2600(1.15)(1.0)(0.879) 2628 psi ' Fvx 265(1.15)(1.0) 305 psi Chapter 5 Solutions Page 5 of 7 E x' 2.0x10 6 (1.0) 2.0 x10 psi 6 ' c. Fbx 2628(0.8) 2102 psi with CM = 0.8 for Fb ' Fvx 305(0.875) 267 psi with CM = 0.875 for Fv E x' 2.0 x10 6 (0.833) 1.67 x10 6 psi with CM = 0.833 for E Chapter 5 Solutions Page 6 of 7 5.14 a. Same results as Problem 5.13, part a. b. CM = 1.0; λ = 0.8 (for L due to occupancy) 1 C (21)(12)(5.125) 10 0.879 (32)(28.5)(5.125) V For bending about the strong (x-x) axis: F+bnx = F+bx (KF) = 2600 psi (2.54) = 6604 psi = 6.60 ksi F+ + bnx’ = F bnx (Φb)(λ)(CM)(CV) = 6.60 ksi (0.85)(0.8)(1.0)(0.879) = 3.95 ksi Fvnx = Fvx (KF) = 265 psi (2.88) = 763 psi = 0.763 ksi Fvnx’ = Fvnx (Φv)(λ)(CM) = 0.763 ksi (0.75)(0.8)(1.0) = 0.458 ksi Ex’ = Ex (CM) = 2.0x106 psi (1.0) = 2000 ksi c. Wet service conditions: CM < 1.0 F+bnx’ = 3.94 ksi (0.8) = 3.16 ksi with CM = 0.8 for Fb Fvnx’ = 0.458 ksi (0.875) = 0.401 ksi with CM = 0.875 for Fv Ex’ = 2000 ksi (0.833) = 1666 ksi with CM = 0.833 for E (28.5)(5.125) 2 5.15 a. S y 124.8 in 3 vs 124.8 in 3 OK 6 (28.5)(5.125) 3 Iy 319.7 in4 vs 319.7 in4 OK 12 b. CD = 1.15; CM = 1.0; Cfu = 1.1 For bending about the weak (y-y) axis: Fby' (1450)(1.15)(1.0)(1.1) 1834 psi Fvy' (230)(1.15)(1.0) 265 psi E y' (1.6 x10 6 )(1.0) 1.6 x10 6 psi ' c. Fby (1834)(0.8) 1467 psi with CM = 0.8 for Fb ' Chapter 5 Solutions Page 6 of 7 Fvy (265)(0.875) 231psi with CM = 0.875 for Fv E y' (1.6 x10 6 )(0.833) 1.3x10 6 psi with CM = 0.833 for E Chapter 5 Solutions Page 5 of 7 5.16 a. Same results as Problem 5.15, part a. b. CM = 1.0; Cfu = 1.1; λ = 0.8 (for 1.2D + 1.6L, with L due to occupancy) For bending about the weak (y-y) axis: Fbny = Fby (KF) = 1450 psi (2.54) = 3683 psi = 3.68 ksi Fbny’ = Fbny (Φb)(λ)(CM)(Cfu) = 3.68 ksi (0.85)(0.8)(1.0)(1.1) = 2.75 ksi Fvny = Fvy (KF) = 230 psi (2.88) = 662 psi = 0.662 ksi Fvny’ = Fvny (Φv)(λ)(CM) = 0.662 ksi (0.75)(0.8)(1.0) = 0.397 ksi Ey’ = Ey (CM) = 1.6x106 psi (1.0) = 1600 ksi c. Wet service conditions: CM < 1.0 Fbny’ = 2.75 ksi (0.8) = 2.20 ksi with CM = 0.8 for Fb Fvny’ = 0.397 ksi (0.875) = 0.348 ksi with CM = 0.875 for Fv Ey’ = 1600 ksi (0.833) = 1300 ksi with CM = 0.833 for E 5.17 a. A = (5.125)(28.5) = 146.1 in2 vs. 146.1 in2 OK b. CD = 1.15; CM = 1.0 Ft ' (1100)(1.15)(1.0) 1265 psi Fc' (1600)(1.15)(1.0) 1840 psi ' E axial (1.7 x10 6 )(1.0) 1.7 x10 6 psi ' c. Ft (1265)(0.8) 1012 psi with CM = 0.8 for Ft ' Fc (1840)(0.73) 1343 psi with CM = 0.73 for Fc ' 6 6 E axial (1.7 x10 )(0.833) 1.4x10 psi with CM = 0.833 for E 5.18 a. Same results as Problem 5.17, part a. b. CM = 1.0; λ = 0.8 (for 1.2D + 1.6L, with L due to occupancy) Ftn = Ft (KF) = 1100 psi (2.7) = 2970 psi = 2.97 ksi Ftn’ = Ftn (Φt)(λ)(CM) = 2.97 ksi (0.8)(0.8)(1.0) = 1.90 ksi Fcn = Fc (KF) = 1600 psi (2.4) = 3840 psi = 3.84 ksi Fcn’ = Fcn (Φc)(λ)(CM) = 3.84 ksi (0.9)(0.8)(1.0) = 2.76 ksi E’axial = Eaxial (CM) = 1.7x106 psi (1.0) = 1700 ksi c. Wet service conditions: CM < 1.0 Ftn’ = 1.90 ksi (0.8) = 1.52 ksi with CM = 0.8 for Ft Fcn’ = 2.76 ksi (0.73) = 2.02 ksi with CM = 0.73 for Fc E’axial = 1700 ksi (0.833) = 1400 ksi with CM = 0.833 for E Chapter 5 Solutions Page 6 of 7 (5.0)(33.0) 2 5.19 a. S x 907.5 in3 vs 907.5 in3 OK 6 (5.0)(33.0) 3 4 4 Ix 14,974 in vs 14,974 in OK 12 b. CD = 1.15; CM = 1.0 1 C (21)(12)(5.125) 20 0.932 (32)(33)(5) V For bending about the strong (x-x) axis: Fbx ' (2600)(1.15)(1.0)(0.932) 2787 psi ' Fvx (300)(1.15)(1.0) 345 psi E x' (1.9x10 6 )(1.0) 1.9x10 6 psi ' c. Fbx (2787)(0.8) 2229 psi with CM = 0.8 for Fb ' Fvx (345)(0.875) 302 psi with CM = 0.875 for Fv E x' (1.9x10 6 )(0.833) 1.6x10 6 psi with CM = 0.833 for E 5.20 a. Same results as Problem 5.19, part a. b. CM = 1.0; λ = 0.8 (for 1.2D + 1.6L, with L due to occupancy) 1 C (21)(12)(5.125) 20 0.932 (32)(33)(5) V For bending about the strong (x-x) axis: F+bnx = F+bx (KF) = 2600 psi (2.54) = 6604 psi = 6.60 ksi F+ + bnx’ = F bnx (Φb)(λ)(CM)(CV) = 6.60 ksi (0.85)(0.8)(1.0)(0.932) = 4.19 ksi Fvnx = Fvx (KF) = 300 psi (2.88) = 864 psi = 0.864 ksi Fvnx’ = Fvnx (Φv)(λ)(CM) = 0.864 ksi (0.75)(0.8)(1.0) = 0.518 ksi Ex’ = Ex (CM) = 1.9x106 psi (1.0) = 1900 ksi c. Wet service conditions: CM < 1.0 F+bnx’ = 4.19 ksi (0.8) = 3.35 ksi with CM = 0.8 for Fb Fvnx’ = 0.518 ksi (0.875) = 0.454 ksi with CM = 0.875 for Fv Ex’ = 1900 ksi (0.833) = 1600 ksi with CM = 0.833 for E 5.21 Higher quality laminations are sometimes placed on the tension face. These higher quality laminations provide increased Fc┴ values. 5.22 CM < 1.0 applies for glulams when MC 16% Chapter 5 Solutions Page 7 of 7 5.23 Fb: CM = 0.8 Fc┴: CM = 0.53 Ft: CM = 0.8 Fc: CM = 0.73 Fv: CM = 0.875 E and Emin: CM = 0.833 Design of Wood Structures-ASD/LRFD 7th Edition Breyer Solutions manual Download more: https://solutionsmanualbank.com/download/design-of-wood-structures- asdlrfd-seventh-7th-edition-solutions-manual (6th edition also avaiable link) design of wood structures breyer 7th edition pdf design of wood structures 7th edition solutions design of wood structures 6th edition pdf design of wood structures 6th edition solutions manual design of wood structures 7th edition solutions pdf design of wood structures breyer pdf design of wood structures-asd/lrfd pdf design of wood structures 7th edition solution manual design of wood structures 6th edition solutions manual pdf design of wood structures solutions manual free design of wood structures 7th edition solutions design of wood structures 7th edition solutions pdf design of wood structures solutions manual pdf design of wood structures 7th edition solution manual design of wood structures breyer 7th edition pdf design of wood structures 6th edition pdf