Structural Wood Design 2nd Edition Solution Manual

Structural Wood Design 2nd Edition By Abi Aghayere, Jason Vigil

Chapter 1

1-10

Solutions

S1-1

Determine the total shrinkage across the width and thickness of a green triple 2x4 Douglas Fir Larch top plates loaded perpendicular to grain as the moisture content decreases from an initial value of 30% to a final value of 12%.

For a 2 x 4 sawn lumber, the actual thickness = 1.5 in.

For a 2 x 4 sawn lumber, the actual width, d1 = 3.5 in.

M1 = 30 and M2 = 12 (a) Shrinkage across the width of the 2 x 4 continuous blocking: The shrinkage parameters from Table 1-3 for shrinkage across the width of the 2x4 are a = 6.031, and b = 0.215 The final width d2 is given as,  6.031- 0.215 (12)   1 100  d2 = 3.5  6.031- 0.215(30)   1  100

= 3.365 in.

Thus, the total shrinkage across the width of the (triple) 2x4 is d1 – d2 = 3.5 in. – 3.365 in. = 0.135 in. (b) Shrinkage across the thickness of the (triple) 2x 4 top plates: The shrinkage parameters from Table 1-3 for shrinkage across the thickness of the 2x4 plates are a = 5.062, and b = 0.181

The final thickness d2 of each plate is given as,  5.062 - 0.181(12)   1 100  d2 =1.5   5.062 - 0.181(30)  1  100

= 1.451 in.

The total shrinkage across the thickness of the triple top plates is the sum of the shrinkage in each of the individual wood member calculated as 3 plates x (d1 – d2) = 3(1.5 in. – 1.451 in.) = 0.147 in.

Chapter 1

1-11

Solutions

S1-2

Determine the total shrinkage over the height of a 2-story building with the exterior wall cross-section shown below as the moisture content decreases from an initial value of 25% to a final value of 12%.

For a 2 x 6 sawn lumber, the actual thickness = 1.5 in.

For a 2 x 12 sawn lumber, the actual width, d1 = 11.25 in.

M1 = 25 and M2 = 12 (a) Shrinkage across the width of the 2 x 12 header joist: The shrinkage parameters from Table 1-3 for shrinkage across the width of the 2x12 are a = 6.031, and b = 0.215 The final width d2 is given as,  6.031- 0.215 (12)   1 100  d2 =11.25  6.031- 0.215(25)   1  100

= 10.93 in.

Thus, the total shrinkage across the width of the 2- 2x12 header joist is 2(d1 – d2)= 2(11.25 in. – 10.93 in.) = 0.64 in. (b) Shrinkage across the thickness of a 2x 6 top plate: The shrinkage parameters from Table 1-3 for shrinkage across the thickness of the 2x6 plates are a = 5.062, and b = 0.181

The final thickness d2 of each plate is given as,  5.062 - 0.181(12)   1 100  d2 =1.5   5.062 - 0.181(25)  1  100

= 1.465 in.

In the section shown in Figure 1.24, there are a total of 7- 2x6 top and sole or sill plates, and 2 2x12 header joists. The total shrinkage of the building cross-section will be the sum of the shrinkage across the thickness of the 2x6 top and sole/sill plates plus the shrinkage across the width of the 2x12 header joists. That is,

Chapter 1

Solutions

S1-3

7, 2x6 plates x (d1 – d2) = 7(1.5 in. – 1.465 in.) = 0.245 in.

The longitudinal shrinkage or shrinkage parallel to grain in the 2x6 wall studs is negligible.

Therefore, the total shrinkage across over the height of the two-story building, which is the sum of the shrinkage of all the wood members at the floor level, is 0.245 in. + 0.64 in. = 0.89 in.

1-12

How many board feet (bf) are there in a 4 x 16 x 36 ft long wood member? How many Mbf are in this wood member? Determine how many pieces of this wood member would amount to 4.84 Mbf or 4840 bf?

# of bf = 4x16x(36’x12)/144 = 192 bf = 192/1000 Mbf = 0.192 Mbf # of pieces = 4840 bf/192 bf = 25.2 or 25 pieces

Chapter 2

2-1.

Solutions

S2-1

Calculate the total uniformly distributed roof dead load in psf of horizontal plan area for a sloped roof with the design parameters given below.

• • • • • •

2x8 rafters at 24” on centers Asphalt shingles on ½” plywood sheathing 6” insulation (fiberglass) Suspended Ceiling Roof slope: 6-in-12 Mechanical & Electrical (i.e. ducts, plumbing etc) = 5 psf

Solution: 2x8 rafters at 24” on-centers = 1.2 psf Asphalt shingles (assume ¼” shingles) = 2.0 psf ½” plywood sheathing = (4 x 0.4 psf/1/8” plywood) = 1.6 psf 6” insulation (fiberglass) = 6 x 1.1 psf/in. = 6.6 psf Suspended Ceiling = 2.0 psf Mechanical & Electrical (i.e. ducts, plumbing etc) = 5.0 psf Total roof dead load, D (psf of sloped roof area) = 18.4 psf The total dead load in psf of horizontal plan area will be:  2 2 w DL = D  6 +12  , psf of horizontal plan area 12   



=18.4 psf (1.118 ) = 20.6 psf of horizontal plan area

2-2.

Given the following design parameters for a sloped roof, calculate the uniform total load and the maximum shear and moment on the rafter. Calculate the horizontal thrust on the exterior wall if rafters are used.

• • • • •

Roof dead load, D= 20 psf (of sloped roof area) Roof snow load, S = 40 psf (of horizontal plan area) Horizontal projected length of rafter, L2 =14 ft Roof slope: 4-in-12 Rafter or Truss spacing = 4’ 0

Solutions:  2 2 Sloped length of rafter, L1 =  4 +12  (14 ft ) =14.8 ft 12   



Chapter 2

Solutions

S2-2

Using the load combinations in section 2.1, the total load in psf of horizontal plan area will be: L  wTL = D  1  + (Lr or S or R), psf of horizontal plan area L   2   = 20 psf  14.8'  + 40 psf  14'  = 61.1 psf of horizontal plan area

The total load in pounds per horizontal linear foot (Ib/ft) is given as, wTL (Ib/ft) = wTL (psf) x Tributary width (TW) or Spacing of rafters = 61.1 psf (4 ft) = 244.4 lb/ft. h = (4/12) (14 ft) = 4.67 ft The horizontal thrust H is,

H=

 L2    244.4 Ib/ft 14'  14'   2 2    =

wTL ( L2 )  h

( )

4.67'

= 5129 Ib.

The collar or ceiling ties must be designed to resist this horizontal thrust.

L2 = 14’ The maximum shear force in the rafter is, L    Vmax = wTL  2  = 244.4  14'  = 1711 Ib  2   2    The maximum moment in the rafter is,

w (L ) M max = TL 2 8

2-3.

2

244.4 (14' ) = = 5989 ft-Ib = 5.9 ft-kip 8 2

Determine the tributary widths and tributary areas of the joists, beams, girders and columns in the panelized roof framing plan shown below. Assuming a roof dead load of 20 psf and an essentially flat roof with a roof slope of ¼” per foot for drainage, determine the following loads using the IBC load combinations. Neglect the rain load, R and assume the snow load, S is zero: a. b. c. d.

The uniform total load on the typical roof joist in Ib/ft The uniform total load on the typical roof girder in Ib/ft The total axial load on the typical interior column, in Ib. The total axial load on the typical perimeter column, in Ib

Chapter 2

Solutions

S2-3

.

Solution: The solution is presented in a tabular format as shown below: Tributary Widths and Tributary Areas of Joists, Beams and Columns Structural Member Purlin Glulam girder Typical interior Column Typical perimeter Column

Tributary Width (TW) 10' + 10' =10' 2 2 20' + 20' = 20' 2 2

Tributary Area (TA) 10’ x 20’ = 200 ft

2

20’ x 60’ = 1200 ft

2

 20' 20'   60' 60'  + + =  2   2 2   2

 20' 20'   60'  + =  2   2   2

1200 ft

600 ft

2

Since the snow and rain load are both zero, the roof live load, Lr will be critical.

With a roof slope of ¼” per foot, the number of inches of rise per foot, F = ¼ = 0.25 Purlin: The tributary width TW = 10 ft and the tributary area, TA = 200 ft2 < 200 ft2 From section 2.5.1, we obtain:R1 = 1.0 and R2 = 1.0, Using equation 2-4 gives the roof live load, Lr = 20 x 1 x 1 = 20 psf The total loads are calculated as follows: wTL (psf) = (D + Lr) = 20 + 20 = 40 psf wTL (Ib/ft) = wTL (psf) x tributary width (TW) = 40 psf x 10 ft = 400 Ib/ft Glulam Girder: 2 The tributary width, TW = 20 ft and the tributary Area, TA = 1200 ft Thus, TA > 600, and from section 2.4, we obtain: R1 = 0.6, and R2 = 1.0 Using equation 2-4 gives the roof live load, Lr = 20 x 0.6 x 1 = 12 psf The total loads are calculated as follows: wTL (psf) = (D + Lr) = 20 + 12 = 32 psf wTL (Ib/ft) = wTL (psf) x tributary width (TW) = 32 psf x 20 ft = 640 Ib/ft Typical Interior Column:

2

Chapter 2

Solutions

The tributary tributary area of the typical interior column, TA = 1200 ft Thus, TA > 600, and from section 2.4, we obtain:

S2-4

2

R1 = 0.6, and R2 = 1.0 Using equation 2-4 gives the roof live load, Lr = 20 x 0.6 x 1 = 12 psf The total loads are calculated as follows: wTL (psf) = (D + Lr) = 20 + 12 = 32 psf 2 The Column Axial Load, P = 32 psf x 1200 ft = 38, 400 Ib = 38.4 kips Typical Perimeter Column: 2 The tributary tributary area of the typical perimeter column, TA = 600 ft Thus, from section 2.5.1, we obtain: R1 = 0.6, and R2 = 1.0 Using equation 2-4 gives the roof live load, Lr = 20 x 0.6 x 1 = 12 psf The total loads are calculated as follows: wTL (psf) = (D + Lr) = 20 + 12 = 32 psf 2 The Column Axial Load, P = 32 psf x 600 ft = 19, 200 Ib = 19.2 kips 2-4.

A building has sloped roof rafters (5:12 slope) spaced at 2’ 0” on centers and is located in Hartford, Connecticut. The roof dead load is 22 psf of sloped area. Assume a fully exposed roof with terrain category “C”, and use the ground snow load from the IBC or ASCE 7 snow map (a) Calculate the total uniform load in lb/ft on a horizontal plane using the IBC. (b) Calculate the maximum shear and moment in the roof rafter.

Solution: The roof slope,  for this building is 22.6, Roof Live Load, Lr: From Section 2.4, the roof slope factor is obtained as, F=5  R2 = 1.2 – 0.05 (5) = 0.95 Assume the tributary area (TA) of the rafter < 200 ft2, The roof live load will be, Lr = 20R1R2 = 20(1.0)(0.95) = 19 psf

 R1 = 1.0

Chapter 2

Solutions

Snow Load: Using IBC Figure 1608.2 or ASCE 7 Figure 7-1, the ground snow load, Pg for Hartford, Connecticut is 30 psf.

Assuming a building with a warm roof and fully exposed, and a building site with terrain category “C”, we obtain the coefficients as follows: Exposure coefficient, Ce = 0.9 (ASCE 7 Table 7-2) The thermal factor, Ct = 1.0 (ASCE 7 Table 7-3) The Importance Factor, I = 1.0 ASCE Table 7-4 The slope factor, Cs = 1.0 (ASCE Figure 7-2 with roof slope,  = 22.6 and a warm roof) The flat roof snow load,

Pf = 0.7 Ce Ct I Pg = 0.7 x 0.9 x 1.0 x 1.0 x 30 = 18.9 psf

Minimum flat roof snow load, Pm = 20Is = 20 (1.0) = 20 psf (governs) Thus, the design roof snow load,

Ps = Cs Pf = 1.0 x 20 = 20 psf

Therefore, the snow load, S = 20 psf The total load in psf of horizontal plan area is given as, L  wTL = D  1  + (Lr or S or R) , psf of horizontal plan area L   2 Since the roof live load, Lr (18 psf) is smaller that the snow load, S (20 psf), the snow load is more critical and will be used in calculating the total roof load.  2 2 wTL = 22 psf  5 +12  + 20 psf 12   

=



43.83 psf of horizontal plan area

The total load in pounds per horizontal linear foot (Ib/ft) is given as, wTL (Ib/ft) = wTL (psf) x Tributary width (TW) or Spacing of rafters = 43.83 psf (2 ft) = 87.7 lb/ft.

Assume L2 = 14’ The maximum moment in the rafter is,

w (L ) M max = TL 2 8

2

87.7 (14' ) = = 2149 ft-Ib = 2.15 ft-kip 8 2

S2-5

Chapter 2

2-5.

Solutions

S2-6

A 3-story building has columns spaced at 18 ft in both orthogonal directions, and is subjected to the roof and floor loads shown below. Using a column load summation table, calculate the cumulative axial loads on a typical interior column with and without live load reduction. Assume a roof slope of ¼” per foot for drainage.

Roof Loads: Dead Load, Droof = 20 psf Snow Load, S = 40 psf 2nd and 3rd Floor Loads: Dead Load, Dfloor = 40 psf Floor Live Load, L = 50 psf

Solution: At each level, the tributary area (TA) supported by a typical interior column is 18’ x 18’ =

324 ft2

Roof Live Load, Lr: From section 2.4, the roof slope factor is obtained as, F = ¼ = 0.25  R2 = 1.0 Since the tributary area (TA) of the column = 324 ft2,  R1 = 1.2 – 0.001 (324) = 0.88 The roof live load will be, Lr = 20R1R2 = 20(0.88)(1.0) = 17.6 psf < Snow load, S = 40 psf The governing load combination from Section 2.1.1 for calculating the column axial loads is D + L + (Lr or S or R). Since the snow load is greater than the roof live load, the critical load combination reduces to D + L + S. The reduced or design floor live load for the 2nd and 3rd floors are calculated using the table below: Reduced or Design Floor Live Load Calculation Table AT Unreduced Member Levels (summation Floor live supported of floor KLL load, Lo tributary (psf) area)

Live Load Reduction Design Factor floor live load, L 0.25 + 15/(KLL AT)

Chapter 2

Solutions

3rd floor Column Roof only (i.e. column below roof)

Floor live load reduction NOT applicable to roofs!!!

2nd floor column (i.e. column below 3rd floor)

Ground or 1st floor column (i.e. column below 2nd floor)

-

S2-7

-

-

50 psf

0.25 + 15/(4 x 324) = 0.67

4 1 floor + roof

1 floor x 324 ft2 = 324 ft2

2 floors + roof

2 floors x 324 ft2 = 648 ft2

KLL AT = 1296 > 400 ft2  Live Load reduction allowed

4 50 psf KLL AT = 2592 > 400 ft2  Live Load reduction allowed

0.25 + 15/(4 x 648) = 0.54

40 psf (Snow load)

0.67 x 50 = 33.5 psf ≥ 0.50 Lo = 25 psf

0.54 x 50 = 27 psf ≥ 0.40 Lo = 20 psf

The column axial loads with and without floor live load reduction are calculated using the column load summation tables below:

Chapter 2

Solutions

S2-8

Column Load Summation Table

Design Live Load

Level

Tributar y area, (TA)

Dead Load, D

(ft2 )

(psf)

Live Load, Lo (S or Lr or R on the roof)

Roof: S or Lr or R Floor: L

Roof: D Floor: D + L

Roof: D+0.75S Floor: D+0.75L

Unfactore d Column Axial Load at each level, P= (TA)(ws1) or (TA)(ws2)

(psf)

(psf)

(kips)

Unfactored total load at each level, ws1

Unfactored total load at each level,ws2

(psf) (psf)

Cumulative Unfactored Axial Load, PD+L (kips)

Maximum Cumulative Cumulative Unfactored Unfactored Axial Axial Load, Load, P PD+0.75L+0.

75S

(kips) (kips)

Roof 3rd Flr

324 324

20 40

40 50

40 33.5

2nd Flr

324

40

50

27

Roof Third floor Secon d floor

324 324

20 40

40 50

40 50

324

40

50

50

With Floor Live Load Reduction 20 50 6.5 or 16.2 73.5 65.1 23.8 or 21.1 67 60.3 21.7 or 19.5 Without Floor Live Load Reduction 20 50 6.5 or 16.2 90 77.5 29.2 or 25.1 90 77.5 29.2 or 25.1

6.5 30.3

16.2 37.3

16.2 37.3

52

56.8

56.8

6.5 35.7

16.2 41.3

16.2 41.3

64.9

66.4

66.4

Chapter 2

2-6.

Solutions

S2-9

A 2-story wood framed structure 36 ft x 75 ft in plan is shown below with the following given information. The floor to floor height is 10 ft and the truss bearing (or roof datum) elevation is at 20 ft and the truss ridge is 28 ft 4” above the ground floor level. The building is “enclosed” and located in Rochester, New York on a site with a category “C” exposure. Assuming the following additional design parameters, calculate:

Floor Dead Load = Roof Dead Load = Exterior Walls = Snow Load (Pf) =

30 psf 20 psf 10 psf 40 psf

Site Class = Importance (Ie)= SS = S1 = R=

D 1.0 0.25% 0.07% 6.5

(a) The total horizontal wind force on the main wind force resisting system (MWFRS) in both the transverse and longitudinal directions. (b) The gross vertical wind uplift pressures and the net vertical wind uplift pressures on the roof (MWFRS) in both the transverse and longitudinal directions. (c) The seismic base shear, V, in kips (d) The lateral seismic load at each level in kips

Solution: (a) Lateral Wind Roof Slope: Run = 18’, Rise = 8’-4”,  = 25 Assuming a Category II building V = 115mph (ASCE 7 Table 26.5-1A) Wind Pressures (from ASCE 7, Figure 28.6-1): Transverse ( = 25): Horizontal Zone A: 26.3 psf Zone B: 4.2 psf Zone C: 19.1 psf Zone D: 4.3 psf

Vertical Zone E: -11.7 psf Zone F: -15.9 psf Zone G: -8.5 psf Zone H: -12.8 psf

Chapter 2

Solutions

S2-10

Longitudinal: ( = 0): Horizontal Zone A: 21 psf Zone B: N/A Zone C: 13.9 psf Zone D: N/A

Vertical Zone E: -25.2 psf Zone F: -14.3 psf Zone G: -17.5 psf Zone H: -11.1 psf

End Zone width: a:

 0.1 x least horizontal dimension of building  0.4 x mean roof height of the building and  0.04 x least horizontal dimension of building  3 feet

a

 0.1 (36’) = 3.6’ (governs)  20'+28.33'   (0.4)  = 9.67’ 2    0.04 (36’) = 1.44’  3 feet

Therefore the Edge Zone = 2a = 2 (3.6’) = 7.2’ Average horizontal pressures: Transverse:

 (end zone)(end zone pressure) + (bldg width − end zone)(int erior zone pressure)   q avg =  (bldgwidth )  

 (7.2 ')(26.3 psf ) + (75'− 7.2 ')(19.1 psf )  qavg ( wall ) =   = 19.8 psf (Zones A, C) (75')    (7.2 ')(4.2 psf ) + (75'− 7.2 ')(4.3 psf )  qavg (roof ) =   = 4.3 psf (Zones B, D) (75')  

Chapter 2

Solutions

Longitudinal:

 (7.2 ')(21 psf ) + (36 '− 7.2 ')(13.9 psf )  qavg ( wall ) =   = 15.32 psf (Zones A, C) (36 ')   Design wind pressures: Height and exposure coefficient:  20'+28.33'  Mean roof height =   = 24.2’ 2  

 = 1.35 (ASCE 7 Figure 28.6-1, Exposure = C, h  25’)

Transverse wind: P = qavg Pwall = (19.8 psf)(1.35) = 26.73 psf Proof = (4.3 psf)(1.35) = 5.81 psf Longitudinal wind: Pwall = (15.32 psf)(1.35) = 20.7 psf Total Wind Force: Transverse wind:

PT =  (26.73 psf )(10 '+ 10 ') + (5.81 psf )(8.33') (75') = 43.7 kips (base shear, transverse) Longitudinal wind: 8.33'   PT =  20 '+  (20.7 psf )(36 ') = 18.0 kips (base shear, longitudinal) 2  

S2-11

Chapter 2

Solutions

S2-12

(b) Wind Uplift Average vertical pressures: P = qavg From Part (a), base uplift pressures: Transverse: Zone E: -11.7 psf Zone F: -15.9 psf Zone G: -8.5 psf Zone H: -12.8 psf

Longitudinal: Zone E: -25.2 psf Zone F: -14.3 psf Zone G: -17.5 psf Zone H: -11.1 psf

Transverse:

  36 '   36 '   Pu ,avg = (−11.7 psf − 15.9 psf )(7.2 ')   + (−8.5 psf − 12.8 psf )(75'− 7.2 ')    (1.35)  2   2   = - 39,922 lb.

qu ,avg =

−39,922 lb = - 14.8 psf (gross uplift, transverse) (75')(36 ')

Longitudinal:

  75'   75'   Pu ,avg = (−25.2 psf − 14.3 psf )(7.2 ')   + (−17.5 psf − 11.1 psf )(36 '− 7.2 ')    (1.35)  2   2   = - 56,097 lb.

qu ,avg =

−56, 097 lb = - 20.8 psf (gross uplift, longitudinal) (75')(36')

Net factored uplift (Longitudinal controls): qnet = 0.9D+W =(0.9)(20psf) + (-20.8psf) = -2.8psf (net uplift)

Chapter 2

Solutions

S2-13

(c) Seismic base shear Calculate “W” for each level

Level

Area (ft.2)

Roof

75’ x 36’ = 2700 ft.2

Trib. Height (ft.)

Wt. Level (kip)*

Wt. Walls (kip)

Wtotal (kip)

(10’/2)= 5’

(2700 ft2) x [20psf + (0.2x40psf)] = 75.6k

(5’) x (10psf) x (2) x (75’ + 36’) = 11.1k

75.6k + 11.1k = 86.7k

(10’) x (10psf) x 81.0k + 2 (2) x (75’ + 36’) = 22.2k = 22.2k 103.2k W = 86.7k + 103.2k = 189.9k * Note: Where the flat roof snow load, Pf, is greater than 30psf, then 20% of the flat roof snow load shall be included in “W” for the roof (ASCE 7 Section 12.14.8.1 ) nd

75’ x 36’ = 2700 ft.2

(10’/2) + (10’/2) = 10’

(2700 ft2) x (30psf) = 81.0k

Seismic Variables: Fa = Fv =

1.6 (ASCE 7 Table 11.4-1) 2.4 (ASCE 7 Table 11.4-2) )

SMS = Fa SS = (1.6) ( 0.25) = 0.40 SM1 = Fa S1 = (2.4) ( 0.07) = 0.168 SDS = (2/3) SMS = (2/3) ( 0.40) = 0.267 SD1 = (2/3) SM1 = (2/3) ( 0.168) = 0.112

Base Shear: V=

F S DS W R

V=

(1.1) (0.267) (189.9) = 8.58k (6.5)

Chapter 2

Solutions

(d) Seismic Forces at each level:

....

Fx =

F S DS Wx R

FR =

(1.1) (0.267) (86.7) = 3.92k (6.5)

F2 =

(1.1) (0.267) (103.2) = 4.67k (6.5)

S2-14

Chapter 2

Solutions

2-7 (see framing plan and floor section) a) Determine the floor dead load in PSF b) Determine the service dead and live loads to J-1 and G-1 in PLF c) Determine the maximum factored loads in PLF to J-1 and G-1 d) Determine the factored maximum moment and shear in J-1 and G-1 e) Determine the maximum service and factored load in kips to C-1

S2-15

Chapter 2

Solutions

S2-16

Chapter 2

Solutions

S2-17

Chapter 2

2-9. w = 500plf; L1 = 20ft Case 1: continuous over support

Solutions

S2-18

Chapter 2

Solutions

Case 2: hinged over support

Case 1 would have less deflection Case 2 is easier to build; 40ft section might be hard to get or ship/handle on-site

S2-19

Chapter 2

2-10

Solutions

S2-20

Chapter 2

Solutions

a) List each truss member in a table (B1, B2, T1, T2, W1 to W6) and list the following: size, length, species, grade, density, weight). b) Calculate the total weight of the truss using the table in (a).

S2-21

Chapter 2

Solutions

S2-22

Chapter 2

Solutions

S2-23

c) Draw a free-body diagram of the truss and indicate the uniformly distributed loads to the top and bottom chords in pounds per lineal foot (plf) and indicate the supports. d) Calculate the maximum possible reaction using the controlling load case Dead + Snow.

Trib width = 2 ft.

Top chord: wD = (2)(10) = 20plf wLr = (2)(20) = 40plf wS = (2)(30.8) = 61.6plf Bot. chord: wD = (2)(10) = 20plf wD = 20+20 = 40 plf wS = 61.6 plf Rmax = ( 20 + 20 + 61.6)(19.41) =986 lb.

2

d) What are the maximum compression loads to W2, W5, and W6 and what is the purpose of the single row of bracing at midpoint?

W2 – 860 lb W5 – 899lb W6 – 354 lb The bracing limits the unbraced length of the members being braced and prevents buckling under compression.

Chapter 2

2-11: Given Loads: Uniform load, w D = 500plf L = 800plf S = 600plf Beam length = 25 ft.

Solutions

S2-24

Concentrated Load, P D = 11k S = 15k W = +12k or -12k E = +8k or - 8k

Do the following: a) Describe a practical framing scenario where these loads could all occur as shown. b) Determine the maximum moment for each individual load effect (D, L, S, W, E) c) Develop a spreadsheet to determine the worst-case bending moments for the code-required load combinations.

Chapter 2

Solutions

a) Transfer beam that has loads transferred from the roof down to a floor level.

S2-25

Chapter 2

Solutions

S2-26

Load Combinations LB = 25ft

Uniform Loads

Concentrated Loads

wD = 500plf

PD = 11kips

wL = 800plf

PS = 15kips

wS = 600plf

PW = 12kips

PWup = −12kips

PE = 8kips

PEup = −8kips

2

wDLB PDLB MD = + = 108 ft kips 8 4

MW =

PW LB = 75 ft kips 4

MWup =

PWup LB = −75 ft kips 4

MEup =

PEupLB = −50 ft kips 4

2

wLLB ML = = 62 ft kips 8

ME =

PELB 4

= 50 ft kips

2

MS =

wS LB PS LB + = 141 ft kips 8 4

(

)

(

) (

) (

)

(

) (

) (

)

(

) (

) (

LC1 = 1.4 MD = 151 ft kips LC2 = 1.2 MD + 1.6 ML + 0.5 MS = 300 ft kips LC3a = 1.2 MD + 1 ML + 1.6 MS = 417 ft kips

)

LC3b = 1.2 MD + 0.5 MW + 1.6 MS = 392 ft kips

( ) ( ) ( ) ( ) LC5 = ( 1.2 MD) + ( ME) + ( ML) + ( 0.2 MS) = 270 ft kips

LC4 = 1.2 MD + 1.6 MW + ML + 0.5 MS = 382 ft kips

(

) (

(

) (

)

LC6 = 0.9 MD + 1.0 MWup = 22 ft kips

)

LC7 = 0.9 MD + MEup = 47 ft kips Mmax = max( LC1  LC2  LC3a  LC3b  LC4  LC5) = 417 ft kips MmaxUp = min( LC6  LC7) = 22 ft kips

Chapter 2

Solutions

S2-27

2-12 (see framing plan) Assuming a roof dead load of 25 psf and a 25 degree roof slope, determine the following using the IBC factored load combinations. Neglect the rain load, R and assume the snow load, S is zero: e. Determine the tributary areas of B1, G1, C1, and W1 f. The uniform dead and roof live load and the factored loads on B1 in PLF g. The uniform dead and roof live load on G1 and the factored loads in PLF (Assume G1 is uniformly loaded) h. The total factored axial load on column C1, in kips i. The total factored uniform load on W1 in PLF (assume trib. length of 50 ft.)

Chapter 2

Solutions

S2-28

Chapter 2

Solutions

S2-29

Chapter 2

Solutions

S2-30

2-13. A 3-story building has columns spaced at 25 ft in both orthogonal directions, and is subjected to the roof and floor loads shown below. Using a column load summation table, calculate the cumulative axial loads on a typical interior column. Develop this table using a spreadsheet. Submit a hard copy that is properly formatted with your HW and submit the XLS file by e-mail.

Roof Loads: 2nd & 3rd floor loads Dead, D = 20psf Dead, D = 60psf Snow, S = 45psf Live, L = 100psf All other loads are 0

Chapter 2

Solutions

S2-31

2-14 Using only the loads shown and the weight of the concrete footing only (conc = 150pcf), determine the required square footing size, BxB using the appropriate load combination to keep the footing from overturning about point A (i.e. - either load combination 6 or 15 Chapter 2 of the text). Loads shown are service level (Mw = 0.6W = 45k-ft)

PD = 10kips

B = 7.67ft

 conc = 150pcf

MW = 45ft kips

H = 1.333ft

Pftg = BBH conc = 11.8 kips

Overturning Moment

Resisting Moment

OM = MW = 45 ft kips

B RM = PD + Pftg  = 83.5 ft kips 2

(

ASD Load Comb UnityASD =

( 0.6 RM) = 1.113 OM

Use B=7.28ft for ASD and 7.67ft for LRFD

)

LRFD Load Comb UnityLRFD =

( 0.9 RM) = 1.002  OM     0.6 

must be greater than 1.0

Chapter 2

Solutions

2-15.

Given: Location - Massena, NY; elevation is less than 1000 feet Total roof DL = 25psf Ignore roof live load; consider load combination 1.2D+1.6S only Use normal occupancy, temperature, and exposure conditions Length of B-1, B-2 is 30 ft.

Find: a) Flat roof snow load and sloped roof snow load b) Sliding snow load c) Determine the depth of the balanced snow load and the sliding snow load on B-1 and B-2 d) Draw a free-body diagram of B-1 showing the service dead and snow loads in PLF e) Find the factored Moment and Shear in B-1.

S2-32

Chapter 2

Solutions

S2-33

Chapter 2

Solutions

S2-34

2-16.

Given: Location - Pottersville, NY; elevation is 1500 feet Total roof DL = 20psf Ignore roof live load; consider load combination 1.2D+1.6S only Use normal occupancy, temperature, and exposure conditions Find: a) Flat roof snow load b) Depth and width of the leeward drift and windward drifts; which one controls the design of J1? c) Determine the depth of the balanced snow load and controlling drift snow load d) Draw a free-body diagram of J-1 showing the service dead and snow loads in PLF

Chapter 2

Solutions

S2-35

Chapter 2

Solutions

Problem 2-16 Part (e)

Factored Reactions and Maximum Moment: RA = 63.18 kips RB = 46.85 kips Mu, max = 1207.6 ft-kips (occurs at 51.55 ft from B)

S2-36

Chapter 3

3-3

Solutions

S3-1

Determine the load duration factors for the following ASCE 7 ASD load combinations.

1.

2.

3.

4.

5.

6.

7.

8.

9.

D D+H+L D + H + (Lr or S or R) D+ H + 0.75(L) + 0.75 (Lr or S or R) D + H + (0.6W or 0.7E) D + H + 0.75(0.6W or 0.7E) + 0.75L + 0.75 (Lr or S or R) D + H + 0.75(0.7E) + 0.75L + 0.75(S) 0.6D+0.6W+H 0.6D+ 0.7E+H

Chapter 3

Solutions

S3-2

Solution:

D D+H+L D+H+ (Lr or S or R) D+H+ (Lr or S or R) D+H+ + 0.75L + 0.75 (Lr or S or R) D+H+ + 0.75L + 0.75 (Lr or S or R) D+H+ (0.6W or 0.7E)

D

L

F

H

T

W

E

Lr

S

R

0.9 0.9 0.9

1.0 -

-

-

-

-

-

-

-

-

-

-

1.25

-

-

CD* for Load combo 0.9 1.0 1.25

0.9

-

-

-

-

-

-

-

1.15

-

1.15

0.9

1.0

-

-

-

-

-

1.25

-

-

1.25

0.9

1.0

-

-

-

-

-

-

1.15

-

1.15

0.9

-

-

-

-

1.6

1.6

-

-

-

1.6

Chapter 3

Solutions

S3-3

D

L

F

H

T

W

E

Lr

S

R

CD* for Load combo

0.9

1.0

-

-

-

1.6

1.6

1.25

1.15

-

1.6

1.0

-

-

-

-

1.6

-

1.15

-

1.6

-

-

-

-

1.6 -

1.6

-

-

-

1.6 1.6

D+H+ 0.75(0.6W or 0.7E) + 0.75L + 0.75 (Lr or S or R) D+H+ 0.9 0.75(0.7E) + 0.75L + 0.75(S) 0.6D+0.6W+H 0.9 0.6D+ 0.7E+H 0.9

* The largest CD value in the load combination governs

Chapter 3

3-8

Solutions

S3-4

Given a 2 x 12 Douglas Fir Larch Select Structural wood member that is fully braced laterally and subject to dead load + wind load + snow load, and exposed to the weather under normal temperature conditions, determine all the applicable allowable stresses.

Solution: Step 1 -

Size classification = Dimension Lumber  Use NDS-S Table 4A

Step 2 -

From NDS-S Table 4A, we obtain the design values: Bending stress, Fb = 1500 psi Tension stress parallel to grain, Ft = 1000 psi Horizontal shear stress parallel to grain, Fv = 180 psi Compression stress perpendicular to, Fc⊥ = 625 psi Compression stress parallel to grain, Fc = 1700 psi Pure bending modulus of elasticity, E = 1.9 x 106 psi Buckling modulus of elasticity, Emin = 0.69 x 106 psi

Step 3 -

Determine the Applicable Stress Adjustment or “C factors From Chapter 3, for dead load plus wind load snow load (D + W + S) combination, the governing load duration factor, CD is 1.6 (i.e. the largest CD value in the load combination governs) From the Adjustment factors section of NDS-S table 4A, we obtain the following “C” factors: CM (Fb) CM (Fv) CM (Fc⊥) CM (Fc) CM (E, E’) CF (Fb) CF (Ft) CF (Fc) CP

= = = = = = = = =

0.85 0.97 0.67 0.8 0.9 1.0 1.0 1.0 1.0 (column stability factor needs to be calculated,

Chapter 3

Solutions

S3-5

but assume 1.0 for this problem) Ct = 1.0 CL, Cfu, Ci, Cb = 1.0 Cr = 1.0

Step 4 -

Calculate the allowable stresses using the adjustment factors applicability table (Table 3.1)

Allowable bending stress: Fb’ = Fb CD CM Ct CL CF Cfu CiCr =

1500 x 1.6 x 0.85 x 1 x 1 x 1 x 1x1x1

=

2040 psi

Allowable tension stress parallel to grain: Ft’ = Ft CD CM Ct CFCi = 1000 x 1.6 x 1 x 1 x 1 x 1

=

1600 psi

Allowable horizontal shear stress: Fv’ = FV CD CM CtCi = 180 x 1.6 x 0.97 x 1x 1

=

279 psi

Allowable bearing stress or compression stress perpendicular to grain: F’c⊥ = Fc⊥ CM CtCiCb = 625 x 0.67 x 1x1x1 = 418 psi Allowable compression stress parallel to grain: Fc’ = Fc CD CM Ct CF CiCp (Assume CP = 1) = 1700 x 1.6 x 0.8 x 1 x 1 x 1x1 =

2176 psi

Allowable pure bending modulus of elasticity: E’ = E CM CtCi = 1.9x 106 x 0.9 x 1x1

=

1.71 x 106 psi

Allowable buckling modulus of elasticity: E’min = Emin CM CtCi = 0.69x 106 x 0.9 x 1x1

=

0.621 x 106 psi

Chapter 3

3-9

Solutions

S3-6

Given 10 x 20 Douglas Fir Larch Select Structural roof girder that is fully braced for bending and subject to dead load + wind load + snow load. Assuming dry service and normal temperature conditions, calculate the allowable bending stress, shear stress, modulus of elasticity, and bearing stress perpendicular to grain.

Solution: Step 1

Size classification is Beam & Stringer (B & S).  Use NDS-S Table 4D For 10x 20 sawn lumber, b = 9.5”, and d = 19.5”

Step 2

From NDS-S Table 4D (for D-F-L Select Structural) we obtain the design values: Bending stress, Fb = 1600 psi Horizontal shear stress, Fv = 170 psi Bearing stress or Compression stress perpendicular to grain, Fc⊥ = 625 psi Pure bending modulus of elasticity, E = 1.6 x 106 psi

Step 3 -

Determine the Applicable Stress Adjustment or “C factors From Chapter 3, for dead load plus wind load plus snow load (D + W + S) combination, the governing load duration factor, CD = 1.6 (i.e. the largest CD value in the load combination governs) From the adjustment factors section of NDS-S table 4D, we obtain the following Stress Adjustment or “C” factors: CM = 1.0 (dry service) 1

 9 CF =  12   1.0  d  1

 9 =  12  = 0.95 < 1.0 19.5  

OK

Ct = 1.0 (normal temperature condition applies) CL = 1.0 (member is fully braced laterally for bending) Cr does not apply to Timbers, therefore ignore or set equal to 1.0.

Chapter 3

Solutions

S3-7

Cfu, Ci, Cb = 1.0 Step 4

-

Calculate the allowable stresses using the Adjustment Factors Applicability table (Table 3-1)

Allowable bending stress: Fb’ = Fb CD CM Ct CL CFCfuCiCr = 1600 x 1.6 x 1 x 1 x 1 x 0.95x1x1x1

=

2432 psi

=

272 psi

Allowable horizontal shear stress: Fv’ = FV CD CM CtCi = 170 x 1.6 x 1 x 1x1

Allowable bearing stress or compression stress perpendicular to grain: F’c⊥ = Fc⊥ CM CtCiCb = 625 x 1x 1x1x1

=

625 psi

Allowable pure bending modulus of elasticity: E’ = E CM CtCi = 1.6 x 106 x 1x 1x1 = 1.6 x 106 psi

3-10

A 5-1/8” x 36” 24 F – V8 DF/DF (i.e. 24F-1.8E) simply supported Glulam roof girder spans 50 ft and is fully braced for bending, and supports uniformly distributed dead load + snow load + wind load combination. Assuming dry service and normal temperature conditions, calculate the allowable bending stress, shear stress, modulus of elasticity, and bearing stress perpendicular to grain.

Solution: Step 1: Size classification = Glulam in bending (i.e. bending combination glulam)  Use NDS-S Table 5A (Expanded) For this glulam beam, the width, b = 5.125” and depth, d = 36”, The distance between points of zero moments, Lo = L = 50 ft (simply supported beam) Step 2: From NDS-S Table 5A (Expanded), obtain the design values: Bending stress with tension lams stressed in tension, Fbx+ = 2400 psi

Chapter 3

Solutions

S3-8

Bending stress with compression lams stressed in tension, Fbx = 2400 psi

Bearing stress or compression perpendicular to grain on tension lam, Fc⊥xx, t = 650 psi Bearing stress or compression perpendicular to grain on tension lam, Fc⊥xx, c = 650 psi Horizontal shear stress parallel to grain, Fvxx

= 265 psi 6

Pure bending modulus of elasticity, Exx = 1.8 x 10 psi Step 3: Determine the Applicable Stress Adjustment or “C” factors From Chapter 3, for Dead load plus wind load plus snow load (D + W + S) combination, CD = 1.6 (check if local code allows CD value of 1.6 to be used for wind loads) From the Adjustment factors section of NDS-S table 5A, we obtain the following Stress Adjustment or “C” factors: CM = 1.0 Ct = 1.0 Cfu, Cc = 1.0 CL = 1.0 From Equation 3-3, we calculate the volume factor as, 1 1 1 1  1291.5  x  21  x  12  x  5.125  x CV =       < 1.0  =   Lo   d   b   (b) (d) (Lo )  1

=

 10 1291.5 = 0.82 < 1.0    (5.125") (36") (50 ft) 

OK

Recall that in calculating the allowable bending stress for glulam, the smaller of the CV and CL factors are used. Since CV = 0.82 is less than CL = 1.0, the CV value of 0.82 will govern for the calculation of the allowable bending stress.

Step 4:

Calculate the Allowable Stresses

Allowable bending stress with tension laminations stressed in tension: F'+ = the smaller of F+ CDCMCtCLCfuCc or F+ CDCMCtCVCfuCc bx

bx

bx

Chapter 3

Solutions

=

S3-9

2400 x 1.6 x 1 x 1 x 1 x1x1 = 3840 psi or

=

2400 x 1.6 x 1 x 1 x 0.82x1x1 = 3149 psi 

F'+bx

=

Governs 3149 psi

Allowable bending stress with compression laminations stressed in tension: = the smaller of Fbx- CDCMCtCLCfuCc or Fbx- CDCMCtCVCfuCc F'bx=

2400 x 1.6 x 1 x 1 x 1x1x1 = 3840 psi or

=

2400 x 1.6 x 1 x 1 x 0.82x1x1 = 3149 psi 

F'bx-

=

Governs 3149 psi

Allowable bearing stress or compression perpendicular to grain in the tension lamination: F’c⊥xx,t = Fc⊥xx,t CM CtCb =

650 x 1 x 1 x1 = 650 psi

Allowable bearing stress or compression perpendicular to grain in the compression lamination: F’c⊥xx,c

=

Fc⊥xx,c CM CtCb

=

650 x 1 x 1x1 = 650 psi

Allowable horizontal shear stress: F’v xx = F’v xx CD CM Ct =

265 x 1.6 x 1 x 1 = 424 psi

Pure bending modulus of elasticity for bending about the strong or x-x axis: E’xx = Exx CM Ct = 1.8 x 106 x 1 x 1 = 1.8 x 106 psi

3-11

A 6-3/4” x 36” 24 F – V8 DF/DF (i.e. 24F-1.8E) simply supported Glulam floor girder spans 64 ft between simple supports and supports uniformly distributed dead load + floor live load combination. The compression edge of the beam is laterally braced at 8-ft on centers. Assuming dry service and normal temperature conditions, calculate the allowable bending stress, shear stress, modulus of elasticity, and bearing stress

Chapter 3

Solutions

S3-10

perpendicular to grain.

Solution: Step 1: Size classification = Glulam in bending (i.e. bending combination glulam)  Use NDS-S Table 5A (Expanded) For this glulam beam, the width, b = 6.75” and depth, d = 36”, The distance between points of zero moments, Lo = L = 64 ft (simply supported beam) Step 2: From NDS-S Table 5A (Expanded), obtain the design values: Bending stress with tension lams stressed in tension, Fbx+ = 2400 psi Bending stress with compression lams stressed in tension, Fbx = 2400 psi

Bearing stress or compression perpendicular to grain on tension lam, Fc⊥xx, t = 650 psi Bearing stress or compression perpendicular to grain on tension lam, Fc⊥xx, c = 650 psi Horizontal shear stress parallel to grain, Fvxx

= 265 psi 6

Pure bending modulus of elasticity, Exx = 1.8 x 10 psi Step 3: Determine the Applicable Stress Adjustment or “C” factors From Chapter 3, for Dead load plus wind load plus snow load (D + W + S) combination, CD = 1.6 (check if local code allows CD value of 1.6 to be used for wind loads) From the Adjustment factors section of NDS-S table 5A, we obtain the following Stress Adjustment or “C” factors: CM = 1.0 Ct = 1.0 CL = 1.0 Cfu, Cc = 1.0 From Equation 3-3, we calculate the volume factor as, 1 1 1 1  1291.5  x  21  x  12  x  5.125  x CV =       < 1.0  =   Lo   d   b   (b) (d) (Lo ) 

Chapter 3

Solutions

S3-11

1

=

 10 1291.5 = 0.78 < 1.0   (6.75") (36") (64 ft)  

OK

Lu = 8 ft = 96” (that is the distance between lateral supports) Le = 2.06 Lu = 2.06 (96”) = 198”

Ld e b2

RB =

=

198(36") = 12.5 < 50 ( 6.75")2

Ey, min’ = Ey, min CM Ct = 0.83x 106 x 1.0 x 1.0 = 0.83x 106 psi

FbE =

1.20  0.83x106    = 6374 psi min =

1.20 E' R2

(12.5)2

B

*+ = F+ Fbx bx, NDS-S x CD CM Ct Cfu Cc = 2400 x 1.0 x 1.0 x 1.0 x 1.0 x 1.0 = 2400 psi Fbe / Fb = *

6374 = 2.656 2400

1+  FbE Fb*  

CL =

1.9

-

  *  1+  FbE Fb        1.9    

2

  FbE -

Fb* 



0.95

1 + 2.656 1 + 2.656  2.656 = 0.972 = −  − 1.9 0.95  1.9  2

Recall that in calculating the allowable bending stress for glulam, the smaller of the CV and CL factors are used. Since CV = 0.78 is less than CL = 0.972, the CV value of 0.78 will govern for the calculation of the allowable bending stress.

Step 4:

Calculate the Allowable Stresses

Allowable bending stress with tension laminations stressed in tension: F'+ = the smaller of F+ CDCMCtCLCfuCc or F+ CDCMCtCVCfuCc bx

bx

bx

Chapter 3

Solutions

= =

S3-12

2400 x 1.0 x 1 x 1 x 0.972x1x1 = 2232 psi or 2400 x 1.0 x 1 x 1 x 0.78x1x1 = 1872 psi  = F'+ bx

Governs 1872 psi

Allowable bending stress with compression laminations stressed in tension: = the smaller of Fbx- CDCMCtCLCfuCc or Fbx- CDCMCtCVCfuCc F'bx= =

2400 x 1.0 x 1 x 1 x 0.972x1x1 = 2232 psi or 2400 x 1.0 x 1 x 1 x 0.78x1x1 = 1872 psi  = F'bx-

Governs 1872 psi

Allowable bearing stress or compression perpendicular to grain in the tension lamination: F’c⊥xx,t = Fc⊥xx,t CM CtCb = 650 x 1 x 1x1 = 650 psi Allowable bearing stress or compression perpendicular to grain in the compression lamination: F’c⊥xx,c = Fc⊥xx,c CM CtCb = 650 x 1 x 1x1 = 650 psi Allowable horizontal shear stress: F’v xx = F’v xx CD CM Ct = 265 x 1.0 x 1 x 1 = 265 psi Pure bending modulus of elasticity for bending about the strong or x-x axis: E’xx = Exx CM Ct = 1.8 x 106 x 1 x 1 = 1.8 x 106 psi

3-12

For a roof beam subject to the following loads, determine the most critical load combination using the normalized load method.

D = 20 psf, Lr= 20 psf, S = 35 psf, W = 10 psf (downwards). Assume the wind load was calculated according to the ASCE 7 load standard

Solution: Dead load, D = 20 psf Roof live load, Lr = 20 psf Snow load, S = 35 psf Wind load, W = +10 psf (since there is no wind uplift, W = 0 in load combination 5) All other loads are neglected.

Chapter 3

Solutions

S3-13

Note: Loads with zero values are not considered in the determination of the load duration factor for that load combination, and have been excluded from the load combinations..

Since all the loads are uniformly distributed loads with the same units of pounds per square foot (psf), the load type or patterns are similar. Therefore, the normalized load method can be used to determine the most critical load combination. The applicable load combinations are shown in Table 3-12 below.

Table 3-12 Applicable and Governing Load Combination Load Combination

Value of Load Combination (psf), w

CD factor for Load Combination+ 0.9

Normalized Load w CD 20 = 22.2 D 20 psf 0.9 40 = 32 D + Lr 20 + 20 = 40 psf 1.25 1.25 55 = 47.8  D+S 20 + 35 = 55 psf 1.15 1.15 (governs) 39.5 = 24.7 D + 0.75(0.6W) + 20 + 0.75(0.6*10 + 20)= 1.6 1.6 0.75(Lr ) 39.5 psf 26.0 D + 0.6W 20 + 0.6*10 = 26 psf 1.6 = 16.3 1.6 50.8 = 31.7 D + 0.75(0.6W) + 20 + 0.75(0.6*10 + 35) = 1.6 1.6 0.75(S) 50.8 psf ++ 6 = 3.8 0.6D + 0.6W 0.6(10) + (0) = 6 psf 1.6 1.6 ++In ASD load combinations 15 and 16, the wind load, W and seismic load, E are opposed by the dead load, D. Therefore, in these combinations, D takes on a positive number while W and E take on negative values only.

Summary: • •

3-13

The Governing Load Combination is D + S (since the floor live load, L = 0 and the seismic load, E = 0) because it has the highest normalized load The roof beam should be designed for total uniform load of 55 psf with a load duration factor, CD of 1.15.

For a column subject to the axial loads given below, determine the most critical load combination using the normalized load method. Assume the wind loads have been calculated according to the ASCE 7 load Standard.

Roof Dead Load, Droof = 8 kip

Floor Dead Load, Dfloor = 15 kip

Chapter 3

Solutions

S3-14

Roof Live Load, Lr = 15 kip, Floor Live Load, L = 26 kip Snow Load, S = 20 kip Wind Load, W = 8 kip, Earthquake or Seismic Load, E = 10 kip .

Solution: In the load combinations, the dead load D is the sum of the dead loads from the all levels of the building. Thus, D = Droof + Dfloor = 8 kip + 15 kip = 23 kips Since all the loads on this column are concentrated axial loads and therefore similar in pattern or type, the normalized load method can be used!

Applicable and Governing Load Combination Load Combination

Value of Load Combination (k), P (kip)

CD factor for Load Combination+ 0.9

D

23

D+L

23 + 26 = 49

1.0

D + Lr

23 + 15 = 38

1.25

D+S

23 + 20 = 43

1.15

D + 0.75(L) + 0.75(Lr)

23 + 0.75(26 + 15)= 53.8

1.25

D + 0.75(L) + 0.75(S)

23 + 0.75(26 + 20) = 57.5

1.15

D + 0.6W

23 + 0.6(8) = 27.8

1.6

D + 0.7E

23 + 0.7(10) = 30.0

1.6

D + 0.75(0.6W) + 0.75L + 0.75(Lr ) D + 0.75(0.6W) + 0.75L + 0.75(S) D + 0.75(0.7E) + 0.75L + 0.75(Lr) D + 0.75(0.7E) + 0.75L + 0.75(S)

23 + 0.75(0.6*8 + 26 + 15 )= 57.4 23 + 0.75(0.6*8 + 26 + 20) * = 61.1 23 + 0.75[(0.7(10) + 26 + 15] = 59 23 + 0.75[(0.7(10) + 26 + 20] = 62.8

1.6 1.6 1.6 1.6

Normalized Load P CD 23 = 25.6 0.9 49 = 49 1.0 38 = 30.4 1.25 43 = 37.4 1.15 53.8 = 43.04 1.25 57.5 = 50.0  1.15 (governs) 27.8 = 17.4 1.6 30.0 = 18.8 1.6 57.4 = 35.9 1.6 61.1 = 38.2 1.6 59 = 36.9 1.6 62.8 = 39.3 1.6

Chapter 3

Solutions

0.6(23) + 0.6(-8) ++ = 9

S3-15

9.0 = 5.6 1.6 6.8 = 4.3 0.6D + 0.7E 0.6(23) + 0.7(-10) ++ = 1.6 1.6 6.8 ++ In ASD load combinations 15 and 16, the wind load, W and seismic load, E are opposed by the dead load, D. Therefore, in these combinations, D takes on a positive number while W and E take on negative values only.

0.6D + 0.6W

1.6

Summary: • •

The Governing load combination is dead load plus floor live load plus snow load (D + 0.75L + 0.75S) because it has the highest normalized load.

The column should be designed for a total axial load, P = 57.5 kip with a load duration factor, CD = 1.15

Chapter 3

Solutions

S3-16

Chapter 3

Solutions

S3-17

Chapter 3

Solutions

S3-18

Chapter 3

Solutions

S3-19

Chapter 3

Solutions

S3-20

Chapter 3

Solutions

S3-21

Chapter 3

Solutions

S3-22

Chapter 3

Solutions

S3-23

Chapter 3

Solutions

S3-24

Chapter 3

3-17.

Alternate Check:

Solutions

S3-25

Chapter 3

Solutions

S3-26

Chapter 3

3-18.

Solutions

S3-27

Chapter 3

3-19: Given Loads: Uniform load, w D = 500plf L = 800plf S = 600plf Beam length = 25 ft.

Solutions

S3-28

Concentrated Load, P D = 11k S = 15k W = +12k or -12k E = +8k or - 8k

Do the following: a) Describe a practical framing scenario where these loads could all occur as shown. b) Determine the maximum moment for each individual load effect (D, L, S, W, E) c) Develop a spreadsheet to determine the worst-case bending moments for the code-required load combinations.

Chapter 3

Solutions

a) Transfer beam that has loads transferred from the roof down to a floor level.

S3-29

Chapter 3

Solutions

S3-30

3.20. A 3-story building has columns spaced at 24 ft in both orthogonal directions, and is subjected to the roof and floor loads shown below. Using a column load summation table, calculate the cumulative axial loads on a typical interior column. Develop this table using a spreadsheet. Submit a hard copy that is properly formatted with your HW and submit the XLS file by e-mail.

Roof Loads: 2nd & 3rd floor loads Dead, D = 20psf Dead, D = 60psf Snow, S = 45psf Live, L = 100psf All other loads are 0

Chapter 3

Solutions

S3-31

Chapter 4

4-1.

Solution

S4-1

Determine the maximum uniformly distributed floor live load that can be supported by a 10 x 24 girder with a simply supported span of 16 ft. Assume wood species and stress grade is Douglas Fir Larch Select Structural (SS), normal duration loading, normal temperature conditions, and continuous lateral support is provided to the compression flange. The girder has a tributary width of 14 feet and a floor dead load of 35 psf in addition to the girder self-weight.

Solution: Density of Douglas fir larch = 31.2 pcf For 10x24 Beam & Stringer, Sxx = 874.4 in3 Selfweight = 48.4 Ib/ft Fb (NDS-S table 4D) = 1600 psi Fv = 170 psi CD = 1.0 (D + L) Ct = Ci = CM = 1.0 Since all the adjustment or C- factors are 1.0, the allowable stresses are as follows: F’b = 1600 psi F’v = 170 psi F’c⊥ = 625 psi E’ = 1.6 x 106 psi As determined from the text, bending typically governs over shear, bearing or deflections.

Therefore, using equation 4-11, the maximum allowable total load,  F' S    b xx 8 12    8 ( M max )  Ib/ft wTL = =  2 2 L L

( )

(

)

 1600 psi 874.4in 3    8  12     = 3643.3 Ib/ft (governs) = 2 (16 ft)

Chapter 4

Solution

S4-2

Check Shear Stress

F' ( A xx ) The allowable maximum shear, Vmax or V’max = v , Ib 1.5 170 psi  223.3 in 2  =





1.5

= 25,307.3 Ib V = Maximum shear at the centerline of the joist, beam or girder bearing support V’ = Maximum a distance “d” from the face of the support Use V’ to calculate the applied shear stress only for beams or girders with no concentrated loads within a distance “d” from the face of support and where the bearing support is subjected only to confining compressive stresses. (i.e. use V’ only for beams or girders subject to compression at the support reactions) Vmax = wTL(16 ft)/2 V’max = (16 ft/2 – 23.5/12)/(16 ft/2)] wTL(16 ft)/2 = 25,307.3 Therefore, wTL = 4189 Ib/ft

Check Deflection From equation 4-15, the allowable maximum live load is 384 E'I  L    Ib/in 5L4  360  384 E'I  L  =   (12 ) Ib/ft 5L4  360  384 (1.6 x 106 )(10,270 in 4 )  16 ft x 12  12 = 5943 Ib/ft =  360  ( ) 5(16 ft x 12)4 

wLL =

Similarly, for a uniformly loaded beam or girder, the maximum allowable total load is obtained from equation 4-16 as, 384 E'I  L    Ib/in 5L4  240  384 E'I  L  =   (12 ) Ib/ft 5L4  240 

w k (DL) + LL =

Chapter 4

Solution

=

S4-3

384 (1.6 x 106 )(10,270 in 4 )  16 ft x 12  12 = 8915 Ib/ft  240  ( ) 5(16 ft x 12)4 

The smallest wTL value governs, therefore, wTL (due to bending stress) = 3643.3 Ib/ft.

Knowing the allowable total load, wTL, and the applied dead load, wDL on the member, the allowable live load wLL based on bending stress alone can be obtained using the load combination relationships (see Section 2.1.1). wTL

= wDL + wLL

3643.3 Ib/ft = 48.4 Ib/ft + 35 psf (14’) + wLL (14’) Therefore, the allowable maximum live load, wLL = 222 psf Maximum Live Load, wLL = 222 psf

4-2.

For the girder in problem 4-1, determine the minimum bearing length required for a live load of 100 psf assuming the girder is simply supported on columns at both ends.

Solution: The maximum reaction, R1 = [48.4 Ib/ft + 35 psf (14’) + 100psf (14’)] (16 ft/2) = 15,507 Ib The minimum required bearing length, lb from equation 4-9 is

lb reqd =

4-3.

R1 bF

c⊥

=

15,507 Ib = 2.7” (9.5") (625 psi)

Determine the joist size required to support a dead load of 12 psf (includes the selfweight of joist) and a floor live load of 40 psf assuming a joist spacing of 24” and a joist span of 15 ft? Assume Southern Pine wood species, normal temperature and dry service conditions.

Solution: Wood specie = SOUTHERN PINE Joist span = 15 ft Live load, L = 40 psf Dead load = 12 psf Tributary area of joist = 24" (15') = 30 ft2 12 From Chapter 2, we obtain the following parameters:

Chapter 4

Solution

AT = tributary area = 30 ft2 and KLL = 2 (interior beam or girder) KLL AT = (2) (30 ft2) = 60 ft2 < 400 ft2  No live load reduction is permitted Tributary width (TW) of joist or beam B1 = 24” = 2.0’ The total uniform load on the joist that will be used to design the joist for bending, shear and bearing is, wTL = (D + L) x TW = (12 psf + 40 psf) x 2’ = 104 lb/ft

(104 Ib/ft ) 15'  w L TL Maximum Shear, Vmax = Rmax = = = 780 lb 2 2 2 104 Ib/ft ) 15'  ( 2   = 2925 ft-lb = 35,100 in lb Maximum Moment, Mmax = w TL L = 8 8 The following loads will be used for calculating the joist deflections.

The uniform dead load is, wDL= 12 psf x 2’ = 24 lb/ft = 2 lb/in The uniform live load is, wLL = 40 psf x 2’ = 80 lb/ft = 6.67 lb/in

Check Bending Stress: Summary of load effects: Maximum Shear, Vmax = 780 lb Maximum Reaction, Rmax = 780 lb Maximum Moment, Mmax = 35,100 in lb Uniform dead load is, wDL= 2.0 lb/in Uniform live load is, wLL = 6.67 lb/in

The wood specie is Southern Pine For the stress grade, assume 2x10 Non-Dense Select Structural.

The size classification for the joist is Dimension Lumber Use NDS-S Table 4B From NDS-S Table 4B, we find Fb, NDS-S = 1850 psi Assume initially that F’b = Fb, NDS-S = 1850 psi From equation 4-1, the required approximate section modulus of the member is given as, Sxx req’d  M max = 35100 = 18.97 in3 1850 Fb, NDS-S

S4-4

Chapter 4

Solution

S4-5

From NDS-S Table 1B, the trial size with the least area that satisfies the section modulus requirement of step 3: Try 2 x 10 SP Non-Dense Select Structural b = 1.5” and d = 9.25” Size is still dimension lumber as assumed in step 1.

Sxx provided = 21.39 in3 > 18.97 in3 Area, A provided = 13.88 in2 Ixx provided = 98.93 in4

OK OK

The NDS-S tabulated stresses are, Fb = 1850 psi Fv = 175 psi Fc⊥ = 480 psi E = 1.7 x 106 psi The Adjustment or “C” factors are: Adjustment Factors for Joist* Adjustment Factor Adjustment factor Symbol Beam stability factor CL

Value 1.0

Rational for the chosen value

The compression face is fully braced by the floor sheathing Size factor CF (Fb) Already included in tabulated values Moisture or wet CM 1.0 Equilibrium moisture content (EMC) service factor is  19% Load duration factor CD 1.0 The largest CD value in the load combination of dead load plus floor live load (i.e. D + L) Temperature factor Ct 1.0 Insulated building, therefore normal temperature condition applies Repetitive member Cr 1.15 All the three required conditions are factor satisfied: • The 2 x 10 trial size is dimension lumber • Spacing = 16”  24” • Plywood floor sheathing nailed to joists Bearing stress factor Cb 1. 0 Cb = 1.0 for bearings at the ends of a member (See Section 3-1 of this text) *Other applicable C-factors not listed default to 1.0 or are neglected Using equation 4-2, the allowable bending stress is,

Chapter 4

Solution

S4-6

Fb’ = Fb NDSS x (Product of Applicable Adjustment or “C” factors) = Fb NDSS CD CM Ct CLCF CfuCi Cr = 1850 x 1.0x1.0x1.0x1.0x1.0x1.0x1.15 = 2128 psi The actual applied bending stress is, fb = M max = 35100 Ib -3in = 1641 psi 21.39 in Sxx < Fb’ = 2128 psi

OK

 The beam is adequate in bending

Check Shear Stress Vmax = 780 lb The beam cross-sectional area, A = 13.88 in2 The applied shear stress in the wood beam at the centerline of the beam support is, 1.5( 780 Ib ) fv = 1.5V = = 84.3 psi A 13.88 in 2

The allowable shear stress is, F’v = Fv CD CM Ct Ci = 175 x 1 x 1 x 1 x 1 = 175 psi Thus, fv < F’v

 The beam is adequate in shear

Check Deflection The allowable pure bending modulus of elasticity is, E’ = E CM Ct = 1.7 x 106 x 1 x 1 = 1.7 x 106 psi The moment of inertia about the strong axis Ixx = 98.93 in4 The uniform dead load is, wDL= 2.0 lb/in The uniform live load is, wLL = 6.67 lb/in

Chapter 4

Solution

Table 4-3 Joist Deflection Limit (see Table 4-1) Deflection Live load deflection, LL Incremental long term deflection due to dead load plus live load (including creep effects), TL = k (DL) + LL

S4-7

Deflection Limit L = 15' x 12 =0.50” 360 360 L = 240 15' x 12 =0.75” 240

The dead load deflection is, 4 4 DL = 5wL = 5 (2.0 Ib/in) (15' x 12) = 0.16” 384EI 384(1.7x106 psi)(98.93in 4 )

The live load deflection is, 4 4 LL = 5wL = 5 (6.67 Ib/in) (15' x 12) = 0.53” > L Not Good 384EI 360 6 4 384(1.7x10 psi)(98.93in )

Since seasoned wood in a dry service condition is assumed to be used in this building, the creep factor, k = 0.5 The total incremental dead plus floor live load deflection is, TL = k (DL) + LL = 0.5 (0.16”) + 0.53” = 0.61” < L OK 240 Alternatively, the required moment of inertia can be calculated using Equations 4-5 and 4-6 as follows: I required =

3 L3 360 ) in4 = 5(6.67)(15'x12) ( 360 ) = 107.3 in4 ( 384 (1.7x106 ) 384 E'

5w

LL

Similarly, the required moment of inertia based on total load is obtained for a uniformly loaded beam or girder as,

5w I required =

k(DL) + LL

384 E'

L3

( 240 ) in4

=

5(0.5x2.0 + 6.67)(15'x12)3 384 (1.7x106 )

( 240 ) = 82.2 in4

Therefore, the required moment of inertia = 107.3 in4, therefore TRY 2 x 12

Check Bearing Stress or Compression Stress Perpendicular (⊥) to Grain

Chapter 4

Solution

S4-8

Maximum reaction at the support, R1 = 780 Ib Thickness of 2 x 12 sawn lumber joist, b = 1.5” The allowable bearing stress or compression stress parallel to grain is, F’c⊥ = Fc⊥ CM Ct Cb = 480 x 1 x 1 x 1 = 480 psi From equation 4-7, the minimum required bearing length, lb is,

lb reqd 

R1 bF

c⊥

=

780 Ib = 1.1” (1.5")(480 psi)

The floor joists will be connected to the floor girder using Face Mount Joist Hangers with the top of the joists at the same level as the top of the girders. A Face Mount Joist Hanger should be selected from a Connector Manufacturer’sCatalog that provides the reaction capacity as well as the required bearing length.

Use 2 x 12 Southern Pine Non-Dense Select Structural Joist

4-4.

Determine the joist size required to support a dead load of 10 psf (includes the selfweight of joist) and a floor live load of 50 psf assuming a joist spacing of 16” and a joist span of 15 ft? In addition, the joists support an additional 10 ft high partition wall weighing 10 psf running perpendicular to the floor joists at the mid-span of the joists.

Assume Spruce Pine Fir (S-P-F) Select Structural wood species, normal temperature and dry service conditions.

Solution: Wood specie = Spruce-Pine-Fir Joist span = 15 ft Live load, L = 50 psf Dead load = 10 psf Tributary area of joist = 16" (15') = 20 ft2 12 From Chapter 2, we obtain the following parameters: AT = tributary area = 20 ft2 and KLL = 2 (interior beam or girder) KLL AT = (2) (20 ft2) = 40 ft2 < 400 ft2  No live load reduction is permitted Tributary width (TW) of joist = 16” = 1.33’ The total uniform load on the joist that will be used to design the joist for bending, shear and bearing is,

Chapter 4

Solution

S4-9

wTL = (D + L) x TW = (10 psf + 50 psf) x 1.33’ = 80 lb/ft Concentrated partition wall load = 10 psf (1.33’)(10 ft) = 133 Ib (80 Ib/ft ) 15' 133 Ib w L TL Maximum Shear, Vmax = Rmax = = = 667 lb + 2 2 2 2 ' 80 Ib/ft 15 ( ) 2 133 Ib (15 ft ) Maximum Moment, Mmax = w TL L = = 2749 ft-lb = 32,988 in lb + 8 8 4

( )

( )

The following loads will be used for calculating the joist deflections.

The uniform dead load is, wDL= 10 psf x 1.33’ = 13.33 lb/ft = 1.11 lb/in The uniform live load is, wLL = 50 psf x 1.33’ = 66.7 lb/ft = 5.55 lb/in Concentrated dead load, PDL = 133 Ib

Check Bending Stress: Summary of load effects: Maximum Shear, Vmax = 667 lb Maximum Reaction, Rmax = 667 lb Maximum Moment, Mmax = 32,988 in lb Uniform dead load is, wDL= 1.11 lb/in Uniform live load is, wLL = 5.55 lb/in Concentrated dead load, PDL = 133 Ib (at midspan)

The wood specie is S-P-F (Spruce Pine Fir); assume Dimension Lumber Therefore, use NDS-S Table 4A From NDS-S Table 4B, we find Fb, NDS-S = 1250 psi Assume initially that F’b = Fb, NDS-S = 1250 psi From equation 4-1, the required approximate section modulus of the member is given as, Sxx req’d  M max = 32,988 = 26.4 in3 1250 Fb, NDS-S From NDS-S Table 1B, the trial size with the least area that satisfies the section modulus requirement of step 3: Try 2 x 12 Select Structural b = 1.5” and d = 11.25”

Chapter 4

Solution

S4-10

Size is still dimension lumber as assumed in step 1.

Sxx provided = 31.64 in3 > 26.4 in3 Area, A provided = 16.88 in2 Ixx provided = 178 in4

OK OK

The NDS-S tabulated stresses are, Fb = 1250 psi Fv = 135 psi Fc⊥ = 425 psi E = 1.5 x 106 psi The Adjustment or “C” factors are: Adjustment Factors for Joist* Adjustment Factor Adjustment factor Symbol Beam stability factor CL

Value

Rational for the chosen value

1.0

The compression face is fully braced by the floor sheathing Size factor CF (Fb) 1.0 NDS-S Table 4A Moisture or wet CM 1.0 Equilibrium moisture content (EMC) service factor is  19% Load duration factor CD 1.0 The largest CD value in the load combination of dead load plus floor live load (i.e. D + L) Temperature factor Ct 1.0 Insulated building, therefore normal temperature condition applies Repetitive member Cr 1.15 All the three required conditions are factor satisfied: • The 2 x 10 trial size is dimension lumber • Spacing = 16”  24” • Plywood floor sheathing nailed to joists Bearing stress factor Cb 1. 0 Cb = 1.0 for bearings at the ends of a member (See Section 3-1 of this text) *Other applicable C-factors not listed default to 1.0 or are neglected

Using equation 4-2, the allowable bending stress is, Fb’ = Fb NDSS x (Product of Applicable Adjustment or “C” factors) = Fb NDSS CD CM Ct CL CF CfuCi Cr = 1250 x 1.0x1.0x1.0x1.0x1.0x1.0x1.15 = 1438 psi

Chapter 4

Solution

S4-11

1) The actual applied bending stress is, fb = M max = 32,988 Ib 3- in = 1043 psi 31.64 in Sxx < Fb’ = 1438 psi

OK

 The beam is adequate in bending

Check Shear Stress Vmax = 667 lb The beam cross-sectional area, A = 16.88 in2 The applied shear stress in the wood beam at the centerline of the beam support is, 1.5( 667 Ib ) fv = 1.5V = = 59.3 psi A 16.88 in 2

The allowable shear stress is, F’v = Fv CD CM Ct Ci = 135 x 1 x 1 x 1 x 1 = 135 psi Thus, fv < F’v

 The beam is adequate in shear

Check Deflection The allowable pure bending modulus of elasticity is, E’ = E CM Ct = 1.5 x 106 x 1 x 1 = 1.5 x 106 psi The moment of inertia about the strong axis Ixx = 178 in4 Uniform dead load is, wDL= 1.11 lb/in Uniform live load is, wLL = 5.55 lb/in Concentrated dead load, PDL = 133 Ib (at midspan) Table 4-3 Joist Deflection Limit (see Table 4-1) Deflection Deflection Limit L Live load deflection, LL = 15' x 12 =0.50” 360 360 L = Incremental long term deflection due to dead load 240 plus live load (including creep effects), 15' x 12 =0.75” TL = k (DL) + LL 240

Chapter 4

Solution

S4-12

The dead load deflection is, 4 5 (1.11 Ib/in) (15' x 12) 4 133 Ib (15' x 12)3 DL = 5wL = = 0.12” + 384EI 384(1.5x106 psi)(178 in 4 ) 48(1.5x106 psi)(178 in 4 )

The live load deflection is, 4 5 (5.55 Ib/in) (15' x 12)4 LL = 5wL = = 0.28” < L OK 384EI 360 6 4 384(1.5x10 psi)(178 in )

Since seasoned wood in a dry service condition is assumed to be used in this building, the creep factor, k = 0.5 The total incremental dead plus floor live load deflection is, TL = k (DL) + LL = 0.5 (0.12”) + 0.28” = 0.34” < L 240

OK

Check Bearing Stress or Compression Stress Perpendicular (⊥) to Grain Maximum reaction at the support, R1 = 667 Ib Thickness of 2 x 12 sawn lumber joist, b = 1.5” The allowable bearing stress or compression stress parallel to grain is, F’c⊥ = Fc⊥ CM Ct Cb = 425 x 1 x 1 x 1 = 425 psi From equation 4-7, the minimum required bearing length, lb is,

lb reqd 

R1 bF

c⊥

=

667 Ib = 1.05” (1.5")(425 psi)

The floor joists will be connected to the floor girder using Face Mount Joist Hangers with the top of the joists at the same level as the top of the girders. A Face Mount Joist Hanger should be selected from a Connector Manufacturer’sCatalog that provides the reaction capacity as well as the required bearing length.

Use 2 x 12 S-P-F Select Structural Joist

Chapter 4

Solution

S4-13

4.5a DL = 15psf

TW = 12in

LL = 50psf

wLL = LLTW = 50 plf

TL = DL + LL = 65 psf

Lb = 16ft

wTL = TLTW = 65.0 plf

2

wTLLb Mb = = 2080 ft lb 8

wTLLb = 520 lbf 2

Vb =

Hem Fir #2 Fb = 850psi

CD = 1.0

Fv = 150psi

CM = 1.0

Cfu = 1.0

E = 1300ksi

Cr = 1.15

Ci = 1.0

Emin = 470ksi

b = 1.5in

Ct = 1.0

CFb = 1.0

CT = 1.0

d = 11.25in

use 2x12

2

Sx =

Ax = b d = 16.875 in

b d 6

2

CL = 1.0

3

3

5

b d 4 = 177.979 in 12

Ix =

= 31.641 in

E'min = EminCM Ct Ci CT = 4.7  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 978 psi

F'b = Fbs CL = 978 psi

Mmax = F'b Sx = 2577.4 ft lb

F'v = Fv CDCM Ct Ci = 150 psi

Vmax =

Mb fb = = 788.9 psi Sx fv =

1.5 Vb Ax

= 46.222 psi

(

F'v Ax 1.5

= 1687 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1300 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.319 in

 LLmax =

Lb = 0.4 in 480

if  LL   LLmax "OK"  "NG" = "OK"

= 0.414 in

 TLmax =

Lb = 0.8 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

(

)

(

)

Chapter 4

Solution

S4-14

4.5b DL = 20psf

TW = 16in

LL = 60psf

wLL = LLTW = 80 plf

TL = DL + LL = 80 psf

Lb = 15ft

wTL = TLTW = 106.7 plf

2

wTLLb Mb = = 3000 ft lb 8

wTLLb = 800 lbf 2

Vb =

Hem Fir #2 Fb = 850psi

CD = 1.15

Fv = 150psi

CM = 1.0

Cfu = 1.0

E = 1300ksi

Cr = 1.15

Ci = 1.0

Emin = 470ksi

b = 1.5in

Ct = 1.0

CFb = 0.9

CT = 1.0

d = 13.25in

use 2x14

2

Sx =

Ax = b d = 19.875 in

b d 6

2

CL = 1.0

3

3

= 43.891 in

5

b d 4 = 290.775 in 12

Ix =

E'min = EminCM Ct Ci CT = 4.7  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1012 psi

F'b = Fbs CL = 1012 psi

Mmax = F'b Sx = 3700.4 ft lb

F'v = Fv CDCM Ct Ci = 172.5 psi

Vmax =

Mb fb = = 820.2 psi Sx fv =

1.5 Vb Ax

= 60.377 psi

(

F'v Ax 1.5

= 2286 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1300 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.241 in

 LLmax =

Lb = 0.375 in 480

if  LL   LLmax "OK"  "NG" = "OK"

= 0.321 in

 TLmax =

Lb = 0.75 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

(

)

(

)

Chapter 4

Solution

S4-15

4.5c DL = 10psf

TW = 24in

LL = 125psf

wLL = LLTW = 250 plf

TL = DL + LL = 135 psf

wTL = TLTW = 270.0 plf

2

wTLLb Ps Lb Mb = + = 3375 ft lb 8 4

Vb =

Lb = 10ft

Nj = 2

PD = 0lb

PL = 0lb

Ps = PD + PL = 0

wTLLb Ps + = 1350 lbf 2 2

Hem Fir #2 Fb = 850psi

CD = 1.0

Fv = 150psi

CM = 1.0

Cfu = 1.0

E = 1300ksi

Cr = 1.15

Ci = 1.0

Emin = 470ksi

b = 1.5in

Ct = 1.0

CFb = 1.1

CT = 1.0

d = 9.25in

CL = 1.0

use (2)-2x10 @ 24 in. O.C.

2

Sx =

Ax = Nj b d = 27.75 in

Nj b d 6

2 3

= 42.781 in

5

Ix =

Nj b d

3

12

4

= 197.863 in

E'min = EminCM Ct Ci CT = 4.7  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1075 psi

F'b = Fbs CL = 1075 psi

Mmax = F'b Sx = 3833.4 ft lb

F'v = Fv CDCM Ct Ci = 150 psi

Vmax =

Mb fb = = 946.7 psi Sx fv =

1.5 Vb Ax

(

 LL =

384 E'Ix

 TL =

384 E'Ix

)

(

)

 LLmax =

Lb = 0.25 in 480

 TLmax =

Lb = 0.5 in 240

3

+

4

5 wTLLb

= 2775 lbf

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1300 ksi

4

1.5

if fb  F'b  "OK"  "NG" = "OK"

= 72.973 psi

5 wLLLb

F'v Ax

PLLb

48 E'Ix

= 0.219 in

(

)

(

)

if  LL   LLmax "OK"  "NG" = "OK"

3

+

Ps Lb

48 E'Ix

= 0.236 in

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-16

4.5d DL = 25psf

TW = 12in

Lb = 14ft

Nj = 2

LL = 50psf

wLL = LLTW = 50 plf

PD = 250lb

PL = 500lb

wTL = TLTW = 75.0 plf

Ps = PD + PL = 750 lbf

TL = DL + LL = 75 psf 2

wTLLb Ps Lb Mb = + = 4462 ft lb 8 4

Vb =

wTLLb Ps + = 900 lbf 2 2

Hem Fir #2 Fb = 850psi

CD = 1.0

Fv = 150psi

CM = 1.0

Cfu = 1.0

E = 1300ksi

Cr = 1.15

Ci = 1.0

CT = 1.0

Emin = 470ksi

b = 1.5in

Ct = 1.0

CFb = 1.0

d = 11.25in

use (2)-2x12 @ 12 in. O.C.

2

Nj b d

Sx =

Ax = Nj b d = 33.75 in

6

CL = 1.0

2 3

Ix =

= 63.281 in

5

Nj b d

3

12

4

= 355.957 in

E'min = EminCM Ct Ci CT = 4.7  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 978 psi

F'b = Fbs CL = 978 psi

Mmax = F'b Sx = 5154.8 ft lb

F'v = Fv CDCM Ct Ci = 150 psi

Vmax =

Mb fb = = 846.2 psi Sx fv =

1.5 Vb Ax

(

 LL =

384 E'Ix

 TL =

384 E'Ix

)

(

)

 LLmax =

Lb = 0.35 in 480

 TLmax =

Lb = 0.7 in 240

3

+

4

5 wTLLb

= 3375 lbf

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1300 ksi

4

1.5

if fb  F'b  "OK"  "NG" = "OK"

= 40 psi

5 wLLLb

F'v Ax

PLLb

48 E'Ix

= 0.200 in

(

)

(

)

if  LL   LLmax "OK"  "NG" = "OK"

3

+

Ps Lb

48 E'Ix

= 0.300 in

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-17

4.6a wTL = 950plf

Ps = 0lb

Lb = 19ft 2

wTLLb Ps Lb Mb = + = 42869 ft lb 8 4

wTLLb Ps Vb = + = 9025 lbf 2 2 4

Mb 3 SxMin = = 381.056 in Fb

b = 5.5in

DFL #1

IxMin =

d = 21.5in

5 wTLLb 384 E

Fb = 1350psi Fv = 170psi

E = 1600ksi Emin = 580ksi

 240  = 1833 in4   Lb 



use 6x22 1 9

CD = 1.15

 12in  = 0.937   d 

CFb = 

Ct = 1.0

CT = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 CL = 1.0

2

Sx =

Ax = b d = 118.25 in

b d 6

2

3

3

Ix =

= 424 in

b d 4 = 4555 in 12

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1455 psi

E'min = EminCM Ct Ci CT = 580 ksi F'b = Fbs CL = 1455 psi F'v = Fv CDCM Ct Ci = 195.5 psi

Mb fb = = 1214 psi Sx fv =

1.5 Vb Ax

(

(

= 114.482 psi

4

 TL =

384 E'Ix

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

 TLmax =

E' = ECM Ct Ci = 1600 ksi

5 wTLLb

)

if fb  F'b  "OK"  "NG" = "OK"

Lb = 0.95 in 240

3

+

Ps Lb

48 E'Ix

= 0.382 in

(

)

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-18

4.6b wTL = 0plf

Ps = 4000lb

Lb = 24ft 2

wTLLb Ps Lb Mb = + = 32000 ft lb 8 3

Fv = 170psi

E = 1600ksi

 5 w L 4 P L 3  TL b s b  240  4 IxMin =  +  = 1777 in   28 E   Lb   384 E

d = 19.5in

use 6x20

Cfu = 1.0

Cr = 1.0

Ci = 1.0

2

 12in  = 0.947   d 

CFb = 

Ct = 1.0

CM = 1.0

Ax = b d = 107.25 in

Sx =

Emin = 580ksi

1 9

CT = 1.0

CD = 1.0

Fb = 1350psi

wTLLb Ps Vb = + = 4000 lbf 2 1

Mb 3 SxMin = = 284 in Fb

b = 5.5in

DFL #1

b d 6

2

3

3

Ix =

= 349 in

b d 4 = 3398 in 12

Lb Lu = = 8 ft 3

Le = 1.68 Lu = 161.28 in

Lu S1 = = 4.923 d RB =

Le d b

2

= 10.196

E'min = EminCM Ct Ci CT = 580 ksi

FBE =

1.2 E'min 2

= 6694.5 psi

RB

E'min = EminCM Ct Ci CT = 580 ksi F'b = Fbs CL = 1264 psi F'v = Fv CDCM Ct Ci = 170 psi

Fbs = Fb CDCM Ct CFb CfuCi Cr = 1279 psi

 FBE  1+   Fbs   CL = − 1.9

2

  FBE   FBE 1 +     Fbs   Fbs = 0.988   − 1.9 0.95  

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1279 psi

Chapter 4

Solution

S4-19

4.6b contd Mb fb = = 1101.7 psi Sx fv =

1.5 Vb Ax

(

(

= 55.944 psi

4

 TL =

384 E'Ix

 TLmax =

3

+

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1600 ksi

5 wTLLb

)

if fb  F'b  "OK"  "NG" = "OK"

Ps Lb

48 E'Ix

= 0.366 in

(

Lb = 1.2 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-20

4.6c Ps = 8000lb

Lb = 20ft

wTL = 0plf 2

wTLLb Ps Lb Mb = + = 40000 ft lb 8 4

 5 w L TL b

IxMin = 



+

 12in  = 0.937   d 

CFb = 

Ct = 1.0 Cfu = 1.0

Cr = 1.0

Ci = 1.0

2

Sx =

Emin = 580ksi

 240  = 1440 in4  48 E   Lb   

1 9

CM = 1.0

Ax = b d = 118.25 in

E = 1600ksi

Ps Lb 

use 6x22

CT = 1.0

CD = 1.0

384 E

Fv = 170psi

3

4

d = 21.5in

b = 5.5in

Fb = 1350psi

wTLLb Ps Vb = + = 4000 lbf 2 2

Mb 3 = 356 in Fb

SxMin =

DFL #1

b d 6

2

3

3

Ix =

= 424 in

b d 4 = 4555 in 12

Lb Lu = = 10 ft 2

Le = 1.11 Lu = 133.2 in

Lu S1 = = 5.581 d RB =

Le d b

2

= 9.73

E'min = EminCM Ct Ci CT = 580 ksi

FBE =

1.2 E'min 2

= 7351.8 psi

RB

E'min = EminCM Ct Ci CT = 580 ksi F'b = Fbs CL = 1252 psi F'v = Fv CDCM Ct Ci = 170 psi

Fbs = Fb CDCM Ct CFb CfuCi Cr = 1265 psi

 FBE  1+  Fbs   CL = − 1.9

2

  FBE   FBE 1 +     Fbs   Fbs = 0.99   − 1.9 0.95  

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1265 psi

Chapter 4

Solution

S4-21

4.6c cont’d Mb fb = = 1132.8 psi Sx fv =

1.5 Vb Ax

(

(

= 50.74 psi

4

 TL =

384 E'Ix

 TLmax =

3

+

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1600 ksi

5 wTLLb

)

if fb  F'b  "OK"  "NG" = "OK"

Ps Lb

48 E'Ix

= 0.316 in

(

Lb = 1 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-22

4.6d Ps = 5000lb

Lb = 18ft

wTL = 700plf 2

wTLLb Ps Lb Mb = + = 50850 ft lb 8 4

 5 w L TL b

IxMin = 



+

 12in  = 0.928   d 

CFb = 

Ct = 1.0 Cfu = 1.0

Cr = 1.0

Ci = 1.0

2

Sx =

Emin = 580ksi

 240  = 1877 in4  48 E   Lb   

1 9

CM = 1.0

Ax = b d = 129.25 in

E = 1600ksi

Ps Lb 

use 6x24

CT = 1.0

CD = 1.0

384 E

Fv = 170psi

3

4

d = 23.5in

b = 5.5in

Fb = 1350psi

wTLLb Ps Vb = + = 8800 lbf 2 2

Mb 3 = 452 in Fb

SxMin =

DFL #1

b d 6

2

3

3

Ix =

= 506 in

b d 4 = 5948 in 12

Lb Lu = = 9 ft 2

Le = 1.11 Lu = 119.88 in

Lu S1 = = 4.596 d RB =

Le d b

2

= 9.65

E'min = EminCM Ct Ci CT = 580 ksi

FBE =

1.2 E'min 2

= 7473.4 psi

RB

E'min = EminCM Ct Ci CT = 580 ksi F'b = Fbs CL = 1241 psi F'v = Fv CDCM Ct Ci = 170 psi

Fbs = Fb CDCM Ct CFb CfuCi Cr = 1253 psi

 FBE  1+  Fbs   CL = − 1.9

2

  FBE   FBE 1 +     Fbs   Fbs = 0.99   − 1.9 0.95  

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1253 psi

Chapter 4

Solution

S4-23

4.6d cont’d Mb fb = = 1205.4 psi Sx fv =

1.5 Vb Ax

(

(

= 102.128 psi

4

 TL =

384 E'Ix

 TLmax =

3

+

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1600 ksi

5 wTLLb

)

if fb  F'b  "OK"  "NG" = "OK"

Ps Lb

48 E'Ix

= 0.284 in

(

Lb = 0.9 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-24

4.7 DL = 16psf

TW = 12in

LL = 50psf

wLL = LLTW = 50 plf

TL = DL + LL = 66 psf

Lb = 17ft

wTL = TLTW = 66.0 plf

2

wTLLb Mb = = 2384 ft lb 8

wTLLb = 561 lbf 2

Vb =

Hem Fir #1 Fb = 975psi

Fcp = 405psi

CD = 1.0

Fv = 150psi

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.15

Ci = 1.0

E = 1500ksi Emin = 550ksi

b = 1.5in

Ct = 1.0

CFb = 1.1

CL = 1.0

CT = 1.0

d = 9.25in

use 2x10

2

Sx =

Ax = b d = 13.875 in

b d 6

2

3

3

5

b d 4 = 98.932 in 12

Ix =

= 21.391 in

E'min = EminCM Ct Ci CT = 5.5  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1233 psi

F'b = Fbs CL = 1233 psi

Mmax = F'b Sx = 2198.6 ft lb

F'v = Fv CDCM Ct Ci = 150 psi

Vmax =

Mb fb = = 1337.5 psi Sx fv =

1.5 Vb Ax

= 60.649 psi

(

F'v Ax 1.5

= 1387 lbf

)

if fb  F'b  "OK"  "NG" = "NG"

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1500 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.633 in

 LLmax =

Lb = 0.567 in 360

if  LL   LLmax "OK"  "NG" = "NG"

= 0.836 in

 TLmax =

Lb = 0.85 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

F'cp = Fcp CM Ct Ci Cb = 405 psi

(

)

(

)

Vb LbMin = = 0.923 in b F'cp

Chapter 4

Solution

S4-25

4.8a DL = 12psf

TW = 24in

LL = 50psf

wLL = LLTW = 100 plf

TL = DL + LL = 62 psf

Lb = 16ft

wTL = TLTW = 124.0 plf

2

wTLLb Mb = = 3968 ft lb 8

wTLLb = 992 lbf 2

Vb =

DFL, Select Fb = 1500psi

Fcp = 625psi

CD = 1.0

Fv = 180psi

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.15

Ci = 1.0

E = 1900ksi Emin = 690ksi

b = 1.5in

Ct = 1.0

CFb = 1.0

CL = 1.0

CT = 1.0

d = 11.25in

use 2x12

2

Sx =

Ax = b d = 16.875 in

b d 6

2

3

3

5

b d 4 = 177.979 in 12

Ix =

= 31.641 in

E'min = EminCM Ct Ci CT = 6.9  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1725 psi

F'b = Fbs CL = 1725 psi

Mmax = F'b Sx = 4548.3 ft lb

F'v = Fv CDCM Ct Ci = 180 psi

Vmax =

Mb fb = = 1504.9 psi Sx fv =

1.5 Vb Ax

= 88.178 psi

(

F'v Ax 1.5

= 2025 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1900 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.436 in

 LLmax =

Lb = 0.533 in 360

if  LL   LLmax "OK"  "NG" = "OK"

= 0.541 in

 TLmax =

Lb = 0.8 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

F'cp = Fcp CM Ct Ci Cb = 625 psi

(

)

(

)

Vb LbMin = = 1.058 in b F'cp

Chapter 4

Solution

S4-26

4.8b Lb = 12ft

DL = 12psf

TW = 16ft

LL = 50psf

wLL = LLTW = 800 plf

TL = DL + LL = 62 psf

PL = 0lb

wTL = TLTW = 992.0 plf

Ps = 0lb

DFL, SS Fb = 1500psi Fv = 180psi

E = 1900ksi Emin = 690ksi

2

wTLLb Ps Lb Mb = + = 17856 ft lb 8 4

Vb = 4

Mb 3 SxMin = = 142.848 in Fb

b = 5.5in

d = 13.5in

IxMin =

5 wTLLb 384 E

wTLLb Ps + = 5952 lbf 2 2

 240  = 406 in4   Lb 



use 6x14 1 9

 12in  = 0.987   d 

CFb = 

Ct = 1.0

CT = 1.0

CD = 1.0 CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 CL = 1.0

2

Ax = b d = 74.25 in

Sx =

b d 6

2

3

3

Ix =

= 167 in

b d 4 = 1128 in 12

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1480 psi

E'min = EminCM Ct Ci CT = 690 ksi F'b = Fbs CL = 1480 psi F'v = Fv CDCM Ct Ci = 180 psi

E' = ECM Ct Ci = 1900 ksi

Mb fb = = 1282.6 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

= 120.242 psi

(

(

 LL =

384 E'Ix

= 0.174 in

 LLmax =

Lb = 0.3 in 480

if  LL   LLmax "OK"  "NG" = "OK"

= 0.216 in

 TLmax =

Lb = 0.6 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

4

5 wLLLb

)

(

)

(

)

Chapter 4

Solution

4.8c Use Simpson LUS210

4.8d Column Load:

Pc = ( DL + LL) ( 12ft 16ft ) = 11.904 kips

Use Simpson CCQ66 column cap

S4-27

Chapter 4

Solution

S4-28

4.9 DL = 20psf

Lb = 17ft

TW = 19.2in wDL = DLTW = 32 plf

DFL, Select Fb = 1500psi

Fcp = 625psi

CD = 1.0

Fv = 180psi

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.15

Ci = 1.0

E = 1900ksi Emin = 690ksi

b = 1.5in

Ct = 1.0

CFb = 1.0

CL = 1.0

CT = 1.0

d = 11.25in

use 2x12

2

Sx =

Ax = b d = 16.875 in

b d 6

2

3

3

5

b d 4 = 177.979 in 12

Ix =

= 31.641 in

E'min = EminCM Ct Ci CT = 6.9  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1725 psi

F'b = Fbs CL = 1725 psi

Mmax = F'b Sx = 4548.3 ft lb

F'v = Fv CDCM Ct Ci = 180 psi

Vmax =

 w L 2  DL b MLL = Mmax −   = 3392 ft lb 8  

WLL =

8MLL

 L 2 TW  b 

F'v Ax 1.5

= 2025 lbf

= 58.691 psf

use office occupancy

wLL = WLLTW = 93.906 plf E' = ECM Ct Ci = 1900 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.522 in

 LLmax =

Lb = 0.567 in 360

(

)

if  LL   LLmax "OK"  "NG" = "OK"

Chapter 4

Solution

4.10 DFL, Select

Lb = 15ft

Fb = 1500psi Fv = 180psi

E = 1900ksi

Fcp = 625psi

CD = 1.0

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0

Emin = 690ksi

b = 1.5in

S4-29

Ct = 1.0

CFb = 1.0

CT = 1.0

d = 11.25in

use 2x12

2

Sx =

Ax = b d = 16.875 in

b d 6

2

3

3

Ix =

= 32 in

b d 4 = 178 in 12

Lb Lu = = 7.5 ft 2

Le = 1.11 Lu = 99.9 in

Lu S1 = =8 d

RB =

b

E'min = EminCM Ct Ci CT = 690 ksi E' = ECM Ct Ci = 1900 ksi

FBE =

1.2 E'min 2

Le d

= 1657.7 psi

RB

2

= 22.349

Fbs = Fb CDCM Ct CFb CfuCi Cr = 1500 psi

 FBE  1+   Fbs   CL = − 1.9

2

  FBE   FBE 1 +     Fbs   Fbs = 0.854   − 1.9 0.95  

F'b = Fbs CL = 1282 psi

Mmax = F'b Sx = 3379.2 ft lb

F'v = Fv CDCM Ct Ci = 180 psi

Vmax =

Pmax1 =

F'v Ax 1.5

= 2025 lbf

4Mmax = 901.131 lbf Lb

Pmax2 = 2 Vmax = 4050 lbf

Pmax3 =

48 E'Ix  Lb   = 2087 lbf  3  240  Lb

(

)

Pmax = min Pmax1 Pmax2 Pmax3 = 901.131 lbf

Chapter 4

Solution

S4-30

4.12 DL = 100psf

TW = 12in

LL = 150psf

wLL = LLTW = 150 plf

TL = DL + LL = 250 psf

wTL = TLTW = 250.0 plf

2

wTLLb Ps Lb Mb = + = 4500 ft lb 8 4

Vb =

Lb = 12ft

Nj = 1

PD = 0lb

PL = 0lb

Ps = PD + PL = 0

wTLLb Ps + = 1500 lbf 2 2

So. Pine #2 Fb = 925psi

CD = 1.15

Fv = 175psi

CM = 1.0

Cfu = 1.0

E = 1400ksi

Cr = 1.15

Ci = 1.0

Emin = 510ksi

b = 1.5in

CT = 1.0

d = 7.25in

Sx =

Ax = Nj b d = 10.875 in

Nj b d 6

2 3

= 13.141 in

4

384 E'Ix 3

Lmax =

384E'Ix

Ix =

Nj b d 12

 TLmax =

E' = ECM Ct Ci = 1400 ksi

5 wTLLb

CL = 1.0

2x8

2

 TL =

Ct = 1.0

CFb = 1.0

3 4

= 47.635 in Lb = 0.8 in 180

3

+

Ps Lb

48 E'Ix

(

= 1.749 in

 1  = 8.701 ft   5 wLL  360 

)

if  TL   TLmax "OK"  "NG" = "NG"

part C

Chapter 4

Solution

S4-31

4.13 15ft 2

DL = 10psf

TW =

LL = 40psf

wLL = LLTW = 300 plf

TL = DL + LL = 50 psf

wTL = TLTW = 375.0 plf

2

wTLLb Mb = = 3000 ft lb 8

Vb =

Lb = 8ft Nj = 3

wTLLb = 1500 lbf 2

DFL, Select Fb = 1500psi

Fcp = 625psi

CD = 1.0

CFb = 1.3

Ct = 1.0

Fv = 180psi

Cb = 1.0

CM = 1.0

CT = 1.0

Cfu = 1.0

E = 1900ksi

CL = 1.0

Cr = 1.0

Ci = 1.0

Emin = 690ksi

b = 1.5in

d = 5.5in

2x6

2

Ax = Nj b d = 24.75 in

Sx =

Njb d 6

2 3

Ix =

= 23 in

Nj b d

3

12

4

= 62 in

Fbs = Fb CDCM Ct CFb CfuCi Cr = 1950 psi

E'min = EminCM Ct Ci CT = 690 ksi E' = ECM Ct Ci = 1900 ksi F'b = Fbs CL = 1950 psi

Mmax = F'b Sx = 3686.7 ft lb

F'v = Fv CDCM Ct Ci = 180 psi

Vmax =

Mb fb = = 1586.8 psi Sx fv =

1.5 Vb Ax

= 90.909 psi

(

 LL =

384 E'Ix

 TL =

384 E'Ix

= 2970 lbf

)

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

= 0.233 in

 LLmax =

Lb = 0.267 in 360

if  LL   LLmax "OK"  "NG" = "OK"

= 0.292 in

 TLmax =

Lb = 0.4 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

5 wTLLb

1.5

if fb  F'b  "OK"  "NG" = "OK"

4

5 wLLLb

F'v Ax

F'cp = Fcp CM Ct Ci Cb = 625 psi

Vb LbMin = = 1.6 in b F'cp

(

)

(

)

(

)

if LbMin  1.5in  "OK"  "NG" = "NG"

Chapter 4

Solution

S4-32

4.14 a = 12ft

Lb = 18ft

Mb =

b = Lb − a = 6 ft

Ps = 7000lb

Hem Fir #2 Fb = 675psi

Ps a b

= 28000 ft lb

Lb

Fv = 140psi

E = 1100ksi d = 19.5in

b = 7.5in CD = 1.25

8x20

 12in  = 0.947   d 

CFb = 

Ct = 1.0

CT = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 2

Sx =

Ax = b d = 146.25 in

Emin = 400ksi

1 9

b d 6

2

Ix =

b d 4 = 4634 in 12

Le1 = 2.06 Lu = 296.64 in

Le2 = 1.63Lu + 3 d = 293.22 in

Lu S1 = = 7.385 d

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 293.22 in

RB =

Le d b

( (

= 10.082

2

5

FBE =

2

= 12 ft

Le3 = 1.84 Lu = 264.96 in

))

 FBE  1+  Fbs   CL = −

= 4722.1 psi

1.9

Mb fb = = 706.9 psi Sx

4

(

)

if fb  F'b  "OK"  "NG" = "OK"

 TLmax =

E' = ECM Ct Ci = 1100 ksi

Ps a b ( a + 2 b )  3 a ( a + 2 b )

2

  FBE   FBE 1 +     Fbs   Fbs = 0.99   − 1.9 0.95  

Mmax = F'b Sx = 3.1  10 ft lb

F'b = Fbs CL = 791 psi

27E'IxLb

3

Fbs = Fb CDCM Ct CFb Ci Cr = 799 psi

RB

 TL =

Lu =

RB must be less than 50

E'min = EminCM Ct Ci CT = 4  10 psi

1.2 E'min

2 Lb

3

3

= 475 in

= 0.011 in

(

Lb = 0.9 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

Solution

S4-33

4.15 DL = 10psf

TW = 24in

LL = 50psf

wLL = LLTW = 100 plf

TL = DL + LL = 60 psf

wTL = TLTW = 120.0 plf

2

wTLLb Ps Lb Mb = + = 3375 ft lb 8 4

Vb =

Lb = 15ft

Nj = 1

PD = 0lb

PL = 0lb

Ps = PD + PL = 0

wTLLb Ps + = 900 lbf 2 2

Hem Fir #1 Fb = 975psi

CD = 1.25

Fv = 150psi

CM = 1.0

Cfu = 1.0

E = 1500ksi

Cr = 1.15

Ci = 1.0

Emin = 550ksi

CT = 1.0

d = 7.25in

b = 1.5in

Ct = 1.0

CFb = 1.2

CL = 1.0

2x8

2

Sx =

Ax = Nj b d = 10.875 in

Nj b d 6

2 3

Ix =

= 13.141 in

5

Nj b d

3

12

4

= 47.635 in

E'min = EminCM Ct Ci CT = 5.5  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 1682 psi

F'b = Fbs CL = 1682 psi

Mmax = F'b Sx = 1841.7 ft lb

F'v = Fv CDCM Ct Ci = 187.5 psi

Vmax =

Mb fb = = 3082 psi Sx fv =

1.5 Vb Ax

= 124.138 psi

(

Lmax =

384E'Ix

 1  = 10.978 ft  5 wTL  240  

1.5

= 1359 lbf

)

if fb  F'b  "OK"  "NG" = "NG"

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1500 ksi

3

F'v Ax

part C

Chapter 4

Solution

S4-34

4.16 Ps = 5000lb

Lb = 24ft

wTL = 0plf 2

wTLLb Ps Lb Mb = + = 30000 ft lb 8 4

d = 15.5in

b = 5.5in

use 6x16

Cr = 1.0

Ci = 1.0

Sx =

E = 1300ksi Emin = 470ksi

 12in  = 0.972   d 

CFb = 

Ct = 1.0 Cfu = 1.0

2

Fv = 125psi

1 9

CM = 1.0

Ax = b d = 85.25 in

Fb = 1100psi

wTLLb Ps Vb = + = 2500 lbf 2 2

CT = 1.0

CD = 1.15

SPF Select

b d 6

2

3

3

Ix =

= 220 in

b d 4 = 1707 in 12

Lb Lu = = 12 ft 2

Le = 1.11 Lu = 159.84 in

S1 =

Lu = 9.29 d

RB =

Le d b

2

= 9.05

Fbs = Fb CDCM Ct CFb CfuCi Cr = 1230 psi

E'min = EminCM Ct Ci CT = 470 ksi

FBE =

1.2 E'min 2

 FBE  1+   Fbs   CL = −

= 6886.3 psi

1.9

RB

2

  FBE   FBE 1 +     Fbs   Fbs = 0.989   − 1.9 0.95  

F'b = Fbs CL = 1216 psi

Mb fb = = 1634.7 psi Sx

(

 TLmax =

E' = ECM Ct Ci = 1300 ksi 4

 TL =

5 wTLLb

384 E'Ix

)

if fb  F'b  "OK"  "NG" = "NG"

Lb = 1.2 in 240

3

+

Ps Lb

48 E'Ix

= 1.121 in

(

)

if  TL   TLmax "OK"  "NG" = "OK"

Chapter 4

4.17

Solution

S4-35

Chapter 4

Solution

S4-36

Joists: 2650 - 1.9E DL = 26.7psf

Lb = 16ft

LL = 125psf

TW = 16in

Fv = 285psi

TL = DL + LL = 151.7 psf

wLL = LLTW = 166.7 plf

E = 1900ksi

N = 1

Fb = 2650psi

Emin = 690ksi

wTL = TLTW = 202.3 plf

d = 11.875in

b = 1.75in

CD = 1.0 2

Le = 0.01in

Mb =

2

A = N b d = 20.781 in Sx = Ix =

N b d 6 N b d 12

2

wTLLb Vb = = 1618.133lbf 2

3

= 41.13 in

3

wTLLb = 6473 ft lb 8

4

Iy =

= 244.207 in

N d b 12

3

 12in  CFb =    d 

4

= 5.304 in

0.136

= 1.001

CM = 1.0 Cr = 1.0 CL = 1.0 CFt = 1.0 CFc = 1.0 CT = 1.0 Ct = 1.0 Cfu = 1.0

5

E'min = EminCM Ct Ci CT = 6.9  10 psi

Fbs = Fb CDCM Ct CFb Ci Cr = 2654 psi

F'b = Fbs CL = 2654 psi

Mmax = F'b Sx = 9096 ft lb

F'v = Fv CDCM Ct Ci = 285 psi

Vmax =

Mb fb = = 1888.4 psi Sx

fv =

1.5 Vb A

(

F'v A 1.5

= 3948 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

= 116.798 psi

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1900 ksi 4

4

 LL =

5 wLLLb

384 E'Ix

 LLmax =

(

 TL =

= 0.530 in

Lb = 0.533 in 360

5 wTLLb

384 E'Ix

 TLmax =

)

if  LL   LLmax "OK"  "NG" = "OK"

(

= 0.643 in

Lb = 0.8 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Ci = 1.0

Chapter 4

Solution

S4-37

Girder: 2650 - 1.9E DL = 29.362psf

Lb = 13ft

LL = 125psf

TW = 8ft

Fv = 285psi

TL = DL + LL = 154.362 psf

wLL = LLTW = 1000 plf

E = 1900ksi

N = 3

Fb = 2650psi

Emin = 690ksi

wTL = TLTW = 1234.9 plf

b = 1.75in

CD = 1.0 2

wTLLb Mb = = 26087 ft lb 8

d = 14in 2

A = N b d = 73.5 in Sx = Ix =

N b d 6 N b d 12

2

wTLLb Vb = = 8026.824lbf 2

3

= 171.5 in

3

3

4

Iy =

= 1.2  10 in

N d b 12

3

= 18.758 in

 12in  CFb =    d 

4

0.136

= 0.979

CM = 1.0 Cr = 1.0 CL = 1.0 CFt = 1.0 CFc = 1.0 CT = 1.0 Ct = 1.0 Cfu = 1.0

5

E'min = EminCM Ct Ci CT = 6.9  10 psi

Fbs = Fb CDCM Ct CFb Ci Cr = 2595 psi

F'b = Fbs CL = 2595 psi

Mmax = F'b Sx = 37087 ft lb

F'v = Fv CDCM Ct Ci = 285 psi

Vmax =

Mb fb = = 1825.3 psi Sx

fv =

1.5 Vb A

(

F'v A 1.5

= 13965 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

= 163.813 psi

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1900 ksi 4

4

 LL =

5 wLLLb

384 E'Ix

 LLmax =

(

 TL =

= 0.282 in

Lb = 0.433 in 360

5 wTLLb

384 E'Ix

 TLmax =

)

if  LL   LLmax "OK"  "NG" = "OK"

(

= 0.348 in

Lb = 0.65 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Ci = 1.0

Chapter 4

Solution

4.18 DL = 15psf

Lb = 15ft

LL = 80psf

TW = 12in

TL = DL + LL = 95 psf

wLL = LLTW = 80 plf

N = 1

wTL = TLTW = 95 plf 2

wTLLb Mb = = 2672 ft lb 8

Vb =

9.5", RED-I45

w = 80plf

L = 15 ft.

d = 9.5 in.

EI = 250x106 in2-lb 22.5wL4 2.67wL2 = + EI dx105 22.5(80)(15)4 2.67(80)(15)2 = + 250 x106 9.5x105 = 0.364"+0.05" = 0.414 in" L/480 = (15)(12)/480 = 0.375 in. < 0.414 in., NG Joist Hanger Use Simpson LBV

wTLLb = 712.5 lbf 2

S4-38

Chapter 4

Solution

S4-39

Girder: 2600 - 1.9E Lb = 10ft 15ft TW = 2

DL = 15psf LL = 80psf

Fb = 2600psi Fv = 285psi

E = 1900ksi

TL = DL + LL = 95 psf

wLL = LLTW = 600 plf

N = 4

wTL = TLTW = 712.5 plf

b = 1.75in

Emin = 690ksi CD = 1.0 2

d = 9.5in

Mb =

2

A = N b d = 66.5 in Sx = Ix =

N b d 6 N b d 12

2

wTLLb Vb = = 3562.5 lbf 2

3

= 105.292 in

3

wTLLb = 8906 ft lb 8

4

Iy =

= 500.135 in

N d b 12

3

= 16.971 in

 12in  CFb =    d 

4

0.136

= 1.032

CM = 1.0 Cr = 1.0 CL = 1.0 CFt = 1.0 CFc = 1.0 CT = 1.0 Ct = 1.0 Cfu = 1.0

5

E'min = EminCM Ct Ci CT = 6.9  10 psi

Fbs = Fb CDCM Ct CFb Ci Cr = 2684 psi

F'b = Fbs CL = 2684 psi

Mmax = F'b Sx = 23550 ft lb

F'v = Fv CDCM Ct Ci = 285 psi

Vmax =

Mb fb = = 1015 psi Sx

fv =

1.5 Vb A

(

F'v A 1.5

= 12635 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

= 80.357 psi

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1900 ksi 4

4

 LL =

5 wLLLb

384 E'Ix

 LLmax =

(

 TL =

= 0.142 in

Lb = 0.333 in 360

5 wTLLb

384 E'Ix

 TLmax =

)

if  LL   LLmax "OK"  "NG" = "OK"

(

= 0.169 in

Lb = 0.5 in 240

)

if  TL   TLmax "OK"  "NG" = "OK"

Ci = 1.0

Chapter 4

Solution

S4-40

4.20 A 4 x14 Hem Fir Select Structural lumber joist which is part of an office floor framing is notched 2.5 inches on the tension face at the end supports because of headroom limitations.

Determine the maximum allowable end support reaction.

Solution: CD (D + L) = 1.0 CM = Ct = Ci = 1.0 F’v = Fv CDCMCtCi = 150 (1)(1)(1)(1) = 150 psi d = 13.25” dn = 13.25 – 2.5 = 10.75” The NDS Code equation from Equation 4-20 for calculating the adjusted applied shear stress when the notch is on the tension face is given as,

fv

3 V  2 2  d   allowable shear stress, F’v = bd  d  n n

fv

3V 2  13.25"  2 =  150 psi   (3.5")(10.75")  10.75" 

Therefore, V = 2477 Ib Assuming a simply supported member, therefore, the maximum allowable end support reaction, V = 2477 Ib Check the maximum notch depth: dn > 0.75d (from Figure 4.21) dn = (0.75)(13.25) = 9.934”, OK

Chapter 4

Solution

S4-41

4.21 A simply supported 6 x 12 girder of Hem Fir No. 1 species supports a total uniformly distributed load (dead plus floor live load) of 600 Ib/ft over a span of 24 ft. Assuming normal moisture and temperature conditions and a fully braced beam, determine if the beam is structurally adequate for bending, shear, and deflection.

Solution:

( )

( 600 Ib/ft ) 24 w L Maximum Shear, Vmax = TL = = 7200 lb 2 2 Maximum Reaction, Rmax = 7200 lb 2 Maximum Moment, Mmax = w TLL 8 '

600 Ib/ft ) ( 24' ) ( = = 43,200 ft-lb = 518,400 in lb 2

8

The following loads will be used for calculating the girder deflections.

The uniform total load is, wTL= 600 lb/ft = 50 lb/in From NDS-S Table 1B, the properties of a 6 x 12 girder (Hem-fir No. 1 species) are: b = 5.5” and d = 11.5” Sxx provided = 121.2 in3 Area, A provided = 63.25 in2 Ixx provided = 697.1 in4 Since 6 x 12 Hem-fir No. 1 is a Beam & Stringer, the tabulated stresses from NDS-S Table 4D are: F'b = 1050 psi Fv = 140 psi E = 1.3 x 106 psi Check Bending Stress (Girders): Summary of load effects: Maximum Shear, Vmax = 7200 lb Maximum Reaction, Rmax = 7200 lb Maximum Moment, Mmax = 518,400 in lb Uniform TOTAL load is, wTL= 50 lb/in Adjustment factors for Girder * Adjustment Factor Adjustment

Value

Rational for the chosen value

Chapter 4

Solution

factor Symbol CL Cfu CM

1.0 1.0 1.0

CD

1.0

Temperature factor

Ct

1.0

Repetitive member factor Incision factor

Cr

-

Cr

1.0

Buckling stiffness factor Bearing stress factor

CT

1.0

Beam stability factor Flat use factor Moisture or wet service factor Load duration factor

Fully braced girder Girder is bending about its strong x-x axis Equilibrium moisture content (EMC) is < 19% The largest CD value in the load combination of dead load plus snow load (i.e. D + L) Insulated building, therefore normal temperature condition applies Does not apply to Timbers. Neglect or enter a value of 1.0 1.0 for wood that is not incised, even if pressure treated Does not apply to Timbers

Cb = lb + 0.375 for lb  6” lb = 1.0 for lb > 6” = 1.0 for bearings at the ends of a member *Other applicable C-factors not listed default to 1.0 or are neglected Cb

1. 0

Fb' = Fb CD CM Ct Cfu CL CF Cr = 1050 x 1.0 x 1.0 x 1.0 x 1.0 x 1.0x1.0x1.0 = 1050 psi

Using Equation 4-2, the actual applied bending stress is, fb = M max = 518,400 Ib3 - in = 4277 psi >>> Fb’ = 1050 psi NOT GOOD 121.2 in Sxx Therefore, the 6x12 Girder is not adequate in bending!

Check Shear Stress Vmax = 7200 lb The girder cross-sectional area, A = 63.25 in2 The applied shear stress in the wood beam at the centerline of the girder support is, fv = 1.5V = A

1.5( 7200 Ib ) 63.25 in 2

S4-42

= 171 psi

Using the Adjustment factor Applicability Table from Table 3-1, we obtain the

Chapter 4

Solution

S4-43

allowable shear stress as, F’v = Fv CD CM Ct = 140 x 1 x 1 x 1= 140 psi Thus, fv > F’v ,

 The girder is not adequate in shear

Check Deflection The allowable pure bending modulus of elasticity for strong (i.e. x-x) axis bending of the girder was previously calculated, The allowable pure bending modulus of elasticity, E’ is calculated as: E’ = E CMCt = 1.3 x 106 x 1.0 x 1.0 = 1.3 x 106 psi The moment of inertia about the strong axis Ixx = 697.1 in4 The uniform TOTAL LOAD is, wDL= 50 lb/in

Chapter 4

Solution

Deflection Limits (see Table 4-1) Deflection Incremental long term deflection due to dead load plus live load (including creep effects), TL = k (DL) + LL

S4-44

Deflection Limit

L = 24' x 12 = 1.0” 240 240

The TOTAL load deflection is, 4 4 DL = 5wL = 5 (50 Ib/in) (24' x 12) = 4.94” >> L/240 384EI 384(1.3x106 psi)(697.1in 4 )

6x12 Girder is not Adequate

NOT GOOD

Chapter 4

Solution

S4-45

4.22 A 5½” x 30” 24F-1.8E Glulam beam spans 32 ft and supports a concentrated moving load of 5000 Ib that can be located anywhere on the beam, in addition to its self weight. Normal temperature and dry service conditions apply, and the beam is laterally braced at support sand at midspan. Assuming a wood density of 36 lb/ft3, a load duration factor, CD of 1.0, and the available bearing length, lb of 4”, is the beam adequate for bending, shear, deflection, and bearing perpendicular to grain?

Solution: Concentrated moving live load, P = 5000 Ib Glulam Girder self-weight = [(5.5”)(30”)/144](36 Ib/ft3) = 41.3 Ib/ft Using the free body diagram of the girder, the maximum load effects are calculated as follows.

Note that the maximum moment occurs when the concentrated load is at the midspan whereas the maximum shear and reaction occurs when the concentrated load is near the supports.:

( )

( 41.3 Ib/ft ) 32 w L Maximum Shear, Vmax = TL + P = + 5000 Ib = 5661 lb 2 2 Maximum Reaction, Rmax = 5661 lb 2 Maximum Moment, Mmax = w TLL + PL 8 4 '

41.3 Ib/ft ) (32' ) (5000 Ib ) (32' ) ( + = = 45,286 ft-lb = 543,432 in lb 2

8

4

The following loads will be used for calculating the girder deflections.

The uniform dead load is, wDL= 41.33 lb/ft = 3.44 lb/in Concentrated Live Load, P = 5000 Ib

From NDS-S Table 1C (for Western species Glulam), the properties of the GIVEN size are: 5½” x 30” b = 5.5” and d = 30” Sxx provided = 825 in3 Area, A provided = 165 in2 Ixx provided = 12,380 in4 For Glulam used primarily in bending, use NDS-S Table 5A or Table 5A- Expanded. Since the glulam stress class is given, we will use NDS-S Table 5A. For 24F – 1.8E Glulam, the tabulated stresses from NDS-S Table 5A are:

Chapter 4

Solution

S4-46

+ = 2400 psi Fbx Fvxx = 265 psi Fc⊥xx, tension lam = 650 psi Ex = 1.8 x 106 psi Ey = 1.6 x 106 psi Eymin = 0.83 x 106 psi (Eymin , and not Exmin , is required for lateral buckling of the girder about the weak axis) Check Bending Stress (Girders): Summary of load effects: Maximum Shear, Vmax = 5661 lb Maximum Reaction, Rmax = 5661 lb Maximum Moment, Mmax = 543,432 in lb Uniform dead load is, wDL= 3.44 lb/in Concentrated Live Load, P = 5000 Ib

Chapter 4

Solution

S4-47

Adjustment factors for Glulam Girder Adjustment Factor

Adjustment factor Symbol CL CV Cc Cfu CM

Value

Rational for the chosen value

0.955 0.87 1.0 1.0 1.0

CD

1.0

Temperature factor

Ct

1.0

Repetitive member factor Incision factor

Cr

-

Cr

1.0

Buckling stiffness factor Bearing stress factor

CT

1.0

See calculation below See calculation below Glulam girder is straight Glulam is bending about its strong x-x axis Equilibrium moisture content (EMC) is < 16% The largest CD value in the load combination of dead load plus snow load (i.e. D + L) Insulated building, therefore normal temperature condition applies Does not apply to Glulams. Neglect or enter a value of 1.0 1.0 for wood that is not incised, even if pressure treated Does not apply to Glulams

Cb

1. 0

Beam stability factor Volume factor Curvature factor Flat use factor Moisture or wet service factor Load duration factor

Cb = lb + 0.375 for lb  6” lb = 1.0 for lb > 6” = 1.0 for bearings at the ends of a member

From the Adjustment factor Applicability Table for Glulam from Table 3-2, we obtain the allowable bending stress of the Glulam girder with CV and CL equal to 1.0 as,

*+ = F+ Fbx bx, NDS-S x CD CM Ct Cfu Cc = 2400 x 1.0 x 1.0 x 1.0 x 1.0 x 1.0 = 2400 psi The allowable pure bending modulus of elasticity, Ex’ and the bending stability modulus of elasticity Ey, min’ are calculated as: Ex’ = Ex CMCt = 1.8 x 106 x 1.0 x 1.0 = 1.8 x 106 psi Ey, min’ = Ey, min CM Ct = 0.83x 106 x 1.0 x 1.0 = 0.83x 106 psi

Calculate the beam stability factor, CL : From equation 3-2, the beam stability factor is calculated as follows:

Chapter 4

Solution

S4-48

The unsupported length of the compression edge of the beam or distance between points of lateral support preventing rotation and/or lateral displacement of the compression edge of the beam is, Lu = 32 ft/2 = 16 ft = 192” L u = 192" = 6.4 30" d

Since the predominant load is the concentrated moving load, the effective length of the girder is obtained from Table 3-9 (or NDS Code Table 3.3.3) for Lu/d < 7 as, Le = 2.06 Lu = 2.06 (192”) = 396” RB =

FbE =

Ld e b2

396"(30") = 19.8 < 50 ( 5.5")2

=

OK

1.20  0.83 x106    = 2541 psi min =

1.20 E' R2

(19.8)2

B

F F* = 2541psi = 1.06 bE b 2400 psi

From Equation 3-2, the beam stability factor is calculated as,

CL =

1+  FbE Fb*  

-

1.9

  *  1+  FbE Fb        1.9    

2

  FbE -

Fb* 



0.95

(1+1.06) -  (1+1.06)  - 1.06 = 0.84 = 2

1.9

 

1.9

 

0.95

Calculate the Volume Factor, CV: LO = length of beam in feet between points of zero Moment (Conservatively, assume LO = span of beam, L) d = depth of beam = 30” b = width of beam, inches ( 10.75”) = 5.5” x = 10 (for Western species Glulam) From Equation 3-3,

=

32 ft

Chapter 4

Solution

1

CV =

1

1

1

 1291.5  x  21  x  12  x  5.125  x        =  L d b     o   (b) (d) (Lo ) 

S4-49

1

 10 1291.5 =   (5.5") (30") (32 ft)  

= 0.87 < 1.0

The smaller of CV and CL will govern, and is used in the allowable bending stress calculation. Since CV = 0.87 > CL= 0.84,  Use CL = 0.84 Using Table 3-2 (i.e. Adjustment factor Applicability Table for Glulam), we obtain the allowable bending stress as, Fb’ = Fb NDSS CD CM Ct CF Cr (CL or CV ) = Fb* (CL or CV ) = 2400 (0.84) = 2016 psi

Using Equation 4-2, the actual applied bending stress is, - in fb = M max = 543,432 Ib = 659 psi 825 in 3 Sxx

<< the allowable bending stress, Fb’ = 2016 psi

OK

Therefore, the 5½” x 30” 24F – 1.8E Glulam Girder is adequate in bending.

Check Shear Stress Vmax = 5661 lb The girder cross-sectional area, A = 165 in2 The applied shear stress in the wood beam at the centerline of the girder support is, fv = 1.5V = A

1.5( 5661 Ib ) 165 in 2

= 51.5 psi

Using the Adjustment factor Applicability Table for Glulam from Table 3-2, we obtain the allowable shear stress as, F’v = Fv CD CM Ct = 265 x 1 x 1 x 1= 265 psi Thus, fv << F’v ,

Check Deflection

 The girder is adequate in shear

Chapter 4

Solution

S4-50

The allowable pure bending modulus of elasticity for strong (i.e. x-x) axis bending of the girder was previously calculated, Ex’ = Ex CMCt = 1.8 x 106 x 1.0 x 1.0 = 1.8 x 106 psi The moment of inertia about the strong axis Ixx = 12,380 in4 The uniform dead load is, wDL= 3.44 lb/in Concentrated Live Load, P = 5000 Ib Deflection Limits (see Table 4-1) Deflection Live load deflection, LL Incremental long term deflection due to dead load plus live load (including creep effects), TL = k (DL) + LL

Deflection Limit L = 32' x 12 = 360 360 1.07” L = 32' x 12 = 1.6” 240 240

The dead load deflection is, 4 4 DL = 5wL = 5 (3.44 Ib/in) (32' x 12) = 0.04” 384EI 384(1.8x106 psi)(12,380 in 4 )

The live load deflection is, 3 5000 Ib (32' x 12)3 LL = PL = = 0.26” << L = 1.07” 48EI 360 6 4 48(1.8x10 psi)(12,380 in )

OK

Since seasoned wood in a dry service condition is assumed to be used in this building, the creep factor, k = 0.5 The total incremental dead plus floor live load deflection is, TL = k (DL) + LL = 0.5 (0.04”) + 0.26” = 0.28” << L = 1.6” 240 Check Bearing Stress or Compression Stress Perpendicular (⊥) to Grain Maximum reaction at the support, R1 = 5661 Ib Width of 5½” x 30” 24F – 1.8E Glulam Girder, b = 5.5” The allowable bearing stress or compression stress parallel to grain is,

OK

Chapter 4

Solution

S4-51

F’c⊥ = Fc⊥ CM Ct Cb = 650 x 1 x 1 x 1 = 650 psi From equation 4-7, the minimum required bearing length, Lb is,

lb reqd 

R1 bF

c⊥

=

5661 Ib = 1.6” << 4” bearing provided (5.5")(650 psi)

5½” x 30” 24F – 1.8E Glulam Girder is Adequate

OK

Chapter 4

Solution

S4-52

4.23 Lb = 12ft

16F-E6 DF/DF Fb = 1600psi Fv = 265psi

E = 1600ksi

Fcp = 625psi

CD = 1.0

Ct = 1.0

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0

Emin = 790ksi

b = 3.5in

CT = 1.0

d = 12in 2

Sx =

Ax = b d = 42 in

b d 6

2

3

3

Ix =

= 84 in

b d 4 = 504 in 12

Lu = Lb = 12 ft

0.1   1291.5   = 1   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le = 1.44 Lu + ( 3 d ) = 243.36 in

Lu S1 = = 12 d

RB =

b

E'min = EminCM Ct Ci CT = 790 ksi E' = ECM Ct Ci = 1600 ksi

FBE =

1.2 E'min 2

Le d

= 3976.6 psi

RB

(

)

2

= 15.44

Fbs = Fb CDCM Ct CfuCi Cr = 1600 psi

 FBE  1+   Fbs   CL = − 1.9

2

  FBE   FBE 1 +     Fbs   Fbs = 0.969   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 1550 psi

Mmax = F'b Sx = 10853 ft lb

F'v = Fv CDCM Ct Ci = 265 psi

Vmax =

Pmax1 =

F'v Ax 1.5

= 7420 lbf

Mmax = 904.434 lbf Lb

Pmax2 = Vmax = 7420 lbf

Pmax3 =

3 E'Ix  Lb   = 292 lbf  3 400  Lb 

(

)

Pmax = min Pmax1 Pmax2 Pmax3 = 291.667 lbf

Chapter 4

Solution

S4-53

4.24 Lb = 36ft

wTL = 450plf

Ps = 0lb

2

wTLLb Mb = = 72900 ft lb 8

wTLLb = 8100 lbf 2

Vb =

20F-1.5E Fb = 2000psi Fv = 195psi

E = 1500ksi

CD = 1.15

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0

Emin = 630ksi

b = 5.5in

Ct = 1.0

Fcp = 425psi

CT = 1.0

d = 15in 2

Sx =

Ax = b d = 82.5 in

b d 6

2

3

3

Ix =

= 206 in

Lb Lu = = 9 ft 4

b d 4 = 1547 in 12

0.1   1291.5   = 0.92   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le2 = 1.63Lu + 3 d = 221.04 in

Le1 = 2.06 Lu = 222.48 in

S1 =

Lu = 7.2 d Le d

RB =

b

( (

RB must be less than 50

5

E'min = EminCM Ct Ci CT = 6.3  10 psi

FBE =

1.2 E'min 2

))

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 221.04 in

= 10.469

2

Le3 = 1.84 Lu = 198.72 in

= 6897.4 psi

RB

Fbs = Fb CDCM Ct Ci Cr = 2300 psi

 FBE  1+   Fbs   CL = − 1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.976   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 2116 psi

F'v = Fv CDCM Ct Ci = 224.25 psi

Mb fb = = 4241.5 psi Sx

if fb  F'b  "OK"  "NG" = "NG"

fv =

1.5 Vb Ax

= 147.273 psi

E' = ECM Ct Ci = 1500 ksi

(

)

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

 TLmax =

Lb = 1.8 in 240

4

 TL =

5 wTLLb

384 E'Ix

= 7.329 in

(

)

if  TL   TLmax "OK"  "NG" = "NG"

Chapter 4

Solution

S4-54

4.25 Lb = 18ft

b1 = Lb − a1 = 9 ft

a1 = 9ft

24F-E4 SP/SP Fb = 2400psi

d = 12.375in

b = 5.125in

Fv = 300psi CD = 1.0

CT = 1.0

Ct = 1.0

CM = 1.0

Cb = 1.0

Cfu = 1.0

E = 1900ksi Emin = 950ksi Fcp = 805psi

Ci = 1.0

Cr = 1.0 2

Sx =

Ax = b d = 63.422 in

b d 6

2

3

3

= 131 in

Lb Lu = = 9 ft 2

b d 4 = 809 in 12

Ix =

0.1   1291.5   = 1   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le1 = 2.06 Lu = 222.48 in

Le2 = 1.63Lu + 3 d = 213.165 in

Lu S1 = = 8.727 d

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 213.165 in

RB =

Le d b

( (

= 10.022

2

5

FBE =

2

4

= 1.1  10 psi

RB

(

))

RB must be less than 50

E'min = EminCM Ct Ci CT = 9.5  10 psi 1.2 E'min

Le3 = 1.84 Lu = 198.72 in

Fbs = Fb CDCM Ct Ci Cr = 2400 psi 2

 FBE  1+   Fbs   CL = −

  FBE   FBE 1 +     Fbs   Fbs = 0.987   − 1.9 0.95  

1.9

)

4

F'b = min Fbs CL  Fbs CV = 2369 psi

Mmax = F'b Sx = 2.6  10 ft lb

F'v = Fv CDCM Ct Ci = 300 psi

Vmax =

F'cp = Fcp CM Ct Ci Cb = 805 psi

Pmax1 =

4Mmax = 5738 lbf Lb

F'v Ax 1.5

= 12684 lbf

Lbr = 5in Pmax2 = 2Vmax = 25369 lbf

Pmax3 = 2Fcp b Lbr = 41256 lbf

(

)

Pmax = min Pmax1 Pmax2 Pmax3 = 5738 lbf b2 = Lb − a2 = 6 ft

a2 = 12ft

Mb =

Ps a2 b2 Lb

= 28000 ft lb

Ps = 7000lb

Mb fb = = 2568.7 psi Sx

(

)

if fb  F'b  "OK"  "NG" = "NG"

Chapter 4

Solution

S4-55

4.26 Lb = 26ft

wTL = 420plf

Ps = 0lb

2

wTLLb Mb = = 35490 ft lb 8

wTLLb = 5460 lbf 2

Vb =

16F-E3, DF/DF Fb = 1600psi Fv = 265psi

E = 1600ksi

Ct = 1.0

Fcp = 560psi

CD = 1.15

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0

Emin = 790ksi

CT = 1.0

d = 16.5in

b = 5.5in

2

Sx =

Ax = b d = 90.75 in

b d 6

2

3

3

Ix =

= 250 in

Lb Lu = = 13 ft 2

b d 4 = 2059 in 12

0.1   1291.5   = 0.942   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le1 = 2.06 Lu = 321.36 in

Le2 = 1.63Lu + 3 d = 303.78 in

Lu S1 = = 9.455 d

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 303.78 in

Le d

RB =

b

( (

= 12.872

2

5

FBE =

2

))

RB must be less than 50

E'min = EminCM Ct Ci CT = 7.9  10 psi 1.2 E'min

Le3 = 1.84 Lu = 287.04 in

= 5721.2 psi

RB

Fbs = Fb CDCM Ct Ci Cr = 1840 psi

 FBE  1+   Fbs   CL = − 1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.978   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 1732 psi

F'v = Fv CDCM Ct Ci = 304.75 psi

Mb fb = = 1706.5 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

= 90.248 psi

E' = ECM Ct Ci = 1600 ksi

(

)

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

 TLmax =

Lb = 1.3 in 240

4

 TL =

5 wTLLb

384 E'Ix

= 1.311 in

(

)

if  TL   TLmax "OK"  "NG" = "NG"

Chapter 4

Solution

S4-56

4.27 Lb = 38ft

wTL = 250plf

Ps = 0lb

2

wTLLb Mb = = 45125 ft lb 8

wTLLb = 4750 lbf 2

Vb =

16F-E6, DF/DF Fb = 1600psi Fv = 265psi

E = 1600ksi

CD = 1.25

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0

Emin = 790ksi

b = 6.75in

Ct = 1.0

Fcp = 560psi

CT = 1.0

d = 18in 2

Sx =

Ax = b d = 121.5 in

b d 6

2

3

3

Ix =

= 365 in

Lb Lu = = 19 ft 2

b d 4 = 3281 in 12

0.1   1291.5   = 0.88   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le1 = 2.06 Lu = 469.68 in

Le2 = 1.63Lu + 3 d = 425.64 in

Lu S1 = = 12.667 d

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 425.64 in

Le d

RB =

b

( (

= 12.967

2

5

FBE =

2

))

RB must be less than 50

E'min = EminCM Ct Ci CT = 7.9  10 psi 1.2 E'min

Le3 = 1.84 Lu = 419.52 in

= 5637.7 psi

RB

Fbs = Fb CDCM Ct Ci Cr = 2000 psi

 FBE  1+   Fbs   CL = − 1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.974   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 1761 psi

F'v = Fv CDCM Ct Ci = 331.25 psi

Mb fb = = 1485.6 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

= 58.642 psi

E' = ECM Ct Ci = 1600 ksi

(

)

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

 TLmax =

Lb = 1.9 in 240

4

 TL =

5 wTLLb

384 E'Ix

= 2.235 in

(

)

if  TL   TLmax "OK"  "NG" = "NG"

Chapter 4

Solution

S4-57

4.28 Ps = 4000lb

Lb = 40ft

wTL = 400plf 2

wTLLb Mb = = 80000 ft lb 8

Vb =

b = 6.75in

24F-E11, HF/HF Fb = 2400psi Fv = 215psi

E = 1800ksi

wTLLb Ps + = 10000 lbf 2 2

d = 24in Ct = 1.0

Fcp = 500psi

CD = 1.25

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0

Emin = 790ksi 2

Sx =

Ax = b d = 162 in

b d 6

2

CT = 1.0

3

3

Ix =

= 648 in

Lb Lu = = 20 ft 2

b d 4 = 7776 in 12

0.1   1291.5    = 0.851 CV = min 1     Lb   b   d                 1ft   1in   1in   

Le2 = 1.63Lu + 3 d = 463.2 in

Le1 = 2.06 Lu = 494.4 in

S1 =

Lu = 10 d Le d

RB =

b

( (

RB must be less than 50 5

E'min = EminCM Ct Ci CT = 7.9  10 psi

FBE =

1.2 E'min 2

))

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 463.2 in

= 15.62

2

Le3 = 1.84 Lu = 441.6 in

E' = ECM Ct Ci = 1800 ksi Fbs = Fb CDCM Ct Ci Cr = 3000 psi

 FBE  1+   Fbs   CL = −

= 3885.4 psi

RB

1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.898   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 2553 psi

F'v = Fv CDCM Ct Ci = 268.75 psi

Mb fb = = 1481.5 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

(

(

= 92.593 psi

4

 LL =

Vb LbMin = = 2.963 in b F'cp

384 E'Ix 4

 TL =

5 wTLLb

384 E'Ix

3

+

0.6Ps Lb

48 E'Ix

= 1.383 in

 LLm =

Lb = 1.333 in 360

if  LL   LLm "OK"  "NG" = "NG"

 TLm =

Lb = 2 in 240

if  TL   TLm "OK"  "NG" = "NG"

3

+

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

F'cp = Fcp CM Ct Ci Cb = 500 psi

0.6 5 wTLLb

)

Ps Lb

48 E'Ix

= 2.305 in

(

)

(

)

Chapter 4

Solution

S4-58

4.30a wTL = 1200plf

Lb = 36ft

Ps = 0lb

2

wTLLb Ps Lb 5 Mb = + = 2  10 ft lb 8 4

wTLLb Ps 4 Vb = + = 2.16  10 lbf 2 2 4

Mb 3 SxMin = = 1166 in Fb

b = 6.75in

IxMin =

d = 33in

CD = 1.15

2000F-1.5E

5 wTLLb 384 E

 240  = 16796 in4   Lb 

Fb = 2000psi Fv = 195psi

E = 1500ksi Emin = 630ksi



Fcp = 425psi

use 6.75x33 Ct = 1.0

CT = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 2

Sx =

Ax = b d = 222.75 in

b d 6

2

3

3

Ix =

= 1225 in

b d 4 = 20215 in 12

Lu = 0.01ft

0.1   1291.5   = 0.833   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le1 = 2.06 Lu = 0.247 in

Le2 = 1.63Lu + 3 d = 99.196 in

Lu −3 S1 = = 3.636  10 d

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 0.247 in

Le d

RB =

b

( (

= 0.423

2

RB must be less than 50 5

E'min = EminCM Ct Ci CT = 6.3  10 psi

FBE =

1.2 E'min 2

6

= 4.2  10 psi

RB

))

E' = ECM Ct Ci = 1500 ksi Fbs = Fb CDCM Ct Ci Cr = 2300 psi

 FBE  1+  Fbs   CL = − 1.9

(

Le3 = 1.84 Lu = 0.221 in

)

2

  FBE   FBE 1 +     Fbs   Fbs =1   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 1916 psi

F'v = Fv CDCM Ct Ci = 224.25 psi

Mb fb = = 1904.1 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

5 wTLLb

384 E'Ix

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

3

+

)

(

= 145.455 psi

4

 TL =

(

Ps Lb

48 E'Ix

= 1.496 in

 TLm =

Lb = 1.8 in 240

(

)

if  TL   TLm "OK"  "NG" = "OK"

Chapter 4

Solution

S4-59

4.30b Ps = 6000lb

Lb = 33ft

wTL = 0plf 2

wTLLb Ps Lb Mb = + = 66000 ft lb 8 3

Fv = 195psi

E = 1500ksi

3

IxMin =

d = 24in

Ps Lb

 240  = 5377 in4  28 E  Lb 

Emin = 630ksi



Fcp = 425psi

use 5.125x24 Ct = 1.0

CT = 1.0

CD = 1.0

Fb = 2000psi

wTLLb Ps Vb = + = 6000 lbf 2 1

Mb 3 SxMin = = 396 in Fb

b = 5.125in

2000F-1.5E

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 2

Sx =

Ax = b d = 123 in

b d 6

2

3

3

Ix =

= 492 in

Lb Lu = = 11 ft 3

b d 4 = 5904 in 12

0.1   1291.5   = 0.892   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le = 1.68 Lu = 221.76 in

Le d

RB =

b

= 14.235

2

RB must be less than 50

5

E'min = EminCM Ct Ci CT = 6.3  10 psi

FBE =

1.2 E'min 2

= 3730.9 psi

RB

E' = ECM Ct Ci = 1500 ksi

Fbs = Fb CDCM Ct Ci Cr = 2000 psi

 FBE  1+   Fbs   CL = − 1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.951   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 1784 psi

F'v = Fv CDCM Ct Ci = 195 psi

Mb fb = = 1609.8 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

= 73.171 psi

(

(

 TL =

28 E'Ix

= 1.503 in

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

3

Ps Lb

)

 TLm =

Lb = 1.65 in 240

(

)

if  TL   TLm "OK"  "NG" = "OK"

Chapter 4

Solution

S4-60

4.30c Ps = 8000lb

Lb = 28ft

wTL = 0plf 2

wTLLb Ps Lb Mb = + = 56000 ft lb 8 4

Fv = 195psi

E = 1500ksi

3

IxMin =

d = 21in

Ps Lb

 240  = 3011 in4  48 E  Lb 

Emin = 630ksi



Fcp = 425psi

use 5.125x21 Ct = 1.0

CT = 1.0

CD = 1.0

Fb = 2000psi

wTLLb Ps Vb = + = 4000 lbf 2 2

Mb 3 SxMin = = 336 in Fb

b = 5.125in

2000F-1.5E

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 2

Sx =

Ax = b d = 107.625 in

b d 6

2

3

3

Ix =

= 377 in

Lb Lu = = 14 ft 2

b d 4 = 3955 in 12

0.1   1291.5   = 0.919   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le = 1.11 Lu = 186.48 in

Le d

RB =

b

= 12.21

2

RB must be less than 50

5

E'min = EminCM Ct Ci CT = 6.3  10 psi

FBE =

1.2 E'min 2

= 5070.6 psi

RB

E' = ECM Ct Ci = 1500 ksi

Fbs = Fb CDCM Ct Ci Cr = 2000 psi

 FBE  1+  Fbs   CL = − 1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.97   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 1838 psi

F'v = Fv CDCM Ct Ci = 195 psi

Mb fb = = 1784 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

= 55.749 psi

(

(

 TL =

48 E'Ix

= 1.066 in

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

3

Ps Lb

)

 TLm =

Lb = 1.4 in 240

(

)

if  TL   TLm "OK"  "NG" = "OK"

Chapter 4

Solution

S4-61

4.30d Ps = 6000lb

Lb = 24ft

wTL = 900plf 2

wTLLb Ps Lb Mb = + = 100800 ft lb 8 4

wTLLb Ps Vb = + = 13800 lbf 2 2

 P L 3 5 w L 4  s b TL b  240  4 IxMin =  +  = 5391 in   384 E   Lb   48 E

Mb 3 SxMin = = 605 in Fb

b = 6.75in

d = 24in

CD = 1.15

2000F-1.5E Fb = 2000psi Fv = 195psi

E = 1500ksi Emin = 630ksi Fcp = 425psi

use 6.75x24 Ct = 1.0

CT = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.0

Ci = 1.0 2

Sx =

Ax = b d = 162 in

b d 6

2

3

3

Ix =

= 648 in

Lb Lu = = 12 ft 2

b d 4 = 7776 in 12

0.1   1291.5   = 0.896   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le = 1.11 Lu = 159.84 in

Le d

RB =

b

= 9.176

2

RB must be less than 50

5

E'min = EminCM Ct Ci CT = 6.3  10 psi

FBE =

1.2 E'min 2

E' = ECM Ct Ci = 1500 ksi

Fbs = Fb CDCM Ct Ci Cr = 2300 psi

 FBE  1+   Fbs   CL = −

= 8979.1 psi

RB

1.9

(

)

2

  FBE   FBE 1 +     Fbs   Fbs = 0.983   − 1.9 0.95  

F'b = min Fbs CL  Fbs CV = 2060 psi

F'v = Fv CDCM Ct Ci = 224.25 psi

Mb fb = = 1866.7 psi Sx

if fb  F'b  "OK"  "NG" = "OK"

fv =

1.5 Vb Ax

Ps Lb

48 E'Ix

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

4

+

)

(

= 127.778 psi

3

 TL =

(

5 wTLLb

384 E'Ix

= 0.832 in

 TLm =

Lb = 1.2 in 240

(

)

if  TL   TLm "OK"  "NG" = "OK"

Chapter 4

Solution

S4-62

4.31a DL = 20psf

TW = 19.2in

LL = 40psf

wLL = LLTW = 64 plf

TL = DL + LL = 60 psf

wTL = TLTW = 96.0 plf

2

Mb =

Lb = 14.5ft

wTLLb = 2523 ft lb 8

wTLLb = 696 lbf 2

Vb =

Hem Fir, #2 CFb = 1.0

Fb = 850psi

Fcp = 405psi

CD = 1.0

Fv = 150psi

Cb = 1.0

CM = 1.0

Cfu = 1.0

Cr = 1.15

Ci = 1.0

E = 1300ksi Emin = 470ksi

b = 1.5in

Ct = 1.0

CL = 1.0

CT = 1.0

d = 11.25in

use 2x12

2

Sx =

Ax = b d = 16.875 in

b d 6

5

2

3

3

b d 4 = 177.979 in 12

Ix =

= 31.641 in

E'min = EminCM Ct Ci CT = 4.7  10 psi

Fbs = Fb CDCM Ct CFb Ci CfuCr = 978 psi

F'b = Fbs CL = 978 psi

Mmax = F'b Sx = 2577.4 ft lb

F'v = Fv CDCM Ct Ci = 150 psi

Vmax =

Mb fb = = 956.9 psi Sx fv =

1.5 Vb Ax

= 61.867 psi

(

F'v Ax 1.5

= 1687 lbf

)

if fb  F'b  "OK"  "NG" = "OK"

(

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1300 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.275 in

 LLmax =

Lb = 0.483 in 360

if  LL   LLmax "OK"  "NG" = "OK"

= 0.413 in

 TLmax =

Lb = 0.725 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

F'cp = Fcp CM Ct Ci Cb = 405 psi

(

)

(

)

Vb LbMin = = 1.146 in b F'cp

Chapter 4

Solution

S4-63

4.31b 14.5ft + 13ft = 13.75 ft 2

DL = 20psf

TW =

LL = 40psf

wLL = LLTW = 550 plf

TL = DL + LL = 60 psf

wTL = TLTW = 825.0 plf

2

wTLLb Mb = = 59400 ft lb 8

4

IxMin =

d = 20in

5 wTLLb 384 E

24F-1.5E Fb = 2400psi Fv = 195psi

E = 1500ksi Emin = 630ksi

wTLLb Vb = = 9900 lbf 2

Mb 3 SxMin = = 297 in Fb

b = 5.5in

Lb = 24ft

Fcp = 425psi

 240  = 3421 in4   Lb 



use 5.5x20

CD = 1.0

Cb = 1.0

Ct = 1.0

CM = 1.0

CT = 1.0

Cfu = 1.0 Ci = 1.0

Cr = 1.0

b d Sx = 6

2

Ax = b d = 110 in

2

3

b d 4 Ix = = 3667 in 12

3

= 367 in

Lu = 0.01ft

0.1   1291.5   = 0.931   Lb   b   d                 1ft   1in   1in   

CV = min1  

Le2 = 1.63Lu + 3 d = 60.196 in

Le1 = 2.06 Lu = 0.247 in

S1 =

Lu −3 = 6  10 d

RB =

Le d b

( (

RB must be less than 50

5

E'min = EminCM Ct Ci CT = 6.3  10 psi

FBE =

1.2 E'min 2

6

= 4.6  10 psi

RB

(

))

Le = if S1  7  Le1  if S1  14.3  Le3  Le2  = 0.247 in

= 0.404

2

Le3 = 1.84 Lu = 0.221 in

Fbs = Fb CDCM Ct Ci Cr = 2400 psi

 FBE  1+  Fbs   CL = − 1.9

)

F'b = min Fbs CL  Fbs CV = 2234 psi

2

  FBE   FBE 1 +     Fbs   Fbs =1   − 1.9 0.95  

F'v = Fv CDCM Ct Ci = 195 psi

Chapter 4

Solution

S4-64

4.31b cont’d Mb fb = = 1944 psi Sx fv =

1.5 Vb Ax

(

)

if fb  F'b  "OK"  "NG" = "OK"

(

= 135 psi

)

if fv  F'v  "OK"  "NO GOOD" = "OK"

E' = ECM Ct Ci = 1500 ksi 4

 LL =

5 wLLLb

384 E'Ix

= 0.746 in

 LLmax =

Lb = 1.2 in 240

if  LL   LLmax "OK"  "NG" = "OK"

= 1.120 in

 TLmax =

Lb = 1.2 in 240

if  TL   TLmax "OK"  "NG" = "OK"

4

 TL =

5 wTLLb

384 E'Ix

F'cp = Fcp CM Ct Ci Cb = 425 psi

Need (4) jack studs, (4)(1.5) =6in. > 4.2in

Vb LbMin = = 4.235 in b F'cp

(

)

(

)

Chapter 4

Solution

S4-65

4.32 d = 5.5in

b = 1.5in

run = 12

rise = 3

 run  = 75.964 deg   rise 

 v = atan

P = 0lb

lb = 3.5in

2

Ab = b lb = 5.25 in

P f = =0 Ab

P fcp = =0 Ab

Doug Fir SS Fc = 1700psi

CD = 1.15

CFb = 1.3

CT = 1.0

Ci = 1.0

Fcp = 625psi

CM = 1.0

CFt = 1.3

Ct = 1.0

Cb = 1.0

Cr = 1.15

CFc = 1.1

Cfu = 1.0

F'cp = Fcp CM Ct Ci Cb = 625 psi Fcs = Fc CDCM Ct CFc Ci = 2150.5 psi

F' =

Fcs F'cp

F  sin  2 + F'  cos  2  cs ( ( v) )   cp ( ( v) ) 

( (

Pmax = Ab  min F'cp  F'

) ) = 3281 lbf

= 652.2 psi

part a

part b

Chapter 4

Solution

S4-66

4.33 d = 9.5in

b = 5.5in

rise = 4.5

 run  = 69.444 deg   rise 

 v = atan

run = 12

P = 0lb

lb = 5.5in

2

Ab = b lb = 30.25 in

P f = =0 Ab

P fcp = =0 Ab

SPF #1 Fc = 625psi

CD = 1.15

CFb = 1.0

CT = 1.0

Ci = 1.0

Fcp = 425psi

CM = 1.0

CFt = 1.0

Ct = 1.0

Cb = 1.0

Cr = 1.0

CFc = 1.0

Cfu = 1.0

= 447.6 psi

part a

F'cp = Fcp CM Ct Ci Cb = 425 psi Fcs = Fc CDCM Ct CFc Ci = 718.8 psi

F' =

Fcs F'cp

F  sin  2 + F'  cos  2  cs ( ( v) )   cp ( ( v) ) 

( (

Pmax = Ab  min F'cp  F'

) ) = 12856 lbf

part b

Chapter 4

Solution

S4-67

4.34 d = 9.25in

b = 1.5in

P = 800lb

run = 12

rise = 10

 run  = 50.194 deg   rise 

 v = atan

lb = 2.5in

2

Ab = b lb = 3.75 in

Hem Fir #2 Fc = 1300psi

CD = 1.15

CFb = 1.0

CT = 1.0

Ci = 1.0

Fcp = 405psi

CM = 1.0

CFt = 1.0

Ct = 1.0

Cb = 1.0

Cr = 1.0

CFc = 1.0

Cfu = 1.0

F'cp = Fcp CM Ct Ci Cb = 405 psi

P f = = 213.3 psi Ab

P fcp = = 213.3 psi Ab

(

Fcs = Fc CDCM Ct CFc Ci = 1495 psi

F' =

Fcs F'cp

F  sin  2 + F'  cos  2  cs ( ( v) )   cp ( ( v) ) 

( (

Pmax = Ab  min F'cp  F'

) ) = 1519 lbf

= 577.6 psi

)

if fcp  F'cp  "OK"  "NG" = "OK"

part a

(

part b

)

if f  F'  "OK"  "NG" = "OK"

part c

Chapter 4

Solution

S4-68

4.36 d = 9.25in

b = 1.5in

rise = 7.5

 run  = 57.995 deg   rise 

 v = atan

P = 1600lb

run = 12 lb = 5.5in

2

Ab = b lb = 8.25 in

Hem Fir, select Fc = 1500psi

CD = 1.15

CFb = 1.0

CT = 1.0

Ci = 1.0

Fcp = 405psi

CM = 1.0

CFt = 1.0

Ct = 1.0

Cb = 1.0

Cr = 1.0

CFc = 1.0

Cfu = 1.0

F'cp = Fcp CM Ct Ci Cb = 405 psi

P f = = 193.9 psi Ab

P fcp = = 193.9 psi Ab

(

Fcs = Fc CDCM Ct CFc Ci = 1725 psi

F' =

Fcs F'cp

F  sin  2 + F'  cos  2  cs ( ( v) )   cp ( ( v) ) 

( (

Pmax = Ab  min F'cp  F'

) ) = 3341 lbf

= 515.9 psi

)

if fcp  F'cp  "OK"  "NG" = "OK"

part a

(

part b

)

if f  F'  "OK"  "NG" = "OK"

part d

Chapter 4

Solution

S4-69

4.37 The sloped 2x10 rafter shown in Figure 4-59 is supported on a ridge beam at the ridge line and on the exterior stud wall. The reaction from dead load plus snow load on the rafter at the exterior stud wall is 2400 Ib. Assuming No. 1 Hem-Fir, normal temperature and dry service conditions, determine the following: a. The allowable bearing stresses at an angle to grain, F' and the allowable bearing θ

stress or allowable stress perpendicular to grain, F'

c⊥

b. The bearing stress at an angle to grain, f, in the rafter c. The bearing stress perpendicular to grain, fc⊥, in the top plate

Solution: 2 x 10 sloped rafter: b = 1.5”; d = 9.25” 2 x 4 top plates on top of stud wall: d = 3.5” Since 2 x 8 rafter and 2 x 4 top plates are Dimension lumber,

Use NDS-S Table 4A

Using NDS-S Table 4A, we obtain the tabulated stresses for No. 1 Hem-Fir as, Bearing stress perpendicular to grain, Fc⊥ = 405 psi Compression stress parallel to grain, Fc = 1350 psi The stress adjustment or “C” factors are: CM = 1.0 (dry service conditions) Ct = 1.0 (normal temperature conditions) Ci = 1.0 (wood is not incised) CF (Fc) = 1.0 CD = 1.15 (dead load plus snow load) The bearing length in the 2 x 4 top plate measured parallel to grain is the same as the thickness of the sloped rafter. l = brafter = 1.5” b

Since l = 1.5” < 6”, and bearing is not nearer than 3” from the end of the top plate and end of b

rafter (see chapter 3) Cb =

l + 0.375" 1.5"+ 0.375" b = = 1.25 1.5" l b

The bearing area at the rafter-stud wall connection is,

Chapter 4

Solution

S4-70

Abearing = (thickness, b of rafter) x (width, d of the 2 x 4 top plates) = 1.5” x 3.5” = 5.25 in2

(a) Calculate the allowable bearing stresses:

F'

c⊥

= Fc⊥ CM CtCiCb = 405 x 1 x 1 x 1 x 1.25 = 506.3 psi

F* = FcCDCMCtCFCi = 1350 x 1.15 x 1 x 1 x 1 x 1 = 1552.5 psi c

 = angle between the direction of the reaction or bearing stress in the sloped member and the direction of grain in the other wood member  tan  = 12/4 and  = 71.6 Using Equation 4-24, the allowable stress at an angle to grain is given as,

F' = θ

F* F' (1552.5)(506.3) c c⊥ = = 543 psi F*sin2θ + F' cos2θ 1552.5sin 2 71.6 + 506.3cos2 71.6 c⊥

c

(b)

Applied bearing stress at angle  to grain of Rafter: = 2400 Ib 5.25 in 2 bearing

P

f =

A

= 457 psi < F' = 543 psi θ

(c)

OK

Applied bearing stress perpendicular to the grain of the top plate: fc⊥ =

= 2400 Ib 5.25 in 2 bearing

P

A

= 457 psi < F'

c⊥

= 506.3 psi

OK

Chapter 4

4.38

Solution

S4-71

Chapter 4

4.39

Solution

S4-72

Chapter 4

Solution

S4-73

4.41 Design a DF-L tongue and grooved roof decking to span 14 ft between roof trusses. The decking is laid out in a pattern such each deck sits on two supports. Assume a dead load of 15 psf including the self-weight of the deck and snow load of 40 psf on a horizontal projected are.

Assume dry service and normal temperature conditions apply. .

Solution: Assume 4” x 6” decking (actual size = 3.5” x 5.5”) Douglas-fir-larch Commercial Dex decking with Simple-span Decking layout Type 1 in Table 4-11.

Section Properties for 4” x 6” decking: Axx = 19.25 in2 (12”/5.5”) = 42 in2 per ft width of deck Syy = 11.23 in3 (12”/5.5”) = 24.5 in3 per ft width of deck Iyy = 19.65 in4 (12”/5.5”) = 42.87 in4 per ft width of deck Loads: Total dead load, D Snow load, S

= 15 psf = 40 psf

The governing load combination for this roof deck will be dead load plus snow load (D + S).  Total load = D + S = 15 + 40 = 55 psf Bending stress:

w L2 Maximum moment, Mmax = = 8

( 55 psf )(14 ft ) 8

2

= 1348 ft-Ib per ft width of deck = 16,170 in-Ib per ft width of deck

The tabulated stress values and appropriate adjustment factors from NDS-S Table 4E for 4” x 6” DF-L Commercial Dex decking are: (Fb)(Cr) = 1650 psi (repetitive member) Fc⊥= 625 psi E = 1.7 x 106 CF = 1.0 (for 3” decking) Ci = 1.0 (lumber is not incised) The allowable stresses are calculated using the sawn lumber adjustment factors applicability table (Table 3-1) as follows, Fb’ = (FbCr) CD CM Ct CL CF Ci = 1650 x 1 x 1 x 1 x 1x 1 x 1 = 1650 psi

Chapter 4

Solution

S4-74

E’ = E CM Ct Ci = 1.7 x 106 x 1 x 1 x 1 = 1.7 x 106 psi The applied bending stress for bending about the weak (y-y) axis is, fby = M max = 16,170 Ib3- in = 660 psi << the allowable stress, Fb’ = 1650 psi 24.5 in Syy

OK

Deflection: Dead load, D = 15 psf = 15 Ib/ft per ft width of deck = 1.25 Ib/in per ft width of deck Live load (snow), S = 40 psf = 40 Ib/ft per ft width of deck = 3.33 Ib/in per ft width of deck The dead load deflection is, DL =

5wL4 384EI

=

yy

5 (1.25 Ib/in) (14' x 12)4 384(1.7x106 psi)(42.87 in 4 )

= 0.18”

The live load deflection is, LL =

5wL4 384EI

=

yy

5 (3.33 Ib/in) (14' x 12)4 384(1.7x106 psi)(42.87 in 4 )

= 0.47”

(14' x 12in/ft)  L = = 0.47” 360 360

OK

Since seasoned wood in a dry service condition is assumed to be used in this building, the creep factor, k = 0.5 The total incremental dead plus floor live load deflection is, TL = k (DL) + LL (14' x 12in/ft) = 0.5 (0.18”) + 0.47” = 0.56” < L = = 0.7” 240 240

Use 4” x 6” DF-L Commercial Dex with Simple Span Deck Layout

OK

Chapter 4

Solution

S4-75

4.42 Determine if a floor framed with 2x12 sawn lumber (Hem Fir, No. 1.) spaced at 24” O.C. and a 15’-0” simple span with ¾” plywood glued and nailed to the framing is adequate for walking vibration. Assume a residential occupancy. Assume 2x solid blocking at 1/3 points.

Solution: Floor Stiffness: E = 1.5x106 psi Am = 16.88 in.2 Im = 178 in.4 EItop = EIwpar = 0.016x106 psi/in. (Table 4-14) (EIwpar) (S) = (0.016x106)(24) = 0.384 x106 psi EAflr = EAwpar = 0.46x106 lb./in. (Table 4-14) (EAwpar) (S) = (0.46x106)(24) = 11.04 x106 lb Sflr = 50,000 lb./in. /in. (Table 4-15) Cfn = 1.0 (Simple span) Lflr = 48”  11.25"   3 / 4"  h top =   +  = 6"  2   2 

EA top =

y=

1+

EA flr = 10EA flr S flr L flr

EA top h top EA m + EA top

2

=

(11.04x10 6 ) = 5.63x10 6 lb.

6 (10) (11.04x10 ) 1+ (50,000) (48) 2

(5.63x10 6 ) (6" ) = 1.09" (1.5x10 6 ) (16.88in.2 ) + (5.63x10 6 )

EI = EIm + EItop + EAmy2 + EAtop(htop – y)2 EI = (1.5x106)(178) + (0.384x106) + (1.5x106)(16.88)(1.09)2 + (0.384x106 )(6 – 1.09)2 EI = 306x106 lb.-in2

Chapter 4

Solution

=

14.4 E (14.4) (1.5x10 6 ) = = 0.070 2 6 2  16.88  (0.1x10 ) L G (15x12)      178  r

r=

Im 1 A 16.88 , 2= m= Am r Im 178

EI (306x10 6 ) EI eff = = = 283x10 6 lb. − in.2 6 EI (0.070) (306x10 ) 1+ 1+ C fn EI m (1.0)(1.5x10 6 )(178) Kj =

EI eff 283x10 6 = 48.5 = L3 (15x12) 3

K bi =

K1 =

0.585EI bi L (0.585)(0.040x10 6 )(15x12) = 304.7 = s3 243

Kj K j + K bi

=

(37.46) = 0.0376 (37.46) + (960)

(2a / L)1.71 E v A K vi = s a=

(15x12) = 60” (blocking at 1/3 points) 3

[(2)(60) /(15x12)]1.71 (2000)(11.25)(1.5) = 703 lb./in K vi = 24 K2 =

K vi 703 = = 2.31 K bi 304.7

DFb = 0.0294 + 0.536K11/4 + 0.516K11/2 – 0.31K13/4 DFb = 0.0294 + (0.536) (0.0376)1/4 + (0.516)( 0.0376)1/2 – (0.31)( 0.0376)3/4 DFb = 0.339 DFv = -0.00253 - 0.0854K11/4 + 0.0797K21/2 – 0.00327K2

S4-76

Chapter 4

Solution

DFv = -0.00253 – (0.0854)( 0.0376)1/4 + 0.0797(2.31)1/2 – 0.00327(2.31) DFv = - 0.0724 N eff =

p =

1 1 = = 3.75 DFb − DFv (0.339) − (−0.0724)

C pd N eff

PL3 (1.0) (225)(15x12) 3 = = 0.026" 48EI eff (3.75) (48)(283x10 6 )

p  0.024 + 0.1e-0.18(L–6.4)  0.08 in. p  (0.024) + (0.1)e-0.18(15–6.4) = 0.045” > 0.026”, OK

S4-77

Chapter 5

5-1

Solutions

S5 - 1

A 4 x 8 wood member is subjected to an axial tension load of 8000 lb. caused by dead plus snow plus wind load. Determine the applied tension stress, ft, and the allowable tension stress, F’t and check the adequacy of this member. The lumber is Douglas Fir Larch No. 1, normal temperature conditions apply and the moisture content is greater than 19%. Assume the end connections are made with a single row of ¾” diameter bolts.

Solution: The gross area of 4x8, Ag The area of bolt holes,

= 25.38 in2 Aholes = 1 hole x (3/4” + 1/8”) x (3.5”) = 3.06 in.2

The net area of the member at the critical section is, An = Ag - Aholes = 25.38 – 3.06 = 22.32 in.2 The applied axial tension force, T = 8000 Ib. Therefore, the applied tension stress is, ft = T = 8000 Ib = 358.4 psi An 22.32 in 2

Since 4x8 is Dimension Lumber, therefore the applicable table is NDS-S Table 4A The tabulated design tension stress and the adjustment factors from NDS-S Table 4A for Douglas-fir-larch No. 1 are: Ft CD

= =

CM(Ft) = Ct = CF = Ci =

675 psi 1.6 (the CD value for the shorter duration load in the load combination is used, i.e. wind load.)

1.0 (wet service condition since MC > 19%) 1.0 (normal temperature) 1.2 1.0 (assumed since no incision or preservative treatment is prescribed)

Using the NDS Applicability table (Table 3.1), the allowable tension stress is as, F’t

= =

Ft CDCMCtCFCi 675 x 1.6x1x1x1.2x1 = 1296 psi

xft = 358.4 psi < F’t = 1296 psi Therefore, the 4x8 Doulas-fir-larch No. 1 is adequate.x

OK

given

Chapter 5

5-2

Solutions

S5 - 2

A 2 x 10 Hem-fir No. 3 wood member is subjected to an axial tension load of 6000 lb. caused by dead plus snow load. Determine the applied tension stress, ft, and the allowable tension stress, F’t and check the adequacy of this member. Normal temperature and dry service conditions apply, and the end connections are made with a single row of ¾” diameter bolts

Solution: The gross area for a 2x10, Ag = 13.88 in2 The area of bolt holes, Aholes = 1 hole x (3/4” + 1/8”) x (1.5”) = 1.31 in.2 The net area of the member at the critical section is, An = Ag - Aholes = 13.88 – 1.31 = 12.57 in.2 The applied axial tension force, T = 6000 Ib. Therefore, the applied tension stress is, ft = T = 6000 Ib = 477.3 psi An 12.57 in 2

Since 2x10 is Dimension Lumber, therefore the applicable table is NDS-S Table 4A The tabulated design tension stress and the adjustment factors from NDS-S Table 4A for Hem-fir No. 3 are: Ft CD CM Ct CF Ci

= 300 psi = 1.15 (D+S) = 1.0 (dry service condition) = 1.0 (normal temperature) = 1.1 = 1.0 (assumed since no incision or preservative treatment is prescribed)

Using the NDS Applicability table (Table 3.1), the allowable tension stress is given as, F’t

= =

Ft CDCMCtCFCi 300 x 1.15 x 1 x 1 x 1.1 x 1 = 379 psi

xft = 477.3 psi > F’t = 379 psi

NOT GOOD

Since the applied tension stress, ft is greater than the allowable tension stress, Ft’, the 2x10 Hem Fir No. 3 is NOT adequate for this tension member.x

Chapter 5

5-3

Solutions

S5 - 3

A 2 ½ x 9 (6 lams) 5DF Glulam Axial Combination member is subjected to an axial tension load of 20,000 lb. caused by dead plus snow load. Determine the applied tension stress, ft, and the allowable tension stress, F’t and check the adequacy of this member.

Normal temperature and dry service conditions apply, and the end connections are made with a single row of 7/8” diameter bolts.

Solution: The gross area, Ag for 2 ½” x 9” Glulam = 22.5 in2 (Note: Glulam is specified using the actual size) The area of bolt holes,  Aholes = 1 hole x (7/8 + 1/8) x 2 ½” = 2.5 in2 The net area at the critical section is, An = Ag -  Aholes = 22.5 – 2.5 = 20.0 in2 The applied tension stress, ft is, ft = T = 20,000 Ib = 1000 psi An 20.0 in 2

For 5DF Glulam Axial Combination, Use NDS-S Table 5B The tabulated design tension stress and the applicable adjustment factors from NDS-S Table 5B are obtained as follows: Ft CM CD Ct

= 1600 psi for 5 DF = 1.0 = 1.15 (D+S) = 1.0 (normal temperature condition)

Using the NDS Applicability table (Table 3.3), the allowable tension stress is given as, Ft’

= Ft CD CMCt = 1600 x 1.15x1x1 = 1840 psi > ft’

OK

Therefore, the 2½ x 9 5DF Glulam Tension Member is adequate.

5-4

For the roof truss elevation shown in Figure 5-21, determine if a 2 x 10 Spruce Pine Fir No. 1 wood member is adequate for the bottom chord of the typical truss assuming the roof dead load is 20 psf, the snow load is 40 psf, and the ceiling dead load is 15 psf of

Chapter 5

Solutions

S5 - 4

horizontal plan area. Assume normal temperature and dry service conditions apply and the members are connected with a single row of ¾” dia. bolt. The trusses span 36 ft and are spaced at 2 ft on centers.

Solution: Calculate the Joint Loads The given loads in psf of horizontal plan area are as follows: Roof dead load, D (roof) = 20 psf Ceiling dead load, D (ceiling) = 15 psf Snow load, S = 40 psf Since the tributary area for a typical roof truss is 72 ft2 (i.e. 2 ft tributary width x 36 ft span), the roof live load, Lr from Equation 2-4 will be 20 psf, which is less than the snow load, S. Therefore, the controlling load combination from Chapter 2 will be dead load plus snow load (i.e. D + S).

Total load on roof, wTL = (20 + 40) psf x (2 ft tributary width) =120 lb/ft Total ceiling load = (15 psf) x (2 ft) = 30 lb/ft The concentrated gravity loads at the truss joints are calculated as follows: The concentrated gravity loads at the truss joints are calculated as follows: Roof Dead Load + Snow load: PA (top) = (9’/2) x (120 lb/ft) PE = (9’/2 + 9’/2) x (120 lb/ft) PF = (9’/2 + 9’/2) x (120 lb/ft) PG = (9’/2 + 9’/2) x (120 lb/ft) PA (top) = PD (top)

= 540 lb = 1080 lb = 1080 lb = 1080 lb = 540 lb

Ceiling Dead Loads: PA (bottom) = (12’/2) x 30 (lb/ft) = 180 lb PD (bottom) = (12’/2) x 30 (lb/ft) = 180 lb PB = PC = (12’/2 + 12’/2) x (30 lb/ft) = 360 lb

Analyzing the truss in Figure 5-21 for the loads above using the method of joints or computer analysis software, we obtain the maximum tension force in the bottom chord member (member AB, BC or CD) as,

Chapter 5

Solutions

S5 - 5

T = 4750 Ib In addition to the tension force on member AC, there is also a uniform ceiling load of 30 Ib/ft acting on the truss bottom chord.

Analysis of 2x10 S-P-F No. 1 wood member in truss bottom chord: Ag = 13.88 in2 Sxx = 21.4 in3 Since given member is dimension lumber, therefore NDS-S Table 4A is applicable From NDS-S Table 4A, we obtain the tabulated design stresses and stress adjustment factors.

Fb = 875 psi (tabulated bending stress) Ft = 450 psi (tabulated tension stress) CF (Fb) = 1.1 (size adjustment factor for bending stress) CF(Ft)= 1.1 (size adjustment factor for tension stress) CD = 1.15 (for design check with Snow Load) CD = 0.9 (for design check with Dead Load only) CL = 1.0 (assuming lateral buckling is prevented by the ceiling and bridging) Cr = 1.15 (all 3 repetitive member requirements are met, see NDS Code 4.3.9) Design Check #1: The applied tension force, T = 4750 Ib (caused by dead load + snow load) The net area, An = Ag - Aholes = 13.88 – 1 hole x (3/4”+1/8”) x (1.5”) =12.57 in2 From Equation 5-4, the applied tension stress at the supports, ft = T = 4750 Ib2 = 378 psi 12.57 in An Using the NDS Applicability table (Table 3.1), the allowable tension stress is given as, Ft’ = (Ft) x (CD) x (CM) x (Ct) x (CF) x (Ci) = (450) x (1.15) x (1.0) x (1.0) x (1.1) x (1.0) = 569 psi > ft Design Check #2: From Equation 5-5, the applied tension stress at the midspan,

OK

Chapter 5

Solutions

S5 - 6

ft = T = 4750 Ib2 = 342 psi < Ft’ = 569 psi 13.88in Ag

OK

Design Check #3: The maximum moment which for this member occurs at the midpsan is given as,

M

2

= 30 Ib/ft (12 ft) = 540 ft-lb = 6480 in-lb max 8

This moment is caused by the ceiling dead load only (since no ceiling live load specified), therefore, the controlling load duration factor CD is 0.9.

From Equation 5.7, the applied bending stress (i.e. tension stress due to bending) is, fbt =

M

max =

Sx

6480 in - Ib = 303 psi 21.4in 3

Using the NDS Applicability table (Table 3.1), the allowable bending stress is given as, Fb’ = (Fb) x (CD) x (CM) x (Ct) x (CL) x (CF) x (Ci) x (Cr ) = (875) x (0.9) x (1.0) x (1.0) x (1.0) x (1.1) x (1.0) x (1.15) = 996 psi > fbt

OK

Design Check #4: For this case, the applicable load is dead plus snow load, therefore, the controlling load duration factor, CD is 1.15 (See Chapter 3) Using the NDS Applicability table (Table 3.1), the allowable axial tension stress and the allowable bending stress are calculated as follows: Ft’ = (Ft) x (CD) x (CM) x (Ct) x (CF) x (Ci) = (450) x (1.15) x (1.0) x (1.0) x (1.1) x (1.0) = 569 psi

Fb* = (Fb) x (CD) x (CM) x (Ct) x (CF) x (Ci) x (Cr ) = (875) x (1.15) x (1.0) x (1.0) x (1.1) x (1.0) x (1.15) = 1273 psi The applied axial tension stress, ft at the midspan is 342 psi as calculated for Design Check #2, while the applied tension stress due to bending, fbt is 303 psi as calculated in Design Check #3.

Chapter 5

Solutions

S5 - 7

From Equation 5.11, the combined axial tension plus bending interaction equation for the stresses in the tension fiber of the member is given as:

f

f

t + bt Ft' Fb*

 1.0

Substitution into the interaction equation yields,

342 psi + 303 psi = 0.84 < 1.0 569 psi 1273 psi

OK

Design Check #5: The applied compression stress due to bending is, fbc =

M

6480 in - Ib = 303 psi (= f since this is a rectangular cross-section) bt 21.4in 3

max =

Sx

It should be noted that fbc and fbt are equal for this problem because the wood member is rectangular in cross-section.

The load duration factor for this case, CD = 1.15 (combined dead + snow loads) Using the NDS Applicability table (Table 3.1), the allowable compression stress due to bending is, Fb** = (Fb) x (CD) x (CM) x (Ct) x (CL) x (CF) x (Ci) x (Cr ) = (875) x (1.15) x (1.0) x (1.0) x (1.0) x (1.1) x (1.0) x (1.15) = 1273 psi From Equation 5.15, the interaction equation for this design check is given as,

f -f bc

Fb'

t

 1.0

Substituting in the interaction equation yields,

303psi - 342 psi = 1273 psi

-0.03 <<< 1.0

OK

From all of the above steps, we find that all the five design checks are satisfied, x2 x 10 S-P-F No. 1 is adequate for the Bottom Chord

therefore,

Chapter 5

5-5

Solutions

S5 - 8

For the roof truss in problem 5-4, determine if a 2 x 10 Spruce Pine Fir No. 1 is adequate for the top chord.

Solution: The maximum axial compression force in the truss top chord is, Pmax (from computer analysis, method of joints, or method of sections) = 5150 lb The maximum moment in member AD is, Mmax = (120lb/ft ) x (9’)2/8 = 1215 ft.-lb. = 14,580 in-lb.

Analysis of given 2 x 10 S-P-F No. 1: 2 x 10 is dimension lumber, therefore, use NDS-S Table 4A From NDSS Table 1B, we obtain the section properties for the trial member size: Gross cross-sectional area, Ag = 13.88 in2 Section modulus, Sx = 21.4 in3 dx = 9.25” dy = 1.5”

From NDS-S Table 4A, we obtain the tabulated design stress values and the stress adjustment or C-factors as follows: Fc = 1150 psi Fb = 875 psi Emin = 0.51 x 106 psi CF(Fc) = 1.0 CF(Fb) = 1.1 CM = 1.0 (normal moisture conditions) Ct = 1.0 (normal temperature conditions) Cr = 1.15 (the condition for repetitiveness is discussed in Chapter 3) CD

= 1.15 (snow load controls for the load combination D + S)

Design Check #1: Compression on net area,

Chapter 5

Solutions

S5 - 9

This condition occurs at the ends (i.e. supports) of the member For this load case, the axial load P is caused by dead load plus snow load (D + S), therefore, the load duration factor, CD from Chapter 3 is 1.15.

The applied compression stress, fc = P  Fc’ An Net area, An = Ag - Aholes = 13.88 - 1 hole x (3/4” + 1/8”) x (1.5”)  An = 12.56 in2 The applied compressive axial stress, fc =

P = 5150 Ib = 410 psi An 12.56 in 2

At support locations, there can be no buckling  CP =1.0 The allowable compression stress parallel to grain is, Fc’ = (Fc) x (CD) x (CM) x (Ct) x (CF) x (Ci) x (Cp) = (1150) x (1.15) x (1.0) x (1.0) x (1.0) x (1.0) x (1.0)  Fc’ = 1323 psi > fc = 410 psi

OK

Design Check #2: Compression on gross area This condition occurs at the midspan of the member. The axial load P for this load case is also caused by dead load plus snow load (D + S), therefore, the load duration factor, CD from Chapter 3 is 1.15.

Applied Axial Compressive stress, fc = P = 5150 Ib = 371 psi Ag 13.88 in 2 Unbraced length: For x-x axis buckling, the unbrace length, Lux = 9.75 ft For y-y axis buckling, the unbrace length, Luy = 0 ft (plywood sheathing braces top chord)

Effective length:

Chapter 5

Solutions

S5 - 10

For building columns that are supported at both ends, it is usual practice to assume and effective length factor, Ke of 1.0.

For x-x axis buckling, effective length, Lex = Ke Lux = (1.0) x (9.75’) = 9.75’ For y-y axis buckling, effective lengthy, Ley = Ke Luy = (1.0) x (0) = 0’ Slenderness ratio: For x-x axis buckling, the effective length is, Lex/dx = (9.75’ x 12)/(9.25”) = 12.7 < 50 (larger slenderness ratio controls) For y-y axis buckling, the effective length is, Ley/dy = 0/1.5” = 0 < 50  (Le/d)max = 12.7 This slenderness ratio will be used in the calculation of the column stability factor, CP and the allowable compression stress parallel to grain, F’c, Buckling modulus of elasticity, Emin’ = (Emin) x (CM) x (Ct) x (Ci) = (0.51 x 106) x (1.0) x (1.0) = 0.51 x 106 psi c = 0.8 (visually graded lumber) The Euler critical buckling stress about the weaker axis (i.e. the axis with the higher slenderness ratio), which for this problem happens to be the x-x axis is, FcE

= 0.822 Emin’ = 0.822 x 0.51 x 106 = 2599 psi (Le/d)max2 (12.7)2

Fc* = (Fc) x (CD) x (CM) x (Ct) x (CF) x (Ci) = (1150) x (1.15) x (1.0) x (1.0) x (1.0) x (1.0) = 1323 psi

2599 psi FcE Fc* = = 1.96 1323 psi From Equation 5.29, the column stability factor is calculated as, 1+ (1.96 ) CP = 2(0.8)

(

)

 1+ 1.96     2(0.8)   

2 -

(1.96 ) = 0.864 (0.8)

Chapter 5

Solutions

S5 - 11

The allowable compression stress parallel to grain is, Fc’ = (Fc*) x (CP) = (1323) x (0.864) = 1143 psi > fc = 371 psi

OK

Design Check #3: Bending only This condition occurs at the point of maximum moment (i.e. at the midspan of the member), and the moment is caused by the uniformly distributed gravity load on the top chord and since this load is dead load plus snow load, the controlling load duration factor (See Chapter 3) is 1.15.

The maximum moment in the top chord, Mmax =

w

D+S

L2

8 2

= (120 Ib/ft) (9 ft) = 1215 ft-Ib = 14,580 in-lb.

8 The applied compression stress in member AD due to bending, fbc = M = 14,580 in 3- Ib = 681 psi Sx 21.4 in The allowable bending stress, Fb’ = (Fb) x (CD) x (CM) x (Ct) x (CL) x (CF) x (Ci) x (Cr ) = (875) x (1.15) x (1.0) x (1.0) x (1.0) x (1.1) x (1.0) x (1.15) Fb’

= 1273 psi > fb = 681 psi

OK

Design Check #4: Bending plus axial compression force This condition occurs at the midspan of the member. For this load case, the loads causing the combined stresses are dead load plus snow load (D + S), therefore, the controlling load duration factor, CD is 1.15. It should be noted that because the load duration factor, CD for this combined load case is the same as for load cases #2 and #3, the parameters to be used in the combined stress interaction equation can be obtained from design check #2 and #3. The reader is cautioned to be aware that the CD value for all the four design checks are not necessarily always equal.

The interaction equation for combined concentric axial load plus uniaxial bending is obtained from Equation 5.48 as,

Chapter 5

Solutions

S5 - 12

 f  bx   2   ' f  Fbx   c  +    1.0  '  f F   c c    1 F  cEx  

fc = 371 psi (occurs at the midspan; See design check #2) F’c = 1143 psi (occurs at the midspan; See design check #2) fbx = 681 psi (occurs at the midspan; See design check #3) F’bx = 1273 psi (occurs at the midspan; See design check #3) The reader should note that in the interaction equation above, it is the Euler critical buckling stress about the x-x axis, FcEx that is required since the bending of the truss top chord member is about that axis.

The effective length for x-x axis buckling is, Lex/dx = (9.75’ x 12)/(9.25”) = 12.7 < 50 The Euler critical buckling stress about the x-x axis is, FcEx

= 0.822 Emin’ = 0.822 x 0.51 x 106 = 2599 psi (Lex/dx) 2 (12.7)2

Substituting the above parameters into the interaction equation yields,  371     1143 

2

 681    +  1273  371    12599  

= 0.73 < 1.0

OK

x2 x 10 S-P-F No. 1 is adequate for the truss top chordx

5-6

Determine if a 10 ft high 2 x 6 Hem Fir No. 1 interior wall studs spaced at 16” on centers is adequate to support a dead load of 300 Ib/ft and a snow load of 600 Ib/ft. Assume the wall is sheathed on both sides, and normal temperature and dry service conditions apply.

Chapter 5

Solutions

Solution: First, we calculate the gravity and lateral loads acting on a typical wall stud.

Tributary width of wall stud = 1.33 ft Dead load, D = 300 Ib/ft (1.33 ft) = 400 Ib Roof snow load, S = 600 Ib/ft (1.33 ft) = 800 Ib Gravity loads: PD = 400 lb PS = 800 Ib Considering the load combinations in Chapter 2, the most critical load combination is D + S = 1200 Ib, and CD = 1.15 Given member size: 2x6 Hem-fir No. 1 2 x 6 is dimension lumber, therefore, use NDS-S Table 4A From NDSS Table 1B, we obtain the section properties for the trial member size: Gross cross-sectional area, Ag = 8.25 in2 Section modulus, Sx = 7.56 in3 dx = 5.5” dy = 1.5” (fully braced by sheathing)

From NDS-S Table 4A, we obtain the tabulated design stress values and the stress adjustment or C-factors as follows: Fc = 1350 psi Fb = 975 psi Fc⊥ = 405 psi Emin = 0.55 x 106 psi

CF(Fc) = 1.1 CF(Fb) = 1.3 CM = 1.0 (normal moisture conditions) Ct = 1.0 (normal temperature conditions) Cr = 1.15 (the condition for repetitiveness is discussed in Chapter 3) CL = 1.0 CD = 1.15 (D + S)

S5 - 13

Chapter 5

Solutions

S5 - 14

The beam stability factor, CL will be 1.0 for bending of the wall stud about the x-x (or strong axis) because the wall stud is braced against lateral torsional buckling by the plywood sheathing.

Design Check #1: Compression on net area This condition occurs at the ends (i.e. supports) of the member. The controlling load combination is D + S for which P = 1200 lb., and CD = 1.15 Since the ends of the wall studs are usually connected to the sill plates and top plates with nails rather than bolts, the net area will be equal to the gross area since there are no bolt holes to consider.

Net area, An = Ag - Aholes = Ag = 8.25 in2 The applied compressive axial stress, fc =

P = 1200 Ib = 145.5 psi An 8.25 in 2

At support locations, there can be no buckling  CP =1.0 The allowable compression stress parallel to grain is, Fc’ = Fc CD CM Ct CF Ci Cp = 1350 x 1.15x 1 x 1 x1.1 x 1 x 1  Fc’ = 1708 psi > fc = 145.5 psi

OK

Design Check #2: Compression on gross area This is also a pure axial load case that occurs at the mid-height of the wall stud, and the critical load combination is D + S for which P = 1200 lb., and CD = 1.15 The axial compression stress, fc = P = 1200 Ib = 145.5 psi Ag 8.25 in 2 The compressive stress perpendicular to grain in the sill plate is, f c⊥ = P = 1200 Ib = 145.5 psi Ag 8.25 in 2 Unbraced length: For x-x axis buckling, the unbrace length, Lux = 10 ft

Chapter 5

Solutions

S5 - 15

For y-y axis buckling, the unbrace length, Luy = 0 ft (plywood sheathing braces y-y axis buckling)

Effective length: Wall studs are usually supported at both ends, therefore, it is usual practice to assume an effective length factor, Ke of 1.0.

For x-x axis buckling, effective length, Lex = Ke Lux = (1.0) x (10’) = 10’ For y-y axis buckling, effective lengthy, Ley = Ke Luy = (1.0) x (0) = 0’ Slenderness ratio: For x-x axis buckling, the effective length is, Lex/dx = (10’ x 12)/(5.5”) = 21.8 < 50 (larger slenderness ratio controls) For y-y axis buckling, the effective length is, Ley/dy = 0/1.5” = 0 < 50  (Le/d)max = 21.8 This slenderness ratio will be used in the calculation of the column stability factor, CP and the allowable compression stress parallel to grain, F’c, Buckling modulus of elasticity, Emin’ = Emin CMCtCi = 0.55 x 106 x 1 x 1 = 0.55 x 106 psi c = 0.8 (visually graded lumber) The Euler critical buckling stress about the “weaker” axis (i.e. the axis with the higher slenderness ratio), which for this problem happens to be the x-x axis is, FcE

= 0.822 Emin’ = 0.822 x 0.55 x 106 = 951 psi (Le/d)max2 (21.8)2

Fc* = Fc CDCMCtCFCi = 1350 x 1.15x 1 x 1 x1.1 x 1 x 1 = 1708 psi

951 psi FcE Fc* = = 0.557 1708 psi

Chapter 5

Solutions

S5 - 16

From Equation 5.29 the column stability factor is calculated as, 1+ ( 0.557 ) CP = 2(0.8)

(

)

1+ 0.557     2(0.8)   

2 -

( 0.557 ) = 0.472 (0.8)

The allowable compression stress parallel to grain is, Fc’ = (Fc*) x (CP) = (1708) x (0.472) = 806 psi > fc = 145.5 psi

OK

NOTE: From design aid B.41, the allowable axial load for each wall stud is at least 6,420 Ib (since CD = 1.0 for this chart) at M = 0. The applied axial load for this stud is 1200 Ib < 6,420 Ib, OK The allowable compression stress perpendicular to grain in the sill plate is, Fc⊥ = Fc⊥ CMCtCi = 625 x 1 x 1 x 1 Fc⊥ = 625 psi > fc⊥ = 145.5 psi 5-7

OK

Determine the axial load capacity of (4) - 2x6 nailed built-up column with a 12 ft unbraced height. Assume Hem Fir No. 2, normal temperature and dry service conditions, and load duration factor CD of 1.0

Solution: For (4)-2x6’s: dx = 5.5”, dy = 4 x 1.5” = 6” (see NDS-S Table 1B) Since 2 x 6 is dimension Lumber, NDS-S Table 4A is applicable. From NDS-S Table 4A, the tabulated design stresses Hem-fir No. 2 are obtained as, Fc = 1300 psi Emin = 0.47 x106 psi From NDS-S Table 4A, the size factor for axial compression stress parallel to grain is, CF (Fc) = 1.1 (size factor for axial compression stress parallel to grain) The following stress adjustment or C- factors were specified in the problem: CM = 1.0 (normal moisture conditions) Ct = 1.0 (normal temperature conditions) CD = 1.0 (given)

Chapter 5

Solutions

S5 - 17

Unbraced length of column, Lx = Ly = 12 ft Ke = 1.0 (building columns are typically assumed to be pinned at both ends) Lex/dx = Ke Lx/dx = (1.0) x (12 ft x 12)/(5.5”) = 26.2 < 50

OK

Ley/dy = Ke Ly/dy = (1.0 x 12 ft x 12)/(6”) = 24 < 50

OK

Since Lex/dx > Ley/dy, two conditions will need to be checked: Ley/dy with Kf = 0.60 and Lex/dx with Kf = 1.0 c = 0.8 (visually graded lumber) Emin’ = (Emin) x (CM) x (Ct) x (Ci) = (0.47 x 106)x (1.0) x (1.0) x (1.0) = 0.47 x 106 psi

FcExx =

0.822E min ' 0.822E min ' and FcEyy = 2 (L e / d x ) (L e / d y ) 2

FcExx =

0.822(0.47x10 6 ) = 563psi (26.2) 2

and

FcEyy =

0.822(0.47x10 6 ) = 671 psi (24) 2

Fc* = (Fc) x (CD) x (CM) x (Ct) x (CF) x (Ci) = (1300) x (1.0) x (1.0) x (1.0) x (1.1) x (1.0) = 1430 psi

FcEx 563 = = 0.393 and Fc * 1430

FcEy Fc *

=

671 = 0.47 1430

From Equation 5.37, the column stability factor is calculated as, 2  1 + (0.393)  (0.393)   1 + (0.393) C Px =1.0 −  − = 0.354 2(0.8)  (0.8)   2(0.8)    2  1 + (0.47)  (0.47)   1 + (0.47) C Py = 0.6 −  − = 0.247  controls 2(0.8)  (0.8)   2(0.8)   

The allowable compression stress is given as, Fc’ = (Fc*) x (Cp) = (1430) x (0.247) = 353 psi

Chapter 5

Solutions

S5 - 18

The axial load capacity of the built-up column is calculated as, Pallowable = Fc’ Ag =

(353) x (6.0” x 5.5”) = 11,600 lb

From Section 15.3.3 of the NDS Code (ref. 5-1), the minimum nailing requirement is 30d nails spaced vertically at 8” on centers and staggered 2½” with an edge distance of 1½” and end distance of 3½” (See Figure 5.11).

5-8

Determine if a 2 x 6 Douglas Fir Larch Select Structural studs spaced at 2 ft on centers is adequate a the ground floor level to support the following loads acting on the exterior stud wall of a 3-story building. The floor-to-floor height is 10 ft and the wall studs are spaced at 2 ft on centers, and assume the stud is sheathed on both sides.

Roof: Dead load = 300 Ib/ft; Snow load = 800 Ib/ft ; Roof live load = 400 Ib/ft 3rd Floor: Dead load = 400 Ib/ft; Live load = 500 Ib/ft 2nd Floor: Dead load = 400 Ib/ft; Live load = 500 Ib/ft Lateral wind load acting perpendicular to the face of the wall = 15 psf

Solution: First, we calculate the gravity and lateral loads acting on a typical wall stud. Note that the wall selfweight is assumed to have already been included in the given dead loads.

Gravity loads: The total dead Load on the Ground floor studs is: PD = PD(roof) + PD(floor) + PD(wall) = (300 Ib/ft)(2 ft) + (400 Ib/ft + 400 Ib/ft)(2 ft) = 2200 lb The total snow load on the Ground floor studs is: PS = (800 Ib/ft)(2 ft) = 1600 Ib The total floor live load on the Ground floor studs is: PL = (500 Ib/ft + 500 Ib/ft)(2 ft) = 2000 Ib Note that the roof live load, Lr has been neglected in this example since it will not govern because its normalized load (i.e. Lr/1.25 = 400 plf/1.25) is less than the normalized snow load (i.e. S/1.15 = 800 plf/1.15).

Lateral loads:

Chapter 5

Solutions

S5 - 19

The wind load acts perpendicular to the face of the stud wall causing bending of the stud about the x-x (or strong) axis (See Figure 5-18).

The lateral wind load, wwind = (15psf) x (2’ tributary width) = 30 lb./ft The maximum moment due to wind load is calculated as, Mw = wL2/8 = (30) x (102)/8 = 375 ft.-lb = 4500 in.-lb The most critical axial load combination and the most critical combined axial and bending load combination are now determined following the normalized load procedure outlined in Chapter 3.

The axial load, P in the wall stud will be caused by the gravity loads, D, L, and S, while the bending load and moment will be caused by the lateral wind load, W. The load values to be used in the load combinations are as follows: D = 2200 Ib S = 1600 Ib L = 2000 Ib W = 4500 in-Ib. All other loads are assumed to be zero and are therefore neglected in the load combinations. The applicable load combinations with all the zero loads neglected are as shown in in the table below.

Applicable and Governing Load Combinations Load Comb.

Axial Load, P (lb.)

Moment, M (in.-lb.)

CD

D

2200

0

D+L

2200 + 2000 = 4200

D+S D + 0.75(L) + 0.75(S) D + 0.75(0.6W) + 0.75(L) + 0.75(S) 0.6D + 0.6W

2200 + 1600 = 3800 2200 + 0.75(2000 + 1600)= 4900 2200 + 0.75(2000+ 1600)= 4900 0.6(2200) = 1320

Normalized Load & Moment P/CD

M/CD

0.9

2444

0

0

1.0

4200

0

0

1.15

3304

0

1.15

4261

0

1.6

3063

1266

1.6

825

1688

0 0.75(0.6)(4500)

= 2025 (0.6)(4500)=

2700

Chapter 5

Solutions

S5 - 20

It will be recalled from Chapter 3 that the necessary condition to use the normalized load method is that all the loads must be similar and of the same type. We can separate the load cases in the above table into two types of loads: pure axial load cases and combined load cases, the most critical load combinations are the load cases with the highest normalized load or moment.. Since not all the loads on this wall stud are pure axial loads only or bending loads only, the normalized load method discussed in Chapter 3 can then only be used to determine the most critical pure axial load case, but not the most critical combined load case. The most critical combined axial load plus bending load case would have to be determined by carrying out the design (or analysis) for all the combined load cases, with some of the load cases eliminated by inspection.

Pure Axial Load Case: For the pure axial load cases, the load combination with the highest normalized load (P/CD) from the table above is, D + 0.75(L + S),

P = 4900 lb., with CD = 1.15

Combined Load Case: For the combined load cases, the load combinations with the highest normalized load (P/CD) and normalized moment (M/CD) from the table above are, D + 0.75(L + S + 0.6W),

P = 4900 lb., M = 2025 in.-lb, CD = 1.6

or 0.6D + 0.6W,

P = 1320 lb., M = 2700 in.-lb, CD = 1.6

By inspection, it would appear as if the load combination D + (L + S + W) will control for the combined load case, but this should be verified through analysis, and both of these combined load cases will have to be investigated.

Analysis of given 2x6 DF-L Select Structural Stud: 2 x 6 is dimension lumber, therefore, use NDS-S Table 4A From NDSS Table 1B, we obtain the section properties for the trial member size: Gross cross-sectional area, Ag = 8.25 in2 Section modulus, Sx = 7.56 in3 dx = 5.5” dy = 1.5” (fully braced by sheathing)

Chapter 5

Solutions

S5 - 21

From NDS-S Table 4A, we obtain the tabulated design stress values and the stress adjustment or C-factors as follows: Fc = 1700 psi Fb = 1500 psi Fc⊥ = 625 psi Emin = 0.69 x 106 psi

CF(Fc) = 1.1 CF(Fb) = 1.3 CM = 1.0 (normal moisture conditions) Ct = 1.0 (normal temperature conditions) Cr = 1.15 (the condition for repetitiveness is discussed in Chapter 3) CL = 1.0 CD = To be determined for each design check The beam stability factor, CL will be 1.0 for bending of the wall stud about the x-x (or strong axis) because the wall stud is braced against lateral torsional buckling by the plywood sheathing.

The load duration factor, CD will be determined for each design check, and for each design check, the controlling load duration factor value will correspond to the CD value of the shortest duration load in the load combination for that design case. Thus, the CD value for the different design cases may vary.

Design Check #1: Compression on net area This condition occurs at the ends (i.e. supports) of the member. Note that for this design check which is a pure axial load case, the load combination with the highest normalized axial load (P/CD) will be most critical. This controlling load combination from the table above is D + 0.75(L + S) for which P = 4900 lb., and CD = 1.15 Since the ends of the wall studs are usually connected to the sill plates and top plates with nails rather than bolts, the net area will be equal to the gross area since there are no bolt holes to consider.

Net area, An = Ag - Aholes = Ag = 8.25 in2 The applied compressive axial stress, fc =

P = 4900 Ib = 594 psi An 8.25 in 2

At support locations, there can be no buckling  CP =1.0 The allowable compression stress parallel to grain is,

Chapter 5

Solutions

S5 - 22

Fc’ = (Fc) x (CD) x (CM) x (Ct) x (CF) x (Ci) x (Cp) = (1700) x (1.15) x (1.0) x (1.0) x (1.1) x (1.0) x (1.0)  Fc’ = 2151 psi > fc = 594 psi

OK

Design Check #2: Compression on gross area This is also a pure axial load case that occurs at the mid-height of the wall stud, and the most critical load combination will be the load combination with the highest normalized axial load (P/CD). This controlling load combination from the table above is D + 0.75(L + S) for which P = 4900 lb., and CD = 1.15 The axial compression stress, fc = P = 4900 Ib = 594 psi Ag 8.25 in 2 The compressive stress perpendicular to grain in the sill plate is, f c⊥ = P = 4900 Ib = 594 psi Ag 8.25 in 2 Unbraced length: For x-x axis buckling, the unbrace length, Lux = 10 ft For y-y axis buckling, the unbrace length, Luy = 0 ft (plywood sheathing braces y-y axis buckling)

Effective length: Wall studs are usually supported at both ends, therefore, it is usual practice to assume an effective length factor, Ke of 1.0.

For x-x axis buckling, effective length, Lex = Ke Lux = (1.0) x (10’) = 10’ For y-y axis buckling, effective lengthy, Ley = Ke Luy = (1.0) x (0) = 0’ Slenderness ratio: For x-x axis buckling, the effective length is, Lex/dx = (10’ x 12)/(5.5”) = 21.8 < 50 (larger slenderness ratio controls) For y-y axis buckling, the effective length is,

Chapter 5

Solutions

S5 - 23

Ley/dy = 0/1.5” = 0 < 50  (Le/d)max = 21.8 This slenderness ratio will be used in the calculation of the column stability factor, CP and the allowable compression stress parallel to grain, F’c, Buckling modulus of elasticity, Emin’ = (Emin) x (CM) x (Ct) x (Ci) = (0.69 x 106) x (1.0) x (1.0) = 0.69 x 106 psi c = 0.8 (visually graded lumber) The Euler critical buckling stress about the “weaker” axis (i.e. the axis with the higher slenderness ratio), which for this problem happens to be the x-x axis is, FcE

= 0.822 Emin’ = 0.822 x 0.69 x 106 = 1193 psi (Le/d)max2 (21.8)2

Fc* = (Fc) x (CD) x (CM) x (Ct) x (CF) x (Ci) = (1700) x (1.15) x (1.0) x (1.0) x (1.1) x (1.0) = 2151 psi

1193 psi FcE Fc* = = 0.555 2151 psi From Equation 5.29 the column stability factor is calculated as, 1+ ( 0.555 ) CP = 2(0.8)

(

)

1+ 0.555     2(0.8)   

2 -

( 0.555 ) = 0.471 (0.8)

The allowable compression stress parallel to grain is, Fc’ = (Fc*) x (CP) = (2151) x (0.471) = 1013 psi > fc = 594 psi OK NOTE: From design aid B.37, the allowable axial load for each wall stud is at least 8,060 Ib (since CD = 1.0 for this chart) at M = 0. The applied axial load for this stud is 4900 Ib < 8,060 Ib, OK The allowable compression stress perpendicular to grain in the sill plate is, Fc⊥ = (Fc⊥) x (CM) x (Ct) x (Ci) = (625) x (1.0) x (1.0) x (1.0) Fc⊥ = 625 psi > fc⊥ = 594 psi

OK

Chapter 5

Solutions

S5 - 24

Design Check #3: Bending only This condition occurs at the mid-height of the wall stud, and the bending moment is caused by the lateral wind load. Only the two controlling combined load cases in the table above need be considered for this design check. And the load combination with the maximum normalized moment, (M/CD) is 0.6D + 0.6W for which the maximum moment M = 2700 in.-lb, P = 1320 lb., and CD = 1.6.

The applied compression stress in wall stud due to bending, fbc = M = 2700 in -3Ib = 357 psi Sx 7.56 in The allowable bending stress, Fb’ = (Fb) x (CD) x (CM) x (Ct) x (CL) x (CF) x (Ci) x (Cr ) = (1500) x (1.6) x (1.0) x (1.0) x (1.0) x (1.3) x (1.0) x (1.15) Fb’

= 3588 psi > fb = 357 psi

OK

Design Check #4: Bending plus axial compression force This condition occurs at the mid-height of the wall stud, and only the two controlling combined load cases in the table above need be considered for this design check, and these are: D + 0.75(L + S + 0.6W),

P = 4900 lb., M = 2025 in.-lb, CD = 1.6

and 0.6D + 0.6W,

P = 1320 lb., M = 2700 in.-lb, CD = 1.6

We will investigate both of these load combinations separately to determine the most critical.

Load combination D + 0.75(L + S + 0.6W) Pmax = 4900 lb., Mmax = 2025 in.-lb. CD = 1.6 The applied bending stress is,

Chapter 5

Solutions

fbx =

S5 - 25

M max = 2025 in - Ib = 268 psi (lateral wind load causes bending about x-x axis) Sxx 7.56 in3

The applied axial compression stress at the mid-height is, fc = 594 psi (from Design Check #2) Allowable bending stress, Fb’ = (Fb) x (CD) x (CM) x (Ct) x (CL) x (CF) x (Ci) x (Cr ) = (1500) x (1.6) x (1.0) x (1.0) x (1.0) x (1.3) x (1.0) x (1.15) Fb’

= 3588 psi > fb = 268 psi

OK

Note that the beam stability factor, CL is 1.0 because the compression edge of the stud is braced laterally by the wall sheathing for bending due to loads acting perpendicular to the face of the wall. We will now proceed to calculate the column stability factor, CP. Fc* = (Fc) x (CD) x (CM) x (Ct) x (CF) x (Ci) = (1700) x (1.6) x (1.0) x (1.0) x (1.1) x (1.0) = 2992 psi The Euler critical buckling stress about the “weaker” axis (i.e. the axis with the higher slenderness ratio), which for this problem happens to be the x-x axis is, FcE(max) = 0.822 Emin’ = 0.822 x 0.69 x 106 = 1193 psi (Le/d)max2 (21.8)2

1193 psi FcE Fc* = = 0.399 2992 psi From Equation 5.29, the column stability factor is calculated as, 1+ ( 0.399 ) CP = 2(0.8)

(

)

 1+ 0.399     2(0.8)   

2 -

( 0.399 ) = 0.359 (0.8)

The allowable compression stress parallel to grain is, Fc’ = (Fc*) x (CP) = (2992) x (0.359) = 1074 psi > fc = 594 psi

OK

The interaction equation for combined concentric axial load plus uniaxial bending is obtained from Equation 5.48 as,

Chapter 5

Solutions

2 

f  c   '   Fc 

S5 - 26

 f  bx     ' Fbx  +    1.0 f  c   1FcEx    

The Euler critical buckling stress about the x-x axis (i.e. the axis of bending of the wall stud due to lateral wind loads) is, FcEx

= 0.822 Emin’ = 0.822 x 0.69 x 106 = 1193 psi (Lex/dx) 2 (21.8)2

Substituting the above parameters into the interaction equation yields,  594     1074 

2

 268    +  3588  594    1  1193 

= 0.45 < 1.0

OK

NOTE: From design aid B.21, the allowable axial load for each wall stud is about 6,800 Ib at M = 281 ft-Ib (3375 in-Ib). The applied axial load for this stud is 4900 Ib < 6,800 Ib, OK

Load combination 0.6D + 0.6W Pmax = 1320 lb., Mmax = 2700 in.-lb. CD = 1.6 The applied bending stress is, fbx =

M max = 2700 in - Ib = 357 psi (lateral wind load causes bending about x-x axis) Sxx 7.56 in3

The axial compression stress, fc = P = 1320 Ib = 160 psi Ag 8.25 in 2 Except for the applied bending and axial stresses, all the other parameters used in the interaction equation for the previous combined load case would also apply to this load case. Therefore, the interaction equation for this load case will be,  160     1074 

2

 357    +  3588  160    1 1193  

= 0.14 <<< 1.0

OK

Chapter 5

Solutions

S5 - 27

NOTE: From design aid B.21, the allowable axial load for each wall stud is about 6,400 Ib at M = 375 ft-Ib (4500 in-Ib). The applied axial load for this stud is 1320 Ib < 6,400 Ib, OK

Chapter 5

Solutions

S5 - 28

Chapter 5

Solutions

S5 - 29

Chapter 5

Solutions

S5 - 30

Chapter 5

Solutions

S5 - 31

Chapter 5

Solutions

S5 - 32

Chapter 5

Solutions

S5 - 33

Chapter 5

Solutions

S5 - 34

Chapter 5

Solutions

S5 - 35

Chapter 5

Solutions

S5 - 36

Chapter 5

Solutions

S5 - 37

Chapter 5

Solutions

S5 - 38

Chapter 5

Solutions

S5 - 39

Chapter 5

Solutions

S5 - 40

Chapter 5

Solutions

S5 - 41

Chapter 5

Solutions

S5 - 42

Chapter 5

Solutions

S5 - 43

Chapter 5

Solutions

S5 - 44

Chapter 5

Solutions

S5 - 45

Chapter 5

Solutions

S5 - 46

Chapter 5

Solutions

S5 - 47

Chapter 5

Solutions

S5 - 48

Chapter 5

Solutions

S5 - 49

Chapter 5

Solutions

S5 - 50

Chapter 5

Solutions

S5 - 51

Chapter 5

Solutions

S5 - 52

Chapter 5

Solutions

S5 - 53

Chapter 5

Solutions

S5 - 54

Chapter 5

Solutions

S5 - 55

Solutions

Chapter 6

6-1

S6-1

For a total roof load of 60psf on 2x framing members at 24” O.C., determine an appropriate plywood thickness and span rating assuming panels with edge support and panels oriented perpendicular to the framing. Specify the minimum fasteners and spacing.

Solution: From IBC Table 2304.7(3), select 15/32” with span rating of 32/16. Total load capacity is 40 psf at a span rating of 32”. Adjusting for actual load capacity: 2

L   32  WTL =  max  WTL =   (40psf ) = 71 psf > 60 psf, OK  24   La  2

From Table 6-3, minimum fastening: 8d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing)

Chapter 6

6-2

Solutions

S6-2

Repeat Problem 6-1 for panels oriented parallel to the supports.

Solution: From IBC Table 2304.7(5), select 19/32” Structural I. Total load capacity is 80 psf at a span rating of 24”.

From Table 6-3, minimum fastening: 8d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing)

Chapter 6

6-3

Solutions

S6-3

For a total floor load of 75psf on 2x floor framing at 24” O.C., determine an appropriate plywood thickness and span rating assuming panels with edge support and panels oriented perpendicular to the framing. Specify the minimum fasteners and spacing.

Solution: From IBC Table 2304.7(3), select 23/32” with span rating of 48/24. Total load capacity is 100 psf at a span rating of 24”. (Single floor grade with a span rating of 24” O.C. could also be selected).

From Table 6-3, minimum fastening: 6d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing)

Solutions

Chapter 6

6-4

S6-4

A one-story building is shown below. The roof panels are 15/32” thick CD-X, with a span rating of 32/16. The gravity loads are: Dead = 20psf and Snow = 40psf. Assume the diaphragm is unblocked, and framing members are Spruce-Pine-Fir. Determine the following for the transverse direction: a.) Unit shear in the diaphragms b.) Required fasteners c.) Maximum Chord Forces d.) Maximum Drag Strut Forces (construct Unit Shear, Net Shear, and Drag Strut Force diagrams)

Note: The applied wind load will either have to be given, or can be calculated based on local conditions. In this case, assume W = 25psf.

Solution:

a)

Wind load to roof diaphragm:

 16'   w T = (25psf )   + 2' = 250plf  2   VT =

w T L (250)(88) = =11,000 lb.

2 2

Unit shear in diaphragm:

vT = b)

VT 11,000 lb = = 234 plf b 47'

Required fasteners: Adjusting for the 40% increase in capacity for wind loads: v r eqd =

234plf = 168plf 1.4

Since 15/32” sheathing is given, IBC Table 2306.3.1 specifies the minimum fasteners to be: 8d nails at 12” O.C. field nailing and 6” O.C. edge nailing. The diaphragm capacity is (240)(0.92) = 220 plf > vT = 168 plf (assuming load case #1).

Solutions

Chapter 6

c)

Chord forces:

wL2 (250)(88) 2 = 242,000 ft.-lb.

M= = 8 8 T=C=

d)

M 242,000 = = 5149 lb. b 47

Drag Strut Forces The starting point for shearwall 1 (25 ft long) is ‘A’, endpoint is ‘B’.

The starting point for shearwall 2 (11 ft long) is ‘C’, endpoint is ‘D’.

The drag strut goes from points ‘B’ to ‘C’.

vW =

VT 11,000 = = 305.6 plf L W 25' + 11'

vd = 234 plf Net shear (A-B) = 305.6 – 234 = 71.5 plf Net shear (B-C) = 234 plf Net shear (C-D) = 305.6 – 234 = 71.5 plf Drag Strut Force (Pt. A) = 0 lb.

Drag Strut Force (Pt. B) = (71.5 plf)(25’) = 1788 lb Drag Strut Force (Pt. C) = (-1788) + (234 plf)(11’) = 786 lb Drag Strut Force (Pt. D) = (786) - (71.5 plf)(11’) = 0 lb

S6-5

Solutions

Chapter 6

6-5

S6-6

Repeat Problem 6-4 for the Longitudinal Direction.

Solution:

a)

Wind load to roof diaphragm:

 16'   w L = (25psf )   + 2' = 250plf  2   VL =

w L L (250)(47) = = 5875 lb.

2 2

Unit shear in diaphragm:

vL = b)

VL 5875 lb = = 66.8 plf b 88'

Required fasteners: Adjusting for the 40% increase in capacity for wind loads: v r eqd =

66.8plf = 48plf 1.4

Since 15/32” sheathing is given, IBC Table 2306.3.1 specifies the minimum fasteners to be: 8d nails at 12” O.C. field nailing and 6” O.C. edge nailing. The diaphragm capacity is (180)(0.92) = 165 plf > vL = 48 plf (assuming load case #3).

c)

Chord forces:

M=

wL2 (250)(47) 2 = 69,031 ft.-lb. = 8 8

T=C=

M 69,031 = = 785 lb. b 88

Solutions

Chapter 6

d)

Drag Strut Forces The starting point for drag strut 1 (13’-6” ft long) is ‘D’, endpoint is ‘E’.

The starting point for shearwall 3 (18 ft long) is ‘E’, endpoint is ‘F’.

The starting point for drag strut 2 (25 ft long) is ‘F’, endpoint is ‘G’.

The starting point for shearwall 4 (18 ft long) is ‘G’, endpoint is ‘H’.

The starting point for drag strut 3 (13’-6” ft long) is ‘G’, endpoint is ‘I’.

vW =

VL 5875 = 163.2 plf = L W 18' + 18'

vd = 66.8 plf

Net shear (D-E) = 66.8 plf Net shear (E-F) = 163.2 – 66.8 = 96.4 plf Net shear (F-G) = 66.8 plf Net shear (G-H) = 163.2 – 66.8 = 96.4 plf Net shear (H-I) = 66.8 plf Drag Strut Force (Pt. D) = 0 lb.

Drag Strut Force (Pt. E) = (66.8 plf)(13.5’) = 901 lb Drag Strut Force (Pt. F) = (901) - (96.4 plf)(18’) = 834 lb Drag Strut Force (Pt. G) = (-834) + (66.8 plf)(25’) = 834 lb Drag Strut Force (Pt. H) = (834) -(96.4 plf)(18’) = 901 lb Drag Strut Force (Pt. I) = (-901) + (66.8 plf)(13.5’) = 0 lb

S6-7

Chapter 6

6.6

Solutions

S6-8

Chapter 6

Solutions

S6-9

From IBC Table 2304.7(3), select 19/32” with span rating of 40/20. Total load capacity is 100 psf at a span rating of 20”.

From Table 6-3, minimum fastening: 6d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing)

Chapter 6

b)

Solutions

S6-10

Chapter 6

6.7

Solutions

S6-11

Solutions

Chapter 6

S6-12

a)

With edge support: From IBC Table 2304.7(3), select 15/32” with span rating of 32/16. Total load capacity is 40 psf at a span rating of 32”. Adjusting for actual load capacity: 2

2

2

2

L   32  WTL =  max  WTL =   ( 40psf ) = 71 psf > 60 psf, OK for total loads  24   La  L   32  WLL =  max  WTL =   (30psf ) = 53 psf > 50 psf, OK for total loads  24   La  Without edge support: From IBC Table 2304.7(3), select 19/32” with span rating of 40/20.

From Table 6-3, minimum fastening: 8d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing)

Chapter 6

Solutions

VT =

b)

S6-13

w T L ( 275)(100) = = 13,750 lb.

2 2

West side

East side

Unit shear in diaphragm:

Unit shear in diaphragm:

vT =

VT 13,750 lb = = 327.4 plf b 42'

V 13,750 lb vT = T = = 275 plf b 50'

Chapter 6

Solutions

S6-14

c)

Vcap = (1.4)(285) = 399plf > 327plf, OK

d) West side

Unit shear in SW1 & 2: v SW =

VT 13,750 lb = = 458.3 plf LSW 15'+15'

East side

Unit shear in SW3: v SW =

VT 13,750 lb = = 458.3 plf LSW 30'

Chapter 6

Solutions

S6-15

e)

f) Load Path: Wind to wall sheathing→wall studs→top/bottom plates→diaphragm→shearwalls→foundation

Solutions

Chapter 6

S6-16

6.8 a)

b)

a) Unit shear in diaphragm:

VT =

w T L ( 427)( 40) = = 8533lb.

2 2

V 8533lb vT = T = = 356 plf b 24'

Chapter 6

Solutions

vd = 356plf < vcap = 425plf, OK 15/32”, Struct I, 10d nails @ 12” o.c. (field) and 4” o.c. (edge) c) Load path: Soil lateral load→CMU wall→footing @ bot / floor framing @ top→diaphragm→shearwalls→foundation

S6-17

Chapter 6

Solutions

S6-18

Chapter 6

6.9 V=

w L L (320)(95) = = 15200 lb.

2 2

vd =

V 15200 lb = = 310.2 plf b 49'

310.2 = 6.53 95' 2 310.2 − 240 = 10.75ft , say 11 ft.

6.53

Solutions

S6-19

Solutions

Chapter 6

S6-20

6.10 LA = 60ft

LB = 50ft

LdA = 23ft

LdB = 36ft

ww = 240plf

LA + LB = 13200 lbf 2

LA P1 = ww = 7200 lbf 2

P2 = ww

Ls1 = 8ft

Ls2 = 13ft

LB P3 = ww = 6000 lbf 2 Ls3 = 12ft

P1 vd1 = = 313 plf LdA

P3 vd2 = = 167 plf LdB

P1 vs1 = = 900 plf Ls1

P2 vs2 = = 1015 plf Ls2

P3 vs3 = = 500 plf Ls3

vnet1 = vs1 − vd1 = 587 plf

vnet2 = vs2 − vd1 − vd2 = 536 plf

vnet3 = vs3 − vd2 = 333 plf

Pds1 = vd1 15ft = 4696 lbf

Pd2 = vd1 + vd2 23ft = 11033 lbf

(

)

vreq’d = (313)/1.4 = 223plf

Select 3/8 STRUCT I, 8d nails @12” o.c. (field) and 6” o.c. (edge)

Pds3 = vd2 12ft = 2000 lbf

Chapter 6

6.11

Solutions

S6-21

Chapter 6

Solutions

S6-22

Solutions

Chapter 6

S6-23

6.12 LA = 100ft

ww = 400plf

LdA = 56ft

LA P1 = ww = 20000 lbf 2

vdi =

vd1 = 7.143 psf  LA 

   2 

Ls1 = 18ft



Ls1   = 9000 lbf  Ls1 + Ls2 

P1 vd1 = = 357 plf LdA

Ldx =

vd1 − 240plf = 16.4 ft vdi

say 17ft, part a

Ls2 = 22ft



Ls2   = 11000 lbf  Ls1 + Ls2 

Ps1 = P1 

Ps2 = P1 

Ps1 vs1 = = 500 plf Ls1

Ps2 vs2 = = 500 plf Ls2

vnet1 = vs1 − vd1 = 143 plf

vnet2 = vs2 − vd1 = 143 plf

Pds1 = vnet1 Ls1 = 2571 lbf

Pds2 = vnet2 Ls2 = 3143 lbf

part b

Chapter 6

6.13

Solutions

S6-24

Chapter 6

6.14

Solutions

S6-25

Solutions

Chapter 7

7-1

S7-1

For the building plan shown below, and based on the following given information:

• Wall framing is 2x6 @ 16” O.C., Hem-Fir, Select Structural • Normal temperature and moisture conditions apply • Roof Dead Load = 15psf, Wall Dead Load = 10 psf, Snow Load, Pf = 40 psf • Floor-to-Roof height = 18’ • Consider loads in the N-S direction only Determine the following: a) For the 18’ and 20’ shear walls on the East and West face only, select the required plywood thickness, fastening, and edge support requirements. Assume the plywood is applied directly to the framing and is on one side only. Give the full specification. b) Determine the maximum tension force in the shear wall chords of the 20’ long shear wall. c) Determine the maximum compression force in the shear wall chords of the 20’ long shear wall Solution: a)

Unit shear in the shearwalls: VT =

w T L (300)(90) = =13,500 lb.

2 2

vW =

VT 13,500 = = 355.3 plf L W 20' + 18'

Since Hem-Fir framing is used, the allowable shear capacity is decreased.

SGAFH-F = 1-(0.5-) = 1 – (0.5-0.43) = 0.93 hem-fir = 0.43 (NDS Table 11.3.2A) Required shear capacity: v r eqd =

vW 355.3plf = 382plf = SGAF 0.93

Adjusting for the 40% increase in capacity for wind loads:

v r eqd =

382plf = 273plf 1.4

Solutions

Chapter 7

S7-2

From IBC Table 2306.4(1), select 15/32” Structural I sheathing, vcap = 280 plf > 273 plf Required fasteners: 8d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing)

b)

Check shearwall aspect ratio: (SW1 is 20’ long) h  3.5 (2.0 for seismic loads) w 18' = 1.0  3.5  SW1 meets aspect ratio requirements for wind loads 18'

Lateral load to shearwall - there are 2 shearwalls in the N-S direction:

  length of SW1  (Fr ( N −S) ) FSW1 =    of all shear wall lengths  =

 20'   (13,500)  20' + 18' 

FSW1 = 7106 lb.

Gravity loads on SW1: Direct load on wall:  90'  RD = (15 psf )   (20' )  2 

= 13,500 lb.

 90'  RS = (40 psf )   (20' )  2 

= 36,000 lb.

WD = (10psf)(18’)(20’)

= 3,600 lb.

Reaction from header (load to other end is 0):  90'  12'  PD = (15 psf )    2  2 

= 4,050 lb.

 90'   12'  PS = (40 psf )     2  2 

= 10,800 lb.

Solutions

Chapter 7

Tension Chord Force, SW1: OM1 = (7106) (18’) = 127,908 ft.-lb.

  20'  RM D1 = (13,500 + 3,600)  = 171,000  2  

c)

T1 =

OM1 − 0.6RM D1 w

T1 =

(127,908) − (0.6)(171,000) = 1,265 lb 20'

Compression chord force

  1.33'  RMT1 = (10,800)(20’) + (36,000)  = 240,000 ft. lb.  2     1.33'  RMD1 = (4,050)(20’) + (13,500 + 3,600)  = 92,400 ft. lb.  2   C1 =

(0.75OM1 + RM D1 + 0.75RM T1 ) (eq. 7-8) w

C1 =

(0.75)(127,908) + (92,400) + (0.75)(240,000) = 18,417 lb.

20'

S7-3

Solutions

Chapter 7

7-2

S7-4

For the building plan shown below for a one story building and based on the following given information:

Floor-to-Roof height = 18’

• Wall framing is 2x6 @ 16” O.C., DF-L, No. 1 • Normal temperature and moisture conditions apply • Roof Dead Load = 15psf, Wall Dead Load = 10 psf, Roof Live Load = 20 psf • Floor-to-Roof height = 18’ • Applied wind load is 250plf to the diaphragm

Determine the following: a) For the typical 19’ shear wall, select the required plywood thickness, fastening, and edge support requirements. Assume the plywood is applied directly to the framing and is on two sides. Give the full specification. b) Determine the maximum tension force in the shear wall chords for the typical shear wall. c) Determine the maximum compression force in the shear wall chords for the interior shear wall. d) Select an appropriate anchor rod layout to resist the applied lateral loads..

Solution: a)

Unit shear in the shearwalls:

VT =

w T L (250)(132) = =16,500 lb.

2 2

vW =

VT 16,500 = = 435 plf L W 19' + 19'

Required shear capacity (sheathing on 2 sides):

v r eqd =

v W 435plf = 218plf = 2 2

Adjusting for the 40% increase in capacity for wind loads: v r eqd =

218plf = 156plf 1.4

From IBC Table 2306.4(1), select 3/8” Structural I sheathing, vcap = 230 plf > 156 plf Required fasteners: 8d nails at 12” O.C. (field nailing) and 6” O.C. (edge nailing).

Note: 3/8” is a minimum practical size for wall sheathing.

Solutions

Chapter 7

b)

S7-5

Check shearwall aspect ratio: (SW1 is 19’ long) h  3.5 (2.0 for seismic loads) w 18' = 0.95  3.5  SW1 meets aspect ratio requirements for wind loads 19'

Lateral load to shearwall - there are 2 shearwalls in the N-S direction:

  length of SW1  (Fr ( N −S) ) FSW1 =    of all shear wall lengths  =

 19'   (16,500) 19 ' + 19 '  

FSW1 = 8,250 lb.

Gravity loads on SW1: Direct load on wall:  12'  RD = (15 psf )   (19' )  2 

= 1,710 lb.

 12'  RLr = (20 psf )   (19' )  2 

= 2,280 lb.

WD = (10psf)(18’)(19’)

= 3,420 lb.

Reaction from header (load to other end is 0):  12'  12'  PD = (15 psf )    2  2 

= 180 lb.

 12'   12'  PLr = (20 psf )     2  2 

= 240 lb.

Solutions

Chapter 7

Tension Chord Force, SW1: OM1 = (8,250) (18’) = 148,500 ft.-lb.

  19'  RM D1 = (180)(19) + (1,710 + 3,420)  = 52,155 ft.-lb.  2  

c)

T1 =

OM1 − 0.6RM D1 (eq. 7-4) w

T1 =

(148,500) − (0.6)(52,155) = 6,168 lb ( net uplift) 19'

Compression chord force

  1.33'  RMT1 = (240)(19’) + (2,280)  = 6,080 ft. lb.  2     1.33'  RMD1 = (180)(19’) + (1,710 + 3,420)  = 6,840 ft. lb.  2   C1 =

(0.75OM1 + RM D1 + 0.75RM T1 ) (eq. 7-8) w

C1 =

(0.75)(148,500) + (6,840) + (0.75)(6,080) = 6,462 lb.

19'

S7-6

Solutions

Chapter 7

d)

S7-7

Sill anchors

The base shear is V1 = 8,250 lb.; the allowable shear parallel to the grain in the sill plate is: Z’ = (Z)(CD)(CM)(Ct)(Cg)(C)(Ceg)(Cdi)(Ctn) CD = 1.6 (wind) CM = 1.0 (MC  19% ) Ct = 1.0 (T  100 F ) Cg, C, Ceg, Cdi, Ctn = 1.0 Zll = 650 Ib (1/2” bolts - NDS Table 11E) The Allowable Shear per bolt Z’ = (Z) (CD) (CM) (Ct) (Cg) (C) (Ceg) (Cdi) (Ctn) = (650) (1.6) (1.0) (1.0) (1.0) (1.0) (1.0) (1.0) (1.0) = 1040 lb per bolt Number of sill anchor bolts required = (8,250 lb)/(1040 lb/bolt)  8 bolts Max bolt spacing:

=

(19' wall length − 1' edge dist − 1' edge dist ) (8 bolts −1)

Use ½” dia anchor bolts @ 2’ – 4” o.c.

= 2.43’  2’-4”

Chapter 7

7-3

Solutions

S7-8

Determine the allowable unit shear capacity for a shear wall panel under both wind and seismic loads assuming the following design parameters: Exterior face of wall: 15/32” plywood with 8d nails @ 4” o.c.(EN); 12” o.c.(FN) Interior face of wall: GWB with allowable unit shear of 175 plf

Solution: From IBC Table 2306.4.1, vcap = 380 plf (Struct I not specified) Adjusting for the 40% increase in capacity: vcap = (1.4)(380 plf) = 532 plf (Note: the 40% increase only applies to plywood panels) For wind loads, the total shear capacity is the direct summation of the capacities on both sides. vcap = (532 plf) + (175 plf) = 707 plf (wind loads) For seismic loads, the total shear capacity is the larger of the shear capacity on each face since the shear capacity of dissimilar materials cannot be summed. vcap = 380 plf (seismic loads) Note that the seismic response or ‘R’ value for wood and GWB walls is different and thus the seismic force to the wall for each is also different. However, wood shearwalls have a higher ‘R’ value and thus a lower seismic force.

Solutions

Chapter 7

7-4

S7-9

Determine the allowable unit shear capacity for a shear wall panel under both wind and seismic loads assuming the following design parameters: Exterior face of wall: 15/32” plywood with 8d nails @ 4” o.c.(EN); 12” o.c.(FN) Interior face of wall: 3/8” plywood with 8d nails @ 4” o.c.(EN); 6” o.c.(FN)

Solution: From IBC Table 2306.4.1 (Struct I not specified) vcap = 380 plf (15/32”) vcap = 320 plf (3/8”) Adjusting for the 40% increase in capacity: vcap = (1.4)(380 plf) = 532 plf (15/32”) vcap = (1.4)(320 plf) = 448 plf (3/8”) For wind and seismic loads, the total shear capacity is the capacity of the stronger side, or twice the capacity of the weaker side, whichever is greater. vcap = 532 plf or vcap = 448 plf + 448 plf = 896 plf (controls for wind)

vcap = 380 plf or vcap = 320 plf + 320 plf = 640 plf (controls for siesmic)

Solutions

Chapter 7

7-5

S7-10

Design a plywood lap splice between floors using sawn lumber with 2x6 wall studs and 7/16” wall sheathing for a net uplift of 600 plf. Stud Spacing is 16” and lumber is DF-L No. 1.

Solution: Ft = 675psi CD = 1.6 CF = 1.3 A = 8.25 in.2 F’t = FtCDCMCtCFCi F’t = (675)(1.6)(1.0)(1.0)(1.3)(1.0) = 1404 lb.

Tmax = F’tA = (1404)(8.25) = 11,583 lb.

Uplift force on each stud:  16"  T = (600plf )  = 800 lb. < 11,583 lb, OK  12 

Fastener Design: 8d nails will be assumed since they are the smallest nail size allowed for 7/16” shearwall panels (see IBC Table 2306.4.1).

From NDS Table 11Q: Z = 73 lb/nail Z’ = ZCD = (73)(1.6) = 116.8 lb/nail Req’d fasteners each side of the splice:

n reqd =

800 = 6.85 → use 7 nails each side of the splice 116.8

Minimum spacing: (L = 2.5”, D = 0.131” for 8d common nail) End distance = 15D = (15)(0.131) = 1.96 in.

Edge distance = 2.5D = (2.5)(0.131) = 0.33 in. c-c spacing = 15D = (15)(0.131) = 1.96 in. min. penetration = 10D = (10)(0.131) = 1.31” (provided: 2.5” – 7/16” = 2.06”)

Chapter 7

7.6

Solutions

S7-11

Chapter 7

Solutions

S7-12

Solutions

Chapter 7

S7-13

7.7 LA = 60ft

LB = 50ft

LdA = 23ft

LdB = 36ft

ww = 240plf

LA + LB = 13200 lbf 2

LA P1 = ww = 7200 lbf 2

P2 = ww

Ls1 = 8ft

Ls2 = 13ft

LB P3 = ww = 6000 lbf 2 Ls3 = 12ft

P1 vd1 = = 313 plf LdA

P3 vd2 = = 167 plf LdB

P1 vs1 = = 900 plf Ls1

P2 vs2 = = 1015 plf Ls2

P3 vs3 = = 500 plf Ls3

vnet1 = vs1 − vd1 = 587 plf

vnet2 = vs2 − vd1 − vd2 = 536 plf

vnet3 = vs3 − vd2 = 333 plf

Pds1 = vd1 15ft = 4696 lbf

Pd2 = vd1 + vd2 23ft = 11033 lbf

(

)

Pds3 = vd2 12ft = 2000 lbf

vreq’d = (900)/1.4 = 643plf (SW2), 15/32 STRUCT I, 8d nails @12” o.c. (field), 4” o.c. (edge), each side, (2)(430)=860> 643, OK vreq’d = (1015)/1.4 = 725plf (SW2) 15/32 STRUCT I, 8d nails @12” o.c. (field), 4” o.c. (edge), each side, (2)(430)=860> 725, OK vreq’d = (500)/1.4 = 358plf (SW2) 15/32 STRUCT I, 8d nails @12” o.c. (field), 4” o.c. (edge), one side, 430 > 358, OK

Chapter 7

7.8

Solutions

S7-14

Chapter 7

Solutions

b) G = 0.42, SGAF = 1-(0.5-0.42) = 0.92 vreq’d = (400)/[(1.4)(0.92)] = 311plf

Select 15/32 STRUCT I, 10d nails @12” o.c. (field) and 6” o.c. (edge) c) Loads to each wall are the same since it is a flexible diaphragm.

S7-15

Chapter 7

7.9

Solutions

S7-16

Chapter 7

Solutions

a) Chord BD, worst case tension Direct load on wall: RD = (300plf + 150plf + 150plf ) (16' ) = 9600 lb.

WD = 1200lb+1200lb+1500lb= 3,900 lb.

Tension Chord Force: OMA = (3000)(39’) + (4000)(27’) + (4000)(15’) = 285,000 ft.-lb.

  16'  RM D1 = (9600 + 3900)  + (300 + 150 + 150)(16)= 117,600  2   TB =

OM A − 0.6RM D1 w

TB =

( 285,000) − (0.6)(117,600) = 13403 lb 16'

b) Chord GE, worst case compression; assume studs @ 16” o.c.

OMF = (3000)(12) = 36,000 ft.-lb

  1.33'  RMT1 = (1600)(16’) + ( 400)(16)  = 29867 ft. lb.  2     1.33'  RMD1 = (1200)(16’) + 1200 + (300)(16)  = 23200 ft. lb.  2   C1 =

(0.75OM1 + RM D1 + 0.75RM T1 ) w

C1 =

(0.75)(36000) + ( 23200) + (0.75)( 29867) = 4538 lb.

16'

S7-17

Solutions

Chapter 7

S7-18

7.10 Fw = 10kips

h1 = 18ft

OMA = Fwh1 = 180 ft kips

PDmin =

OMA = 30 kips 0.6 bwall

wwall = 3000lb

bwall = 10ft

Solutions

Chapter 7

S7-19

7.11 Fw = 8kips

h1 = 13ft

Fw Vsw = = 666.667 plf bwall

wwall = 2000lb

bwall = 12ft

use 7/16" Struct I, 8d nails, 1 3/8" penetration Va = (255plf)(1.4)(2) = 714plf > 667plf spacing" 6" o.c. at edges, 12" o.c. at field

OMA = Fwh1 = 104 ft kips

Tmax =

 wwallbwall    + PDL( dx1 + dx2) 2    = 6.717 kips

OMA − 0.6

bwall

dy2 = 9ft

dy1 = 6ft

PLL = 1000lb

Cmax =

dx2 = 6ft

dx1 = 3ft

PDL = 3000lb

(

)

(

)

 wwallbwall   + PDL( dy1 + dy2) 2   = 12.188 kips

( 0.75) OMA + ( 0.75) PLL dy1 + dy2  +  bwall

Solutions

Chapter 7

S7-20

7.12 Fw = 10kips

h1 = 20ft

Fw Vsw = = 666.667 plf bwall

Tmax =

use 7/16" Struct I, 8d nails, 1 3/8" penetration Va = (505plf)(1.4) = 707plf > 667plf spacing 3" o.c. at edges, 12" o.c. at field (1 side)

dx2 = 10ft

dx1 = 5ft

 wwallbwall    + PDL( dx1 + dx2) 2    = 9.883 kips

OMA − 0.6

bwall

dy1 = bwall − dx1 = 10 ft

PLL = 2000lb

Cmax =

bwall = 15ft

use 7/16" Struct I, 8d nails, 1 3/8" penetration Va = (255plf)(1.4)(2) = 714plf > 667plf spacing" 6" o.c. at edges, 12" o.c. at field (2 sides)

OMA = Fwh1 = 200 ft kips PDL = 4000lb

wwall = 3500lb

(

)

(

)

 wwallbwall   + PDL( dy1 + dy2) 2   = 17.25 kips

( 0.75) OMA + ( 0.75) PLL dy1 + dy2  +  bwall

dy2 = bwall − dx2 = 5 ft

Solutions

Chapter 8

8-1

S8-1

Calculate the nominal withdrawal design values, W for the following and compare the results with the appropriate table in the NDS Code. Lumber is Douglas Fir-Larch a.) Common Nails: 8d, 10d, 12d b.) Wood Screws: #10, #12, #14 c.) Lag Screws: ½”, ¾” Solution: G = 0.50 (DF-L → NDS Table 11.3.2A) a)

Nails: W = 1380G5/2D (eq. 8-23) From NDS Table L4: D = 0.131” (8d) = 0.148” (10d) = 0.148” (12d) W8d = (1380)(0.50)5/2(0.131) = 32.0 lb./in.

W10d = (1380)(0.50)5/2(0.148) = 36.1 lb./in.

W12d = (1380)(0.50)5/2(0.148) = 36.1 lb./in.

Values agree with NDS Table 11.2C

b)

Wood Screws: W = 2850G2D (eq. 8-22) From NDS Table L3: D = 0.19” (#10) = 0.216” (#12) = 0.242” (#14) W#10 = (2850)(0.50)2(0.19) = 135.3 lb./in.

W#12 = (2850)(0.50)2(0.216) = 153.9 lb./in.

W#14 = (2850)(0.50)2(0.242) = 172.4 lb./in.

Values agree with NDS Table 11.2B

Solutions

Chapter 8

c)

Lag Screws: W = 1800G3/2D3/4 (eq. 8-21) W1/2” = (1800)(0.50)3/2(1/2”)3/4 = 378.4 lb./in.

W3/4” = (1800)(0.50)3/2(3/4”)3/4 = 512.9 lb./in.

Values agree with NDS Table 11.2A

S8-2

Solutions

Chapter 8

8-2

S8-3

Calculate the group action factor for the following connection and compare the results with 10.3.6A and explain why the result are different. Lumber is Hem-Fir Select Structural.

Solution: Es = Em = 1.6x106 psi As = Am = 8.25 in.2 R EA = the lesser of :

EsAs E A or m m = 1.0 EmAm EsAs

D = 0.625” s = 5”  = (180,000)(D1.5) = (180,000) (0.6251.5) = 88,939

s 1 1  u =1 +   +  2  E m A m EsAs  u = 1 + (88,939)

 (5" )  1 1 +   = 1.034 6 6 2  (1.6 x10 )(8.25) (1.6 x10 )(8.25) 

m = u − u 2 −1 m = (1.034) − (1.034) 2 −1 = 0.771   1 + R EA  m(1 − m 2 n ) Cg =   n 2n   n[(1 + R EA m )(1 + m) − 1 + m )]   1 − m     1 + (1.0)  (0.771)(1 − 0.771( 2 )(6) ) Cg =   = 0.904 6 ( 2 )( 6 )   )]  1 − (0.771)   (6)[(1 + (1.0)(0.771) )(1 + 0.771) − 1 + 0.771 From NDS Table10.3.6A with As/Am = 1.0 and with As = 8.25in2, the group action factor Cg = 0.82. This table is based on values of D = 1” and = 1.4x106, and s = 5” – which is conservative for most cases.

Solutions

Chapter 8

8-3

S8-4

For the following connectors, calculate the dowel bearing strength parallel to the grain (Fe), perpendicular to the grain (Fe⊥) and at an angle of 30 degrees to the grain Fe).

Lumber is Spruce-Pine-Fir. Compare the results with Table 11.3.2 of the NDS Code. a.) 16d common nail b.) Wood screws: #14, #18 c.) Bolts: 5/8”, 1” Solution: G = 0.55 (S-P-F → NDS Table 11.3.2A) a) 16d common nail D = 0.162” (NDS Table L4) Fe = Fe⊥ = Fe = 16600G1.84 = (16600)(0.55)1.84 = 5,525 psi Agrees with NDS Table 11.3.2 (Fe = 5550 psi) b) Wood Screws From NDS Table L3: D = 0.242” (#14) D = 0.294” (#18) #14 wood screw: Fe = Fe⊥ = Fe = 16600G1.84 = (16600)(0.55)1.84 = 5,525 psi Agrees with NDS Table 11.3.2

#18 wood screw: Fe = 11200G = (11200)(0.55) = 6160 psi

Fe ⊥ =

6,100G 1.45 D

=

(6,100)(0.55)1.45 0.294

= 4728 psi

Agrees with NDS Table 11.3.2

Fe =

Fe Fe ⊥ Fe sin  + Fe ⊥ cos  2

2

=

(6160)(4728) = 5726 psi (6160) sin 2 30 + (4728) cos 2 30

Solutions

Chapter 8

S8-5

c) Bolts 5/8” bolt: Fe = 11200G = (11200)(0.55) = 6160 psi

Fe ⊥ =

6,100G 1.45 D

=

(6,100)(0.55)1.45 0.625

= 3242 psi

Agrees with NDS Table 11.3.2 Fe =

Fe Fe ⊥ Fe sin  + Fe ⊥ cos  2

2

=

(6160)(3242) = 5028 psi (6160) sin 2 30 + (3242) cos 2 30

1” bolt: Fe = 11200G = (11200)(0.55) = 6160 psi

Fe ⊥ =

6,100G 1.45 D

=

(6,100)(0.55)1.45 1

= 2563 psi

Agrees with NDS Table 11.3.2

Fe =

Fe Fe ⊥ Fe sin  + Fe ⊥ cos  2

2

=

(6160)(2563) = 4560 psi (6160) sin 2 30 + (2563) cos 2 30

Solutions

Chapter 8

8-4

Using the yield limit equations, determine the nominal lateral design value, Z, for the connection shown below. Compare the results with the appropriate table in the NDS Code. Lumber is Douglas Fir-Larch.

Solution G = 0.50 (NDS Table 11.3.2A) D = 0.625” lm = 1.5” (2x main member) Fyb = 45,000 psi (Table 8-4) Fe = 11200G = (11200)(0.50) = 5600 psi (agrees with NDS Table 11.3.2) Re = Fem/Fes = 5600/5600 = 1.0 Rt = lm/ls = 1.5”/ 1.5” = 1.0 = 0 K = 1+0.25(/90) = 1+0.25(0/90) = 1.0

k 3 = − 1+

2(1 + R e ) 2Fyb (2 + R e )D + 2 Re 3Fem l s

k 3 = − 1+

2(1 + 1) (2)(45,000)(2 + 1)(0.625) 2 + = 1.606 1 (3)(5600)(1.5) 2

2

Calculate Z for each mode of failure:

Mode Im : (Rd = 4K)

Z=

D l m Fem (0.625)(1.5)(5600) = = 1313 lb.

Rd (4)(1)

Mode Is : (Rd = 4K)

Z=

2D l s Fes (2)(0.625)(1.5)(5600) = = 2625 lb.

Rd (4)(1)

Mode II: N/A

S8-6

Solutions

Chapter 8

S8-7

Mode IIIm : N/A Mode IIIs : (Rd = 3.2K)

Z=

2k 3 D l s Fem (2)(1.606)(0.625)(1.5)(5600) = 1756 lb. = (2 + R e )R d (2 + (1))(3.2)(1.0)

Mode IV:

2D 2 Z= Rd

(2)(0.625) 2 = 3(1 + R e ) (3.2)(1) 2Fem Fyb

(2)(5600)(45,000) = 2238 lb.

3(1 + (1))

The lowest value was 1313 lb. (Mode Im), therefore Z = 1313 lb., which agrees with Table 11F of the NDS code (Z = 1310 lb).

Chapter 8

8-6

ZII = 1800lb/bolt = 3600lb (6x6) Zperp = 540lb/bolt = 1080lb (2x12) Rmax = 1080lb

Solutions

S8-8

Chapter 8

8.7

Solutions

S8-9

Chapter 8

Solutions

S8-10

Solutions

Chapter 8

Tread: A = 16.88 in.2 Sy = 4.219 in.3 (16.88)(0.43)(62.4)/144 = 3.15 plf Fb = 850psi Fv = 150psi Cfu = 1.2 CF = 1.0

Iy = 3.164 in.4

S8-11

self wt =

Fc = 405psi E = E' = 1300ksi G = 0.42 (1020)( 4.219) F'b = (850)(1.2) = 1020psi Mmax = = 358 ft-lb (4303 in.-lb.) ← controls design 12 (150)(16.88) F'v =150psi Vmax = = 1688 lb (Lmax is practically unlimited) 1.5 F'c =405psi Rmax = (405)(1.5)(11.25) = 6834 lb (Lmax is practically unlimited) Load cases:

Case 1: Lmax =

8M = w

(8)(358) = 5.43 ft.

97.15

Case 2:

(3.15) L2 (300) L + 8 4 Solving for L = 4.66 ft. (controls) Mmax = 358 =

5wL4 PL3 (5)(3.15 / 12)( 4.66x12) 4 (300)( 4.66) 3 + = + = 0.008+0.266 = 0.274 in.

384EI 48EI (384)(1.3E6)(3.164) ( 48)(1.3E6)(3.164) (L/204) b)  =

Chapter 8

Solutions

c)

Rmax = (0.5)(3.08)(4.66) + 300 = 308 lb. < (2)(280) = 560 lb., OK

S8-12

Solutions

Chapter 8

S8-13

8.8 G = 0.43

CD = 0.9

p = 2in

D = 0.216in W = 2850

lb 2  D  lb G   = 113.824 in in  1in 

WT = W p = 227.649 lbf W' = WTCD = 204.884 lbf Pmax = Nb W' = 410 lbf

per inch

spa = 24in

Nb = 2

Solutions

Chapter 8

S8-14

8.9 use #12 wood screw and 20d nail

T = 800lb

Ds = 0.216in

CD = 1

Dn = 0.192in

T lb Wreqd = = 160 p in

Gs =

Wreqd  in   1   1    = 0.51 lb  Ds   2850  1 in

2

Gn = 5

Wreqd  in  1  1  lb  Dn  1380 1 in

Wreqd  in   1   1   lb  Dn   1380  1 in

= 0.86

use southern pine

none worked

p = 5in

Chapter 8

8.10 Nail Capacity:

Solutions

Vmax = 120lb/nail : (6)(120) = 720lb at splice

Screw capacity: Vmax = 101 lb/screw, D = 0.19in Req’d number of screws: 720/101 = 7.1 – say 8 screws Spacing: Edge = 2.5D = 0.475in < 1.25 in, OK Centers = 15D = 2.85in < 3in, OK Row = 5D = 0.95in < 3in, OK End = 15D = 2.85in > 2in, NG, need to increase to 3in

S8-15

Chapter 8

8.11

Solutions

S8-16

Chapter 8

Solutions

S8-17

Solutions

Chapter 8

8-12

S8-18

For the sill plate connection shown below, calculate the design shear capacity, Z’, parallel to the grain. The concrete strength is f’c = 3,000 psi and the lumber is SprucePine-Fir. Loads are dues to Seismic (E), and normal temperature and moisture conditions apply.

Solution: G = 0.42 (Table 11.3.2A NDS Code) D = 0.625” (Table L4 NDS Code) l = 7” (Table L4 NDS Code) Fyb = 45,000 psi (Table 8-4) Fes = Fe = 11200G = (11200)(0.42) = 4704 psi (agrees with NDS Table 11.3.2) Fem = 7500 psi (see NDS Table 11E) Re = Fem/Fes = 7500/4704 = 1.594 Rt = lm/ls = 7”/ 1.5” = 4.67 = 0 K = 1+0.25(/90) = 1+0.25(0/90) = 1.0 2

k1 =

k1 =

(

)

R e + 2R e 1 + R t + R t + R t R e − R e (1 + R t ) 2

2

3

(1 + R e )

(

)

1.594 + (2)(1.594) 2 1 + 4.67 + (4.67) 2 + (4.67) 2 (1.594) 3 − (1.594)(1 + 4.67) ) =2.35 (1 + (1.594))

k 2 = − 1+ 2(1 + R e ) +

2Fyb (1 + 2R e )D 2

k 2 = − 1 + 2(1 + (1.594)) +

3Feml m

2

(2)(45,000)(1 + (2)(1.594) )(0.625) 2 = 1.31 (3)(7500)(7) 2

2(1 + R e ) 2Fyb (2 + R e )D k 3 = − 1+ + 2 Re 3Fem l s

2

2(1 + (1.594)) (2)(45,000)(2 + (1.594) )(0.625) 2 k 3 = − 1+ + = 1.40 (1.594) (3)(7500)(1.5) 2

Solutions

Chapter 8

S8-19

Calculate Z for each mode of failure: Mode Im : (Rd = 4K)

Z=

D l m Fem (0.625)(7)(7500) = 8203 lb = Rd (4)(1.0)

Mode Is : (Rd = 4K)

Z=

D l s Fes (0.625)(1.5)(4704) = = 1102 lb Rd (4)(1.0)

Mode II: (Rd = 3.6K)

Z=

k 1 D l s Fes (2.35)(0.625)(1.5)(4704) = = 2878 lb.

Rd (3.6)(1.0)

Mode IIIm : (Rd = 3.2K)

Z=

k 2 D l m Fem (1.31)(0.625)(7)(7500) = = 3207 lb.

(1 + 2R e )R d (1 + (2)(1.594))(3.2)(1.0)

Mode IIIs : (Rd = 3.2K)

Z=

k 3 D l s Fem (1.40)(0.625)(1.5)(7500) = (2 + R e )R d (2 + (1.594))(3.2)(1.0)

= 855 lb.

Mode IV: (Rd = 3.2K)

D2 Z= Rd

(0.625) 2 (2)(7500)(45,000) = = 1136 lb.

3(1 + R e ) (3.2)(1.0) 3(1 + (1.594)) 2Fem Fyb

The lowest value was 855 lb. (Mode IIIs), therefore Z = 855 lb., which agrees with Table 11E of NDS Table 11E (Z = 850 lb).

Solutions

Chapter 8

8-13

S8-20

Design a 2x6 lap splice connection using: a.) Wood Screws b.) Bolts Applied load is 1,800 lb. due to wind. Lumber is Southern Pine.

Solution: A lap splice length of 6’-0” is assumed. a) Wood Screws Try #12 wood screws, 3” long D = 0.216” From NDS Table 11L, Z = 161 lb.

Minimum spacing (see Table 8-1): end distance = 15D = (15)(0.216) = 3.24” use 4”min. edge distance = 2.5D = (2.5)(0.216) = 0.54” use 1”min. center-center spacing = 15D = (15)(0.216) = 3.24” use 4”min. row spacing = 5D = (5)(0.216) = 1.08” use 3.5” (see edge distance calc.)

Penetration: 6D < p < 10D (see Table 8-2) (6)(0.216) = 1.30” (minimum) (10)(0.216) = 2.16” (maximum) Since the provided penetration p = 1.5” is less than 10D, the allowable lateral load must be reduced by p/10D (see footnote 3 in NDS Table 11L).

Reduction: = 1.5”/2.16” = 0.694 Z = (161 lb.)(0.694) = 111.8 lb.

Allowable lateral resistance per screw, Z’ = ZCDCMCtCgCCegCdiCtn CD = 1.6 (wind) CM = 1.0 (MC  19% ) Ct = 1.0 (T  100 F ) Cg, C, Ceg, Cdi, Ctn = 1.0

Solutions

Chapter 8

S8-21

The allowable shear per screw, Z’ = ZCDCMCtCgCCegCdiCtn = (111.8)(1.6)(1.0)(1.0)(1.0)(1.0)(1.0)(1.0)(1.0) = 179 lb. per screw No. of screws req’d = T/ Z’ = 1,800 Ib / 179 Ib = 10.1, use 12 screws.

Use (12) - #12, 2 rows each side of splice. Row spacing is 3.5”, c-c spacing is 4”, end distance is 4”.

b) Bolts Assume ½” through bolts.

From NDS Table 11A, Z = 530 lb.

Minimum spacing (see Table 8-3, assume full design value): End Distance: 7D = (7)(0.5”) = 3.5” Edge Distance: 1.5D = (1.5)(0.5”) = 0.75” l/D = 1.5”/0.5” = 3 c-c spacing: 4D = (4)(0.5”) = 2” row spacing: 1.5D = N/A → bolts provided in single row Allowable lateral resistance per bolt, Z’ = ZCDCMCtCgCCegCdiCtn CD = 1.6 (wind) CM = 1.0 (MC  19% ) Ct = 1.0 (T  100 F ) Cg, C, Ceg, Cdi, Ctn = 1.0 The allowable shear per screw, Z’ = ZCDCMCtCgCCegCdiCtn = (530)(1.6)(1.0)(1.0)(1.0)(1.0)(1.0)(1.0)(1.0) = 848 lb. per bolt No. of bolts req’d = T/ Z’ = 1,800 Ib / 848 lb = 2.2, use 3 bolts.

Use (3) – ½” bolts each side of splice in a single row, centered in splice. C-C spacing is 2”, end distance is 3.5”.