Chapter 1 Solutions S 1 - 1 1 - 10 Determine the total shrinkage across the width and thickness of a green triple 2x4 Douglas Fir Larch top plates loade d 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, d 1 = 3.5 in. M 1 = 30 and M 2 = 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 d 2 is given as, 6.031- 0.215 (12) 1- 100 d = 3.5 2 6.031- 0.215(30) 1- 100             = 3.365 in. Thus, the total shrinkage across the width of the (triple) 2x4 is d 1 – d 2 = 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 d 2 of each plate is given as, 5.062 - 0.181(12) 1- 100 d =1.5 2 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 (d 1 – d 2 ) = 3(1.5 in. – 1.451 in.) = 0.147 in. Chapter 1 Solutions S 1 - 2 1 - 11 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, d 1 = 11.25 in. M 1 = 25 and M 2 = 1 2 (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 d 2 is given as, 6.031-0.215 (12) 1- 100 d =11.25 2 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( d 1 – d 2 ) = 2( 11.25 in. – 10.93 in. ) = 0. 6 4 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 d 2 of each plate is given as, 5.062 - 0.181(12) 1- 100 d =1.5 2 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 S 1 - 3 7, 2x6 plates x (d 1 – d 2 ) = 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. 6 4 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 Solutions S 2 - 1 2 - 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 T he total dead load in psf of horizontal plan area will be: DL 2 2 6 +12 w = D 12           , psf of horizontal plan area ( ) =18.4psf 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, L 2 = 14 ft • Roof slope: 4 - in - 12 • Rafter or Truss spacing = 4’ 0 Solutions: Sloped length of rafter, L 1 = ( ) 2 2 4 +12 14 14.8 12           = ft ft Chapter 2 Solutions S 2 - 2 Using the load combinations in section 2.1, the total load in psf of horizontal plan area will be: TL L 1 w = D L 2         + (L r or S or R), psf of horizontal plan area 14.8' = 20 psf 14'         + 40 psf = 61.1 psf of horizontal plan area The total load in pounds per horizontal linear foot (Ib/ft) is given as, w TL (Ib/ft) = w TL (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, ( ) 2 TL 2 L w L 2 H = h         = ( ) 14' 244.4 Ib/ft 14' 2 4.67'         = 5129 Ib. The collar or ceiling ties must be designed to resist this horizontal thrust. L 2 = 1 4 ’ The maximum shear force in the rafter is, 2 max TL L V = w 2         = 14' 244.4 2         = 1711 Ib The maximum moment in the rafter is, ( ) 2 TL 2 max w L M = 8 = ( ) 2 244.4 14' 8 = 5989 ft - Ib = 5.9 ft - kip 2 - 3. 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. The uniform total load on the typical roof joist in Ib/ft b. The uniform total load on the typical roof girder in Ib/ft c. The total axial load on the typical interior column, in Ib. d. The total axial load on the typical perimeter column, in Ib Chapter 2 Solutions S 2 - 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 Tributary Width (TW) Tributary Area (TA) Purlin 10' 10' + =10' 2 2 10 ’ x 20 ’ = 200 ft 2 Glulam girder 20' 20' + = 20' 2 2 20 ’ x 60’ = 1200 ft 2 Typical interior Column 20' 20' 60' 60' + + = 2 2 2 2                 1200 ft 2 Typical perimeter Column 20' 20' 60' + = 2 2 2                 600 ft 2 Since the snow and rain load are both zero, the roof live load, L r 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 ft 2 < 200 ft 2 From section 2.5.1, we obtain:R 1 = 1.0 and R 2 = 1.0, Using equation 2 - 4 gives the roof live load, L r = 20 x 1 x 1 = 20 psf The total loads are calculated as follows: w TL (psf) = (D + L r ) = 20 + 20 = 40 psf w TL (Ib/ft) = w TL (psf) x tributary width (TW) = 40 psf x 10 ft = 400 Ib/ft Glulam Girder: The tributary width, TW = 20 ft and the tributary Area, TA = 1200 ft 2 Thus, TA > 600, and from section 2.4 , we obtain: R 1 = 0.6, and R 2 = 1.0 Using equation 2 - 4 gives the roof live load, L r = 20 x 0.6 x 1 = 12 psf The total loads are calculated as follows: w TL (psf) = (D + L r ) = 20 + 12 = 32 psf w TL (Ib/ft) = w TL (psf) x tributary width (TW) = 32 psf x 20 ft = 640 Ib/ft Typical Interior Column: Chapter 2 Solutions S 2 - 4 The tributary tributary area of the typical interior column, TA = 1200 ft 2 Thus, TA > 600, and from section 2.4 , we obtain: R 1 = 0.6, and R 2 = 1.0 Using equation 2 - 4 gives the roof live load, L r = 20 x 0.6 x 1 = 12 psf The total loads are calculated as follows: w TL (psf) = (D + L r ) = 20 + 12 = 32 psf The Column Axial Load, P = 32 psf x 12 00 ft 2 = 38 , 4 00 Ib = 38 . 4 kips Typical Perimeter Column: The tributary tributary area of the typical perimeter column, TA = 600 ft 2 Thus, from section 2.5.1, we obtain: R 1 = 0.6, and R 2 = 1.0 Using equation 2 - 4 gives the roof live load, L r = 20 x 0.6 x 1 = 12 psf The total loads are calculated as follows: w TL (psf) = (D + L r ) = 20 + 12 = 32 psf The Column Axial Load, P = 32 psf x 600 ft 2 = 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 , L r : From Section 2.4, the roof slope factor is obtained as, F = 5  R 2 = 1.2 – 0.05 (5) = 0.95 Assume the tributary area (TA) of the rafter < 200 ft 2 ,  R 1 = 1.0 The roof live load will be, L r = 20R 1 R 2 = 20(1.0)(0.9 5 ) = 1 9 psf Chapter 2 Solutions S 2 - 5 Snow Load : Using IBC Figu re 1608.2 or ASCE 7 Figure 7 - 1 , the ground snow load, P g 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, C e = 0.9 (ASCE 7 Table 7 - 2) The thermal factor, C t = 1.0 (ASCE 7 Table 7 - 3) The Importance Factor, I = 1.0 ASCE Table 7 - 4 The slope factor, C s = 1.0 (ASCE Figure 7 - 2 with roof slope,  = 2 2 .6  and a warm roof) The flat roof snow load, P f = 0.7 C e C t I P g = 0.7 x 0.9 x 1.0 x 1.0 x 30 = 18.9 psf Minimum flat roof snow load, Pm = 20I s = 20 (1.0) = 20 psf (governs) Thus, the design roof snow load, P s = C s P f = 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, TL L 1 w = D L 2         + (L r or S or R) , psf of horizontal plan area Since the roof live load, L r (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. TL w 2 2 5 +12 = 22 psf 12           + 20 psf = 4 3 .8 3 psf of horizontal plan area The total load in pounds per horizontal linear foot (Ib/ft) is given as, w TL (Ib/ft) = w TL (psf) x Tributary width (TW) or Spacing of rafters = 4 3 .83 psf ( 2 ft) = 8 7 . 7 lb/ft. Assume L 2 = 14 ’ The maximum moment in the rafter is, ( ) 2 TL 2 max w L M = 8 = ( ) 2 87.7 14' 8 = 2149 ft - Ib = 2.1 5 ft - kip Chapter 2 Solutions S 2 - 6 2 - 5. 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, D roof = 20 psf Snow Load, S = 40 psf 2 nd and 3 rd Floor Loads: Dead Load, D floor = 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 ft 2 Roof Live Load , L r : From section 2.4 , the roof slope factor is obtained as, F = ¼ = 0.25  R 2 = 1.0 Since the tributary area (TA) of the column = 324 ft 2 ,  R 1 = 1.2 – 0.001 ( 324 ) = 0.8 8 The roof live load will be, L r = 20R 1 R 2 = 20(0.8 8 )(1.0) = 1 7.6 psf < Snow load, S = 40 psf The governing load combination from Section 2.1.1 for calculating the column axial loads is D + L + (L r 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 Member Levels supported A T (summation of floor tributary area) K LL Unreduced Floor live load, Lo (psf) Live Load Reduction Factor 0.25 + 15/  (K LL A T ) Design floor live load, L Chapter 2 Solutions S 2 - 7 3 rd floor Column (i.e. column below roof) Roof only Floor live load reduction NOT applicable to roofs!!! - - - 40 psf (Snow load) 2 nd floor column (i.e. column below 3 rd floor) 1 floor + roof 1 floor x 324 ft 2 = 324 ft 2 4 K LL A T = 1 296 > 400 ft 2  Live Load reduction allowed 5 0 psf 0.25 + 15/  (4 x 324) = 0.67 0.67 x 5 0 = 33.5 psf ≥ 0.50 Lo = 2 5 psf Ground or 1 st floor column (i.e. column below 2 nd floor) 2 floors + roof 2 floors x 324 ft 2 = 648 ft 2 4 K LL A T = 2592 > 400 ft 2  Live Load reduction allowed 5 0 psf 0.25 + 15/  (4 x 648) = 0.54 0.54 x 5 0 = 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 S 2 - 8 Column Load Summation Table Level Tributar y area, (TA) (ft 2 ) Dead Load, D (psf) Live Load, L o ( S or L r or R on the roof) (psf) Design Live Load Roof: S or L r or R Floor: L (psf) Unfactored total load at each level, w s1 Roof: D Floor: D + L (psf) Unfactored total load at each level,w s2 Roof: D +0.75 S Floor: D +0.75 L (psf) Unfactore d Column Axial Load at each level, P = (TA)(w s1 ) or (TA)(w s2 ) (kips) Cumulative Unfactored Axial Load,  P D + L (kips) Cumulative Unfactored Axial Load,  P D +0.75 L +0. 75 S (kips) Maximum Cumulative Unfactored Axial Load,  P (kips) With Floor Live Load Reduction Roof 324 20 40 40 20 50 6.5 or 16.2 6.5 16.2 16.2 3 rd Flr 324 40 50 33.5 73.5 65.1 23.8 or 21.1 30.3 37.3 37.3 2 nd Flr 324 4 0 5 0 2 7 67 6 0.3 21.7 or 19.5 52 56.8 56.8 Without Floor Live Load Reduction Roof 324 20 40 40 20 50 6.5 or 16.2 6.5 16.2 16.2 Third floor 324 4 0 50 50 90 77.5 29.2 or 25.1 35.7 41.3 41.3 Secon d floor 324 4 0 50 5 0 90 77.5 29.2 or 25.1 64.9 66.4 66.4 Chapter 2 Solutions S 2 - 9 2 - 6. 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 = 30 psf Roof Dead Load = 20 psf Exterior Walls = 10 psf Snow Load (P f ) = 40 psf Site Class = D Importance (I e )= 1.0 S S = 0.25% S 1 = 0.07% R = 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 = 115 mph ( ASCE 7 Table 26.5 - 1A ) Wind Pressures (from ASCE 7, Figure 28.6 - 1 ): Transverse (  = 25  ): Horizontal Vertical Zone A: 26.3 psf Zone E: - 11.7 psf Zone B: 4.2 psf Zone F: - 15.9 psf Zone C: 19.1 psf Zone G: - 8.5 psf Zone D: 4.3 psf Zone H: - 12 .8 psf Chapter 2 Solutions S 2 - 10 Longitudinal: (  = 0  ): Horizontal Vertical Zone A: 21 psf Zone E: - 2 5. 2 psf Zone B: N/A Zone F: - 14.3 psf Zone C: 13.9 psf Zone G: - 17.5 psf Zone D: N/A 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)        + 2 ' 33 . 28 ' 20 ) 4 . 0 ( = 9.67’  0.04 (36’) = 1.44’  3 feet Therefore the Edge Zone = 2a = 2 (3.6’) = 7.2’ Average horizontal pressures: Transverse:         − + = ) width bldg ( ) pressure zone erior )(int zone end width bldg ( ) pressure zone end )( zone end ( q avg (7.2')(26.3 ) (75' 7.2')(19.1 ) ( ) (75') avg psf psf q wall   + − =     = 19.8 psf (Zones A, C) (7.2')(4.2 ) (75' 7.2')(4.3 ) ( ) (75') avg psf psf q roof   + − =     = 4.3 psf (Zones B, D) Chapter 2 Solutions S 2 - 11 Longitudinal: (7.2')(21 ) (36' 7.2')(13.9 ) ( ) (36') avg psf psf q wall   + − =     = 15.32 psf (Zones A, C) Design wind pressures: Height and exposure coefficient: Mean roof height =       + 2 ' 33 . 28 ' 20 = 24.2’  = 1.35 (ASCE 7 Figure 28.6 - 1 , Exposure = C, h  25’) Transverse wind: P = q avg  P wall = ( 19.8 psf)(1.35) = 26.73 psf P roof = (4.3 psf) (1.35) = 5.81 psf Longitudinal wind: P wall = ( 15.32 psf) (1.35) = 2 0 .7 psf Total Wind Force: Transverse wind:   (26.73 )(10' 10') (5.81 )(8.33') (75') T P psf psf = + + = 43.7 kips (base shear, transverse) Longitudinal wind: 8.33' 20 ' (20.7 )(36 ') 2 T P psf   = +     = 1 8.0 kips (base shear, longitudinal) Chapter 2 Solutions S 2 - 12 (b) Wind Uplift Average vertical pressures: P = q avg  From Part (a), base uplift pressures: Transverse: Longitudinal: Zone E: - 11.7 psf Zone E: - 25.2 psf Zone F: - 15.9 psf Zone F: - 14.3 psf Zone G: - 8.5 psf Zone G: - 17.5 psf Zone H: - 12.8 psf Zone H: - 11.1 psf Transverse: , 36' 36' ( 11.7 15.9 )(7.2') ( 8.5 12.8 )(75' 7.2') (1.35) 2 2 u avg P psf psf psf psf       = − − + − − −             = - 39,922 lb. , 39,922 (75')(36') u avg lb q − = = - 14.8 psf (gross uplift, transverse) Longitudinal: , 75' 75' ( 25.2 14.3 )(7.2') ( 17.5 11.1 )(36' 7.2') (1.35) 2 2 u avg P psf psf psf psf       = − − + − − −             = - 56 , 097 lb. , 56, 097 (75')(36') u avg lb q − = = - 20.8 psf (gross uplift, longitudinal) Net factored uplift (Longitudinal controls): q net = 0.9 D+W =(0.9)(20psf) + ( - 20 .8psf) = - 2 .8psf (net uplift)