Tutorial 1 civi 454 -2024

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Concordia University *

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CIVI 454

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Mechanical Engineering

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Feb 20, 2024

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1 Tutorial 1 – CIVI454 1 . The plan given below is the floor plan of a 2-storey residential building located in Montreal. There are 3 bays in the E-W direction and 2 bays in the N-S direction. The building height is 7m and both storey have the same height. The composite steel deck is expanded 300 mm from the grid lines. At typical floor and roof, dead load is: DL floor = 3.6 kPa and DL roof = 3.0 kPa, respectively. The live load at roof and floor is LL roof =1 kPa and LL floor =1.9 kPa, respectively. Cladding dead load is 1.2 kPa. In this example, the wind load combination is neglected. a) Calculate the snow load and design the secondary beams located at roof between grid lines 2 and 3 and grid lines A and B . Repeat the calculation for the floor level. b) Design the girder (main beam) located in grid line 2 between A and B , as well as that between B and C located at the floor level, as well as the roof level. Design also the girder located in grid line 1 between A and B at the floor level. 2 . Consider the same building as described in Example 1 but change the occupancy type from residential to school building. Consider the roof level and calculate the typical secondary beam and the girder in grid line 2 between A and B. A B C D 1 2 3

2 Snow load will be discussed in Tutorial 2. In this example, we consider the following parameters: C b = 0.8, C w = 1.0, C s = 1.0 and C a = 1.0.  Question 1 Snow load Calculation: For a building of normal importance category, I s = 1.0. For a building of high importance category (e.g. schools), I s = 1.15 It is noted that these values for I s are in accordance with the ultimate limit state (ULS). In both cases, I s =0.9 when serviceability limit state (SLS) is considered in order to check the member’s deflection. According to Table C-2 of Appendix C on NBCC 2015, for Montreal, S s = 2.6 kPa and S r = 0.4 kPa. Thus: - for a normal importance category building, S = 1[2.6(0.8x1x1x1) + 0.4] = 2.48 kPa - for a high importance category building, S = 1.15[2.6(0.8x1x1x1) + 0.4] = 2.85 kPa

3 Design the secondary beam between grid lines 2 and 3 and A and B located at roof. Load combinations: Case 2: (1.25DL+1.5LL+1.0SL) Case 3: (1.25DL+1.0LL+1.5SL) For roof: w DL = 2m x 3 kPa = 6 kN/m, where 2.0 m is the tributary width for the secondary beam. w SL = 2m x 2.48 kPa = 4.96 kN/m w LL = 2m x 1.0 kPa = 2 kN/m w CASE 2 = 1.25 x 6 + 1.5 x 2 + 1.0 x 4.96 = 15.46 kN/m w CASE 3 = 1.25 x 6 + 1.0 x 2 + 1.5 x 4.96 = 16.94 kN/m Now, M f = 𝑊 𝐿 2 8 ⁄ = 16.94 x 6 2 /8 = 76.5 kNm Select BEAM section based on M f: W310 x 21 which has M r = 89.1 kN-m > M f O.K. The nominal mass is 21 kg/m or 0.21 kN/m  Compute moment due to beam self – weight , M f = 𝑊 𝐿 2 8 ⁄ = 1.25 x 0.21x 6 2 /8 = 1.2 kN-m = 16.94 kN/m = 6.0 m

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4 M f (total) = 76.5 + 1.2 = 77.7 kN-m < 89.1 ( M r) O.K .  Check Beam Deflection Consider SL and LL The Specified snow load is 0.9 * 2.48 = 2.23 kPa. For W310 x 21, I = 37 x 10 6 mm 4 Δ maxSL = 5x (2.23x2) x 6000 4 /(384x200000x37x10 6 ) = 10.2 mm < L/360 O.K For Live load, Δ maxSL = 5x (1x2) x 6000 4 /(384x200000x37x10 6 ) = 4.6 mm < L/360 O.K Total deflection, Δ max = 10.2 mm + 4.6 mm = 14. 8 mm > L/360 = 6000/360 = 16.6 mm O.K. Selected BEAM section is O.K. Calculate reaction of W310 X 21 beam located at the roof: Considering only service load for girder deflection calculation at roof Design the secondary beam between grid lines 2 and 3 and A and B located at floor. Case 2: (1.25DL+1.5LL+1.0SL) For floor: w DL = 2m x 3.6 kPa = 7.2 kN/m w LL = 2m x 1.9 kPa = 3.8 kN/m; w SL =0 = 16.94 + .21*1.25 =17.2 kN/m = 6 m 51 kN = 6 m = (2.23*2) + (1*2) =6.46 19.4 KN 51 kN 19.4 kN <

5 Consider the same beam at floor that was considered at roof as W 310 x 21 . W CASE 2 = 1.25 x (7.2+ 0.21) + 1.5 x 3.8 = 15.0 kN/m (W*L)⁄2= (15*6)⁄2=45 KN M f = ? ∗ 𝐿 2 8 ⁄ = 15 x 6 2 /8 = 67.5 kN-m < 89.1 kN-m , O.K  Check Deflection of secondary beam of floor W 310X21 for which w LL = 1.9 x 2 = 3.8 kN/m (service load). For this beam I = 37 x 10 6 mm 4 Δ maxSL = 5x (3.8) x 6000 4 /(384x200000x37x10 6 ) = 8.7 mm < L/360 O.K Design the FLOOR girder (main beam) located in grid line 2 between A and B M max =90x2 = 180 kN-m Select BEAM W 360 x 39 For Lu = 2000 mm (unbraced length), M r = 193 kNm = 15 KN/m = 6 m 45 KN w= 1.9 x 2 =3.8 KN/m(LL service) 11.4 kN 45 * 2 = 90 KN 90 KN 90 KN =2m = 6 m 45 * 2 = 90 KN 2m =2m 11.4 KN LL

6  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25x0.39x6 2 /8 = 1.8 kN-m M f (total) = 180 + 1.8 = 181.8 KN-m < 193 ( M r) So, selected Beam W360x39 is O.K  Check Deflection For tW360x39, I x = 102 x 10 6 mm 4 Δ max = 𝑃𝑎 24𝐸𝐼 ( 3 𝑙 2 − 4 𝑎 2 ) = 22 . 8 𝑥 1000 𝑥 2000 24 𝑥 200000 𝑥 102 𝑥 10 6 (3 x 6000 2 – 4x2000 2 ) = 8.36 mm < 16.6mm O.K. Select W 360x39 beam is O.K Design the FLOOR girder (main beam) located in grid line 2 between B and C M f= 135X4-90X2 = 360 kN-m Select W 460x60 beam which has M r= 393 kNm  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.6 x 8 2 /8= 6 KN-m M f (total) = 360 + 6 = 366 KN-m < 396 ( M r) O.K. So, selected Beam W460x60 is O.K 11.4 * 2 = 22.8 KN 11.4 * 2 = 22.8 KN 45 X 2 = 90 KN 270/2=135 KN 270/2=135 KN 8 m 2 m 2 m 2 m 2 m =2m =2m L=6m 90 KN 90 KN 2m 8.57

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7  Check Deflection for girder W 460x60 (I = 255 x 10 6 mm 4 ) Δ max(1) 𝑃𝑎 24𝐸𝐼 ( 3 𝑙 2 − 4 𝑎 2 ) = ( 22 . 8 𝑥 1000 𝑥 2000 24 𝑥 200000 𝑥 255 𝑥 10 6 (3x8000 2 – 4x2000 2 ) = 6.6 mm Δ max (2) = 𝑃 𝐿 3 48𝐸𝐼 = 22 . 8 𝑥1000𝑥 8000 3 48 𝑥200000 𝑥255𝑥 10 6 = 4.8 mm Total deflection Δ max = 6.6 mm + 4.8 mm = 11.4 mm < L/360 = 8000/360 = 22.2 mm So selected beam is O.K. Design the ROOF girder (main beam) located in grid line 2 between A and B Consider, Case 3: (1.25DL+1.0LL+1.5SL) M f=102 * 2 = 204 kN-m Select W 410x39 which has M r=216 kN-m and I = 126x10 6 mm 6  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.39 x 6 2 /8 = 2.2 kN-m 11.4 x 2 = 22.8 KN 22.8 KN 2 m 2 m 4 m 8 m 22.8 KN 4 m 4 m 8 m 102 KN a =2m 2m 102 KN = 6 m a=2 m 51 * 2 = 102 kN 51 * 2 = 102 kN + (1) (2)

8 M f (total) = 204 + 2.2 = 206.2 KN-m < 216 ( M r) O.K. So, selected Beam W410X39 is O.K  Check Deflection ∆ → Δ max = 𝑃𝑎 24𝐸𝐼 ( 3 𝑙 2 − 4 𝑎 2 ) = 38 . 8 𝑥 1000 ) 𝑥 2000 24 𝑥 200000 𝑥 126 𝑥 10 6 ( 3x6000 2 – 4x2000 2 ) = 11.56 mm < 16.6 mm O.K. Selected Beam is O.K. Design the ROOF girder (main beam) located in grid line 2 between B and C M f = 153 * 4 – 102*2 = 408 kN-m Select W 410X74 which has Mr= 448 KN-m  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.74 x 8 2 /8 = 7.4 kN-m M f (total) = 408 + 7.4 = 415.4 kN-m < 448 kN-m ( M r) Select Beam is O.K.  Check Deflection of beam W 410 x74 ( I = 275 x10 6 mm 4 ) 51 * 2 = 102 KN 306/2=153 kN 153 kN 8 m 2 m 2 m 2 m 2 m 102 KN 102 KN a =2m 2m = 6 m a=2 m 19.4X2 KN 19.4X2 KN 38.8 kN 38.8 kN 469

9 (1) (2) ∆ → Δmax( 1 ) = 𝑃𝑎 24𝐸𝐼 ( 3 𝑙 2 − 4 𝑎 2 ) = 38 . 8 𝑥 1000 𝑥 2000 24 𝑥 200000 𝑥 275 𝑥 10 6 (3 x 8000 2 – 4x2000 2 ) = 10.13 mm ∆ → Δmax( 2 ) = 𝑃 𝐿 3 48𝐸𝐼 = 38 . 8 ?1000? 8000 3 48?200000?275? 10 6 Total deflection Δ max = 10.13 mm + 7.37 mm = 17.5 mm < L/360 = 8000/360 = 22.2 mm O.K. So, the selected beam is O.K. Design the girder located in grid line 1 between A and B at the floor level . Case 2: (1.25DL+1.5LL+1.0SL) no snow load for floor level. W CASE 2 = 1.25 x 3.6 + 1.5 x 1.9 + 1.0 x 0 = 7.35 kN/m From floor extension of 300m UDL load from extension is = 7.35 x 0.3 = 2.2 kN/m 19.4 x 2 = 38.8 KN 38.8 KN a=2 m a=2 m 4 m 8 m 38.8 KN 4 m 4 m 8 m 12.8 kN/m 45 kN 2 m 2 m 2 m 45 kN L=6 m 45 +12.8*6/2= 83.4 kN 45+12.8*6/2 = 83.4 kN = 7.37 mm 10.34 7.52 7.52 17.86 17.86 0.34 Pa

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10 Cladding (usually do not consider when storey height is less than 11m). When cladding is considered, the DL on beam at typical floor is associated to half of above storey height + half of the bellow storey height: 1.25 [1.2 x (7/2+7/2)] kN/m= 10.6 kN/m Total UDL load acting on Beam = 10.6 + 2.2 = 12.8 kN/m M f = 83.4x 3 – 45x1 – 12.8x3x1.5 = 147.6 kN-m Select W 360X39 which has Mr= 193 kN-m  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.39 x 6 2 /8 = 2.2 kN-m M f (total) = 147.6 + 2.2 = 149.8 kN-m < 193 ( M r) So, the selected Beam is O.K Deflection is within the accepted limit. Design the secondary beam in grid line A between 2-3 at the floor level . Case 2: (1.25DL+1.5LL+1.0SL) no snow load for floor level. W CASE 2 = 1.25 x 3.6 + 1.5 x 1.9 + 1.0 x 0 = 7.35 kN/m From floor extension of 300m UDL load from extension is = 7.35 x (1+0.3) = 9.6 kN/m Cladding (usually do not consider when storey height is less than 11m). When cladding is considered, the DL on beam at typical floor is associated to half of above storey height + half of the bellow storey height: 1.25 [1.2 x (7/2+7/2)] kN/m= 10.5 kN/m Total UDL load acting on Beam = 10.5 + 9.6 = 20.1 kN/m 20.1 kN/m L=6 m 20.1*6/2= 40.2 kN 20.1*6/2 = 40.2 kN Pa

11 M f = 20.1 x6 2 /8= 90.5 kN-m Select W 310X24 which has Mr= 102 kN-m  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.24 x 6 2 /8 = 1.4 kN-m M f (total) = 90.5 + 1.4 = 92 kN-m < 102 ( M r) So, the selected Beam is O.K Deflection is within the accepted limit. Location of beams Beam Section Secondary beam between grid lines 2-3 and A - B located at roof W 310X21 Secondary beam between grid lines 2-3 and A- B located at Floor. W 310X21 Girder (main beam) located in grid line 2 between A and B at floor W 360X39 Girder (main beam) located in grid line 2 between B and C at floor W 460X60 Girder (main beam) located in grid line 2 between A and B at roof W 410X39 Girder (main beam) located in grid line 2 between B and C at roof W 410X74 Girder(main beam) located in grid line 1 between A and B at floor W 360X39 Secondary beam between grid lines A between 2-3 located at Floor. W310x24

12 Floor plan W310x24 W360x39 W360x39 W360x57 W310x21 W310x21 W360x39 W460x60 W360x39 W360x39 W360x39 W360x57 W310x24 W310x24 W310x24

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13 Roof floor W410x39 W310x21 W310x21 W410x39 W410x39 W410x39 W410x39 W410x39 W410x74 W410x54 W410x54

14  Question 2 School Montreal : S s =2.6 kPa and S r = 0.4 kPa [ subsection 1.1.3] C w= C a= C s=1 I s = 1.15 S = 1.15[2.6(0.8x1.0x1.0x1.0) +0.4] =2.8 kPa Design Secondary beam at roof located in grid line 2 between A and B W DL = 2m x 3 kPa = 6 kN/ m W SL = 2m x 2.8 kPa = 5.6 kN/ m W LL = 2m x 1 kPa = 2 kN/ m W CASE 3 = 1.25 x 6 + 1.0 x 2 + 1.5 x 5.6 = 18 kN/m M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 18 x 6 2 /8 = 81 kN-m Check W 310x21 from previous design with Mr=89.1 kN-m.  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.21 x 6 2 /8 = 1.2 kN-m M f (total) = 81 + 1.2 = 82.2 KN-m < 89.1 kN-m ( M r) So, Selected Beam W310 x 21 is O.K Deflection was checked in example 1. = 18 KN/m = 6 m 18*6/2=54kN

15 Design Girder at roof between grid lines 2-3 and A - B Check W410x39 Considering self weight of the secondary beam = 1.25 x0.21* 6= 1.575 KN M f = 109.6 * 2 = 219.2 kN-m > 213kN-m Not good Increase the size of girder which coming from example -1. Select W410x46 with Mr=265 kN-m  Check moment due to beam self – weight , M f = 𝑊 ∗ 𝐿 2 8 ⁄ = 1.25 x 0.46 x 6 2 /8 = 2.6 kN-m M f (total) = 219.2 + 2.6 = 221.8 KN-m < 265 kN-m ( M r) So, Selected Beam is O.K Calculating secondary beam reaction considering S.L.S For Specified snow load = 0.9 * 2.8 = 2.52 kPa Total service load = 0.9SL+LL = 2.52+1 = 3.52 kPa w(SL)= 3.52* 2= 7.04 kN/m = 7.04 KN/m = 6 m 7.04*6/2=21.12 kN 54*2+ 1.575 = 109.6 109.3 2m 109.6 2m 109.3 = 6 m 2m

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16  Check Deflection ∆ → Δmax = 𝑃𝑎 24𝐸𝐼 ( 3 𝑙 2 − 4 𝑎 2 ) = 42 . 3 ? 1000 ? 2000 24 ? 200000 ? 156 ? 10 6 10.5 mm < 16.6 mm The selected beam is O.K 21.12*2= 42.3 KN 2m 42.3 KN 2m = 6 m 2m (3 x 6000 2 – 4x 2000 2 ) =