CEE 377 HW4 SOLUTION

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Dec 6, 2023

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Homework 4 CEE 377 - Introduction to Structural Design Problem 1 Use the basic building layout shown below and determine the requested loads using the ASCE 7 tables provided on the course website. You do not need to consider the self weight of the structural members. 3 @ 30 ′ 0” = 90 ′ 0” 3 @ 40 ′ 0” = 120 ′ 0” A B C D 1 2 3 4 10 ′ 0” (Typ.) 1. The roof of the building is flat. It is composed of 4-ply felt and gravel on 3-inches of reinforced concrete. The ceiling beneath the roof is a suspended steel channel system. The building is a storage facility located on the UW campus in a fully exposed area. a. The roof dead load From ASCE 7 Table C3-1 and C3-2, we have 4-ply felt on stone and gravel: 5.5 psf Suspended steel channel ceiling system: 2 psf 3” reinforced concrete slab: (3”/12)(150 pcf) = 37.5 psf Summing, the roof dead load is 45 psf. b. The flat roof snow load in psf to be applied across the roof. Note all assumptions you make related to various factors associated with Equation 7.3-1. Per ASCE 7-16, the ground snow load is 20 psf, but footnote 3 points us to the SEAW white paper (also on the course website) that provides a ground snow load of 25 psf. Further, we can assume that the UW campus is (1) an urban/suburban area representative of surface roughness B, that the structure has no special considerations for thermal effects, and that we can use Risk Category I for a low impact to human life. Thus, p g = 25 psf (SEAW White Paper) C e = 0 . 9 (ASCE 7-16 Table 7.3-1, Surface Roughness B) C t = 1 . 0 (ASCE 7-16 Table 7.3-2, ”All structures...” I s = 0 . 8 (ASCE 7-16 Table 1.5-1, Risk Category I) Then we define the flat roof snow load as p f = 0 . 7 C e C t I s p g = 12 . 6psf per ASCE 7-16 Equation 7.3-1. However, section 7.3.4 requires a minimum snow load of p m = 20 I s when the ground snow load exceeds 20 psf, thus our snow load is p f = p m = 20(0 . 8) = 16 psf . The assumptions made here are not definite, especially regarding the various factors, and another engineer may have a different opinion. c. The live load in psf to be applied to Column B2 Storage facilities in ASCE 7-16 Table 4.3-1 are defined as either heavy or light storage.

Either could be appropriate here, so the live load is either 125 psf (light storage) or 250 psf (heavy storage). In either case, we cannot reduce the live load because it is greater than 100 psf. 2. The roof of the building is flat and there is a 3-inch parapet wall surrounding the perimeter. Drainage is managed through pipes located flush with the roof, and secondary drainage occurs by overtopping the 3-inch parapet wall. It is composed of 2-inches of reinforced concrete on 18-gauge metal decking. A single-ply waterproof sheet will be used. The ceiling beneath the roof is unfinished, but allowance for mechanical ducts should be provided. a. The roof dead load From ASCE 7 Table C3-1 and C3-2, we have 18-gauge metal decking: 3 psf Single-ply waterproof sheet: 0.7 psf Mechanical duct allowance: 4 psf 2” reinforced concrete slab: (2”/12)(150 pcf) = 25 psf Summing, the roof dead load is 32.7 psf. b. The rain load in psf to be applied across the roof The 3-inch parapet wall will result in ponding of 3 inches across the entire roof. Secondary drainage is overtopping, therefore we cannot build a hydraulic head. This gives us d s = 3 inches and d h = 0 , thus we have R = 5 . 2( d s + d h ) = 5 . 2(3 + 0) = 15 . 4 psf.

3. The dead load and live load for the second floor of a library in which any area can be used for stacks. Assume that there will be a steel channel ceiling system and asphalt tile floors. The floors are 6-inch reinforced concrete. Allowance should be provided for mechanical ducts. From ASCE 7 Table C3-1 and C3-2, we have Steel channel ceiling system: 2 psf Asphalt tile floors: 1 psf Mechanical duct allowance: 4 psf 6” reinforced concrete slab: (6”/12)(150 pcf) = 75 psf Summing, the roof dead load is 85 psf. For the live load, we must consider the worse case scenario between the 1000 lb concentrated load and the 150 psf distributed load. Note that we cannot reduce the live load because it exceeds 100 psf. 4. The dead load and live load for a floor in a light manufacturing warehouse/office complex in which any area can be used for storage. Assume that there will be no ceiling or floor finish and that the floors are 4-inch reinforced concrete. Allowance for mechanical ducts should be provided. From ASCE 7 Table C3-1 and C3-2, we have Mechanical duct allowance: 4 psf 4” reinforced concrete slab: (4”/12)(150 pcf) = 50 psf Summing, the roof dead load is 54 psf. For the live load, the light manufacturing facility will control, thus we will assume that L = 125 psf. Again, we cannot reduce. 5. The dead load and live load for a typical upper floor in an office building with movable partition walls. The ceiling is a steel channel system and the floors have a linoleum finish. the floors are 3-inch reinforced concrete. Allowance should be provided for mechanical ducts. From ASCE 7 Table C3-1 and C3-2, we have Steel channel ceiling system: 2 psf Linoleum floor finish: 1 psf Mechanical duct allowance: 4 psf 3” reinforced concrete slab: (3”/12)(150 pcf) = 37.5 psf Summing, the roof dead load is 44.5 psf. For an office, the live load is 50 psf. We also have to consider the movable partitions. Per ASCE 7-16 Section 4.3.2, movable partitions require a 15 psf live load, thus the total live load is 65 psf. 6. An upper floor in a school with steel stud walls (1/2-inch gypsum on each side), a steel channel ceiling system, and a 3-inch reinforced concrete floor with asphalt tile covering. Allowance should be made for mechanical ducts. From ASCE 7 Table C3-1 and C3-2, we have Steel stud walls: 8 psf Asphalt tile floors: 1 psf Steel channel ceiling system: 2 psf Mechanical duct allowance: 4 psf 3” reinforced concrete slab: (3”/12)(150 pcf) = 37.5 psf Summing, the roof dead load is 52.5 psf. Live load for a school is is 40 psf for the classrooms and 80 psf for the hallways (corridors).

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Problem 2 Find the live load on the following members (from Problem 1), including live load reduction as appropriate. Assume that this is an upper story in a multi-story office building. We can solve this in tabular form. Note that the live load reduction factor for column B3 is listed as 1.0. This is because we cannot reduce any live load over 100 psf. All other loads are reduceable. Member L 0 , psf A T , in 2 K LL A I , in 2 Reduction L, psf Load Column B3 150 1200 4 4800 1.0 150 P L = 180 kips Column A3 75 600 4 2400 0.556 41.7 P L = 25.0 kips Column A1 60 300 4 1200 0.683 41.0 P L = 12.3 kips Interior Beam 75 300 2 600 0.862 64.7 w L = 0.647 klf Girder B2-B3 40 1200 2 2400 0.556 22.2 w L = 0.667 klf

Problem 3 An architect has designed a two-story elevated reading room that will be inside a new downtown library. Both plan and elevation views are shown in the sketches below and the plan view is identical for both floors. The floor of both levels is a 4 inch thick reinforced concrete slab with stone aggregate and has a hardwood floor finish. Below the slab on each story is a suspended metal lath and gypsum plaster ceiling. The railing shown in the elevation can be assumed to completely encircle each level. The weights of the beams and columns are shown and must be accounted for in calculations (recall a W AxB weighs B lbs/ft). Determine: a. The tributary areas for the first story columns: A, B, C, and D. b. The tributary areas for beams A, B, C, D, E, F, and G (they are the same on each story) c. The K LL factors for all columns and beams from ASCE 7 Table 4-2 d. The K LL factors that would be used for columns A, B, C, and D and beam A considering their actual influence areas . Do these differ from ASCE 7? A B C D Girder A W27 × 118 Girder B W27 × 118 Beam C Beam D W18 × 45 (Typ.) Beam E Beam F Beam G 5 ft 12 ft 14 ft 9 ft 7 . 5 ft 7 . 5 ft 15 ft Rigid (Moment Resisting) Column Base (Typ.) Simple Connection (Typ.) Railing (70 lb/ft) 40 ft 15 ft 15 ft

We can answer these questions all at once in a big table. Tributary areas for the columns, girders, and beams are shown below. There are a number of ways that you can solve the area of an arbitrary quadrilateral. I used a tool found online that calculates area by cartesian coordinates. A T K LL A I K LL Element in 2 ASCE 7-16 Table 4.7-1 in 2 A I / A T , Measured Column A 243.75 4 750 3.08 Column B 243.75 4 750 3.08 Column C 131.25 4 750 5.71 Column D 131.25 4 750 5.71 Girder A 375 2 750 2.0 Girder B 375 2 750 2.0 Beam C 73.2 2 143.0 1.95 Beam D 222.7 2 428.7 1.92 Beam E 262.0 2 516.8 1.97 Beam F 152.6 2 321.3 2.11 Beam G 39.5 2 90.3 2.29 Because of the significant differences in the tributary/influence areas caused by the irregular layout, we would not expect to see the same live load reduction factors if we calculate them directly. As the engineer, you would have the option of using the ASCE 7-16 values or using the measured values. In this case, the difference is significant enough (especially for columns) that I would certainly want to use the higher values. A B C D Girder A Girder B Beam C Beam D Beam E Beam F Beam G

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Problem 4 In class, we found the demand on four members of a hotel framing system. In a similar manner, find the factored demands on the following members: a. Exterior Beam X, located on the roof. Use the tributary area method. b. Corner Column Y, located between the ground floor and and first story. Opening 20 ′ 5 ′ 10 ′ 5 ′ 20 ′ 3 @ 9 ′ = 27 ′ (Typical) Room (Typ.) Hallway Hollow Concrete Masonry Wall Full Story Height Opening Girders (Typ.) Beams (Typ.) Column Y Beam X

Beam X Beam X is 20 feet long and has a tributary width of 4.5 feet. From the example problem, for a roof load we have D = 49 psf and L = 20 psf. The tributary area for this beam is (20 ft)(4.5 ft) = 90 ft 2 and the roof is flat, thus we have A T = 90 ft 2 ⇒ R 1 = 1 . 0 F = 0 . 0 ⇒ R 2 = 1 . 0 L = L 0 R 1 R 2 = ( 1 . 0 )( 1 . 0 )( 20 psf ) = 20 psf Multiplying by the tributary width, this gives us w D = (49 psf)(4.5 ft) = 220.5 plf and w L = (20 psf)(4.5 ft) = 90 plf. In addition, we must consider the load from the exterior cement tile as a distributed load on Beam X because we are on the exterior of the building. As sketched out for Column D in the example, the roof receives half of the exterior cement tile load across the third story, which can be reduced to a distributed load as w tile = (16 psf)(7.5 ft) = 120 plf. Then we can apply these loads as uniformly distributed loads on a simply supported beam, as shown below. For a simply supported beam under uniform distributed loading w, we have M max = wL 2 / 8 and V max = wL/2. This gives us Dead Load M max = ( 340 . 5plf )( 20ft ) 2 8 = 17 . 03kipft V max = ( 340 . 5plf )( 20ft ) 2 = 3 . 405kip Live Load M max = ( 90plf )( 20ft ) 2 8 = 4 . 5kipft V max = ( 90plf )( 20ft ) 2 = 0 . 9kip And now we simply apply these in the load combinations. As a general rule, the dead load must be at least 8 times higher than the live load for load combination 1 to control, thus we can confidently say that both moment and shear will be controlled by LC2: M u = 1 . 2M max , D + 1 . 6M max , L = 27 . 6 kip ft V u = 1 . 2V max , D + 1 . 6V max , L = 5 . 53 kip

Column Y For Column Y at the ground level, we must consider the area of all floors and the roof. This will be very similar to the approach for Column D in the example problem. The tributary area for the roof and for each floor is (10 ft)(13.5 ft) = 135 ft 2 with a corresponding influence area of (4)(135) = 540 ft 2 . The dead load for the roof is 49 psf and for the floors is 71 psf. We also must consider the weight of the exterior cement tile based on a unit weight of 16 psf. For the bottom story column, the reduced live load is L = 0 . 25 + 15 √ A I L 0 = 0 . 706 ( 40 psf ) = 28 . 3 psf The tributary area and flat slope of the roof do not allow for live load reduction, thus the live roof load is 20 psf. Roof dead load: P D = (135 ft 2 )(49 psf) = 6.62 kips Floor dead load: P D = (135 ft 2 )(71 psf) = 9.59 kips Cement tile dead load: P D = (10 ft + 13.5 ft)(45 ft)(16 psf) = 16.92 kips Live roof load: P L r = (135 ft 2 )(20 psf) = 2.70 kips Floor live load: P L = (135 ft 2 )(28.3 psf) = 3.82 kips We now have to consider two times the floor dead load and two times the live roof load in our calculations below. We can then define D = 6.62 kips + 2(9.59 kips) + 16.9 kips = 42.7 kips, L = 2(3.82 kips) = 7.64 kips, and L r = 2.70 kips. We apply the load combinations accordingly. P u = 1 . 4D = 1.4(42.7 kips) = 59.8 kips P u = 1 . 2D + 1 . 6L + 0 . 5L r = 1.2(25.8 kips) + 1.6(7.64 kips) + 0.5(2.70 kips) = 64.8 kips P u = 1 . 2D + 1 . 6L r + 0 . 5L = 1.2(25.8 kips) + 1.6(2.70 kips) + 0.5(7.64 kips) = 59.4 kips The design load for Column Y is thus P u = 64.8 kips. Note that I assumed (conservatively) that the corner column at the base of the structure would take the entirety of the cement tile. It is reasonable to assume that the column takes a smaller portion like we did in the example problem as well.

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