1 1–1. The floor of a heavy storage warehouse building is made of 6-in.-thick stone concrete. If the floor is a slab having a length of 15 ft and width of 10 ft, determine the resultant force caused by the dead load and the live load. From Table 1–3 DL = [12 lb  ft 2 # in.(6 in.)] (15 ft)(10 ft) = 10,800 lb From Table 1–4 LL = (250 lb  ft 2 )(15 ft)(10 ft) = 37,500 lb Total Load F = 48,300 lb = 48.3 k Ans. 1–2. The floor of the office building is made of 4-in.-thick lightweight concrete. If the office floor is a slab having a length of 20 ft and width of 15 ft, determine the resultant force caused by the dead load and the live load. From Table 1–3 DL = [8 lb  ft 2 # in. (4 in.)] (20 ft)(15 ft) = 9600 lb From Table 1–4 LL = (50 lb  ft 2 )(20 ft)(15 ft) = 15,000 lb Total Load F = 24,600 lb = 24.6 k Ans. 1–3. The T-beam is made from concrete having a specific weight of 150 lb  ft 3 . Determine the dead load per foot length of beam. Neglect the weight of the steel reinforcement. w = (150 lb  ft 3 ) [(40 in.)(8 in.) + (18 in.) (10 in.)] w = 521 lb  ft Ans. a 1 ft 2 144 in 2 b 26 in. 40 in. 8 in. 10 in. 2 *1–4. The “New Jersey” barrier is commonly used during highway construction. Determine its weight per foot of length if it is made from plain stone concrete. 1–5. The floor of a light storage warehouse is made of 150-mm-thick lightweight plain concrete. If the floor is a slab having a length of 7 m and width of 3 m, determine the resultant force caused by the dead load and the live load. From Table 1–3 DL = [0.015 kN  m 2 # mm (150 mm)] (7 m) (3 m) = 47.25 kN From Table 1–4 LL = (6.00 kN  m 2 ) (7 m) (3 m) = 126 kN Total Load F = 126 kN + 47.25 kN = 173 kN Ans. 12 in. 4 in. 24 in. 6 in. 55 ° 75 ° Cross-sectional area = 6(24) + (24 + 7.1950)(12) + (4 + 7.1950)(5.9620) = 364.54 in 2 Use Table 1–2. w = 144 lb  ft 3 (364.54 in 2 ) = 365 lb  ft Ans. a 1 ft 2 144 in 2 b a 1 2 b a 1 2 b 3 1–7. The wall is 2.5 m high and consists of 51 mm  102 mm studs plastered on one side. On the other side is 13 mm fiberboard, and 102 mm clay brick. Determine the average load in kN  m of length of wall that the wall exerts on the floor. 8 in. 8 in. 4 in. 4 in. 6 in. 6 in. 6 in. 20 in. Use Table 1–3. For studs Weight = 0.57 kN  m 2 (2.5 m) = 1.425 kN  m For fiberboard Weight = 0.04 kN  m 2 (2.5 m) = 0.1 kN  m For clay brick Weight = 1.87 kN  m 2 (2.5 m) = 4.675 kN  m Total weight = 6.20 kN  m Ans. 1–6. The prestressed concrete girder is made from plain stone concrete and four -in. cold form steel reinforcing rods. Determine the dead weight of the girder per foot of its length. 3 4 Area of concrete = 48(6) + 4 (14 + 8)(4) - 4(  ) = 462.23 in 2 Area of steel = 4(  ) = 1.767 in 2 From Table 1–2, w = (144 lb  ft 3 )(462.23 in 2 ) + 492 lb  ft 3 (1.767 in 2 ) = 468 lb  ft a 1 ft 2 144 in 2 b a 1 ft 2 144 in 2 b a 3 8 b 2 a 3 8 b 2 d 1 2 c 2.5 m Ans. 4 1–10. The second floor of a light manufacturing building is constructed from a 5-in.-thick stone concrete slab with an added 4-in. cinder concrete fill as shown. If the suspended ceiling of the first floor consists of metal lath and gypsum plaster, determine the dead load for design in pounds per square foot of floor area. 4 in. cinder fill 5 in. concrete slab ceiling From Table 1–3, 5-in. concrete slab = (12)(5) = 60.0 4-in. cinder fill = (9)(4) = 36.0 metal lath & plaster = 10.0 Total dead load = 106.0 lb  ft 2 Ans. 1–9. The interior wall of a building is made from 2  4 wood studs, plastered on two sides. If the wall is 12 ft high, determine the load in lb  ft of length of wall that it exerts on the floor. From Table 1–3 w = (20 lb  ft 2 )(12 ft) = 240 lb  ft Ans. * 1–8. A building wall consists of exterior stud walls with brick veneer and 13 mm fiberboard on one side. If the wall is 4 m high, determine the load in kN  m that it exerts on the floor. For stud wall with brick veneer. w = (2.30 kN  m 2 )(4 m) = 9.20 kN  m For Fiber board w = (0.04 kN  m 2 )(4 m) = 0.16 kN  m Total weight = 9.2 + 0.16 = 9.36 kN  m Ans. 5 1–11. A four-story office building has interior columns spaced 30 ft apart in two perpendicular directions. If the flat-roof live loading is estimated to be 30 lb  ft 2 , determine the reduced live load supported by a typical interior column located at ground level. Floor load: L o = 50 psf A t = (30)(30) = 900 ft 2 7 400 ft 2 L = L o (0.25 + ) L = 50 (0.25 + ) = 25 psf % reduction = = 50% 7 40% (OK) F s = 3[(25 psf)(30 ft)(30 ft)] + 30 psf(30 ft)(30 ft) = 94.5 k Ans. 25 50 15 2 4(900) 15 2 K LL A T *1–12. A two-story light storage warehouse has interior columns that are spaced 12 ft apart in two perpendicular directions. If the live loading on the roof is estimated to be 25 lb  ft 2 , determine the reduced live load supported by a typical interior column at (a) the ground-floor level, and (b) the second-floor level. A t = (12)(12) = 144 ft 2 F R = (25)(144) = 3600 lb = 3.6 k Since A t = 4(144) ft 2 7 400 ft 2 L = 12.5 (0.25 + ) = 109.375 lb  ft 2 (a) For ground floor column L = 109 psf 7 0.5 L o = 62.5 psf OK F F = (109.375)(144) = 15.75 k F = F F + F R = 15.75 k + 3.6 k = 19.4 k Ans. (b) For second floor column F = F R = 3.60 k Ans. 15 2 (4)(144) 6 1–14. A two-story hotel has interior columns for the rooms that are spaced 6 m apart in two perpendicular directions. Determine the reduced live load supported by a typical interior column on the first floor under the public rooms. Table 1–4 L o = 4.79 kN  m 2 A T = (6 m)(6 m) = 36 m 2 K LL = 4 L = L o (0.25 + ) L = 4.79 (0.25 + ) L = 3.02 kN  m 2 Ans. 3.02 kN  m 2 7 0.4 L o = 1.916 kN  m 2 OK 4.57 2 4(36) 4.57 2 K LL A T 1–13. The office building has interior columns spaced 5 m apart in perpendicular directions. Determine the reduced live load supported by a typical interior column located on the first floor under the offices. From Table 1–4 L o = 2.40 kN  m 2 A T = (5 m)(5 m) = 25 m 2 K LL = 4 L = L o (0.25 + ) L = 2.40 (0.25 + ) L = 1.70 kN  m 2 Ans. 1.70 kN  m 2 7 0.4 L o = 0.96 kN  m 2 OK 4.57 2 4(25) 4.57 2 K LL A T 7 *1–16. Wind blows on the side of the fully enclosed hospital located on open flat terrain in Arizona. Determine the external pressure acting on the leeward wall, which has a length of 200 ft and a height of 30 ft. V = 120 mi  h K zt = 1.0 K d = 1.0 q h = 0.00256 K z K zt K d V 2 = 0.00256 K z (1.0)(1.0)(120) 2 = 36.86 K z 1–15. Wind blows on the side of a fully enclosed hospital located on open flat terrain in Arizona. Determine the external pressure acting over the windward wall, which has a height of 30 ft. The roof is flat. V = 120 mi  h K zt = 1.0 K d = 1.0 q z = 0.00256 K z K zt K d V 2 = 0.00256 K z (1.0)(1.0)(120) 2 = 36.86 K z From Table 1–5, z K z q z 0–15 0.85 31.33 20 0.90 33.18 25 0.94 34.65 30 0.98 36.13 Thus, p = q G C p – q h ( G C p i ) = q (0.85)(0.8) - 36.13 ( ; 0.18) = 0.68 q < 6.503 p 0–15 = 0.68(31.33) < 6.503 = 14.8 psf or 27.8 psf Ans. p 20 = 0.68(33.18) < 6.503 = 16.1 psf or 29.1 psf Ans. p 25 = 0.68(34.65) < 6.503 = 17.1 psf or 30.1 psf Ans. p 30 = 0.68(36.13) < 6.503 = 18.1 psf or 31.1 psf Ans. 8 1–17. A closed storage building is located on open flat terrain in central Ohio. If the side wall of the building is 20 ft high, determine the external wind pressure acting on the windward and leeward walls. Each wall is 60 ft long. Assume the roof is essentially flat. From Table 1–5, for z = h = 30 ft, K z = 0.98 q h = 36.86(0.98) = 36.13 From the text = = 1 so that C p = - 0.5 p = q GC p - q h ( GC p 2 ) p = 36.13(0.85)( - 0.5) - 36.13( ; 0.18) p = - 21.9 psf or - 8.85 psf Ans. 200 200 L o B V = 105 mi  h K zt = 1.0 K d = 1.0 q = 0.00256 K z K zt K d V 2 = 0.00256 K z (1.0)(1.0) (105) 2 = 28.22 K z From Table 1–5 z K z q z 0–15 0.85 23.99 20 0.90 25.40 Thus, for windward wall p = qGC p – q h ( GC p i ) = q (0.85)(0.8) – 25.40( ; 0.18) = 0.68 q < 4.572 p 0 – 15 = 0.68 (23.99) < 4.572 = 11.7 psf or 20.9 psf Ans. p 20 = 0.68 (25.40) < 4.572 = 12.7 psf or 21.8 psf Ans. Leeward wall = = 1 so that C p = - 0.5 p = q GC p - q h ( GC p i ) p = 25.40(0.85)( - 0.5) - 25.40 ( ; 0.18) p = - 15.4 psf or - 6.22 psf Ans. 60 60 L B 1–16. Continued 9 1–18. The light metal storage building is on open flat terrain in central Oklahoma. If the side wall of the building is 14 ft high, what are the two values of the external wind pressure acting on this wall when the wind blows on the back of the building? The roof is essentially flat and the building is fully enclosed. V = 105 mi  h K zt = 1.0 K d = 1.0 q z = 0.00256 K z K zt K d V 2 = 0.00256 K z (1.0)(1.0)(105) 2 = 28.22 K z From Table 1–5 For 0 … z … 15 ft K z = 0.85 Thus, q z = 28.22(0.85) = 23.99 p = q GC p - q h ( GC p i ) p = (23.99)(0.85)(0.7) - (23.99)( 0.18) p = - 9.96 psf or p = - 18.6 psf Ans. ; 1–19. Determine the resultant force acting perpendicular to the face of the billboard and through its center if it is located in Michigan on open flat terrain. The sign is rigid and has a width of 12 m and a height of 3 m. Its top side is 15 m from the ground. q h = 0.613 K z K zt K d V 2 Since z = h = 15 m K z = 1.09 K zt = 1.0 K d = 1.0 V = 47 m  s q h = 0.613(1.09)(1.0)(1.0)(47) 2 = 1476.0 N  m 2 B  s = = 4, s  h = = 0.2 From Table 1–6 C f = 1.80 F = q h GC f A s = (1476.0)(0.85)(1.80)(12)(3) = 81.3 kN Ans. 3 15 12 m 3 m 10 1–21. The school building has a flat roof. It is located in an open area where the ground snow load is 0.68 kN  m 2 . Determine the snow load that is required to design the roof. p f = 0.7 C c C t I s p g p f = 0.7(0.8)(1.0)(1.20)(0.68) = 0.457 kN  m 2 Also p f = p f = I s p g = (1.20)(0.68) = 0.816 kN  m 2 use p f = 0.816 kN  m 2 Ans. 1–22. The hospital is located in an open area and has a flat roof and the ground snow load is 30 lb  ft 2 . Determine the design snow load for the roof. Since p q = 30 lb  ft 2 7 20 lb  ft 2 then p f = I s p g = 1.20(30) = 36 lb  ft 2 Ans. *1–20. A hospital located in central Illinois has a flat roof. Determine the snow load in kN  m 2 that is required to design the roof. p f = 0.7 C c C t I s p g p f = 0.7(0.8)(1.0)(1.20)(0.96) = 0.6451 kN  m 2 Also p f = I s p g = (1.20)(0.96) = 1.152 kN  m 2 use p f = 1.15 kN  m 2 Ans. 11 2–1. The steel framework is used to support the reinforced stone concrete slab that is used for an office. The slab is 200 mm thick. Sketch the loading that acts along members BE and FED . Take , . Hint : See Tables 1–2 and 1–4. b = 5 m a = 2 m 200 mm thick reinforced stone concrete slab: (23.6 kN > m 3 )(0.2 m)(2 m) = 9.44 kN > m Live load for office: (2.40 kN > m 2 )(2 m) = Ans. Due to symmetry the vertical reaction at B and E are B y = E y = (14.24 kN > m)(5) > 2 = 35.6 kN The loading diagram for beam BE is shown in Fig. b . 480 kN > m 14.24 kN > m Beam FED . The only load this beam supports is the vertical reaction of beam BE at E which is E y = 35.6 kN. The loading diagram for this beam is shown in Fig. c . Beam BE . Since the concrete slab will behave as a one way slab. Thus, the tributary area for this beam is rectangular shown in Fig. a and the intensity of the uniform distributed load is b a = 5 m 2 m = 2.5, A B C D E F b a a 12 2–2. Solve Prob. 2–1 with , . b = 4 m a = 3 m Beam BE . Since , the concrete slab will behave as a two way slab. Thus, the tributary area for this beam is the hexagonal area shown in Fig. a and the maximum intensity of the distributed load is 200 mm thick reinforced stone concrete slab: (23.6 kN > m 3 )(0.2 m)(3 m) = 14.16 kN > m Live load for office: (2.40 kN > m 2 )(3 m) = Ans. Due to symmetry, the vertical reactions at B and E are = 26.70 kN The loading diagram for Beam BE is shown in Fig. b . Beam FED . The loadings that are supported by this beam are the vertical reaction of beam BE at E which is E y = 26.70 kN and the triangular distributed load of which its tributary area is the triangular area shown in Fig. a . Its maximum intensity is 200 mm thick reinforced stone concrete slab: (23.6 kN > m 3 )(0.2 m)(1.5 m) = 7.08 kN > m Live load for office: (2.40 kN > m 2 )(1.5 m) = Ans. The loading diagram for Beam FED is shown in Fig. c . 3.60 kN > m 10 .68 kN > m B y = E y = 2 c 1 2 (21.36 kN > m)(1.5 m) d + (21.36 kN > m)(1 m) 2 720 kN > m 21 .36 kN > m b a = 4 3 6 2 A B C D E F b a a 13 2–3. The floor system used in a school classroom consists of a 4-in. reinforced stone concrete slab. Sketch the loading that acts along the joist BF and side girder ABCDE . Set , . Hint : See Tables 1–2 and 1–4. b = 30 ft a = 10 ft A E b a a a a B C D F Joist BF . Since , the concrete slab will behave as a one way slab. Thus, the tributary area for this joist is the rectangular area shown in Fig. a and the intensity of the uniform distributed load is 4 in thick reinforced stone concrete slab: (0.15 k > ft 3 ) (10 ft) = 0.5 k > ft Live load for classroom: (0.04 k > ft 2 )(10 ft) = Ans. Due to symmetry, the vertical reactions at B and F are B y = F y = (0.9 k > ft)(30 ft) > 2 = 13.5 k Ans. The loading diagram for joist BF is shown in Fig. b . Girder ABCDE . The loads that act on this girder are the vertical reactions of the joists at B , C , and D , which are B y = C y = D y = 13.5 k. The loading diagram for this girder is shown in Fig. c . 0.4 k > ft 0.9 k > ft a 4 12 ft b b a = 30 ft 10 ft = 3 14 *2–4. Solve Prob. 2–3 with , . b = 15 ft a = 10 ft A E b a a a a B C D F Joist BF . Since , the concrete slab will behave as a two way slab. Thus, the tributary area for the joist is the hexagonal area as shown in Fig. a and the maximum intensity of the distributed load is 4 in thick reinforced stone concrete slab: (0.15 k > ft 3 ) (10 ft) = 0.5 k > ft Live load for classroom: (0.04 k > ft 2 )(10 ft) = Ans. Due to symmetry, the vertical reactions at B and G are Ans. The loading diagram for beam BF is shown in Fig. b . Girder ABCDE . The loadings that are supported by this girder are the vertical reactions of the joist at B , C and D which are B y = C y = D y = 4.50 k and the triangular distributed load shown in Fig. a . Its maximum intensity is 4 in thick reinforced stone concrete slab: (0.15 k > ft 3 ) (5 ft) = 0.25 k > ft Live load for classroom: (0.04 k > ft 2 )(5 ft) = Ans. The loading diagram for the girder ABCDE is shown in Fig. c . 0.20 k  ft 0.45 k  ft a 4 12 ft b B y = F y = 2 c 1 2 (0.9 k > ft )(5 ft ) d + (0.9 k > ft )(5 ft ) 2 = 4.50 k 0.4 k > ft 0.9 k > ft a 4 12 ft b b a = 15 ft 10 ft = 1.5 < 2 15 2–5. Solve Prob. 2–3 with , . b = 20 ft a = 7.5 ft A E b a a a a B C D F Beam BF . Since , the concrete slab will behave as a one way slab. Thus, the tributary area for this beam is a rectangle shown in Fig. a and the intensity of the distributed load is 4 in thick reinforced stone concrete slab: (0.15 k > ft 3 ) (7.5 ft) = 0.375 k > ft Live load from classroom: (0.04 k > ft 2 )(7.5 ft) = Ans. Due to symmetry, the vertical reactions at B and F are Ans. The loading diagram for beam BF is shown in Fig. b . Beam ABCD . The loading diagram for this beam is shown in Fig. c . B y = F y = (0.675 k > ft)(20 ft) 2 = 6.75 k 0.300 k > ft 0.675 k > ft a 4 12 ft b b a = 20 ft 7.5 ft = 2.7 7 2