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Structural Analysis, 10th Edition Solution Manual

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    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. S0LUTI0N 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. Ans. F = 48.3 k 1

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    2 1–2. The wall is 15 ft high and consists of 2 × 4 in. studs, plastered on one side. On the other side there is 4-in. clay brick. Determine the average load in lb / ft of length of wall that the wall exerts on the floor. S0LUTI0N Using the data tabulated in Table 1–3, 4 @ in. clay brick: (39 lb / ft 2 )(15 ft) = 585 lb / ft 2 × 4 @ in. studs plastered on one side: (12 lb / ft 2 ) (15 ft) = 180 lb / ft w D = 765 lb / ft Ans. Ans. w D = 765 lb / ft

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    3 1–3. A building wall consists of 12-in. clay brick and 1 2 -in. fiberboard on one side. If the wall is 10 ft high, determine the load in pounds per foot that it exerts on the floor. SOLUTION From Table 1–3, 12 @ in. clay brick: (115 lb / ft 2 )(10 ft) = 1150 lb / ft 1 / 2 @ in. fiberboard: (0.75 lb / ft 2 )(10 ft) = 7.5 lb / ft Total: 1157.5 lb / ft = 1.16 k / ft Ans. Ans. w = 1.16k / ft

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    4 *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. 12 in. 4 in. 24 in. 6 in. 55 ◦ 75 ◦ SOLUTION Cross @ sectional area = 6(24) + ( 1 2 ) (24 + 7.1950)(12) + ( 1 2 ) (4 + 7.1950)(5.9620) = 364.54 in 2 Use Table 1–2. w = 144 lb / ft 3 (364.54 in 2 ) ( l ft 2 144 in 2 ) = 365 lb / ft Ans. Ans. w = 365 lb / ft

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    5 1–5. The precast floor beam is made from concrete having a specific weight of 23.6 kN / m 3 . If it is to be used for a floor of an office building, calculate its dead and live loadings per foot length of beam. 1.5 m 0.15 m 0.3 m 0.15 m SOLUTION The dead load is caused by the self-weight of the beam. w D = [(1.5 m)(0.15 m) + (0.15 m)(0.3 m)](23.6 kN / m 3 ) = 6.372 kN / m = 6.37 kN / m Ans. For the office, the recommended line load for design in Table 1–4 is 2.4 kN / m 2 . Thus, w L = (2.40 kN / m 2 )(1.5 m) = 3.60 kN / m Ans. Ans. w D = 6.37 kN / m w L = 3.60 kN / m

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    6 1–6. 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. SOLUTION 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. Ans. F = 173 kN

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    7 1–7. The precast inverted T-beam has the cross section shown. Determine its weight per foot of length if it is made from reinforced stone concrete and twelve 3 4 -in.-diameter cold-formed steel reinforcing rods. 9 in. 12 in. 36 in. 48 in. SOLUTION From Table 1–2, the specific weight of reinforced stone concrete and the cold-formed steel are γ C = 150 lb / ft 3 and γ H = 492 lb / ft 3 , respectively. Reinforced stone concrete: | ( 48 12 ft ) ( 12 12 ft ) + ( 9 12 ft ) ( 36 12 ft ) − 12 ( π 4 ) ( 0.75 12 ft ) 2 | (150 lb / ft) = 931.98 lb / ft Cold @ formed steel: | 12 ( π 4 ) ( 0.75 12 ft ) 2 | (492 lb / ft 3 ) = 18.11 lb / ft 950.09 lb / ft w D = (950.09 lb / ft) ( 1 k 1000 lb ) = 0.950 k / ft Ans. Ans. w D = 0.950 k / ft

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    8 *1–8. The hollow core panel is made from plain stone concrete. Determine the dead weight of the panel. The holes each have a diameter of 4 in. 7 in. 12 ft 12 in. 12 in. 12 in. 12 in. 12 in. 12 in. SOLUTION From Table 1–2, W = (144 lb / ft 3 )[(12 ft)(6 ft) ( 7 12 ft ) − 5(12 ft)( π ) ( 2 12 ft ) 2 ] = 5.29 k Ans. Ans. W = 5.29 k

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    9 1–9. The floor of a light storage warehouse is made of 6-in.-thick cinder concrete. If the floor is a slab having a length of 10 ft and width of 8 ft, determine the resultant force caused by the dead load and that caused by the live load. SOLUTION From Table 1–3, DL = (6 in.)(9 lb / ft 2 . in.)(8 ft)(10 ft) = 4.32 k Ans. From Table 1–4, LL = (125 lb / ft 2 )(8 ft)(10 ft) = 10.0 k Ans. Ans. DL = 4.32 k LL = 10.0 k

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    10 1–10. 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. SOLUTION From Table 1–3, w = (20 lb / ft 2 )(12 ft) = 240 lb / ft Ans. Ans. w = 240 lb / ft

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    11 1–11. 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 fi ll 5 in. concrete slab ceiling SOLUTION 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. Ans. DL = 106 lb / ft 2

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    12 *1–12. 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. SOLUTION Table 1–4: L o = 4.79 kN / m 2 A T = (6 m)(6 m) = 36 m 2 K LL = 4 K LL A T = 4(36) = 144 m 2 > 37.2 m 2 From Eq. 1-1, LL = L o ( 0.25 + 4.57 1 K LL T A ) LL = 4.79 ( 0.25 + 4.57 1 4(36) ) LL = 3.02 kN / m 2 Ans. 3.02 kN / m 2 > 0.4 L o = 1.916 kN / m 2 OK Ans. LL = 3.02 kN / m 2

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    13 1–13. 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. SOLUTION From Table 1–4, L o = 50 psf A T = (30)(30) = 900 ft 2 K LL A T = 4(900) = 3600 ft 2 > 400 ft 2 From Eq. 1-1, L = L a ( 0.25 − 15 1 K LL A T ) L = 50 ( 0.25 − 15 1 4(900) ) = 25 psf % reduction = 25 50 = 50% > 40% (OK) F = 3[(25 psf)(30 ft)(30 ft)] + 30 psf(30 ft)(30 ft) = 94.5 k Ans. Ans. LL = 94.5 k

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    14 1–14. 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. SOLUTION 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 + 4.57 1 K LL A T ) L = 2.40 ( 0.25 + 4.57 1 4(25) ) L = 1.70 kN / m 2 Ans. 1.70 kN / m 2 > 0.4 L o = 0.96 kN / m 2 OK Ans. LL = 1.70 kN / m 2

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    15 1–15. A hospital located in Chicago, Illinois, has a flat roof, where the ground snow load is 25 lb / ft 2 . Determine the design snow load on the roof of the hospital. SOLUTION C e = 1.2 C t = 1.0 I = 1.2 p f = 0.7 C e C t I pg p f = 0.7(1.2)(1.0)(1.2)(25) = 25.2 lb / ft 2 Ans. Ans. p f = 25.2 lb / ft 2

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