Problem Set 2 1. What is the required base area in ποΏ½π‘οΏ½2 of the foundation to support an engine with specific speed of 1200 ποΏ½ποΏ½ποΏ½, and weight of 12,000 ποΏ½ποΏ½, assume bearing capacity of soil as 48 Kpa. Use c= 0.12. Given: π = 1,200οΏ½πππ ππ = 12,000οΏ½πππ ππ΅πΆ = 48οΏ½ ππ π2 π = 0.12 Required: π΄π οΏ½πποΏ½ππ‘ 2 Solution: πππππ’πππ = ππππ πππ = ππ + ππ π΄π ππ΅πΆ ; πΉπ = 2 → 4οΏ½(π’π ποΏ½4) πΉπ πππππ’πππ = ππππ πππ ππ + ππ ππ΅πΆ = π΄π πΉπ ππ = πππ √π ππ = (0.12)(12,000οΏ½πππ )√1,200οΏ½πππ ππ = 49,883.0632οΏ½πππ ππ΅πΆ = 48οΏ½ πππ ππ 1000οΏ½π 1οΏ½πππ × × = 4,892.9663οΏ½ π2 1οΏ½ππ 9.81οΏ½π π2 πππ 49,883.0632οΏ½πππ + 12,000οΏ½πππ 4,892.9663οΏ½ π2 = π΄π 4 π΄π = 50.589οΏ½π2 × π΄π = 544.256οΏ½ππ‘ 2 3.282 οΏ½ππ‘ 2 1οΏ½π2 Problem Set 2 2. Determine the required speed of an engine having a weight and foundation area to be 24,750 ποΏ½ποΏ½π οΏ½ and 155 ποΏ½π‘οΏ½2, respectively. Assume a soil bearing capacity as 45 ποΏ½ποΏ½/ποΏ½ποΏ½2. Use c =0.11. Given: ππ = 24,750οΏ½πππ π΄π = 155οΏ½ππ‘ 2 ππ΅πΆ = 45οΏ½ πππ ππ2 π = 0.11 Required: πππππππ Solution: πππππ’πππ = ππππ πππ ππ + ππ ππ΅πΆ = ; οΏ½πΉπ = 2 → 4οΏ½(π’π ποΏ½4) π΄π πΉπ πππ 2.205οΏ½πππ 1002 οΏ½ππ2 1οΏ½π2 ππ΅πΆ = 45οΏ½ × × × ππ2 1οΏ½πππ 1οΏ½π2 3.282 οΏ½ππ‘ 2 πππ ππ΅πΆ = 92,230. 2573οΏ½ 2 ππ‘ πππ 24,750οΏ½πππ + ππ 92,230. 2573οΏ½ ππ‘ 2 = 155οΏ½ππ‘ 2 4 ππ = 3,549,172.47οΏ½πππ ππ = πππ √π 3,549,172.47οΏ½πππ = (0.11)(24,750οΏ½πππ )√π π = 1,699,489.287οΏ½πππ 3. The volume of concrete needed for the foundation of an engine is 15 ποΏ½π’οΏ½ποΏ½ποΏ½ποΏ½ ποΏ½ποΏ½π‘οΏ½ποΏ½ποΏ½π οΏ½. The concrete mixture is 1: 3:5 by volume. Calculate the number of 40 ποΏ½ποΏ½ − ποΏ½ποΏ½ποΏ½π οΏ½ of cement needed considering the density of cement as 1500 ποΏ½ποΏ½/ποΏ½3. Given: ππ = 15οΏ½π3 Problem Set 2 πΆππππππ‘ποΏ½πππ₯π‘π’ππ = 1: 3: 5 → 1οΏ½ππποΏ½ππππππ‘, 3οΏ½ππ‘ 3 πποΏ½ππππ, 5οΏ½ππ‘ 3 πποΏ½πΊπππ£ππ πππππππ‘ = 1,500οΏ½ ππππ = 40οΏ½ ππ π3 ππ πππ Required: ππ. πποΏ½ππππ Solution: πππππ = ∑ π΄ππ πππ’π‘ποΏ½π£πππ’πποΏ½πποΏ½πππ‘ππππππ οΏ½ → ππππ’ππ‘οΏ½πποΏ½ππππππ‘ππποΏ½πππ‘πππππ = ππ‘ 3 πππ πΎπππ‘πππππ (ππΊπππ‘πππππ )(πΎπ€ππ‘ππ ) ππ 40οΏ½ πππππππ‘ π3 3.283 οΏ½ππ‘ 3 πππ ππππ’ππ‘οΏ½πποΏ½ππππππ‘ = = = 0.02667 × ππ πππππππ‘ πππ 1οΏ½π3 1,500οΏ½ 3 π ππ‘ 3 ππππ’ππ‘οΏ½πποΏ½ππππππ‘ = 0.941 πππ πππ 110οΏ½ 3 ππ‘ 3 ππ‘ ππππ’ππ‘οΏ½πποΏ½π πππ = × 3οΏ½ lb bag (2.64) (62.4οΏ½ f3 ) ππ‘ ππ‘ 3 ππππ’ππ‘οΏ½πποΏ½π πππ = 2.0032 bag πππ 96οΏ½ 3 ππ‘ 3 ππ‘ ππππ’ππ‘οΏ½πποΏ½ππππ£ππ = × 5οΏ½ lb bag (2.66) (62.4οΏ½ f3 ) ππ‘ ππ‘ 3 ππππ’ππ‘οΏ½πποΏ½ππππ£ππ = 2.8918 πππ πππππππ 1οΏ½ππ‘ 3 × πππ 7.48οΏ½πππππππ 3 ππ‘ ππππ’ππ‘οΏ½πποΏ½π€ππ‘ππ = 0.9358 πππ ππππ’ππ‘οΏ½πποΏ½π€ππ‘ππ = 7οΏ½ πππππ = 0.941 ππ‘ 3 ππ‘ 3 ππ‘ 3 ππ‘ 3 + 2.0032 + 2.8918 + 0.9358 πππ πππ πππ bag Problem Set 2 ππ‘ 3 πππππ = 6.7718οΏ½ πππ ππ. πποΏ½ππππππ‘οΏ½πππ = ππ = 15οΏ½π3 × ππ πππππ 3.283 οΏ½ππ‘ 3 = 529.313οΏ½ππ‘ 3 οΏ½ 1οΏ½π3 529.313οΏ½ππ‘ 3 ππ‘ 3 6.7718οΏ½ πππ ππ. πποΏ½ππππππ‘οΏ½πππ = ππ. πποΏ½ππππππ‘οΏ½πππ = 78.1643οΏ½ππππ ππ. πποΏ½ππππππ‘οΏ½πππ ≈ 79οΏ½ππππ 4. A foundation measure 12′π₯οΏ½14′π₯οΏ½16′. Find the number of sacks of cement needed for 1:2:4 mixture. Given: ππ = 12οΏ½ππ‘ × 14οΏ½ππ‘ × 16οΏ½ππ‘ = 2688οΏ½ππ‘ 3 πππππππ‘ποΏ½πππ‘ππ = 1: 2: 4 Required: ππ’πππποΏ½πποΏ½π ππππ /ππππ Solution: ππ. πποΏ½ππππππ‘οΏ½πππ = ππ πππππ 2688οΏ½ππ‘3 ππ. πποΏ½ππππππ‘οΏ½πππ = ππ‘ 3 πππ ππ. πποΏ½ππππππ‘οΏ½πππ = 530.90947 ππ. πποΏ½ππππππ‘οΏ½πππ ≈ 531οΏ½ππππ 5.06301 5. A machine foundation has a base dimension of 4ποΏ½ π₯οΏ½ 8ποΏ½ and has a weight equal to 4 times the weight of the engine. Find the maximum weight of the engine to be mounted if the safe bearing pressure is 220 πΎοΏ½ποΏ½ποΏ½. Given: π΄πππ‘π‘ππ = 4οΏ½π × 8οΏ½π = 32οΏ½π2 ππ = 4ππ Problem Set 2 ππ΅πΆ = 220 ππ π2 Required: πππ₯.οΏ½οΏ½πππππππ Solution: ππππ πππ = πππππ’πππ ππ΅πΆ ππ + ππ = οΏ½; πΉπ = 2οΏ½ππποΏ½πππ₯. π€πππβπ‘ πΉπ π΄π ππ π2 = 4ππ + ππ 2 32οΏ½π2 ππ 220 2 π = 5ππ 2 32οΏ½π2 ππ = 704οΏ½ππ 220 6. A machine foundation has a trapezoidal cross section with bases 4ποΏ½ ποΏ½ποΏ½ποΏ½ 6ποΏ½. The height is 2.5ποΏ½ and the foundation length of 7ποΏ½. Find the required gravel to be used for 1: 2: 4 ποΏ½ποΏ½π₯οΏ½π‘οΏ½π’οΏ½ποΏ½ποΏ½. Given: π = 4οΏ½π π ′ = 6οΏ½π β = 2.5οΏ½π πΏ = 7οΏ½π πΆππππππ‘ποΏ½π ππ‘ππ = 1: 2: 4 Required: ππππ’ππ‘οΏ½πποΏ½ππππ£πποΏ½πποΏ½ππ‘ 3 Problem Set 2 Solution: π + π′ ππ = [( ) β] πΏ 2 4οΏ½π + 6οΏ½π ππ = [( ) 2.5οΏ½π] 7οΏ½π 2 3.283 οΏ½ππ‘ 3 3 ππ = 87.5οΏ½π × 1οΏ½π3 ππ = 3,087.6608οΏ½ππ‘ 3 ππ πππππ ππ‘ 3 πππππ1:2:4 = 5.06301 οΏ½οΏ½ πππ ππ. πποΏ½ππππππ‘οΏ½πππ = 3,087.6608οΏ½ππ‘ 3 ππ. πποΏ½ππππππ‘οΏ½πππ = ππ‘ 3 5.06301 πππ ππ. πποΏ½ππππππ‘οΏ½πππ = 609.846οΏ½ππππ ππ. πποΏ½ππππππ‘οΏ½πππ ≈ 610οΏ½ππππ 96οΏ½ ππππ’ππ‘οΏ½πποΏ½ππππ£ππ ≈ πππ ππ‘ 3 (2.66) (62.4οΏ½ lbf ) ππ‘ 3 ππ‘ 3 × 4οΏ½ × 610οΏ½ππππ bag ππππ’ππ‘οΏ½πποΏ½ππππ£ππ ≈ 1,411.220οΏ½ππ‘ 3 7. A batch of concrete consisted of 200 ποΏ½ποΏ½π οΏ½ fine aggregate, 350 ποΏ½ποΏ½π οΏ½ coarse aggregate, 94 ποΏ½ποΏ½π οΏ½ cement and 5 ποΏ½ποΏ½ποΏ½ποΏ½ποΏ½ποΏ½π οΏ½ ποΏ½ποΏ½ π€οΏ½ποΏ½π‘οΏ½ποΏ½ποΏ½. The specific gravity of the sand and gravel may be taken as 2.65 and that of cement of 3.10. What was the weight of concrete in place per cubic foot. Given: ππ πππ = 200οΏ½πππ πππππ£ππ = 350οΏ½πππ πππππππ‘ = 94οΏ½πππ ππ€ππ‘ππ = 5οΏ½πππππππ ππΊπ πππ = 2.65 ππΊππππ£ππ = 2.65 ππΊππππππ‘ = 3.1 Required: Problem Set 2 πΎπππππππ‘π → πππ ππ‘ 3 Solution: πππ πππ ) = 165.36 ππ‘ 3 ππ‘ 3 πππ πππ πΎππππ£ππ = ππΊππππ£ππ πΎπ€ππ‘ππ = (2.65) (62.4 3 ) = 165.36 3 ππ‘ ππ‘ πππ πππ πΎππππππ‘ = ππΊππππππ‘ πΎπ€ππ‘ππ = (3.1) (62.4 3 ) = 193.44 3 ππ‘ ππ‘ πππ πΎπ€ππ‘ππ = 62.4 3 ππ‘ πΎπ πππ = ππΊπ πππ πΎπ€ππ‘ππ = (2.65) (62.4 200οΏ½πππ = 1.209οΏ½ππ‘ 3 πππ 165.36 3 ππ‘ πππππ£ππ 350οΏ½πππ πππππ£ππ = = = 2.116οΏ½ππ‘ 3 οΏ½ πππ πΎππππ£ππ 165.36 3 ππ‘ 94οΏ½πππ πππππππ‘ πππππππ‘ = = = 0.485οΏ½ππ‘ 3 οΏ½ πππ πΎππππππ‘ 193.44 3 ππ‘ 7.48οΏ½ππ‘ 3 ππ€ππ‘ππ = 5οΏ½πππππππ × = 37.4οΏ½ππ‘ 3 1οΏ½ππππππ ππ πππ = ππ πππ = πΎπ πππ ππ = ππ πππ + πππππ£ππ + πππππππ‘ + ππ€ππ‘ππ ππ = 1.209οΏ½ππ‘ 3 + 2.116οΏ½ππ‘ 3 + 0.485οΏ½ππ‘ 3 + 37.4οΏ½ππ‘ 3 ππ = 41.21οΏ½ππ‘ 3 ππ = ππ πππ + πππππ£ππ + πππππππ‘ + ππ€ππ‘ππ ππ€ππ‘ππ = ππ€ππ‘ππ πΎπ€ππ‘ππ ππ€ππ‘ππ = (37.4οΏ½ππ‘ 3 ) (62.4 πππ ) ππ‘ 3 οΏ½ππ€ππ‘ππ = 2,333.76οΏ½πππ π = 200οΏ½πππ + 350οΏ½πππ + 94οΏ½πππ + 2,333.76οΏ½πππ π = 2977.76οΏ½πππ πΎπ = ππ 2,977.76οΏ½πππ = ππ 41.21οΏ½ππ‘ 3 Problem Set 2 πΎπ = 72.258 πππ ππ‘ 3 Problem Set 2 π π′
0
advertisement
Download
advertisement
Add this document to collection(s)
You can add this document to your study collection(s)
Sign in Available only to authorized usersAdd this document to saved
You can add this document to your saved list
Sign in Available only to authorized users