Appendix B_ Calculations_ STRUCTURES PORTION (1-17-2015)
- 1. RICE UNIVERISTY ENGINEERS WITHOUT BORDERS Appendix D : Calculations Hydraulic, Structural, and Electrical Project Leaders: Edgar Silva and Rachel Sterling 9/14/2014
- 2. Given Values Value Origins General Total Population (people) 1,200 Census from community Growth Rate 7% UN Nicaraguan growth rate with safety factor of 5 Location Of Tank (Elevation, M) 825 Given by ENACAL, determined using GPS Location Of Pump (Elevation, M) 724 Given by ENACAL, determined using GPS Hydraulics Gallons Per Person (Gallons/Day) 24 Provided by handbook from engineer at ENACAL Needed Tank Capacity (Days) 1 Community request Pump Functioning Hours 5.5 - 9.5 From pressure test completed on assessment trip, only accounting for time that PSI at this elevation is >10 (min pressure for pump is < 3 psi) Number of tanks (22,000L each) Max 8 This number is capped at 8 because it was determined to be Structural Soil Type Clay Determined by assessment trip Psi Strength of Concrete 3000 Determined by mix ratios Psi Strength of Steel 60000 Assuming grade 60 steel is used Mass of pump (lbs) 201 From pump spec sheet Mass of filled tank (kg) 22000 Determined by volume of tank * More values given in structural section Electrical Power Supply Single Phase This is determined from the phase of the Nicaraguan electrical grid
- 3. Structural Calculations Soil Bearing Capacity Calculations To calculate the ultimate bearing capacity of the soil, we used the following equation, the Terzaghi Bearing Capacity equation. The highest pressures will be experienced by the slabs holding the 22000 liter water tanks. 𝑞_𝑢𝑙𝑡 = (𝑐)(𝑁𝑐)(𝑠𝑐) + (ɤ)(𝐷)(𝑁𝑞)(𝑠𝑞) + (0.5)(𝐵)(ɤ)(𝑁𝑔)(𝑠ɤ) The following chart contains important values: Cohesive Strength of Soil (c) 239.4 kPa Unit Weight of Soil (ɤ) 8.568 kN/m3 Depth of Foundation (D) 0 m Width of Foundation (B) 4 m Safety Factor 3 Angle Of Internal Friction 17 degrees Bearing Capacity Factor (cohesion) Nc 12.3 Bearing Capacity Factor (surcharge and friction) Nq 4.8 Bearing Capacity Factor (self weight and friction) Nɤ 1.7 Shape Factor (Sc) 1.3 Shape Factor (Sq) 1 Shape Factor (Sɤ) 0.8 The cohesive strength of soil was determined by using a pocket penetrometer at the surface and 6 inches below the surface. A value of 239.4 kPa (2.5 tons per square foot) was measured at the surface. A value of 430.9 kPa (4.5 tons per square foot) was measured 6 inches below the surface. From these measurements we know the soil strengthens as depth increases. For our calculations we will be on the safe side and use the weakest measurement of 239.4 kPa. The unit weight of the soil was determined by taking a 600 ml sample of soil in a bottle, and measuring its mass. The depth and width of foundation were determined by our design criteria. The slabs need to be at least 4 meters wide to accommodate the water tanks. For simplicity of construction the slabs will just lay on the surface of the ground. The safety factor was a value recommended to us by our mentor. The internal angle of friction and bearing capacities were determined by identifying the type of soil at our construction site. By performing the ribbon test, the soil was identified as high plasticity clay soil. According to Swiss Standard SN 670 010b, this soil has an angle of friction between 17 and 31 degrees. We will use an angle of 17 degrees in our calculations to assume worst case soil conditions. From Terzaghi’s tables, an angle of 17 degrees correspond to Nc = 12.3, Nq = 4.8, and Nɤ = 1.7.
- 4. Shape factors were derived from Terzaghi’s tables by assuming a square foundation. A square foundation was used for simplicity of construction. 𝑄_𝑢𝑙𝑡 = (239.4)(12.3)(1.3)+ (8.568)(0)(4.8)(1)+ 0.5(4)(8.568)(1.7)(0.8) = 3851.3𝑘𝑃𝑎 𝑆𝑎𝑓𝑒 𝐵𝑒𝑎𝑟𝑖𝑛𝑔 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = 𝑄𝑠 = 𝑄𝑢𝑙𝑡 𝑆𝑎𝑓𝑒𝑡𝑦 𝐹𝑎𝑐𝑡𝑜𝑟 = 3851.3 3 = 1283.8 𝑘𝑃𝑎 𝑇ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝐿𝑜𝑎𝑑 𝑠𝑢𝑝𝑝𝑜𝑟𝑡𝑒𝑑 𝑏𝑦 𝑆𝑙𝑎𝑏 = 1000(𝑄𝑠)(𝐵2 ) 𝑔 = 1000(1283.8)(42 ) 9.81 = 2093814 𝑘𝑔 The mass of the 22000 liter tank is 22000 kg. Therefore, the soil is more than strong enough to support the tank.
- 5. Pump Station Cross-Section Calculations a. Design slab thickness: The thickness of the slab will be controlled by the shear forces it needs to resist. From our previous calculations, we know the soil is more than strong enough to support our structures. However, the soil is expansive, and will expand and contract based on precipitation in the area. The the slab will trap moisture in the soil below it, and that soil will remain at a relatively constant moisture content, while the soil at the edges of the slab will quickly lose moisture in the dry season and shrink, and will quickly absorb moisture in the wet season and expand. This leads to two critical conditions for the slab, called dishing and doming, or also called edge list condition and center lift condition. Dishing occurs in the wet season when the slab is only supported by the expanded soil on the edges. Doming occurs in the dry season when the slab is only supported by the still moist soil under the slab. For our calculations, we will treat the slab as a wide beam. Additionally, we will assume worse case soil expansion and contraction. We will assume the slab dimensions of 5.80 x 2.60 due to the amount of spacing needed for the pump and valves. Overhead view of forces:
- 6. Doming force diagram: Dishing force distribution: Table of Constraints Dimensions of Slab 5.80 x 2.60 meters Mass of pump 91.17 kg Dimensions of pump platform 1.00 x 0.40 x 0.50 meters Density of concrete and CMU blocks 2400 kg/m3 Strength of concrete, f’c 13.8 MPa Shear Strength Safety Factor 0.75 Height of CMU wall 2.40 meters Width of CMU wall 0.15 meters Acceleration due to gravity (g) 9.81 m/s2
- 7. i. Doming slab thickness calculation: 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑖𝑛 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 𝑠𝑞𝑢𝑎𝑟𝑒𝑑 = √ 𝑓′ 𝑐2 6 (1000000) The slab will be designed to have no vertical web reinforcement. By ACI code, the design shear strength of concrete without web reinforcement is (𝐷𝑒𝑠𝑖𝑔𝑛 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠) = (0.5)(𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)(𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ) The maximum shear will occur in the center of the slab, where the soil supports the slab. (0.5)( 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)( 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ)( 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 + 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 = (𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑔) 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 = (2.6)(.15)(2.4)(2400)(9.81) = 22037 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒)(𝑔)(𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)/2 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (2.6)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(2400)(9.81)(5.8) 2 = 177522(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 = 2(𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑔)(𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)/2 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 = 2(. 15)(2.4)(2400)(9.81)(5.8) 2 = 49160 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 (0.5)(0.75)( √13.8 2 6 )(1000000)(2.6)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 22037 + 177522(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)+ 49160 Height of slab = .167 meters ii. Dishing slab thickness calculation: The maximum shear will occur at the left and right of the slab, where it is supported by the soil. Calculations will find shear at the left support.
- 8. (0.5)( 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)( 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ)( 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑢𝑚𝑝 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚 + 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 + 𝑟𝑖𝑔ℎ𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 − 𝑟𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 𝐹𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑢𝑚𝑝 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚 = ( 𝑝𝑢𝑚𝑝 𝑤𝑒𝑖𝑔ℎ𝑡)(𝑔) + (𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑝𝑢𝑚𝑝 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒)(𝑔) 𝐹𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑢𝑚𝑝 𝑝𝑙𝑎𝑡𝑓𝑜𝑟𝑚 = (91.17)(9.81)+ (1)(0.4)(0.5)(2400)(9.81) = 5603 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑟𝑖𝑔ℎ𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 = (𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑔) 𝑟𝑖𝑔ℎ𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 = (2.6)(.15)(2.4)(2400)(9.81) = 22037 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒)(𝑔)(𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (2.6)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(2400)(9.81)(5.8) = 355044(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 = 2(𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑜𝑓 𝑤𝑎𝑙𝑙)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑙𝑙)(𝑔)(𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 = 2(. 15)(2.4)(2400)(9.81)(5.8) = 98320 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑐𝑒 2 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 5603 + (2)22037 + 98320 + 355044(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) 2 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 73999 + 177522(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) (0.5)(0.75)( √13.82 6 )(1000000)(2.6)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 5603 + 355044(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)+ 98320 + 22037 − 73999 − 177522(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) Height of slab = .123 meters iii. Final slab thickness Bowling requires a thicker slab, so that is our controlling case for shear resistance. Therefore, the slab must be at least .167 meters thick. For ease of construction, the slab will be designed as 0.18 meters thick
- 9. b. Design moment calculation: The slab is relatively thin, so bending moments in the slab should be considered for the dishing and bowling case. These moments will be used to calculate the needed steel reinforcement. To calculate the design moment, we need to determine what part of the slab will experience the most moment. We will assume the slab dimensions of 5.80 x 2.60 x 0.18 meters as given, due to the amount of spacing needed for the pump and valves. The previous figures and table of constraints will be used. i. doming moment calculation In the doming case, the slab acts like a cantilever, so maximum moment will be at the center of the slab. 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) + (𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) + (𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) From doming shear calculations 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 = 22037 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 = 49160 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = 177522(0.18) = 31954 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (22037)(2.9)+ (49160)( 2.9 2 ) + 31954( 2.9 2 ) 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 181523 𝑛𝑒𝑤𝑡𝑜𝑛 𝑚𝑒𝑡𝑒𝑟𝑠 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 ∗ 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 181523 ∗ 1.2 = 217827 𝑁 ∗ 𝑚 ii. dishing moment calculation In the dishing case, the slab acts like a simply supported beam, so maximum moment will be at the center of the slab. 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = ( 𝑟𝑖𝑔ℎ𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡)( 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) + (𝑟𝑖𝑔ℎ𝑡 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 𝑓𝑜𝑟𝑐𝑒)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) + ( 𝑟𝑖𝑔ℎ𝑡 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒)( 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) − (𝑟𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) From dishing shear calculations 𝑟𝑖𝑔ℎ𝑡 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑤𝑎𝑙𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 = 22037 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑟𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑝𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑤𝑎𝑙𝑙 = 98320 2 = 49160 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑟𝑖𝑔ℎ𝑡 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = 355044(0.18)/2 = 31954 𝑛𝑒𝑤𝑡𝑜𝑛𝑠
- 10. 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 73999 + 177522(.18) = 105953 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (22037)(2.9)+ (49160)( 2.9 2 ) + 31954( 2.9 2 ) − 105953(2.9) 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = −125741 𝑛𝑒𝑤𝑡𝑜𝑛 𝑚𝑒𝑡𝑒𝑟𝑠 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 ∗ 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 125741 ∗ 1.2 = 150889 𝑁 ∗ 𝑚 c. Rebar design calculations For rebar design, Whitney Stress Block theory will be used. We need to solve for the area of steel to be used. Tensile stresses will be at the top of the slab during doming, and tensile stresses will be at the bottom of the slab during dishing. A concrete skirt will extend into the ground around the perimeter of the slab, to prevent the erosion of soil under the slab. It is submerged in the soil, so its weight has been neglected in moment calculations, but it will provide a compressive surface for the doming case. We will assume the bars are placed 5 cm from the faces of the slab Design Inputs Concrete Design Compressive Stress (f'c) 13.8 MPa Yield Stress of steel (fy) 420 MPa Base (b) 0.18 meter Doming Rebar Design Moment 217827 N-m Dishing Rebar Design Moment 150889 N-m Safety Factor (phi) 0.9 Depth (d) 0.13 meter Width 2.60 meter Modulus of elasticity of steel (Es) 200 GPa a to c ratio (β) 0.85 Dishing force diagram
- 11. Doming force diagram Doming force cross section 𝛽𝑐 = 𝑎 Dishing Force Equilibrium: 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝑠𝑡𝑒𝑒𝑙 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑓𝑜𝑟𝑐𝑒 (𝑓′ 𝑐)(𝛽𝑐1)(𝑤𝑖𝑑𝑡ℎ) = (𝐴𝑠)(𝑓𝑦) (13800000)(0.85∗ 𝑐1)(2.6) = (𝐴𝑠)(420000000) (30498000)(𝑐1) = (𝐴𝑠)(420000000)
- 12. Doming Force Equilibrium: 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝑠𝑡𝑒𝑒𝑙 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑓𝑜𝑟𝑐𝑒 (𝑓′ 𝑐)(𝛽𝑐2)(𝑤𝑖𝑑𝑡ℎ) = (𝐴𝑠)(𝑓𝑦) (13800000)(0.85 ∗ 𝑐2)(2 ∗ .15) = (𝐴𝑠)(420000000) (3519000)( 𝑐2) = (𝐴𝑠)(420000000) Dishing Moment Equilibrium: ( 𝑓′ 𝑐)( 𝛽𝑐1)( 𝑤𝑖𝑑𝑡ℎ) (𝑑 − 𝛽𝑐1 2 ) = 𝑑𝑖𝑠ℎ 𝑚𝑜𝑚𝑒𝑛𝑡 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 (13800000)(0.85∗ 𝑐1)(2.6)(. 13 − 0.85 ∗ 𝑐1 2 ) = 150889 0.9 (30498000)(𝑐1)(.13 − .425(𝑐1)) = 167654 Doming Moment Equilibrium: ( 𝑓′ 𝑐)( 𝛽𝑐2)( 𝑤𝑖𝑑𝑡ℎ)(𝑑 − 𝛽𝑐2 2 ) = 𝑑𝑜𝑚𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 (13800000)(0.85∗ 𝑐2)(2∗ .15)(. 63 − 0.85 ∗ 𝑐2 2 ) = 217827 0.9 (3519000)(𝑐2)(.63 − .425(𝑐2)) = 242030 Using the Dishing Moment Equation, c1=.0507 m Using the Dishing Force Equation, As = .0037 m2 The moment capacity in the doming case must be checked for this area of steel. Using the Doming Force Equation, c2 = .440 m The compressive block remains within the concrete skirt, so the assumptions made in the force balance equation are correct. Using the Doming Moment Equation, 685382 n*m>242030 n*m The moment capacity exceeds the expected moment during doming. Therefore, the design is good and As = .0037 m2 𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 = 𝐴𝑠 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = . 0037 . 18 ∗ 2.6 = 0.79% 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑛𝑒𝑒𝑑𝑒𝑑 𝑎𝑙𝑜𝑛𝑔 𝑠ℎ𝑜𝑟𝑡 𝑠𝑖𝑑𝑒 = .0037 𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑛𝑒𝑒𝑑𝑒𝑑 𝑎𝑙𝑜𝑛𝑔 𝑙𝑜𝑛𝑔 𝑠𝑖𝑑𝑒 = .0037 ∗ 5.8 2.6 = .083 𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 #4 𝑟𝑒𝑏𝑎𝑟 = 0.000129032 𝑚2 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #4 𝑟𝑒𝑏𝑎𝑟 𝑠ℎ𝑜𝑟𝑡 𝑠𝑖𝑑𝑒 = .0037 .000129032 = 29 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #4 𝑟𝑒𝑏𝑎𝑟 𝑙𝑜𝑛𝑔 𝑠𝑖𝑑𝑒 = .0083 .000129032 = 65
- 13. 𝐴𝑟𝑒𝑎 𝑜𝑓 #5 𝑟𝑒𝑏𝑎𝑟 = 0.0002 𝑚2 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #5 𝑟𝑒𝑏𝑎𝑟 𝑠ℎ𝑜𝑟𝑡 𝑠𝑖𝑑𝑒 = .0037 .0002 = 19 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #5 𝑟𝑒𝑏𝑎𝑟 𝑙𝑜𝑛𝑔 𝑠𝑖𝑑𝑒 = .0083 .0002 = 42 Calculations such as this can be easily repeated for other rebar sizes. Selection of rebar will depend on local prices of steel. Drawings will be for #4 rebar, due to its known price in Nicaragua. Rebar will be equally spaced, with 10 cm or greater clearance from the edges. The density of steel of 2.05% is greater than the recommended value of 0.1% of steel needed to prevent cracking from ACI committee 360, so this design is good.
- 14. Pump Station Construction Notes a. Concrete skirt The purpose of the concrete skirt is to prevent the slab from being undermined by erosion. The subterranean concrete skirt will be poured before the slab is poured and the slab rebar is placed. Ensure that the rebar that is inside the CMU walls is vertically placed in the concrete skirt and sticking out of the ground and to the proper measurements. This rebar will help secure the skirt and the CMU walls. b. Pump platform Ensure that rebar is sticking up in the middle of the slab before pouring. Also ensure that this rebar sticks out at least 0.45 meters from the surface of the slab. The pump will be placed on a raised concrete platform, a 1.00x0.50x0.50 meter block of concrete. This platform will be poured separately from the underlying slab, so it needs to be secured to the slab through the rebar. This rebar will secure the platform, and therefore the pump, to the slab, and the rebar will minimize cracking so that the pump remains level. c. Thrust block The thrust block is a block of concrete placed around the pipes in front of the discharge end of the pump so that the pipes are protected from hydraulic forces. The thrust block adds mass to the pipes, so for a given force they are deflected a shorter distance. This reduction in deflections reduces the stress on the pipes and reduces the risk of the pipes breaking. It is poured after the slab is poured and after the pipes are set. d. Top of wall slope After the wall is complete, ensure that the top of the wall has a slope. This can be achieved by mixing concrete and placing it on the top of the CMU wall. Make sure the 0.2 meter high section is on the side of the facility facing uphill and that the top of the wall slope is in the same direction as the surrounding terrain. This ensures that rain runs off of the roof and away from the building and not back towards the foundation. e. CMU wall and roof interface Ensure that the rebar that is placed in the CMU walls penetrates through the top of the wall and the sloped concrete layer. This allows the welders to weld metal beams directly onto the structure. These beams go across the roof, which provides an anchor point for the sheet metal roof.
- 15. Storage Facility Concrete Cross Sections a. Design slab thickness: The thickness of the slab will be controlled by the shear forces it needs to resist. From our previous calculations, we know the soil is more than strong enough to support our structures. However, the soil is expansive, and will expand and contract based on precipitation in the area. The slab will trap moisture in the soil below it, and that soil will remain at a relatively constant moisture content, while the soil at the edges of the slab will quickly lose moisture in the dry season and shrink, and will quickly absorb moisture in the wet season and expand. This leads to two critical conditions for the slab, called dishing and doming, or also called edge list condition and center lift condition. Dishing occurs in the wet season when the slab is only supported by the expanded soil on the edges. Doming occurs in the dry season when the slab is only supported by the still moist soil under the slab. For our calculations, we will treat the slab as a wide beam. Additionally, we will assume worse case soil expansion and contraction. We will assume the slab dimensions of 3.15 x 3.15 due to the amount of spacing needed for the water tank. Overhead view of forces:
- 16. Doming force diagram: Dishing force diagram: Table of Constraints Dimensions of Slab 3.15 x 3.15 meters Mass of water tank 22000 kg Strength of concrete, f’c 13.8 MPa Shear Strength Safety Factor 0.75 Acceleration due to gravity (g) 9.81 m/s2
- 17. i. Doming slab thickness calculation: 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ 𝑖𝑛 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑝𝑒𝑟 𝑚𝑒𝑡𝑒𝑟 𝑠𝑞𝑢𝑎𝑟𝑒𝑑 = √ 𝑓′ 𝑐2 6 (1000000) The slab will be designed to have no vertical web reinforcement. By ACI code, the design shear strength of concrete without web reinforcement is (𝐷𝑒𝑠𝑖𝑔𝑛 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠) = (0.5)(𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)(𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ) The maximum shear will occur in the center of the slab, where the soil supports the slab. (0.5)( 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)( 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ)( 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒)(𝑔)(𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)/2 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (3.15)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(2400)(9.81)(3.15) 2 = 116808(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡𝑎𝑛𝑘)(𝑔)/2 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 = (22000)(9.81) 2 = 107910 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 (0.5)(0.75)( √13.8 2 6 )(1000000)(3.15)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 116808(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) + 107910 Height of slab = .176 meters ii. Dishing slab thickness calculation: The maximum shear will occur at the left and right of the slab, where it is supported by the soil. Calculations will find shear at the left support. (0.5)( 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟)( 𝐶𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ)( 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 + 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 − 𝑟𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒)(𝑔)(𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑠𝑙𝑎𝑏) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = (3.15)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)(2400)(9.81)(3.15) = 233615(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏)
- 18. 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 = (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡𝑎𝑛𝑘)(𝑔) 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 = (22000)(9.81) = 215820 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑜𝑤𝑛𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑐𝑒 2 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 233615(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) + 215820 2 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 107910 + 116808(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) (0.5)(0.75)( √13.8 2 6 )(1000000)(3.15)(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) = 233615(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) + 215820 − 107910− 116808(ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏) Height of slab = .176 meters iii. Final slab thickness Both cases result in the same thickness slab. Therefore, the slab must be at least .176 meters thick. For ease of construction, the slab will be designed as 0.18 meters thick b. Design moment calculation: The slab is relatively thin, so bending moments in the slab should be considered for the dishing and bowling case. These moments will be used to calculate the needed steel reinforcement. To calculate the design moment, we need to determine what part of the slab will experience the most moment. We will assume the slab dimensions of 3.15 x 3.15 x 0.18 meters as given, due to the amount of spacing needed for the pump and valves. The previous figures and table of constraints will be used. i. doming moment calculation In the doming case, the slab acts like a cantilever, so maximum moment will be at the center of the slab. 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘)(𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) + (𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) From doming shear calculations 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 = 107910 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡 = 116808(0.18) = 21025 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (107910)(0.6366) + (21025)( 1.575 2 ) 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = 85253 𝑛𝑒𝑤𝑡𝑜𝑛 𝑚𝑒𝑡𝑒𝑟𝑠
- 19. 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 ∗ 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 85253 ∗ 1.2 = 102303 𝑁 ∗ 𝑚 ii. dishing moment calculation In the dishing case, the slab acts like a simply supported beam, so maximum moment will be at the center of the slab. 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = ( 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑙𝑓 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘)( 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) + ( 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑖𝑔ℎ𝑡 𝑠𝑙𝑎𝑏 𝑤𝑒𝑖𝑔ℎ𝑡)( 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) − (𝑟𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒)(𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) From dishing shear calculations 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑖𝑔ℎ𝑡 ℎ𝑎𝑙𝑓 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑡𝑎𝑛𝑘 = 215820 2 = 107910 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑓𝑜𝑟𝑐𝑒 𝑓𝑟𝑜𝑚 𝑟𝑖𝑔ℎ𝑡 𝑝𝑎𝑟𝑡 𝑜𝑓 𝑠𝑙𝑎𝑏 = 233615(0.18) 2 = 21025 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝑅𝑖𝑔ℎ𝑡 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 𝑓𝑜𝑟𝑐𝑒 = 107910 + 116808(0.18) = 128935 𝑛𝑒𝑤𝑡𝑜𝑛𝑠 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = (107910)(0.6366) + (21025)( 1.575 2 ) − 128935(1.575) 𝐵𝑒𝑛𝑑𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 = −117820 𝑛𝑒𝑤𝑡𝑜𝑛 𝑚𝑒𝑡𝑒𝑟𝑠 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 ∗ 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 𝐷𝑒𝑠𝑖𝑔𝑛 𝑚𝑜𝑚𝑒𝑛𝑡 = 117820 ∗ 1.2 = 141384 𝑁 ∗ 𝑚 c. Rebar design calculations For rebar design, Whitney Stress Block theory will be used. We need to solve for the area of steel to be used. Tensile stresses will be at the top of the slab during doming, and tensile stresses will be at the bottom of the slab during dishing. Due to the narrow slab and the existence of positive and negative moments, 2 rows of steel will be used. We will assume the bars are placed 5 cm from the faces of the slab Design Inputs Concrete Design Compressive Stress (f'c) 13.8 MPa Yield Stress of steel (fy) 420 MPa Base (b) 0.18 meter Dishing Rebar Design Moment 141384 N-m Doming Rebar Design Moment 102303 N-m Safety Factor (phi) 0.9 Depth (d) 0.13 meter Width 3.15 meter Modulus of elasticity of steel (Es) 200 GPa a to c ratio (β) 0.85
- 20. Dishing force diagram Doming force diagram 𝛽𝑐 = 𝑎 Dishing force equilibrium: 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 + 𝑠𝑡𝑒𝑒𝑙 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝑠𝑡𝑒𝑒𝑙 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑓𝑜𝑟𝑐𝑒 (𝑓′ 𝑐)(𝛽𝑐1)(𝑤𝑖𝑑𝑡ℎ) + (𝐴𝑠𝑇𝑜𝑝)(𝐸𝑠)(.003)(𝑐1 − .05)/𝑐1 = (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(𝑓𝑦) (13800000)(0.85 ∗ 𝑐1)(3.15) + (𝐴𝑠𝑇𝑜𝑝)(200000000000)(.003)(𝑐1 − .05)/𝑐1 = (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(420000000) (36949500)(𝑐1)+ (𝐴𝑠𝑇𝑜𝑝)(600000000)(𝑐1 − .05)/𝑐1 = (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(420000000) Doming force equilibrium: 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 + 𝑠𝑡𝑒𝑒𝑙 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒 = 𝑠𝑡𝑒𝑒𝑙 𝑡𝑒𝑛𝑠𝑖𝑙𝑒 𝑓𝑜𝑟𝑐𝑒 (𝑓′ 𝑐)(𝛽𝑐2)(𝑤𝑖𝑑𝑡ℎ) + (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(𝐸𝑠)(.003)(𝑐2− .05)/𝑐2 = (𝐴𝑠𝑇𝑜𝑝)(𝑓𝑦) (13800000)(0.85∗ 𝑐2)(3.15)+ (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(200000000000)(.003)(𝑐2 − .05)/𝑐2 = (𝐴𝑠𝑇𝑜𝑝)(420000000) (36949500)(𝑐2)+ (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(600000000)(𝑐2 − .05)/𝑐2 = (𝐴𝑠𝑇𝑜𝑝)(420000000) Dishing Moment Equilibrium: ( 𝑓′ 𝑐)( 𝛽𝑐1)( 𝑤𝑖𝑑𝑡ℎ)(𝑑 − 𝛽𝑐1 2 ) + (𝐴𝑠𝑇𝑜𝑝)(𝐸𝑠)(.003)(𝑐1 − .05)(.05)/𝑐1 = 𝑑𝑖𝑠ℎ 𝑚𝑜𝑚𝑒𝑛𝑡 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟 (13800000)(0.85∗ 𝑐1)(3.15)(. 13 − 0.85 ∗ 𝑐1 2 ) + (𝐴𝑠𝑇𝑜𝑝)(200000000000)(.003)(𝑐1 − .05)(.08)/𝑐1 = 141384 0.9 (36949500)(𝑐1)(.13 − .425(𝑐1))+ (𝐴𝑠𝑇𝑜𝑝)(48000000)(𝑐1− .05)/𝑐1 = 157093 Doming Moment Equilibrium: ( 𝑓′ 𝑐)( 𝛽𝑐2)( 𝑤𝑖𝑑𝑡ℎ)(𝑑 − 𝛽𝑐2 2 ) + (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(𝐸𝑠)(.003)(𝑐2 − .05)(.08)/𝑐2 = 𝑑𝑜𝑚𝑒 𝑚𝑜𝑚𝑒𝑛𝑡 𝑠𝑎𝑓𝑒𝑡𝑦 𝑓𝑎𝑐𝑡𝑜𝑟
- 21. (13800000)(0.85 ∗ 𝑐2)(3.15)(. 13 − 0.85 ∗ 𝑐2 2 ) + (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(200000000000)(.003)(𝑐2 − .05)(.05)/𝑐2 = 102303 0.9 (36949500)(𝑐2)(.13 − .425(𝑐2)) + (𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚)(48000000)(𝑐2 − .05)/𝑐2 = 113670 With four equations and four unknowns, c1, c2, AsTop, and AsBottom, these values can be determined. Using Matlab: AsTop = .0020 m2 AsBottom = .0032 m2 𝑇𝑜𝑡𝑎𝑙 𝐷𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 = 𝐴𝑠𝑇𝑜𝑝 + 𝐴𝑠𝐵𝑜𝑡𝑡𝑜𝑚 𝐶𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑛𝑐𝑟𝑒𝑡𝑒 = . 0020 + .0032 . 18 ∗ 3.15 = 0.917% 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑛𝑒𝑒𝑑𝑒𝑑 𝑎𝑙𝑜𝑛𝑔 𝑡𝑜𝑝 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 = .0020 𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 𝑛𝑒𝑒𝑑𝑒𝑑 𝑎𝑙𝑜𝑛𝑔 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 = .0032 𝑚2 𝐴𝑟𝑒𝑎 𝑜𝑓 #4 𝑟𝑒𝑏𝑎𝑟 = 0.000129032 𝑚2 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #4 𝑟𝑒𝑏𝑎𝑟 𝑡𝑜𝑝 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 = .0020 .000129032 = 16 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #4 𝑟𝑒𝑏𝑎𝑟 𝑏𝑜𝑡𝑡𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 = .0032 .000129032 = 25 𝐴𝑟𝑒𝑎 𝑜𝑓 #5 𝑟𝑒𝑏𝑎𝑟 = 0.0002 𝑚2 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #5 𝑟𝑒𝑏𝑎𝑟 𝑡𝑜𝑝 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 = .0020 .0002 = 10 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 #5 𝑟𝑒𝑏𝑎𝑟 𝑏𝑜𝑡𝑡𝑜𝑚 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 = .0032 .0002 = 16 Calculations such as this can be easily repeated for other rebar sizes. Selection of rebar will depend on local prices of steel. Drawings will be for #4 rebar, due to its known price in Nicaragua. Rebar will be equally spaced, with 10 cm or greater clearance from the edges. The density of steel of 0.917% is greater than the recommended value of 0.1% of steel needed to prevent cracking from ACI committee 360, so this design is good.
- 22. d. Landslide Risk Analysis This storage facility will be placed on a sloped hill. The construction area will be flattened prior to construction, but a mass of soil will sit uphill of the facility. In heavy rains some of the hillside may flow downhill and impact the walls of the storage facility. These calculations determine the resistance of the walls to a mass of fluid soil pushing against it. Specifically these calculations will determine the maximum height of flowing soil that the wall can resist. If soil is determined to be flowing near this height, then the wall must be reinforced and braced at the expected centroid of the fluid force. Calculation Inputs Density of Soil 9270 N/m^3 Yield Stress of Steel (fy) 420 MPa Width of Wall 5.6 meter Height of Wall 2.4 meter Thickness of Wall 0.15 meter Depth of rebar in wall 0.075 meter i. Moment Strength of the CMU wall Cross section of wall: Number of rebar = 12 #4 rebar 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙 = 2.36 𝑖𝑛2 = .00152 𝑚2 Using Whitney Stress Block theory 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑜𝑟𝑐𝑒 𝑖𝑛 𝑠𝑡𝑒𝑒𝑙 ∗ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 (𝑓𝑦)(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙)(𝑑 − 𝑎 2 ) = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑎 = (𝑓𝑦)(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙) (0.85)(𝑓’𝑐)(𝑤𝑖𝑑𝑡ℎ) (𝑓𝑦)(𝐴𝑠)(𝑑 − (𝑓𝑦)(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑡𝑒𝑒𝑙) (1.7)(𝑓’𝑐)(𝑤𝑖𝑑𝑡ℎ) ) = 𝑚𝑜𝑚𝑒𝑛𝑡 (420000000)(.00152)(.075− (420000000)(.00152) (1.7)(21000000)(5.6) ) = 𝑀𝑜𝑚𝑒𝑛𝑡 𝑀𝑜𝑚𝑒𝑛𝑡 = 45844 𝑁 ∗ 𝑚
- 23. ii. Height of fluid soil needed to break the wall Assuming the soil is perfectly fluid, then as it accumulates up against a wall segment it will exert a hydrostatic pressure against the wall. This pressure then exerts a moment on the fixed end of the wall in the ground. Force diagram: 𝐹𝑜𝑟𝑐𝑒 = (𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒) ∗ (𝑎𝑟𝑒𝑎) 𝐹𝑜𝑟𝑐𝑒 = (𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡)(ℎ𝑒𝑖𝑔ℎ𝑡) 2 ∗ (ℎ𝑒𝑖𝑔ℎ𝑡)(𝑊𝑖𝑑𝑡ℎ) 𝑀𝑜𝑚𝑒𝑛𝑡 = 𝐹𝑜𝑟𝑐𝑒 ∗ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑀𝑜𝑚𝑒𝑛𝑡 = (𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡)(ℎ𝑒𝑖𝑔ℎ𝑡) 2 ∗ (ℎ𝑒𝑖𝑔ℎ𝑡)(𝑊𝑖𝑑𝑡ℎ) ∗ ( ℎ𝑒𝑖𝑔ℎ𝑡 3 ) 45844 = (9270)(ℎ) 2 ∗ (ℎ)(5.6)∗ ℎ 3 ℎ = 1.74 𝑚𝑒𝑡𝑒𝑟𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 𝑠𝑜𝑖𝑙 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑓𝑜𝑟𝑐𝑒 = 1.74/3 = 0.58 𝑚𝑒𝑡𝑒𝑟𝑠
- 24. Storage Facility Construction Notes a. CMU wall vertical rebar Unlike the pump station rebar, there will be 2 pieces of rebar in each indicated cell. These walls are longer and taller than the pump station so the extra rebar is to provide greater strength. b. CMU wall horizontal rebar At specific heights horizontal rebar will be placed in the wall. This horizontal rebar prevents the wall from bending around a vertical axis. The CMU blocks must have notches cut into them so that the rebar can fit. Also, concrete must be poured around the rebar to secure it to the CMU wall. c. CMU wall gaps The wall is designed to have periodic vertical gaps where CMU blocks are not mortared together. A long continuous wall will randomly crack due to shrinkage, moisture, and thermal effects. Random cracks could compromise the structural strength of the wall and should be avoided. The designed gaps break the wall up into shorter segments that are less likely to randomly crack. Also, these gaps act like controlled cracks so we know where the weaknesses are located and we know where to place reinforcement. d. Landslide risk The walls as designed can withstand a landslide of perfectly fluid soil that is 1.74 meters tall. If soil accumulates past this height then additional reinforcement of the walls will be needed. This can be accomplished by adding steel reinforcement or steel bracing on the inside of the structure. It must be attached to a point on the wall that is 0.6 meters or higher off the ground to be effective because this is the height where the centroid of the landslide force will occur.