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Main Code: ACI 318 – 14
Reference Code: Load: ASCE 10
How many design methods
- Strength Design Method: the strength design method requires service loads or related internal moments and forces to be increased by specified load factors ( required strength ) and computed nominal strengths to be reduced by specified strength reduction factors ( design strength ) – load factor and strength reduction factors specified in chapter 9 (ACI 318 – 11). Chapter 9 strength and serviceability requirements
6.4.1
For the design of roof or floors to resist gravity loads, it shall be permitted to assume that live load applied only to the level under consideration ( it means we can model only that floor or roof to determine internal moment, but how about dead load?) – convenient to use SAFE
Level Ground: Dorm office, Lobby, stair, toilet, multi-purposes area
Electricity Generator supporter: ???
Dorm office: 2.4 kN/m2
Stair: 4.79 kN/m2
Lobby: 4.79 kN/m2
Toilet (similar to public room): 4.79 kN/m2
Multi-purposes area (Public room): 4.79kN/m2
Second floor: multi-purposes area, roof (?? Do you want to have a swimming pool here), Canopy, corridor, stair, dorm room,
Multi-purposes area (Public room): 4.79kN/m2
Corridor and dorm room: 1.92 kN/m2
Stair: 4.79 kN/m2
Roof (ordinary flat): 0.96 kN/m2
Canopy: 0.24 kN/m2
Third Floor – 16th floor: Multi-purpose area, stair, dorm room, corridor,
Multi-purposes area (Public room): 4.79kN/m2
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Corridor and dorm room: 1.92 kN/m2
Stair: 4.79 kN/m2
Balcony:=1.5*1.92=2.88 kN/m2 (1.5 times the live load for the occupancy served. Not required to exceed 4.79 kN/m)
17th Floor: Pump Room, Water Tank 1, Water Tank 2, Multi-purpose area, dorm room, corridor
Pump Room: ???
Water tank: 2.78x7.7x3x10x1000/(2.78x7.7)=30 kN/m2
Multi-purposes area (Public room): 4.79kN/m2
Corridor and dorm room: 1.92 kN/m2
Stair: 4.79 kN/m2
Lobby: 4.79 kN/m2
18th floor: health club, fitness center, multi-purpose area, toilets, lobby, stair,
Gymnasiums: 4.79 kN/m2
Multi-purposes area (Public room): 4.79kN/m2
Stair: 4.79 kN/m2
Lobby: 4.79 kN/m2
Roof floor: (any assembly purposes)
Ordinary roof: 0.96 kN/m2
Building and other structure, Flexible: Slender buildings and other structures that have a fundamental natural frequency less than 1 Hz.
The natural frequency for each mode of vibration follows this rule:
F = sqrt( K/M)/(2PI) (Hz)
F: Natural frequency
K: the stiffness of the building associated with this mode
M: the mass of the building associated with this mode
It mean at first we don’t need to calculate the wind load to assign to building when we use Etabs to calculate the frequency.
Step:
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Wind load: Cp: table 16H, page 29/545
Components and cladding: elements of the building envelope that do not qualify as part of the MWFRS
Select Concrete and Steel Grade (base on which Code ?)
Determine the load (dead, live, wind, earthquake base on UBS) on each floor (architectural drawing )
Dead load:
Members’ weight: Etabs can calculate.
Wall load: Strip load or uniform load
100 mm – wall
W = 0.1*
Water load:
Machine load
Finishing (tile,motar) + cladding
Live load:
Reduction factor for Column
When we change the reduction factor for beam, it has no effect on the result
On such structures, the minimum roof live load shall be 12 psf (0.58 kN/m2).
Learn more about At
Should we reduce load on roof
Wind load (windward, leeward, side ward, roof)
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Stair load
Roof load
Determine Load Combination (base on Chapter 2 ASCE 10):
Load combination for strength design * ( why is it different from the Myanmar service report )
Load combination for allowable stress design
Choose Type of Slab and beam
How to model the join between the column and beam ( edge area ).
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Rigid diaphragms have infinite in-plane stiffness properties, and therefore they neither exhibit membrane deformation nor report the associated forces, whereas semi-rigid diaphragms simulate actual in-plane stiffness properties and behavior. For most reinforced-concrete slab systems, in which the slab is sufficiently thick and membrane deformation due to lateral loading is negligible, rigid diaphragms produce results nearly identical to those of semi-rigid diaphragms, while taking advantage of faster computation. Semi-rigid diaphragms should be modeled when significant in-plane deformation does occur, or when required by code
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Question:
1. Is it the normal roof ( or roof used for garden, both 2 roofs (2 floor and 18 floor)2. Does the basement cover the area of electricity generator supporter3. Can I change the position of shear wall ( lift pitch) as the drawing, is the result change much
compare with the original design, we change in Etabs, yet we still follow the design in construction
4. Is it necessary to have beam at lift area to connect the shear wall and column5. What is the weight of Pump Machine ( is it dead load or live load ? in the code says: machine
weight is dead load when it is supported by structural members)6. Water load is dead load or live load?7. Water Tank ( slab for 18th – 300 or 150 )?8. Weight of electricity generator?9. Which code to apply for Seismic Design ( most of the project follow UBC not ASCE 7 )10. Where to apply the load for Stairs and lift, ( If apply on Beam? How to calculate? What is the
weight of its self and full loaded weight?)11. Can I add superimposed dead load to dead load?12. Is it necessary to connect the gap between shear wall by beam? – draw the opening in elevation
Longitudinal reinforcement!
1. Which combination is used to design beams, Columns, slabs, foundations, joints?2. Can we design by first select the As then compare Mn and Mu?3. Should we design doubly reinforced beam?4. Can we assume phi = 0.9 at first (tension controlled)?5. Asmin does not take account of reinforcement in compression fiber!6. What is a procedure to calculate reinforcement for beam, do we need to examine the
reinforcement ratio – strain later to make sure that section is tension-controlled?7. Which section we take to design reinforcement?8. What is the optimum reinforcement ratio for beam? 1 - 1.5%
Web reinforcement design!
1. What is the envelop to design web reinforcement ( stirrup) – shear force at the mid-span will be biggest if load is distribute un-symmetrically.
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Etab