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REINFORCE CONCRETE DESIGN
Transcript of REINFORCE CONCRETE DESIGN
Design column c-1 gb groundbase.
Unbraced column,short column <10.
Axial load and biaxial moment.
Mx and My was obtained from esteem,while axial load was determined from ULS calculated from the design project.
1. Determine total dead load, Gk
2. Determine live load, Qk
3. Design load ( ULS and SLS)
4. Determine Vmax and Mmax
5. Determine K value
6. Determine As and As'
7. Shear link test
8. Deflection test
9. Cracking test
10. Reinforcement details
REINFORCED CONCRETE DESIGN
- TWO STOREY BUNGALOW
- LOCATED AT :
LOT 904, JALAN KAMPUNG KU SAYANG, MUKIM BUKIT BARU, DAERAH MELAKA TENGAH,
MELAKA BANDARAYA BERSEJARAH.
SPECIFICATION AND REFERENCES
Based on BS 8110 : 1985 Structure Use of Concrete
BS 6399 : Code of Practice for Dead and
due to differencess of tributery area affect design load.
square area and circle area.
Lots of human error that can't be determined and detected.
Lots of differences between esteem and manual design calculation.
Lacks of experienced.
MUHAMAD HANIS BUANG
AHMAD NAFEK AUGUST FOUZY
AHMAD SUFIAN NUDIN
MUHAMAD ARIF ABDUL AZIZ
a) Dead Loads
Dead loads for the typical area are as follows:
Unit weight of concrete 24.0 kN / m2
Finishes 0.2 kN / m2 (Cement render)
1.0 kN / m2(Homogenous/Ceramic Tiles)
Brickwall 2.6 kN / m2 (115 mm thick with 20 mm thick of plaster on both side)
Glass 0.3 kN / m2
Roof tiles 0.7 kN / m2
Battens 0.05 kN / m2
Ceiling, M&E services, duct, trusses 1.5 kN / m2
Water 9.81 kN /m3
b) Imposed/Live load
Imposed loads for particular areas shall be referred to the usage of the floor space and being taken in accordance with BS 6399 : Part 1
Bedrooms 1.5 kN / m2
Living room 1.5 kN / m2
Kitchen 2.0 kN / m2
Toilet 2.0 kN / m2
Car porch 3.0 kN / m2
Roof 0.75 kN / m2
c) Wind Loads
Wind action is not critical and negligible, notional loads govern
d) Seismic Loads
No significant seismic activity in the project area. Thus, seismic loading is not applicable.
for designing slab, 2 type of slab :1 way slab, 2ways slab.
for this project,design slab was a 2 ways slab.
to design the reinforcement of slab, we always need to find M in both direction.
by using shear coefficient, we can determined both moment in x and y direction.
there was difference area and steel requirement in slab design when compared to esteem. This is because,lots of different value of load was placing on the slab design.
lack of experienced also causes lots of information does not meet the requirement in designing slab.
DESIGN PAD FOOTING
There are 2 type of pada footing; axial load, and moment bending which divide into 2, 1 direction and 2 direction.
We only take into consideration axial load. Same goes to Esteem software.
BILL OF QUANTITIES
Q & A session
choose the most beneficial tributery area.
Carefully and guided from supervisor from experinced people.
En. Mohd Nurazam Bin Husin
Penolong Jurutera JKR Daerah
Pn. Nor Khadijah Binti Ruslan
Juruteknik Kanan JKR Melaka
En. Baharom Bin Hashim
Juruteknik Kanan JKR
Prof Dato' Dr. Ir. Wan Hamidon Wan Badaruzzaman
En. Yazmil Yatim
Mentor Vintage Engineer
DESIGN PAD FOOTING
Thus, we get w x L values.
due to error in calculating load on slab, it produce lots of error that extremely hard to find,thus a guidance can help in designing better slab.
Slab - Fariq
1) Calculate the load that are applied to the pad footing. Load from slab to beam and so on. Also calculate dead and live load.
2) Q allowable (soil investigation)
3) Find punching shear to know the effective depth . Find D first, which is D = d min + cover + diameter of bars/2
4) Take a little bit larger value for D (more safe)
5) Find the footing size. First find the depth of soil = Df - D. Next, calculate total load = (soil height x unit weight) + (concrete high, D x unit weight concrete)
6) Next, find the serviceability limit state = 1.0 Gk + 1.0 Qk. And serviceability/net allowable bearing pressure, Qnet = Area, m2.
7) Find area. Then calculate flexural reinforcement to find the reinforcement size.
x3 = (B-C)/2
P = ULS/area
M = W x (P x X3)^2/2
8) k = M/bd^2(fcu). Must k <0.156. We do not need compression steel. k > 0.156, we need compression. (cut cost)
9) find z = d (0.5 + (0.25 - k/0.9)^1/2). To prevent cracking.
10) S required = M/0.87(fy)(z)
Characteristics strength of concrete, fcu
Beam and slab 30 N / mm2
Pad footing and column 30 N / mm2
Characteristics strength of reinforced, fy 460 N / mm2
Characteristics strength of links, fyv 250 N / mm2
Concrete density 24 KN / m3
Fire resistance 1.5 hours
Exposure condition Light
Ground floor suspended slab
Soil bearing pressure 100 KN / m2
Design staircase as slab
Step to determine Structural Layout:
1. Understand the Architectural Drawing
- Use of Space
2. By 'try and error' placed the structure of:
Beam - placed under brickwall
- secondary beam under long span of slab
Slab - At space need for the use
column - Intersection of beam
3. Find at most severe condition at the structure
- To solve the problem, placed more structure to support the load.
1. Project Site Preparation
2. Project Earth Excavation Costing Tables
earth excavation volume and cost
3. Project Formwork and Concrete Costing Tables
Concrete Volume and Cost
Lean Concrete Volume and Cost
Formwork Area and Cost (Exclude Opening for RC Wall)
4. Project Reinforcement Costing Tables
5. Brickwall cost
6. Roofing cost
7. Equipment renting
Total Project Costing with Pad Foundation