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CASE - 6 ONE LONG EDGE CONTINUOUS
PARAM
Short Side (A): 3 m rebar diameter Long Side (B): 4.5 m concrete cover f'c: (^28) MPa concrete unit weight fy: (^420) MPa φ (^) 0. β (^) 0. assumed thickness (^) 0.013 m LOAD LIVE LOAD = 3.8 kPa DEAD LOAD = 6.94 kPa floor finish: 1. ceiling finish: 0. mech. duct allowance: 0. electrical load: 0 partition load: 0 miscellaneous load: 3 Total Dead Load = 6.94 kPa DETERMINE HEIGHT AND DEPTH OF SLAB d1= d2=
0.667 < 2, DESIGN AS TWO WAY SLAB!
DESIGN OF TWO WAY SLAB
SAY, 100 mm thick slab DETERMINE FACTORED MOMENT FOR THE SLAB MIDDLE STRIP, SHORT SPAN 25.06464 kN.m 40.409037 (^) kN.m COLUMN STRIP, SHORT SPAN 12.53232 kN.m
α𝑡=𝐼𝑡𝑏/𝐼𝑡𝑠 α𝑡=𝐼𝑙𝑏/𝐼𝑙𝑠 α𝑡=𝐼𝑏𝑏/𝐼𝑏𝑠 α𝑡=𝐼𝑟𝑏/𝐼𝑟𝑠
ℎ=ln(0.8+𝑓𝑦/1400)/(36+9β)β)) ℎ= ≥9β)0𝑚𝑚 𝑀𝐴+=𝐶𝑎𝑤𝑢𝐴^2 𝑀𝐴−=𝐶𝑎𝑛𝑒𝑔𝑤𝑢𝐴^2 𝑀𝐴+=𝐶𝑎𝑤𝑢𝐴^2
ρ = ρ = 0.
Since ρmin > ρ, use ρ = c. Compute for Area of Tensile Steel Needed. ρbd
d. Compute for Spacing of Rebars.
200 USE: 170 b. Check for value of ρmin ρmin = ρmin = ρmin = ρmin = Use higher value of ρmin = As = As =
(0.85𝑓^′ 𝑐)/𝑓𝑦 (1− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐)))
√(𝑓^′ 𝑐)/4𝑓𝑦
s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 𝒔= s𝑚𝑎𝑥=2(ℎ) 𝒔=
LONG SPAN, COLUMN STRIP
AT MIDSPAN
- Design = 0. ρ = ρ = 0.
Since ρmin > ρ, use ρ = c. Compute for Area of Tensile Steel Needed.
- Solve for ρmax. ρmax = ρmax = a. Calculate the percentage of tensile reinforcement (ρ). Rn = b. Check for value of ρmin ρmin = ρmin = ρmin = ρmin = Use higher value of ρmin =
(0.85𝑓^′ 𝑐)/𝑓𝑦 (1− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐)))
√(𝑓^′ 𝑐)/4𝑓𝑦
(3(0.85𝑓^′ 𝑐β) 1 ))/ 7𝑓𝑦
Since ρmin > ρ, use ρ = c. Compute for Area of Tensile Steel Needed. ρbd
d. Compute for Spacing of Rebars.
200 USE: 150 LONG SPAN, COLUMN STRIP AT CONTINUOUS EDGE b. Check for value of ρmin ρmin = ρmin = ρmin = ρmin = Use higher value of ρmin = As = As =
- Solve for ρmax. ρmax =
√(𝑓^′ 𝑐)/4𝑓𝑦
s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 𝒔= s𝑚𝑎𝑥=2(ℎ) 𝒔=
- Design = 0 ρ = ρ = 0
Since ρmin > ρ, use ρ = c. Compute for Area of Tensile Steel Needed. ρbd
d. Compute for Spacing of Rebars. ρmax = a. Calculate the percentage of tensile reinforcement (ρ). Rn = b. Check for value of ρmin ρmin = ρmin = ρmin = ρmin = Use higher value of ρmin = As = As =
(0.85𝑓^′ 𝑐)/𝑓𝑦 (1− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐)))
√(𝑓^′ 𝑐)/4𝑓𝑦
s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 (3(0.85𝑓^′ 𝑐β) 1 ))/ 7𝑓𝑦
Since ρmin < ρ < ρmax, ρ = c. Compute for Area of Tensile Steel Needed. ρbd
d. Compute for Spacing of Rebars.
200 USE: 90 SHORT SPAN, COLUMN STRIP AT CONTINOUS EDGE =
Design ρmin = Use higher value of ρmin = As = As =
Solve for ρmax. ρmax = ρmax = a. Calculate the percentage of tensile reinforcement (ρ). s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 𝒔= s𝑚𝑎𝑥=2(ℎ) 𝒔= (3(0.85𝑓^′ 𝑐β) 1 ))/ 7𝑓𝑦
ρ = ρ = Err:
Err: c. Compute for Area of Tensile Steel Needed. ρbd Err: d. Compute for Spacing of Rebars. Err: 200 Rn = b. Check for value of ρmin ρmin = ρmin = ρmin = ρmin = Use higher value of ρmin = As = As =
(0.85𝑓^′ 𝑐)/𝑓𝑦 (1− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐)))
√(𝑓^′ 𝑐)/4𝑓𝑦
s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 𝒔= s𝑚𝑎𝑥=2(ℎ) 𝒔=
c. Compute for Area of Tensile Steel Needed. ρbd Err: d. Compute for Spacing of Rebars. Err: 200 USE: Err: SHORT SPAN, MIDDLESTRIP AT MIDSPAN =
Design = 7. ρ = As = As =
Solve for ρmax. ρmax = ρmax = a. Calculate the percentage of tensile reinforcement (ρ). Rn = s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 𝒔= s𝑚𝑎𝑥=2(ℎ) 𝒔= 𝑀𝑢/(∅𝑏𝑑2) (0.85𝑓^′ 𝑐)/𝑓𝑦 (1− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐))) (3(0.85𝑓^′ 𝑐β) 1 ))/ 7𝑓𝑦
ρ = 0.
Since ρmin < ρ < ρmax, ρ = c. Compute for Area of Tensile Steel Needed. ρbd
d. Compute for Spacing of Rebars.
200 USE: 40 b. Check for value of ρmin ρmin = ρmin = ρmin = ρmin = Use higher value of ρmin = As = As =
√(𝑓^′ 𝑐)/4𝑓𝑦
s=(𝐴𝑟𝑒𝑎 𝑜𝑓 𝑜𝑛𝑒 𝑏𝑎𝑟)/𝐴𝑠 𝒔= s𝑚𝑎𝑥=2(ℎ) 𝒔=
OF TWO WAY SLAB
0.103448275862069 m say, 0.11 m = 4. = 3. < 90 mm slab THIS CAN ONLY BE USED IF αave>2. OTHERWISE, COMPUTE MANUALLY. ≥9β)0𝑚𝑚
einforcement (ρ).
Since ρmin > ρ, use ρ = 0. eeded. einforcement (ρ). Compare to values of ρmax and ρmin
− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐)))
s. mm say, (^690) mm mm mm SPACING O.C. m^2 einforcement (ρ). Compare to values of ρmax and ρmin
− √(1−2𝑅𝑛/(0.85𝑓^′ 𝑐)))
