RIICWD533E - REINFORCED CONCRETE DESIGN

Project 2: One-way slab
Part 1: ADOPTED DESIGN PARAMETERS FOR DESIGN
Type of Load Load Set ?
Dead Load ? kPa
Live Load (q) ? kPa
Additional Load n/a

Site and material parameters Group ?
Site Exposure classification A1
Unit weight (concrete) ? kN/m3
Concrete Grade (Slab) - MPa 32
Concrete Ec - MPa 30,100
Reinforcing bar (diameter) N12
Compression bars (Asc) - % ?
Grid, deflection, cracking control, ? & k parameters
Group ?
Grid length (mm) ?
Total ? (maximum) L/250
Incr. ? (maximum) L/500
Cracking control ?
multip. ?s & ?l 0.7 & 0.4
factor k3 1.0 (for one-way slab)
factor k4 1.75 end span / 2.1 internal span

Working Solution
Slab Deflection requirement
Estimated minimum thickness of slab (using clear span dimension Ln)
using Table 5.1

= D=L_n/?= ???mm
? Try ??? mm minimum slab thickness
Weight of slab (self-weight) =0.??? ×24= ??kN/m^2
Total dead load g=?+??=???kN/m^2
Taking Asc/Ast to be as ??
Therefore, long-term deflection multiplier (due to creep & shrinkage effects) is:
k_cs=2-1.2(A_sc/A_st )=??
For total deflection, the design service load is:
F_(d.ef)=(1+k_cs )g+(f_s+f_l k_cs )q= ????kN/m^2
For incremental deflection, the design service load is:
F_(d.ef)=k_cs g+(f_s+f_l k_cs )q= ????kN/m^2
Now, implementing equation for
the maximum ratio of effective span (Lef) to effective depth (d):
Total deflection L_ef/d=k_3×k_4 ?((?/L_ef ×E_c)/(F_(d.ef)/1000))?^(1/3)= A
Incremental deflection L_ef/d=k_3×k_4 ?((?/L_ef ×E_c)/(F_(d.ef)/1000))?^(1/3)= B

Consider smaller value L_ef/d of the above deflections A or B (as it is critical)!
finding required d L_ef/d=smaller (A or B) ? d=L_ef/(s (A or B))= ??mm
Therefore, the overall slab thickness is determined as:
D=d+cover+ (barØ)/2= ??mm
Strength Design
(for 1.0m wide slab strip)
Accepted the thickness of slab D=??mm
slab self-weight s_sw=D×24=? ×24= ? kN/m^2
dead load (total) g=s_sw+g_table= ??kN/m^2
live load q= ???kN/m?^2
Strength design load w^*=g×1.2+q×1.5= ??kN/m^2 = ??kN/m per meter width
Calculating maximum moments for spans and supports using coefficients.

You will calculate moments, parameters ? and choose N12-spacing for four locations: End support, End span, Interior support & Interior span.
End support (example)
max (-ve) moment M^*=(w^*×L_n^2)/?= ??kNm per m width
Design moment parameter ?=M^*/(b×d^2 )= ??

End span (example)
max (+ve) moment M^*=(w^*×L_n^2)/?= ??kNm per m width
Design moment parameter ?=M^*/(b×d^2 )= ??

reading from the chart C1 N12 @ ?? mm (Bottom layer)
Interior support
max (-ve) moment M^*=(w^*×L_n^2)/?= ??kNm per m width
Design moment parameter ?=M^*/(b×d^2 )= ??

reading from the chart C1 N12 @ ?? mm (Top layer)
Interior span
max (+ve) moment M^*=(w^*×L_n^2)/?= ??kNm per m width
Design moment parameter ?=M^*/(b×d^2 )= ??

reading from the chart C1 N12 @ ?? mm (Bottom layer)
Summarise your calculated outputs in a table.
Location M* ? Bar N12 spacing
End support ?? (Top)
End span ?? (Bottom)
Interior support ?? (Top)
Interior span ?? (Bottom)

Secondary reinforcement (Top & Bottom layers)
Taking minor / moderate (read from your data) control of cracking
due to temperature and shrinkage
Apply appropriate equation for your case:
Minor - A_(s.min)=0.00175×b×D=0.00175×1,000×D= ?? ?mm?^2
Moderate - A_(s.min)=0.0035×b×D=0.0035×1,000×D= ?? ?mm?^2
Chose secondary reinforcing bar & spacing from the table below.
(consider N12 bars for this project, knowing 1N12 = 110 mm2)

Adopted N12 @ ?? mm