MATH: Reinforced Concrete Design 01

Problem 2.20
Design a one-way slab having a simple span of 3 m. the slab is to carry a uniform live load of 7150 Pa. Assume fc=27.6 MPa and fy=276 MPa for main and temperature bars.
SOLUTION
Minimum slab thickness from Table 2.1:
hmin= L200.4+fy700
= 3000200.4+ 276700
hmin=119mm use 120 mm
Effective depth:




d= 120-20 mm (covering) – ½ bar diameter (12 mm)
d= 94 mm
Weight of slab= γconc x h = 23.5 (0.12)
= 2.82 kPa
Weight of slab= 2,820 Pa
Factored floor pressure= 1.4 DL + 1.7LL
= 1.4(2.820) + 1.7(7,150)
Factored floor pressure= 16,103 Pa
Analyzing 1m ( b -1000mm) width of slab:
Wu= 16,103 = 1
Wu= 16,103 N/m
Maximum factored moment, Mu = WuL28
= 16.103(3)38
Maximum factored moment, Mu = 18,115,875 N-m
Mu= ϕRu b d2
18,115.875 x 103 = 0.90 Ru (1000)(94)2
R = 2.278 MPa
ρ= 0.85 f'cfy 1- 1- 2 Ru0.85 f'c
= 0.85 27.6276 1- 1- 2(2.278)0.85(27.6)
ρ=0.0087
ρmin= 1.4fy
ρmin= 1.4276
ρmin=0.0051 < 0.0087 (OK)
[ρmax=0.75 0.85f'cβ1600fy (600+fy)]

ρmax=0.75 0.8527.6(0.85)600276(600+276)
ρmax=0.057>0.0087 OK

As= ρbd
= 0.0087(1000)()
As= 817.8 mm2 per meter
Using 12mm main bars:
Spacing= AbarAsx1000
= π4122817.8 x 1000
Spacing= 138 mm say 135 mm
Maximum spacing required by the code:
a) 3(h)= 3(120) = 360mm
b) 450 mm
Thus, use 12mm main bars at 135mm o.c.
Temperature bars: (Grade 275)
As= 0.002bh
As= 0.002(1000)(120) = 240 mm2
Spacing= AbarAs x 1000
= π4102240 x 1000
Spacing = 327 mm say 325mm
Maximum spacing required by the code:
5h = 5(120) = 600 mm
450 mm
Thus, use 10 mm temeperature bars at 325 mm o.c.







ASSUMPTIONS IN STRENGTH DESIGN IN FLEXURE
(CODE SECTION 5.10.2)
Strain in reinforcement and concrete shall be assumed directly proportional to the distance from the neutral axis. Except for deep flexural members with overall depth to clear span ratio, h/L > 2/5 for continuous spans and h/L > 4/5 for simple spans, a nonlinear distribution of strain shall be considered (See Sec.5.10.7).
Maximum usable strain at extreme concrete compression fiber, εc shall be assumed equal to 0.003.
For fs below fy, fs shall be taken as Es x εs . For εs > εy , fs = fy.
Tensile strength of concrete shall be neglected in axial and flexural calculations.
Relationship between compressive stress distribution and concrete strain may be assumed rectangular, trapezoidal, parabolic, or any other form that results in prediction of strength in substantial agreement with results of comprehensive tests.
For rectangular distribution of stress:
Concrete stress of 0.85 fc shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross-section and a straight line located parallel to the neutral axis at a distance of a=β1c from the fiber of maximum compressive strain.
Distance c from fiber of maximum strain to the neutral axis shall be measured in the direction perpendicular to N.A.
Factor β1 shall be taken as 0.85 for f c 30 MPa and β1 shall be reduced continuously at a rate of 0.008 for each 1 MPa of strength in excess of 30 MPa, but β1 shall not be taken less than 0.65. i.e.

For f c 30 MPa, β1 = 0.85
ii. for f c > 30 MPa,
β1 = 0.85 – 0.008 (f c – 30) but shall not be less than 0.65