Problem: For a RC slab with uniform thickness, select thickness and area of flex

Question

Problem: For a RC slab with uniform thickness, select thickness and area of flexural reinforcement required to resist factored bending moment M Determine the thickness h and required tension reinforcement area A, of a reinforced concrete slab with design compressive strengthf 4,000 psi. Use reinforcement with f, -60,000 psi. The factored bending moment to be designed for is M-11 ft-kip per foot of the slab Note: The slab is long n lateral dimensions, so in order to apply the beam bending design concept, the design of slab for flexure is typically reduced to designing a strip of the slab with a unit width of 1 foot (or a beam with b-12 in.). The result is then applied uniformly across the length of the slab r Guidance: We are trying to determine two related parameters (slab dimension and reinforcement, or d and pi, so the design process starts with assuming cne. For this problem, choose a that is less than te P corresponding to a steel strain 4-0001-000s can be computed as 0.85a + 0ms (see slides #13 and 15 of Handout m2). A starting can be« 0.5 0.6, 7, or 0.3 x Areas mis ensures that steel will have tensde strain greater than 0.005, tusa tension-controlled failure mode and .. 0.9). With the assumed determined With 4R known, compute d using Compute required area of reinforcement A, pod. Choose number of bars (Note: These are number of bars per 12" strip of the slab. Compute corresponding fR(where R,",-o59- 1. . eMed, (choose a round value for d not a fraction). * They are to be replicated for the entire width or length of the slat) Consult with the best practices for reinforcement placement (on first slide) to select the concrete Cover, then compute the slab thickness: h d·4·Cover (gnore diameter of stirrup der on the first slide. Slabs don't need shear reinforcement so there are no stirrups). Provide a design sketch. Good Practices for Reinforcement Placement (S, S', and Concrete Cover) Minimum Cover) Inch Cover Not exposed to weather /contact ground Beams and columns Slabs, walls, and joists with No.11 and smaller bars 3/4 Slabs, walls, and joists with No.14 and 18 bars #2 1 in tl-- Exposed to earth or weather Members with No. 5 and smaller bars Coverl Members with No. 6 through 18 bars Cast against and permanently exposed to earth db S211/3 x S'2 1.0 in. Not yet account for fire protection max, 1 in amax = Maximum aggregate size d, Bar diameter dstr = Stirrup diameter

Explanation / Answer

Rhobalance = 0.85 1 (fc / fy) (ecu /( ecu + ey)

Let concrete compressive strain at failure, ecu = 0.003

Es = 29000 ksi

ey = fy / 29000000 psi

So, Rhobalance = 0.85 1 (fc / fy) (87000 /( 87000 + fy)

1 = 0.85 for fc 4000 psi

Rhomax for underreinforced behaviour = 0.85 1 (fc / fy) (ecu /( ecu + ey)

                                                                     = 0.85 x 0.85 x (4000/60000) x (0.003/(0.003+0.005))

                                                                     = 0.0181

Rhobalance = 0.85 1 (fc / fy) (87000 /( 87000 + fy) = 0.85 x 0.85 x (4000/60000) x (87000/(87000+60000))

                                                                                     = 0.0285

Limit is 0.75xRhobalance = 0.75 x 0.0285 = 0.0213 > Rhomax (OK)

Assume Rho = 0.6 x Rhomax = 0.6 x 0.0181 = 0.0108

Now, Mu/bd2 = øRho fy (1-0.59x(Rho fy / fc))

=> d2 = Mu / (ø b Rho fy (1-0.59x(Rho fy / fc))) = 20.79

=> d = sqrt (20.79) = 4.56 in

Rho = Ast/bd

=> Ast = Rho bd = 0.0108 x 12 x 4.56 = 0.593 in2

Use 4 #4 bars per ft of slab

Bar spacing = 12/4 = 3 in

Diameter of #4 bars = 0.5 in

Ast provided = 4 x pi x 0.52 / 4 = 0.785in2

Rho provided = 0.785/(12 x 4.56) = 0.0144

Cover provided = 1.5 in

Total depth (h) = 4.56 + 0.25 + 1.5 = 6.31 in

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