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Design of One-way Slab using Trial-and-Error Method:

Problem 2) Design a solid one-way slab for flexure using trial-and-error method assuming the following

properties and conditions shown in the figure below.

No. 4 bars for temperature and shrinkage,

spacing=?

No. 4 bars at 10" on-center

Livel load = 100 psf

Dead load = 20 psf + self weight of the slab.

10 ft (simply supported)

fy 60 ksi (Gr. 60 steel)

fe= 4 ksi (Concrete)

1/2/n(a) Estimate the minimum thickness for the slab if deflections are not computed. [Table 4.1 JCM Text]

(b) Design the slab using minimum thickness from part (a) and determine the total factored load per

linear foot (lbs/ft) and also the maximum factored moment (use 150 lbs/ft for the density of the

slab, and design using a 12"-strip).

(c) Determine the minimum cover (cs) for the slab steel if the concrete is not exposed to weather or

ground contact. Then calculate the depth (d) to the primary steel assuming #4 bars will be used.

Hint; d =h-cs-4

(d) Determine the flexural steel required (A) per foot using trial-and-error approach. Let your initial

trial moment arm (d-) equal to 0.9d. Assume reduction factor () equal to 0.9. 'a' converges

after one or two iterations.

(e) Determine the maximum spacing for the slab including the flexural steel requirement from part (d).

S

max

3h

18in

-(12in)

vey

(f) Select 10" as the spacing for the slab (S=10") and calculate the actual steel area provided per foot (

4.) using the following equation:

A₁ = A

12in

S

(g) Check the provided steel ratio against the minimum and maximum steel ratios (Pmin and Pmax).

Calculate the strain (&) in the steel and determine if the reduction factor () is equal to 0.9.

(h) Determine the minimum area for the temperature and shrinkage steel. Select spacing for

temperature and shrinkage reinforcement using No. 4 bars. Remember to check the maximum

spacing requirements,

(i) Calculate the actual design moment capacity, OM.

Fig: 1

Fig: 2