Question

Design a simple-span, one-way slab to carry a uniformly distributed live load of 375 psf plus the self-weight of the slab. The span is 10-ft., center-to-center of the supports. Assume fe' = 4,000 psi and fy = 60,000 psi. Design per foot of width. Use #5 bars for both longitudinal and temperature and shrinkage reinforcement. (30 pts.)

Answer 1

GIVEN:

Span,L = 10 ft

b = 1 ft = 12 in.

Live load, W1 = 375 psf

f'c = 4000 psi 4ksi

fy = 60000 psi = 60 ksi

Unit weight of concrete = 150 pcf

Young's modulus of steel, E = 29000 psi

SOLUTION:

L 10 * 12 minimum thickness of slab to control deflection, h = 6 in 20 20

Take h = 7in.

3 Assume a clear cover of 2 in.

Diameter of Bar used = 0.625 in.

Area of Bar used, Ab = 0.31 in.?

3 Effective depth of slab, d = 7 - 0.625 2 = 5.9375 in 4

CALCULATION OF LOAD:

7 Self weight of the slab, Wd 150 = 87.5 psf 12

Factored Load, Wu = (1.2 *Wd) + (1.6*WI) = (1.2 * 87.5) + (1.6* 375) = 705 psf

Factored Moment, Mu (Wu* L) 0.705 * 102 = 8.8125 K.ft 8 8

CALCULATION OF STEEL AREA:

Ru = Mu b*d2 8.8125 * (12000) 12 * 5.93752 = - 250 psi

0.85 * f'c fy 1- 1 2 * Ru 0.9 * 0.85 *f'c 0.85 * 4 60 1- 2 * 0.25 0.9 * 0.85 * 4 = 0.0048

0.003 + fy E Check Maximum steel ratio,pmax *pb 0.008

87 Balanced steel ratio, pb=0.85 * B1 * f'c fy 87 + fy

B1 = 0.85

87 pb 0.85 * 0.85 * 5* (con 4 60% 0.0285 87 + 60

60 0.003 + 29000 0.008 pmax * 0.0285 0.01805

:: p<pmax ,Hence it is safe

Area of Tension steel , As = p*b* d = 0.0048 * 12 * 5.9375

By using no.5 bars

0.31 Ab Spacing, S = 12 * = = 12 * As 10.87 in. 0.342

:: Hence provide no.5 bars @ 10.5 in.center to center

Ab 0.31 As provided = 12*= 12* 10.5 = 0.354 in.2 S

CHECK FOR MOMENT CAPACITY :

a As* fy 0.85 * f'c*b 0.354 * 60 = 0.52 in. 0.85 * 4* 12

0.52 Moment capacity = 0.9 * As* fy* d- **(d -) = = 0.9 * 0.354 * 60 *(5.9375

ØMn 108.53 K-in.= 108.53 K.ft = 9.04 K.ft > Mu= 8.8125 K.ft 12

SHRINKAGE REINFORCEMENT :

Min.steel will be provided as shrinkage reinforcement, p min= 0.0018

Area of shrinkage reinforcement, A sh 0.0018 *b* h = 0.0018 * 12 * 7 = 0.1512 in.2

Ab Spacing = 12 * Ash = 0.31 12* 0.1512 = 24.60 in.

Hence provide no.5 Bars 24 in.center to center normal to the main reinforcement