Reinforced Concrete Design Design a 6 inch thick oneway rein

Reinforced Concrete Design:

Design a 6 inch thick one-way reinforced concrete slab that is 16 feet wide. The slab is one span and is simply supported. fc 5ksi fy-60 ksi Live load 200psf Dead Load- weight of the slab Design shrinkage and temperature steel transverse to the main reinforcement also. Provide a sketch of your final design

Solution

Solution:-Design of one way slab:-

Given

D = 6 inch

b = 16 feet

f_c = 5ksi

f_y = 60 ksi

Live load = 200 psf

Live load = 200 *16 = 3200 pound/feet

Let span of slab = 12 feet

Let Effective depth(d) = 5 inch

Dead load = 6/12 *16*150

Dead load = 1216.6 pound/feet

Total load = 3200 + 1216.6 = 4416.6 pound/feet

Factored load = 1.5 * 4416.6 = 6624.5 pound/feet

Maximum bending moment = total factored load * (effective span)²/8

B.M = 6624.5 *(12.41)²/8 = 127.5 kilopound – feet

B.M. = 0.87 f_yA_st* ( d – f_y*A_st/(f_ck*b))

127.5*10³*12 = 0.87 *60000 *A_st * ( 5 – 60000*A_st/(5000*16*12))

A_st = 6.36 inch²

Provide 0.7 inch diameter bars @ 6 inch spacing

Shrinkage and temperature reinforcement:-

Let 0.15% Area of concrete =steel in temperature and shrinkage

Reinforcement in temperature and shrinkage = 0.0015*5*(192) = 1.44 inch²

Use 0.5 inch diameter bars @ 4 inch spacing