Calculate the uniform load (in addition to beam weight) that will cause the section in figure 1 to begin to crack if the are used for 28ft simple span.
fc’=4000psi, fr=7.5√fc’ and reinforced concrete weight =150 lb/ft3.
for the above example, we need to calculate the value of the cracking moment. using cracking moment we can calculate the required load.
fr=7.5√4000=474.34 psimoment of inertia will be calculated using parallel axis theorem I=Ic+(A*d^2) I=2*((12*(4^3)/12)+4*12*(13^2))+(4*(22^3)/12)=19901.33in4
y is the distance from the neutral axis to extreme fiber in the tension zone. neutral axis will be located at the center of the beam due to symmetry. therefore, y=15 in
Mcr=(474.34*19901.33)/15=629,333.1248Ib.in=52.44 kips.ftthe maximum moment for simply beam =(w.l^2)/8w=((Mcr*8)/(l^2))=((52.44*8)/(28^2))=0.53514kip/ft=535.147Ib/ft
uniform load=668.93-beam weight=535.147-150*(2*(12/12)*(4/12)+(22/12)*(4/12))=343.48Ib/ft