Question

Roof Given the framing plans in Fig- Roof Loading Loading on All Other Floors D-40 pat ure 2, determine the uniform 4 - 35 pot L-60 psf line load, ww in units of k/ft, R-30 pat 20 (typ.) that acts on beam AB. The area loads for each floor are reported. Please only consider Load Cases 1 through 4 from ACI Table 5.3.1 in. your analysis. (10 points) Beam AB 4 Floor 3 Floor Beam AB 8.33 ft (typ) 2 Floor 25 (typ.) Building Plan View Building Elevation View Figure 2

Answer 1

Length of beam AB L = 20ft width of slab = B=8.33 ft L/B= 20/8.33= 2.4 > 2 The slab is one way slab hence Tributory width of beam AB W, = 8.33 ft T

Given Roof loading (service ) Dead load D = 40 psf Roof live load Lr = 35 psf Snow load S=25 psf Rain load R = 30 psf ACI Table 5.3.1 Load combination Load case 1 U = 1.4*D Load case 2 U = 1.2D +1.6L +0.5(Lr or Sor R) Load case 3 U=1.2D +1.6(Lr or Sor R)+(1.0L or 0.5W)

Load case 4 U=1.2D +1.0L +1.0W +0.5(Lr or Sor R) Let us analyze each case Load case 1 U=1.4 * 40 = 56 psf Load case 2 U=1.2*40+0.5* max(35,25, 30) =1.2* 40+0.5 * 35 = 65.5 psf Load case 3 U =1.2* 40 +1.6* max(35,25,30) + max(0,0) =1.2* 40 +1.6*35 = 104 psf =

Load case 4 U = 1.2* 40+1*0+1*0+0.5* max(35,25, 30) =1.2* 40 +0.5* 35 = 65.5 psf Hence maximum of U from 4 load case U = 104 psf Uniform line load = =W, *U w =W1 =8.33* 104 = 866.32 plf = 0.866 k/ft