TehMightyEngineer
BRB Starting an Engineering Firm
So, in continuing my last minute cramming totally thought out review for the exam next weekend I decided to run through the afternoon problems of the lateral section. Problem 801 has us design a masonry wall using the ASD combinations from the 2006 IBC. If you follow through that then you get the wall as unacceptable for flexure. However, if you do it by LRFD (which we should be allowed to do on the actual exam I hope) then you actually get it being satisfactory. Now, I know that LRFD in ACI 530 gives you more credit than ASD does but I want to make sure I didn't goof somewhere.
Here's my math for 801. c)
Given 4' parapet with 100 PSF wind, #5 at 48" o.c. at centerline of wall. f'm = 1500 psi, fy = 60000 psi. Weight of wall = 60 PSF.
Design in 1 ft widths.
Mu = 0.9D + 1.6W = 0.9*(0) + 1.6*(800 lb-ft) = 1280 lb-ft
As = 0.31 in^2 / 4ft = 0.0775 in^2 / ft
Pu = 0.9D = 216 lb/ft
Ase = (Pu + As*fy) / fy = 0.0811 in^2/ft
a = (Ase*fy) / (0.80*f'm*b) = 0.3379 in
d = 3.82 in
Mn = Ase*fy*(d - a/2) = 17,766 lb-in = 1480.5 lb-ft
ϕMn = 0.9*Mn = 1332 lb-ft > Mu OK!
Check max steel = OK!
The ASD solution given by NCEES shows that the wall is insufficient for the load with the reinforcement given.
Here's my math for 801. c)
Given 4' parapet with 100 PSF wind, #5 at 48" o.c. at centerline of wall. f'm = 1500 psi, fy = 60000 psi. Weight of wall = 60 PSF.
Design in 1 ft widths.
Mu = 0.9D + 1.6W = 0.9*(0) + 1.6*(800 lb-ft) = 1280 lb-ft
As = 0.31 in^2 / 4ft = 0.0775 in^2 / ft
Pu = 0.9D = 216 lb/ft
Ase = (Pu + As*fy) / fy = 0.0811 in^2/ft
a = (Ase*fy) / (0.80*f'm*b) = 0.3379 in
d = 3.82 in
Mn = Ase*fy*(d - a/2) = 17,766 lb-in = 1480.5 lb-ft
ϕMn = 0.9*Mn = 1332 lb-ft > Mu OK!
Check max steel = OK!
The ASD solution given by NCEES shows that the wall is insufficient for the load with the reinforcement given.