Compute the steel stress and resultant tensile force at ultimate load:/ 0.5/?/;\
fsu=fs' 1-^M (61)
V Jc IOr,fsu = 248,000(1 - 0.5 x 0.00137 x 248,000/5000) = 240,00 lb/in^2 (1,654,800 kPa);
TU=AJSU = 1.089(240,000) = 261,400 Ib (1,162,707.2N).
Compute the depth of the compression block. This depth, a, is found from C 11 =
0.85(5000)(72«) = 261,400 Ib (1,162,707.2 N); a = 0.854 in (21.6916 mm); Jd = d-a/2 =
10.66 in (270.764 mm); Mn = </>TJd= 0.90(261,40O)(10.66) = 2,500,000 in-lb (282,450.0
N-m).
Calculate the steel index to ascertain that it is below the limit imposed by the ACI
Code, or q =pfjfc' = 0.00137 (240,000)75000 = 0.0658 < 0.30. This is acceptable.
- Calculate the required ultimate-moment capacity as given
by the ACI Code
Thus, WDL = 329 + 12(6) = 401 Ib/lin ft (5852.2 N/m); WLL = 40(6) = 240 Ib/lin ft (3502.5
N/m); wu = LSw 0 L + L8wLL - 1034 Ib/lin ft (15,090.1 N/m); Mu required =
(%)(1034)(40)^2 (12) = 2,480,000 < 2,500,000 in-lb (282,450.0 N-m). The member is there-
fore adequate with respect to its ultimate-moment capacity. - Calculate the maximum and minimum area of web
reinforcement in the manner prescribed in the ACI Code
Since the maximum shearing stress does not vary linearly with the applied load, the shear
analysis is performed at ultimate-load conditions. Let Av = area of web reinforcement
placed perpendicular to the longitudinal axis; Kc' = ultimate-shear capacity of concrete;
Vp = vertical component of Ff at the given section; K 1 / = ultimate shear at given section;
s = center-to-center spacing of stirrups;^ = stress due to Ff 9 evaluated at the centroidal
axis, or at the junction of the web and flange when the centroidal axis lies in the flange.
Calculate the ultimate shear at the critical section, which lies at a distance d/2 from the
face of the support. Then distance from midspan to the critical section =^1 A(L — d)= 19.54
ft (5.955 m); KJ = 1034(19.54) = 20,200 Ib (89,849.6 N).
Evaluate Vc' by solving the following equations and selecting the smaller value:
Kc;-=1.7Z>'</(/cT^5 (62)where d = effective depth, in (mm); b' = width of web at centroidal axis, in (mm); b' =
2(5 + 1.5 x 10.98/12) = 12.74 in (323.596 mm); V^ = 1.7(12.74)(11.09)(5000)^05 =
17,000 Ib (75,616.0 N). Also,Vc'w = b'd(3.5f™ + 0.3JP + K; (63)where d = effective depth or 80 percent of the overall depth, whichever is greater, in
(mm). Thus, d = 0.80(16) = 12.8 in (325.12 mm); K; = O. From step 4,j£ = 0.85(574) =
+488 lb/in^2 (3364.8 kPa); K^ = 12.74(12.8)(3.5 x 50000 -^5 + 0.3 x 488) = 64,300 Ib
(286,006.4 N); therefore, Kc' = 17,000 Ib (75,616.0 N).
Calculate the maximum web-reinforcement area by applying the following equation:W = W)
A
«=
-^wT~(64)
where d = effective depth at section of maximum moment, in (mm). Use/J, = 40,000 lb/in
2