- Compute the value of ej under the balanced-design
ultimate moment
Compare this value with the strain at incipient yielding. By Eq. 3, cb = l.\$qbd/ki =
1.18(0.432)(20.5)/0.80 = 13.1 in (332.74 mm); e;/ec = (13.1 - 2.5)713.1 = 0.809; e/ =
0.809(0.003) = 0.00243; ey = 50/29,000 = 0.0017 < e/. The compression reinforcement
will therefore yield before the concrete fails, and/' -fy may be used. - Alternatively, test the compression steel for yielding
Apply
_ , 0.85^/^(87,000)
P P //(87,000-JT) < >If this relation obtains, the compression steel will yield. The value of the right-hand mem-
ber is 0.85(0.80)(5/50)(2.5/20.5)(87/37) = 0.0195. From the preceding calculations,/? -p'
= 0.0324 > 0.0195. This is acceptable.- Determine the areas of reinforcement
By Eq. 2, A 3 = A's = qmj)dftlfy = 0.324(14)(20.5)(5/50) = 9.30 in^2 (60.00 cm^2 ); A 3 =
CuAWy) = 74,400/[0.90(50,00O)] - 1.65 in^2 (10.646 cm^2 ); A 3 = 9.30 + 1.65 = 10.95 in^2
(70.649 cm
2
). - Verify the solution
Apply the following equations for the ultimate-moment capacity:
(4,-AMy m,
a== 0.85/'* (13)So a = 9.30(50,000)/[0.85(5000)(14)] = 7.82 in (198.628 mm). Also,Mu = 4>fy\(Aa-4)(d-±\ +A,(d-d')\ (14)So Mn = 0.90(50,000)(9.30 x ( 16.59 + 1.65 x 18) = 8,280,000 in-lb (935,474.4 N-m), as
before. Therefore, the solution has been verified.
DESIGN OF WEB REINFORCEMENT
A 15-in (381-mm) wide 22.5-in (571.5-mm) effective-depth beam carries a uniform ulti-
mate load of 10.2 kips/lin ft (148.86 kN/m). The beam is simply supported, and the clear
distance between supports is 18 ft (5.5 m). Using// = 3000 lb/in
2
(20,685 kPa) and/, =
40,000 lb/in^2 (275,800 kPa), design web reinforcement in the form of vertical U stirrups
for this beam.
Calculation Procedure:- Construct the shearing-stress diagram for half-span
The ACI Code provides two alternative methods for computing the allowable shearing
stress on an unreinforced web. The more precise method recognizes the contribution of
both the shearing stress and flexural stress on a cross section in producing diagonal ten-