169

Assuming that both #8 bars are terminated so that only 4 #9 bars

at the bottom wall (B) of the member are effective in the design zone

5-6, the area of longitudinal steel due to flexure required for the

corner truss bar is 2.82/4 + 0.26 = 0.97 in. 2 , which is once again less

than the area of the #9 bar provided at the corners of the section.

The four #9 bars are then continued into the support and as a

consequence the longitudinal reinforcement requirements of the design

zones 4-5, 3-4, 2-3, and 1-2 would be satisfied. Figure 4.26 shows the

final detailing of the longitudinal steel in the member. The flexure

requirements for the two 118 tension bars are examined next. The area

required for flexure at midspan in the tension face of the member is 5

in.2 Neglecting the excess area of longitudinal steel the distance at

which the two #8 bars could be terminated is X = [(5-4)(156/5)2/5]1/2 =

70 in. Since the bar is going to be terminated without bending it into

the compression zone then the total distance from the centerline of the

span at which the two fl8 bars could be terminated is 70 + 12 db = 70 +

12(1) = 82 in., where db is the bar diameter or 70 + d = 70 + 15.44 =

85.4", whichever is greater. Thus the two #8 bars could be terminated

at 86 inches from the midspan. Since the two bars are going to be

continued up to the section 6, then the distance from the midspan at

which those two #8 bars are going to be terminated is 156 - 65 = 91

inches. Therefore, this satisfies the flexural requirements.

Due to the presence of a vertical shear force the longi tud inal

reinforcement which is terminated in the flexural tension face of the

member (Bottom face [B]) must be provided with an additional embedment

170

13" 13" 13" -,- .... ... ... ..

--1.,13 ..j

1 2 3 4 5 6 7 8 9 10 II 12 13

I #9 #9 #8 #8 #9 #9

.J

65" I 91" I ~ of support

Fig. 4.26 Detailing of the longitudinal reinforcement

171

length Is beyond the theoretic~l cut-off point. The additional

embedment length Is for the case of members subjected to distributed

loading is

(4.26)

where Al is the total area of longitud inal reinforcement to be

terminated, Vu is the factored ultimate shear force at the section, Wu

is the factored distributed load, and ld is the anchorage length

required to develop yielding of the bar. The basic development length

of a #8 bar evaluated in accordance with the ACI Building Code (2) is 24

in. The area of steel to be terminated is that corresponding to two

number 8 bars or 1.58 in. 2 Vu at the section where the bars are no

longer required for flexure ero inches from the midspan section) is 33

kips, coto'equals 1.0 and zL is 12.94 in. Therefore, Is = 2.0 -

[(1.58)(60)/{(33/1.08) + <3.5/2)}] = - 0.9'. The negative value

ind icates that the magni tude of the shear force is such that for the

amount of longitudinal steel to be terminated at that particular zone,

no additional embedment length would be required past the theoretical

cut-off point for flexure located at 70 inches from the midspan section.

Since the 2 #8 bars would be continued up to 91 inches from the midspan

section, all requirements would be satisfied.

Finally, the longitudinal tension reinforcement continued into

the support (4 #9) because of the presence of compression fans at the

support regions has to be provided with an anchorage length such that a

172

force Vu/2cotO' is adequately developed. In this case Vu/2cot<Yis equal

to 45.5/2 = 22.8 ki ps.

The truss model resisting the applied shear and torsion has two

vertical walls (L) and (R). Hence, each one takes 1/2 of the applied

shear force. Thus, the force that needs to be anchored in the truss

chord located at the corners of the wall where the applied bending

moment induces tension is 1/2(22.15) = 11.4 kips. Although 4 f#9 bars

are coming into the support r~gion, only one of them will actually be

located at each of the bottom corners of the truss model. Hence, the

force of 11.4 kips has to be totally taken by the 1 19 bar at the corner

of the section.

From column (4) of Table 4.10, due to the presence of shear and

2 torsion, an area of longitudinal steel of 0.34 in. working at its full

yield strength has to be :if>veloped at each bottom corner of the truss

model. Thus, the force thqt, has ~o be developed in the corner bottom f#9

bar of the section at the support region is (0.34)(60) + (11.4) or

approximately equal to 32 kips. The 119 corner bars have to be provided

with an embedment length such that a force of 32 kips is adequately

developed.

Since all the longitudinal bars anchored into the support region

will be provided with a 6 in. straight embedment length past the support

centerline, it is then necessary to check if this 6 in. straight

embedment length is enough to adequately develop the 32 kip force, or if

a standard hook is necessary for the two 19 bars located at each of the

bottom corners of the member.