Structural Concrete - Hassoun

8.6 Deep Members 307
P u = 768 K
D
6"
60"
A
θ = 36.5°
81" 81"
B
6"
384 K
384 K
Figure 8.13 Example 8.1.
The vertical struts at A, B and D have uniform sections, and therefore β s = 1.0 and
f ce = 0.85 × 1.0 × 4 = 3.4ksi
The nodal zone D has C–C–C forces, and therefore β s = 1.0. The effective strength at nodal
zone D is given as
f cu = 0.85 × 1.0 × 4 = 3.4ksi
Since the struts AD and BD connect to the other nodes, then f cu = 2.55 ksi controls to all nodal
zones.
7. Design of nodal zones:
a. Design the nodal zone at A. Assume that the faces of the nodal zone have the same stress of
2.55 ksi and the faces are perpendicular to their respective forces:
φF n ≥ F u or φf cu A c ≥ F u
where φ equals 0.75 for struts, ties, and nodes.
The length of the horizontal face ab, Fig. 8.14a, is equal to F u /(φ f cu b) = 384/(0.75 ×
2.55 × 18) = 11.2 in.
From geometry, the length ac = 11.2 (518.3/384) = 15.2 in.
Similarly, the length of bc = 11.2 (645/384) = 18.8 in.
The center of the nodal zone is located at 15.2/2 = 7.6 in. from the bottom of the beam,
which is close to 6.0 in., assumed earlier.
b. Design the nodal zone at D (Fig. 8.14b):
The length of the horizontal face de = 768∕(0.75 × 2.55 × 18) =22.3in.
The length of df = ef = 22.3 (645/768) = 18.7 in.
The length of fg = 15.0 in, and the center of the nodal zone is located at 15/3 = 5.0 in. from
the top, which is close to the assumed 6.0 in.