Structural Concrete - Hassoun

312 Chapter 8 Design of Deep Beams by the Strut-and-Tie Method
(160)(40)−0.75(0.85 × 1.0 × 4)(12)w s (40 − 1.125w s )=0
w s = 6.4in. and w t = 1.125w s = 7.2in.
Therefore
jd = 40 − 7.2
2 − 6.4
2 = 33.2in.
5. Calculate the forces in all truss members:
√
length of diagnal struts, AB and CD = 40 2 + 33.2 2 = 52 in.
Let θ be the angle between the strut and the tie,
tanθ = 33.2
40 = 0.831, θ = 39.7∘ > 25 ∘ OK
( ) 40
F u,BC = F u,AD = 160 = 192.5K
33.2
160
F u,AB = F u,CD =
sin 39.7 ∘ = 250K
6. Calculate the effective stress, f ce :
For struts:
Assume that confining reinforcements is provided to resist the splitting forces. Struts, AB, BC,
and CD represent the bottle-shaped compression members, therefore, β s = 0.75
f ce = 0.85β s f c ′ = 0.85 × 0.75 × 4 = 2.55 ksi
For nodes:
The nodal zone B or C has C-C-C forces, therefore β n = 1.0.
f ce = 0.85β n f c ′ = 0.85 × 1 × 4 = 3.4ksi
Since the three struts are connected to the other nodes, then f cu = 2.55 ksi controls all nodes.
7. Design of nodal zones:
a. Design the nodal zone at B or C (Fig. 8.18a):AsshowninFig.8.19a, the width of the top
strut is
w st = 12 sin θ + w s cos θ = 12 sin 39.7 ∘ + 6.4 cos 39.7 ∘ = 11 in.
b. Design the nodal zone at A or D (Fig. 8.19b):
w sb = 16 sin θ + w t cos θ = 16 sin 39.7 ∘ + 7.2 cos 39.7 ∘ = 14.4in.
Struts AB, BC,andCD represent the bottle-shaped compression members, β s = 0.75.
φF ns = φ(0.85β s f c ′)bw st = 0.75(0.85)(0.75)(4)(12)(11) =251.3K
φF ns = φ(0.85β s f c ′ )bw sb = 0.75(0.85)(0.75)(4)(12)(14.4) =329 K
use φF ns = 251.3K
Because φF ns is higher than the required forces, struts AB, BC,andCD are adequate.
φF ns ≥ F uAB,CD
or 251.3 > 250 and φF ns ≥ F uBC
or 251.3 > 250 OK
8. Design of horizontal and vertical reinforcement:
Vertical web reinforcements provided must be at least:
A v = 0.0025bs
And horizontal web reinforcements provided must be at least:
A vh = 0.0025bs