Structural Concrete - Hassoun

8.6 Deep Members 311
V u = 160 K
CL
40"
A
θ
F u,AB
STRUT
TIE
B
STRUT
w s
w t
jd
F u,BC
F u,AD
40" 28"
Figure 8.17
Free-body diagram of the left third of the beam.
Check if
f cu > V u
A c
2500 psi > 666.7psi OK
Therefore, the bearing plate loading point is adequate.
3. Calculate the maximum shear strength of beam cross section:
Let V u at A = R A = 160 K, and assume d = 0.9h = 0.9 × 40 = 36 in.
V n = 10 √ √
f cb ′ w d = 10 × 4000 × 12 × 36 = 273 K
φV n = 0.75 × 273 = 205 > V u = 160K
Therefore, the cross sectional dimensions are adequate.
4. Select strut and tie model and geometry:
A truss model is chosen as shown in Fig. 8.16. Assume that the nodes act at the centerline of
the supports and loading points. Therefore, the horizontal position of A, B, C, andD are defined.
The vertical position of the nodes must be as close to the top and bottom of the beam. To reach
this goal, the lever arm, jd shown in Fig. 8.17, for the coupled forces should be at a maximum, or
w s and w t should be at a minimum.
To minimize w s and w t , the strut and tie should reach their capacity:
For strut BC:
F u,BC = φF nc = φf cu A c = φ(0.85β s f c ′)bw s
where β s = 1.0 (horizontal strut).
For tie AD:
F u,AD = φF nt = φf cu A c = φ(0.85β n f c ′ )bw t
where β n = 0.8 (C-C-T node).
As shown in Fig. 8.17, strut BC and tie AD form a couple, therefore F u, BC = F u, AD or
φ(0.85 × 1.0f ′ c )bw s = φ(0.85 × 0.8f ′ c )bw t ,
w t = 1.25w s
jd = 40 − w s
2 − w t
2 = 40 − 1.125w s
By writing the moment equilibrium about point A we have:
V u (40)−F u,BC (jd) =0
(160)(40)−φ(0.85β s f ′ c )bw s (40 − 1.125w s )=0