Structural Concrete - Hassoun

310 Chapter 8 Design of Deep Beams by the Strut-and-Tie Method
40" 56" 40"
95 K 95 K
B
C
w s
40"
A
STRUT
STRUT
TIE
STRUT
D
w t
16"
10' = 120"
16"
Figure 8.16
Idealized truss model.
Solution
1. Calculate the factored load:
(
Weight of the beam = 10 + 16
12 + 16 )( )( ) 40 12
(0.150) =6.3K
12 12 12
Since the weight of beam is small relative to concentrated loads, add to the concentrated loads.
P u = 1.2D + 1.6L = 1.2(6.3)+1.6 × 95 = 160 K
R A = R D = 160 K
2. Check if beam is deep according to ACI Code, Section 9.9:
Clear span, l n = 10 ft, h = 3.3ft, therefore l n
= 3 < 4, deep beam.
h
Check the bearing capacity at support and loading location:
a. At supports A and D: The area of bearing plate at each support is A c = 16 × 20 = 320 in. 2 The
bearing stresses at each support is:
V u
= 160(1000) = 500 psi
A c 320
The nodal zone over the support is a compression-tension node (C-C-T), therefore:
f cu = 0.85β n f c ′ = 0.85 × 0.8 × 4000 = 2720 psi,φf cu = 0.75(2720) =2040 psi
Check if
φf cu > V u
A c
2040 psi > 500 psi OK
Therefore, the bearing plate at the support is adequate.
b. At loading points B and C: The area of bearing plate at each loading point is A c = 12 × 20 = 240 in. 2
The bearing stress at each loading point is
V u
= 160(1000) = 666.7psi
A c 240
The nodal zone beneath each loading point is a pure compression node (C-C-C), therefore,
f cu = 0.85β n f c ′ = 0.85 × 1.0 × 4000 = 3400 psi φf
cu
= 0.75(3400) =2550 psi