Design of Deep Beams and Joints
Design of Deep Beams and Joints
Design of Deep Beams and Joints
Naveed Anwar
Buddhi S. Sharma
ACECOMS, AIT
Definition of
Deep Members
Strain Profile – The Starting Point
• Section Capacity is represented by Stress
Resultants
• Stress Resultants are based on stress
Distribution
• Stress Distribution is based on Strain
Distribution
• Strain Distribution for a particular
deformation is not known for reinforced
concrete sections
1 n
N z 1 x, y dx dy ...
1
Ai i ( x, y ) ...
1 x y 2 i 1
1 1 n
M x 2 x, y dx dy . y ... Ai i ( x, y ) yi ...
1 x y 2 i 1
1 1 n
M y 3 x, y dx dy . x ... Ai i ( x, y ) xi ...
1 x y 2 i 1
h
f1
f2 s for
sse l
fn e
Str Stee
s for Horizontal
sse and
e
Str crete
n
co R/F
train
S
D B D
• Deep Members:
– Where most of the beam length is “D” Region
• Thick Members:
– Flexural Deformations are Predominant and shear
deformations can be ignored
• Thin Members:
– Shear Deformations are Significant and can not be
ignored
Tension
Compression
Tension
Compression
Tension
Compression
D B D
Beam elements
with rigid ends
Beam elements
in “Truss
Model”
U1, R1 U1, R1
Shell
Fi Ai f i
A n
f1 P Fi
C i 1
f2 n
M Fi xi
i 1
f3 n
x1
f4 V Ai vi
T i 1
x1
f5
t
M P
f1, f2, …..fn are the nodal stresses at
section A-A , obtained from analysis V
“Zipper”
L
Ties
L
Tension
Compression
Tension
Compression
Tension
Compression
Tension
Compression
L/d = 5
L/a = 2.5
L/d = 6
L/a = 3
Angle = 34 De g
Angle 35 - 45 Deg
Gives the most
OK: USed by ACI Code
economical
and realistic design
Angle = 45 De g
Angle > 50 Deg.
Too steep. Requires too
NOT OK: Too Steep and Expensive
much stirrups. Not good.
Angle = 64 De g
• Failure of Struts
• By Longitudinal
Crushing
• Compression failure
of Struts
H C
t t x 2t
a=1.6 a=1.6
P=10,000 kN
d=1.4
L=2.5 L=2.5
1
= tan-1 d/0.5L
= tan-1 d/0.5(L-d1)
= 48 deg = 68.5 deg
T = 0.5P/tan T = 0.5P/tan
T = 1970 kN
T = 4502 kN
a2
a2
P4
P2
P3
d
L2
L1
Main members
Secondary members
T
Ast
f y
M P
– Continuous Beam:
l n /d 2.5
l n /d 5.0 P
• Special Case
Deep
Beam Shallow Beam d
Deep Beam or
Veirendel Girder
Veirendel Girder
Vc 2 f c' bw d 2 ld '
Max. Vn 10 f c bw d when l / d is 2 to 5
3 d
Vu d
Vc 1.9 f c 2500 w
'
bw d Vc 2 f c' bw d
Mu
V d
Vc F 1.9 f c' 2500 w u bw d
Mu
V d
where F 3.5 2.5 u 2.5
Mu
Complete Model