6 SEISMIC BEHAVIOUR OF LIGHTWEIGHT CONCRETE BEAM – NORMAL
6.7 Experiments
6.7.1 Interior beam-column joints
6.7.1.2 Strut-and-tie model and influence of beam reinforcement
In order to study the influence of beam reinforcement type and ratio on the overall behaviour of LWC beam and NC column joint, two types of reinforcement bars with two different ratios were used. The specimens S2 and S3 were reinforced with steel reinforcement bars with 1.75 % and 3.50 % reinforcement ratios respectively. While the specimens S4 and S5 were reinforced with glass-fibre reinforcement bars with 1.75 % and 3.50 % reinforcement ratios respectively.
-80 -60 -40 -20 0 20 40 60
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
Story Drift [%]
Beam Moment [KN.m] .
1.75 % - SRFT
-120 -100 -80 -60 -40 -20 0 20 40 60 80 100
-5 -4 -3 -2 -1 0 1 2 3 4 5
Story Drift [%]
Beam Moment [KN.m] .
3.50 % - SRFT
a. Story drift – beam moment diagram for S2 (left) and S3 (right)
-80 -60 -40 -20 0 20 40 60 80
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
Story Drift [%]
Beam Moment [KN.m] .
1.75 % - GFR
-80 -60 -40 -20 0 20 40 60 80 100
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
Story Drift [%]
Beam Moment [KN.m] .
3.50 % - GFR
b. Story drift – beam moment diagram for S4 (left) and S5 (right)
Figure 6.10: Story drift – beam moment diagram for specimens S2, S3, S4 and S5.
In general, the low reinforcement ratio enhances the overall behaviour of the joint by reducing the flexural strength of the beam (Fig. 6.10) comparable to that of the column and then minimizes the cracks in the column and joint intersection area to a very low limit especially in the case of steel reinforcement bars as shown in Fig. 6.11.
a. Crack pattern for S2 (left) and S3 (right)
b. Crack pattern for S4 (left) and S5 (right)
Figure 6.11: Crack pattern for specimens S2, S3, S4 and S5.
-100 0 100 200 300 400
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
Story Drift [%]
Joint Shear [KN] .
1.75 % - SRFT
-100 0 100 200 300 400 500 600 700 800 900
-5 -4 -3 -2 -1 0 1 2 3 4 5
Story Drift [%]
Joint Shear [KN] .
3.50 % - SRFT
Figure 6.12: Story drift – joint shear diagram for specimens S2 (left) and S3 (right).
The story drift – joint shear diagrams for the specimens S2 and S3 are shown in Figure 6.12.
The joint shear force is reduced in the joint intersection area due to reduction of the longitudinal reinforcement ratio in the beams which coincide with that proved by strut-and-tie model.
T
Cb b Tc
Tb
Tc Cc
-Cb
Cc
-T = C = (C +C )
b b c sb
-T = C = (C +C )c c c sc
-Vc
V c
Vb
Vb
N
N
N = 0 KN T + C - Vb b c
-T + C - Vb b c
-T + C - Vc c b T + C - Vc c b
-Cc -Cc
-Tc
Tc
Tb
Tb Cb
-Cb -Vb
Vb
Vc
Vc
A
f
Th Tv
- - - Compression Tension
Cc- A
Cb -Cd
f
Detail A
Figure 6.13: Strut-and-tie model for interior joint.
Figure 6.13 shows the strut-and-tie model for the studied interior joint, Vc is the horizontal force that is applied to the specimens by horizontal hydraulic jack and represent the story shear force. Vb represents the reaction of the two link elements at the end of the beam arms. Tc
and Tb represent the tensile forces in the longitudinal reinforcement bars in column and beam, respectively. Cc and Cs represent the compression forces that are carried with concrete and
reinforcement bars, respectively. Cc- and Cb
represent the total compression forces in column and beam, respectively. It should be noted that the direction of Vc and Vb will be reversibly changed during the test.
Figure 6.14: Principle stress for the interior joint using finite element program Diana.
The strut-and-tie model shows diagonal compression trajectories in the joint intersection area.
The diagonal compression force Cd and the diagonal stresses are verified by FEM - Diana (Fig. 6.14).
Figure 6.15: Strut-and-tie models for an interior joint with different details [Schlaich J., et. al., 2001].
The different structural details for an interior joint and its corresponding strut-and-tie model according to Schlaich J., et. al., (2001) are shown in Figure 6.15. It is suitable to use reinforcement details that is shown in Fig. 6.15b which simulate the details for frame corner, however using of diagonal reinforcement bars in the joint as shown in Fig. 6.15a is not practical. The details in Fig. 6.15c that is used in our study is more suitable for hysteretic loading of the joint but the concentration of compression stress at the corner A should be taken into account.
According to the rules for design of structural elements by strut-and-tie model, the following equations are used to calculate the joint shear force (Th). This shear force is the horizontal component of the diagonal compression force in the joint intersection area (Fig. 6.13). The middle longitudinal bars in the column are very essential to carry the vertical component (Tv) of the diagonal compression force in the joint intersection area.
Vc Cb Tb
Th = + − −
(6.1)
Vb Cc Tc
Tv = + − − (When N is neglected) (6.2)
where:
Th = joint shear force carried by transverse stirrups in the joint intersection area
Tv = vertical component of the diagonal compression force in the joint intersection area which is carried with the middle vertical bars in the column
Vc = the horizontal force that is applied to the specimens by horizontal hydraulic jack and represents the story shear force
Vb = reaction of the two link elements at the end of the beam arms
Tc and Tb = tensile forces in the longitudinal reinforcement bars in column and beam respectively
Cc- and Cb
= the total compression forces in column and beam respectively
As recommended in strut-and-tie model method; it is important to design the stress concentration at corner A (Fig. 6.13 and Fig. 6.15c). The corner A is considered as three compression components joint (CCC-Joint). The effective design strength fcd,eff for CCC-joint can be calculated as
1 , fcd
eff
fcd =α (6.3)
where:
fcd,eff = The effective design compression strength α = reduction factor and equals1.0 for CCC-joint
c fck fcd
α γ
1= (6.4)
where:
fcd1 = design compression strength
α = reduction factor for long-term effect on concrete strength and equals 0.85 for normal concrete and 0.80 for lightweight concrete according to DIN 1045-1
γc = concrete strength safety factor and equals 1.5
According to the obtained results from the test of interior joints, the effective design strength fcd,eff was not reached. Therefore, it is recommended to reduce α in Equation (6.3) to be 0.85 for two reasons; the first is, the three compression components Cd, Cb- and Cc
in Fig. 6.13 act on two different materials which are normal and lightweight concrete with equal compression strength but with different E-modulus, the second reason is the type of loads which is hysteretic load as in this study, this means that the concrete at this corner A is cracked while it reaches its tensile strength at early cycles of loading.