Overall Stability Analysis of Improved Buckling Restrained Braces

2.1. Design Methodology of the Specimens

2.1.1. Sectional Area of Inner Tube of BRBs

Fig. (1) shows the schematic of a structure, in which the BRBs are designed to resist all lateral loads. The columns and beams are designed for gravity load and the force transferred from brace, so that they remain elastic during the earthquake. It is intended that the maximum roof displacement is almost the same with the given target displacement. The lateral stiffness of BRB of the single story structure is derived as follows:

(2.1)

Thus,

(2.2)

The lateral stiffness of BRB is

Where E is young's modulus, A is the sectional area of inner tube, θ is the slope.

Define nominal rigidity ratio as R_s = K_D/K_f . K_D is the lateral stiffness of BRB. K_f is the lateral stiffness of steel frame and the recommended value of R_s is between 3 to 5. The lateral stiffness of steel frame is obtained by D-value method with K_f = 12αE_cI_c/h³, α is correction coefficient with the derivation in Table 1. E_c is young's modulus of the column, I_c is the moment of inertia of the column. h is the height of column.

Table 1.

Floors	Schematic of the structure	Beam-column stiffness ratio	Correction coefficient of stiffness
The ground floor
The other floors

Notes: i₁, i₂, i₃ and i₄ are the linear stiffness of beam which can be deviated by the formula of i = EI/l. E is young's modulus of the beam, I is the moment of inertia of the beam. l is the length of beam.

From the statement above, the sectional area of inner tube of BRB can be obtained, that is

2.1.3. Sectional Area of Outer Tube of BRBs

The inner tube is combined tightly with outer tube through contact rings. At initial stage, both inner tube and outer tube deformed together under axial compression so that the deformation of inner tube is agreed with outer tube at the midpoint. Assume that contact pressure of central contact ring is Q when inner tube and outer tube bending together. In the real structure, two ends of the brace are articulated with main body structure. The simplified calculation model is shown below (see Fig. 2). The differential equations [12] are based on small deflection theory of mechanics of materials. Balance equation of inner tube is:

(2.3)

E₁ is the Young’s modulus of inner tube.

I₁ is the moment of inertia of an area of inner tube.

v₁ (x) is y axial displacement of inner tube

Q is external force by outer constraints

P is external axial force

Balance equation of constraint is :

(2.4)

E₂ is the Young’s modulus of outer constraints.

I₂ is the moment of inertia of an area of outer constraints.

v₂ (x) is y axial displacement of outer constraints.

Substitute Eq. (2.4) into Eq. (2.3), Eq. (2.5) can be obtained :

(2.5)

Apply boundary conditions to Eq. (2.5), the overall critical load can be obtained as Eq. (2.6):

(2.6)

µ-represents the coefficient of calculation length, l represents the length of inner tube. When µ = 1.0 it illustrates that both ends are articulated, µ = 0.5 are fixed. However, in the real structure, µ is between 0.5 and 1.0.

In recent researches, scholars hold the opinion that the flexural stiffness of inner tube is far less than the flexural stiffness of outer tube. Compare E₂I₂ with E₁I₁, E₁I₁ could be ignored, the formula can be simplified as Eq. (2.7):

(2.7)

To make full use of energy dissipation of inner tube, the members of BRBs are required to bear the maximum axial force, the value is P_max = f_yA₁. It is necessary that P_max ≤ P_cr when considering the safety of stability of BRBs. Define ζ = P_cr/P_max = π²E₂I₂/(πl)²f_yA₁, in theory, if ζ ≥ 1, overall buckling of the member will not occur, however considering the safety, ζ ≥ 1.5 is suggested. The value of I₂ can be obtained, also, the sectional area of outer tube can be acquired.

Table 2.

Specimens	Length of member/cm		Member specification/mm				Stability strength capacity/kN
	Outer tube	Inner tube	Outer tube	Inner tube	Contact ring
					Number	Specification
IBRBs-1	2.4	2.0	Ф89×4.0	Ф60×3.5	2	Ф80×10.0	87.9
IBRBs-2	2.4	2.0	Ф89×4.0	Ф60×3.5	3	Ф80×10.0	129.1
IBRBs-3	2.4	2.0	Ф89×4.0	Ф60×3.5	5	Ф80×10.0	138.5
IBRBs-4	2.4	2.0	Ф76×4.0	Ф60×3.5	5	Ф68×4.0	136.4
IBRBs-5	2.4	2.0	Ф108×4.0	Ф60×3.5	5	Ф100×20	137.2

2.2. Geometrical Parameter

According to the above statement, it indicates that IBRBs in the condition of reasonable design can have good energy dissipation ability. Because of the limitation of equipment, reduced scale test was done by us. Geometrical parameters of IBRBs are shown in Table 2 and Fig. (3).

2.3. Theoretical Analysis

After development of the deformation of inner tube, it will be constrained by outer tube. Assume outer tube is in elastic stage, it provides inner tube with some equivalent force (see Fig. 4). The force depends on the deformation of outer tube and geometric parameters. When the number of the spring approaches to infinite, it equals to the double-tube without clearance. When the number of the spring is finite, it refers to the IBRBs. However, some more assumptions should be considered. 1) Both ends of the brace are articulated and the brace is considered as axially loaded member; 2) the outer tube is considered as elastic model and inner tube is considered as elastoplastic model; 3) the contact rings and outer tube are considered as springs, with the stiffness of k_i.

Assume that each contact ring provides inner tube with stiffness k_i and the stability strength capacity can be obtained according to the energy theory. The derivation can be expressed as Eq. (2.8):

(2.8)

In the formula, E_i is elastic modulus, when σ_x < f_y, E_i equals E, when σ_x ≥ f_y, E_iequals 0;

k_i is the ith stiffness coefficient of the spring

v_i is y axial displacement of the ith spring

u is x axial displacement of inner tube

v is y axial displacement of inner tube

x is the external load

In local system, Eq. (2.8) can be expressed as Eq. (2.9),

(2.9)

Due to the axial displacement has little effect on the whole system, it can be omitted. The formula can be expressed as:

(2.10)

The incremental displacement of the element in local coordinate can be expressed as:

(2.11)

The node incremental displacement and shape function matrix in local coordinate are:

(2.12)

(2.13)

(2.14)

The relationship of the incremental displacement of the element between local coordinate and global coordinate is:

(2.15)

In which,

Discrete Eq. (2.10), Eq. (2.16) can be obtained.

(2.16)

In which,

Consider the relationship between local system and whole system, set

Then, the whole system can be expressed as Eq. (2.17)

(2.17)

For δΠ^k=0, the differential calculus of Eq. (2.17) is Eq. (2.18)

(2.18)

In Eq. (2.18), the last item is the part of spring. This item can also be expressed as a column vector where the nodes k_i with springs has non zero value and the others without springs have zero values. Thus,

(2.19)

The characteristics is that V ≠ 0, Thus | K-pM + k_i^k | = 0, the minimum value of p is the value of stability strength capacity.

According to the above process, a program can be applied by using MATLAB; this program can calculate the values of stability strength capacity of different IBRBs. The values are listed in Table 1.

3.1. Models of the IBRBs

The model is established by using the ANSYS finite element analysis software. Solid45 is applied to the inner tube, outer tube and contact rings of IBRBs. There is a large room of contact area between inner tube and outer tube. Conta174 which has high order quadrilateral element of 8 nodes is adopted. For the displacement of structure will cause nonlinear structure. The comprehensive yield strength of inner tube is 235Mpa. The yield strength of outer tube is 235Mpa, employing steel Q235 with 0.3 elastic modulus. Mapped meshing method and displacement load method are applied, both ends are articulated. The figures of different IBRBs are as Fig. (5).

3.2. Strength Capacity Analysis

Generally, the strength capacity of the member mainly depends on its geometry and material. As for the IBRBs, inner tube is under constraint of outer tube and the strength capacity has been significantly improved. The main factors are: 1) The number of contact rings; 2) The restraining stiffness; 3) Geometric imperfection.

3.2.1. The Effect of the Number of Contact Rings

Through tracking one-way loading of IBRBs-1, IBRBs-2 and IBRBs-3, the load displacement curves (see Fig. 6) and the calculation results (see Table 3) can be obtained.

(1) The curves can be divided into two stages. Namely, elastic stage, which is from the beginning to the first turning point, and plastic stage, which is from the first turning point to the end. In the plastic stage, the curve is smooth, assuming it as a flat line. Therefore, the model can be simplified as an ideal elastoplastic model.

(2) The yield strengths of the members are respectively: 124.8kN, 126.4kN and 128.6kN with the increase of contact rings. It shows that the effect of contact ring is good and can improve the yield strength.

(3) After the development of the four curves reaching to the first turning point, there is a small fluctuation on the curves with the increase of the number of contact rings. The member of IBRBs-1 has the greatest curve fluctuation and the load has a downward trend. However, the curves of IBRBs-2 and IBRBs-3 have a remarkable rising trend. The stability of IBRBs-3 is good and fully meets the standard of ideal plasticity.

Table 3.

Number	Yield strength Pu/kN	Ultimate strength Pu/kN	Clearance/mm	Number of contact ring
IBRBs-1	124.8	131.6	10	2
IBRBs-2	126.4	135.3	10	3
IBRBs-3	128.6	142.5	10	5

3.3. Analysis of Load Displacement Curve

Through the calculation of restraining stiffness of the three members, three curves (see Fig. 6 and Fig. 7) could be obtained. In general; the curve of IBRBs-5 has a poor stability. In contrast, the curves of IBRBs-3 and IBRBs-4 are stable.

(1) The development tendency of IBRBs-3, IBRBs-4 and IBRBs-5 agreed with each other. They are all in the ideal linear period. The initial stiffness of the three members has a little difference, for it mainly reflects the stiffness of inner tube at the initial stage.

(2) The curves of the three members from the yield point to the displacement reaching to 20mm are relatively stable. It means they are all in the plastic stage with mainly axial deformation.

(3) After the displacement of 20mm, the strength capacity of IBRBs-3 and IBRBs-4 have been enhanced. They are all in the strengthening stage. However, the strength capacity of IBRBs-5 is unstable. It means that the member has been affected by the outer tube. For the comprehensive restraining stiffness coefficient η is large and the clearance is small, with deformation of inner tube well constrained by outer tube. Thus, the continuing development is stable. The comprehensive restraining stiffness coefficient η of IBRBs-5 is small and the clearance is large, so the constraint can hardly meet the deformation of inner tube. As a result, the constraint effect is declined and the curve is fluctuated.

(4) After the displacement of 60mm the curve is a horizontal line, IBRBs-3 and IBRBs-4 reach to their ultimate strength.

4.1. The Test Load Program

The loading device consists of a rigid base and loading portal frame (see Fig. 8). The test adopts one way loading and cyclic loading. In order to check the experimental apparatus and data acquisition system, 1.5mm preload should be applied for the specimen before loading. After then, 3 specimens are under one way loading and cyclic loading, the load should be terminated under these following circumstance:

(1) Overall buckling of the brace.

(2) Insufficient rigidity of constraint leads to higher order buckling instability.

(3) Crack of weld seam between flanges and brace.

(4) Fracture of the end of inner tube.

Considering the strength capacity of the loading device and the load program of exiting test, displacement load method is adapted for the whole loading process. Before the value of 100kN, the incremental displacement is 1.5mm, when the value exceeds 100kN; the incremental displacement is 1mm, each stage of load cycling once.

4.2. The Arrangement of Survey Points

There is a load sensor arranged between the upper flanges plate and the load beam (see Fig. 9). The arrangements of displacement meter are:

(1) One is arranged at the upper flanges, mainly to determine the out of plane lateral displacement.

(2) Three of them are distributed along the trisection of circumference of outer tube.

(3) Two of them are arranged at the brace of bottom flanges, mainly to measure axial displacement. Strain gauges (see Fig. 9) are distributed along the position of the trisection of circumference of the steel tube; with a strain gauge set in each position. Thus, three strain gauges are arranged in each section.

4.3. Strength Capacity Analysis

Table 5 lists the main information of the test and Fig. (10) is the load displacement curve. Following conclusions can be obtained.

(1) The yield strengths of IBRBs-3, IBRBs-4 and IBRBs-5 are close to each other. On the whole, the yield strength increases with the increase of restraining stiffness. Because at the initial stage, the main deformation is the axial deformation and the contact ring is closely combined with outer tube. The grater the constraint of outer tube is, the better the strength capacity is. The strength capacity is proportional to the constraining stiffness. Compared with the finite element numerical calculation results, the test results are relatively large, because the value of yield strength in reality is larger than that in theory. However the main principles between them are the same.

(2) In the process of the test, the three members of lateral displacement are very small and the maximum lateral displacement is 2.1mm. The deformation of the member is 1.75‰ of its length. It means overall buckling doesn't occur. The test value of strength capacity has exceeded the calculated value of yield strength (see Table 5). In view of the results, the inner tube is on yield form. That means it reaches to the goal that the buckling restrained brace of the tube is on yield form without buckling.

4.4. Hysteretic Loops Analysis

The hysteretic loops of the specimens under cyclic loading are shown in Fig. (11). The weld in IBRBs-5 was cracked due to the poor quality in the test and the core of the member wasn’t in plastic stage. Thus, the figure of IBRBs-5 hasn’t been presented below. The load and displacement loop is plump and fusiform shaped so that specimen energy dissipation is good. Hysteresis loops show:

(1) The shapes of the hysteresis loops of both IBRBs-3 and IBRBs-4 are similar with each other, they have basically the same envelope area and the same energy dissipation ability.

(2) The hysteresis loops without big fluctuations and strength degradation are plump. The reason is that, with the increment of contact rings, the outer tube plays an important role in constraining the inner tube so that it will yield without buckling, reaching to the effect of same performance in compression and tension.

4.5. The Strength Back-Bone Curves

The strength back-bone curves of IBRBs-3 and IBRBs-4 are shown in Fig. (12). As can be seen from the figure:

(1) The maximum load in tension and compression of IBRBs-3 and IBRBs-4 are similar with each other.

(2) The compression and tension curves of IBRBs-3 and IBRBs-4 are relatively smooth.

(3) The shape of the strength back-bone curves by cyclic loading is similar with the curve of the one way loading, however, the maximum load is slightly lower.