USER BULLETIN 3: DETERMINATION OF UNSUPPORTED LENGTH RATIO L/Db

Size: px

Start display at page:

Download "USER BULLETIN 3: DETERMINATION OF UNSUPPORTED LENGTH RATIO L/Db"

Joshua Hunter
6 years ago
Views:

1 USER BULLETIN 3: DETERMINATION OF UNSUPPORTED LENGTH RATIO L/Db The unsupported length ratio (L/Db) is calculated by the automatic Sectional modeler spreadsheet using Dhakal and Maekawa [1]. This method consists in the calculation of an equivalent stiffness ratio (k eq ) that is used as input in a table that associates this ratio with the number of spaces between ties over the buckling length (n). This number is multiplied by the distance between ties and divided by the diameter of the bar to result in the desired unsupported ratio. Equivalent stiffness ratio (k eq ) The equivalent stiffness ratio is the division of the normalized stiffness of the rebar (k) by the tie stiffness (k t ). The normalized stiffness of the rebar is k eq = k k t k = π4 E r I s 3 Where E r I is the reduced flexural rigidity of the rebar and is E r I = 0.5E s I (f y 400); I is the second moment of inertia of the rebar and is I = πd 4 /64; s is the tie spacing. The tie stiffness is: Rectangular sections Circular Sections k t = E ta t l e n l n b k t = 2E ta t D core where E t is the modulus of elasticity of the tie; A t is the area of the tie; l e is the length of the tie leg; D core is the diameter of the cross section s core; n b is the number of longitudinal bars supported by these tie legs, and n l is the number of tie legs parallel to the lateral load. The length of the tie leg (l e ) is exemplified as follows: n b = 2 ; n l = 2 n b = 3 ; n l = 2 n b = 5 ; n l = 4 n b = 5 ; n l = 4 Figure 1 - Number of bars and ties exemplified. This document was prepared by Rafael Salgado as a part of a project supervised by Dr. Serhan Guner Version 1.1

2 Number of spaces between ties over the buckling length (n) The equivalent stiffness ratio is used as the input for table 1 that associates this ratio to a number of spaces between ties. Unsupported Length Ratio Stable buckling mode, n k eq 1 > Table 1 - Determination of n The buckling length is determined by L = n. s and the unsupported length ratio is calculated as follows: where D b is the diameter of the rebar. Unsupported Length Ratio = L D b = Different methods must be considered based on the purpose of which the spreadsheet is being used for. Therefore, three main cases were considered in this spreadsheet, namely, circular, beam/column and slab. Circular With hoop reinforcement For the circular section with hoop reinforcement, the equivalent stiffness ratio is defined for the entire diameter of the cross section as being k eq = k t /k, as shown previously, and the unsupported length ratio can be calculated following the steps shown previously. No hoop reinforcement In this case, the unsupported length ratio is considered to be D/D b, where D is the diameter of the cross section. n. s D b 2 of 10

3 Beam/Column The beam/column presents two different types of unsupported length ratio. The first one is representative of the buckling contribution on the edge s bars due to the bending effect. The second one considers the buckling contribution on the core s bars due to pure compression (Figure 1). Figure 2 - Bending and pure compression bars With transverse reinforcement For the bending unsupported length ratio, the equivalent stiffness ratio is calculated considering the length of the tie leg (l e ) of the tie that is restraining the edge s bars in the tie stiffness (Figure 2(a), next page). Figure 3 - Tie leg for bending and pure compression cases For the pure compression unsupported length ratio, the equivalent stiffness ratio is calculated considering the length of the tie leg (l e ) of the tie that is restraining the core s bars in the tie stiffness 3 of 10

4 (Figure 2(b)). Since pure compression acts on all core s bars, the number of longitudinal bars supported by the ties (exemplified in Figure 1) are multiplied by 2. For both cases, the unsupported length ratio calculation follows its normal procedure. No transverse reinforcement In this case, the unsupported length ratio is section height divided by bar diameter. Slab With shear reinforcement The unsupported length ratio is due to bending of the section. No shear reinforcement The unsupported length ratio is three times section height divided by bar diameter. Slenderness ratio The slenderness ratio (r b ) indicates approximately the reduction of normalized stress (f i f y ) on the modified compressive stress-strain curve due to buckling [2]. Its calculation follows: r b = L D f y 100 Table 2 associates the value of the slenderness ratio with the buckling effect level. r b f i f y Buckling effect level < 8 - No effect Small High Very High Table 2 - Determination of r b and its relation with normalized intermediate stress ratio 4 of 10

EXAMPLES 1. Circular I = πd4 64 = πx19.544 = 7155.96 mm 4 64 E r I = 0.5E s I (f y 400) = 0.5x200000x7155.96x (400 400) = 715 596 382.2 Nmm 2 k t = 2E ta t D core = k = π4 E r I s 3 = π4 x715 596 382.

Therefore, the unsupported length ratio is: The slenderness ratio is: L n. s = = 1x250 D b D b 19.54 = 12.79 r b = L D f y 100 = 12.79 400 100 = 25.

5 EXAMPLES 1. Circular I = πd4 64 = πx = mm 4 64 E r I = 0.5E s I (f y 400) = 0.5x200000x x ( ) = Nmm 2 k t = 2E ta t D core = k = π4 E r I s 3 = π4 x = N/mm 2E t A t D 2xClearCover TieDiameter = k eq = k t k = = x200000x100 = N/mm 500 2x By entering with k eq in table 1, we get the n = 1. Therefore, the unsupported length ratio is: The slenderness ratio is: L n. s = = 1x250 D b D b = r b = L D f y 100 = = Based on Table 2, the table below gives high buckling effect for all reinforcement layers. N L/Db n r b Buckling Effects High High High High High High High High High 5 of 10

2. Beam/Column I = πd4 64 = πx19.544 = 7155.96 mm 4 64 E r I = 0.5E s I (f y 400) = 0.5x200000x7155.96x (447 400) = 756 469 993.

88 N/mm k t = E ta t l e n l n b = For pure compression unsupported length: k t = E ta t l e n l n b = 200000x100 2 = 40863.03 N/mm 300 2x22 11.28 4 k eq = k t k = 40863.03 9210.

6 2. Beam/Column I = πd4 64 = πx = mm 4 64 E r I = 0.5E s I (f y 400) = 0.5x200000x x ( ) = Nmm 2 For bending unsupported length: k = π4 E r I s 3 = π4 x = N/mm k t = E ta t l e n l n b = For pure compression unsupported length: k t = E ta t l e n l n b = x100 2 = N/mm 300 2x k eq = k t k = = x x k eq = k t k = = = N/mm 2x8 By entering with k eq in table 1, we get n = 1 for the bending ratio and n=2 for the pure compression ratio. Therefore, the unsupported length ratio is: For bending unsupported length L D b = n. s D b = 1x = of 10

For pure compression unsupported length The slenderness ratio is: For bending ratio For pure compression ratio L n. s = = 2x200 D b D b 19.54 = 20.47 r b = L D f y 100 = 10.

26 Based on Table 2, the table below gives high buckling effect for the reinforcement layers with bending unsupported length and very high buckling effect for the

7 For pure compression unsupported length The slenderness ratio is: For bending ratio For pure compression ratio L n. s = = 2x200 D b D b = r b = L D f y 100 = = r b = L D f y 100 = = Based on Table 2, the table below gives high buckling effect for the reinforcement layers with bending unsupported length and very high buckling effect for the reinforcement layers with pure compression unsupported length. 3. Slab N L/Db n rb Buckling Effects High Very High Very High Very High Very High Very High Very High High 7 of 10

I = πd4 64 = πx15.964 = 3184.94 mm 4 64 E r I = 0.5E s I (f y 400) = 0.5x200000x3184.94x (400 400) = 318 494 140.5 Nmm 2 k = π4 E r I s 3 = π4 x318 494 140.5 250 3 = 1985.

8 I = πd4 64 = πx = mm 4 64 E r I = 0.5E s I (f y 400) = 0.5x200000x x ( ) = Nmm 2 k = π4 E r I s 3 = π4 x = N/mm k t = E ta t l e n l = x100 3 = N/mm n b 700 2x30 8 k eq = k t k = = 5.90 By entering with k eq in table 1, we get the n = 1. Therefore, the unsupported length ratio is: The slenderness ratio is: L n. s = = 1x250 D b D b = r b = L D f y 100 = = Based on Table 2, the table below gives high buckling effect for both reinforcement layers. N L/Db n rb Buckling Effects High High SPECIAL CASES 1. Open Stirrups Case This bar layer does not have any buckling protection. Thus, L/Db calculation as if no stirrups for this layer. (section height / bar diameter) For core rebar layers : Proceed as specified in manual. Pure compression case. For bottom rebar layer: Proceed as specified in manual. Bending case. 8 of 10

Top Stirrups Case The stirrups provided can t prevent the bars to buckle.

9 2. Top & Bottom Stirrups Case The stirrups provided can t prevent the bars to buckle. Thus, all bar layers should be calculated as no stirrup case. 3. Top Stirrups Case The stirrups provided can t prevent the bars to buckle. All bar layers should be calculated as no stirrup case. 4. Bottom Stirrups Case The stirrups provided can t prevent the bars to buckle. All bar layers should be calculated as no stirrup case. 9 of 10

10 REFERENCES [1] Dhakal, R.P., and Maekawa, K., 2002c, Path-dependent Cyclic Stress-Strain Relationship of Reinforcing bar including buckling, Engineering Structures, Vol. 24, No. 11, pp [2] Wong, P.S., Vecchio, F.J. and Trommels H., 2013, VecTor2 & FormWorks User s Manual, Second Edition, pp of 10

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram

ε t increases from the compressioncontrolled Figure 9.15: Adjusted interaction diagram CHAPTER NINE COLUMNS 4 b. The modified axial strength in compression is reduced to account for accidental eccentricity. The magnitude of axial force evaluated in step (a) is multiplied by 0.80 in case