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

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

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

Transcription

1 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 of tied columns, and by 0.85 for spirally reinforced columns. c. ACI Code 9... specifies that for sections in which the net tensile strain in the extreme tension steel at nominal strength, ε t is between the limits for compression-controlled and tension-controlled sections,ϕ shall be permitted to be linearly increased from that for compression-controlled sections to 0.90 as ε t increases from the compressioncontrolled strain limit to Figure 9.5: Adjusted interaction diagram 9.. Non-dimensional Design Interaction Diagrams Non-dimensional design interaction diagrams are independent of column dimensions. In this text, ACI 40 prepared diagrams are used in design of rectangular and circular columns ( a sample of which is shown in Figure 9.6). The vertical coordinate K P / ( f' A ) represents the non-dimensional form of the nominal axial load capacity of the section. The horizontal coordinate R M / ( f' A h) n n c g represents the non-dimensional nominal bending moment capacity of the section. These diagrams could be used with any system of units. The strength reduction factor (φ) was considered to be.0 so that the nominal values contained in the interaction diagrams could be used with any set of φ factors. In order to use these diagrams, it is necessary to compute the value of, which is equal to the distance from the center of the bars on one side of the column to the center of the bars on the other side of the column divided by the depth of the column, both taken in the direction of bending. n n c g

2 CHAPTER NINE COLUMNS 5 Figure 9.6: Nominal load-moment strength interaction diagram Example (9.7): Design reinforcement for a 40 cm 50 cm tied column. The column, which is part of a nonsway frame (bent into single curvature), has an unsupported length of.0 m. It is subjected to a factored axial load of 40 tons in addition to a factored bending moment of 50 t.m at both column ends. Use f ' c 80 kg / cm and f y 400 kg / cm. Solution: For the column to be short,

3 CHAPTER NINE COLUMNS 6 k l u r k l u ( M / M ) ( 00) ( 50) / r ( M / M ) 4 ( 50 / 50) > i.e., the column is classified as being short. h (cover ) ( d γ h Assuming 50 γ stirrup φ 5mm for bars, ( 4) ( ) 50 ) d ( 000) / ( 40)( 50) P K n n ( 00000) / 0.65 ( 40)( 50)( 50) P e 50 R n n 0.75 h 80 b Using strength interaction diagram L4-60.7, ρ g 4.67 %., while using interaction diagram L4-60.8, ρ g 4.5 %.. Interpolating between the two values of γ, one gets ρ g 4.46 %. ( 40)( 50) 89.cm A s , use 0 φ 5mm Clear distance between bars in the short direction ( 4) 5( ) 6 (.5) cm >.5(.5) cm 4.0 cm (O.K) Spacing of φ 0 mm ties is the smallest of: 48 () 48 cm 6 (.5) 40 cm 40 cm Use φ 0 mm 40 cm as shown in Figure 9.7.

4 CHAPTER NINE COLUMNS 7 Figure 9.7: Designed cross section Example (9.8): Design a short, square tied column to carry the following service loads: P L 5tons, M D 6.5t. m, and M L.5t. m. Use f ' c 80 kg / cm, f y 400 kg / cm, and % ρ. P D 75tons, Solution: ( 75) +.6 ( 5) 74.8 tons P u. ( 6.5) +.6 (.5).4 tons M u. In the next table, different cross sections are assumed and corresponding reinforcement ratios are then evaluated using strength interaction diagrams L4-60.6, L and L cm x 40 cm 50 cm x 50 cm 60 cm x 60 cm Pn 74.8( 000) / K n 80( 40)( 40) ( 000) / ( )( ) ( 000) / ( 60)( 60) R Pn e h.4 ( 00000) / n 80 ( 40)( 40)( 40) ( 00000) / ( )( )( ).4( 00000) / ( )( )( ) γ 40 ( 4) ( ).5 50 ( 4) ( ).5 60 ( 4) ( ) ρ g

5 CHAPTER NINE COLUMNS 8 Choosing 40 cm 40 cm cross section. ( 40)( 40) 6.0 cm A s 0.0, use 8 φ 6 mm. Clear distance between bars 40 ( 4) ( ) 4 (.6) 7.5 cm >.5 Spacing of φ 0 mm ties is the smallest of: 48 () 48 cm 6 (.6) 5.6 cm 40 cm Use φ 0 mm 5 cm as shown in Figure 9.8. (.6 ) cm > 4.0 cm Figure 9.8: Designed cross section Example (9.9): Design a short spirally reinforced column using minimum reinforcement to support a factored axial load of 4 tons and a factored bending moment of 5 t.m. Use 80 kg / cm, and f c f y 400 kg / cm.. Solution: P u 4.0 tons M u 5.0 t.m D γ ( 4) ( ) D.0

6 CHAPTER NINE COLUMNS 9 In the next table, different cross sections are assumed and corresponding reinforcement ratios are then evaluated using interaction diagrams C4-60.6, C and C n Pn Try a 0 cm diameter cross section. Now, let us check the assumed value of γ, ( / 4)( 0) 7.06 cm A s 0.0 π, use 7 φ mm. Trying φ 8 mm spirals, the pitch S is given by 4( 0.5) ( ) ( π / 4 )( 0 ) ( π / 4)( ) 4a s S. 5 cm Ag f 80 c 0.45D 0.45 c A c f 400 sy Clear pitch of spiral 0 cm 5 cm 40 cm 45 cm K R Pn e h n γ ρ g S c cm Use φ 8 mm spiral with a pitch of.5 cm as shown in Figure 9.9. Figure 9.9: Designed cross section

7 CHAPTER NINE COLUMNS 0 9. Design of Long (Slender) Columns The total moment in along column is the sum of the secondary moment caused by deformations resulting from buckling, and the primary moment caused by transverse loads and column end moments, as shown in Figure 9.0. Once the deflected shape of the column, shown in Figure 9., is determined, the secondary moment can be evaluated at any section. One way to determine the deflected shape is to integrate the basic equation (a) (b) (c) (d) (e) Figure 9.0: (a) beam-column; (b) free body diagram; (c) primary moment; (d) secondary moment; (e) total moment ( x) d y M d x E I (9.7) The stiffness E I can not be determined accurately due to change of the modulus of elasticity of concrete with compressive stress level and an unknown extent of cracking of the concrete sections.

8 CHAPTER NINE COLUMNS ACI Code 0.0. suggests that the moments in long columns be determined using second-order analysis that accounts for the influence of cracking and creep of concrete. If second-order analysis software is not available, approximate analysis using the moment magnification method can be used. The moment magnification method is based on magnifying the primary moments, determined from first-order analysis, by a certain factor in an attempt to reach approximate value of the total moment originally caused by primary and secondary moments. 9.. Moment Magnification In his book Theory of Elastic stability, Timoshenko studied the elastic stability of structures. One of the problems he discussed was a beam-column subjected to an axial load P in addition to a concentrated load Q applied at midspan, shown in Figure 9.. He proved that the maximum deflection following equations ( tanu u) Figure 9.: Deflected shape of column δ max and the maximum bending moment M max are given by the Q L δ max 48EI (9.8) u Q L tan u M max (9.9) 4 u where L u P EI

9 CHAPTER NINE COLUMNS (a) (b) (c) (d) Figure 9.: (a) Beam-column and loads; (b) primary moment (P 0); (c) secondary moment (Q 0); (d) primary + secondary moments Looking carefully at the two previous equations, one can easily notice that the maximum deflection δ max is equal to Q L 48 EI absence of P, multiplied by the factor which is the maximum deflection due to the load Q in the ( tan u u) force P. Similarly, the maximum bending moment u which accounts for the presence of the M max is Q L, which is the maximum 4 moment due to the load Q in the absence of the load P, multiplied by the factor tan u, u which accounts for the presence of the force P. These factors are called amplification factors, and their values are unity when P is equal to zero.

10 CHAPTER NINE COLUMNS Timoshenko derived an approximate expressions for the amplification factor as for P / Pcr less than 0.60, as shown in Figure 9.. P / P cr Eq. (9.8) and (9.9) can be written as Figure 9.: Amplification factors Q L δ max (9.0) 48 EI P / P cr Q L M max (9.) 4 P / P cr where P cr is Euler s critical buckling load given as π EI P (9.) cr ( k L ) u The previous equations can be used for beam-columns bent in single curvature in nonsway frames and with the maximum values of primary and secondary moments coincide near midspan.

11 CHAPTER NINE COLUMNS ACI Moment Magnification Method for Nonsway Frames (a) (b) Figure 9.4: (a) nonsway column bent in single curvature; (b) nonsway column bent in double curvature If a beam-column in nonsway frame is loaded by unequal end moments, without transverse loads, the maximum design moment will occur either at one of the column ends when the secondary moment is small or between the two ends when the secondary moment is large, as shown in Figure 9.4. Since the maximum primary and secondary moments do not coincide, the amplification factor P / P cr can not be applied directly. To deal with this

12 CHAPTER NINE COLUMNS 5 situation, the maximum end moment M is multiplied by an equivalent moment correcting factor C m. This factor is used to convert M into an equivalent uniform moment which gives the same total moment due to actual primary and secondary moments when multiplied by the amplification factor P / P cr, as shown in Figure 9.5. (a) (b) Figure 9.5: (a) Actual moments M max primary moment+secondary moment; (b) Equivalent Uniform moment M max C m M According to ACI Code 0.0.6, slender columns in nonsway frames are designed for the factored axial load curvature, Mc where P u and the factored moment amplified for the effects of member Mc δ ns M δ ns M,min (9.) and M column s larger-end moment, not taken less than ( h) M, min,where M Pu 0, where the units within the brackets are given in millimeters.,min + δ ns moment magnification factor for nonsway frames, given by Cm δ ns.0 (9.4) Pu 0.75 Pc where

13 CHAPTER NINE COLUMNS 6 C m factor relating actual moment diagram to an equivalent uniform moment diagram. For members without transverse loads between the supports, C m is taken as M C m (9.5) M where M / M is positive if the column is bent in single curvature, and negative if the member is bent into double curvature. For columns with transverse loads between supports, which M, min exceeds M, computed end moments M / M. P c Euler s critical buckling load given by Pc π EI ( k l ) u C m is taken as.0. For members for C m is either taken equal to.0, or be based on the ratio of the In the previous equation, the flexural stiffness EI is to account for cracking, and creep given by either of the two following equations ( 0. E I + E I ) c g s se EI (9.6) + βdns EI 0.4 Ec I g + β (9.7) dns E c modulus of elasticity of concrete E s modulus of elasticity of steel reinforcement I g moment of inertia of gross concrete section about centroidal axis, neglecting reinforcement I se moment of inertia of reinforcement about centroidal axis of member cross section β dns creep effect factor equals the ratio of maximum factored axial sustained load to maximum factored axial load associated with the same load combination, but shall not be taken greater than.0.

14 CHAPTER NINE COLUMNS 7 Example (9.0): Design the reinforcement for a 50 cm 55 cm column that carries a factored axial load of 50 tons, a service dead load of 7.9 tons, a smaller-end moment of 0 t.m, and a larger-end moment of 40 t.m, shown in Figure 9.6.a. The column is considered nonsway, its effective length kl u 8.0 m, and is bent into single curvature. Use 80 kg / cm and f c f y 400 kg / cm. Figure 9.6.a: Given cross section Solution: - Check whether the column is short or long: k l u r ( 55) ( M / M ) 4 ( 0 / 40) i.e., the column is classified as being long, thus column s larger-end moment needs to be magnified. - Evaluate the equivalent moment correction factor C m: C m M ( M / ) ( 0 / 40) Evaluate the critical buckling load P c : ( 7.9). β dns E c kg/cm

15 CHAPTER NINE COLUMNS 8 ( ) ( 50)( 55) ( ) 0. 4 EI 59. ( 5.9)( 0) ( 800) ( 000) 0 Pc π tons 4- Evaluate the magnification factor δ ns : 0 ( 0) kg.cm δ ns ( 800.6).54 >.0 5- Evaluate the design moment M c : ( 550) M,min t. m < 40 t. m 000 M c.54 ( 40) 6.6 t. m 6- Design the reinforcement: 55 γ ( 4) ( ) ( 000) ( 0.65) 80 ( 50)( 55) P K n n ( 00000) ( 0.65) 80 ( 50)( 55)( 55) P e R n n 0. h Using strength interaction diagram L4-60.7, one gets, while using strength interaction diagram L , one gets ρ g.%. For γ 0. 78, ρ g.40%. ( 50)( 55) 66.0 cm A s 0.04, use 4 φ 5 mm Clear distance between bars S c 50 ( 4) 4 ( ) 5 (.5) 4 S c is given as 6.75 cm Spacing of φ 0 mm ties is the smallest of: > (.5 (.5).75cm) > 4.0 cm O.K 48 () 48 cm

16 CHAPTER NINE COLUMNS 9 6 (.5) 40 cm 50 cm Use two sets of φ 0 mm 40 cm as shown in Figure 9.6.b. Example (9.): Design a 7.0 m high column that carries a service dead load of 55 tons, and a service live load of 45 tons, shown in Figure 9.7.a. Use 80 kg / cm and f c f y 400 kg / cm. Figure 9.7: (a) Column and eccentricities; (b) end moments

17 CHAPTER NINE COLUMNS 40 Solution: - Compute column end moments M and M : ( 55) +.6 ( 45) 8 tons P u. ( 6/00) 8.8 t m M 8. ( 0/00).8 t m M 8. - Estimate the column size: For an assumed reinforcement ratio of %, A 0.45 P 8( 000) ( ) u g ( fc + ρ g fy ) 0.45 Try a 40 cm 40 cm cross section - Check whether the column is short or long: k l u r ( 40) 58. < 00 ( M / M ) 4 ( 8.8/.8) A g may be assumed as follows: 95.8 cm i.e., the column is classified as being long, thus column s larger-end moment needs to be magnified. 4- Evaluate the equivalent moment correction factor C m: C m M ( M / ) ( 8.8/.8) Evaluate the critical buckling load P c : ( 55). β dns E c kg/cm ( 567.8) ( 40)( 40) ( ) 0.4 EI ( )( 0) ( 700) ( 000) 0 Pc π 9.8 tons 6- Evaluate the magnification factor δ ns : 0 ( 0) kg.cm

18 CHAPTER NINE COLUMNS δ ns.5 > ( 9.8).0 7- Evaluate the design moment M max : ( 400) M,min 8.7 t. m <.8 t. m 000 M c.5 (.8).05 t. m 8- Design the reinforcement: 40 γ ( 4) ( 0.8) ( 000) ( 0.65) 80 ( 40)( 40) P K n n ( 00000) ( 0.65) 80 ( 40)( 40)( 40) P e R n n 0.7 h Using strength interaction diagram L4-60.7, one gets ρ g.80%. The reinforcement ratio is rather high, try a 50 cm 50 cm cross section 9- Evaluate the critical buckling load P cr : E c kg / cm ( 567.8) ( 50)( 50) ( ) 0.4 EI ( )( 0) ( 700) ( 000) 0 Pcr π 77.7 tons 0- Evaluate the magnification factor δ ns : 0 ( 0) kg.cm 0.84 δ ns. > ( 77.7).0 - Evaluate the design moment M c :

19 CHAPTER NINE COLUMNS 4 ( 500) M,min t. m <.8 t. m 000 M c. (.8) 5.59 t. m - Design the reinforcement: 50 γ ( 4) ( 0.8) ( 000) ( 0.65) 80 ( 50)( 50) P K n n ( 00000) ( 0.65) 80 ( 50)( 50)( 50) P e R n n 0.07 h Using strength interaction diagrams L and L4-60.8, one gets ρ g.0%. ( 50)( 50) 5.0 A s 0.0 cm, use 8 φ 0mm. Clear distance between bars S c is given as S c 50 ( 4) ( 0.8) 4(.0) Spacing of φ 8 mm ties is the smallest of: 0.5 cm > 4. 0 cm O.K. 48 (0.8) 8.4 cm 6 (.8) 8.8 cm 50 cm Use two sets of φ 8 mm 5 cm as shown in Figure 9.7.c.

20 CHAPTER NINE COLUMNS 4 Figure 9.7.c: Designed cross section