> 0. 1 f, they are treated as beam-columns.

Transcription

1 223 A- Flexural Members (Beams) of Special Moment Frames Requirements of ACI 21.5 are applicable for special moment frame members proportioned primarily to resist flexure with factored axial forces 0. 1 f. If such members are subjected to axial forces c Ag c A g > 0. 1 f, they are treated as beam-columns. 1- General Requirements: Clear span for the member, l n, shall not be less than four times the effective depth. Width of member, b w, is not to be less than the smaller of 0.3 h and 25 cm, where bw is web width and h is overall thickness of member. Width of member is not to be more than the width of supporting member plus distances on each side of the supporting member equal to the smaller of (a) and (b): (a) Width of supporting member in the direction of the span, C 2, and (b) 0.75 times width of the supporting member in direction perpendicular to C Longitudinal Reinforcement: Minimum amounts of top as well as bottom reinforcement, A, is not to be less than the larger of s, min 0.80 f' f c y b w d and 14 b f w y d

2 224 This requirement needs not be satisfied if the tension reinforcement provided at every section is 1/3 larger than required by analysis. Maximum reinforcement ratio is not to exceed At least two bars are to be provided continuously both top and bottom. Positive moment strength at joint face is not to be less than ½ of the negative moment strength provided at the face of the joint. The negative or positive moment at any section along the member is not to be less than ¼ the maximum moment strength provided at face of either joint. Lap splices of flexural reinforcement are permitted only if hoop or spiral reinforcement is provided over the lap length. Maximum spacing of the transverse reinforcement in the lap region is not to exceed the smaller of d/4 or 10 cm. Lap splices are not to be used within the joints, within a distance of twice the member depth from the face of the joint, and at locations where analysis indicates flexural yielding caused by inelastic lateral displacements of the frame.

3 225 Reinforcement Requirements for Flexural Members of Special Moment Frames 3- Transverse Reinforcement: Hoops are to be provided in the following regions of frame members: (a) Over a length equal to twice the member depth measured from the face of the supporting member toward mid span, at both end of the flexural member; (b) Over lengths equal to twice the member depth on both sides of a section where flexural yielding is likely to occur in connection with inelastic lateral displacements of the frame. The first hoop is to be located at a distance not more than 5 cm from the face of the supporting member. Spacing of such reinforcement is not to exceed the smallest of: d/4, 8d b where d b is the diameter of the smallest longitudinal bars, 24 times the diameter of hoop bars and 30 cm.

4 226 Where hoops are required, they are arranged in away similar to that of column ties. Where hoops are not required, stirrups with seismic hooks at both ends are to spaced at a distance not more than d/2 throughout the length of the member. Hoops in flexural members are permitted to be made up of two pieces of reinforcement: a stirrup having seismic hooks at both ends and closed by a crosstie. Consecutive crossties engaging the same longitudinal bar shall have their 90 deg hooks at opposite sides of a flexural member. If the longitudinal reinforcing bars secured by the crossties are confined by a slab on only one side of the flexural frame member, the 90-degree hooks of the crossties shall be placed on that side. Transverse Reinforcement for Flexural Members of Special Moment Frames

5 227 Splices and Hoop Reinforcement for Flexural Members of Special Moment Frames 3- Shear Strength Reinforcement: The design shear force, V e, is to be determined from consideration of the statical forces on the portion of the member between faces of the joint. It is assumed that moments of opposite sign corresponding to probable flexural moment strength, M, act at the joint faces and that the member is loaded with the factored tributary gravity load along its span. For calculation of M it is assumed pr pr

6 228 that tensile strength in the longitudinal bars is 1.25 f y and a strength reduction factor φ of 1.0. M pr A where A ( 1.25 f ) a s 0.85 s ( 1.25 f )( d a / 2) f' c y b y Transverse reinforcement over the lengths identified in 3(a) and 3(b) shall be proportioned to resist shear assuming V c 0 when both of the following conditions occur: (a) The design shear force, V e, represents ½ or more of the maximum required shear strength within these lengths; (b) The factored axial compressive force, P u, including earthquake effects is less than 0.05 A f'. g c Design Shear Forces For Flexural Members of Special Moment Frames

7 229 Example (8): Design the transverse reinforcement for the potential hinge regions of the earthquake resisting beam in a monolithic reinforced concrete frame shown in the figure. The beam which is part of a special moment resisting frame is subjected to a service dead load of 3.0 t/m and a service live load of 2.0 t/m. 2 2 Note that f ' 300 Kg / cm and f 4200 Kg / cm. c y Solution: In this example requirements of section 21.3 of ACI are to be satisfied. A- ACI "Scope": Based on ACI , factored axial compressive force acting on the member < 0.1f ' c A. (O.K) g

8 230 Based on ACI , clear span of beam is not to be less than four times its effective depth. d cm > 4.0 (O.K) Based on ACI , the width-to-depth ratio is not to be less than Based on ACI , width of beam is not to be less than 25 cm. (O.K) > 0.30 (O.K) Width of beam is not to be more than column width plus three-fourths depth of beam on each side of the column. Width of beam width of column. (O.K) B- ACI "Longitudinal Reinforcement": Based on ACI , minimum ratio of top as well as bottom reinforcement is not to be less than the larger of: and ρ min ( provided) > (O.K) 45( 53.75) - Maximum reinforcement ratio is not to exceed ρ max ( provided) < (O.K) 45( 53.75) - At least two bars are to be provided continuously top and bottom. 2 φ 25mm bars are provided throughout the length of the beam on the top side, while 4 φ 25mm bars are provided continuously on the bottom side. (O.K) Based on ACI , positive moment strength at joint face is not to be less than 1/2 of the negative moment strength provided at the face of the joint. Positive moment strength at face of joint is evaluated as follows: M n ( ve) ( A ) f ( d a / 2) + s, + ve y From equilibrium of forces, C ( + ve) T ( ve) and ( 300)( a)( 45) 19.63( 4200) n n and a 7.18 cm

9 231 ( + ve) ( 19.63)( 4200) [ / 2] t. m M n 5 10 Negative moment strength at face of joint is evaluated as follows: M n ( ve) ( A ) f ( d a / 2) s, ve y From equilibrium of forces, C ( ve) T ( ve) and ( 300)( a)( 45) 29.45( 4200) 0.85 and a cm ( + ve) ( 29.45)( 4200) [ / 2] t. m Mn 5 10 M ( ve) Thus, M ( ve) n + > at face of joint. (O.K) n 2 n - The negative or positive moment at any section along the member is not to be less than 1/4 the maximum moment strength provided at face of either joint. At section of least reinforcement moment strength is evaluated as follows: From equilibrium of forces, ( 300)( a)( 45) 9.817( 4200) n Tn n C and 0.85 and a 3.59 cm ( 9.817)( 4200) [ ] / t.m t. m Mn > (O.K) Based on ACI , lap splices of flexural reinforcement are permitted only if hoop or spiral reinforcement is provided over the lap length. Maximum spacing of the transverse reinforcement in the lap region is not to exceed the smaller of d/4 or 10 cm. Thus, maximum spacing is not to exceed 10 cm within the lap length. - Lap splices are not to be used (a) within the joints; (b) within a distance of twice the member depth from the face of the joint and (c) at locations where analysis indicates flexural yielding caused by inelastic lateral displacements of the frame. Development length of top bars (in tension):

10 f y αβ γ λ l d db C + K tr f ' c db α 1.3, β 1, γ 1, and λ 1 C cm or C [(45 4 (2) 2 (1) 2.5]/ (2) cm i.e., C is taken as 6.25 cm ( 4200) ( 2) Atr f yt (2)(0.785) K tr 3.13 cm s n (10) C + Ktr > 2.5, taken as 2.5. db fy αβ γ λ 0.283( 4200)( 1.3) l db 2.5 C K tr f ' c d b Required development length l d 90 cm d Development length of bottom bars (in tension): 0.283f y αβ γ λ l d db C + K tr f ' c db α 1.0, β 1, γ 1, and λ 1 C cm or C [(45 4 (2) 2 (1) 2.5]/ (6) 5.42cm i.e., C is taken as 5.42 cm ( 4200) ( 4) cm Atr f yt (2)(0.785) Ktr cm s n (10) C + K tr db > 2.5, taken as 2.5.

11 f y αβ γ λ 0.283( 4200)( 1) ld db cm C K tr f ' c d b Required development length l d 70 cm C- ACI "Transverse Reinforcement": Based on ACI , hoops are to be provided in the following regions of frame members: (c) Over a length equal to twice the member depth measured from the face of the supporting member toward mid span, at both end of the member; (d) Over lengths equal to twice the member depth on both sides of a section where flexural yielding is likely to occur in connection with inelastic lateral displacements of the frame. Based on ACI , the first hoop is to be located at a distance not more than 5 cm from the face of the supporting member. Maximum spacing of such reinforcement is not to exceed the smallest of: d/4, 8db where d b is the diameter of the smallest longitudinal bars; 24 times the diameter of hoop bars, and 30 cm. Hoops are to be provided over a distance of 2 h 120 cm from faces of joints. Maximum hoop spacing d / / 4 8db 8 24dh ( 2.5) 24() cm 20cm, taken as 12.5 cm. 24cm 30cm Based on ACI , where hoops are required they are arranged in away similar to that of column ties (ACI ).

12 234 Based on ACI , where hoops are not required, stirrups with seismic hooks at both ends are to spaced at a distance not more than d/2 throughout the length of the member. Maximum spacing d/ / cm, taken as 25 cm. D- ACI "Shear Strength Requirements": Based on ACI , the design shear force V e is to be determined from consideration of the static forces on the portion of the member between faces of the joint. It is assumed that moments of opposite sign corresponding to probable flexural moment strength M act at the joint faces and that the member is loaded with the factored tributary gravity load along its span. For calculation of M it is assumed that tensile strength in the longitudinal bars pr is 1.25 f y and a strength reduction factor φ of 1.0. pr

13 235 ( 3) + 0.5( 2) 4.6 t / m w u 1.2 w u l c / M + ve 1.25 A + ve pr ( ) t ( ) s ( ) f y ( d a / 2) ( 300)( a)( 45) 1.25( 19.63)( 4200) a ( ) ( 19.63)( 4200) + ve [ / 2] t. m 0.85 and cm Mpr 5 10 M pr ( ve) 1.25 As ( ve) f y ( d a / 2) ( 300)( a)( 45) 1.25( 29.45)( 4200) a ( ) ( 29.45)( 4200) ve [ / 2] t. m 0.85 and cm Mpr 5 10 M pr V, max ( + ve) + M pr ( ve) lc t e For sway to the right side for sway to the left. Ve, max t occurs at the right side, while it occurs at the left Based on ACI , transverse reinforcement over the lengths identified in 3(a) and 3(b) shall be proportioned to resist shear assuming V c 0 when both of the following conditions occur: (b) The design shear force represents ½ or more of the maximum required shear strength within these lengths;

14 236 (c) The factored axial compressive force including earthquake effects is less than 0.05 A f '. g c Seismic induced shear 13.27tons < 34.66/ 2 tonsand the above-mentioned requirement is not applicable. ( 45)( 53.75) / tons Vc 0.53 f ' c bd V Vs Vn Vc and V u s Vc Φ V s tons 0.75 For two-legged 10 mm transverse reinforcement, A v f y d 2( 0.785)( 4200)( 53.75) Vs 24.01(1000) and S cms S S Use two-legged 10 mm 12.5 cm, S cms Stirrups at other locations: cm At the end of the hoop region, V u and V u V Vs Vn Vc and u Vs Vc Φ V s tons 0.75 For two-legged 10 mm transverse reinforcement, A v f y d 2( 0.785)( 4200)( 53.75) Vs 16.65(1000) and S cms< 53.75/2 S S Use 10 mm 20 cm. tons

15 237

16 238 B- Special Moment Frame Members Subjected to Bending and Axial Load Requirements of ACI 21.4 are applicable for special moment frame members proportioned to resist axial forces > 0. 1 f. 1- General Requirements: c A g The shortest cross-sectional dimension, measured on a straight line passing through the geometric centroid, shall not be less than 30 cm. The ratio of shortest cross-sectional dimension to the perpendicular dimension shall not be less than Minimum Flexural Strength of Columns: The flexural strengths of the columns shall satisfy the following equation: M nc 1.2 Mnb Where M nc sum of nominal flexural strengths of columns framing into the joint, evaluated at the faces of the joint. Column flexural strength shall be calculated for the factored axial force, consistent with the direction of the lateral forces considered, resulting in the lowest flexural strength. M nb sum of nominal flexural strengths of the beams framing into the joint, evaluated at the faces of the joint. Flexural strengths shall be summed such that the column moments oppose the beam moments. The intent of the above equation is to reduce the likelihood of inelastic action. If columns are not stronger than beams framing into a joint, flexural yielding can occur at both ends

17 239 of all columns in a given story, resulting in a column failure mechanism that cal lead to collapse. Columns not satisfying the previous equation shall be ignored in determining the calculated strength and stiffness of the structure, and shall conform to ACI (frame members not proportioned to resist forces induced by earthquake motions). Strong Column-Weak Beam Requirements for Special Moment Frames 3- Longitudinal Reinforcement: The reinforcement ratio ρ shall not be less than 0.01 and shall not exceed Lap splices are permitted only within the center half of the member length, and shall be designed as tension lap g

18 240 splices and enclosed within transverse reinforcement conforming to ACI and Typical Lap Splice Details of Columns in Special Moment Frames 4- Transverse Reinforcement: Transverse reinforcement shall be provided over a length o l from each joint face and on both sides of any section where

19 241 flexural yielding is likely to occur as a result of inelastic lateral displacements of the frame. The length l o shall not be less than the largest of: (a) The depth of the member at the joint face or that section where flexural yielding is likely to occur; (b) 1/6 of the clear span of the member; and (c) 45 cm. Transverse reinforcement shall be provided by either single or overlapping hoops, spirals, circular hoops or rectilinear hoops, with or without crossties. Crossties of the same or smaller bar size as the hoops shall be permitted. Each end of the crossties shall engage a peripheral long reinforcing bar. Consecutive crossties shall be alternated end for end and along the longitudinal reinforcement. Spacing of cross ties or legs of rectilinear hoops, h x, within a cross section of the member shall not exceed 35 cm on center. Example of Transverse Reinforcement in Columns

20 242 Spacing of transverse reinforcement along the length l o of the member shall not exceed the smallest of (a), (b) and (c): (a) one-quarter of the minimum member dimension; (b) six times the diameter of the smallest longitudinal bar, and (c) s o h 3 x, where s o shall not exceed 15 cm and need not be taken less than 10 cm. In the same expression h x is maximum horizontal spacing of hoop or crosstie legs on all faces of the column. The volumetric ratio of spiral or circular hoop reinforcement ρ s shall not be less than the larger value evaluated from the following equations: ρ ρ s s 0.12 f ' f yt c A 0.45 A g ch 1 f' f c yt where f yt yield stress of the transverse reinforcement A gross cross-sectional area of concrete section g A ch cross-sectional area of a structural member measured to the outside edges of transverse reinforcement. The total cross-sectional area of rectangular hoop reinforcement shall not be less than that required by the following equations:

21 243 A sh s b f' A 0.30 c c g 1 fyt Ach 0.09 s b A sh f yt c f' c Where s center-to-center spacing of transverse reinforcement measured along the longitudinal axis of the structural member b c cross-sectional dimension of column core measured to the outside edges of the transverse reinforcement composing A sh A ch cross-sectional area of a structural member measured to the outside edges of transverse reinforcement Beyond the length l o, the column shall contain spiral or hoop reinforcement with center-to-enter spacing, s, not exceeding the smaller of six times the diameter of the smallest longitudinal column bars and 15 cm. Columns supporting reactions from discontinued stiff members, such as walls, shall satisfy (a) and (b): (a) Transverse reinforcement as required in 4 shall be provided over their full height at all levels beneath the discontinuity if the factored axial compressive force in these members, related to earthquake effect, exceeds 0.1 f' c Ag. Where design forces have been magnified to account for the over strength of the vertical elements of the seismic-force-resisting system, the limit of 0.1 f' c Ag shall be increased to 0.25 f' c Ag.

22 244 (b) The transverse reinforcement shall extend into the discontinued member at least l d of the largest longitudinal column bar, where l d is determined in accordance with ACI Where the lower end of the column terminates on a wall, the required transverse reinforcement shall extend into the wall at least l d of the largest longitudinal column bar at the point of termination. Where the column terminates on a footing or mat, the required transverse reinforcement shall extend at least 30 cm into the footing or mat. Confinement Requirements at Column Ends (a) Spiral hoop reinforcement

23 Confinement Requirements at Column Ends (b) Rectangular hoop reinforcement 245

24 246 Columns Supporting Discontinued Stiff Members 5- Shear Strength Requirements: The design shear force, V e, is to be determined from consideration of maximum forces that can be generated at the faces of the joint at each end of the member. These joint forces shall be determined using the maximum probable moment strengths, M pr, of the member associated with the range of factored axial loads, P u, acting on the member. The member shears need not exceed those determined from joint strengths based on the probable moment strength M of the transverse members framing into the joint. In no case shall V e be less than the factored shear determined by analysis of the structure. Transverse reinforcement over the length l o shall be proportioned to resist shear assuming V c 0 when both (a) and (b) occur: (a) The earthquake-induced shear force represents ½ or more of the maximum required shear strength within l o ; pr

25 247 (b) The factored axial compressive force, earthquake effects is less than 0.05 A f'. g c P u, including Loading Cases for Design of Shear Reinforcement in Columns of Special Moment Frames

26 248 Example (9): For the column shown in the figure, check the requirements of ACI 21.4 in relation to columns which are part of special moment frames. Note that design column loads are: P u 337 tonsand M u 84.4 tons. 2 2 Use f ' 300 Kg / cm and f 4200 Kg / cm. c y

27 249 Solution: A- ACI "Scope": Based on ACI ,.1 f ' A 0.1( 300)( 45)( 70) / tons 337 tons 0 c g <. Thus, requirements of section ACI 21.4 apply. Based on ACI , the shortest cross-sectional dimension, measured on a straight line passing through the geometric centroid shall not be less than 30 cm. This requirement is satisfied since shortest cross-sectional dimension 45 cm. Based on ACI , width of beam is not to be less than 25 cm. (O.K) The ratio of the shortest cross-sectional dimension to the perpendicular dimension shall not be less than Ratio 0.64 > (O.K) 70 B- ACI "Minimum Flexural Strengths of Columns": Based on ACI , the flexural strengths of the columns shall satisfy the following equation: M c 1.2 Mg Considering the columns on both sides of the joint are of equal flexural strengths, the flexural strength of each of the columns is determined using strength interaction diagrams.

28 ( 4) 2( 1) ρ g , f ' c 4ksi, γ 0.821, P u 337 tons e M u 84.4 tons, e m, m 337 h 0.7 Φ M Using strength interaction diagram E , n 0.48 ksi Ag h Φ M Using strength interaction diagram E , n 0.52 ksi Ag h Φ M Interpolating for γ , n ksi and M n t. m Ag h Note that Ksi Kg/cm 2 From example (8), M ( + ve) M ( + ve) t. mand M ( ve) M ( ve) t. m nr nl nr nl M c t.m, M g t. m Mc Mg > 1.2 (O.K) C- ACI "Longitudinal Reinforcement": Based on ACI , the reinforcement ratio ρ g shall not be less than 0.01 and shall not exceed ρ g (O.K) ( 45)( 70) Based on ACI , lap splices are only permitted within the center half of the member length and shall be designed as tension lap splices enclosed within transverse reinforcement conforming to ACI and Length of lap splice of longitudinal bars (in tension): 250 For Class "B" lap splice, 0.283f y αβ γ λ l d db C + K tr f ' c db l 1.3 l sp d

29 251 α 1.0, β 1, γ 1, and λ 1 C cm or C [(45 4 (2) 2 (1) 2.5]/ (8) cm i.e., C is taken as cm Ignoring the effect of transverse reinforcement, K tr 0 C + Ktr < 2.5 (O.K) db f y αβ γ λ 0.283( 4200)( 1) ld db cm C K tr f ' c d b Required splice length l sp 1.3( ) cm, taken as 140 cm. Based on ACI , transverse reinforcement shall be spaced at a distance not exceeding (a) one-quarter of the minimum member dimension, (b) six times the diameter of the longitudinal reinforcement, and (c) Sx hx, where S x is maximum longitudinal spacing of transverse reinforcement, shall not exceed 15 cm and need not be taken less than 10 cm. In the same expression h x is maximum horizontal spacing of hoop or crosstie legs on all faces of the column. Maximum vertical spacing of transverse reinforcement is not to exceed the smallest of : i.45/ cm ii.6 (2.5) 15 cm iii.s x h x cm 70 2 ( 4) 1 h x 30.5 cm and S x ( 30.5) 19.41cm, taken as 15 2 cm (maximum permitted). Thus maximum spacing is limited to 10 cm (based on the minimum of a, b and c). Based on ACI , crossties or legs of overlapping hoops shall not be spaced more than 35 cm on center-to-center in the direction perpendicular to the longitudinal axis of a structural member. Two cross ties are added to the present φ 10 mm hoops to satisfy this requirement (maximum spacing of 35 cm).

30 252 D- ACI "Transverse Reinforcement": Based on ACI (b), the total cross-sectional area of rectangular hoop reinforcement shall not be less than that required by ACI equations (21-3) and (21-4). For shear in the direction of longer side of the column: sh1 ( 10)( 36)( 300) 45( 70) ( 62) ( 10)( 36)( 300) A 0.09 A sh1' 2.31 cm i.e., A 2.88 cm sh cm 2 Use φ 12 mm tie plus one 12 mm 2 φ cross tie ( A 3( 1.13) 3.39 cm sh ) For shear in the direction of shorter side of the column: 0.3( 10)( 61)( 300) 45( 70) 2 A sh cm ( 62) 0.09( 10)( 61)( 300) 2 A sh2' 3.92 cm i.e., A 4.88 cm sh2 Use φ 12 mmtie plus three 12 mm 2 φ cross ties ( A 5( 1.13) 5.65 cm sh )

31 253 Based on ACI , transverse reinforcement in amount specified before shall be provided over a length l o from each joint face and on both sides of any section where flexural yielding is likely to occur as a result of inelastic lateral displacements of the frame. The length l o shall not be less than the largest of: (d) The depth of the member at the joint face 70 cm (e) 1/6 of the clear span of the member 400/ cm (f) 45 cm. i.e., l o 70 cm. Based on ACI , where transverse reinforcement as specified before is not provided throughout the full length of the column, the remainder of the column length shall contain spiral or hoop reinforcement with center-tocenter spacing not exceeding the smaller of six times the diameter of the longitudinal column bars or 15 cm. S max the larger of 6 (2.5) cm and 15 cm 15 cm E- ACI "Shear Strength Reinforcement": The design shear force V e is to be determined from consideration of maximum forces that can be generated at the faces of the joint at each end of the member. These joint forces shall be determined using the maximum probable moment strengths M of the member associated with the range of pr factored axial loads on the member. The member shears need not exceed those determined from joint strengths based on the probable moment strength M pr of the transverse members framing into the joint. In no case shall V e be less than the factored shear determined by analysis of the structure.

32 254 1/ 2( ) + ( ) tons(see Example 8 for M pr V e 4 values) d cm ( 45)( 63.55) / tons(neglecting effect of axial force) V c V Vs Vn Vc and u Vs Vc Φ V s tons 0.75 A v f y d Vs S and Av V 14.9( 1000) s S f y d 4200( 63.55) Av 3.5( 45) < (O.K) S 4200 min 2 For S 10 cms, A v cm Available (within the length l o ) 3 (1.13) cm 2 > (O.K). A v

33 255

34 256 C- Joints of Special Moment Frames Requirements of ACI 21.7 are applicable for joints of special moment frames. 1- General Requirements: Forces in longitudinal beam reinforcement at the joint face shall be determined by assuming that the stress in the flexural tensile reinforcement is 1.25 f. Beam longitudinal reinforcement terminated in a column shall be extended to the far face the confined column core and anchored in tension according 4 and in compression according to chapter 12. Where longitudinal beam reinforcement extends through abeam-column joint, the column dimension parallel to the beam reinforcement shall not be less than 20 times the diameter of the largest longitudinal bar. 2- Transverse Reinforcement: Transverse reinforcement as discussed in B shall be provided within the joint, unless the joint is confined by structural members as shown below. Within the depth of the shallowest framing member, transverse reinforcement equal to at least ½ the amount shown in B shall be provided where members frame into all four sides of the joint and where each member width is at least ¾ the column width. At these locations spacing is permitted to be increased to 15 cm. Transverse reinforcement as required in B shall be provided through the joint to provide confinement for longitudinal beam reinforcement outside the y

35 257 column core if such confinement is not provided by a beam framing into the joint. Effective Area of Joint 3- Shear Strength: The nominal shear strength of the joint shall not be taken greater than the values specified below: - For joints confined on all four sides 5.3 f c A j - For joints confined on three faces or on two opposite faces f c A j - For others 3.18 f c Aj A member that frames into a face is considered to provide confinement to the joint if at least ¾ of the face of the joint is covered by the framing member. A joint is considered to be confined if such members frame into all faces of the joint.

36 Development length of bars in tension: The development length l dh for a bar with a standard 90 degree hook shall not be less than the largest of 8 db, 15 cm, and the length required by the following equation which is applicable to bar diameters ranging from 10 mm to 36 mm. f y db ldh fc The 90-degree hook shall be located within the confined core of a column. For bar diameters 10 mm through 36 mm, the development length l d for a straight bar shall not be less than (a) and (b): (a) 2.5 times the length required by the previous equation if the depth of the concrete cast in one lift beneath the bar does not exceed 30 cm, and (b) 3.5 times the length provided by the same equation if the depth of the concrete cast in one lift beneath the bar exceeds 30 cm.

37 Horizontal Shear in Beam-Column Connection 259

38 260 Example (10): Determine the transverse reinforcement and shear strength requirements for the interior beam-column connection shown in Example (9). Solution: A- ACI "General Requirements" Based on ACI , forces in longitudinal beam reinforcement at the joint face shall be determined as assuming that the stress in the flexural tensile reinforcement is 1.25f y. Based on ACI , where longitudinal beam reinforcement extends through a beam-column joint, the column dimension parallel to the beam reinforcement shall not be less than 20 times the diameter of the larger longitudinal bar. 20( 2.5) 50 cm 70 cm (O.K) 20d b < B- ACI "Transverse Reinforcement": Based on ACI , transverse reinforcement in shall be provided within the joint.

39 261 A 2.88 cm sh 2 C- ACI "Shear Strength": V col tons 4.6 V u, jo int T + C 1 2 V col tons b j bb + 2x 45 cm ( )cm 2 ( 70)( 45) 3150 cm A j b j hcol ( 3150) / tons V n Φ V n 0.75 Vu Φ V n ( ) tons > and column dimension in the direction of shear force needs to be increased. Column's longer cross-sectional dimension may be increased to 0.45/ cm without violating the specified ratio of For Vu, 0.75( ) 300( 45)( h col )/ tons and h col Φ Vn Increase column cross sectional dimension to 45 cm x 90 cm.