3. AXIALLY LOADED MEMBERS

3 AXIALLY LOADED MEMBERS 31 Reading Assignment: Section 19 and Sections 81 and 82 of text Most axially loaded structural members carry some moment in addition to axial load -- for this discussion, restrict consideration to axial load only 32 Reinforcement of Compression Members 321 Plain concrete columns prohibited: possibility of bending is always present: ACI 109: 001 A s /A g 008 ACI 1091: where A s = Area of longitudinal reinforcement; A g = Total area of column cross section; 322 Possible column configuration a Tied -- Deformed bars or wires placed normal to column axis Different tied arrangements ACI 7105 Longitudinal Rods and lateral ties 36

3221 Spacing of Ties to Prevent Longitudinal Bar Buckling A Tied column may fail prior to steel yield if shell spalls and longitudinal bars buckle; B Insure that bar buckling load is greater than yield load (σ cr >f y) Assume that bar buckling load is greater than yield load -- Assume a pin--pin bar between ties: P cr = π2 EI L 2 The moment of inertia of a circular bar is: I = π D4 64 and for σ cr = P cr /A = P cr /(πd 2 /4) L Example: σ cr = π 2 E 16 (L D) 2 For f y =40ksi = σ cr σ cr = π 2 E 16 (L D) 2 40 = π 2 E 16 (L D) 2 Solving for critical buckling condition L=21D So, to prevent buckling, space ties more closely than this ACI Code requires (ACI-02, Sect 71052) that spacing not to be greater than 16 D (D = Diameter of longitudinal bar); 48 tie bar diameter; Least member dimension Other Code requirements are given in ACI-02 Sections 7105 37

b Spiral: Circular arrangement of longitudinal bars confined by a continuous wire which spirals around the bars for the entire length of the member; Longitudinal Rods and spiral hooping c Composite and combination: Structural steel member encased in concrete or steel pipe filled with concrete ACI 1092: at least 4 bars in tied columns at least 6 bars in spiral columns at least 3 bars in triangular ties Note-- Tied or spiral columns are used in order: to prevent buckling of longitudinal bars to prevent movement of longitudinal bars during construction Bundles of steel bars are sometimes used to prevent congestion It is shown that they act as a unit with area as the same as all of the bundle bars In buildings columns generally have proportions with the ratio of length to cross section width (L/h) in the range from about 8 to 12 (use of high strength, more slender column becoming more popular) 38

33 Design Assumptions: A Strain compatibility between steel and concrete (to prefect bonding; no slip; mechanical interlocking) B Axially loaded only-- Uniform strain over cross section; Axial load plus moment--strains vary linearly; Moment only-- Strains vary linearly C Concrete tensile strength usually zero D The internal forces must be in equilibrium with applied external loads E Plane cross section remains plane after application of loading F Theory is based on real strain--stress relationships 39

34 Elastic behavior of column -- Example See ACI section 102 40 Á compressive stress 30 ksi 20 10 steel Á y = 00138 (f y = 40 ksi) Strains the same, stresses much different concrete 0 0001 0002 0003 Strain (in/in) 1 Up to f c 1 2 f c concrete stress--strain approximately linear This is known as the working or service load range: 2 For strain equality in this range: letting Á = Á c = Á s = f c E c = f s E s or f s = E s E c f c f s =nf c Note: n is generally rounded to the nearest whole no 40

Adopting the following notation: A g = gross section area, (in 2 ) A s =areaofsteel(in 2 ) A c = net concrete area = A g -- A s P = Axial force on column Then P=A c f c +A s f s = A c f c +nf c A s = f c (A c +na s ) Transformed area A s 2 na s 2 (n 1)A s 2 Actual Size (a) Transformed Section A t =A c +na s (b) Transformed section in axial compression Transformed Section A t =A g +(n--1)a s (c) The three bars along each of the two faces are thought of as being removed and replaced, at the same distance from the axis of the section, with added areas of fictitious concrete of total amount of na s Alternatively, as shown in figure c, we can think of the area of the steel bars as replaced with concrete in which case one has to add to the gross concrete area A g so obtained only (n --1)A s in order to obtain the same total transformed area So, knowing A c =A g -- A s P= f c (A g + (n--1)a s ) 41

341 Example 1 12 12 Given 4#8bars Assume: f c = 4000 psi f y =40ksi area of steel = 4(area of # 8 bars) 4(079) = 316in 2 (see ACI 318 -- bar dimension table, or page iii of class notes) A s A g = 316 144 = 0022 OK (ACI-02 Sect 109 0010 < 0022 < 0080) What axial load will cause concrete to be at its maximum working stress? Solution f c = (4000)/2 = 2000 psi f s =(E s /E c )f c =nf c E c = 57, 000 4000 = 36 10 6 psi E s = 29x10 6 psi (always) Therefore n = (29,000,000/3,600,000) = 804 8 P= f c (A g + (n--1)a s ) = 2000[144 + (8--1)x316] = 332,000 lb = 332 kips both steel and concrete behaved elastically 42

342 Example 2 For the previous example find the axial load P which produces ε c= ε s = 0001 Solution Á s < 0001379 f s = E s Á s = 29, 000, 000(psi) 0001 = 29, 000 psi See the slow rate curve of text on page 26, read stress in concrete for given strain of 0001 f c = 2, 400 psi therefore P = f c A c + f s A s = 2, 400 (144 316) + 29, 000 316 = 338, 016 + 91, 640 = 429, 655 lbs = 430 kips stress Steel 40 24 000139 strain 35 Nominal axial load of column P n ;(P u = ΦP n ) -- Greatest calculated load A Should occur when concrete stress peaks, steel reaches yield -- assume this condition B Concrete stress will not be f c : f c based on test of standard cylinder; ends confined; f c depends on the rate of loading; Strength of actual column varies over length -- water migrates to top, causing top to be slightly weaker use f c =085f c at nominal load condition then P N = A c f c +A s f s = A c (085f c )+A s f y for column of previous example: P N = (144 -- 316)(085)(4000) + 316(40,000) = 605 kips 43

351 Example 3 Consider a rectangular column subjected to axial compression The material stress--strain relationships have been idealized as shown below P 12--#14 bars 24 24 P CONCRETE 60 ksi STEEL 5000 psi E c = 57, 000 f c psi E s = 29, 000 (ksi) Á 0 Á y 1 Determine the stress in the concrete and stress in the steel if the applied load is equal to 3100 kips 2 Determine the stress in the concrete and stress in the steel if the applied load is equal to 4050 kips See the solution on the next page 44

Solution: f c =5ksi 5000 E c = 57,000 = 4030 ksi ( Sect 851 of ACI) 12#14 bars A s =27in 2 (from table A2) E s = 29,000 ksi 1 Assume ε < ε 0 or ε < 000124 Assume elastic behavior for ε < ε y or ε < 000207 both steel and concrete P=f c A c +f s A s where A c =A g -- A s =24x24 -- 27 = 549 in 2 P = 3100 = E c ε c A c +E s ε s A s 3100 = (4030)ε(549) + (29000)ε(27) Note: ε = ε c= ε s (perfect bonding) Therefore we have: ε = 3100/(2,212,470 + 783,000) = 000103 < 000124 ok Assumption was correct f s = εe s = 000103 x 29,000 = 299 ksi f c = εe c = 000103 x 4,030 = 42 ksi 2 Assume ε > ε 0 or ε > 000124 Assume elastic behavior for ε < ε y or ε < 000207 steel and inelastic behavior for concrete P=f c A c +f s A s 4050 = (500) x (549) + (29,000) ε (27) ε = 0001667 > 000124 ok ε = 0001667 < 000207 ok Assumption was correct (Concrete behaves inelastically) Assumption was correct ( Steel behaves elastically) f s = εe s = 0001667 x 29,000 = 483 ksi f c = 5000 ksi 45

36 Behavior of Spirally Reinforced Columns P A P B after spalling Behavior of Spirally Reinforced and Tied Columns 37 Confinement A ACI spiral reinforcement ratio based on tests by Richart, Brandtzeg and Brown -- 1928; (Univ of Illinois experimental bulletin no 185) Using 6 x 12 cylinders, they related lateral confining pressure to axial capacity; Axial capacity Psi f c f * =085f c +40f 2 Lateral Confining Pressure where f * c =Compressive strength of spirally confined core concrete 085 f c = compressive strength of concrete if unconfined f 2 = lateral confinement stress in core concrete produced by spiral 46

Spiral Column 47

48 B What sort of lateral confinement can a given spiral provide? Consider a length of a spiral--wrapped circular section: for a length S : volume of spiral = A sp π D (approximately) volume of concrete = (πd 2 /4)S Let à s = volume of spiral volume of concrete = 4A sp DS Calculate equivalent confinement: f 2 DS =2A sp f ys or f 2 =(ρ s f ys )/2 from previous research: f * c = 085f c + 40f 2 = 085f c + 40 à sf ys 2 f * c = 085f c + 20à s f ys D A sp f ys A sp f ys f 2 S D ACI objective is to insure that P B >P N Therefore, make sure spiral increases capacity of core enough to make up for loss of shell Before shell spalls: P N =A s f y + 085f c (A g -- A s ) After shell spalls: P B =A s f y +(A core -- A s )(085f c +2ρ s f ys ) s A sp D

Set P B =P N, calculate like terms, expand: A s f y + 085f c A g 085f c A s = A s f y + A core (085f c +2à s f ys ) A s (085f c ) 2A s à s f ys Small Ignore the last term -- very small then; 085f c (A g A core ) = A core (2à s f ys ) solve for spiral reinforcement ratio we have: à s = 085f c (A g A core ) A core (2f ys ) or à s = 0425f c f ys A g A core 1 conservatively, change 0425 to 045 to get Eq 10--6 of ACI-02: à s = 045f c f ys A g A core 1 Eq 10--6 which says that the ratio of spiral reinforcement shall not be less than the value given by the equation above; where f y is the specified yield strength of spiral reinforcement but not more than 60,000 psi 38 Maximum Loads for Spiral Column Prior to spalling of shell: same as tied column or P A =P N =A s f y + 085f c (A g -- A s ) after spalling of shell: P B =A s f y +(A core -- A s )(085f c +2ρ s f ys ) P B =A s f y +(A core -- A s )085f c +(A core -- A s )(2ρ s f ys ) ---------------------------------- 49

The underlined term is the added capacity of the core resulting from the presence of the spiral where will ρ s be critical? High strength concrete (shell carries large loads); Small columns or square columns; Columns with large cover 381 Example -- Ultimate Strength of Spiral Column Choose a column with areas equivalent to those of previous example (page 37) A g =144in 2 f c = 4000 psi A core =(π(106) 2 /4) = 874 in 2 f y =40ksi A s =316in 2 s=15in f ys =50ksi Use (3/8) diameter spiral: A sp =011 106 A s A g = 316 144 = 0022 OK Ã s = 4A sp DS = 4 011 106 15 = 0028 Check spiral ratio against ACI requirements 135 Ã s = 045f c f y A g A core 1 = 045 144 874 1 4 50 = 0023 Since 0028 > 0023, the column satisfies the minimum spiral reinforcement requirements Load prior to shell spalling: P N =A s f y + 085f c (A g -- A s ) = 316(40) + 085(4)(144--316) = 605 kips -- same as tied column (compare this with the axial capacity we found in page 37) After spalling of shell: P B =A s f y +(A core -- A s )(085f c +2ρ s f ys ) = 085(4)(874--316) + 2(0028)(50)(874--316) + 316(40) = 6478 kips or 648 kips 50

39 ACI Provisions for Spiral Columns ACI 10351 φp n(max) =085φ [A st f y +085f c (A g -- A st )] ACI (10--1) 310 ACI Provisions for Tied Columns ACI 10352 φp n(max) =080φ [A st f y +085f c (A g -- A st ) ] ACI (10--2) 311 Strength and Serviceability 1 ACI assigns two types of safety factors for design: a Load factor -- increase design loads (ACI 92) Dead load -- 12 Live load -- 16 b Strength reduction factor -- reduce calculated design capacity -- variability of material and construction 2 Columns are assigned the following strength reduction factor (ACI 9322) Tied column -- 065 Spiral column -- 070 spiral column allowed more because of ductility 51

312 Example -- Design of Axial Members under Axial Loads Design a rectangular tied column to accept the following service dead and live loads Ignore length effects Given: P D = 142 kips P L = 213 kips Solution: f c = 4000 psi f y =60ksi b=? A s =? We need to find b, h, and A s Sometimes architectural considerations limit allowable column width and height sizes to a set of given dimensions In this case one needs to determine the reinforcement area and detail the column For this problem assume that we do not have any architectural limitations h=? Calculate design loads: P u = 12 P d + 16 P L = 12 142 + 16 213 = 5112 kips For the first trial use a 12 by 12 column (b =12inandh =12in) FromACI 10351 we have P u = φp n(max) =080φ [A st f y + 085f c (A g -- A st )] P u = 080 065 [A st 60 + 085 4 (144 A st )] = 278A st + 2546 therefore 5112 = 278A st + 2546 A st = 923 in 2 Therefore, use 6--#11 bars with A s = 937 in 2 Check ACI 109 A s = A 937 = 0065 OK g 144 ACI Code requires (ACI-02, Sect 71052) that spacing not to be greater than 12 16 (bar diameter) = 16 x 141 = 2256 in 48 tie bar diameter = 48 x 05 = 24 in (05 diameter of #4 ties) Least member dimension = 12 in Use#4barsfortieswith12inchesofspacing #4 Striuup 6--#11 12 52