Introduction to Reinforced Concrete PDF

CHAP ERR TE: INTRODUCTION TO REINFORCED CONCRETE 1.0 NOTATIONS & SYMBOLS Ay area of bar, mm? D Dead Load dana 7 diameter of aggregates, mam dj,...

CHAP ERR TE: INTRODUCTION TO REINFORCED CONCRETE 1.0 NOTATIONS & SYMBOLS Ay area of bar, mm? D Dead Load dana 7 diameter of aggregates, mam dj, > diameter of bar, mum d, = diameter of stirrups, mm = Earthquake Load E. = modulus of elasticity of concrete, MPa E, = modulus of elasticity of steel = 200,000 MPa ke = specified compressive stress of concrete, MPa fh = specified yield strength of steel, MPa L = Live Load W = Wind Load 1.1 CONCRETE AND REINFORCED CONCRETE COMPONENTS CONCRETE - is a mixture of sand, gravel, crushed rock, or other aggregates held together in a rocklike mass with a paste of cement and water. Sometimes one or more admixtures are added to change certain characteristics of concrete such as its workability, durability, and time of hardening. As with most rocklike substances, concrete has a high compressive strength and a very low tensile strength. REINFORCED CONCRETE - is a combination of concrete and steel wherein the steel reinforcement provides the tensile strength lacking in the concrete. Steel reinforcing is also capable of resisting compression forces and is used in columns as well as in other situations. AGGREGATES The aggregates used in concrete occupy about fhree-fourths of the concrete volume. Since they are less expensive than the cement, it is desirable to use as much of them as possible, Both fine aggregates (usually sand) and coarse aggregates (usually gravel or crushed stone) are used, Any aggregate that passes a No. 4 sieve (about 6mm in size) is suid to be fine aggregates, Material of a larger size is course aggregate. CamScanner a7 ’ CHAPTER I: INTRODUCTION TO REINFORCED CONCRETE WATER Water used in mixing concrete shall be clean and free from injurious amounts of oils, acids, alkalis, salts, organic matcrials, or other substances that may be deleterious to concrete or reinforcement. Non-potable (non -drinkable) water shall not be used in concrete unless selection of concrete proportions shall be basey on concrete mixes using water from the same sources and mortar test cubes made with non-potable mix Ing water shall have 7-day and 28-day strength equal to at least 90% of strength of similar specimens made with potable water. REINFORCING STEEL BARS Reinforcement used for concrete structures maybe in the form of bars or welded wire fabric. Reinforcing bars are referred to as plane or deformed bars. The deformed bars ribbed projections rolled into their surfaces to provide better bonding between the concrete and steel. Plane bars are not used very often except for wrapping around longitudinal bars primarily in columns. Reinforcing steel bars are commercially available in lengths of 6m, 7.5m, 9m, 10.5 m and even longer lengths. 6m bars are always available, 7.5m bars are sometimes readily available, but longer lengths, when needed may be ordered. STEEL REINFORCEMENT INFORMATION “ASTM STANDARD © ©) J) PHILIPPINE STANDARD (SH) Yr j Reeninel : Nominal - scipn | Nominal beet me aie meter. Area ee ils Arca fo on Designation neh | mm[n® | mm® mm? “g/m _| 3 0.375 9.5 0.11 71 79 0.618 4 0.500 | 12.7 | 0.20 129 113 0.890 5 0.625 | 15.9 | 0.31 199 201 1.580 6 0.750 19.1 0.44 294 314 2.465 7 0.875 | 22.2 | 0.60 387 N.A, N.A. 8 1.000 | 25.4 | 0.79 510 491 3.851 9 1.128 | 28.7 | 1.00 645 616 4.831 10 1.270 | 32.3 | 1.27 819 804 6.310 11 1.410 | 35.8 1.57 1006 1019 7.986 14 1.693 | 43.0 | 2.25 1452 1257 10.861 18 2.257 | 57.3 | 4.00 | 2581 2642 20.729 1.2 ADVANTAGES OF REINFORCED CONCRETE AS A STRUCTURAL MATERIAL Reinforced concrete may be a dominant structural material available for construction in every country. It is used in one form or another for almost all structures such as buildings, bridges, dams and so on. Some of the numerous advantages of reinforced concrete are the following: 1. Concrete is strong in compression compared with most other materials. 2. Concrete building has a great resistance to the actions of fire and water. me. 3. Reinforced concrete structures are very rigid. 689 CamScanner  CHAPTER 1: INTRODUCTION TO REINFORCED CONCRETE 4. It requires less maintenance than a structural steel or thnber, §. Whasa very long service life us compared to other materials, Under proper conditions, reinforced concrete structures can be used indefinitely without reduction of their load-carrying abilities. 6 Conerete material is very widely available, and the reinforcing steel can be transported easily to the construction site than a structural steel. In tost remote areas, reinforced concrete is the preferred constriction materials, 7. Conerete has its ability to be cast into any varlety of shapes and dimenstons of desired structures. 1.3 DISADVANTAGES OF REINFORCED CONCRETE AS A STRUCTURAL MATERIAL Some of the factors that may cause to select a material other than reinforced concrete are the following: a. Concrete has a very low tensile strength requiring the usc of tension stcel reinforcement. b. The construction of cast-in-place structures requires forms and shoring to hold the concrete in place and support its weight until it hardens and until its strength is adequate. ce. Ithas a relatively low strength per unit weight or volume. It means concrete members require a larger volume and a greater weight of materials which has a great effect on bending moments. d. The properties of concrete vary widely because of variations in its proportioning and mixing 1.4 DESIGN CODES AND SPECIFICATIONS — THE NATIONAL STRUCTURAL CODE OF THE PHILIPPINES 2015 Design and construction of buildings are regulated by building codes to protect the public’s health and safety nationwide. The latest design specifications for the reinforced concrete buildings in the Philippines is THE NATIONAL STRUCTURAL CODE OF THE PHILIPPINES (NSCP 2015) by the Association of Structural Engineers of the Philippines, Inc. (ASEP) as adopted in ACI 3/8M-14 “Building Code Requirements for Structural Concrete and Commentary” generally referred to as the ACI Code. Minimum design requirements for various types of reinforced concrete members are presented in the code. Frequent references are made in this text and section numbers are provided. It should be realized that codes provide only a general guide for design. NSCP 2015 Code allowed design of concrete structures either by strength design or by working stress design. Busding Coge Requirements ® | 18 for Stuctural Cantrete : (ACI 313404) ana NATIONAL Cor (ACI 378240 44) ntary re hop. STRUCTURAL ome CODE OF THE st PHILIPPINES a. 2015 0 ~m oO 1.5.2 MODULUS OF ELASTICITY (NSCP 2015 SECTION 419.2.2) It is the ratio of the normal stress to corresponding strain for tensile or compressive stresses below the proportional limit of the material. Concrete has a modulus of elasticity which varics with the different concrete strength, concrete age, type of loading and the proportions of cement and aggregates. * 419.2.2.1 Modulus of Elasticity, E-, for concrete shall be permitted to be calculated as (a) or (b): a. For values of w, between 1440 and 2560 kg/m3 E, = w.75 0.043,/f= (in MPa) b. For normal weight concrete 689 CamScanner CHAPTER 1: INTRODUCTION TO REINFORCED CONCRETE 1.5.3 MINIMUM SPACING OF REINFORCEMENT (NSCP 2015 SECTION 425.2) 425.2.1 For parallel non-prestressed reinforcement in a horizontal layer, clear spacing shall be at least the greatest of 25mm, d,,, and (4/3)dagg 425.2.2 For parallel non-prestressed reinforcement placed in two or more horizontal layers shall be placed directly above reinforcement in the bottom layer with a clear spacing between layers of at least 25mm. 425.2.3 For longitudinal reinforcement in columns, pedestals, struts, and boundary elements in walls, clear spacing between bars shall be at least the greatest of 40mm, 1. 5d, and (4/3)dagg. 1.5.4 BUNDLED REINFORCEMENT (NSCP 2015 SECTION 425.6) 425.6.1 Non-Prestressed Reinforcement 425.6.1.1 Groups of parallel reinforcing bars bundled in contact to act as a unit shall be limited to four in any one bundle. 425.6.1.2 Bundled bars shall be enclosed within transverse reinforcement. Bundled bars in compression members shall be enclosed by transverse reinforcement at least 12mm @ in size. 425.6.1.3 Bars larger than a 36mm @ shall not be bundled in beams. 425.6.1.4 Individual bars within a bundle terminated within the span of flexural members shall terminate at different points with at least 40d, stagger. 425.6.1.5 Development length for individual bars within a bundle, in tension or compression, shall be that of the individual bar, increased 20 percent for a three-bar bundle, and 33 percent for a four-bar bundle. 425.6.1.6 A unit bundled bars shall be treated as a single bar with an area equivalent to that of the bundle and a centroid coinciding with that of the bundle. The diameter of equivalent bar shall be used for d,, in (a) through (e): Spacing limitations based on d,; Cover requirements based on dy; Spacing and cover values in Section 425,4,2.2; Confinement term in Section 425.4.2.3; , factor in Section 425.4.4. i eco $f & o 8 &- © Fig: Bundled-bar Arrangement Fig: Equivalent Diameter, D 689 CamScanner | CotAPTER 1: INTRODUCTION TO REINFORCED CONCRETE > 1.5.5 SPECIFIED CONCRETE COVER REQUIREMENTS (NSCP 2015 SECTION 420.6.1.3) © = $20,6.13.1 Non-prestressed cast-in-place concrete members shall have specified concrete cover for reinforcement at least that given in Table 420.6.1.3.1. Table 420.6.1.3.1 Specified Concrete Cover for Cast-in-Place Non-Prestressed Concrete Members me ~ Reinforcement — : Cast against and 5 permanently in All an contact with —r | f a | fround weatheorin ay [Temmg bar, MW200 40 ground or MD200 wire, and smaller Slabs, joists, 40mm ¢? = 58mmm 40 36mm bar and smaller 20 Not exposed 10 ane walls contact with brine Primary reinforcement, ground ? stirrups, ties, spirals, 40 pedestals, and tension ties and hoops " 420,6.1.3.2 Cast-in-place prestressed concrete members shall have specified concrete cover for reinforcement, ducts and end fittings at least that given in Table 420.6.1.3.2. Table 420.6.1.3.2 Specified Concrete Cover for Cast-in-Place Prestressed Concrete Members | Concrete | Member "| Reinforcement he __Exposure ae | eee ee on | eee Cast against and permanently in contact with All All 75 ground Exposed to Slabs, joists, All 25 weather or in and walls contact with =) ground 1 others All 40 ; Slabs, joists, 7 Not cr poset to and walls All 20 vet a or th Beams, Primary reinforcement 40 contac "i columns, and Stirmups, ties, spirals, 25 groun tension ties and hoops “ 689 CamScanner  CHAPTER 1: INTRODUCTION TO REINFORCED CONCRETE * 420.6.1.3.4 For bundled bars, specified concrete cover shall be at least the smaller of (a) and (5): a. The equivalent diameter of the bundle b. SOmm and for concrete cast against and permanently in contact with ground, the specified cover shall be 75mm. > 1.5.6 STRENGTH REDUCTION FACTORS (NSCP 2015 SECTION 421.2) * 21.2.1 Strength reduction factor @ shall be in accordance with Table 421.2.1 except as modified by Sections 421.2.2, 421.2.3, and 421.2.4. Table 421.2.1 Strength Reduction Factors @ [Action or StructuralElement [Ss Exceptions Near ends of pretensioned Moment, axial force, or combined 0.65 10 0.90 in members where strands are (a) moment and axial force accordance with | not fully developed, @ shall Section 421.2.2 be in accordance with Section 421.2.3 Additional requirements are given in Section 421.2.4 for (b) Shear 0.75 structures designed to resist earthquake effects. (c) Torsion 0.75 - (d) Bearing 0.65 - (e) | Post-tensioned anchorage zones 0.85 - (f) Brackets and Corbels 0.75 - Struts, ties, nodal zones, and bearing areas designed in _ (8) accordance with strut-and-tie os method in Section 423 Components of connections of precast members controlled by. (h) yielding of steel elements in oon tension i Plain concrete elements 0.60 - 0.45 to 0.75 in (j) Anchors in concrete elements accordance with - Section 417 = 421.2.2 Strength reduction factor for moment, axial force, or combined moment and axial force shall be in accordance with Table 421.2.2. 689 CamScanner 8 CHAPTER 1: INTRODUCTION TO REINFORCED CONCRETE Table 421.2.2 Strength Reduction Factors, @, for Moment, Axial Force, or Combined Moment and Axial Force oe = ae aE ™ iF mean 7 7 ee - m " — @ detec. evening sina h - ansverse Reinforcement 7 Other a 0.65 (b) controlled Ey) | (€ 4 — —~ dt) Ey 0.005 a 0.90 (c) 0.90 (f 11 For sections classified as transition, it shall be permitted to use © corresponding to compression-controllcd sections. * 421.2.2.1 For deformed reinforcement, &y shall be fy/E;. For Grade 280 deformed reinforcement, it shall be permitted to take &,, equal to 0,002. 1.6 LOADS AND LOAD COMBINATIONS Perhaps the most important and most difficult task faced by the structural designer is the accurate estimation of the loads that may be applied to a structure during its life. No loads that may reasonably be expected to n occur may be overlooked. After loads are estimated, the next problem is to decide the worst possible p combinations of these loads that might occur at one time. Loads are classified as being dead, live, or environmental. » DEAD LOADS —are loads of constant magnitude that remain in one position. They include the weight of the structure under consideration as well as any fixtures that are permanently attached to it. For a reinforced concrete building, some dead loads are frames, wall, floors, ceilings, stairways, roofs, and plumbing. Unit weight of concrete = 23.54 kN/m? * LIVE LOADS — are loads that can change in magnitude and position. They include occupancy loads, warehouse materials, construction loads, overhead service cranes, equipment operating loads, and many others. In general, they are induced by gravity. * ENVIRONMENTAL LOADS - are loads caused by the environment where the structure is located. For buildings, they are caused by rain, wind, temperature change, and earthquake. In fact, these are also live loads, but they are the result of environment where the structure is located. Although they do vary with time, they are not all caused by gravity or operating conditions, as is typical with other live loads. NSCP 2015 SECTION 203.3 presents the load factors and combinations that are to be used in design of reinforced concrete structures using Strength Design Mcthod. The required strength must at least equal to the largest value obtained by substituting into NSCP Equations 203-1 to 203-7. On the other hand, NSCP 2015 SECTION 203.4 presents the load factors and combinations using Working Stress Design and must at least equal to the largest value obtained by substituting into NSCP Equations 203-8 to 203-12. 689 CamScanner  CHAPTER ft: INTRODUCTION TO REINFORCED CONCRETE Vv) > 16.1 LOAD COMBINATIONS USING STRENGTIEDESIGN OR LOAD AND RESISTANCE FACTOR DESIGN (NSCP 2015 SECTION 203,3) ® 2033.1 Where strength design or load and resistance factor design is used, structures and all portions thercof shall resist the most critical effects from the following combinations of factored loads: 14(D + F) (203-1) 1.2(D + F +T) + 1.6(L +H) + 0.5(L, or R) (203-2) 1.2D + 1.6(L, or R) + (fil or 0.5W) (203-3) 1.2D + 1.0W + f,L + 0.5(L, or R) (203-4) 12D+10E+f,L (203-5) 0.9D + 1.0W + 1.6H (203-6) 0.9D +1.0E +1.6H (203-7) where; D = Dead Load L = Live Load W = Wind Load E = Earthquake Load L, = Roof Live Load T = Temperature Load F = Fluid Load > 1.6.22 LOAD COMBINATIONS USING ALLOWABLE STRESS OR ALLOWABLE STRENGTH DESIGN (NSCP 2015 SECTION 203.4) * 203.4.1 Where allowable stress or allowable strength design is used, structures and all portions thereof shall resist the most critical effects from the following combinations of loads: D+F (203-8) D+H+F+L+T (203-9) D+H+F+(L,orR) (203-10) D+H+F +0,75[L+T +(L, or R)] (203-11) D+H+F +(0.6W or +) (203-12) 689 CamScanner |’ CHAPTER I: INTRODUCTION TO REINFORCED CONCRETE The following are the most common load combinations for most applications: a LAD b. 12D + 1.6L ce 12D+1.0L+1.0W d. 1.2D+4 1.0L + 1.0£ e 0.9) + 1.0W f. 0.9D + 1.0E 1.7 ARRANGEMENT OF REINFORCING BARS FOR MOST EFFICIENT RECTANGULAR BEAM SECTION For rectangular beams to be the most efficient, reinforcing bars should be arranged in such a way that the effective depth of the reinforcement is the maximum. Minimum cover, which for beams not exposed to weather or in contact with the ground is 40mm, and minimum spacing between parallel bars which is the greater between 25mm or diameter of bar, must be followed. The minimum width of a rectangular beam, Dyin: Bru { - if Dinin = 2c + 2d, + ndp + (n— 1)S, ws 5 where: ° ; C; = concrete cover vay @ A a ‘ d, = diameter of stirrup ders ; 7 114. h a |i d, = diameter of bar 4 ¢ _ S, = clear spacing between the longitudinal bars | ae a n= number of longitudinal bars - e e [25mm ‘ 4 “ cL J ‘4 ¢ d Se —T } Colo} CamScanner  ILLUSTRATIVE PROBLEMS: Solution: From NSCP 2015 Section 425.2.1: Clear spacing shall be at least the greatest of: a 25mm- use b. d,=16mm c. (4/3) dagg 360mm Try single layer: 250 — 2(40) — 2(10) — 6(16) Sc= 5 S_ = 10.8mm < 25mm not ok! Try double layer (4 bars @ Ist layer & 2 bars @ 2nd layer): 250 — 2(40) — 2(10) — 4(16) Se= 3 Sc = 28.67mm > 25mm ok! 689 CamScanner  CHAPTER 2: ANALYSIS AND DESIGN OF BEAMS FOR FLEXURE (STRENGTH DESIGN METHOD) 2.0 NOTATIONS & SYMBOLS: a = depth of equivalent stress block, mm A, = area of tension reinforcement, mm? b = width of compression face of member, mm = distance from extreme compression fiber to neutral axis, mm C. = compressive force of concrete, N Cp = balanced c, mm d = distance from extreme compression fiber to centroid of tension reinforcement, mm d’ = distance from extreme compression fiber to centroid of compression reinforcement E. = modulus of elasticity of concrete, MPa E, = modulus of elasticity of steel = 200,000 MPa fe = specified compressive stress of concrete, MPa fe = calculated stress in reinforcement at service loads, MPa fy = specified yield strength of steel, MPa h = overall thickness of member, mm M,, = nominal moment, Nmm Mnp = nominal balanced moment, Nmm Mane = nominal moment for tension-controlled section, Nmm Mnmax = maximum nominal moment, Nmm My = ultimate moment at section, Nmm T = tensile force of steel reinforcement, N By = factor depending on the value of f? Es = strain in steel below yield point = f[/E, Ec = strain in concrete, (maximum = 0.003) Ey = strain in steel at yield point = f,/E; ) = strength reduction factor OMn = ultimate moment capacity, Nmm @Mnmax = maximum ultimate moment, Nmm Pb = balanced steel ratio p = ratio of tension reinforcement Ww = reinforcement index ratio CamScanner 2.1. INTRODUCTION Beams are structural members that are primarily subjected to flexure or bending. There arc two philosophies of design that are used to analyze the beam sections — (1) Working Stress Method and (2) Strength Design Method. In the Working Stress Method, a structural clement is so designed that the stresses resulting from the action of service loads/working loads and computed by the mechanics of clastic members do not exceed some predesignated allowable values. In the Strength Design Method (formerly called Ultimate Strength Design), the service loads are increased by factors to obtain the load at which failure is considered to be imminent. This load is called the factored load or factored service load. The Strength Design Method requires the conditions of static equilibrium and strain compatibility across the depth of the section to be satisfied. The cross-sectional dimensions of a flexural member and the required amount of flexural reinforcement at critical sections are determined using the strength and serviceability requirements of the Code. For flexure, design strength for at all sections shall satisfy OM,, = M,, where @M,, is the design strength of the member at a particular section and M,, is the moment due to the factored loads, which commonly is referred to as the factored design moment. 2.2 DESIGN ASSUMPTIONS FOR CONCRETE (NSCP 2015 SECTION 422.2.2) The following are the assumptions for Strength Design method: = 422.2.2.1 Maximum strain at the extreme concrete compression fiber shall be assumed equal to 0.003 = 422.2.2.2 Tensile strength of concrete shall be neglected in flexural and axial strength calculations. = 422.2.2,3 The relationship between concrete compressive stress and strain shall be represented by a rectangular, trapezoidal, parabolic, or other shape that results in prediction of strength in substantial agreement with results of comprehensive tests. = 422.2.2.4.1 Concrete stress of 0.85f¢ shall be assumed uniformly distributed over an equivalent compression zone bounded by edges of the cross section and a line parallel to the neutral axis located a distance a from the fiber of maximum compressive strain, as calculated by: a= fBi4c « 422,2.2.4.2 Distance from the fiber of maximum compressive strain to the neutral axis, c, shall be measured perpendicular to the neutral axis. s 422.2,2.4.3 Values of B, shall be in accordance with Table 422,2.2.4.3 CamScanner  Table 4$22.2.2.4.3 Values of B, for Equivalent Rectangular Concrete Stress Distribution i7sfe Fa =0; C,=T 0.85 fab = Af, a=— ~ 2, (&q2.2) A 0.85/15 d Multiplying Eq (2.2) by 7’ _ Achy od a= 0.B5fib d _ As | fyd °= bd 0.85f'. 689 CamScanner  A.. The term 4 is called the ratio of steel reinforcement and Is denoted as p ( pfyd , =o.nsp, de) fy. Leta = Par — Reinforcement Index Ratlo e wd a= 7957 (Eq 2.5) Nominal Moment Capacity: From the stress diagram, a Mn = C, (4-5) M,, = 0.85f2.ab (d — 5) a —_ t wd 1 wd My,n = 0. BS fe ob 0.85 (4-5) b\a 20.85 M,, = fi wbd?(1 — 0.59) > (Eq 2.6) Ultimate Moment Capacity (Design Strength): M,, = OM, M,, = Of.wbd?(1 — 0.59w) > (Eq 2.7) Coefficient of Resistance: R, = f-w(1 — 0.59w) — (Eq 2.8) Then, M,, = OR,,bd? — (Eq 2.9) Solving for w in Eq 2.8 and substitute w = p 2, Cc CamScanner y 23 TYPES OF FAILURE AND STRAIN LIMIT Three types of flexural failure ofa structural member can be expected depending on the percentage of steel used in the section. l. Steel may reach its yield strength before the concrete reaches its maximum strength, in this case, the failure is duc to the yielding of stecl reaching a high strain equal to or greater than 0.005. The section contains a relatively small amount of stecl lesser than what is required for balance condition and is called a TENSION-CONTROLLED SECTION. The maximum strain at the extreme concrete compression fiber just reaches the crushing strain ty £, = 0.003 at the same time as the tension steel reaches a strain €, = f/E, causing them to fail simultaneously, the section is called a BALANCED SECTION. Concrete may fail before the yield of steel, due to the presence of high percentage of steel in the section more than what is required for balanced condition. In this case, the concrete strength and its maximum strain of 0.003 are reached, but the steel stress is less than the yield strength. that is, f; < fy. The strain in the steel is equal to or less than 0.002. This section is called a COMPRESSION-CONTROLLED SECTION. E, = 0.003 E, = 0.003 E, = 0.003 —,- = 0.003 s oa e e c - a) t Cmax / Ce Cy a a“ * a *_ €, 20.005 £, = 0.004 £, < 0.002 &,=f,/Es TENSION MAXIMUM COMPRESSION BALANCED CONTROLLED CONDITION CONTROLLED CONDITION 0 _= 0.75 + 0.287005 f.—fy d-c where f, = 600—— f; = 800MPa for maximum condition f; = 1000M Pa for tension — controlled @ = 0.90 for Tension-centrolled 0 /0-4- @ = 0.65 for Compression-controlled Strength Reduction Factor a = » 23.11 THE BALANCED CONDITION Considering the case of balanced section which implies that at ultimate load, the strain in concrete Ec = 0,003 and that of steel e, = fy /Es. b E, = 0.003 - s = 4 GY ras ay a cp J Seow. 4 ae _ 1 _ _ _ _ pe 7 aa i” on al | r Agy*: -:4. T a - @— * > [ _ 4%: ‘) Es = f,/E; By Ratio & Proportion: Ch _ d— Ch 0.003 — fy] s Cp _ d— Cp 0.003 f,/200000 c 600d Bal d b =600 +f, - alanced c a Since a = Byc 9c => By Asfy pbaf, ox O.85fb O85f/b _—paf, A, By 0.85 f¢ 8; C=C; pdf, 600d 0.85//2, 600+f, Py = 0.858, —fyfe| ——— 600 ——- } > Balanced Steel Ratio \600+ fy CamScanner  > 2.3.1.2 THE MAXIMUM CONDITION For maximum condition, €, = 0.004 b E, = 0.003 ~—-=— c es _ wl ! net! ane d—a/2 T, =Asify Ta = Aude  Ultimate Moment Capacity: M, = Mui -+- Mu Balanced condition @ = 0.65 600d > 600 + fy es = Maximum Condition 3 Cmax = 3d 800 — fy ® = 0.65 + 0.25 1000 — f, 3.3.1 STEPS TO DETERMINE A, AND A, OF A DOUBLY-REINFORCED RECTANGULAR 4 a BEAM WITH KNOWN REQUIRED MOMENT M,, AND OTHER BEAM PROPERTIES. Given: b,d, fe, fy, and My I. Solve for OMpmax 3 St * =) ¥ 3 5 Cmax = ad Mrmax = Tao’? fod” U2 = 14 @, } a= By emer © = 0.6540 oe oe Z vy B00 — fy OR oe 1000 = A D = 0.65 + 0,25 ———-2-1000 ™ ly OM mar = Se ees il OMiamaa = OC, (d 7 =) ee men It OMama, 2 My design as Singly Reinforced WOMpmas < My design as Doubly Reinforeed (Step1)  II. For My; 0.85fab = Asify Ag, = —_ Note: Ag; = Asmax Il]. For M2 a7 i,t __ - iff,2f,; tension steel yields f, = fy = iff,f,; compression steel yields f; = f, = iff{ Asfy = Ag f; As = Asz = Ag, = As — Asz =  Solve fora and ¢: C.=T, 0.85 fab =: Asify a= ——————— C= Cheek assumptions: d-c f; = 600 ; = = iff,2f,; _ tension steel yieldsf, = f, = iff; 5.2.1 SHEAR STRENGTH PROVIDED BY CONCRETE V, (NSCP 2015 SECTION 422.5.5) » 422,5,5.1 For non-prestressed members without axial force, V; shall be calculated by; 1 t V.= 6 fc b,d Unless a more detailed calculation is made in accordance with Table 422.5.5.1  » 5: SUEEAROEN TEAMS CHAP ERR Vible AQAA Detntled Method for Caleulating V, aay, (0.toayf: +17 Val ir) byl mn} (a) ofa, | ; — sedbiibicaiteats ) b), 7 mas | (0.t6ay hi + 170) ae a (b) (ec); ——s = ny ie = Tere pre ct ee 0.29AVf! hyd (c) UY Af, occurs simultaneously with Vy, at the section considered © = $22,5.6.1 Kor non-prestressed members with axial compression, V_ shall be calculated by; 1 Ny ; Vo= 5.2.2 SHEAR STRENGTH PROVIDED BY STIRRUPS V, (NSCP 2015 SECTION 422.5.16, © 422.5.10.1 At each section where V,, > OV, transverse reinforcement shall be prov ided such equation is satisfied V V,>——-/e 0 = 422,5.10.5.3 V, for shear reinforcement shall be calculated by; _ Avfyd Vs : where: s = spiral pitch or the longitudinal spacing of shear reinforcement, mm A, = effective area of all bar legs or wires within spacing s, mm? fyt = specified yield strength of hoop, tie or spiral reinforcement, MPa pa pn4 oe SBF Of Ay = 2Ay Ay = 2Ap Ay = 4A, > 5.2.3 SPACING LIMITS OF SHEAR REINFORCEMENT (NSCP 2015 SECTION 409.7.6.2} 409.7.6.2.1 If required, shear reinforcement shall be provided using stirrups, hoops, oF » longitudinal bent bars. = 409.7.6.2.2 Maximum spacing of shear reinforcement shall be in accordance with Tabi 409.7.6.2.2. Table 409.7.6.2.2 Maximum Spacing of Shear Reinforcement | d/2/ 3h/4 / < 0.33V fe byd Lesser of: 600 Lesser of: d/4 3h/8 > 0.33/f! byd 300 }  75 CHAPTER 5: SUEAR IN BEAMS >» §2.4 MINIMUM SHEAR REINFORCEMENT (NSCP 2015 SECTION 409.6.3) © £096.30 Ammimun area of shear reinforcement, Apmins Shall be provided in aff regions where V,, > 0. 50V, except fo the case in Table 409,6.3,1, where at least A ymin Shall be provided where V,, > OV, Table 409,6.3.1 Cases where A is not Required if 0.50V, < V,, < OV Shallow Depth h< 250mm h S greater of 2.5t, or 0.5b,, Integral with Slab and. h < 600 mm Constructed with steel fibered reinforced normal- weight concrete h = 600 mm conforming to Sections and 426.4.1.5(a), 426.4.2.2(d), V, < 0.170,/f! byd and 426.12.5.1(a) and with f2 < 40 MPa is In accordance with Section One-way joist system 409.8 409.6.3.3 If shear reinforcement is required and torsional effects can be neglected according to Section 409.5.4.1, Amin Shall be in accordance with Table 409.6.3,3. Table 409.6.3.3 Required Aymin b 0.062,/ f) — a Non-prestressed and prestressed — fye with Ansfse < 0.40(A,, + fou + _ A, + fy) 0.35 Pw b hye b 0.062) 6 — c fye Greater of: b * Prestressed with Ap. fee 2 Lesser of 0.35 Fn w d 0.40(Aps + fou +45 + fy) y 126) CHAPTER 5: SHEAR IN BEAMS > §,2.5 CRITICAL SECTION FOR BEAM SHEAR (NSCP 2015 SECTION 409.43) © 409.4,3.1 For beams built integrally with supports, Vy at the support shall be permitted to +, calculated at the face of support. * 409.4.3.2 Sections between the face of support and a critical section located d from the face «) support for non-prestressed beams and h/2 from the face of support for prestressed beams hz be permitted to be designed for V,, at that critical section if (a) through (c) are satisfied: a. Support reaction, in direction of applied shear, introduces compression into the er region of the beam; b, Loads are applied at or near the top surface of the beam; ¢. Noconcentrated load occurs between the face of support and critical section P p ) w |. I eo ew i : “ — ' ih. i iphiss) et Gam el ae! es ce sO oe | i t iey } , >d | {4 / >hW2 a = For Non-Prestressed beams for Prestressed beams 5.3 STEPS IN VERTICAL STIRRUP DESIGN 1, Calculate the factored shear force V,, at critical sections or at any section you want the spacing to® determined : Il. Calculate the shear strength provided by concrete, V, 1 Ve= awh b,d if ¥,, > OV,, stirrups are necessary, proceed to Step III 1 if¥, < OV, but ¥, > 3 ON. proceed to Step V I ify, Ov ,beam section is capable to carry the required shear. CHAPTER 5: SHEAR IN BEAMS II]. Calculate the shear strength V, to be provided by the stirrup. o%,=O0K%+K)=h Vi Vs = Oo Vy * 2 f ifV, avi b,,d, adjust the size of the beam IV. Spacing of stirrups: A spacing s = Avf yd Vs Maximum spacing, s (a) WhenV, Ss svfe 1 7 b,d, Smax = d/2 or 600mm (b) WhenV,>sJfebwd, Sax = d/4 or 300mm V. ifYi < OV, but %, > 50% Minimum area of stirrup: Avmin = 0.062,/f: 7 bes 5g agile! yt 0 CHAPTER 6: TORSION CHAPTER 6: TORSION 6.0 NOTATIONS AND SYMBOLS: area enclosed by outside perimeter of concrete cross-section, mm? gross area of section, mm? additional area of longitudinal reinforcement due to torsion, mm #. 7.. am ? zZ minimum area of longitudinal torsional reinforcement, mm *. *. re. area enclosed by centerline of the outermost closed transverse torsior: reinf., mm? area of one leg of closed stirrup resisting torsion, mm? area of shear reinforcement within a distance s, mm* minimum area of stirrup, mm? width of compression face of member, mm distance from extreme compression fiber to centroid of tension reinforcement, mm specified compressive stress of concrete, MPa specified yield strength of steel, MPa specified yield strength of hoop, tie, or spiral reinforcement, MPa overall depth of section, mm factored moment, kNm factored axial load normal to cross section occurring simultaneously with ¥, to be taken positive for compression and negative for tension, and to includ: | effects of tension due to creep and shrinkage, N outside perimeter of concrete cross-section, mm perimeter of centerline of outermost closed transverse torsional reinforcement, mm spacing of stirrups, mm Cracking torsion, kNm nominal torsional moment, kNm Threshold torsion, kNm factored torsional moment, kNm shear strength provided by concrete, N nominal shear force, N shear strength provided by the reinforcement, N factored shear force, N strength reduction factor ll modification factor (1 for normal weight concrete)  CHAPTER 6: TORSION Al ALINTRODUCTION Romford comctete bean are offen subjected to torsional moments in addition to bending moments and soe) ot Shear forces. Ht must be designed for the efleets from torsional loads wherever applicable Torsion Soyo Ops i strvctural members as a result of asymmetrical loading (eccentric loadings), peometry of the spammer, or structural framing. Torstonal moment tends to twist the structural member around its Some rmadoaal anrs, inducing shear stresses. However, structural members are rarely subjected to torsional omont alone, Usually, torstonal moment usually occurs in combination with bending moment and shear o2 NSCP 2015 CODE REQUIREMENTS > 62.1 TORSIONAL STRENGTH (NSCP 2015 SECTION 422.7) = $22.°.1.J. This section shall apply to members if T,, = OT, where @ is given in Section 421 and threshold torsion T;,, is given in Section 422.7.4. If T,, < OT;,, it shall be permitted to neglect torsional effects. « 422.7.4.1 Threshold torsion, T,,, shall be calculated in accordance with Table 422.7.4.| Table 422.7.4.1 Threshold Torsion for Solid Cross Sections 1 As foe Prestressed member aa AN fe (=) at (b) 12 *"° \ Py { 0334/2 Non-prestressed 1 aff ( ae!) Ny ©) member subjected to ssrA fi |= Ly c axial force 1200 VEcp 0.33A,a/f Where: Ac = area enclosed by outside perimeter of concrete cross section, mm? P., = outside perimeter of concrete cross section, mm A = modification factor (1 for normal weight concrete) = 422.7,3.11fT,, > OT,, and T,, is required to maintain equilibrium, the member shall be designed to resist T,,. _ e  CHAPTER 6; TORSION © 422742 Cracking torsion, T,, shall be calculated it accordance with Table 422.7.5.1 for so id and hollow cross-sections, where N,, is positive for compression and negative for tension. Table 422.7.5.1 Cracking Torsion ah cs Non-prestressed 1 a ie Acy ee | member 3 AV fe ( Fop ) a | | 1 An foc | Prestressed member —A, fe =) 1+—_—— (b) 3 e (# p 0.33AV f Non-prestressed La Ae a N,, member subjected to ile (= 2) 1+ (c) axial force Kop 0.334, afc « 422.7.6.1 For non-prestressed and prestressed members, T,, shall be the lesser of (a) and (b): 2A,A J) Y., = : Sy cote s b. T, = 2A,A a ify P;, cot@ whereA, shall be determined by analysis, 8 shall not be taken less than 30 degrees nor greater than 60 degrees, A, is the area of one leg of a closed stirrup resisting torsion, A, is the area of longitudinal torsional reinforcement, and P,, is the perimeter of the centerline of the outermost closed stirrup. * — 422.7.6.1.1 In Eq 422.7.6.1a and 422.7.6.1b, it shall be permitted to take A, equal to 0. B5Aaa * — 422.7.6.1,2 In Eq 422.7.6.1a and 422.7.6.1b, it shall be permitted to take @ equal to (a) or (b): a. 45 degrees for non-prestressed members or members with Aste < 0, 4(Ays foa + Agfy} b. 37.5 degrees for prestressed members with Ansfze 2 0.4(A pelon ¥ ely) » 6.2.2 CROSS-SECTIONAL LIMITS (NSCP 2015 SECTION 422.7.7) * 422.7.7.1 Cross-sectional dimensions shall be selected such that (a) or (b) is satisfied:  6: TORSION 143 CHAPTER a, For solid sections | ( m ay + for ) 6.2.3 DETAILS OF TORSIONAL REINFORCEMENT (NSCP 2015 SECTION 409.7.6.3) « 409.7.6.3.1 If required transverse torsional reinforcement shall be closed stirrups. = 409,7.6,3.2 Transverse torsional reinforcement shall extend a distance of at least (b, + d) beyond the point required by analysis. = 409,7.6.3.3 Spacing of transverse torsional reinforcement shall not exceed the lesser of P;,/8 and 300mm. > 6.2.4 MINIMUM TORSIONAL REINFORCEMENT (NSCP 2015 SECTION 409.6.4) = 409.6.4.1 A minimum area of torsional reinforcement shall be provided in all regions where T,, = OT; accordance with Section 422.7. =» 409.6.4.2 If torsional reinforcement is required, minimum transverse reinforcement (Ay + 2At)min/S Shall be the greater of (a) and (b): a, 0.062,/Fr— oF yt by b. 0.35— fy = $409.6.4.3 If torsional reinforcement is required, minimum area of longitudinal reinforcement Aimin Shall be the lesser of (a) and (b): 0.42 Fedep - (a40) py fyt fy “Ty 0.42) fA _ (0-17 )Ps frye fy fy "fy > 6.2.5 ADDITIONAL AREA OF LONGITUDINAL REINFORCEMEN DUE TO TORSION  CHAPTER 7: SHORT COLUMNS 7.0 NOTATIONS AND SYMBOLS: A. = net area of conerete, mm? Ag = gross area of concrete, mm? Ags = area of structural steel shape of strength F,, mm? Ag - total area of steel reinforcement, mm? = width of compression face of member, mm d = distance from extreme compression fiber to centroid of tenso- dagg = nominal maximum size of coarse aggregate, mm dy = nominal diameter of bar, mm e = eccentricity, mm ep = balanced eccentricity, mm i = specified compressive stress of concrete, MPa ly = specified yield strength of steel, MPa Mon = nominal balanced moment capacity, kNm M,, = nominal moment capacity, kNm M, = ultimate moment capacity, kNm Pie = nominal balanced load, N P, = nominal axial compressive strength of column, V P, = ultimate axial capacity of column, N s = spacing of stirrups, mm a) = strength reduction factor  CHAPTER 7: SHORT COLUMNS 155. 7.1 INTRODUCTION A column is a vertical structural clement which is subjected to axial compressive forces with or without moments, Columns are generally referred to as compression members because the compression forces dominate their behavior, They support vertical loads from the upper floors and transmit these loads to the lower floors and then to the foundation. TYPES OF COLUMNS Reinforced concrete column can be classified on the basis of the form and types of reinforcement, the ed oe position of the load on the cross-section, and the length of the column in relation to its lateral dirsens:oms > BASED ON FORM AND TYPES OF REINFORCEMENT: (a) PLAIN CONCRETE PEDESTAL -— this may be used only if the height does not exceed three times the least lateral dimension. (b) TIED COLUMNS - A column in which the main longitudinal bars are braced within closely spaced lateral ties (c) SPIRAL COLUMNS —a column in which the main longitudinal bars are enclosed within closely spaced helix or spiral. (d) COMPOSITE COLUMNS - these columns may contain a structural steel sections or pipes. with or without additional longitudinal reinforcement. BASED ON THE POSITION OF THE LOAD ON CROSS-SECTION: W (a) AXIALLY LOADED COLUMNS, where loads are assumed acting at the center of the column section (b) ECCENTRICALLY LOADED COLUMNS, where loads are acting at a distance e from the center of the column section along the x or y — axis, causing moments either about x or y axis BASED ON THE LENGTH OF THE COLUMN IN RELATION TO ITS LATERAL Vv DIMENSIONS: (a) SHORT COLUMNS, where the column's failure is due to the crushing of concrete or the yielding of the steel bars under the full load capacity of the column (b) LONG COLUMNS, where buckling effect and slenderness ratio must be taken into consideration in the design, thus reducing the load capacity of the column relative to that of a short column. 6 = CHAPTER 7: SHORT COLUMNS ™~. Composite Columns Tied Column Spiral Column Figure: Types of Columns — Tied, Spiral and Composite 7.2 AXIALLY LOADED COLUMNS > 7.2.1 AXIAL LOAD CAPACITY OF COLUMNS = Nominal load Capacity P, Py = 0.85f (Ag — Ase) + fyAse = Ultimate Load Capacity P,, of Tied Column OP, = Py = 00.80(0.85f4(Ay — Agr) + fyAse] = Ultimate Load Capacity P,, of Spiral Column OP» = Py = 00.85[0.85f+(A, — Ast) + fyAst] « — Ultimate Load Capacity P,, of Composite Columns OP, = P,, = 00.80[0.85f.A, + fyAg + FyAss| Tied OP, = Py = 00.85[0.85fA, + fyAg + FyAss| Spiral where; ® = 0.65 for Tied Column % = 0.75 for Spiral Column  CHAPTER 7: SHORT COLUMNS 357 >» 7.2.2 LIMIT OF REINFORCEMENT (NSCP 2015 SECTION 416.6) 410.611 For non-prestressed columns and for prestressed columns with average fpe < 1.60MPa, area of longitudinal reinforcement shall be at least 0.01A, but shall not exceed 0.084, 4170,7.3.1 For non-prestressed columns and for prestressed columns with average fpe < 1.60MPa, the minimum number of longitudinal bars shall be (a), (b), or (c): Three within triangular ties Pf Four within rectangular or circular ties a Six enclosed by spirals or for columns of special moment frames enclosed by circular G hoops. ™, - 7.2.3 MINIMUM SPACING OF MAIN REINFORCEMENT (NSCP 2015 SECTION 425.2) 425. 2.3 For longitudinal] reinforcement in columns, pedestals, struts, and boundary elements in walls, clear spacing between bars shall be at least the greatest of 40mm, 1.5d,, and (4/3)dagg 7.2.4 SIZES AND SPACING OF TIES (NSCP 2015 SECTION 425.7.2) 425.7.2.1 Ties shall consist of a closed loop of deformed bar or deformed wire with spacing in accordance with (a) and (b): a. Clear spacing of at least (4/3)dagg b. Center-to-center spacing shall not exceed the least of 16d, of longitudinal bar, 48d, of tie diameter, and smallest dimension of member. 425.7,2.2 Diameter of tie bar or wire shall be at least (a) or (b): a. 10mm 9 enclosing 32mm @ or smaller longitudinal bars; b. 12mm @ enclosing 36mm @ or larger longitudinal bars or bundled longitudinal bars. 425. 7.2.2.1 Rectilinear ties shall be arranged to satisfy (a) and (b): a. Every corner and alternate longitudinal bars shall have lateral support provided by the corner of a tie with an included angle of not more than 135 degrees; b. No unsupported bar shall be farther than 150m clear on each side along the tie from a laterally supported bar. 425.7.2.4 Circular ties shall be permuted where longitudinal bars are located around the penmeter of a circle.  150mm 150mm 150mm 150mm max max max max — a 150mm may 150mm May > 150mm > 150mm > 150mm > 150mm > 150mm > 150mm f (a p } A a ¥ > 150mm > 150mm f 1 = < 135° > 150mm > 150mm i ao __|i e/ ‘S oo. ee A 150mm 150mm 150mm 150mm 150mm 150mm max max max max max max ‘caatice Ul cee = (@ | rr > 150mm fr _}150mm max 1 __$150mm max { \ > 150mm ~4150mm max leis , a (e_le So) _9) \ lea g) Figure: Typical Tie Arrangements  — CHAPTER 7: SHORT COLUMNS 7.4 SHORT COLUMNS SUBJECTED TO AXIAL LOAD AND BENDING All columns are subjected to some bending as well as axial forces, and they need to be proportioned to resist both, The axial load capacity given are only applicable to an eccentricity @ = 0.10h for tied colurmn and e = 0.05h for spiral column, Columns will bend under the action of moments, and those moments will tend to produce compression on one side of the columns and tettsion on the other side. Depending on the relative magnitudes of the moments and axial loads, there are several ways in which the sections might fail. P (a) Large axial load with negligible moment - For this situation, failure will occur by the crushing of concrete, with all reinforcing bars in the column having reached their yield stress in compression (b) Large axial load and small moment such that the entire cross section is in compression — When a column is subject to a smal] bending moment (i.e., when the eccentricity is small), the entire column will be in compression, but the compression will be higher on one side than the other. The maximum compressive stress in the column will be 0.85f/ and failure will occur by the crushing of the concrete with ail the bars in compression. (c) Eccentricity larger than in case (b) such that tension begins to develop on one side of the column — If the eccentricity is increased somewhat from the preceding case, tension will begin to develop en one side of the column, and the steel on that side will be in tension but less than the yield stress. On the other side, the steel will be in compression. Failure will occur as a result of the crushing of the concrete on the compression side. A balanced loading condition — As we continue to increase the eccentricity, a condition will be reached in which the reinforcing bars on the tension side will reach their yield stress at the same time that the concrete on the opposite side reaches its maximum compression, 0.85 f-. This situation is called the balanced loading condition. Large moment with small axial load — \t the eccentricity is further increased, failure will be initiated by the yielding of the bars on the tensile side of the column prior to concrete crushing. cel Large moment with no appreciable axial load — For this condition, failure will occur as it does in a beam, » 17.3.1 ANALYSIS OF ECCENTRICALLY LOADED COLUMNS The following procedure and formulas may be applied on determining the value of P,, for columns with four rows of reinforcement perpendicular to the axis of bending (along side h) h d' d’ Be Ey ee kd _@ @., @ , & ra) «4 ‘| et ‘, © die “@. a ee 2 8 |. o 0: | Steel stress: far = 600——* fsx = 600-—— fi. m600——— fos = 600-— T =Agifsr Cy = Asafs2 C3 =Assfs3 Cy = Asafsa C, = 0.85ftab yee ; P, =C,+0C€3;+0C,-T ». My =0; CoX2 43x + Cyx4 + C.x, — Pax = 0  > 7.3.2 THE PLASTIC CENTROID The eccentricity of the column load is the distance from the load to the plastic centroid of the column. The plastic centroid represents the location of the resultant force produced by the steel and the concrete. It is the point in the column cross-section through which the resultant column load must pass to produce uniform strain at failure. For locating the centroid, all concrete is assumed to be stressed in compression to 0. 85f7 and all steel to f,, in compression. For symmetrical sections, the plastic centroid coincides with the centroid of the column cross section, while for nonsymmetrical sections, it can be located by taking moments. > 7.3.3 BALANCED LOADING CONDITION Columns normally fail by either tension or compression. In between the two lies the so-called balanced load condition where failure may be of either type. Balanced loading occurs when the tension steel just reached its yield strain f,/E, and the concrete is strained to 0.003. For every column there is always a balanced loading situation where an ultimate load Ps, placed at an eccentricity e, will produce a moment Mpp. if e < e, — the column fails in compression (compression controls fi, < i) if e > ey — the column fails in tension (tension controls f,, = fy) For any column, the balanced "c" is: 600d “b ~ 600+fy >» 7.3.4 COLUMN INTERACTION DIAGRAM It can be seen that the analysis of column with statics equations is very tedious and complicated. As a result, designers resort to diagrams or computer in designing columns, Interaction Diagram is a plot the axial load P,, that the column can carry versus its moment capacity My. This diagram is very useful for studying the strengths of columns with varying proportions of loads and moments. Any combination of loading that falls inside the curve is satisfactory, whereas any combination that falls outside represents failure. The diagram is made by plotting the axial load capacity of the column at A, then the balanced loading B, then the bending strength of the column if it is subjected to pure moment only at C. In between the points A and C, the column fails due to combination of axial and bending. Point B is called the balanced point. In reference to point D, the vertical and horizontal dotted lined represents the particular load combination of axial load and moment. =—R 7: SHORT COLUMNS 100% axial load ad pure bending strength of member Figure: Column Interaction Diagram (McCORMAC “Design of Reinforced Concrete”)  CHAPTER 38: LONG/SLENDER COLUMNS 8.0 NOTATIONS AND SYMBOLS: Ay area of gross section of column, mumé (EDs; effective flexural stiffness of member, Ninm4 moment of inertia of gross section of column, mm* effective length factor unsupported length of the member, mm factored moment used for design of columns, kNm smaller factor end moment, kNm larger factor end moment, kNm i] factored end moment on a compression member at the end at which M- ii acts, due to loads that cause no appreciable sidesway, calculated using 2 first-order elastic frame analysis, Nmm. factored end moment on a compression member at the end at which M, Il acts, due to loads that cause appreciable sidesway, calculated using a first- order elastic frame analysis, Nmm. factored end moment on a compression member at the end at which Mz acts, due to loads that cause no appreciable sidesway, calculated using = first-order elastic frame analysis, Nmm. factored end moment on a compression member at the end at which M, acts, due to loads that cause appreciable sidesway, calculated using a first- order elastic frame analysis, Nmm. critical buckling load, N maximum factored axial load, N radius of gyration, mm the ratio of maximum factored sustained shear within a storey to the maximum factored shear in that storey associated with the same load combination ratio used to account for reduction of stiffness of columns due to sustained axial loads (creep factor) moment magnification factor used to reflect effects of member curvature Il between ends of compression member magnification factor for frames not braced against sideways, to reflect lateral drift resulting from lateral and gravity loads, 8.1 INTRODUCTION A column is said to be slender if its cross-sectional dimensions are émall compared with pte leneth Most columns are subjected to bending moments producing lateral deflection of a member between rts ends and may result in relative lateral displacement of joints associated with secondary moments that are added to the primary moments and that may become very large for slender colurnns leading to failure. A slender column may be detined as a column that has significant reduction in its axtal load capacity duc to more=ts resulting from lateral deflections of the column, When a column bends or deflects laterally an amount A, its axial load will cause additional morment aon ved to the column equal to PA. This moment is called secondary moment or P-Delta mormert. This moment will be superimposed onto any moments already in the column. $8.2 SLENDERNESS EFFECTS lt is important to determine whether the effects of slenderness need to be considered or to be neglected in the design of any column because second-order effects can have a significant influence on design strength. The degree of slendermess is generally expressed in terms of the slenderness ratio KL,,/r , where L, is the unsupported length of the member and r is the radius of gyration of its cross section. If slendermess effects are considered small, then columns can be considered “short” and can be designed as short columns. However, if they are “s/ender’, the moment for which the column must be designed is imcreased or magnified. Once the moment is magnified, the column is then designed using the increased moment. An approximate method based on moment magnification may be used. > SLENDERNESS EFFECTS ON COLUMNS (NSCP 2015 SECTION 406.2.5) = 406.2.5 Slenderness effects shall be permitted to be neglected if (a) or (b) is satisfied: a. For columns not braced against sidesway ls 22 T b. For columns braced against sidesway Where M,/Mz is negative if the column is bent in single curvature, and positive for doubl: curvature. M, = smaller factored end moment M2 = larger factored end moment  _ _—_—— eet es US If bracing elements resisting lateral movement of a storey have a total stiffness of at least 12 times the Bross lateral stiffness of the columns in the direction considered, it shall be permitted to consider columns within the Storey to be braced against sidesway. hy) P, 4 i ‘ i \ i \ i j f | i i i / ' é ' \ ' f \ f \ / \ | \ \ \ \ \ \ u M, M, P, P, Single Curvature Double Curvature * 406.2.5.1 The radius of gyration, r, shall be permitted to be calculated by (a), (b), or (c): b. 0.30 times the dimension in the direction stability is being considered for rectangular columns; c. 0.25 times the diameter of circular columns 8.3 ALIGNMENT CHART The preliminary procedure used for estimating effective lengths involves the use of the alignment charts. The primary design aid to estimate the effective length factor k is the Jackson and Moreland Alignment Charts. The chart of part (a) of the figure is applicable to braced frames, whereas the one of part (5) is applicable to unbraced frames. To use the alignment charts for a particular column, factors are computed at each end of the column. The Y factor at one end of the column equals the sum of the stiffness [}}(EI/L)] of the columns meeting at that joint, including the column in question, divided by the sum of all the stiffness of the beams meeting at the joint. Should one end of the column be pinned, ® is theoretically equal to 09, and if fixed, YW = 0. Since a perfectly fixed end 1s practically impossible to have, ¥ is usually taken as 1.0 instead of 0 for assumed fixed ends. When column ends are supported by, but not rigidly connected to a footing, ‘P is theretically infinity but usually is taken as about 10 for practical design. One of the two ¥ values is called 4 and the other is called Pp. After these values are computed, the effective length factor k is obtained by placing a straightedge between 'P, and ‘Wg. The point where the straightedge crosses the middle nomograph is kK  FOO © =) GBeEeo00a0 oO oO 833 2 Ee OUMwOW f O ray YE a Sway frames ACI 318-14 Effective Length Factor k