## Info

Negative m 14.132 x 9421 Positive mvf 14,132 X j 4711 It should be emphasized that the loads shown are psf and would be multiplied by 2 to obtain loads per foot acting on the strong bands. Correspondingly, the moments just obtained are per foot width and must be multiplied by 2 to give the support and span moments for the 2 ft wide strong band. The moments for the Y direction middle strip of the basic case may be used w ithout change thus, in Fig. 15.13g, Moments for the Y direction edge strips...

## Function and Types of Retaining Walls

Retaining walls are used to hold back masses of earth or other loose material where conditions make it impossible to let those masses assume their natural slopes. Such conditions occur when the width of an excavation, cut, or embankment is restricted by conditions of ownership, use of the structure, or economy. For example, in railway or highway construction the width of the right of way is fixed, and the cut or embankment must be contained within that width. Similarly, the basement walls of...

## Introduction

Chapter 3 dealt with the flexural behavior and flexural strength of beams. Beams must also have an adequate safety margin against other types of failure, some of which may be more dangerous than flexural failure. This may be so because of greater uncertainty in predicting certain other modes of collapse, or because of the catastrophic nature of some other types of failure, should they occur. Shear failure of reinforced concrete, more properly called diagonal tension failure. is one example....

## E Heel Slab

The heel slab, too, acts as a cantilever, projecting in this case from the back face of the stem and loaded by surcharge, earth fill, and its own weight. The upward reaction of the soil will be neglected here, for reasons given earlier. Applying appropriate load factors, the moment to be resisted is Mu 1.2 X 225 X 4.672 X 1 2 + 1.6 400 x 4.672 XI 2 + 1620 X 4.672 XI 2 38,200 ft-lb Thus, Mh 38,200 X 12 bd2 0.90 X 12 X 14.52 Interpolating from Graph AAb, the required reinforcement ratio is...

## Table

Strength reduction factors in the ACI Code Tension-controlled sections 0.90 Compression-controlled sections Members with spiral reinforcement 0.70 Post-tensioned anchorage zones 0.85 u Chapter 3 contains a discussion of the linear variation of between tension and compression-controlled sections. Chapter 8 discusses the conditions that allow an increase in for spirally reinforced columns. b StruKuul-lie models are described in Chapter 10. In addition to the values given in Table 1.3, ACI Code...

## Prestressed Concrete

Relaxation that will occur later at a lower stress level. The relaxation rate can thus be artificially accelerated by temporary overtensioning. This technique is the basis for producing low-relaxation steel. c. Loss Estimation by the Time-Step Method The loss calculations of the preceding paragraphs recognized the interdependence of creep, shrinkage, and relaxation losses in an approximate way, by an arbitrary reduction of 10 percent of the initial prestress force P, to obtain the force for...

## Cracking in Flexural Members

All reinforced concrete beams crack, generally starting at loads well below service level, and possibly even prior to loading due to restrained shrinkage. Flexural cracking due to loads is not only inevitable, but actually necessary for the reinforcement to be used effectively. Prior to the formation of flexural cracks, the steel stress is no more I Nilson-Darwin-Dolan I 6. Serviceability I Text I I The McGraw-Hill Design ol Concrete Camples. 2C04 204 DESIGN OF CONCRETE STRUCTURES Chapter 6...

## Flexural Design Based on Concrete Stress Limits

As in reinforced concrete, problems in prestressed concrete can be separated generally as analysis problems or design problems. For the former, with the applied loads, the concrete cross section, steel area, and the amount and point of application of the prestress force known, Eqs. (19.1) to (19.4) permit the direct calculation of the resulting concrete stresses. The equations in Section 19.7 will predict the flexural strength. However, if the dimensions of a concrete section, the steel area...

## A Monolithic Beamand Girder Floors

A beam-and-girder floor consists of a series of parallel beams supported at their extremities by girders, which in turn frame into concrete columns placed at more or less regular intervals over the entire floor area, as shown in Fig. 18.3. This framework is covered by a one-way reinforced concrete slab, the load from which is transmitted first to the beams and then to the girders and columns. The beams are usually spaced Nilson-Darwin-Dolan Design of Concrete Structures, Thirteenth Edition...

## B Equivalent Loads

The effect of a change in the vertical alignment of a prestressing tendon is to produce a vertical force on the concrete beam. That force, together with the prestressing force acting at the ends of the beam through the tendon anchorages, can be looked upon as a system of external loads. In Fig. 19.2a. for example, a tendon that applies force P at the centroid of the concrete section at the ends of a beam and that has a uniform slope at angle between the ends and midspan introduces a transverse...

## ACI Code Provisions for Torsion Design

The basic principles upon which ACI Code design provisions arc based have been presented in the preceding sections. ACI Code 11.6.3.5 safety provisions require that where Tu - required torsional strength at factored loads Tn nominal torsional strength of member The strength reduction factor 0.75 applies for torsion. Tn is based on Eq. (7.4) with A0 substituted for Aofr thus In accordance with ACI Code 11.6.2, sections located less than a distance d from the face of a support may be designed for...

## Torsion Plus Shear

Members are rarely subjected to torsion alone. The prevalent situation is that of a beam subject to the usual flexural moments and shear forces, which, in addition, must also resist torsional moments. In an uncracked member, shear forces as well as torque produce shear stresses. In a cracked member, both shear and torsion increase the forces in the diagonal struts (Figs. 4.2CW and 7.8 > ), they increase the width of diagonal cracks, and they increase the forces required in the transverse...

## B Stresses Elastic and Section Cracked

When the tensile stress ., exceeds the modulus of rupture, cracks form, as shown in Fig. 3.2d. If the concrete compressive stress is less than approximately i - and the steel stress has not reached the yield point, both materials continue to behave elasti-cally, or very nearly so. This situation generally occurs in structures under normal service conditions and loads, since at these loads the stresses are generally of the order of magnitude just discussed. At this stage, for simplicity and with...

## D Toe Slab

The toe slab acts as a cantilever projecting outward from the face of the stem. It must resist the upward pressures shown in Fig. l.lb or c and the downward load of the toe slab itself, each multiplied by appropriate load factors. The downward load of the earth fill over the toe will be neglected because it is subject to possible erosion or removal. A load factor of 1.6 will be applied to the service load bearing pressures. Comparison of the pressures of Fig. 17.7 and c indicates that for the...

## Figure

Fixed portal frame, laterally braced. 3. In members in frames not braced against sidesway, the maximum moments of both kinds. M0 and Py almost always occur at the same locations, the ends of the columns they are fully additive, regardless of the presence or absence of an inflection point. Here, too, other things being equal, the additional deflections and the corresponding moments increase with increasing kl r. This discussion is a simplified presentation of a fairly complex subject. The...

## Example Design of a Gravity Retaining Wall

A gravity wall is to retain a bank 11 ft 6 in. high whose horizontal surface is subject to a live load surcharge of 400 psf. The soil is a sand and gravel mixture with a rather moderate amount of fine, silty particles. It can. therefore, be assumed to be in class 2 of Table 17.1 with the following characteristics unit weight w 120 pcf, 30 (with adequate drainage to be provided), and base friction coefficient 0.5. With sin 30 0.5, from Eqs. (17.4) and (17.5), the soil pressure coefficients are...

## Figure Pi

Only the girders frame into two opposite faces, as before. All reinforcement is the same as for the joint of Problem 11.1. Design and detail the joint, specifying bar placement, cutoff points, and details such as bar hook dimensions. 11.3. The precast columns of a proposed parking garage will incorporate symmetrical brackets to carry the end reactions of short girders that, in turn, carry long-span precast, prestressed double T floor units. The girder reactions will be applied 6 in. from the...

## D Minimum Bonded Reinforcement

To control cracking in beams and one-way prestressed slabs with unbonded tendons. some bonded reinforcement must be added in the form of nonprestressed reinforcing bars, uniformly distributed over the tension zone as close as permissible to the extreme tension fiber. According to ACI Code 18.9.2, the minimum amount of such reinforcement is where A is the area of that part of the cross section between the flexural tension face and the centroid of the gross concrete cross section. Exceptions are...

## D ACI Code Provisions for Underreinforced Beams

While the nominal strength of a member may be computed based on principles of mechanics, the mechanics alone cannot establish safe limits for maximum reinforcement ratios. These limits are defined by the ACI Code. The limitations take two forms. First, the Code addresses the minimum tensile reinforcement strain allowed at nominal strength in the design of beams. Second, the Code defines strength reduction factors that may depend on the tensile strain at nominal strength. Both limitations are...

## Slender Columns

Are usually wind or earthquake loads of short duration, so creep effects are minimal. In general, the use of sway frames to resist sustained lateral loads, e.g., from earth or liquid pressures, is not recommended, and it would be preferable to include shear walls or other elements to resist these loads. 9.1. S. P. llinoshenko and J. M. Gere. Theory of Elastic Stability. 3d ed McGraw-Hill. New York. 1969. 9.2. T. V. ( alambos (ed.). Guide to Stability Design Criteria for Metal Structures. 5th...

## Bar Cutoff and Bend Points in Beams

Chapter 3 dealt with moments, flexural stresses, concrete dimensions, and longitudinal bar areas at the critical moment sections of beams. These critical moment sections are generally at the face of the supports (negative bending) and near the middle of the span (positive bending). Occasionally, haunched members having variable depth or width arc used so that the concrete flexural capacity will agree more closely with the variation of bending moment along a span or series of spans. Usually,...

## Short Columns

(e) From the calculations just completed, plus similar repetitive calculations that will not be given here, the strength interaction curve of Fig. 8. Od is constructed. Note the characteristic shape, described earlier, the location of the balanced failure point as well as the small eccentricity and large eccentricity points just found, and the axial load capacity. (f) In the process of developing a strength interaction curve, it is possible to select the values of steel strain as done in step...

## F Examples of Rectangular Beam Analysis and Design

Flexural problems can be classified broadly as analysis problems or design problems. In analysis problems, the section dimensions, reinforcement, and material strengths are known, and the moment capacity is required. In the case of design problems, the required moment capacity is given, as are the material strengths, and it is required to find the section dimensions and reinforcement. Examples 3.5 and 3.6 illustrate analysis and design, respectively. EXAMPLE 3.5 Flexural strength of a given...

## Strip Method For Slabs

The Final arrangement of bar reinforcement is shown in Fig. 15.1and . Negative bar cutoff locations arc as indicated, and development by embedded lengths into the supports will be provided. All positive bars in the slab and strong band will be carried 6 in. into the support faces. A design problem commonly met in practice is that of a slab supported along three edges and unsupported along the fourth, with a distributed load that increases linearly from zero along the free edge to a maximum at...

## Integrated Beam Design Example

In this and in the preceding chapters, the several aspects of the design of reinforced concrete beams have been studied more or less separately first the flexural design, then design for shear, and finally for bond and anchorage. The following example is presented to show how the various requirements for beams, which are often in some respects conflicting, are satisfied in the overall design of a representative member. EXAMPLE 5.3 Integrated design of T beam. A floor system consists of single...

## Analysis And Design Of Slabs 479

< > The McGraw-Hill Compares. 2004 13.1. A footbridge is to be built, consisting of a one-way solid slab spanning 16 ft between masonry abutments, as shown in Fig. PI3.1. A service live load of 100 psf must be carried. In addition, a 2000 lb concentrated load, assumed to be uniformly distributed across the bridge width, may act at any location on the span. A 2 in. asphalt wearing surface will be used, weighing 20 psf. Precast concrete curbs are attached so as to be nonstructural. Prepare a...

## Bond Anchorage and Development Length

If the reinforced concrete beam of Fig. 5.1 were constructed using plain round reinforcing bars, and, furthermore, if those bars were to be greased or otherwise lubricated before the concrete were cast, the beam would be very little stronger than if it were built of plain concrete, without reinforcement. If a load were applied, as shown in Fig. 5.1 , the bars would tend to maintain their original length as the beam deflects. The bars would slip longitudinally with respect to the adjacent...

## Flexural Analysis and Design of Beams

The fundamental assumptions upon which the analysis and design of reinforced concrete members are based were introduced in Section 1.8. and the application of those assumptions to the simple case of axial loading was developed in Section 1.9. The student should review Sections 1.8 and 1.9 at this time. In developing methods for the analysis and design of beams in this chapter, the same assumptions apply, and identical concepts will be used. This chapter will include analysis and design for...

## A Concrete Stress Control by Prestressing

Many important features of prestressed concrete can be demonstrated by simple examples. Consider first the plain, unreinforced concrete beam with a rectangular cross section shown in Fig. 19.1 . It carrics a single concentrated load at the center of its span. (The self-weight of the member will be neglected here.) As the load W is gradually applied, longitudinal flexural stresses arc induccd. If the concrete is stressed only within its clastic range, the flexural stress distribution at midspan...

## C I I I

Tensile and compressive forces place nodes in compression because tensile forces are treated as if they pass through the node and apply a compressive force on the far side, or anchorage face. Thus, within the plane of a strut-and-tie model truss, nodal zones arc considered to be in a state of hydrostatic compression, as shown in Fig. 10.6 . Nodal zone dimensions wnl, w,l2, and wn3 are proportional to the applied compressive forces. The dimension of one side of a nodal zone is often determined...

## 60

0 0.85 X 3 X 10 m' Mn2 3.03 x 60 26 - 3.56 4080 in-kips The depth to the neutral axis is c a , 7.13 0.85 8.39 and d, 27.5 in. to the lowest bar. The c d, ratio is 8.39 27.5 0.305 0.325, so the , 0.005 requirement is met and 0.90. When the ACI strength reduction factor is incorporated, the design strength is Mn 0.90(6330 + 4080) 9370 in-kips EXAMPLE 3.15 Determination of steel area for a given moment. A floor system consists of a 3 in. concrete slab supported by continuous T beams with a 24 ft...

## Analysis of Indeterminate Beams and Frames

The individual members that compose a steel or timber structure are fabricated or cut separately and joined together by rivets, bolts, welds, or nails. Unless the joints are specially designed for rigidity, they are too flexible to transfer moments of significant magnitude from one member to another. In contrast, in reinforced concrete structures, as much of the concrete as is practical is placed in one single operation. Reinforcing steel is not terminated at the ends of a member but is...

## B Beams with Constant Eccentricity

The design method presented in the previous section was based on stress conditions at the maximum moment section of a beam, with the maximum value of moment M0 resulting from the self-weight immediately being superimposed. If Pt and e were to be held constant along the span, as is often convenient in pretensioned prestressed construction, then the stress limits and fci would be exceeded elsewhere along the span, where M(, is less than its maximum value. To avoid this condition, the constant...

## 0966

Four No. 4 (No. 13) hoops will be used, providing an area of 1.60 in2. I Nilson-Darwin-Dolan I 4. Shear and Diagonal I Text I I The McGraw-Hill Design ol Concrete Tension ol Beams Compares. 2004 160 DESIGN OF CONCRETE STRUCTURES Chapter 4 The maximum shear force that can be transferred, according to the ACI Code limits, will be based conservatively on a horizontal plane 32 in. long. No strength reduction factor need be included in the calculation of this maximum value because it was already...

## Strip Method For Slabs 523

The value of k may be selected so as to make use of the minimum steel in the X direction required by ACI Code 7.12. In choosing mys to be used in Eq. (15.18) for calculating k2, one should again recognize that the deflection of the strong band along the free edge will tend to increase the Y direction moment at the supported edge above the propped cantilever value based on zero deflection. A value for ntyi of about half the free cantilever moment may be appropriate in typical cases. A high ratio...

## B Balanced Strain Condition

A reinforcement ratio h producing balanced strain conditions can be established based on the condition that, at balanced failure, the steel strain is exactly equal to v when the strain in the concrete simultaneously reaches the crushing strain of u 0.003. Referring to Fig. 3.6, Nilson-Darwin-Dolan I 3. Flexural Analysis and I Text I I < > The McGraw-Hill I Design ol Concrete Design ol Beams Company. 2C04 DESIGN OF CONCRETE STRUCTURES Chapter 3 which is seen to be identical to Eq. (3.23)....

## Figure 114

Q (b) strengths, 5 and (c) safety margin. M. Strength of the entire structure or of a population of repetitive structures, e.g., highway overpasses, can also be considered a random variable with a probability density function of the type shown in Fig. 1.14 . As in the case of loads, the exact form of this function cannot be known but can be approximated from known data, such as statistics of actual, measured materials and member strengths and similar information....

## C Examples of Analysis and Design of Beams with Tension and Compression Steel

As was the case for beams with only tension reinforcement, doubly reinforced beam problems can be placed in one of two categories analysis problems or design problems. For analysis, in which the concrete dimensions, reinforcement, and material strengths are given, one can find the flexural strength directly from the equations in Section 3.7a or Section 3.7b. First, it must be confirmed that the tensile reinforcement ratio is less than h given by Eq. (3.52), with compression steel stress from...

## A Stresses Elastic and Section Uncracked

As long as the tensile stress in the concrete is smaller than the modulus of rupture, so that no tension cracks develop, the strain and stress distribution as shown in Fig. 3.2c is essentially the same as in an elastic, homogeneous beam (Fig. 3.1 > ). The only difference is the presence of another material, the steel reinforcement. As shown in Section 1.9a, in the elastic range, for any given value of strain, the stress in the steel is n times that of the concrete Eq. (1.6) . In the same...

## E C

Where Ecs modulus of elasticity of slab concrete c2 size of rectangular column, capital, or bracket in direction l2 C cross-sectional constant see Eq. (13.6) The summation applies to the typical case in which there are slab beams (with or without edge beams) on both sides of the column. The length l2 is measured center-to-center of the supports and, thus, may have different values in each of the summation terms in Eq. (13.10), if the transverse spans are unequal. If a panel contains a beam...

## A

Where all terms are as previously defined. The strength reduction factor is to be taken equal to 0.75 for shear. The additional conservatism, compared with the value of 0.90 for bending for typical beam designs, reflects both the sudden nature of diagonal tension failure and the large scatter of test results. For typical support conditions, where the reaction from the support surface or from a monolithic column introduces vertical compression at the end of the beam, sections located less than a...

## Engineering Drawings for Buildings

Design information is conveyed to the builder mainly by engineering drawings. Their preparation is therefore a matter of the utmost importance, and they should be carefully checked by the design engineer to ensure that concrete dimensions and reinforcement agree with the calculations. Engineering drawings for buildings usually consist of a plan view of each floor showing overall dimensions and locating the main structural elements, cross-sectional views through typical members, and beam and...

## Problems

A rectangular beam made using concrete with 4000 psi and steel with . 60.000 psi has width b - 24 in., total depth h 18 in., and effective depth d 15.5 in. Concrete modulus of rupture 475 psi. The elastic modulus of the steel and concrete are, respectively. 29.000,000 psi and 3.600,000 psi. The tensile steel area is As five No. 11 (No. 36) bars. (a) Find the maximum service load moment that can be resisted without stressing the concrete higher than 0.45 ' or the steel above 0.40 . (b)...

## Yield Line Analysis For Slabs

Orthotropic Reinforcement and Skewed Yield Lines Generally slab reinforcement is placed orthogonally, i.e in two perpendicular directions. The same reinforcement is often provided in each direction, but the effective depths will be different. In many practical cases, economical designs are obtained using reinforcement having different bar areas or different spacings in each direction, in such cases, the slab will have different moment capacitics in the two orthogonal directions and is said to...

## I

If members framing into a joint in a direction perpendicular to the plane of the frame under analysis have sufficient torsional stiffness, and if their far ends are fixed or nearly so. their effect on joint rigidity should be included in the computations. The torsional stiffness of a member of length L is given by the expression GJ L, where G is the shear modulus of elasticity of concrete (approximately to Ec 2.2) and J is the torsional stiffness factor of the member. For beams with rectangular...

## T

When the concrete reaches its limit strain of 0.003. the strain distribution is that shown in Fig. 8.11 b. the strains at the locations of the four bar groups arc found from similar triangles, after which the stresses are found by multiplying strains by E, 29.000 ksi applying the limit value 0.00258 tl 75.0 ksi compression s2 0.00142 fsl 41.2 ksi compression 0.00025 fa 7.3 ksi compression 0.00091 f< A 26.4 ksi tension For j' - 6000 psi. , 0.75 and the depth of the equivalent...

## Figure 145

Alternative mechanisms for a slab supported on three sides. Positive yield lines form along the lines of intersection of the rotating segments of the slab. A rectangular two-way slab on simple supports is shown in Fig. 14.4 . The diagonal yield lines must pass through the comers, while the central yield line is parallel to the two long sides (axes of rotation along opposite supports intersect at infinity in this case). With this background, the reader should have no difficulty in applying the...

## Figure 1212

Three-span continuous beam after the formation of plastic hinges at the interior supports. b. Plastic Hinges and Collapse Mechanisms If a short segment of a reinforced concrete beam is subjected to a bending moment, curvature of the beam axis will result, and there will be a corresponding rotation of one face of the segment with respect to the other. It is convenient to express this in terms of an angular change per unit length of the member. The relation between moment and angle change per...

## Table 102

Nodal Zone Condition Classification Bounded by struts or bearing area C-C-C l .0 Anchoring one tie C-C-T 0.80 Anchoring two or more lies C-T-T or 7-7-7' 0.60 The effective width of a tie w, depends on the distribution of the tie reinforcement. If the reinforcement in a tie is placed in a single layer, the effective width of a tie may be taken as the diameter of the largest bars in the tie plus twice the cover to the surface of the bars. Alternatively, the width of a tie may be taken as the...

## E Two Way Edge Supported Slabs

Two-way solid slabs supported by beams on the column lines on all sides of each slab panel have been discussed in detail in Chapter 13. The perimeter beams are usually concrete cast monolithically with the slab, although they may also be structural steel, often encased in concrete for composite action and for improved fire resistance. For monolithic concrete, both the beams and the slabs are designed using the direct design method or the equivalent frame method described in Chapter 13. Two-way...

## Strutandtie Models 339

2 X 1.5 4 X 0.625 4 spaces 2 x 1.41 5 bars x 1.41 23.8 in., which fits within the 24 in. beam thickness. Design details and minimum reinforcement requirements ACT Code 11.8.6 requires that shear reinforcement in deep beams satisfy a both Eqs. 10.10 and 10.11 or b Eq. 10.5 . Using Eq. 10.10 , the minimum required vertical steel is K 0.0025 0.0025 x 24 x 12 0.72 in2 ft. This is satisfied by No. 5 No. 16 bars at 10 in. placed on each face, giving a total area of reinforcement equal to 0.74 in2...

## Strutandtie Models 333

To design transverse reinforcement for bottle-shaped struts, AC1 Code A.3.3 permits the assumption that the compressive force in the struts spreads at a slope of two longitudinal to one transverse along the axis of the strut, as shown in Fig. 10.3 . Forf'c 6000 psi. the ACI Code considers the transverse reinforcement requirement to be satisfied if the strut is crossed by layers of reinforcement that satisfy where Asi is the total area of reinforcement at spacing .v, in a layer of reinforcement...

## Reinforcement Of Concrete Slab To Resist Shear

Analysis and Dosign oi I Text I I The McGraw-Hill Design of Concrete Slabs Campiw s. 2C01 462 DESIGN OF CONCRETE STRUCTURES Chapter 13 satisfactorily above the actual Vu. When shear is resisted by combined action of concrete and bar reinforcement, the concrete contribution is reduced to V 0.75 X 2 4000 X 72 X 6 41.0 kips No. 3 No. 10 vertical closed hoop stirrups will be used since d must be 16 times the stirrup diameter d 16 f in. and arranged along four integral...

## Biaxially Loaded Column

a Find the required column reinforcement for the condition that the full live load acts. b Check to ensure that the column is adequate for the condition of no live load on the roof. Material strengths are. .' 4000 psi and v 60,000 psi. Solution. a The column will be designed initially for full load, then checked for adequacy when live load is partially removed. According to the AG safety provisions, the column must be designed for a factored load Pu 1.2 X 222 1.6X 333 799 kips and a factored...

## Seismic Design 733

29 kips in the upper column and 31 kips in the lower column. Minimum factored axial loads are 21 and 25 kips below the forces specified in Problem 20.1 for the upper and lower columns, respectively. 20.3. In Example 20.1, the columns are spaced 28 ft on center in the direction of the spandrel beams. The total dead load on the spandrel beam including self-weight is 2 kips ft and the total live load is 0.93 kips ft. Design the spandrel beam transverse reinforcement for a building subject to high...

## Figure 819

Splice details at typical interior column. Special ties to resist outward thrust Special ties to resist outward thrust through the depth of the joint to a level not more than one-half the usual spacing s below the lowest reinforcement in the slab. Analogous requirements are found in ACI Code 7.10.4 and are illustrated in Ref. 8.1 for spirally reinforced columns. As discussed in Section 5.11, in frames subjected to lateral loading, a viable alternative to splicing bars just above the floor is to...

## Figure 916

Cross section of column C3, Example 9.2. Two more checks are required to complete the design. First, in accordance with Eq. 9.21 , a higher magnified moment must he calculated using the values of Mx and M2 and Eqs. 9.11 , 9.13 , and 9.9 if lu r 35 P feAr In this case. 13 12 0.3 x 18 29 compared to 35 459 4 x 324 59, indicating that the current analysis and design are satisfactory. A second check is needed to protect against sidesway instability of the entire story under gravity loads. When Q is...

## A Stress in the Prestressed Steel at Flexural Failure

When a prestressed concrete beam fails in flexure, the prcstressing steel is at a stress fps that is higher than the effective prestress. p , but below the tensile strength fpir If the effective prestress J - Pe Aps is not less than 0.50fpu, ACI Code 18.7.2 permits use of certain approximate equations for fp . These equations appear quite complex as they are presented in the ACI Code, mainly because they are written in general form to account for differences in type of prcstressing steel and to...

## Figure 142

Fixed-end, uniformly loaded one-way slab. mn is calculated by the usual equations. For design purposes, mp would be taken equal to w , with typically equal to 0.90, since is well below max for most slabs. For a statically determinate slab like that in Fig. 14.1, the formation of one yield line results in collapse. A mechanism forms, i.e the segments of the slab between the hinge and the supports are able to move without an increase in load. Indeterminate structures, however, can usually sustain...

## Jl

A panel of the vertical wall between two counterforts is a slab acted upon by horizontal earth pressure and supported along three sides, i.e., at the two counterforts and the base slab, while the fourth side, the top edge, is not supported. The earth pressure increases with distance from the free surface. The determination of moments and shears in such a slab supported on three sides and nonuniformly loaded is rather involved. It is customary in the design of such walls to disregard the support...

## Figure 1416

Yield fan geometry at concentrated load a yield fan b moment vectors acting on fan segment c resultant of positive-moment vectors d edge view of fan segment. Nilson-Darwin-Dolan I 14. Yield Line Analysis for I Text I I lt S gt The McGraw-Hill Design ol Concrete Slabs Companies. 2C04 Structures, Thirteenth Edition DESIGN OF CONCRETE STRUCTURES Chapter 14 If the positive resisting moment per unit length is m and the negative resisting moment m', the moments per unit length acting along the edges...

## Control of Deflections

In addition to limitations on cracking, dcscribcd in the preceding sections, it is usually necessary to impose certain controls on deflections of beams to ensure serviceability. Excessive deflections can lead to cracking of supported walls and partitions, ill-fitting doors and windows, poor roof drainage, misalignment of sensitive machinery and equipment, or visually offensive sag. It is important, therefore, to maintain control of deflections, in one way or another, so that members designed...

## Strip Method For Slabs 517

Fig. 15.7c, to the maximum positive moment section is chosen as b. It follows that the maximum positive moment is Applying the same methods as used for the X direction shows that the negative support moment in the Y direction middle strips is It is easily confirmed that the moments in the Y direction edge strips are just one-eighth of those in the Y direction middle strip. With the above expressions, all of the moments in the slab can be found once a suitable value for is chosen. From Kq. 15.12...

## 12 X 575

For which 0.0033, and the required positive steel area per strip is As 0.0033 x 12 x 5.75 0.23 in2 ft to be provided by No. 4 No. 13 bars on 10 in. centers. Note that slight adjustments downward and upward have been made in the steel required at negative and positive bending sections, as permitted by ACI Code 8.4, to arrive at practical bar spacings. Note also that all bar I Nilson-Darwin-Dolan 15. Strip Method for Slabs I Text I I The McGraw-Hill Design of Concrete Companies. 2C04 520 design...

## Slender Columns 299

In accordance with ACI Code 10.11.1, the section properties of the frame members used to calculate Q must take into account the effects of axial loads, cracked regions along the length of the member, and the duration of the loads. Alternately, the section properties may be represented using the modulus of elasticity Ec given in Eq. 2.3 and the following section properties Flat plates and flat slabs 0.251 g The moments of inertia must be divided by 1 d when sustained lateral loads act or for...

## Minimum beam depths for compression reinforcement to yield

Maximum for ' 2.5 in Maximum for ' 2.5 in., 40.000 0.23 10.8 0.20 12.3 60.000 0.13 18.8 0.12 21.5 75.000 0.06 42.7 0.05 48.8 This is a quadratic equation in c, the only unknown, and is easily solved for c. The nominal flexural strength is found using the value of fs from Eq. 3.55 , and a ,c in the expression Mn 0.85fcab d - y AJS d - d 3.57 This nominal capacity is reduced by the strength reduction factor to obtain the design strength. If compression bars arc used in a flexural member,...

## Figure P89

P8.8 must be designed for a factored axial load of 130 kips. Material strengths are f'c - 4000 psi and J - 60,000 psi. a Select the longitudinal and transverse reinforcement for an eccentricity ey 2.7 in. b Select the longitudinal and transverse reinforcement for the same axial load with ex ey 2.7 in. c Construct the strength interaction diagram and design strength curves for the column designed in part b , given that the column will be subjected to biaxial...

## Partial Prestressing

Early in the development of prestressed concrete, the goal of prestressing was the complete elimination of concrete tensile stress at service load. This kind of design, in which the service load tensile stress limit fls 0. is often referred to as full prestressing. While full prestressing offers many advantages over nonprestressed construction, some problems can arise. Heavily prestressed beams, particularly those for which full live load is seldom in place, may have excessively large upward...

## J J J1

Thus, according to the Bresler method, the design load of Pu 0.65 x 410 267 kips can be applied safely. b By the load contour method, for K axis bending with Pu Ag 255 0.65 x 4 X 240 0.41. The average from Graphs A.6 and A.7 of Appendix A is Hence. M 0.224 x 4 x 240 x 20 4300 in-kips. Then for X axis bending, with Pu f'c 0.41, as before, from Graph A.5, Nilson-Darwin-Dolan I 8. Short Columns I Text I I The McGraw-Hill Design ol Concrete Compares, 2C04 Structures,...

## Figure 814

AO safety provisions superimposed on column strength interaction diagram. , fy- Es at the balanced condition. The effect of the transition in lower right end of the design strength curve. The design of eccentrically loaded columns using the strain compatibility method of analysis described requires that a trial column be selected. The trial column is then investigated to determine if it is adequate to carry any combination of Pu and Mu that may act on it should the structure be overloaded,...

## Figure 417

Comparison of equations for V, for members subject to axial loads. where Ag is the gross area of the concrete and Nu-A amp is expressed in psi units. As an alternative to the rather complicated determination of V . using Eqs. 4.12 , 4.16 , and 4.17 . ACI Code 11.3.1.2 permits the use of an alternative simplified expression Figure 4.17 shows a comparison of Vc calculated by the more complex and simplified expressions for beams with compression load. Equation 4.18 is seen to be generally quite...

## Figure 1114

Failure due to lack of support for diagonal compression in beam-girder joint. Courtesy of M. I'. Collins, University of Toronto. 11.9 . These stirrups serve as tension ties to transmit the reaction of the beam to the compression zone of the girder, where it can he equilibrated by diagonal compression struts in the girder. The hanger stirrups, which are required in addition to the normal girder stirrups required lor shear, can be designed based on equilibrating part or all of Nilson-Darwin-Dolan...

## F Beamless Flat Slabs with Drop Panels or Column Capitals

By suitably proportioning and reinforcing the slab, it is possible to eliminate supporting beams altogether. The slab is supported directly on the columns. In a rectangular or square region centered on the columns, the slab may be thickened and the column tops flared, as shown in Fig. 18.8. The thickened slab is termed a drop panel and the column flare is referred to as a column capital. Both of these serve a double purpose they increase the shear strength of the floor system in the critical...

## Deflections Due to Long Term Loads

Initial deflections are increased significantly if loads are sustained over a long period of time, due to the effects of shrinkage and creep. These two effects are usually combined in deflection calculations. Creep generally dominates, but for some types of members, shrinkage deflections are large and should be considered separately see Section 6.8 . It was pointed out in Section 2.8 that creep deformations of concrete are directly proportional to the compressive stress up to and beyond the...

## Figure 1335

Flat slab span with variable moment of inertia. direction see Fig. 13.35 . It is suggested in Rcf. 13.29 that a weighted average moment of inertia be used in such cases Iin. 2l-flr 2ljlll l-fl, 13.22 where I moment of inertia of slab including both drop panel and capital I l - moment of inertia of slab with drop panel only , moment of inertia of slab alone Span distances are defined in Fig. 13.35. Next it is necessary to correct for the rotations of the equivalent frame at the supports, which...

## Example 142

If a slab is reinforced in orthogonal directions so that the resisting moment is the same in these two directions, the moment capacity of the slab will be the same along any other line, regardless of direction. Such a slab is said to be isotropicaily reinforced. If, however, the strengths are different in two perpendicular directions, the slab is called orthogonally anisotropic, or simply orthotropic. Only isotropic slabs will be discussed in this section. Orthotropic reinforcement, which is...