Design of Reinforced Concrete

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Design Of Reinforced Concrete

ACI 318-11 Code Edition

Anas G. Dawas

Hashemite University

ANAS

Preface

 This textbook presents an introduction to reinforced concrete design. I hope the material is written in such a manner as to interest students in the subject and to encourage them to continue its study in the years to come.

 This textbook covers the following topics :

 Design One Way Ribbed Slab  Design Two Way Slabs  Serviceability  Design For Torsion  Design Footings  Design Columns

 Sample Exams :  Examples of sample exams are included for most topics in the text. Problems in the back of each chapter are also suitable for exam questions

 About the Author  I am currently a third year student in the CIVIL ENGINEERING at Hashemite University

2 كلمة بسى هللا انشدًٍ انشدٍى خٍش كالو أبذأ فٍّ سسانخً , ٔانذًذ هللا انزي ٔفق ٔقذس نُا اكًال ْزا انًٕضٕع انًخٕاضع نعهّ ٌسٓم انكثٍش عهى انضيالء األفاضم ٔأسأل هللا أٌ ٌكٌٕ عهًا َافعا ٔنٕجّٓ خانصا .

يا أٔد انخأكٍذ عهٍّ ْٕ أٌ ْزا انعًم ْٕ يٍ صُع بشش ٌخطئ ٌٔصٍب , ٔأٌ انًشاجع راحٓا انخً اعخًذث عهٍٓا فً جًع انًٕضٕع فٍٓا اخطاء دسابٍت يخخهفت ْٔزِ عهى أٌذي كباس انعهًاء انغشبٍٍٍ , نزنك أحًُى يٍ صيالئً أٌ ٌخذشٔا انًعهٕيت أٌا كاَج فانخطأ ٔاسد يًٓا كاَج دسجت انخشكٍض ٔأٌ ٌسايذًَٕ اٌ أخطأث فً أيش يا ٔأٌ ٌكَٕٕا عَٕا نً عهى حذقٍق ْزا انطشح قذس االيكاٌ .

أخٍشاٌ , أحًُى يٍ هللا أٌ ٌغفش نضيالئً انزٌٍ ٔافخٓى انًٍُت خالل دٍاحً انجايعٍت ٔانزٌٍ كإَا بٍُُا ٌطًذٌٕ نًا َطًخ انٍّ َذٍ اٌَ ٌٔشٌٔ فً أَفسٓى يسخقبال يششقاً نكٍ هلل عاقبت األيٕس , ٔأحًُى يٍ كم شخص اَخفع بٓزا انًٕضٕع أٌ ٌذعٕا نٓى ٔنجًٍع انًسهًٍٍ بانًغفشة ٔانشدًت , فُذٍ صائشٌٔ نًا صاسٔا

ٔفقكى هللا أدبخً

ONE WAY JOIST SLAB

Figure 1 Types of slabs

One-Way Joist Floor System :

Long-span floors for relatively light live loads can be constructed as a series of closely spaced, cast-in-place T-beams (or joists) with a cross section as shown in Fig. The joists span one way between beams. Most often, removable metal forms referred to as fillers or pans are used to form the joists. Occasionally, joist floors are built by using clay-tile fillers, which serve as forms for the concrete in the ribs that are left in place to serve as the ceiling (ACI Code Section 8.13.5).

When the dimensions of the joists conform to ACI Code Sections 8.13.1 to 8.13.3, they are eligible for less cover to the reinforcement than for beams (ACI Code Section 7.7.2(c)) and for a 10 percent increase in the shear, carried by the concrete (ACI

5 Code Section 8.13.8). The principal requirements are that the floor be a monolithic combination of regularly spaced ribs and a top slab with 1. ribs not less than 100 mm. in width, 2. depth of ribs not more than 3.5 times the minimum web width, and 3. clear spacing between ribs not greater than 750 mm.

without filler  40mm   50mm h f    with filler   1  S  12

Ribbed slabs not meeting these requirements are designed as slabs and beams.

Although not required by the ACI Code, load-distributing ribs perpendicular to the joists are provided at the midspan or at the third points of long spans. These have at least one continuous No. 4 (13 mm) bar at the top and the bottom. The CRSI Handbook [10-4] suggests no load-distributing ribs in spans of up to 6 m, one at midspan for spans of 6 to 9 m , and two at the third points for spans over 9 m .

For joist floors meeting the requirements of ACI Code Section 8.3.3, the ACI moment and shear coefficients can be used in design, taking (Ln) as the clear span of the joists themselves. For uneven spans, it is necessary to analyze the floor. The negative moments in the ends of the joists will be underestimated if this is not done.

Reinforcing Details

9.5.2.1 — Minimum thickness stipulated in Table 9.5(a) shall apply for one-way construction not supporting or attached to partitions or other construction likely to be damaged by large deflections, unless computation of deflection indicates a lesser thickness can be used without adverse effects.

7 gbaiJ tgioJingiseD

Figure 2 Design Example 0.52 m

Design Example :

Design the joist solid slab for the floor system shown below .

2 Live load = 2KN/m , fc=25 Mpa and fy=420 Mpa. Cross section is shown in fig.2 above .

1. Calculating minimum depth according to ACI code (Table 9.5a) l h  n (one end continuous) min 18.5

4400  400   216.216mm  use typical slab thickness  250mm 18.5

2. Calculating ultimate load : a) Dead load for 1 m2 :

Slab weight = 0.07×1×25 1.75 KN/m2 Rib weight =0.14×0.18×25/0.52 1.2 KN/m2 Block weight =5×0.18/0.52 1.73 KN/m2 Plastering 0.025×22 0.55 KN/m2 Mortar 0.025×22 0.55 KN/m2 Sand fill 0.1×13 1.3 KN/m2 Tiles 0.025×22 0.55 KN/m2 Total Dead Load 7.63 KN/m2

2 b) Ultimate load = wu=1.2(7.63)+1.6(2)=12.35 KN/m

9 Notes :

 Ultimate load on slab = 12 .35 KN/m2 but the Ultimate load on Joist = 12.35×0.52=6.422 KN/m .  Usually the number of blocks per meter in joist slab equal to 5 blocks with thickness equal to 20 cm and weight =0.18 KN per one .

 Typical thickness depend on the depth of the blocks , for example , for 24 cm blocks ,the typical thickness equal to 240+70=310 mm

 Ln used to calculate minimum thickness for joist slab is in the direction of the joist .

Primary Beam

3. Calculating shear and moment subjected to the Joist :

6.42 42 M Ve   4.28 KN.m 24 For negative moment the rib will be 2 Ve 6.42 4 designed as rectangular beam and M   7.33 KN.m 14 for positive moment will be 2 6.42 4 designed as T-beam M Ve  11.41 KN.m 9

Effective depth =250-20-10-10/2=215 mm

11 A. Design for positive moment :

bw 120mm ,bf  520mm d  215mm

7.331000 2 100.5420 As   95 mm  a   3.62mm 0.904200.95215 0.8525520 7.331000 2 As   91mm 3.61 0.90420(215 ) 2 1.4bw d 1.4(120)(215) 2    86mm   fy 420 As(min)    use 210 0.25 fc bw d   mm 2  76  fy

B. Design for negative moment :

12 11.441000 157420 As  148.2 mm2  a   24.4 mm 0.904200.95215 0.8525120 11.441000 As  149.2 mm2 25.8 0.90420(215  ) 2 1.4bwd 1.4(120)(215)   86mm2   f  y 420 A    use 210 s(min) . f b d 0 25 c w  mm2  76  fy 4. Shear Design: W l V 1.15 ( u n ) 14.76 KN u 2

fc 25 V 1.1 b d 1.1 0.75 120 215 17.73KN c 6 w 6 Vc  Vu  OK 5. Solid Slab Design : we can design this part as fixe fixebeam but also we can design it as simple beam. w l 2 12.34 0.42 M  u n  u 8 8  0.2468 KN.m 0.24681000 40cm A  s 0.9 420 0.95 45 15mm2 2 As(min)  0.0018(1000)(70) 126mm Use meshor 10 / block

14 Problems:

1-1 The one way joist slab shown below consists of 011 mm wide joists spaced at 510 mm , the web width of the spandrel beams is 400 mm and 500 for interior beams . Column dimension (400×400 mm). A B C D

 The minimum required depth of the slab upon the ACI code requirement is most nearly …………  Calculate the Ultimate load /rib assuming that the Live load = 3 KN/m2,  25 mm Covering material with unit weight is 20 KN/m3 .  blocks are 40×25×17 cm in dimension , each 17 Kg in weight . Use

fc =30 Mpa , 3  fy=420 Mpa and 25 mm plaster with unit weight = 21 KN/m .

 The maximum ultimate negative moment in the joist ………  The maximum ultimate shear in the joist ………  The ultimate distributed load /m on the beam between column A and B is most nearly ……….  The ultimate load carried by column A is most nearly……..  For the overhang slab , The minimum required depth of the slab upon the ACI code requirement is…………  The minimum required depth of the exterior beam upon the ACI code requirement is …….  The maximum ultimate negative moment in the overhang joist is most nearly ………….  The maximum ultimate positive moment in the joist is most nearly ……….  The largest ultimate load carried by the intermediate column is most ………  The minimum area of steel required to resist the maximum ultimate positive moment in beam between column C2 and D2 upon ACI code is most nearly (assume the depth of the beam is 600 mm ) …………  Show by suitable sketch the reinforcements required to resist the maximum ultimate positive moment for any interior joist.  The minimum area of steel must be provided to solid slab part upon ACI code requirements is most nearly …….. ANAS DAWAS All Rights Reserved to Icivil-Hu uploaded by icivil-hu.com

1-2 For Simple supported square slab shown below is apart of a floor in a typical 2 residential building and the live load = 2 KN/m ,fc=28 Mpa and fy=420 Mpa .use 12mm bars

 The minimum required depth of the slab upon the ACI code requirement is most nearly………..  Calculate the Ultimate load /rib based on the typical section shown above…….  Draw shear and moment diagram for the simple supported Joist ……  Calculate the shear strength capacity for the joist upon ACI code………  Design the reinforcement in the joist to carry the positive moment, show suitable sketch ………..  The minimum required depth of the primary beam upon the ACI code requirement is most nearly………..

17  Draw shear and moment diagram for the primary beam based on the minimum depth calculated above ………

1-3 The one way joist slab shown below consists of 150 mm wide joists spaced at 550 mm , the web width of the spandrel beams is 400 mm and 500 for interior beams . Column dimension (400×400 mm).

7m 4m

Assume that the dead load and live load on the slab is 7KN/m2 and 3 KN/m2:

18  The minimum required depth of the slab upon the ACI code requirement is most nearly …………

 The maximum ultimate negative moment in the joist ………  The maximum ultimate shear in the joist ………  The ultimate distributed load /m on the beam between column A3 and B3 is most nearly ………. ( neglect self weight of beam )

 The minimum required depth of the exterior beam upon the ACI code requirement is …….

 The maximum ultimate positive moment in the joist is most nearly ……….  Show by suitable sketch the reinforcements required to resist the maximum ultimate positive moment for any interior joist.  The minimum area of steel must be provided to solid slab part upon ACI code requirements is most nearly ……..  Determine the design ultimate bending moment , ultimate shear force at the most critical section for each for the simple supported joist ………..

TWO WAY SLAB

Two Way Slabs

Behavior , Analysis and Design

Two-way slabs are a form of construction unique to reinforced concrete among the major structural materials. It is an efficient, economical, and widely used structural system. In practice, two-way slabs take various forms. For relatively light loads, as experienced in

apartments or similar buildings, flat plates are used. such a plate is sim ply a slab of uniform thickness supported on columns. In an apartment building, the top of the slab would be carpeted, and the bottom of the slab would be finished as the ceiling for the story below. Flat plates are most economical for spans from 4.5 to 6 m .

21 In general, slabs are classified as being one way or two way. Slabs that primarily deflect in one direction are referred to as one way slab When slabs are supported by columns arranged generally in rows so that the slabs can deflect in two directions, they are usually referred to as Two way slabs .

22 Two-way slabs can be strengthened by the addition of beams between the columns, by thickening the slabs around the columns (drop panels), and by flaring the columns under the slabs (column capitals). These situations are shown in Figure 16.1 and discussed in the next several paragraphs.

Flat plates present a possible problem in transferring the shear at the perimeter of the columns. In other words, there is a danger that the columns may punch through the slabs. As a result, it is frequently necessary to increase column sizes or slab thicknesses or to use Shearheads .

Analysis of Two-Way Slabs:

A theoretical elastic analysis for such slabs is a very complex problem because of their highly indeterminate nature. Numerical techniques such as finite difference and finite elements are required, but such methods require sophisticated software to be practical in design. The methods described in this chapter can be done by hand or with simple spreadsheets, and are sufficiently accurate for most design problems.

the design of two-way slabs is generally based on empirical moment coefficients, which, although they might not accurately predict stress variations, result in slabs with satisfactory overall safety factors. In other words, if too much reinforcing is placed in one part of a slab and too little somewhere else, the resulting slab behavior will probably still be satisfactory. The total amount of reinforcement in a slab seems more important than its exact placement.

23 DESIGN OF SLABS: Two procedures for the flexural analysis and design of two-way floor systems are presented in detail in the ACI Code:  Direct-design method—considered in the following section: The calculation of moments in the direct-design method is based on the total statical moment  Equivalent-frame design method: Here, the slab is divided into a series of two-dimensional frames (in each direction), and the positive and negative moments are computed via an elastic-frame analysis.

THE DIRECT-DESIGN METHOD

Limitations on the Use of the Direct-Design Method:

The direct design method was developed from theoretical procedures for determination of moments in slabs, requirements for simple design, construction procedures, and performance of existing slabs. Therefore, the slab system, to be designed using the direct design method, should conform to the following limitations as given by ACI Code 13.6.1:

1. There must be three or more spans in each direction. 2. Slab panels must be rectangular with a ratio of longer to shorter span, center-to-center of supports, not greater than 2.0. 3. Successive span lengths, center-to-center of supports, in each direction must not differ by more than one-third of the longer span. 4. Columns must not be offset more than 10 % of the span in the direction of offset from either axis between centerlines of successive columns. 5. Loads must be due to gravity only and uniformly distributed over the entire panel. The live load must not exceed 2 times the dead load. 6. For a panel with beams between supports on all sides, the relative stiffness of beams in two perpendicular directions is not less than 0.20 and not greater than 5.0.

2  f 1l2 0.20   5 2  f 2l1

24  f  ratio of flexural stiffness of beam section to flexural stiffness of a width of slab bounded laterally by centerlines of adjacent panels , if any, on each side of the beam

Ecb Ib  f  Ecs Is

Ib  moment of inertia about centroidal axis of gross section of beam

Is  moment of inertia about centroidal axis of gross section of slab      in direction of l  f 1 f 3 f 1 f 1 2      in direction of l  f 2 f 4 f 2 f 2 2

25 Design Procedure:

1) Determination of the total factored static moment: Total factored static moment for a span is determined in a strip bounded laterally by centerline of panel on each side of centerline of supports, as shown in Figure.

Absolute sum of positive and average negative factored moments in each direction is not to be less than :

wu  factored load per unit area w l l 2 M  u 2 n where l  transverse width of the strip o 8 2 ln  clear span between columns

2) Distribution of the total factored static moment to negative and positive moments:

3) Distribution of the positive and negative factored moments to the column and middle strips: - Column strips : Column strip is a design strip with a width on each side of a column centerline equal to 0.25 l1 or 2 0.25 l2 whichever is less - Middle strip : Middle strip is a design strip bounded by two column strips .

27  Factored moments in column strips: - According to ACI Code 13.6.4, column strip moments, as percentages of total factored positive and negative moments

Aspect Ratio L2/L1  f 1l2 / l1 0.5 1.0 2.0 Negative moment at interior support 0 75 75 75 ≥1.0 90 75 45 positive moment near mid span 0 60 60 60 ≥1.0 90 75 45

Aspect Ratio L2/L1  f 1l2 / l1 t 0.50 1.0 2.0 Negative moment at exterior support 0 0 100 100 100 ≥2.5 75 75 75 ≥ 1.0 0 100 100 100 ≥2.5 90 75 45 Positive moment near mid span 0 60 60 60 ≥ 1.0 90 75 45 Negative moment at interior 0 75 75 75 ≥1.0 90 75 45

t  is defined as ratio of torsional stiffness of edge beam section

to flexural stiffness of a width of slab equal to span length of beam, centerto -

EcbCmax center of supports  t  2Ecs Is

 x  x3 y is the smallest dimension , y is the longest one  C   1 0.63   where: x  y  3 

 Factored moments in beams caused by slab loads

Depth Limitations and Stiffness Requirements

It is obviously very important to keep the various panels of a two-way slab relatively level (i.e., with reasonably small deflections). Thin reinforced two-way slabs have quite a bit of moment resistance, but deflections are often large. As a consequence, their depths are very carefully controlled by the ACI Code so as to limit these deflections. This is accomplished by requiring the designer to either (a) compute deflections and make sure they are within certain limitations or (b) use certain minimum thicknesses as specified in Section 9.5.3 of the code. Deflection computations for two-way slabs are rather complicated, so the average designer usually uses the minimum ACI thickness values, presented in the next few paragraphs of this chapter.

29 1) Slabs without Interior Beams For a slab without interior beams spanning between its supports and with a ratio of its long span to short span not greater than 2.0, the minimum thickness can be taken from [Table 9.5(c) in the code]. The values selected from the table, however, must not be less than the following values (ACI 9.5.3.2):  Slabs without drop panels 125 mm .  Thickness of those slabs with drop panels outside the panels 100 mm

Very often slabs are built without interior beams between the columns but with edge beams running around the perimeter of the building. These beams are very helpful in stiffening the slabs and reducing the deflections in the exterior slab

panels. The stiffness of slabs with edge beams is expressed as a function of αf , which follows.

2) Slabs with Interior Beams

To determine the minimum thickness of slabs with beams spanning between their supports on all sides, Section 9.5.3.3 of the code must be followed. Involved in the expressions presented there are span lengths, panel shapes, flexural stiffness of beams if they are used, steel yield stresses, and so on. In these equations, the following terms are used:

 ln  the clear span in the long direction, measured face to face, of (a) columns for slabs without beams and (b) beams for slab with beams    the ratio of the long to the short clear span

fm  the average value of the ratios of beam- to- slab stiffness on all sides of a panel

 For fm  0.2 , the minimum thicknesses are obtained as they were for slabs without interior beams spanning between their supports.(table 16.1 )

 For 2  fm  0.20 ,

f l ( .  y ) n 0 8 h  1400 125mm 365 (fm  0.20)

For fm  2.0,

fy ln (0.8  ) h  1400 90mm 36 9

31 SHEAR STRENGTH OF TWO-WAY SLABS A shear failure in a beam results from an inclined crack caused by flexural and shearing stresses. This crack starts at the tensile face of the beam and extends diagonally to the compression zone . One-way shear or beam-action shear (Fig. 4) involves an inclined crack extending across the entire width of the structure. Two-way shear or punching shear involves a truncated cone or pyramid-shaped surface around the column, as shown schematically in Fig. 3.

Figure 4 Beam shear / one way Figure 3 Two way shear / punching shear

Once a punching-shear failure has occurred at a slab–column joint, the shear capacity of that particular joint is almost completely lost. In the case of a two-way slab, as the slab slides down, the column load is transferred to adjacent column- slab connections, thereby possibly overloading them and causing them to fail. Thus, although a two-way slab possesses great ductility if it fails in flexure, it has very little ductility if it fails in shear

One Way Shear : The critical section for one way shear is located at “d” from the face of the support or at “d” from the face of the drop panel or other change in thickness . The shear strength on the critical section is computed as for beams . 1 V  f b d  should be  V at "d" c 6 c w u Two Way Shear:

Location of the Critical Perimeter: Two-way shear is assumed to be critical on a vertical section through the slab or footing extending around the column. According to ACI Code Section 11.11.1.2, this section is

chosen so that it is never less than d/2 from the face of the column and so that its length ,bo , is a minimum.

34 Tributary Areas for Shear in Two-Way Slabs

In most slab Vs  0 Vc is taken as the smallest of (a),(b) and (c) where :   ratio of long side of column to the short side c   f b d 4 c o bo  is the perimeter of the critical section (a) Vc  2    c  12 40 for interior column   d  f b d  (b) V  2  s  c o s  30 for edge column c  b  12   o  20 for corner column 1 (c) V  f b d c 3 c o

Shear Reinforcement for Two-Way Slabs

elemasa1

Design the two-way flat plate with no edge (spandrel) beams, shown in Figure , given the following: interior columns are 40 cm × 40 cm, exterior columns are 30 cm × 30 cm, covering materials weigh 2.29 KN/m2 2 and the live load is 3 KN/m . Use fc= 28 Mpa and Fy= 420 Mpa .

37 Solution: 1) Evaluate slab thickness

For corner panel Ln=7-0.3=6.7 m

For Edge panel Ln=7-0.15-0.2=6.65 m

For interior panel Ln=7-0.40=6.6 m For flat plates with no edge beams, minimum slab thickness = Ln/30 = 6700/30 = 223.33 mm Take it as 30 cm .

2) Check limitations for slab analysis by the direct design method: The first five conditions are satisfied, while the sixth condition does not apply due to the nonexistence of beams.

3) Calculate the factored load on the slab:

Wu= 1.2 ( 0.3×25+2.29)+1.6(3)=16.55

4) Check slab thickness for shear: a) Interior column :

davg = 300 – 20 -20 (use ø20 bar ) = 26 cm

- Punching shear

bo  4(260 400) 2640mm 2 Vu 16.55((67)0.660 ) 687.89KN  1.0  4 282640260 2  1816KN  1  12  40260 282640260 Vc  smaller of   1192.35KN   2640  12 1  2826402601210.7KN 3

Vc 1192.350.75  894.3 KN Vc Vu OK

- Beam shear For section 1-1 : 7000 400 x    260  3040 mm 2 2 Vu 16.55(3.040 6)  301.871 KN 1 V  0.75 28  6000 260 1031.8 KN  OK c 6 For section 2-2: 6000 400 y    260  2540mm 2 2 Vu 16.55 (7 2.54)  294.3 KN

1 V  0.75 28  7000 260 1203 KN c 6 OK

b) Edge column :

Figure 5 Edge column in the long direction

Figure 6 Edge column the short direction

You have to check the edge column as we did for interior column , by yourself.

c) Corner column Select tributary area as shown below then check beam and punching shear .

When you select thickness for slab you have to check the shear strength , and you can increase thickness when “Not Ok”

I have assumed that the shear strength for edge and corner column is ok so h=30 cm

5) Calculate the factored static moment:

41  Strip in shorter direction

Width of intermediate strip = 7.0 m Width of the column strip is ( 6/4 +6/4)=3.0 m >>> Width of middle=7-3=4m ( 2m per half) Clear span for exterior panels = 600 – 15 – 20 = 565 cm Clear span for interior panels = 600 – 20 – 20 = 560 cm The larger of the two values will be used in moment calculations, or you can calculate static moment for each panel separately . 16.5575.652 M   463 KN.m o 8 6) Distribute the total factored static moment into positive and negative moments:

 For the interior panel : positive moment  0.35 Mo  0.35 463 162.05 KN.m

negative moment  0.65  463  300.95 KN.m  For exterior panel : positve moment  0.52  463  240.76 KN.m ANAS D AWAS negative moment on the exterior support  0.26  463  120.38 KN.m All negative Rights moment on Reserved the interior support to 0.70 Icivil-Hu463  324.1 KN.m uploaded by icivil-hu.com

300.95 120.38 324.1 300.95 324.1

162.05 240.76 240.76

Figure 7 Total Static moment

243

144 97

Figure 8 Column strip moment

96 64.8

Figure 9 Middle Strip moment

7) Distribute the positive and negative moments to the column and middle strips:

Slab Moment End span Interior span Exterior Positive Interior Positive Negative negative negative Total moment 120.38 240.76 324.1 162.05 300.95

Column 120.38 0.6×240.76= 0.75×324.1= 0.6×162.05= 0.75×300.95= moment 144.45 243.075 97.23 225.7 Middle strip 0 240.76-144.45= 324.1-243.075= 162.05- 300.95-225.7= moment 96.31 81.025 97.23=64.82 75.25

8) Design the reinforcement: Column strip reinforcement and Half middle strip reinforcement: Design sections at maximum positive and negative moments as rectangular section :

Column Strip Half Middle Strip Negative Positive Negative Positive Moment(KN.m) 243 144 81/2= 40.5 96/2= 48 b (mm) 3000 3000 2000 2000 d (mm) 260 260 260 260 h (mm) 300 300 300 300 Fy ( Mpa) 420 420 420 420 Fc (Mpa) 28 28 28 28 As (mm2) As-min (mm2) 0.0018bh 0.0018bh 0.0018bh 0.0018bh Bar Dim Ø12 Ø12 Ø12 Ø12 No of bar As Provided

øMn (KN.m)

Sample Of Calculation To Design Column Strip :

For Positive Moment

44 1441000 A   1542.3mm2 s 0.9 420 0.95 260 1542.3 420 a   9.1mm 0.85 28 3000 1441000 A   1491.3mm2 s 9.1 0.9 420 (260  ) 2 2 As(min)  0.0018 (3000) (300)  1620mm (use this) 1620 No.Bar   14.3  use 1512 bars (1696mm2 )  (12)2 4  3000 (12)2 b  Ab 4 S    200mm  check Smax As 1696 a 1696 420 M   A f  (d  )  a   9.97mm n s y 2 0.85 28 3000 9.97 M  0.901696 420 (260  )  163.5  144 (OK ) n 2

Strip in the long direction Width of intermediate strip = 600 cm and width of column strip is the smaller of (L1 / 2) and (L2 / 2), taken as (600/2) = 300 cm. Total factored static moment: Clear span for exterior panels = 700 – 15 – 20 = 665 cm Clear span for interior panels = 700 – 20 – 20 = 660 cm The larger of the two values will be used in moment calculations. 16.55 6 6.652 M   549 KN.m o 8

You can design this strip by yourself now

elemasagE

For the two-way solid slab with beams on all column lines, shown in Figure , evaluate the moments acting on any of the internal beams, using the direct design method. All columns are 30 cm × 30 cm in cross section, all beams are 30 cm × 60 cm in cross section, slab thickness is equal to 14 cm, covering materials weigh 1.83 KN/m2 and the live load is 4 kN/m2. Use

Use fc= 28 Mpa and Fy= 420 Mpa .

1) Evaluate the slab thickness  For internal beams :

47 140 460 (1220140)( )  (300 460)( 140) y  2 2  204.1 mm (1220140)  (300 460) 3 (1220)(140) 140 2  Ib    (1220140)(204.1 )    12 2  3 (300)(460) 2  9 9 4   (300 460)(370  204.1)   3.35  6.2310  9.5810 mm  12 

For Edge Beam

460 (760)(140)(70)  (460)(300)( 140) y  2  239.4mm (760)(140)  (460)(300)

3 (760)(140) 2  Ib    (760140)(239.4  70)    12  3 (300)(460) 2  9 9 4   (300 460)(370  239.4)   (3.23  4.8)10  810 mm 12  

48  For Interior Slab :

3 (6000)(140) 9 2 I   1.37210 mm s 12

 For Exterior Slab: 3.0m (3000 150)(140)3 I s  12  0.72109 mm4

 Torsional Constant

4 C A  305347.3 cm 4 CB  403907.3 cm 4 Cmax  403907.3 cm 403907.3 t   1.472 (units in cm) 2(600)(14)3 /12

 Calculating α :

4 9850 cm  f 1   6.98 (For interior beams) 1372 cm4

8000    11.11 (For exterior beams) f 2 720 11 11  6.9  6.9    8.95 Figure 10 Alpha Values f m1 4 6.9  6.9  6.9  6.9    6.9 f m2 4 all values  2.0

f y ln (0.8  )  h  1400 90mm 36  9 420 5.7 (0.8  ) 6. 0.30    1.0  h  1400  139.33 mm (140  139.33) Ok 6  0.30 36  9(1)

Generally , you have to calculate αm for each span then calculate thickness for each one to determine which values will control ……

2) Factored load on the slab 2 Wu=1.2 ( 0.14×25+1.83)+1.6(4)=12.8 KN/m 3) Check slab thickness for shear Note : No need to check two way shear for two way slabs with beam , and you should check one way shear for interior and exterior beams but in this example they are the same so we will check for the interior one. x

d  140  20 12  108 mm 6000 300 Width of critical section  x   108  2742 mm 2 2 Vu  12.8 (2.742 6)  210.5 KN 0.75 V  28  6000108  428.6 KN  V  Ok c 6 u

51 4) Total factored static moment

12.8 6 (5.7)2 M   312 KN.m o 8 l2 6  f 1  6.9( )  6.90 l1 6

t  1.472 (calculated befor)

Figure 11 Positive / Negative Moment

Figure 12 Column Strip Moment

Figure 14 Beam’s Moment

Figure 13 Middle Strip Moment

For t  0.0 100

For t  1.472  x

For t  2.5  75% by inerpolation  the ratio  85%

l2  f 1( )  1  Beams resist 85% of the column strip l1

Finally , you should design middle strip , column strip and beam between supports based on the maximum moment for each one .

Section across the strip : 1220 mm

6/4=1.5 m 6/4=1.5m

Middle Strip Column Strip Beam

Column Strip Middle Strip Beam Negative Positive Negative Positive Negative Positive Moment(KN.m) 139.8 133.35 54.9 44.49 139 113.34 b (mm) 1780 1780 3000 3000 300 300 d (mm) 108 108 108 108 108 544 h (mm) 140 140 140 140 600 600 Fy ( Mpa) 420 420 420 420 420 420 Fc (Mpa) 28 28 28 28 28 28 As (mm2) As-min (mm2) 0.0018bh 0.0018bh 0.0018bh 0.0018bh Bar Dim Ø12 Ø12 Ø12 Ø12 Ø12 Ø12 No of bar As Provided

øMn (KN.m)

Notes :

 Design the beam as a rectangle for negative moment and as T beam for a positive moment .  As (min) for a beam is not as slab ( column and middle strip )  You can design middle strip as a one part or you can design half middle strip based on half value of the moment .

55 gbeoag etas

Drop panels are thicker portions of the slab adjacent to the columns, as shown in Fig.

1. The minimum thickness of slab required to limit deflections may be reduced by 10 percent if the slab has drop panels conforming to ACI Code Section 13.2.5. The drop panel stiffens the slab in the region of highest moments and hence reduces the deflection.

56 2. A drop panel with dimensions conforming to ACI Code Section 13.2.5 can be used to reduce the amount of negative-moment reinforcement required over a column in a flat slab. By increasing the overall depth of the slab, the lever arm, jd, used in computing the area of steel is increased, resulting in less required reinforcement in this region.

3. A drop panel gives additional slab depth at the column, thereby increasing the area of the critical shear perimeter.

Example : Design a drop panel according to the ACI code requirements for the floor system shown below in the fig. Check two way shear . Interior columns are ( 500×300 mm) and exterior columns are (300×300 mm ), Live load = 2 KN/m2 and super imposed Dead load = 1.2 KN/m2.

Min imum slab thickness to control deflection 5.5m 6m No beams (  0.0) 0.3 0.5 l  5.5    5.1 m n1 2 2 4.5m l 5.1 min h  n   155 mm 33 33

ln2  6  0.5  5.5 m l 5.5 min h  n   153 mm 36 36 use h  160 mm

h The drop panel thickness  h  200mm 4

58 The half width is at least ( L/6) in length  6000/6 =1000 mm  4500/6 =750 mm This is minimum dimension for the panel for the interior column ,so you can increase this to 2500×1800 area with 200 thickness .

Calculating the ultimate load based on the slab’s thickness : 2 Wu= 1.2( 0.16×24+1.2)+1.6(2)=9.25 KN/m Calculating the ultimate load based on the drop panel’s thickness : 2 Wu= 1.2( 0.2×24+1.2)+1.6(2)=10.4 KN/m 2 1 2 Avg.Ultimate Load  W  (9.25)  (10.4)  9.63 KN / m u 3 3

Now you have to check two way shear :  For interior column :

d  200  20 12 168 mm

Vu  9.63 (5.75 4.5  0.468 0.668)  246.2 KN 500   1.67 , b  2 (468  668)  2272mm c 300 o

4 28  2272 168 (a) V  (2  )   739.8 KN c 1.67 12 1000 40168 28  2272 168 (b)Vc  (  2)   834.4 KN 2272 12 1000 1 168 (c)V   28  2272  673.248 KN c 3 1000 Vc  0.75 673.248  505 KN  Vu OK

60  The punching shear at drop panel d  160  20 12  128mm

Vu  9.63 (5.75 4.5  2.6281.928)  200.4 KN 500    1.67 c 300 bo  2 (2628 1928)  9112 mm

(a) Vc 

(b) Vc  1 128 (c) V   28 9112   2057 KN c 3 1000 Vc  0.75 2057  1543 KN Vu

By the same way , check the two way shear for edge and corner columns.

g MELBeRe

2-1 flat plate floor system with panel is 6.0 × 8.0 m (on centers of column ) is supported on 0.5 m interior columns and 0.40 m square exterior columns . Edge beams are not used

along the exterior floor edges . Use fc=28 Mpa , fy=420 Mpa .

(a) Using ACI code , determine the minimum slab thickness required for panels 1 and 3 based on deflection criterion . (b) Assume the floor is to support a service dead load ( including self-weight ) Of 10 KN/m2 and service live load of 4.0 KN/m2 . If slab thickness is 160 mm and ( d= 134 mm ) , then check punching shear for corner column . (c ) Use direct design method to calculate column and middle in the edge frame spanning 6.0 m in panel 2 . Use ultimate load of 15 KN/m2.

2-2 30 × 30 m a flat plate with no drop panel is shown below . (fy=420 Mpa) . Based on ACI code :

Edge beam with (311×211) mm

a) Determine the minimum permissible total thickness required for slabs in panel 3 . b) Determine the minimum permissible total thickness required for slabs in panel 2

c) The ACI code states that for slabs with beams between column along exterior

edges , the value αf for the edge beam shall not less than 0.8 . Determine αf if the slab is 200 mm in thickness and show if the provision is satisfied .

2-3 Consider the floor system shown below , columns 60×60 cm2 and beams as shown. Answer the following problems . Assume beams and slabs are monolithic

And fy = 420 Mpa .

a) Compute α for interior beam in the short direction b) Compute α for exterior beam in the short direction c) Compute α for exterior beam in the long direction d) Compute βt for the exterior beam . e) Use the direct design method to calculate column and middle strip moments at the shaded 2 edge frame spanning . Use the ultimate load of 15 KN/m . f) Check one way shear for the beam between columns 6 and 7 .

2-4 The concrete for the slab and the beam shown below was placed in one pour . ( all dimensions in mm ):

1) Compute αf for the beam . 2) Compute the torsional constant for the edge beam shown .

2-5 For the flat plate shown in the figure below ( edge beam are not used ). Answer the

following questions . The columns dimensions are (40×40 cm ) , fc=28 Mpa and

fy=420 Mpa .

6 m 5.5 m 6 m

6 m

5 m

5 m 41×41 cm

a) The minimum required depth of the slab based upon ACI code requirement Is most nearly:

2 If the ultimate load on the slab is Wu= 15 KN/m , slab thickness h= 15 cm and effective depth d = 12 cm :

b) The maximum one way shear Vu in the area around column A2 ………

c) The Ultimate punching shear Vu for the corner column W3. ………

d) The Ultimate punching shear Vu for the interior column A2. ………

For the shaded area ( column and middle strip) shown in the figure e) The Total static moment in the W-E direction (hatched frame ) is ………

If the total static moment in W-E direction is 300 KN.m : f) The positive moment in the frame ( hatched frame ) is : ……… g) The column strip positive moment in the frame (hatched frame ) is : ……… h) Design a drop panel for the interior column A2 ….. i) The ultimate punching shear Vu for the 2000 by 2000 mm interior drop panel ( use ø12 bars and 20 mm cover ) .

2-6 A flat plate floor system with panels 7.0 by 8.0 m (on centers of columns ) is supported on 0.6 square interior columns and 0.40 m square exterior columns , all interior

beams are 0.6 × o.6 m and exterior beams are 0.40 × 0.60 m , Use fc=21 Mpa

Fy=420 Mpa and answer the following question :

8.0m 8.0m

7.0m

a) For the exterior beam in the long direction , the span of the slab considered in α calculation is most nearly : :……….

68 b) For interior panel C if the average α is equal to 2.3 , then the minimum depth of the slab is most nearly : :………. c) For the corner panel (A) , if the floor system is assumed to be without beams , then the minimum depth of slab is most nearly : :………. d) If the ultimate slab load is 20 KN/m2, and effective depth d =190 mm , the one way shear in panel C is most nearly : :………. e) For the exterior frame spanning in the long direction (hatched frame ) , the clear span

Ln of the of the frame in panel A is most nearly : :………. f) For the exterior frame spanning in the long direction (hatched frame ) , the transvers

length L2 of the of the frame in panel A is most nearly: :………. g) If the total static moment in panel (A) for the exterior frame spanning in the long direction (hatched frame ) is 300 KN.m , the exterior negative moment in the frame is most nearly : :………. h) If the total static moment in panel (A) for the exterior frame spanning in the long direction (hatched frame ) is 300 KN.m , the exterior positive moment in the frame is most nearly : :………. i) Assume the floor has no beams and the ultimate load is 20 KN/m2 , if d=190 mm then

the ultimate punching shear Vu for an interior column is most nearly :………. j) Assume the floor has no beams and the ultimate load is 20 KN/m2 , if d=190 mm then

the ultimate punching shear Vu for an the corner column is most nearly :……….

Notes :

Serviceability

70 yaelJbaeDJsJnS

Today the structural design profession is concerned with a limit states philosophy. The term limit state is used to describe a condition at which a structure or some part of a structure ceases to perform its intended function. There are two categories of limit states: strength and serviceability.

Strength limit states are based on the safety or load-carrying capacity of structures and include buckling, fracture, fatigue, overturning, and so on. with the bending limit state of various members.

Serviceability limit states refer to the performance of structures under normal service loads and are concerned with the uses and/or occupancy of structures. Serviceability is measured by considering the magnitudes of deflections, cracks, and vibrations of structures, as well as by considering the amounts of surface deterioration of the concrete and corrosion of the reinforcing. You will note that these items may disrupt the use of structures but do not usually involve collapse.

CRACKING This section presents a few introductory comments concerning some of the several types of cracks that occur in reinforced concrete beams. The remainder of this chapter is concerned with the estimated widths of flexural cracks and recommended maximum spacing of flexural bars to control cracks.

Flexural cracks are vertical cracks that extend from the tension sides of beams up to the region of their neutral axes. Cracking starts when the tensile stress in the concrete reaches the tensile strength of the concrete at some point in the bar. When this occurs, the prism cracks.

Figure 15 types of cracks

72 Control of Flexural Cracks Although cracks cannot be eliminated, they can be limited to acceptable sizes by spreading out or distributing the reinforcement. In other words, smaller cracks will result if several small bars are used with moderate spacing rather than a few large ones with large spacing.

ACI Code Provisions Concerning Cracks In the ACI Code, Sections 10.6.3 and 10.6.4 require that flexural tensile reinforcement be well distributed within the zones of maximum tension so that the center-to-center spacing of the reinforcing closest to a tension surface is not greater than the value computed with the following expression: 280 280 S  380   2.5Cc 300   fs   fs  S  bar spacing in mm (center to center ) 2 f  f  service load bar stress in MPa s 3 y

Cc  Least distance from surface of the reinforcement to the tension face

73 elpmaxE

Is the spacing of the bars shown in the figures within the requirements of the ACI code from

the standpoint of the cracking ? If fy= 420 Mpa.

db 28.7 Cc  75   75   60.65mm 2 2 280 280 S  380 ( )  2.5(60.65)  300( ) 2 2  420  420 3 3

 228mm  300 mm Since the actual bar spacing of 75 mm is less than 228 mm, this spacing is acceptable.

74 Skin Reinforcement points above the reinforcement than they are at the level of the steel, as shown in Fig. 9-3b. To control the width of these cracks, ACI Code Section 10.6.7 requires the use of skin reinforcement that is distributed uniformly along both faces of the beam web for a distance of d/2 measured from the centroid of the longitudinal tension reinforcement toward the neutral axis.

If h ≥ 900 mm , Skin reinforcement shall be used

Askin=0.015bw S2

S2= smaller of ( d/6 or 300 mm )

Total Askin ≤ As/2

280 280 S  380   2.5Cc 300   fs   fs 

75 Deflections

Control of Deflections One of the best ways to reduce deflections is by increasing member depths—but designers are always under pressure to keep members as shallow as possible. (As you can see, shallower members mean thinner floors, and thinner floors mean buildings with less height, with consequent reductions in many costs, such as plumbing, wiring, elevators, outside materials on buildings, and so on.) Reinforced concrete specifications usually limit deflections by specifying certain minimum depths or maximum permissible computed deflections.

Minimum Thicknesses Table 9.5(a) of the ACI Code, provides a set of minimum thicknesses for beams and one-way slabs to be used, unless actual deflection calculations indicate that lesser thicknesses are permissible. These minimum thickness values, which were developed primarily on the basis of experience over many years, should be used only for beams and slabs that are not supporting or attached to partitions or other members likely to be damaged by deflections.

Maximum Deflections If the designer chooses not to meet the minimum thicknesses given in Table 9.5(a) , he or she must compute deflections. If this is done, the values determined may not exceed the values specified in Table 6.1, which is Table 9.5(b) of the ACI Code.

Camber The deflection of reinforced concrete members may also be controlled by cambering.

Common cases

Effective Moments of Inertia Regardless of the method used for calculating deflections, there is a problem in determining the moment of inertia to be used. The trouble lies in the amount of cracking that has occurred.

Figure 16 : As moment subjected to the beam increases , the moment of inertia decreases because of cracks .

 for M a  M cr Use I  I g

 M a  M cr Use I  Ie M  M   I  ( cr )3 I  1 ( cr )3 I e M g  M  cr a  a  f I r g  M cr  yt

 Ec  4700 fc

 fr  0.70 fc

M cr  cracking moment

I g  gross moment of inertia

fr  Modulus of rupture

yt  distance from centroid to the extreme tension fiber

M a  maximum moment in the member

at the loading stage for which the momentof inertia is being computed or at any previous loading stage

Continuous-Beam Deflections The following discussion considers a continuous T beam subjected to both positive and negative moments. As shown in Figure below , the effective moment of inertia used for calculating deflections varies a great deal throughout the member.

Figure 17 This formula to calculate deflection along continuous beam

81 Critical section shall be permitted to obtain deflection in ACI code :

 Mid span for simple supported beam  At support for cantilever  Critical positive and negative moment sections for continuous beam

ACI code :( Ig )is the moment of inertia of the gross concrete section neglecting area of tension steel .

Transformed Section

82 Elastic Theory For Flexure This theory helps us to calculate deflection with several assumptions must be taken in considered :

1) Plan sections remain plane after bending . 2) Linear stress – strain curves for steel and concrete . 3) Perfect bond between steel and concrete . 4) Concrete tension capacity is neglected .

 This theory cannot be used when concrete stress are higher than 0.6 fc .

elpmaxE

For the beam shown below , check if we can use Theory of elasticity or not at the given moments :

1- 28 KN.m 2- 113 KN.m

2 As=3000 mm , n=10

 Transform section (un-cracked ) 2 n 10 ,(n 1) As  (10 1)  3000  27000 mm  Determine the centroid 600 (600  300)( ) (27000)(500) 2 y   326.1mm t (600  300)  (27000)

83  Determine the moment of inertia

3 300  600 2  2 I g    (300  600) (326.1 300)   (27000) (500  326.1)   12   6.34 109 mm4

 Determine Cracking moment 2.8 6.34 109 M cr    65 KN.m 600  326.1 106  For M = 28 KN.m

M  28  M cr  65

28326.1 6 fc  9 10 1.44 MPa  0.45 fc 6.3410 28 (500  326.1) 6 f 10 10  7.64 MPa  0.5 f s 6.34109 y 28 (600  326.1) f  106 1.21 MPa  f t 6.34109 r

all checks are ok

 For M = 113 KN.m you have to determine the location of the N.A based on the transformed cracked section .

84  Ai yi  0.0 about N.A y (300  y )  10  3000  (500  y)  y  231.7 mm 2 300  231.73  I   30000(500  231.7)2  3.4 109 mm4 cr 3

113  231.7 6   fc  10  7.7 MPa  0.5 fc 3.4 109

113  (500  231.7) 6  fs 10  10  89.17  0.5 f y 3.4 109 all checks are ok  you can use ElasticTheory

Calculate The Deflection  Instantaneous Deflection : When a concrete beam is loaded, it undergoes a deflection referred to as an instantaneous deflection.

 Sustained Load Deflection Under sustained loads, concrete undergoes creep strains and the curvature of a cross section increases.

If compression steel is present, the increased compressive strains will cause an increase in stress in the compression reinforcement, thereby shifting some of the compressive force from the concrete to the compression steel. As a result, the compressive stress in the concrete decreases, resulting in reduced creep strains. A The greater ratio of compression steel   s , bd the greater reductionincreep.

The Total Long Time Deflection

 LT   L    D  t  SL

 L  Immediate live load deflection    Immediate dead load deflection   D   SL  Sustained live load deflection    Time factor for infinite duration of sustained load      t  Time factor for limited load duration 

   1 50 

elpmaxE : Simple Supported Beam

A simple supported beam with the cross section shown in the figure , has a span of 6 m

And supports un-factored dead load of 30 KN/m2 including its self-weight plus an un- factored live load of 20 KN/m2 .

fc  28 MPa

f y  420 MPa

DL  30 KN / m LL  20 KN / m 2 As  2000 mm

6.0 m

2000 mm2

1- Calculate the properties of the section a) Un-cracked section :

4006003  I   7.2109 mm2 g 12  y  600/ 2  300 mm t Ec  4700 28  25000 Mpa 200 Gpa n   8.0 25000Mpa  f  0.7 28  3.7 Mpa r 3.77.2 109 M cr    88.8 KN 300 106

88 b) Cracked section y2  400  8 (2000)(525  y)  y 168.8 mm 2

400168.83 I  8(2000)(525 168.8)2  2.67109 mm4 cr 3

Section Y before after crack I g I Cr M cr properties crack (Transformed)

At mid span 9 4 9 4 300 mm 168.8 mm 7.210 mm 2.67 10 mm 88.8 KN.M

2- Calculate ultimate moment at the critical section (mid span ) 30  62 M  135 KN.m  M  I  I DL 8 cr e

(30  20)  62 M   225 KN.m  M  I  I DLLL 8 cr e

3- Calculate effective moment of inertia 3 3 88.8  88.8  9 4  Ie (Based on M DL )    7.2  1     2.67  3.9510 mm  135    135   3 3 88.8  88.8  9 4  Ie (Based on M DLLL )    7.2  1     2.67  2.9510 mm  225    225  

4- Calculate instantaneous deflection : 5wl 4 5 30  60004 DL   9  5.126 mm 384Ec IeDL 384  25000  3.9510

89 5wl 4 5 50  60004  DLLL   9  11.44 mm 384Ec IeDL 384  25000  2.9510

 LL   DLLL   DL  11.44  5.126  6.314 mm

5- Calculate long term deflection 2    2 1 50  (0)

 LT  6.314 2  5.126  0.0 16.56 mm

6- Compare with ACI code limits : this just an example span 6000     16.7 mm  6.31mm OK iLL 360 360 span    so and so LT 480

7- Summary Load Mm  I (KN) (KN.m) mid ( mm) 9 4 10 mm DL 30 135 3.95 5.126

DL+LL 50 225 2.95 11.44 DL+SL 30+0 No sus.L ------

Duration    LL   DL t  SL  LT

5 YEARS 2 2 11.44- (2)(5.126)= ______16.566 5.126= 10.252 mm 6.314 mm

elpmaxE : Continuous Beam

3ø25

6ø25

9.0 m

Determine the long term deflection at the mid span of the continues T beam shown above , The member supports a dead load including its self-weight of 16 KN/m and live load of 14

KN/m , Fc=28 Mpa , Fy = 420 Mpa . The beam cross section is shown below . Assume that 30% of the live load is sustained .

Notes :

Firstly , you can check minimum thickness according to Table 9.5 (a) in ACI code Then , you can fill the following tables to get perfect solution :

1) You should calculate the properties for the sections at support and at mid span .

I Section Y before after crack g I Cr M cr properties crack (Transformed) At mid span At supports

2) Calculate the moment caused by DL , (DL+LL) at the two edges of the member and at the mid span .

M1 M2

3) Compare the moments from previous step with the cracking moment to determine which value of moment of inertia should be used . 4) Calculate Average value for moment of inertia using :

5) Calculate deflection using the following formula :

 In this formula , you should substitute moments value with its sign

93 You can use these tables :

Load M1 Mm M2 I1 Imid I2 Iavg  (KN) (KN.m) (KN.m) (KN.m) ( mm) DL DL+LL DL+SL

Duration          LL  DL t SL LT

Note : In case of negative moment , consider the section as a rectangular beam and all calculation based on this consideration as shown below :

PROBLEMS

6-1 A simple supported beam with cross section shown below has a span of 6.0 m . The beam supports un-factored dead load of 24 KN/m , including its own self-weight plus an un-factored live load of 16 KN/m . The concrete compressive strength is 21 Mpa .

 Compute the following : 1) Gross moment of inertia , cracking moment of inertia and cracking moment . 2) Immediate dead load deflection . 3) Immediate live load deflection . 4) Long term deflection at an age of 3 years , if 30% of live load is sustained .

6-2 A Cantilever beam with cross section shown below has a span of 5.0 m . The beam supports an un-factored concentrated live load of 60 KN and un-factored dead load of 15 KN/m . The concrete compressive strength is 28 Mpa and effective depth = 800 mm

6-3 A simple supported beam with cross section shown below has a span of 6.0 m . The beam supports un-factored dead load of 20 KN/m , including its own self-weight plus an un-factored concentrated live load of 40 KN. The concrete compressive strength is 28 Mpa.

6-4 For the cross sections shown below , determine whether the reinforcement satisfies the ACI code requirements for crack width control .  For section C design skein reinforcement Upon ACI code requirements .

97 A B

C D

6-5 For 3.6 m span cantilever beam shown in the figure below , (Ignore weight in your calculation ) : 1) Determine instantaneous live load deflection at the free end of the beam due to the load condition shown below . 2) Determine the Total long term deflection . Assume that only dead load is Sustained. What code deflection criteria it meets and what limitations, if any , have to be placed on its use ?

99 6-6 The beam cross section shown is on a 8000 mm simple span and carries uniformly

distributed service dead load of wD=15 kN/m and concentrated service live load of

PL=30 kN. Use fc’ = 28 MPa, fy = 420 MPa and assume n = 9 (Ignore beam self-weight). Compute : 1. Cracked moment of inertia, 2. Cracking Moment, 3. Immediate deflection due to live load only, 4. Ultimate long-term deflection due to dead load

6-7 Continues beam with cross section shown below has a span of 9.0 m. The beam supports un-factored dead load of 22 KN/m , including its own self weight , plus un- factored live load of 36 KN/m . The concrete compressive strength is 21 Mpa . Compute the following :

1) Gross moment of inertia for positive moment section . 2) Cracking moment for positive moment moment section . 3) Cracked moment of inertia for positive moment section

100

4) Using the following assumption , Answers the followings :

I Section Y before after crack g I Cr M cr properties crack (Transformed) At mid span - - 2×1010 mm4 8×109 mm4 106 KN.m At supports - - 3×1010 mm4 7.4×109 mm4 76 KN.m

101 A. Immediate dead load deflection B. Immediate live load deflection . C. Ultimate long term deflection of 30 % live load sustained .

102

Torsion

103 gnoeiJot

The average designer probably does not worry about torsion very much. He or she thinks almost exclusively of axial forces, shears, and bending moments, and yet most reinforced concrete structures are subject to some degree of torsion. Until recent years, the safety factors required by codes for the design of reinforced concrete members for shear, moment, and so forth were so large that the effects of torsion could be safely neglected in all but the most extreme cases. Today, however, overall safety factors are less than they used to be and members are smaller, with the result that torsion is a more common problem.

Appreciable torsion does occur in many structures, such as:

1) In the main girders of bridges, which are twisted by transverse beams or slabs. 2) In buildings where the edge of a floor slab and its beams are supported by a spandrel beam running between the exterior columns. 3) Earthquakes can cause dangerous torsional forces in all buildings. 4) In curved bridge girders, spiral stairways, and balcony girders

104

 It should be realized that if the supporting member is able to rotate, the resulting torsional stresses will be fairly small. If, however, the member is restrained, the torsional stresses can be quite large.

Torsion Cracks Should a plain concrete member be subjected to pure torsion, it will crack and fail along o 45 spiral lines because of the diagonal tension corresponding to the torsional stresses. For a very effective demonstration of this type of failure, you can take a piece of chalk in your hands and twist it until it breaks. Although the diagonal tension stresses produced by twisting are very similar to those caused by shear, they will occur on all faces of a member. As a result, they add to the stresses caused by shear on one side and subtract from them on the other.

105 Strength of Material Review

 In a bar with a rectangular cross section, however, the torsional stresses vary from a maximum at the middle of the long sides of the rectangle to zero at the corners.

Torsional Reinforcing Tests have shown that both longitudinal bars and closed stirrups (or spirals) are necessary to intercept the numerous diagonal tension cracks that occur on all surfaces of members subject to appreciable torsional forces.

106

 The normal -shaped stirrups are not satisfactory. They must be closed either by welding their ends together to form a continuous loop, as illustrated in Figure (a), or by bending their ends around a longitudinal bar, as shown in part (b) of the same figure.

107  ACI Code Section 11.5.4.3 requires that longitudinal reinforcement for torsion be developed at both ends of a beam. Because the maximum torsions generally act at the ends of a beam, it is generally necessary to anchor the longitudinal torsional reinforcement for its yield strength at the face of the support. This may require hooks or horizontal U-shaped bars lap spliced with the longitudinal torsion reinforcement.

108  The strength of closed stirrups cannot be developed unless additional longitudinal reinforcing is supplied. Longitudinal bars should be spaced uniformly around the insides of the stirrups, not more than 300 mm apart. There must be at least one bar in each corner of the stirrups to provide anchorage for the stirrup legs (Code 11.6.6.2); otherwise, if the concrete inside the corners were to be crushed, the stirrups would slip and the result would be even larger torsional cracks. These longitudinal bars must have diameters at least equal to 0.042 times the stirrup spacing. Their size may not be less than 10 mm .

DESIGN METHODS FOR TORSION  Skew bending theory developed: This theory assumes that some shear and torsion is resisted by the concrete, the rest by shear and torsion reinforcement.  Thin-walled tube/plastic space truss model Torsion is assumed to be resisted by shear flow, q , around the perimeter of the member as shown in Fig.. The beam is idealized as a thin-walled tube. After cracking, the tube is idealized as a hollow truss consisting of closed stirrups, longitudinal bars in the corners, and compression diagonals approximately centered on the stirrups, as shown in Fig.

109 Definitions To understand the subject , we should define some terms used a lot in our calculations :

1) Acp and Pcp

2) Aoh and Ph : represents the gross area enclosed by the shear flow path around the perimeter of the tube.

110

Location of Critical Section for Torsion The critical section for shear was found to be located at a distance d away from the face of the support. For an analogous reason, ACI Code Section 11.5.2.4 allows sections located at less than d from the support to be designed for the same torque, that exists at a distance d from the support. This would not apply if a large torque were applied within a distance d from the support.

Types Of Torsion

 Equilibrium torsion For a statically determinate structure, there is only one path along which a torsional moment can be transmitted to the supports. This type of torsional moment, which is referred to as equilibrium torsion or statically determinate torsion, cannot be reduced by a redistribution of internal forces or by a rotation of the member. The edge beam must be designed to resist the full calculated torsional moment.

111

 Compatibility torsion The torsional moment in a particular part of a statically indeterminate structure may be substantially reduced if that part of the structure cracks under the torsion and “gives,” or rotates. The result will be a redistribution of forces in the structure. This type of torsion, which referred to as statically indeterminate torsion or compatibility torsion.

112

113

Design Procedure For Members Subjected to Bending Moment , Shear and Torsion 1) . Draw the shear force, bending moment, and torque diagrams. 2) Select cross-sectional dimensions “b” and “h” based on factored bending moment, and determine the required area of reinforcement. 2 M u d  c  bd  ,  0.90  s  0.003 ( )  Kn c c  d Kn  fc (1 0.59)   0.003 ( ) s c      ( f y / fc) fc   0.01 (Economic)  As (max)  0.319  bd f y

3) Check if torsion may be neglected. Torsion may be neglected if the torsion less than : 2 A cp 0.083 fc ( ) Without axial force Pcp 2 A cp NU 0.083 fc ( )( 1 ) With axial force Pcp 0.33Ag fc

 If this is the case, proceed on with shear design and choose flexural and shear reinforcement.

 Note : The critical section for torsion is located at distance d from the face of the support if no torques are applied within this distance. If torques are applied within distance d from face of support, critical torsion is located at face of the support.

114

4) .Determine whether the case involves equilibrium or compatibility torsion. A. For equilibrium torsion , use full Tu .

B. For compatibility torsion , reduce Tu to the following : 2 A cp T  0.33 f  ( )  Without axial force u c P cp 2 A cp N T  0.33 f  ( )( 1  U )  With axial force u c P 0.33 A f  cp g c

5) Check the adequacy of the size of the cross section in terms of preventing brittle mode of failure resulting from diagonal compressive stresses due to shear and torsion combined. V 2 Tu P 2 V   ( u )  ( h )  ( c  0.66 f  )  Solid Section max 2 c bwd 1.7A oh bwd

V 2 Tu P V   ( u )  ( h )  ( c  0.66 f  )  Hollow Section max 2 c bwd 1.7A oh bwd

115  Note : If Eq. ( in step 5 ) is not satisfied, cross sectional dimensions need to be increased.

6) Determine the area of stirrups required for shear. To facilitate the addition of stirrups for shear and torsion, the area of shear reinforcement is expressed in terms: Av Vs   Av  2 stirrups one leg area s f yv d

Vu 1 but V  V and V  f  b d s  C c 6 c w

if Vs  4VC the cross Section needs to be enlarge.

 Also, determine maximum stirrup spacing based on shear and if Vu less than 0.5Vc No need for stirrups .

7) Determine the required area of stirrups for torsion in terms of :

At Tn   A0  0.85Aoh s 2 f yv Ao

Besides, compute maximum stirrup spacing based on torsion :  p  h Smax  smaller of  8 300 mm

116

8) Determine combined area of stirrups required for shear and torsion:

0.35bw  A f Avt Av 2At (vt)min  yv    But  max of  s s s s 0.062 f  b  c w   f yv

9) Select stirrup size, and compute stirrup spacing based on the amount determined in step 7. A for selected stirrups size S  v A ( vt calculated s )

 NOTE Maximum stirrup spacing must not exceed the smaller of the two values evaluated in steps 5 and 6.

10) Calculate the longitudinal reinforcement required for torsion.

At f yv 2 Al  Ph cot  s f y

 0.42 fc Acp At f yv A t bw Al min   ( )Ph Where  0.175 f y s f y s f yv

117

11)

2 f  A cp  Cracking Torsion  c  3 Pcp

 Ao  0.85 Aoh

elpmaxE

The cantilever beam shown below supports its own weight plus concentrated load . The beam is 1500 mm and concentrated load at a point 150 mm from the end of the beam and 150 mm away from the centroidal axis of the member . The un-factored concentrated load consists of a 85 KN dead load and a 85 KN live load . Use normal weight of concrete with compressive strength of 20 MPa and yielding strength for bars and stirrups equal to 420 Mpa .

118

1) Assume a dimension for the section to calculate own weight ( d = 535 mm ) wt for major beam  0.4  0.6  24  5.76 wt for 150 mm part  0.4  0.6  24  5.76 2 wt factored 1.2  5.76  6.912 KN / m

Pult 1.2(85) 1.6(85)  238 KN

2) Draw Shear , Moment and Torsion Diagram 600mm

400 mm

 Vu  6.91 0.15  238  6.911.5  249.4 KN

1.52  M  (6.91 0.15) 1.35  6.91  238 1.35  330.5 KN .m u 2 0.15 T  (6.91 0.15)   238  0.15  35.8 KN .m u 2

119

Moment Load ø Moment 0.90 Shear 0.75 Torsion 0.75 Shear

Tortion

3) Design for flexure .

Mu 330.51000 2 As   1000 1815.86 mm   f y jd 0.90 4200.90535

A  f 1815.86 420 a  s y  106.81mm  A 1831mm2 s 0.85 fc b 0.85 20 400

check t , As (max) , As (min) OK

120

4) Check if torsion may be neglected:

2  Acp  b  h  400  600  240000 mm

 Pcp  2 (b  h)  2 (400  600)  2000 mm 2 2 A cp 240000 6  Tth  0.083 fc ( )  0.75 0.083 20 ( ) 10  8.1 KN.m Pcp 2000

Tu  35.8  8.1  must be considered

5) The Torsion is needed for equilibrium , so design for full of 35 .8 KN.m

6) Check the adequacy of the size of the cross section (assume 40 mm cover and ø10 stirrups )

 xo  400  2(40) 10  310 mm

 yo  600  2(40) 10  510 mm 2  Aoh  xo  yo  310510 158100 mm

 Ph  2 (xo  yo )  2 (310  510) 1640 mm 2 2  6  V 2 TuP 2  245.7 1000   35.8 10 1640     ( u )  ( h )     max 2  400  535   2  bwd 1.7A oh    1.7 158100  max  1.8 MPa f  Vc c (  0.66 fc )  (  0.66 fc )  2.77 Mpa b d 6 w  1.8  2.77  Ok

121

 Note : Vu used above to check section size is at “d” distance from the support

V u @d  249.4  6.91(0.535)  245.7 KN

7) Determine the area of stirrups required for shear 1 1 V  f  b d  20  400  535103 159.5 KN c 6 c w 6

Vu 245.7 V  V  159.5 168.1 KN s  c 0.75 A V 168.11000  v  s   0.748 mm2 / mm s f yv d 420  535

8) Determine the required area of stirrups

35.8103 At Tn 0.75 3 2   10  0.422 mm / mm s 2 f yv Ao 2  420  (0.85158100)

9) Determine combined area of stirrups required for shear and torsion: A A 2A 2  vt  v  t  0.748  2(0.422)  1.59 mm / mm s s s 0.35b 2 w  0.333 mm  f A(vt)min  yv   max of  s 0.062 fc bw 2   0.264 mm   f yv

122

10) Calculate the longitudinal reinforcement required for torsion

At f yv 2 2  Al  Ph cot   0.422116401 692 mm s f y

 0.42 fc Acp At f yv At bw  Al min   ( )Ph Where  0.422  0.175  0.167 f y s f y s f yv

0.42 20  240000 2  Al min   0.42216401 381 mm 420

11) Spacing between stirrups :  2  (10)2 A for selected stirrups size  S  v  4  99 mm A ( vt calculated 1.59 s )

 p 1640  h   205 mm   Smax  smaller of  8 8 300 mm

 You can increase the diameter of the stirrups to increase the spacing (s).

12) Provide 3ø10 at the top , 3ø10 at the middle and the rest to the flexure steel:  610  471mm2  692  471 221mm2

1831 221 2052 mm2  use 5 25 at bottom.

123

124 elpmaxE

The one way joist system shown in figure , supports a total factored dead load of 7.5 KN/m2 and factored live load of of 8 KN/m2. Totaling 15.5 KN/m2 . Design the end span AB , of the exterior spandrel beam on grid line 1 . The factored dead load of the beam (i.e , self-weight) and the factored loads applied directly to it total 16 KN/m . The spans and loadings are such that the moments and shears can be calculated by using the moment coefficients from ACI

section 8.3.3 . Use fy =420 MPa and fc = 30 MPa .

125 1) Initial choice of beam size . The beam overhangs the inside face of the columns by 50 mm. Effective length 405 mm and slab thickness is 110 mm .

650 mm

110 mm Spandrel Column

470 mm 470 beam

600 mm

2) Assume that the joists behave as a one unit with uniform load for simplicity , calculate the loads now :  Total load on the beam =load from the slab + load on it self

wln 15.5 9.3  16  16  88.1 KN 2 2  Compute the moment along the edge beam .

126 wl 2 88.1 6.62  Exterior end negative:  M  n   239.9 KN u 16 16 2 2 wln 88.1 6.6  Mid span positive:  M    274.1 KN u 14 14 wl 2 88.1 6.62  Interior end negative:  M  n   383.8 KN u 10 10

3) Check adequacy of the section size based on the maximum moment subjected to the edge beam. The section has adequate size for flexure . 4) Calculate area of steel for flexure :

127 Area of steel required (mm2) Exterior end negative 1791

Mid-span positive 2046 First interior negative 2865

5) Draw final Shear and Torsion diagram for the edge beam .

88.1 6.6  Shear Force: A) at exterior edge Vu   290.7 KN 2

B) at int . edge Vu 1.15 290.7  334.3 KN

128  Torsion:Moment at the joist is Transferred to beam as a torsion w l 2 15.5 9.32  M  n   55.9 KN.m u 24 24 w l 15.5 9.3  n   72.1 KN 2 2

tu  55.9  72.1(0.325)  79.3 KN.m/ m t  l 79.3 6.6  T  n   261.7 KN.m u 2 2

55.9 KN.M

325 mm 325 72.1 KN

Vu @ d  334.7  88.07 (0.405)  298.7 KN

129 6) Should torsion be considered ?

 A  (650  470)  (360 110)  345100 mm2 cp  Pcp  2(650)  470 110  360  2(360)  2960 mm

2 2 A cp 345100 6  Tth  0.083 fc ( )  0.75 0.083 30 ( ) 10  13.7 KN.m Pcp 2960

Tu  261.7  13.7  must be considered

7) Equilibrium or Compatibility ?

 The torque resulting from the moments at the ends of the joists exists only because the joint is monolithic and the edge beam has a torsional stiffness.  The torque resulting from the 25 mm. offset of the axes of the beam and column necessary for the equilibrium of the structure and hence is equilibrium torque.

Note : You should calculate reduced value and equilibrium torsion at offset the design must be based on the maximum value .

Reduced Value at d from support  2 A cp 345100 2 T  0.33 f  ( )  0.75  0.33  30  ( )  55 KN.m u c P 2960 cp

 Check equilibrium torsion 6.6 T  88.1 0.025  7.3 KN.m 2

130

88.1 KN

25 mm

88.1 KN

 Design should be based on T=55.1 KN

8) Check the adequacy of the section >>>>> OK

Ph Aoh Tn  Ao 2 1868 mm 209989 mm 73.5 KN.m 45 0.85 Aoh

131

Table 1 These values are from book ( copy and paste ), Check it by yourself

 max 1.781 MPa Vs 2 Av / s 1.2076 mm /mm Vc 240 KN 2 At / s 0.4902 mm /mm 2 ( Avt / s) 2.188 mm /mm 2 Al 916 mm 2 Al,min 670 mm Stirrups , bars steel diameter

132 Redistribution for the moment along joist

T 55.1 t  u  19 KN.m ln  (2d)/ 2 6.6  2(0.405)/ 2

 Moment along the joist before and after redistribution : wl 2 15.5 9.32  M at ext. edge  n   55.85 KN.m u 24 24 2 2 wln 15.5 9.3  M at midspan.    95.7 KN.m u 14 14 wl 2 15.5 9.32  M at int . edge  n   134.1 KN.m u 10 10

134.1 KN.m 55.85 KN.m

95.7 KN.m

36.85 18.4

133 g MELBeRy

4-1 A simple supported precast T beam as shown in the sketch below . Use yielding strength of 420 MPa for all steel , compressive strength of concrete of 28 MPa . Assume 40 mm cover from all sides , ø12 stirrups then determine the following :

1200 mm

200 mm

800 mm

300 mm

1) Compute the following section properties : 2  Acp = …………………………. mm 2  Aoh = …………………………. mm 2  Ao = …………………………. mm

 Pcp = …………………………. mm

 Ph …………………………. mm

2) If the ultimate torque Tu @ d = 55 KN.m and Ultimate shear force at “d” distance from face of the support is 285 KN , estimate the maximum shear stress in the section .

134 3) Is the section adequate for shear design ? ( Yes/No) 4) Compute the cracking torqe for given section . 5) Compute the total area required for shear and torsion based on the values assumed in (2).

Avs 2 (mm / mm) s

A t 2 6) Compute required area of longitudinal reinforcement for torsion if  0.654 mm . s

4-2 The 8 m span beam shown below is a part of a continuous frame that carries a cantilever slab of 1.5 m width . The beam supports a service live load of 20 KN/m along the beam centerline plus 2 KN/m2 over the slab surface . Assume the effective depth of the beam is 530 mm and the clear cover to the stirrups is 40 mm . Including self-weight of the beam and slab in your calculation .

 Draw the shear force and torsion moment diagrams for this beam .  Determine the shear force and torsion moment at the most critical section .  Design the reinforcement in the beam to carry the forces in previous part .  Show a layout of the beam’s cross section with the reinforcement obtained . From previous part . f  420 Mpa y f   28 MPa c

135

4-3 The figure below shows a cantilever beam subjected to a point load . Indicate the type of torsion in this problem . Explain your answer .

4-4 A) Draw the Torsion moment diagram for part CD given the bending moment diagram for part AB . Determine this type of torsion with explanation .

136 B) Draw torsion moment diagrams for following cantilever .

4-5 The following figure shows a partial plan of precast roof system . Roof members are double tee units simple supported on precast beams .Continuity of spandrel beams is not provided . The roof is supporting a service live load of 1.5 KN/m2 and service dead load of 3.0 KN/m2 . Assume effective depth d =730 mm and ignore the self-weight in your calculation . a) Determine the factored load that will carried by the spandrel beam . b) Draw bending moment , shear force and torsion diagrams for the spandrel beam . c) Determine the design ultimate bending moment , ultimate shear and ultimate torsion at most critical sections . d) Design shear reinforcement for the beam based on values from ( c ) .

137

4-6 Design the T beam shown in the figure below ,considering the contribution of the flanges for torsion if its subjected to a ultimate torsional moment of 18 KN.m .

compressive strength of concrete is 30 MPa and f y =360 MPa for all steel .Assume concrete cover of 40 mm to the centerline of the stirrups al around the cross section .

138

4-7 Architectural and clearance requirements call for the use of transfer girder , shown in the figure below , spanning 6 m between supporting columns. The girder must carry from above an ultimate concentrated column load of 90 KN at midspan , applied with eccentricity 600 mm from girder centerline . (Load factors are already included , as is an allowance for girder self-weight ). Flexural rigidity at the ends of the span can be assumed to develop 40 percent of the maximum moment that would be obtained if the girder were simple supported . Design both transvers and longitudinal steel for the

beam . fc  28 MPa and f y  420 MPa

The member is to have dimensions , b=250 mm , h=500 mm , xo = 163 , yo=413 mm , d =450 mm .

139

4-8 Joint A is monolithically built with cross beam C-D. Beam A-B spans 8.0 meters between supports A and B while beam C-D spans 6.0 meters between supports C and D. Beam A- B carries a factored load of 120 kN/m and beam C-D carries a factored load of 85 kN/m (both are including self-weights). Use fy = 420 MPa for all steel and fc’ = 26 MPa and calculate the following:

140 1) Compatibility torsion applied to beam C-D: 2) Compatibility torsion that beam C-D should be designed for: 3) Design moments for beam A-B are :

4-9 The one way joist system shown in figure , supports a total factored dead load of 9 KN/m2 and factored live load of of 10 KN/m2. Totaling 19 KN/m2 . Design the end span AB , of the exterior spandrel beam on grid line 1 . The factored dead load of the beam (i.e , self-weight) and the factored loads applied directly to it total 18 KN/m . The spans and loadings are such that the moments and shears can be calculated by using the moment

coefficients from ACI section 8.3.3 . Use fy =420 MPa and fc = 30 MPa .

141  it was found that joists with overall depth of 470 mm would be required .The slab thickness is 110 mm . The spandrel beam was made the same depth . The column supporting the beam are 600 mm square , the beam overhangs the inside face of the columns by 50 mm .Beam size is h= 470 mm , b=650 mm and d= 405 mm. Answer the following :

1) Draw bending moment and shear forces diagrams for the beam. 2) Draw bending moment and shear forces diagrams for joist before redistribution. 3) Calculate the torque resulting from the moments at the ends of the joist if the joint is monolithic .Indicate this type of torsion and explain your answer. 4) Calculate the torque resulting from the 25 mm offset of the axis of the beam and column . Indicate this type of torsion and explain your answer . 5) Calculate the design ultimate torsion at most critical section 6) Calculate all section properties : Acp , Aoh ,Pcp ,Ph , Ao

 For the hollow section shown below , calculate the following : Assume the cover = 40 mm

1) Gross moment if inertia 2) Cracking moment of inertia

3) Acp

4) Pcp

5) Aoh

6) Ph

7) Ao

142

Footings

143 FOOTINGS

 Footings are structural members used to support columns and walls and transmit their loads to the underlying soils. Reinforced concrete is a material admirably suited for footings and is used as such for both reinforced concrete and structural steel buildings, bridges, towers, and other structures.

Types of Footings  Among the several types of reinforced concrete footings in common use are the wall, isolated, combined, raft, and pile-cap types. These are briefly introduced in this section; the remainder of the chapter is used to provide more detailed information about the simpler types of this group.

144 1) A wall footing  Wall footings are normally used around the perimeter of a building and perhaps for some of the interior walls.

2) An isolated or single-column footing  is used to support the load of a single column. These are the most commonly used footings, particularly where the loads are relatively light and the columns are not closely spaced.

3) Combined footings  Are used to support two or more column loads. A combined footing might be economical where two or more heavily loaded columns are so spaced that normally designed single-column footings would run into each other.

145

4) A mat or raft or floating foundation  Is a continuous reinforced concrete slab over a large area used to support many columns and walls. This kind of foundation is used where soil strength is low or where column loads are large but where piles or caissons are not used. For such cases, isolated footings would be so large that it is more economical to use a continuous raft or mat under the entire area. The cost of the formwork for a mat footing is far less than is the cost of the forms for a large number of isolated footings. If individual footings are designed for each column and if their combined area is greater than half of the area contained within the perimeter of the building, it is usually more economical to use one large footing or mat. The raft or mat foundation is particularly useful in reducing differential settlements between columns—the reduction being 50% or more. For these types of footings, the excavations are often rather deep. The goal is to remove an amount of earth approximately equal to the building weight. If this is done, the net soil pressure after the building is constructed will theoretically equal what it was before the excavation was made. Thus, the building will float on the raft foundation.

146

SOIL PRESSURE UNDER FOOTINGS The soil pressure at the surface of contact between a footing and the soil is assumed to be uniformly distributed as long as the load above is applied at the center of gravity of the footing. The distribution of soil pressure under a footing is a function of the type of soil and the relative rigidity of the soil and the foundation pad.

147

Allowable Soil Pressures  The allowable soil pressures to be used for designing the footings for a particular structure are preferably obtained by using the services of a geotechnical engineer.

Limit States for the Design of Foundations 1) Limit States Governed by the Soil:  a bearing failure of the soil under the footing  a serviceability failure in which excessive differential settlement between adjacent footings  excessive total settlement.

2) Limit States Governed by the Structure  flexural failure of the portions of the footing that project from the column or wall,  shear failure of the footing,  bearing failure at member interfaces, and  inadequate anchorage of the flexural reinforcement in the footing.

148 Elastic Distribution of Soil Pressure under a Footing P M y Generally , the stress under footing given by q   A I P  Vertical load A  Area of contact surface between the soil and the footing

I  Moment of inertia of this area M  Moment about the centroidal axis of the footing area y  The distance from the centroidal axis to the point where the stress are being calculated

 The loads within Kern distance :  Loads applied within the “ kern “ , shaded area in the figure alongside, will cause compression over the entire area of the footing .

149

150  Loads out of the “kern” distance  This occurs when eccentricity in rectangular footing exceeds the shaded area of the kern distance . A triangular stress distribution will develop over part of the base as shown below , applying the equilibrium equation gives : 1 f  b  (3a)  P 2 max u

2Pu  f  max 3 a  b L  a   e 2

151 Net Soil Pressures(qn)

qn  qa  soil wt  concrete wt  Surcharge wt

Factored Net Soil Pressure

pcolumnultimate q  factored net load  nu A

DESIGN PROCEDURE 1) Assume thickness of the footing then calculate area required .  it is necessary to guess a thickness for a first trial. Generally, the thickness will be 1 to 1.5 the wall thickness or to 2 times column thickness .  Thicknesses of wall footings are chosen in 25 mm. increments, widths in 50- or 75-mm. increments.

Service load(DL  LL) A  qn for soil

qn  qa  soil wt  concrete wt  Surcharge

2) Calculate factored net pressure . pcolumnultimate qnu  factored net load  A

3) Check shear capacity A. Punching shear

152

 4 fcbod   (2  )   c 12   sd fcbod  Vc  smaller of  (  2)    bo 12   1   fcbod  3   if Vc  Vu  no need to increase depth

change CHECK SHEAR if you change depth , you should change qn  A 

B. One way shear  Calculate avarege effective length in the two directions

153 effectivedepth  d  h  cover (75 mm) - db 1 Vc  0.75( fcbwd ) 6 if Vc  Vu  OK otherwise change thickness

4) Design flexural reinforcement

 The deflection for the spread footing shown is in two directions , so the reinforcement should be provided in both the short and long directions .

154

2 d  c f    0.003 ( )  c  a , if   0.005    0.90  Mu  qnu  b  t  t 2 c 1 A f ? Mu S y  check  M n   As f y jd  Mu  As  , a   fy jd 0.85 fcb  find NO.of bars should be used and the spacing " S "

 Asmin  0.0018bh  S should be Smax 2h Temperature steel  Smax  smaller of  500mm 5h  As  0.0018bh but Smax  smallest  500

155

5) Design column footing joint

156

  0.65

6) Area Of dowels  The purpose of the dowels is the same as the keyway. But dowels provide stronger connection .  You should place dowels in the footing while the concrete is still wet .  The number of dowel bars needed is four these may be placed at the four corners of the column. The dowel bars are usually extended into the footing, bent at the ends, and tied to the main footing reinforcement. The dowel diameter shall not exceed the diameter of the longitudinal bars in the column .

157  According to ACI code we can determine required area of dowels by

 Asmin  0.005A1 WhenPnb  Pu pu Pnb  Asreq  ;  0.65When Pu  Pnb  fy

 A1  Column section area

7) Check development length for dowels

158

159 8) Check development length for flexural reinforcement

160 elpmaxE (Design wall Footing )

Design a plain concrete footing to support a 400 mm thick concrete wall . The on the wall consist of 230 KN/m dead load (including self-weight) and a 146 KN/m live load KN/m .

The base of the footing is 1200 mm below final grade . fc  21 MPa , fy  420 MPa, the gross allowable soil pressure = 240 KN/m2 , and the soil density is 18 KN/m3 .

1) Estimate size of the footing and the factored net pressure.  Assume depth of the footing =(1-1.5)×wall thickness =1.25×400=500 mm 2  qn  240  0.5(25)  18(1.2  0.5)  214.9 KN / m 230  146 2  A   1.75 m assume 1 m strip  width of the footing  1.75 m 214.9  Use width = 1.80 m

2) Compute factored net soil pressure 1.2(230)  1.6(146) 2  qnu   283.1 KN / m 1.80 3) Check one way shear :

Estimate d  500  75  12.5  412.5 mm 1.80  0.40  Vu @ d  283.1   0.4125 1m  81.4 KN  2  1  Vc  0.75   21 1000  412.5  241 KN 6 Vu   Vc  OK

161

 No need to check two way shear in the wall footings . 4) Design for flexure :

283.1 1.8  0.4 2 Mu   ( )  69.4 KN .m 2 2

69.4  1000 As    0.9  420  0.95  412.5 2  A  451 mm s 2 As min  0.0018  412.5  1000  742.5 mm

1000 Use 514 / m  S   200 mm 5  As,min=0.0018 bh not 0.0018bd (there is an error above , but its difficult to edit it )

162 5) Check development length

9 f d y b  l  d 10 f  c

9  420  14   1155 mm 10  21 1800  400  l   75  625 mm 2  Use 90 hook 0.24 f d y b  l  dh f  You should check this length c 0.24  420  14   308 mm 21

163 elpmaxE (Square footing )

Design a square footing to support a 450 mm square tied interior column reinforced with 8 25bars . The column carries an un-factored axial dead load of 1000 KN and an axial live load of 900 KN. The base of the footing is 1200 mm below final grade and allowable soil 2 pressure is 240 KN/m . Use f y 420 MPa , fc 28 MPa

1) Estimate the footing size and factored net soil pressure . Assume footing depth h  600 mm 2 qn  240  0.6(15)  (1.2  0.6)(18)  214.2 KN / m 1000  900 2 A   8.87 m 214.5 2 Use 3.0  3.0 m 1.2(1000)  1.6(900) 2 qnu   293.3 KN / m 3  3

2) Check two way shear davg  600  75  1.5(25)  487.5 mm

bo  4(450  487.5)  3750 mm Vc  the smaller value of the three formulas a , b and c

Vc  2394.2 KN 2 Vu  2640  393.2(0.9375)  2382.3 KN Vu   Vc OK

164

3) Check one way shear 3 0.45 Vu  293.3  (   0.4875)  3  692.9 KN 2 2 1  Vc  0.75   28  3000  487.5  986.7 KN 6

Vu   Vc OK

165

4) Design for flexure 3  0.45 Mu  293.3  ( )  3  1122 KN .m 2 2  As  6325 mm 2 Asmin  0.0018  3000  600  3240 mm 2 3000  2(75) Use 13 25  6370 mm  S   237.5 mm 13  1

166

5) Design column footing connection 2 A1  0.45  0.45  0.203 mm 2 2 A2  (1.2  0.45  1.2)  8.123 mm N1  1.7  0.65  28  0.203  6281 KN 8.123 N2  0.85  0.65  28  0.203    19865 KN 0.203  N  6281 KN

Pu  2640  N OK

6) Design area of dowels 2 Adowels  0.005  (450  450)  1013 mm 2 use 420  1256 mm

7) Check development length of dowels: 0.24  db  fy ldh   0.044db fy fc

0.24  20  420   381 mm  370mm 28

8) Check development length of reinforcement of footing 9 fydb 9  420  25 ld    1786 mm 10 fc 10  28

167 3000  450 Available l   75  1200  1786 mm 2

0.24  25  420 use 90 hook  ldh   476 mm  1200mm 28

450 mm

4ø20

600-75 mm

75 mm

3000 mm

13ø25

168

ggomDJta geoonJt

elpmaxE

2 2 600  400 mm 600  600 mm   COL 1  DL  890 KN COL2  DL  1300 KN   LL  650 KN LL  1000 KN All loads are service loads 2 fc  20 MPa , fy  420 MPa , qa  240 KN / m

169

400 mm 600 mm

6.0 m 125 mm

1.2 m

Basement floor with thickness = 125 mm supporting an ultimate live load of 5 KN/m

1) Determine required area Take 22 as the average density for soil and concrete just for simplification, then : 2 qn  240 -1.2(22) - 5  208.6 KN/m 890  650  1300  1000 2  Area requuired   18.4 m 208.6

2) Locate the point of application of column loads on the footing. Q1 Q2 Q2  L x  3  Q1 Q2 x Q1  890  650 1540KN Q 2 13001000 2300KN

2300 6 x   3.594m 23001540

170 3) Determine the dimension of the footing       L 2(x L2) 2(3.594 0.2) 7.588m 7.6 m A 18.4 B    2.5 m L 7.6

L2 L3 L1

Q1+Q2

L/2

Q1+Q2

4) Factored net pressure and factored loads 1.2(890 1300) 1.6(650 1000) 2 qnu   277.3 KN /m 7.6  2.5    Q1u 1.2(890) 1.6(650) 2108KN    Q2u 1.2(1300) 1.6(1000) 3160KN

171 5) Draw moment and shear diagrams  Diagrams should be based on 277.3×2.5=693.25 KN/m

2108 KN 3160 KN

693.25 KN/m

2782.6 KN.m

13.8 KN.m 679.4 KN.m

1969.35 KN

969.85 KN

138.65 KN

2190 KN

6) Check two way shear h 1000mm Try  d  900mm  For interior column

Vu  3160 277.3  (1.5 1.5)  2536KN length of critical shear perimeter bo  41.5  6 m c 1

172

 4 fc bo d a)(2  )  9056KN  c 12   s d fc bo d Vc  smallest of b )(2  ) 12074KN  bo 12  1 c )  fc bo d  6037.4 KN  3  Vc Vu OK

 For interior column

Vu  2108 277.3  (1.5  0.85) 1754.4 KN bo  2(0.85) 1.5  3.2 m 600 c  1.5 400 a)3756.6 KN  Vc  smalles of b ) 8402KN  c )3220KN  Vc Vu OK

7) Check one way shear One way shear is critical at d from the face of the interior column ( d  300 mm from center of the column )

Vu  2190 693.25(0.3  0.9) 1358.1KN 1 1 Vc    fc bw d  0.75  20 2500900 1257.7 KN  increase the thickness 6 6 h 1100mm Try  Vu  2190 693.25(0.3 1.0) 1288.8 KN d 1000mm 1 Vc  0.75  20 250010001397.5 KN OK 6

173

8) Design the flexure reinforcement in the longitudinal direction  3  Mu  2782.6 10  2 As 6 3 8179mm  f y (d  a /2) 0.9  42010  0.9 100010 2 after many iterations  As  7651mm       2 As min 0.0018 bh 0.0018 2500 1100 4950mm Use 1725  8340mm 2

174 For interior column subjected to positive moment

6791000 2 As  1995.9 mm  A 0.9  420 0.9 1000 s min Use 1125  5401mm 2

9) Design for transverse beam

B+1.5 d =2.1 m 0.75 d

3160 KN

wl 2 1264 0.952 Mu    570.4 KN /m 2 2 570.41000 As   0.9  420 0.9 1000 0.95 m after many iterations  As 1588.4 mm2     2 As min 0.0018 2100 1100 4158mm Use 925  4418mm 2

175 2108 KN

For edge beam : 843.2  0.952 Mu   380.4 KN /m 2     2 As min 0.0018 1150 1100 2277mm Use 525  2454.4 mm 2

2018/2.5=843.2 KN/m

 Check development length for all reinforcements required

176 NOTES

 You can use these equations to calculate the ultimate bending moment , ultimate shear force and ultimate load subjected to footings with eccentricity at any point .

qmin

qmax

2 2  X  X  Mu  (q B )( )  (qmax q )(B )( ) 2 3   L x but q  (qmax q )  q L min min   X  Vu  (q B )(X )  (qmax q )(B )( ) 2  L   Pu  (q L)  (qmax q )( )B  min min 2 

177  Moment transfer from columns to footings depends on how the column–footing connection is constructed. Many designers treat the connection between columns and footings as a pinned connection. Others treat it as fixed, and still others treat it as somewhere in between. If it is truly pinned, no moment is transferred to the footing, and this section of the text is not applicable. If, however, it is treated as fixed or partially fixed, this section is applicable. If a column–footing joint is to behave as a pin or hinge, it would have to be constructed accordingly. The reinforcing in the column might be terminated at the column base instead of continuing into the footing. Dowels would be provided, but these would not be adequate to provide a moment connection. To provide continuity at the column–footing interface, the reinforcing steel would have to be continued into the footing. This is normally accomplished by embedding hooked bars into the footing and having them extend into the air where the columns will be located. The length they extend into the air must be at least the lap splice length; sometimes this can be a significant length. These bars are then lap spliced or mechanically spliced with the column bars, providing continuity of tension force in the reinforcing steel.

178 SMELBeRP

5-1 Design a square footing to support a 500 x 500 mm-square tied interior column reinforced with 825 bars. The column carries unfactored axial dead load of 1100 kN and unfactored axial live load of 750 kN. The base of the footing is 1200 mm below 2 final grade and allowable NET soil pressure is 195 kN/m . Use fc’ = 28 MPa and fy = 420 MPa. Assume d = h – 100mm.

5-2 A 4.0 x 4.0 m-square footing is subjected to an ultimate (factored) axial

load Pu = 2400 kN due to eccentric column as shown below (ignore footing self-weight). Assume effective depth d = 0.535 m.

 Calculate maximum and minimum ultimate soil pressure.  Calculate maximum ultimate flexural moment in the footing  Calculate maximum one-way shear in the footing (Vu)  Calculate the punching shear exerted by the 400 mm-square column on the footing (Vu).

179 5-3 A spread footing is subjected to 500×500 mm square tied interior column The base of the footing is 1500 mm below final grade and allowable soil pressure is 200 kN/m2, effective depth equal to (h-100 mm ). Answer the followings :

 If the column carries unfactored axial dead load of 1300 kN and unfactored axial live load of 800 kN,. Assume thickness of footing = 700 mm , answer the following : a) Determine the minimum dimension of the square footing required to support the loads . b) Calculate the ultimate design bending moment and ultimate design shear force at the most critical sections for the 3.5×3.5 m2 footing . 2 c) Calculate the maximum one way shear capacity(Vc ) for 3.5×3.5 m square footing. 2 d) Calculate the maximum two way shear capacity (Vc ) for the 3.5×3.5 m square footing. e) Check development length for the 3.5×3.5 m2 square footing .

 For a 4×6 m rectangular footing . Assume that the ultimate design bending moment in the long direction is 900 KN.m , the ultimate design bending moment in the short direction is 600 KN.m and the ultimate one way shear force is 700 KN , answer the following :

180 f) Calculate the area of steel required in the long direction . g) Calculate the area of steel required in the long direction . h) Calculate the minimum area of steel in the long direction.

i) If we have to use 30ø12in the short direction , make a simple sketch showing the distribution of the bars in this direction .

5-4 The rectangular footing shown below is subjected to a concentrated factored column 2 load Pu=3000 KN and having an area 3.00×4.00 m . Given the column size is 350×450 mm2 and effective depth of the footing is 610 mm :

a) Determine the value of the moment at the critical section(s) . b) Check the adequacy of the footing depth for one way shear . c) Check the adequacy of the footing depth for punching shear . d) Assume that the main reinforcement in the footing consists of 18-mm diameter bars at 150 mm O.C and the dowels in the footing are 14- mm bars . Sketch the distribution of the reinforcement in the short direction .

181 5-5 For the wall footing shown below , draw the ultimate bearing stress distribution for the following cases and determine the values of ultimate shear and ultimate moment at their critical sections . Given effective depth of the footing is 310 mm .

a) Pu= 800 KN

b) Pu= 600 KN , Mu =100 KN.m

c) Pu= 300 KN , Mu =100 KN.m

d) Pu= 300 KN , Mu =200 KN.m

5-6 For the rectangular spread footing shown in the figure below , service dead load of 800 KN and service live load of 600 KN and the eccentricities along x-axis and y-axis equal to 0.20 m .Unit weight of the soil is 18 KN/m3. a) Calculate the soil pressure at the four corners of the footing . b) Calculate the maximum 1.2 m bearing stress capacity at the 0.6 m column-footing joint . c) If the allowable soil pressure is 260 KN/m2 , show 6.0 m whether footing size is adequate or not ?

182 5-7 For the combined footing shown in the figure below . Allowable soil pressure is 250 2 3 KN/m at 1.5 m below surface , use avg  22 KN/m for all materials (soil and

concrete ) ,fc =21 MPa and fy =420 MPa . Column size Service dead load Service live load 1 450×450 mm 700 KN 500 KN 2 600×600 mm 1000 KN 700 KN

0.8m 6.0m

COL1 COL2 = 0.60 m

183

a) Calculate the area required , and find the dimensions B and L .  Assume that the dimensions of the footing are 2.5×9 m : b) Draw ultimate bending and shear diagrams for the footing . c) Calculate the required area of steel in the long direction for positive and negative moments . d) Calculate the minimum area of steel in the long direction .

e) Calculate the maximum one way shear (Vu) at the most critical section (s) . f) Calculate the required area of steel in the transverse direction .

g) Prepare neat design drawings showing footing dimensions and provided reinforcement.

5-8 A wall footing with 700 mm thickness and 4.0 m width , carries a service dead load of 250 KN/m and service live load of 180 KN/m . The base of the footing is 1300 mm under the soil with unit weight equal to 20 KN/m3 . Gross soil pressure is 240 KN/m2 ,

concrete compressive strength is 23 MPa and fy for all steel is 420 MPa .

184 a) Calculate the maximum bending moment at the most critical section(s) . b) Calculate the required area of steel in the short direction. c) Calculate the required area of steel in the long direction. d) How many bars per meter needed in the long and short directions . e) Calculate the ultimate shear force (Vu ) at the most critical section . f) Calculate the maximum shear strength capacity . g) Calculate the development length of the reinforcement used in the short direction . Use ø25 M

5-9 Why it’s better to put the reinforcement in the long direction in a rectangular footing under the reinforcement in the short direction .

185

Biaxial Columns

186

Columns :Combined Axial Load and Bending

The majority of reinforced concrete columns are subjected to primary stresses caused by flexure, axial force, and shear. Secondary stresses associated with deformations are usually very small in most columns used in practice. These columns are referred to as "short columns." Short columns are designed using the interaction diagrams presented in this chapter. The capacity of a short column is the same as the capacity of its section under primary stresses, irrespective of its length.  Many columns are subjected to biaxial bending, that is, bending about both axes, such as : 1) Corner columns in buildings where beams and girders frame into the columns from both directions are the most common cases. 2) where columns are cast monolithically as part of frames in both directions . 3) where columns are supporting heavy spandrel beams. Bridge piers are almost always subject to biaxial bending.

 Most columns are subjected to significant bending in one direction, while subjected to relatively small bending moments in the orthogonal direction. These columns are designed by using the interaction diagrams discussed in RC1 course for uniaxial bending and if required checked for the adequacy of capacity in the orthogonal direction. However, some columns, as in the case of corner columns, are subjected to equally significant bending moments in two orthogonal directions. These columns may have to be designed for biaxial bending.

187

 A circular column subjected to moments about two axes may be designed as a uniaxial column acted upon by the resultant moment; (Why ?? )  Circular columns have polar symmetry and, thus, the same ultimate capacity in all directions. The design process is the same, therefore, regardless of the directions of the moments. If there is bending about both the x- and y-axes, the biaxial moment can be computed by combining the two moments or their eccentricities as follows: 2 2 Mu  (Mux )  (Muy )

2 2 e  (ex )  (e y )

 For shapes other than circular ones, it is necessary to consider the three-dimensional interaction effects.

 Whenever possible, it is desirable to make columns subject to biaxial bending circular in shape.

 Several methods used to determine the nominal strength of the columns subjected to moments in the two directions :

1) static equations : - Such a procedure will lead to the correct answer, but the mathematics involved is so complicated because of the shape of the compression side of the column that the method is not a practical one.

188 2) Computer programs 3) The strain-compatibility method 4) The equivalent eccentricity method. 5) Method based on 45 slice through interaction surface. 6) Bresler reciprocal load method : this will be discussed in this part .

Bresler reciprocal load method

The Bresler equation works rather well as long as Pn is at least as large as 0.10Po .

Should Pn be less than 0.10Po it is satisfactory to neglect the axial force completely and design the section as a member subject to biaxial bending only. This procedure is a little on the conservative side. For this lower part of the interaction curve, it will be remembered that a little axial load increases the moment capacity of the section. The Bresler equation does not apply to axial tension loads. Professor Bresler found that the ultimate loads predicted by his equation for the conditions described do not vary from test results by more than 10%.

 ACI Commentary Sections 10.3.6 and 10.3.7 give the following equation, originally presented by Bresler [11-18], for calculating the capacity under biaxial bending: 1 1 1 1     p  p  p  p n nx ny n

189

Procedure Used To Design Biaxial Columns

1) Select a trial section:

(1  2)%for Tied Column assume that    g (2.5  5)%for spiral Column   p u for Tied Column    0.4(fc f y g ) A   g (trial ) p  u for spiral Column    0.5(fc f y g ) Min. dimension for a tied column  250 mm and 300 mm for spiral .

2) Select appropriate number of bars to be subjected to the column : A bar is adequately supported against lateral movement if its located at a corner of a tie and if the dimension “x” shown in the figure below is less than 15 cm .

190

Figure 18 Ties shown dashed may be omitted if x < 150 mm

 3) Calculate Gamma x h x

b  2cov 2d  d   s b  cov  40mm b b

191

4) Compute ex , ey and eo

Compute e x ,e y and eo :

Muy Mux e x  ,e y  then use the diagramto find Pnx Pu Pu e x     , g Pny l x and Pny depending on :  e y , g  Pnx l  y

5) Compute  Po

ComputePn :

Pn  (0.85 fc (Ag  As )  As f y )

s  0.75 and T  0.65 0.8Pn for Tied Max. axial load   0.85Pn for spiral

6) Solve for  Pn

1  1  1  1 p n p nx p ny p n p should be  p n u

192 Interaction Diagrams

 The ACI column interaction diagrams are used in Examples to design or analyze columns for different situations. In order to correctly use these diagrams, it is necessary to compute the value of γ (gamma), which is equal to the distance from the center of the bars on one side of the column to the center of the bars on the other side of the column divided by h, the depth of the column (both values being taken in the direction of bending). Usually the value of γ obtained falls in between a pair of curves, and interpolation of the curve readings will have to be made.

 Caution 1) Be sure that the column picture at the upper right of the interaction curve being used agrees with the column being considered. In other words, are there bars on two faces of the column or on all four faces? If the wrong curves are selected, the answers may be quite incorrect.

193

elpmaxE

 Select column section size and reinforcement for a rectangular tied column with bars distributed along four faces, subject to biaxial bending. Given :

Pu Mux Muy fc fy 1900 KN 100 KN.m 190 KN.m 28 MPa 420 MPa

1) Estimate column size

Pu 19001000 Ag  assume gross steel ratio  0.015 A g  0.4(fc  g f y ) 0.4(28  0.015 420) 138483.96 mm2 Try 500 400mm 2  200000mm 2 You can select smaller size

2) Select appropriate number of ø25 bars

194

3) Compute Pnx (at ex=0.0)

Mux 100 e y 52.6 mm e e y    52.6 mm    0.132  Pu 1900 l y 400mm h  2 10  25 AS 4  g    0.025 0.040OK A 500 400 g 400  2(40)  2(10)  25    0.6875 400   P      n    for 0.60 2.2 Ksi   bh    Pn for   0.6875  ???  by interpolation  2.26Ksi  2.26 6.9 MPa 15.60   bh     Pn for   0.75  2.3 Ksi   bh 

3  Pnx 15.60500 40010  3120KN note : I have used interaction diagrams with Ksi unit so its necessary to takecare of units

2.2 Ksi

195

4) Compute Pny ( at ey =0.0)

M 190 e 100mm e e  uy  100mm  x   0.2  x Pu 1900 l x 500mm h  2 10  25 AS 4  g    0.025 0.040OK A 500 400 g 500  2(40)  2(10)  25    0.75 500  Pn for   0.75 1.9 Ksi bh  Pn 1.9  6.9 MPa 13.11MPa bh      3  Pny 13.11 500 400 10 2622KN note : I have used interaction diagrams with Ksi unit so its necessary to takecare of units

5) Compute Pno  You can calculate it by equation or by interaction diagrams ( independent of Gamma value ) Using Diagrams

  g  0.025    Pn  e    2.5 Ksi 17.25 MPa   0.0  bh  h        3  Pno 17.25 500 400 10 3450KN Using Equation

   2    2   Pn  0.65 0.85 28500 400 10  25   10  25  420  4357KN o   4   4 

note that , there is a large difference in values of  Pn , may be because of units or due to less accuracy of using diagrams .

196 6) Solve for Bresler equation 1 1 1 1      4.7238510 4  Pn 3120 2622 4357

 Pn  2117KN  Pu 1900KN OK

In this chapter, Pn values were obtained only for rectangular tied columns. The same theory could be used for round columns, but the mathematics would be somewhat complicated because of the circular layout of the bars, and the calculations of distances would be rather tedious. Several approximate methods have been developed that greatly simplify the mathematics. Perhaps the best known of these is the one proposed by Charles Whitney, in which equivalent rectangular columns are used to replace the circular ones. This method gives results that correspond quite closely with test results.

197 SMELBeRP

For all problems :

 Cover Bars diameter Ties diameter fc f y 40 mm 25 mm 10 mm 28 MPa 420 MPa

6-1 The column shown below is to carry a factored load , Pu ,of 800 KN with eccentricities

of ey = 90 mm and ex = 300 mm . Check the adequacy of the trial design by using Bresler method .

6-2 Check the adequacy of the circular column for the following axial load and biaxial bending moments. Use and procedure you feel is most suitable .

Pu Mux Muy 2200 KN 100 KN.m 175 KN.m

198

6-3 For square column shown in the figure below , determine the maximum ultimate

biaxial bending moment ( Mux and Muy ) that can be applied simultaneously with a load

Pu = 3000 KN to column section . Assume ex = ey and  =0.80

6-4 The section of a short tied column is 400×400 mm and is reinforced with 8ø25 bars as

shown. Determine the allowable ultimate load on the section  Pn if its acts at ex = 108

mm and ey = 270 mm. Use Reciprocal Load Method: Bresler’s Formula and the interaction diagram given on Design Aids . Fill the table below .

Pno

Pnx

Pny

 Pn

199

200

Slender Columns

Reference Chapter12: Reinforced Concrete MECHANICS AND DESIGN 6th edition Maher Fakhoury

Slender Columns

Definition: A slender column is defined as a column that has a significant reduction in its axial-load capacity due to moments resulting from lateral deflections of the column. In the derivation of the ACI Code, “a significant reduction” was arbitrarily taken as anything greater than about 5 percent.

 Most building columns fall in the short-column category. Exceptions occur in industrial buildings and in buildings that have a high first-floor story for architectural or functional reasons. An extreme example is shown in Fig. 12-7. The left corner column has a height of 50 times its least thickness. Some bridge piers and the decks of cable stayed bridges fall into the slender-column category.  Less than 10 % of columns in “braced” or “nonsway” frames and less than half of columns in “unbraced” or “sway” frames would be classified as “slender” following ACI Code Procedure.

Fig. 12-7 Bank of Brazil building, Porto Alegre, Brazil. Each Floor extends out over the Floor below it. (Photograph Courtesy of J. G. MacGregor.)

Concept of Slenderness Effect

An eccentrically loaded, pin-ended column is shown in Fig the moments at the ends of the column are when the loads P are applied, the column deflects laterally by an amount as shown. Me = Pe For equilibrium, the internal moment at mid height Mc = P(e + δ) The deflection increases the moments for which the column must be designed. In the symmetrical column shown here, the maximum moment occurs at mid height.

The slenderness of columns is based on their geometry and on their lateral bracing. As their slenderness increases, their bending stresses increase, and thus buckling may occur. If they are “slender,” the moment for which the column must be designed is increased or magnified. Once the moment is magnified, it’s very simple the column is then designed as a short column using the increased moment.

The line O–A is referred to as a load–moment curve for the end moment, while the line O–B is the load– moment curve for the maximum column moment. Failure occurs when the load–moment curve O–B for the point of maximum moment intersects the interaction diagram for the cross section. Thus the load and moment at failure are denoted by point B in Fig. Because of the increase in maximum moment due to deflections, the axial-load capacity is reduced from A to B. This reduction in axial-load capacity results from what are referred to as slenderness effects.

Buckling of Axially Loaded Elastic Columns

Recall Leonhard Euler Equation from strength of materials laboratory,

Where EI = flexural rigidity of column cross section. L = length of the column. n = number of half-sine waves in the deformed shape of the column.

*The effective length factor is K= 1/n

Computation of Effective Length Factor K:

 In buildings, columns are restrained by beams or footings which always allow some rotation of the ends of the column. The effective lengths will be greater than the values for completely fixed ends. The actual value of k for an elastic column is a function of the relative stiffnesses , of the beams and columns at each end of the column, where is

Where b and c refer to beams and columns, respectively, and the lengths and are measured center-to-center of the joints. Where Ic=0.70 Ig , Ib=0.35 Ig , Ec=4700

 After Computing , the Effective Length Factor K is computed using Nomographs Fig12-6 (NEXT PAGE) A and B are top and bottom factors of columns. For a hinged end is infinite or 10 and for a fixed end is zero or 1. -Assumptions for nomographs: 1. Symmetrical rectangular frames

2. Equal load applied at top of columns 3. Unloaded beams. 4. All columns buckle at the same moment

 Calculation of, Column Footing Joints:

Where is the If moment of inertia of the contact area

between the bottom of the footing and the soil and Ks is the coefficient of subgrade reaction which can be taken from Fig. 12-27

 Calculation of k from Tables (another way but NOT according to ACI CODE) Table 12-2 can be used to select values of k for the design of nonsway frames. The shaded areas correspond to one or both ends truly fixed. Because such a case rarely, if ever, occurs in practice, this part of the table should not be used. The column and row labeled “Hinged”, “elastic” through to “fixed” represent conservative practical degrees of end fixity.

Limiting Slenderness Ratios for Slender Columns

Most columns in structures are sufficiently short to be unaffected by slenderness effects. To avoid checking slenderness effects for all columns, ACI Code Section 10.10.1 allows slenderness effects to be ignored in the case of columns in sway frames if,

*IF Then a second order analysis must be done.

Unsupported Lengths (Lu): The length used for calculating the slenderness ratio of a column, Lu, is its unsupported length. This length is considered to be equal to the clear distance between slabs, beams, or other members

1. Length of column. The unsupported length, Lu, is defined as the clear distance between members capable of giving lateral support to the column. For a pin- ended column it is the distance between the hinges. 2. Effective length. States that the effective length factor, k, can be computed from nangoraphs. 3. Radius of gyration. For a rectangular section and for a circular section, 4. Consideration of slenderness effects. For columns in nonsway frames, allows slenderness to be neglected if satisfies Eq. (12-20b). The sign convention for M1/M2 is given in Fig. 12-13. 5. Minimum moment. Requires that the maximum end moment on the column, not be taken less than

M2,min = Pu(15 + 0.03h2) where h is in mm

6. Moment-magnifier equation. ACI Code Section 10.10.6 states that the columns shall be designed for the factored axial load, Where M2 is the larger end moment, and the magnified moment Mc, defined by:

*Extra Information

Example 1: Design of a Slender Pin-Ended Column (Nonsway)

Design a 6m-tall column to support an unfactored dead load of 400 KN and an unfactored live load of 334 KN. The loads act at an eccentricity of 76 mm. at the top and 50 mm. at the bottom, ’ as shown in Fig. Use f c= 28 MPa and Fy=420 MPa

1. Compute the factored loads Pu and moments M1/M2 and

Pu= 1.2 DL+1.6LL = 1.2 X 400+1.6 X 334 = 1014.4 KN

The Moment at the top = 1014.4 X 0.076 = 77.1 KN-m The Moment at the bottom = 1014.4*0.05= 50.72 KN-m

By definition, M2 is the larger end moment in the column. Therefore, M2= 77 KN-m and M1= 51.7 KN-m the ratio M1/M2 is taken to be positive, because the column is bent in single curvature Thus M1/M2=0.658

2. Estimate the column size. Use ρ=1.5%

2 = = 73935.86mm

use bXh = 300 X 300 mm2

3. Is the column slender? a column in a nonsway frame is short if

K=1 because the column is pin ended, where r=0.3*300= 90 mm

For M1/M2=0.658 34 - 12 = 34-12 X 0.658= 26

Because 66.7 exceeds 26, the column is quite slender. This suggests that the Section 300mm by 300mm probably is inadequate. We shall select a 400mm-by-400mm.-section for the first trial.

4. Check whether the moments are less than the minimum. a braced column be designed for a minimum eccentricity of 15+0.03 X 350 = 25.5 Because the maximum end eccentricity exceeds this, design for the moments from step 1.

5. Compute EI.

Ec=4700 f’c= 4700 28=24870 MPa Ig= bh3 /12 = 2.13X 109 mm4

The term Bdnsis the ratio of the factored sustained (dead) load to the total factored axial load:

Bdns = 1.2 X 400 / 1014.4 = 0.473

EI = = 1.44 X 1013 N-mm2

6. Compute the magnified moment.

= 0.6 + 0.4 x 0.658 =0.8632

= KN

1.3

Magnified moment (Mc) =1.3 X 77= 100.1 KN-m

7. Select the column reinforcement. We will use the tied-column interaction diagrams assuming an equal distribution of longitudinal bars in two opposite faces of the column. The parameters required for entering the interaction diagrams are:

Cover 50 mm and assume Ø25

From both and the required value for is less than ρ= 0.01. Therefore, to satisfy the minimum column longitudinal-reinforcement ratio from ACI Code Section 10.9.1, use ρ= 0.01Thus,

As’ rqd = Ag * 0.015 = 400*400 * 0.01 = 1600 mm2

Use 8Ø16 with As= 1609mm2 for the 400x400 mm2 section. This section design would be very conservative if we were designing a short column, but the slenderness of the column has required the use of this larger section.