COURSE OUTLINE

FCE 431 – STRUCTURAL DESIGN 1A (60 hrs)

Reinforced concrete:

Elements of concrete constructionPrinciples of reinforced concrete designElastic and ultimate limit statesDesigning of reinforced concrete elements for bending shear and compression loading.Fire protection.Bond and anchorageReinforced concrete slabs and yield lime analysis.Detailing of beams, slabs, columns, staircase and walls.Foundations and Retaining walls.Codes of PractiseDesign Exercise.

Prestressed Concrete:-

Elements of precast prestressed concrete constructionPrinciple of prestressed concrete design.Tendon ProfilesDesign of statically determinate prestressed concrete element.Applications.Code of practice.Design Exercise.

1. Elements of Concrete Construction.

Concrete is a composite material made up of :-(a) A binding medium, cement + water which forms the cement paste.(b) A relatively inert filter which is the aggregate.The mix design is by proportions to ensure the following properties.

(a) Workability(b) Strength(c) Durability(d) Economy

Portland Cement is made from clinker at 1450⁰C. Its constituents are Lime (Cao) 60%-65%, Silica Sio2 (18 – 25 %), Alumina (Al2O3)- (3%-8%) and Iron Oxide (Fe2O3)- 0.5% - 5%There are various forms of cement : : Ordinary Portland cement , Rapid Hardening Cement, Pozollana Cement etc.

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The cement paste normally sets after hydration to give concrete its strength. Cubes can be cast to measure the strength of the resulting concrete. The specific gravity of cement is 3.15 and has a bulk density of 1450 kg/m3 .

Aggregate which is inert is divided into two: Coarse Aggregate and Fine Aggregate. Coarse aggregate is any aggregate of size greater than 5 mm. Fine aggregate has a size less than 5 mm. The aggregate can be obtained from crushed stone or crushed gravel. The coarse aggregate is of size 40, 20, 14 or 10 mm. Fine aggregate fill the voids between the coarse aggregate. The cement fills voids within the fine aggregate. Voids depend on grading. This can be obtained from sieve analysis. The bulk density of sand and aggregate is approximately 1700 kg/m3.Concrete can be denoted as Class 20/20 or Class 20/40. The first item is its strength whereas the second number is the maximum aggregate size.

1.1 Properties of Concrete

1.1.1 Compressive Strength

The strength: of concrete is denoted by its compressive strength at 28 days, This is obtained by obtaining the cube strength. The strength of concrete increases with age. It may have a strength at 7 days of 13.5 N/mm2 and at 28 days of 20.0 N/mm2. Normally cubes of size 150 mm are used. Cylinders of size 300x150 can also be used to obtain the strength. Normally the cylinder strength is 70-90% of the cube strength.

1.1.2 Tensile Strength :

Indirect tensile strength is obtained by the splitting cylinder test .

ft = 2F/πdl

The flexural strength can also be obtained by the Modulus of Rupture test ft = M/z

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The strength of concrete is primarily dependent on water- cement ratio. Curing must be done for a minimum of 7 days. Moist hessian or sand can be used to cover the concrete. This will allow hydration to continue with increase in strength.

1.1.3 Elasticity and Poisson Ratio

The initial tangent modulus or secant modulus depends on the rate of loading. The load is normally applied at a rate of application of a 0.5 N/mm2 per second. The value of the modulus of elasticity for Grade 20 concrete is 18,0000 N/mm2

1.1.4 Workability

This is the ease with which concrete can be mixed placed, and compacted. Compactibility, mobility, stability all depends on the workability. It is measured by slump test. 3 layers of concrete are tamped 25 times each.. Slump detects variation of concrete while construction is going on. Both the workability and strength depends on the water content.

2. DESIGN PHILOSOPHY 3

The strength of concrete if obtained from a number of samples follows a normal distribution. Characteristic strength is defined as the strength below which only 5% of the cubes are expected to fall. It is 1.64 standard deviations from the mean.

fk = fm – 1.64 sd

Limit state design philosophy is one in which the structure is designed to satisfy different limit states. The states are the ultimate limit state and the serviceability limit state. The ultimate limit state deals with strength of the structure. The serviceability limit state deals with cracking, deflections and vibrations etc. Design is done for ultimate limit state whereupon it is checked for the serviceability limit state. The probability of collapse should also be made small compared to the serviceability limit states being reached.

The characteristic load should not be exceeded during the life of the structure. The design load is the characteristic load multiplied by the partial safety factor for loads. The load factor for dead loads is typically 1.4 whereas the load factor for live load is 1.6. The design strength is the characteristic strength divided by the material factor. The material factor for steel reinforcement is 1.15 , and that for concrete 1.5.

2.1 Stress Strain Characteristic of Steel and Concrete

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Hot rolled steel has a definite yield strength whereas cold worked steel does not have a definite yield stress. Instead we use the 0.2% proof stress as the yield stress. 0.43% strain can be used as the ultimate strength.

Elastic theory is affected by the effects of shrinkage, creep, and cracking in the tension zone, hence the move to ultimate strength as a design philosophy.

2.2 Bending Theory2.2.1. Assumptions:

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i). Strains in concrete and reinforcing steel are directly proportional to the distance from the neutral axis, at which the strain is zero.ii) Ultimate limit state of collapse reached when concrete strain at extreme compression value reaches a specified value of εcu i.e. 0.0035iii) Tensile strength of concrete is ignored.iv) Stresses in reinforcement delivered from appropriate stress/strain graph.

εs = d-x/x, εcu

εs1 = x- d/x . εcu

Note balanced failure occurs when steel fails at 0.87 fy and concrete at a strain of 0.0035 .

In the simplified stress block BS 8110, the value of . x/d < 0.5 i.e. x/d should not be greater than 0.5 . The steel fail at a stress of 0.87 fy. The simplified stress block has an intensity of 0.45 fcu to depth of 0.9x.

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C=0.405 fcubx T= 0.87 fyAs

M=(0.405 fcubx) (d-0.45x)

= (0.405x/d)(1-0.45x/d) fcubd2

= Kfcubd2

If x/d = 0.5 F= 0.405fcu .0.5d = 0.2fcubdz = d- 0.45x d/0.5 = 0.775d.M= 0.2fcubd x 0.775d = 0.156 fcubd2 = moment capacity ( Mu )If M > Mu - A’s = compression reinforcement.

M – Mu = 0.87 fyA’s(d-d’)0.87fyAs = 0.87fyA’s + 0.2fcubdAs = A’s + Mu/z(0.87fy)oZ= 0.775d.

2.2.2 Proof of equation for lever arm

M= 0.45fcubx(0.9)(d-0.45x)Defn K = M/bd2fcu M= Kbd2fcu

Z= d-0.45x0.0405xd – 0.18225x2 = Kd2

0.18225x2 – 0.0405xd +Kd2 = 0x= 0.405d ±√(0.0405d)2 – 4 (0.18225)kd2/ 2(0.18225)x= d{1.11 ± √3 – k(5.4869)}Z= d- 0.45 { d [ 1.11 ± √1.23 – k(5.4869)}Z = d-0.4995d ± √{0.55 – K/0.9}Z= d { 0.5 ± √(0.25 – K/0.9) }

If K> K’ compression reinforcement is requiredK’ = 0.156K = M/bd2fcu

Z = d {0.5 + √(0.25 – K’/0.9)}x = (d-z)/0.45A’s = (K-K’)fcubd2/ 0.87 fy (d-d’)A’s = K’fcubd2/0.87fyz +A’s

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2.2.3 FLANGED BEAMS

b = breadth of flange

T beam: bw + 0.2lz = b lz = 0.7L for a continuous beam.L beam: bw+ 0.1 lz = b

Method 1: Quick Design Method

Assumption: Depth of BS 8110 rectangular stress block is not less than flange thickness. i.e. 0.9x > hf.Compressive force in web below flange is neglected.M = 0.87 fy As [d – hf/2] if M< Mu

and Mu = 0.45 fcu bhf [d- hf/2]

Method 2

K = M/fcubd2 ; obtain z and x.a. If 0.9x <hf As = M / 0.87fyz beam width b.b. if 0.9x > hf Muf = 0.45fcu (b-bw).hf (d - 0.5hf). Kt = M-Muf/fcubwd2

AS = M – Muf / 0.87 fyz + Muf / 0.87 fy (d-0.5 hf).

If Kt > 0.156 redesign the section

DESIGN METHODOLOGY - BS 8110K’ = 0.156K = M/bd2fcu

If K ≤ K’ no compression reinforcement required.Z = lever arm.Z = d {0.5 + √(0.25 – K/0.9)}, but Z is not greater than 0.95dX = (d-z) / 0.45X = depth to neutral axisZ = Lever armAs = M/0.87fy Z

If K > K’ , compression reinforcement required.Z = d{0.5+√(0.25 – K’/0.9)} X = (d-z)/0.45A’s = (K-K’) fcubd2/0.87fy(d-d’)As = K’fcubd2/0.87fyZ + A’s

3. SHEAR

-Failure preferable to bending . Shear failure is small deflections and lack of ductility.-Ultimate state design.-Nominal Shear stress. v = V/bvd = design shear stress.

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Mechanisms of shear transfer.

Vd= dowel action ,Va = aggregate interlock , Vcz = Compression zone shearVcz = 20 – 40%, Vd = 15 – 25%, Va = 35 – 50%-Shear reinforcement is necessary to increase shear capacity and to increases ductility.- One form of shear reinforcement is by Stirrups.

Vs = Shear force carried by web bars.

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0.87 fy As. d/sv = VAsv/sv = V –Vc/ 0.87fyd = V- Vcb/bd0.87fy

= (v-vc)b/0.87fyvAlso bent up bars can be used.BS 8110 design methodology

(i) Design shear stress v = V/bd.(ii) If v> 0.8√fcu or 5 N/mm2 increase bvd. ie the dimensions of the beam.(iii) If v< 0.5vc - nominal shear reinforcement is required. Asv/sv > 0.4b/0.87fyv

Table showing values of vc

Effective depth d (mm)

100As/bd 150 175 200 225 225 300 >400<0.15 0.50 0.48 0.47 0.45 0.44 0.42 0.400.25 0.60 0.57 0.55 0.54 0.53 0.50 0.470.50 0.75 0.73 0.70 0.68 0.65 0.63 0.590.75 0.85 0.83 0.80 0.77 0.76 0.72 0.671.00 0.95 0.91 0.88 0.85 0.83 0.80 0.741.50 1.08 1.04 1.01 0.97 0.95 0.91 0.842.00 1.19 1.15 1.11 1.08 1.04 1.01 0.94>3.00 1.36 1.31 1.27 1.23 1.19 1.15 1.07

As = Area of tension steel for at least a distance d past supportMaximum link spacing Sv should not be greater than 0.75d

If 0.5vc < v < (vc + 0.4) then minimum links should be provided for the whole lengthIf v> (vc+0.4) Asv/sv > (v – vc) bv/ 0.87fyvLinks should go round a bar of at least the same size for an angle of, 90⁰. The ends should have an additional length equal to 8Ф

4.BOND AND ANCHORAGE

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Bond stress is the shear stress acting parallel to the reinforcement bar on the interface between the bar and the concrete. The strain in reinforcement = strain in concrete. It is caused by adhesion , friction and bearing.

Anchorage bond stress fb = Fs/ πФl < fbu

fbu = Ultimate anchorage bond stress = β√fcu.

β= bond co-efficient.

VALUES OF β

Tension ComprPlain Bar 0.28 0.35Deformed Bar 0.50 0.63

fb = fsπФ2/4 / πФL = fsФ / 4L

The anchorage length L = fsФ/ 4β√fcu

Ultimate anchorage bond length = 0.87fyФ/ 4β√fcu

LENGTH OF BEND OR HOOK FOR DIFFERENT REINFORCEMENT fy = 250 fy = 460 bend hook bend hook

Ф 8Ф 16Ф 12Ф 20Ф

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r > 2Ф for mild steel r > 3Ф for high yield up to 20 mm and 4Ф for larger sizes

fy = 250 fy = 460

Ф bend hook bend hook8 15Ф 15Ф 10 13Ф 15Ф12 11Ф 16Ф 14Ф 24Ф16 9Ф 12Ф20 8Ф 12Ф25 8Ф 12Ф32 8Ф

5. COLUMNS

1) Elastic behaviorN = fcAc + fsc Asc Equilibriumfc / Ec = fs /Es Compatibility

fc= N / {Ac + αeAs } fs = αeN / {Ac + αeAs}

α = modular ratio

N the load capacity for short columns = 0.67 fcu / ᵧm . Ac + fy/ ᵧm Asc

= 0.45fcu Ac + 0.87fy Asc

To take into account the eccentricity we utilize only 90% of the capacity of the column. N=0.40 fcuAc + 0.75 fyAsc

6% ≤ As ≤ 0.4% of the total concrete cross section.

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Links ¼ Ф and 12 Ф spacing

Buckling should be considered in dealing with columns.

Ncr = π2EI/L2

and Le = β l where the value of β depends on the end conditions. The end conditions are pertinent at the top of the column and at the bottom of the column. These ends could be fixed or partially fixed, or free.

Interaction curves are used to cater for the effect of axial load on the column together with bending moment coming into the column. All columns have an initial eccentricity eadd especially if it is a slender column.

Madd = Neadd.Mt = Mi + M add.

VALUES OF β FOR DIFFERENT END CONDITIONS OF THE COLUMN.

Bottomfixed Partially

restrainedPinnedTop

Fixed 0.75 0.80 0.90Partially restrained 0.80 0.85 0.95Pinned 0.90 0.95 1.00

i) Short uniaxial - chartii) Short biaxial Mx/h1 ≥ My/b’M’x = Mx + β h’/b’ My.Or else M’y = My + βb’/h’. Mx

N/fcubh 0 0.1 0.2 0.3 0.4 0.5 ≥0.6β 1.00 0.88 0.77 0.65 0.53 0.42 0.30

Slenderness. A slender column is one which Le/b or Le/h › 15Mt = Mi + Ne add

Bending about minorMt = Mi + MaddMt = M2

Mt = M1 + ½ Madd

Mi = 0.4M1 + 0.6M2 ≥ 0.4 M2

Mt = Nemin emin = 0.05h or 20 mmMadd = Nβa kh.

Le/b 12 15 20 25 30 35βa 0.07 0.11 0.20 0.31 0.45 0.61

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Bending about major axis.h/b < 3 le/h < 20 - as minor axis.Or biaxial with o initial moment about minor axis.Slender biaxial

K= Nu2-N/Nu2 – Nbal <1Nu2 = 0.45fuAc + 0.82fy As

Nbal 0.0035 core

6. PRESTRESSED CONCRETE

Class 1: No tensile stress.Class 2: No. visible cracks - tensile stress limited.Class 3: Cracks limited to between 0.1 mm, and 0.2 mm depend environments-Ordinary elastic beam theory.-Sign convention.(a) Sagging M= +ve, hogging M= -ve(b) Conc stress comp = +ve tensile = -ve.(c) Tendon stress: tensile = +ve(d) Tendon eccentricity: downward +ve, upwards –ve.Force tendons.

(a) 7 wire strands – fpu – 1770N/mm2

(b) Cold drawn wires = fpu = 1570 N/mm2.(c) Hign tensile alloys bars fpu = 1030 N/mm2.

Initial presstress = 70-80% fpu.Diff: pretensioning and post tensioning.Advantages:-

a) Whole concrete section utilized to resist load. Greatly reduced deflections.b) Curved profiles – some of the shear taken by the cables.c) Use of lighter sections – more clearance and longer span to be used.d) No cracks in the beams at all.

STRESS IN SERVICE: ELASTIC THEORYLosses: Effective prestressing force Pe = Pα . α= 0.8f1 = Pe/A + Pees/Z1

f2=Pe/A - Pees/Z2

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f1 – (Mimax + Md)/Z1 ≥ famin

f1 – (Mimin + Md )/Z1 ≥ famax

f2 + (Mimax + Md ) /Z2 ≥ famax

f2 + (Mimin + Md ) /Z2 ≥ famin

Z1 ≥( Mimax – Mimin )/ (famax - famin )Z2

Pe = (f1Z1 + f2Z2)A / Z1 + Z2

es = (f1 – f2) Z1Z2/ (f1Z1 + f2Z2)A

The limits of the permissible tendon zone are in fact given by eqns (9.2-4) to (9.2-7) if f1 and f2 in these equations are expressed in terms of Pe and es . The reader should verify that:

es ≥ ( Mimax + Md) / Pe – Z1/A +Z1famin/Pe

es ≥ ( Mimax + Md) / Pe – Z2/A +Z2famax/Pe

es ≤ ( Mimin + Md) / Pe – Z1/A +Z1famax/Pe

es ≤ ( Mimin + Md) / Pe – Z2/A +Z2famin/Pe

STRESSES AT TRANSFER

f1t – Md / Z1 ≥ fa min tf1t – Md / Z1 ≥ fa max tf2t + Md / Z2 ≤ fa max tf2t + Md / Z2 ≥ fa min tCriticalP= Pe / αfa max t = 0.5 fcc - extreme fibre 0.4 fci – uniform.

fa min t = Class 1 = - 1 N/mm2

Class 2 = - 2.4 N/mm2

LOSS OF PRESTRESS

(a) Relaxation of tendons - 5%(b) Elastic deformation of concrete - 5%(c) Shrinkage and Creep of concrete – 4%(d) Slip of tendons during anchoring(e) Friction between tendon and duct in post tensioned beam . – 3%

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