## Dhj

Radius of gyration, i (-f- ) j ( J 86.6mm Slendemess ratio A 1.77 x 103 86.6 20.4 For a braced column the minimum limiting value of A will be given by NuI(AJ'ca) 1280 x 103 (400 x 300 x 25 1.5) 0.64 thus A,im 26.2 v 0lS4 32.7 (> 20.4) Hence, compared with the minimum limiting value of A the column is short and second order moment effects would not have to be taken into account. Short columns usually fail by crushing but a slender column is liable to fail by buckling. The end moments on a...

## A J

Shear resistance of section Vpi, rj - - ''y For the steel section the web depth, d 407.6 mm and the web thickness I 9 mm. Using the conservative value of Shear resistance of section - Vpi rj - - -- -x 10 From the calculations for bending and shear it can be seen the loading on the beam during construction is relatively low compared to the strength of the beam. Also, the steel decking with the corrugations at right angles to the span gives lateral and torsional restraint to the steel beam. For...

## Wau

Examples of good and poor elevations tied together with continuity reinforcement so that the loading can be redistributed and alternative structural actions may develop if necessary. The principles discussed in section 6,7 are relevant to this. Slabs can provide rigid diaphragms to transfer loads at the roof and each floor Figure 6.21 shows how. in a one-storey building, a rigid horizontal slab or bracing at roof level enables the structure to act as a closed box giving more rigidity and...

## Chapter Introduction

Reinforced concrete is a strong durable building material that can be formed into many varied shapes and sizes ranging from a simple rectangular column, to a slender curved dome or shell. Its utility and versatility are achieved by combining the best features of concrete and steel. Consider some of the widely differing properties of these two materials that are listed below.

## Contents

1 Properties of reinforced concrete 1 1.2 Stress-strain relations 3 1.3 Shrinkage and thermal movement 6 1.6 Specification of materials 11 2.2 Characteristic material strengths and characteristic loads 17 2.3 Partial factors of safety 18 2.4 Combination of actions 23 2.5 Global factor of safety 27 3 Analysis of the structure at the ultimate limit state 28 3.2 Load combinations and patterns 30 3.5 Shear wall structures resisting horizontal loads 48 3.6 Redistribution of moments 53 4 Analysis of...

## Example

Design of shear reinforcement for a beam Shear reinforcement is to be designed for the one-span beam of example 7.6 as shown in figures 7.13 and 7.16. The total ultimate load is 108 kN metre and the characteristic strengths of the concrete and steel are fa 30 N mm and fa 500N mnr. 9 - H8 200 H8 links 350 9-H8 200 300

## General design approach

Although EC7 presents three alternative design approaches the UK National Annex allows for only the lirst of these. In this design approach, two sets of load combinations (referred to as combinations 1 and 2 in table 10.1) must be considered at the ultimate limit state. These two combinations will be used for consideration of both structural failure. STR (excessive deformation, cracking or failure of the structure), and geotechnical failure, GEO (excessive deformation or complete failure of the...

## H

Analysis example - singly reinforced section For equilibrium of the compressive and tensile forces on the section Fee F* therefore 0.567fobs 0.87 yk 4, 0.567 x 25 x 300 x i 0.87 x 500 x 1470 This value of x is less than the value of 0.617d derived from section 4.2, and therefore the steel has yielded and 0.87 U as assumed. Moment of resistance of the section is 0.87 x 500 x 1470(520 - 150 2) x 10 6 284 kNm

## Bhw

Slope length of stairs s (32 + 5 ) 3.35 m Consider a I m width of stairs Weight of waist plus steps (0.14 x 3.35 + 0.26 x 1.5 2)25 16.60 kN Variable load 3.0 x 3 9.0 kN Ultimate load, F 1.35 x 16.60+ 1.5 x 9.0 35.91 kN With no effective end restraint H 35.91x3.0 13 46kNm From the lever-arm curve, figure 4.5. ., - 0.95 (the maximum normally adopted in practice), therefore m z 0.87 x 500 x0.95 x 115 283 mm'm Maximum allowable spacing is 3It 3 x 140 - 420mm with an upper limit of 400mm. Provide...

## Cem

39 4kN 34.0kN passive ,800 400 2200 39 4kN 34.0kN passive ,800 400 2200 Therefore ihe horizontal force on I m length of wall is given by kicanh 0.5 V 0.5 x 27.0 x 4.9 66.1 kN from the active earth pressure and kisur) PJ> 3.3 x 4.9 I6.2kN from the surcharge pressure wall (0.4 4- 0.3) x 4.5 x 25 39.4 base 0.4 x 3.4 x 25 34.0 earth 2.2 x 4.5 x 1700 x 10 3 x 9.81 165.1 surcharge 2.2 x 10 22.0 kN The partial factors of safety as given in table 10.1 will be used. (i) Overturning taking moments...

## Ezl

3* -71-9- -IT- -B-0- Kisi , t_ ffr- _ . _a_ 1 H10-200 H10-400 H10-250 H10-250 From figure 6.3 this corresponds to n basic span-effective depth ratio in excess of 32 x 1.3 (for an end span) 41. The actual ratio 4500 140 32.1 hence the chosen effective depth is acceptable. Similar calculations for the supports and the interior span give the steel areas shown in figure 8.5. At the end supports there is a monolithic connection between the slab and the beam, therefore top steel should be provided...

## Vno

(I) First design solution Estimate of slab depth Try a basic span-depth ratio of 27 (approx. 40 above value from figure 6.3) 27 x correction factors (c.f.) 4500 167 27 x c.f. cZ As high yield steel is being used and the span is less than 7 m ihe correction factors can be taken as unity. Try an effective depth of 170 mm. For a class XC'-l exposure the cover 25 mm. Allowing, say. 5 mm as half the bar diameter of the reinforcing bar overall depth of slab - 170 + 25 f 5 200 mm Slab loading...

## Wck

Note the link spacing is reduced to 0.60 x these values lor 400mm above and below each floor level and at laps below 1 st floor level Cover for the reinforcement is taken as 50 mm and < l' lt - 80 400 0.2. The minimum area of reinforcement allowed in the section is given by As 0.002W 0.002 x 300 x 400 240 mm2 and the reinforcement provided is within these limits. Although EC2 permits the use of 12 mm main steel. 16 mm bars have been used to ensure adequate rigidity of the reinforcing cage. A...

## Hi

Cheek if concrctc section is adequate IKd. mux 7 77 7 TT (see equation 5.31) v, 0.6(1 - A 250) 0.6( I - 30 250) 0.528 Therefore 1.33 x 0.528 x 30 x 100 x 100 x 10 3 Rd' x (2.5+4.0) Therefore the concrete section is adequate. 4. Calculate the additional link reinforcement required to resist torsion. (Note that Ay is for one leg only) 5. Therefore for shear plus torsion and based on the area of two legs For 8 mm stirrups at 125 mm centres Asw s 0.805 (see Appendix table A4) Spacing s 125 mm (<...

## I

9.7 Design of slender columns 275 10 Foundations and retaining walls 281 10.2 Combined footings 291 10.6 Piled foundations 300 10.7 Design of pile caps 304 11 Prestressed concrete 319 11.1 Principles of prestressing 321 11.2 Methods of prestressing 322 11.3 Analysis of concrete section under working loads 324 11.4 Design for the serviceability limit state 329 11.5 Analysis and design at the ultimate limit state 353 12 Composite construction 369 12.1 The design procedure 372 12.2 Design of the...

## Eii

Hence at mid-span the resultant of the tendon force must lie at an eccentricity in the range of 47.8 to 66 mm. Provided that the tendons can be arranged so that their resultant force lies within the calculated limits then the design will be acceptable. If a Magne diagram for the stress condition at mid-span had been drawn, as in example 11.5. then the eccentricity range could have been determined directly from the diagram without further calculation. For tendons with a combined prestress force...

## Fbv

Figure 2.4 illustrates how the factors in table 2,2 and 2.4 can be applied when considering the stability of the office building shown for overturning about point 13. Figure 2.4(a) treats the wind load (Wk) as the leading variable action and the live load ((2t) on 'lie roof as the accompanying variable action. Figure 2.4(b) considers the live load as ihe leading variable action and the wind as the accompanying variable action. 2.4.1 Design values of actions at the ultimate limit state In...

## Fzo

Serviceability, durability and stability requirements 153 6.5 Thermal and shrinkage cracking Thermal and shrinkage effects, and the stresses developed prior to cracking of the concrete, were discussed in chapter 1. The rules for providing minimum areas of reinforcement and limiting bar sizes to control thermal and shrinkage cracking were discussed in sections 6.1.5. 6.1.7 and 6.1.8. In this section, further consideration will be given to the control of such cracking and the calculations that...

## Lbx

B.M Diagram (all loads including the point loads shown) Distribution of shear connectors with concentr loads 12.5 Transverse reinforcement in the concrete flange Transverse reinforcement is required to resist the longitudinal shear in the concrete flange. This shear acts on vertical planes either side of the shear connectors as shown in figure 12.14. Transverse reinforcement i the concrete flanges The analysis and design for the transverse reinforcement to resist the longitudinal shear in a...

## If

Non-continuous beam-she reinforcement (a) Check maximum shear at face of support Maximum design shear wu x effective span 2 - 108 x 6.0 2 324kN Design shear at face of support Vu 324 - 108 x 0.15 308 kN Crushing strehgth VRd mas of diagonal strut, assuming angle 0 22 , cot 9 2.5 is 0.124 x 300 x 540(1 - 30 250) x 30 x 10 3 530 kN (> Ve< , 308 kN) Therefore angle 0 22 and col 0 2.5 as assumed. At distance < 1 from face of support the design shear is Vrj 308 - wucl 308 - 108 x 0.54 250 kN

## Ij

Analysis equations for a singly reinforced section The following equations may be used to calculate the moment of resistance of a given section with a known area of steel reinforcement. For equilibrium of the compressive force in the concrete and the tensile force in the steel in ligure 4.4 0.567fab x s 0.87 viA Therefore depth of stress block is Therefore the moment of resistance of the section is These equations assume the tension reinforcement has yielded, which will be the case if .v <...

## Info

Analysis example with triangular stress block 120 x 106 1 x 3000 x 197 x cc U60 - cc 10.3 N mm2 Froin equation 4.46 4.10.2 Triangular stress block - uncracked section The concrete may be considered to resist a small amount of tension. In this ease a tensile stress resultant acts through the centroid of the triangular stress block in the tension zone as shown in figure 4.31. For equilibrium of the section F 0.5 b h-x and F , Asx s, Taking moments about Fcc, the moment of resistance of the...

## Info Cqp

If .c < (Ii - lip) then the neutral axis is within the flange and the concrete is cracked at service and these equations cannot apply. Deflection at service due to the permanent and variable loads The deflections are calculated for the unfactorcd actions. The second moment of area of the composite section is used in the calculations. For the unpropped case the total deflection is tolal constr f* composite corwr is the deflection of the steel beam due to the permanent load at construction...

## Its

These are used in conjunction with permissible stresses appropriate to the type of member and covering the following conditions 1. Initial transfer of prestress force with the associated loading (often just the beam's self-weight) 2. At service, after prestress losses, with minimum and maximum characteristic loading 3. At service with the quasi-permanent loading. The loadings must encompass the full range that the member will encounter during its life, and the minimum values...

## Iv

The points of conlrallexure occur at M 0, that is Vab* - + WAB 0 where x the distance from support A. Taking the roots of this equation gives _ VAB J VAH2 + 2U',WAB) A similar analysis can be applied to beams that do not support a uniformly distributed load. In manual calculations it is usually not considered necessary to calculate the distances u . < h and < u which locate the points of conlrallexure and maximum moment -a sketch of the bending moment is often adequate - but if a computer...

## J

The use of partial factors of safety on materials and actions offers considerable flexibility, which may be used to allow lor special conditions such as very high standards of construction and control or. at the other extreme, where structural failure would be particularly disastrous. The global factor of safety against a particular type of failure may be obtained by multiplying the appropriate partial factors of safety. For instance, a beam failure caused by yielding of tensile reinforcement...

## Ass

Continuous beams with approximately equal spans and uniform loading The ultimate bending moments and shearing forces in continuous beams of three or more approximately equal spans without cantilevers can be obtained using relevant coefficients provided that the spans differ by no more than 15 per cent of the longest span, that the loading is uniform, and that ihe characteristic variable action does not exceed the characteristic permanent action. The values of these coefficients are shown in...

## Dpj

The methods of analysis for continuous beams may also be applied to continuous slabs which span in one direction. A continuous beam is considered to have no fixity with the supports so that the beam is free to rotate. This assumption is not strictly true for beams framing into columns and for that type of continuous beam it is more accurate to analyse them as part of a frame, as described in section 3.4. A continuous beam should be analysed for the loading arrangements which give the maximum...

## Enl

4.6.2 Flanged section - the depth of the stress block extends below the flange, s > hf For the design of a flanged section, the procedure described in section 4.6.1 will check if the depth of the stress block extends below the flange. An alternative procedure is to calculate the moment of resistance. Mr, of the section with s lif, the depth of the - ge (see equation 4.22 of example 4.6 following). Hence if the design moment. MA, is jch that .n the stress block must extend below the flange,...

## Jdr

Beam doubly reinforced to resist a hogging moment 0.87 yk(d - d') (165 x I0fi - 0.167 x 25 x 230 x 3302) Provide two H20 bars for A , area 628 mm2, bottom steel. Provide three H25 bars for area 1470 mm2, top steel. 4. Check equation 7.5 for the areas of steel required and provided for the compression and tension reinforcement to ensure ductility of the section ( s.prov s.req) (As.prov s.rcq) 628 - 496 ( 132) > 1470 -1384 ( - 86) mm2 5. The bar areas provided arc within the upper and lower...

## Qex

7.2.2 Rectangular sections with tension and compression reinforcement, no moment redistribution Compression steel is required whenever the concrete in compression, by itself, is unable to develop the necessary moment of resistance. Design charts such as the one in figure 4.9 may be used to determine the steel areas but the simplified equations based on the equivalent rectangular stress block are quick to apply. The arrangement of the reinforcement to resist a sagging moment is shown in figure...

## Stk

7.9 Curtailment and anchorage of reinforcing bars As the magnitude of the bending moment on a beam decreases along its length so may the area of bending reinforcement be reduced by curtailing the bars since they are no longer required, as shown in figure 7.25. It should be recognised though that because of the approximations and assumptions made for the loading, the structural analysis and the behaviour of the reinforced concrete, the curtailment cannot be a particularly precise procedure. In...

## Vbt

( EXAMPLE 4.4 Analysis of a doubly reinforced rectangular section Determine the ultimate moment of resistance of the cross-section shown in figure 4.11 given that the characteristic strengths are yk 500N mnr for the reinforcement and ,k 25N mnr for the concrete. For equilibrium of the tensile and compressive forces on the section Assuming initially that the steel stresses sl and fx are the design yield values, then 0.87 ykAs 0.567 ck + 0.87 yk< Analysis example, doubly reinforced section 0.87...

## Vkr

Design a waffle slab for an internal panel of a floor system, each panel spanning 6.0 m in each direction. The characteristic material strengths are fa 25 N mnr and yk 500N mm The section as used in example 8.8, figure 8.16 is to be tried with characteristic permanent load including self-weight of 6.0 kN nr and characteristic variable load of 2.5 kN nr. (1.35 x 6.0) + (1.5 x 2.5) 11.85 kN nr As the slab has the same span in each direction the moment coefficients. 3m, Jsy arc taken from tabic...

## Wxy

4.6 Flanged section in bending at the ultimate limit state - r singly reinforced section it is necessary to consider two conditions 4 re - ress block lies within the compression flange, and X 1. block extends below the flange. - Flanged section - the depth of the stress block lies within the s < hf (figure 4.12) .pth of stress block, the beam can be considered as an equivalent rectangular breadth b equal to the flange width. This is because the non-rectangular ciosv (lie neutral axis is in...

## Yuj

Note d is the distance between the fillets of the steel section It is important to note in the figures that the stress block for the concrete extends to the depth of the neutral axis as specified in EC4 for composite design. There are three possible locations of the neutral axis as shown in figure 12.8. These are (a) The neutral axis in the concrete flange (b) The neutral axis in the steel flange (c) The neutral axis in the steel web.

## Cqy

260 x 4902 x 25 0.146 > Km 0.116 Therefore compression steel is required. 0.87 fyv(d-d') _ (0.146 - 0.116)25 x 260 x 4902 Using the UK Annex of EC2 and applying the equations developed in section 4.7 > m the UK Annex of EC2 clause 5.5 k 0.4 and 1.0 Km 0.454((5 - kl) k2 - 0.182 (< 5 - k ) fc2f 0.454(0.8 - 0.4) 1.0 - 0.182 (0.8 - 0.4) 1,0 2 0.182 - 0.029 0.153 i- nich agrees with the value given in table 4.2. M 260 x 4902 x 25 0.147 < Kbal 0.153 erefore compression steel is not required,...

## Wgy

End deflection 01 (if a 1 then k 0.25) Although the derivation has been on the basis of an uncracked section, the linal expression is in a form that will deal with a cracked section simply by the substitution of the appropriate curvature. Since the expression involves the square of the span, it is important that the true effective span as defined in chapter 7 is used, particularly in the case of cantilevers. Deflections of cantilevers may also be increased by rotation of the supporting member,...

## L

The beam should not be too narrow if it is much less than 200 mm wide there may be difficulty in providing adequate side cover and space for the reinforcing bars. Suitable dimensions for b and d can be decided by a few trial calculations as follows 1. For no compression reinforcement Km 0.167 for dt < C50 With compression reinforcement it can be shown that approximately, if the area of bending reinforcement is not to be excessive. 2. The maximum design shear force VKd.max should not...

## M

143 x I0ft 230 x d1 x 25 Rearranging, d > 386 mm. Assume a concrete cover of 25 mm to the reinforcing steel. So for 10 mm links and. say, 32 mm bars Overall beam depth h d + 25 + 10 + 32 2 d + 51 Therefore make h 525 mm as an integer number of brick courses. So that d 525 - 51 474 mm Maximum shear resistance is 0.18 x 230 x 474 x (I - 25 250) x 25 x 10 3 446 kN > Vru 107 kN Basic span-effective depth 8.35 < 20 (for a lightly stressed beam in C25

## Mhmhh

On a sloping site the foundations should be constructed on a horizontal bearing and stepped where necessary. At the steps the footings should be lapped as shown in figure 10.12. The footings are analysed and designed as an inverted continuous beam subjected to the ground bearing pressures. With a thick rigid footing and a firm soil, a linear distribution of bearing pressure is considered. If the columns are equally spaced and equally loaded the pressure is uniformly distributed but if the...

## N

Maximum pressure p ---(10.4)* A typical arrangement of the reinforcement in a pad footing is shown in figure 10.4. With a square base the reinforcement to resist bending should be distributed uniformly across the full width of the footing. For a rectangular base the reinforcement in the short direction should be distributed with a closer spacing in the region under and near the column, to allow for the fact that the transverse moments must be greater nearer the column. It is recommended that at...

## P

In cases where the basic ratio has been modified for spans greater ilian 7 m, maximurr deflections are unlikely to cxceed span 500 after construction of partitions and finishes When another deflection limit is required, the ratios given should be multiplied b 500 where a is the proposed maximum deflection.

## R

Calculate the stability ties required in an eight-storey building of plan area shown in figure 6.17 Clear storey height under beams 2.9 m Characteristic permanent load 6kN nr Characteristic variable load (qy) 3kN nr Characteristic steel strength (f ) 500N mtrr F, (20 + 4 x number of storeys) 20 + 4 x 8 52 kN < 60 kN (a) Peripheral ties Force to be resisted F, 52 kN

## T

Prestress force after losses 2590 kN 145 106 x 106 mm4 A 500 x 103 mm2 Ap 3450 mm2 yk 500 N mnr for the shear links ctk 2.2N mnr The calculations will he presented for a section at the support and then repeated and tabulated at 3 m intervals along the span. (1) Calculate shear force at the section Although the maximum shear force can be taken at the face of the support, in this example we will, for illustrative purposes, take the section at the middle of the support itself. Hence (2) Check if...

## U E

8.5.1 Simply supported slab spanning in two directions A slab simply supported on its four sides will deflect about both axes under load and the comers will tend to lift and curl up from the supports, causing torsional moments. When no provision has been made to prevent this lifting or to resist the torsion then the moment coefficients of table 8.4 may be used and the maximum moments are given b MiX and Msv are the moments at mid-span on strips of unit w idth with spans x and I. respectively n...

## Ull

(i) Calculate the neutral axis depth of the cracked section Taking moments about the neutral axis (ii) Calculate the stress in the tension steel, rrs Taking moments about the level of the compressive force in the concrete trs M (d x 3)A, 650 x l()h (930 - 45773)3770 222N mm2 A', 0.4 assuming long-term loading fa,at dm (from table 6.11) 2.6N mnr Es 200 ** 2.5(1000-930)400 0' 53S> e 222 - '4x ('+645 x 00539),Q6 222 sm cm - -i 0.0

## W

According to EC2, kt 0.44 and kj 1.25 0 10 15 20 25 30 According to EC2, UK Annex, k, - 0.4 and k2 1.0 0 10 15 20 25 30b Maximum permitted redistribution (or class A normal ductility steel Maximum permitted redistribution (or class B and C higher ductility steel, see section 1,6.2 Iii this chapter the examples will be based on the UK Annex's equation 7.6b. but. because many of the designs in the UK arc for projects overseas which may require the use of the 1 C2 specifications, example 4.9 part...

## Lbc

Columns and their connections to beams are critical parts of a structure. Failure of a column in a building can be catastrophic leading to a progressive collapse, and the formation of plastic hinges in columns above the base of a building should be avoided. Horizontal hoops of helical reinforcing bars have been found to give a stronger containment to the longitudinal vertical bars than that provided by rectangular links and at a beam-to-column joint horizontal steel reinforcement hoops not less...

## Y

Where i'y and e, are the first-order eccentrieitics in the direction of the section dimensions b and h respectively. Where these conditions are not fulfilled biaxial bending must be accounted for and EC2 presents an interaction equation, relating the moments about the two axes to the moment of resistance about the two axes, which must be satisfied. However, the given formula cannot be used directly to design a column subject to biaxial bending but rather to check it once designed. In the...

## Zibs

Shear walls with openings can be idealised into equivalent plane frames as shown in figure 3.25. In the plane frame the second moment of area c of the columns is equivalent to that of the wall on either side of the openings. The second moment of area i, of the beams is equivalent to that part of the wall between the openings. The lengths of beam that extend beyond the openings as shown shaded in figure 3.25 are given a very large stiffnesses so that their second moment of area would be say The...

## Nonrectangular sections

Design charts are not usually available for columns of other than a rectangular or a circular cross-section. Therefore the design of a non-rectangular section entails either < I) an iterative solution of design equations. (2) a simplified form of design, or (3) construction of M-N interaction diagrams. For a non-rectangular section it is much simpler to consider the equivalent rectangular stress block. Determination of the reinforcement areas follows the same procedure as described for a...

## Ct

Note All slab reinforcement perpendicular to a free edge transferring moment to the column should be concentrated within the width be The reinforcement for a flat slab should generally be arranged according to the rules illustrated in ligure 8.2, but at least 2 bottom bars in each orthogonal direction should pass through internal columns to enhance robustness. Important features in the design of the slabs are the calculations for punching shear at the head of the columns and at the change in...

## M D

So that for p2 always to be positive. M N - or the effective eccentricity, e - must never be greater than D 6. In these cases the eccentricity of loading is said to lie within the 'middle third' of the base. 3. When the eccentricity, e is greater than D 6 there is no longer a positive pressure along the length D and the pressure diagram is triangular as shown in figure 10.3(c). Balancing the downward load and the upward pressures

## A P

- - 0.0017 ( o. 17 < 2 ) hence from table 8.2 iRt ,c 0.4 N mnr. Therefore the shear resistance of the concrete, VR() c is given by VVc - VRU.Cud -40 x 8134 x 520 x 10 3 1691 kN o VEd 626 kN) 7. Maximum Shear Force - see figure 10.7(b) At the critical section for shear. 1IW from the column face Design shear V& , 239 x 2.8 x 0.68 455 kN 0.40 x 2800 x 520 x 10_i 582 kN (> Vm 455 kN) Therefore no shear reinforcement is required. Instead of assuming a footing weight of 150 kN at the start of...

## Notation

Notation is generally in accordance with EC2 and the principal symbols are listed below. Other symbols arc defined in the text where necessary. The symbols e for strain and for stress have been adopted throughout, with the general system of subscripts such that the first subscript refers to the material, c - concrete, s - steel, and the second subscript refers to the type of stress, c - compression, t - tension. (I effective depth of tension reinforcement d' depth to compression reinforcement h...

## Kj

The location of the neutral axis is determined from the equilibrium equation of the resistance forces R at the section. The moment of resistance at the section is then obtained by taking moments about a convenient axis such as the centreline of the steel section, so that where z is the lever arm about a chosen axis for the resistance R. For cases (b) and (c) the analysis is facilitated by considering an equivalent system of the resistance forces as shown in the relevant diagrams. (a) Neutral...

## JL x

BcPfa 400 x 160- x 25 From the lever-arm curve, figure 4.5 l 0.95. Thus * 0.87 yiiC 0.87 x 500 x 0.95 x 160 mm Provide two H10 bars in each 0.4 m width of slab. A, 157 mm2. 3. At the section where the ribs terminate the maximum hogging moment of resistance of the concrete ribs is 13.36 kN m. as in the previous example. This is greater than the moment at this section, therefore compression steel is not required. At mid-span p 4< j(j x 760 hence from figure 6.3, limiting basic span depth ratio...

## Short columns resisting moments and axial forces

The area of longitudinal steel for these columns is determined by 1. using design charts or constructing W-A' interaction diagrams as in chapter 4. 2. a solution of the basic design equations, or Design charts are usually used for columns having a rectangular or circular cross-section and a symmetrical arrangement of reinforcement, but interaction diagrams can be constructed for any arrangement of cross-section as illustrated in examples 4.10 and 4.11. The basic equations or the approximate...

## Qwx

Hence equaling internal and external work, the maximum u.d.l. that the slab can sustain is given by 2,2 VK 6 12( ii 2 + '2 2) It is clear that the result will vary according (o the value of 3. The maximum value of ii' may be obtained by trial and error using several values of 3, or alternatively, b> differentiation, lei mi . 'i- 'hen 0 will give the critical value of 3 A negative value is impossible, hence the critical value of 3 for use in the analysis is given by the positive root. This is...

## S

The purpose of design is to achieve acceptable probabilities that a structure will not become unfit for its intended use - that is, that it will not reach a limit stale. Thus, any way in which a structure may cease to be lit for use will constitute a limit state and the design aim is to avoid any such condition being reached during the expected life of the structure. The two principal types of limit state are the ultimate limit state and the serviceability limit slate. This requires that the...

## Pad footings

The footing for a single column may be made square in plan, but where there is a large moment acting about one axis it may be more economical to have a rectangular base. Assuming there is a linear distribution the bearing pressures across the base will lake one of the three forms shown in figure 10.3, according to the relative magnitudes of the axial load N and the moment M acting on the base. 1. In figure 10.3(a) there is no moment and the pressure is uniform N 2. With a moment M acting as...

## N M

As with foundations, the design of bending and shear reinforcement is based on an analysis of the loads for the ultimate limit state (STR). with the corresponding bearing pressures. Gravity walls will seldom require bending or shear steel, while the walls in counterfort and cantilever construction will be designed as slabs. The design of counterforts will generally be similar to that of a cantilever beam unless (hey arc massive. With a cantilever-type retaining wall the stem is designed to...

## A C

Commonly used for reinforcement, may behave in a similar manner or may, on the other hand, not have such a definite yield point but may show a more gradual change from elastic to plastic behaviour and reduced ductility depending on the manufacturing process. All materials have a similar slope of the elastic region with elastic modulus 200 kN mnr approximately. The specified strength used in design is based on either the yield stress or a specified proof stress. A 0.2 per cent proof stress is...

## Mvx

6.6 Other serviceability requirements The two principal other serviceability considerations are those of durability and resistance to fire, although occasionally a situation arises in which some other factor may be of importance to ensure the proper performance of a structural member in service. This may include fatigue due to moving loads or machinery, or specific thermal and sound insulation properties. The methods of dealing with such requirements may range from the use of reduced working...

## Tre

From the Magnel diagram of figure 11.13 it can be seen that for any chosen value of prestress force there is an eccentricity range within which the resultant tendon force must lie. As the force approaches a value corresponding to the top and bottom limits of the diagram the width of the available cable zone diminishes until at the very extremities the upper and lower limits of eccentricity coincide, giving zero width of cable zone. Practically, therefore, a prestress force will be chosen which...

## F

The possibility of hogging moments in any of the spans should not be ignored, even if it is not indicated by these coefficients. For example, a beam of three equal spans may-have a hogging moment in the centre span if Qk exceeds 0.45Gk. In situ reinforced concrete structures behave as rigid frames, and should be analysed as such. They can be analysed as a complete space frame or be divided into a series of plane frames. Bridge deck-type structures can be analysed as an equivalent grillage,...

## Under Which Conditions A Corbel Should Be Designed As Cantilever Beam

The effective span of a cantilever is either (a) the length to the face of the support plus half the beam's overall depth. It or (b) the distance to the centre of the support if the beam is continuous. The moments, shears and deflections for a cantilever beam are substantially greater than those for a beam that is supported at both ends with an equivalent load. Also the moments in a cantilever can never be redistributed to other parts of the structure - the beam must always be capable of...

## Wyw

Good bond conditions in unhatched zone Poor bond conditions in hatched zone Definition of good and poor bond conditions Good bond conditions are considered to be when (a) bars are inclined at an angle of between 45 and 90 to the horizontal or (b) zero to 45 provided that in this second case additional requirements are met. These additional conditions are that bars are 1. either placed in members whose depth in the direction of concreting does not exceed 250 mm or 2. embedded in members with a...

## Uat

A soils survey of a proposed site should be earned out to determine the depth to firm soil and the properties of the soil. This information will provide a guide to the lengths of pile required and the probable safe load capacity of the piles. On a large contract the safe loads are often determined from full-scale load tests on typical piles or groups of piles. Willi driven piles the safe load can be calculated from equations which relate the resistance of the pile to the measured set per blow...

## Column classification and failure modes

The slenderness ratio A of a column benl about an axis is given by 0 is the effective height of the column is the radius of gyration about the axis considered I is the second moment of area of the section about the axis is the cross-sectional area of the column The effective height of a column. . is the height of a theoretical column of equivalent section but pinned at both ends. This depends on the degree of fixity at each end of the column, which itself depends on the relative stiffnesses of...

## Loading and moments

The loading arrangements and the analysis of a structural frame have been described with examples in chapter 3. in the analysis it was necessary to classify the structure into one of the following types 1. braced - where the lateral loads are resisted by shear walls or other forms of bracing capable of transmitting all horizontal loading to the foundations, and 2. unbraced - where horizontal loads arc resisted by the frame action of rigidly connected columns, beams and slabs. With a braced...

## Ukri

Where ', is the perpendicular distance between the axis of each wall and the centre of rotation. Determine the distribution of the lOOkN horizontal force F into the shear walls A, B, C. D and E as shown in figure 3.24. The relative stiffness of each shear wall is shown in the figure in tenns of multiples of k. Unsymmetrical arrangement of shear walls 20 5 5 30 Taking moments for kx about YY at wall A E M _ 20 x 0 5 x 32 5 x 40 v E 30 12.0 metres X gt , 6 4 10 Taking moments for ky about XX at...

## Biaxial bending of short columns

For most columns, biaxial bending will not govern the design. The loading patterns necessary to cause biaxial bending in a building's internal and edge columns will not usually cause large moments in both directions. Corner columns may have to resist significant bending about both axes, but the axial loads are usually small and a design similar to the adjacent edge columns is generally adequate. A design for biaxial bending based on a rigorous analysis of the cross-section and the strain and...

## Why Is Strut Angle Between 22 Degrees And 45 Degrees Used In Shear Design In Ec2

0.5 sin 20 see proof in the Appendix which alternatively can be expressed as fl O-Ssin-'i ' 1 lt 45 5.8b where Vnr is the shear force at the face of the support and the calculated value of the angle 9 can then be used to determine col 9 and calculate the shear reinforcement AiW s from equation 5.9 below when 22 lt lt 45 . 2 The vertical shear reinforcement As previously noted, all shear will be resisted by the provision of links with no direct contribution from the .shear capacity of the...

## Reinforced Concrete

Bill Mosley, John Bungey and Ray H lse Other titles of interest to civil engineers Civil Engineering Materials, fifth edition edited by n. iackson amp r. k. dhir Civil Engineering Quantities, sixth edition l.H. seeley Design of Structural Elements W. M.C. McKenzie Design of Structural Timber to EC5, second edition w. M. c. Mckenzie amp b. zhang Design of Structural Masonry w. m. c. Mckenzie Design of Structural Steelwork w. m.c. Mckenzie Engineering Hydrology, fourth edition e. m. wilson...

## Hkf

Curtailment described in section 7.9. The application of these nilcs establishes the cutoff points beyond which the bars must extend at least a curtailment anchorage length. It should be noted that at the external columns the reinforcement has been bent to give a full anchorage bond length. The shear-force envelope and the arrangement of the shear reinforcement for the same continuous beam are shown Is figure 7.18. On the shear-force envelope the resistance of the minimum stirrups has been...

## Ray Hulse

Formerly faculty of engineering and computing c W.H. Hosley and J.H. Bungey 1976, 1982, 1987, 1990 i W.H. Mosley, J.H. Bungey and R. Hulse 1999, 2007 All rights reserved. No reproduction, copy or transmission of this publication may be made without written permission. No paragraph of this publication may be reproduced, copied or transmitted save with written permission or in accordance with the provisions of the Copyright, Designs and Patents Act 1988, or under the terms of any licence...

## 1

As the coefficients of thermal expansion of steel and concrete a-r.s and qt,c are similar, differential movement between the steel and concrete will only be very small and is unlikely to cause cracking. The differential thermal strain due to a temperature change T may be calculated as and should be added to the shrinkage strain lt S if significant. The overall thermal contraction of concrete is, however, frequently effective in producing the first crack in a restrained member, since the...

## Yield line and strip methods

For cases which are more complex as a result of shape, support conditions, the presence i openings, or loading conditions it may be worthwhile adopting an ultimate analysis ethod. The two principal approaches are the yield line method, which is particularly litable for slabs with a complex shape or concentrated loading, and the strip method hich is valuable where the slab contains openings. These methods have been the subject of research, and are well documented although hey are of a relatively...

## Ribbed and hollow block floors

Cross-sections through a ribbed and hollow block floor slab are shown in ligure 8.15. The ribbed floor is formed using temporary or permanent shuttering while the hollow block floor is generally constructed with blocks made of clay tile or with concrete containing a lightweight aggregate. If the blocks are suitably manufactured and have adequate strength they can be considered to contribute to the strength of the slab in the design calculations, but in many designs no such allowance is made....

## Ksg

So that the roof's external connection Ma 0.54 x - x 60 1.6 kN m As a check at each joint, Mr Y Mq. The bending moments due to characteristic wind loads in all the columns and beams of this structure are shown in figure 3.21. 3.5 Shear wall structures resisting horizontal loads A reinforced concrete structure with shear walls is shown in figure 3.22 . Shear walls are very effective in resisting horizontal loads such as Fz in the figure which act in the direction of the plane of the walls. As...

## Stair slabs

The usual form of stairs can be classified into two types 1 those spanning horizontally in the transverse direction, and 2 those spanning longitudinally. Stairs of this type may be supported on both sides or they may be cantilevered from a supporting wall. Figure 8.17 shows a stair supported on one side by a wall and on the other by a stringer beam. Each step is usually designed as having a breadth b and an effective depth of d D 2 as shown in the ligurc a more rigorous analysis of the section...

## Concrete Drop Panel

Moments in a continuous two-way slab The panel considered is an edge panel, as shown in figure 8.9 and he uniformly distributed load, n 1.35 k I.5i k 10kN nr. The moment coefficients are taken from table 8.5. Ma lianl 0.045 x 10 x 52 11.25 kN m in direction , WS gt . 0.028 x 10 x 52 7.0 kN m in direction ly Support ad. A x 0.059 x 10 x 52 14.75 kN m Supports ab and dc. ,Wy 0.037 x 10 x 52 9.25 kN m The moments calculated are for a metre width of slab. The design of reinforcement to resist these...

## Lan

Strip footing with bending reinforcement In tlie transverse direction the maximum moment can he calculated on the assumption that the 2.2 m wide looting is acting as a 1.1 ni long cantilever for the purposes of calculating the design moment 0.87 x 500 x 0.95 x 720 497 mm m Minimum As 0.15 x 1000 x 1080 mnr m Provide H20 bars at 250mm centres, area 1260 mnr m. bottom steel. 5. Normal shear will govern as the punching perimeter is outside the fooling. The critical section for shear is taken l.fW...

## Slx

b Design of main tension reinforcement From equation 10.8, the required area of reinforcement in each truss is _ T 2 N x 's 0.87 yk 4d x 0.87 yk _ 5000 x 103 x 1350 2 4 x 875 x 0.87 x 500 2216mm2 The total area of reinforcement required in each direction 2 x A, 2x2216 4432 mm2. As the piles are spaced at three times ilic pile diameter this reinforcement may be distributed uniformly across the section. Hence provide fifteen H20 bars, area 4710 mm2, at 140 mm centres in both directions Shear...

## Column Design Charts

These equations ure not suitable for direct solution and the design of a column with symmetrical reinforcement in each face is best carried out using design charts as illustrated in figure 9.8. Sets of these charts can be found in the Concise Eurocode rcf. 21 , the Manual for the Design of Concrete Structures rcf. 23 and the website www.eurocode2.info. f EXAMPLE 9.2 Column design using design charts Figure 9.9 shows a frame of a heavily loaded industrial structure for which the centre columns...

## Preface

The purpose of this book is to provide a straightforward introduction to the principles and methods of design for concrelc structures. It is directed primarily at students and young engineers who require an understanding of the basic theory and a concise guide to design procedures. Although the detailed design methods are generally according to European Standards Eurocodes , much of the theory and practice is of a fundamental nature and should, therefore, be useful to engineers in countries...

## Concrete Interaction Diagram

M-N interaction diagram or a non-symmetrical section Non-rectangular M N interaction example EXAMPLE 4.11 M-N interaction diagram for a non-rectangular section Construct the interaction diagram for the equilateral triangular column section in figure 4.23 with i 25N mm2 and gt k 500N mm2. The bending is about an axis parallel to the side AA and causes maximum compression on the corner adjacent to the steel area A'. Non-rectangular M N interaction example

## Pile Cap Truss Analogy

Therefore, substituting for y and .vn P, 166.7 - 35.4 x 1.67 33.3 x 1.0 - 140.9kN P2 166.7 - 35.4 x 1.67 - 33.3 x 1.0 74.3 kN Pi 166.7 35.4 x 0.33 33.3 x 1.0 211.7 kN P4 166.7 35.4 x 0.33 - 33.3 x 1.0 145.1 kN Ps 166.7 35.4 x 1.33 33.3 x 1.0 247.1 kN P6 166.7 35.4 x 1.33 - 33.3 x 1.0 180.5 kN When a pile group is unsymmetrieal about both co-ordinate axes it is necessary to consider the theory of bending about the principal axes which is dealt with in most textbooks on strength of materials. In...

## Combined footings

Where two columns are close together it is sometimes necessary or convenient to combine their footings to form a continuous base. The dimensions of the footing should he chosen so that the resultant load passes through the centroid of the base area. This may be assumed to give a uniform bearing pressure under the footing and help to prevent differential settlement. For most structures the ratios of permanent and variable loads carried by each column are similar so that if the resultant passes...

## Torsional Reinforcement In Euro Code

c Forces acting on one lace of the section c Forces acting on one lace of the section As q is the shear force per unit length of the circumference of the box section, the force produced by the shear How is the product of q and the circumference 0 of the area Ak. Hence, if it assumed that this force is resisted by the truss action of the concrete compressive struts acting at an angle. 0. together with tension in the longitudinal steel, from figure 5.16b the force F in the longitudinal tension...

## Kqm

He the compressive force developed in the concrete and acting through ti. centroid of the stress block l c be the compressive force in the reinforcement area A' and acting through centroid l be the tensile or compressive force in the reinforcement area and actin through its centroid. i Basic equations and design charts The applied force N must be balanced by the forces developed within the cros-section, therefore In this equation, h will be negative whenever the position of the neutral axis is...

## Wlg

4.6.3 Flanged section with compression reinforcement With x 0.45f in figure 4.16 and taking moments about A,, the maximum resistance moment of the concrete is 0.16 Ub d2 0.567 ck fcf - bw d - if 2 4.24 Note that the value of 0.167 was derived in equation 4.10 for the rectangular section. e applied design moment. W gt A bai- compression reinforcement is required. In tte e the area of compression steel can be calculated from 4id . msidering the equilibrium of forces on the section r - cl Fc2 F.C...

## Hhmmm

Common practice to use a raft foundation in conjunction with a more flexible superstructure. The simplest type of raft is a flat slab of uniform thickness supporting the columns. Where punching shears are large the columns may be provided with a pedestal at the base as shown in figure 10.15. The pedestal serves a similar function to the drop panel in a fiat slab floor. Other, more heavily loaded rafts require the foundation to be strengthened by beams to form a ribbed construction. The beams...

## Ireircri ooac

From table 6.11 the cracking strength of the concrete, fam, is given as 2.6 N mnr. Hence from elastic bending theory and considering the uncrackcd concrete section, the moment that will just cause cracking of the section. M , is given by 2.6 x 300 x 7002 6 x 10 6 63.7 kNm i I-P MJM 2 I - 0.5 x 63.7 200 2 0.95 ii Calculate the 'average' curvature 0.95 x 3.08 x Itr6 I - 0.95 x 2.86 x 10 fi 3.07 x I0-6 mm S Ai d-x 2450 600 - 329 664 x I03 mm3 and from table 6.13. ecs 470 x 10'' because 2AJu 210,...

## Counterfort Wall

Such walls are usually required to resist a combination of earth and hydrostatic loadings. The fundamental requirement is that the wall is capable of holding the retained material in place without undue movement arising from deflection, overturning or sliding. Concrete retaining walls may be considered in terms of three basic categories I gravity. 2 counterfort, and 3 cantilever. Within these groups many common variations exist, for example cantilever walls may have additional supporting ties...

## Concrete Design By Dass Euro Code

Lightly stressed when p equals 0.5 per cent, p is given by l 0Aymi l gt d where i4s,rCq i the area of tension reinforcement required in the section. Interpolation between the values of p indicated is permissible. In the case of slabs it is reasonable to assume lha they are lightly stressed. Since the value of allowable span-effective depth ratio is affected by both reinforcement ratio and concrete strength it may be more convenient to use the char in figure 6.3 which is for a simply supported...

## Moment Capacity

Ultimate moment of resistance 253 x 123 x 10 3 31.1 kNm Untensioned steel is therefore required to permit the beam to support an ultimate moment of 40 kNm. Additional moment capacity to be provided 40-31.1 8.9kNm Effective depth of additional steel 245 mm lever arm to additional steel 220 mm and

## Xf

Freeze-thaw resisting aggregates to be specified. 1. Freeze-thaw resisting aggregates to be specified. a gt cno ib lt w 2 Figure 6.1 a gt cno ib lt w 2 Figure 6.1 6.1.2 Minimum member dimensions and cover fire resistance In order that a reinforced concrete member is capable of withstanding Hre for a specified period of time, it is necessary to ensure the provision of minimum dimensions and cover here defined as nominal minimum concrete surface to main bar axis dimension as illustrated in...