Prof. Dias Graded Examples in Reinforced Concrete Design Dias

REINFORCED CONCRETE DESIGN

W.P.s. Dias BSc{Eng), PhD(Lond),DIC, pEng, MIStructE, MIE(SL) Senior Lecturer Department of Civil Engineering University of Moratuwa Moratuwa Sri Lanka

/

Society of Structural Engineers - Sri Lanka

.,. "'~

.... _._.J

PublisheJI. by

ji

Soc. of Structural Engineers - Sri Lanka, Colombo, Sri Lanka, 1995. ,

ISBN 955-9347-00-4

FOREWORD

The Society of Structural Engineers - Sril;-anka was incorporated in July 1993 .

.

Our membership is very small and our fmancial resources are absolutely minimal. Nevertheless, the members of our Committee have contributed a great deal of their time and effort to collect funds from various sources to help advance the knowledge and practice of structural engineering in Sri Lanka through, inter alia, the publication of books on related topics. As the majority of structures in this country are constructed of reinforced concrete, the selection of GRADED EXAMPLES IN REINFORCED CONCRETE DESIGN as the object of the Society's first book publishing effort constitutes an ideal beginning. Dr Priyan Dias is a brilliant young academic and is highly motivated towards training engineers to use a "thinking" approach to solve technical problems. Whilst this book itself is of an immediately practical nature, Dr Dias and others will, no doubt, follow up with more publications which will help our engineers to think laterally so as to come up with innovative solutions to any structural problems they encounter.

I

A.C. Visvalingam MA, PhD, DIC, MICE, MIStructE, MIE(SL), CEng PRESIDENT, Society of Structural Engineers - Sri Lanka

2 March 1995

•.

GRADED EXAMPLES IN REINFORCED CONCRETE DESIGN (with explanatory notes, using Grade 25 concrete to BS 8110)

CONTENTS Introduction

1

Analysis of Beam Sections in Flexure (Examples 1 - 4)

5

Design of Beam Sections in Flexure (Examples 5 - 9)

13

Design of Beams for Shear (Examples 10 - 11)

26

Serviceability Checks and Detailing in Beams (Example 12)

31

Design of Slabs (Examples 13 - 17)

38

Design of Columns (Examples 18 - 21)

58

Design of Foundations (Examples 22 - 24)

66

Design of Staircases (Examples 25 - 26)

76

Design of Wall and Corbel (Examples 27 - 28)

83

Design of Beam for Torsion (Example 29)

90

Frame Analysis and Moment Redistribution (Examples 30 - 32)

94

Design for Stability (Example 33)

104

Serviceability Limit State Calculations (Examples 34 -35)

107

INTRODUCTION

A Case for Worked Examples Educational purists may argue that Worked Examples are detrimental to student learning because there is an element of "spoonfeeding" involved. While acknowledging that there is some truth in this argument, the author would like to contend that Worked Examples do have a place in the educational process. Knowledge can be acquired using two broad approaches - i.e. the deductive approach, having its roots in Greek rationality, and the inductive approach, having its roots in Renaissance empiricism. Learning through worked examples is an inductive approach, and both the format and content of this book reflect that approach. The book has been developed through the author's teaching of a course in Reinforced Concrete Design at the University of Moratuwa. The examples are graded, leading from an appreciation of reinforced concrete behaviour, through the design of structural elements, to the analysis of a reinforced concrete structure. The student's understanding of the calculations is deepened by the "Notes on Calculations" while the Introductory and Concluding Notes set each example in a wider context. Hence, in this book, design principles·are reinforced through practice, with guidance from notes. However, this book caimot and should not be used as a "stand alone" text. It must essentially be complementary to another text or series of lectures that teaches design from a deductive approach - i.e. one .which moves students from principles to practice. It can, of course, be used by practising engineers, who already have a grasp of reinforced concrete fundamentals. In order to equip students for real design practice, the book is very· much code based, with extensive references given in the calculations to clauses in BS 8110 (1985) - "Strueturaluse of concrete". This is another reason for the book's usefulness for Practising engineers. The examples cover most of the reinforced concrete elements and stress states dealt with by Part I of BS 8110. In addition, examples are also given for the de3ign for torsion and the calculation of deflection and cracking, dealt with in Part 2 of BS 8110.

Sections of code are referred to by indicating the relevant clause, table Or equation of BS 8110: Part 1. Where clauses, tables, charts or equations from Parts 2 and 3 of BS 8110 are referenced, the relevant Part is also indicated. One very useful feature of BS 8110 is that each table also gives the equation from which its values· are derived. .This is a clear advantage for computerised design, and even hand calculations. Therefore, although the tables have in fact been referred to in the following calculations, very often it is the corresponding equations that have been used.

A Case for Using Lower Grades of Concrete Table 3.4 in Part 1 of BS 8110 (1985) specifies durability by cover and grade, but also indicates cement contents and water/cement ratios correspondingro the grade specified. The background to this table is given in the paper by Deacon and Dewar ("Concrete durability 1

\,

- specifying more simply and surely by strength. Concrete, February 1982, pp.19-2l), which describes how U.K. concrete strengths vary for given cement contents and water/cement ratios and shows how the grade specified covers the cement content and water/cement ratio requirements 96% of the time.

It must be emphasised here that the index of durability used in BS 8110 is mix proportions. However, it has related these mix proportions to strength, which is a much easier parameter to measure and control. This is clearly evident in the provisions made in the code for reducing the grade if a checking regime establishes that a lower grade of concrete complies with the cement content and water/cement ratio limits (Clauses 3.3.5.2 and 3.3.5.3 of Part 1). Such a relaxation of grade is not allowed, however for concretes using blended cements. Even a cursory glance at Table 3.4 in BS 8110: Part 1 will indicate that at least grade 40 concrete will have to be used for all but mild and moderate exposure conditions, although the corresponding minimum cement content and maximum water/cement ratio are only 325kg/m3 and 0.55 respectively. This seems to be a very stringent condition to be imposed on concreting practice in developing countries, where most concrete specified is still grade 20 to 25. In fact, even in the U.K., the most commonly used grades were grades 20 to 30, even up to the early 19805. The question arises as to whether Table 3.4 in BS 8110: Part 1, developed for the U.K. is applicable in other (especially developing) countries, where materials and practices may be very different. This problem was studied by the author using Sri Lanka as a case in point. The strengths that could be achieved for various cement content and water/cement ratio values were obtained on the basis of a batching plant survey. Specifications based on the above survey are given in TABLE 1. This table is taken from the author's publication "Specifying for Concrete Durability: Part II - The Sri Lankan Context, Engineer, Vol. XX, Nos 1-4, 1992, pp. 4-14". The Notes in TABLE 1 indicate the scope of the specifications, and also conditions under which deviations from the tabulated values can be allowed. In particular, Notes 5 and 6 allow reductions in grade and cover values that bring these recommendations in line with current Sri Lankan practice. In short, these recommendations rationalise satisfactory Sri Lankan practice (especially under mild exposure conditions) with respect to BS 8110, while suggesting improvements to Sri Lankan practice where problem areas (such as concrete exposed to sea spray) are concerned. Although the recommendations in TABLE 1 make it possible to use grade 20 concrete for mild exposure conditions, it was felt that basing the examples on such a low grade would have deviated too much from the provisions of BS 8110, where grade 25 is specified as the lowest grade to be used with normal weight aggregate concrete (Clause 3.1.7.2) and where all tables and charts have grade 25 as the lowest grade. As such, it is grade 25 concrete that is used for all the following examples, except in Examples 28 and 29, where the use of grade 30 concrete is illustrated.

2

TABLE 1 - NOMINAL COVER TO ALL REINFORCEMENT (INCLUDING LINKS) TO MEET DURABILITY REQUIREMENTS - ADAPTED FROM BS 8110: 1985 FOR SRI LANKAN PRACTICE

Exposure Classification

Examples of Exposure

Mild Moderate Severe Very severe Extreme

Indoor Outdoor Driving Rain Sea Spray Abrasive

mm 25

--

Nominal Cover mm mm 20 35

--

20* 30 40 50

mm

mm

20* 25 30 40 60

20* 20 25 30 50

----

--

Maximum free water/cement ratio

. 0.65

0.60

0.55

0.50

0.45

Minimum cement content (kg/m3 )

275 (300)

300 (325)

325 (350)

350 (400)

400 (450)

40

45

Lowest grade of concrete

25

--

30

--

35

.

Note 1

This table applies to normal-weight aggregate OPC concrete of 20 mm nominal maximum aggregate size and river sand fine aggregate. In no case should the cover be less than the maximum aggregate size or diameter of main reinforcement.

Note 2

A minimum of 25 mm cover to all reinforcement should be maintained in beams and columns.

Note 3

Cover values marked with asterisks (*) can be reduced to 15 mm, provided the nominal maximum aggregate size does not exceed 15 mm, subject to the conditions in Notes 1 and 2.

Note 4

The minimum cement content values in parentheses should be maintained if no water-reducing admixtures are used.

Note 5

The grade requirement can be reduced by 5 if a checking regime establishes that the maximum free water/cement ratio and minimum cement cot\tent requirements are met.

Note 6

The above cover values can be reduced by 5 mm, subject to the conditions in Notes 1 and 2 and a minimum of 15 mm, provided a 1:3 cement: sand rendering of 10 mm, 15 mm or 20 mm is applied to concrete made to water/cement ratios of 0.65, 0.6 and 0.55 respectively.

3

EXAMPLE 1 - ANALYSIS OF UNDER-REINFORCED SECTION Determine the lever arm for the beam section shown in the figure; find also its moment of resistance. (

225

)

f eu = 25 N/mm2 f y = 460 N/mm2 3-20

000 (All dimensions in mm)

Introductory Notes 1.

This example is regarding the analysis of an existing beam. The first step in finding the moment of resistance is to find the lever arm.

Area of steel Note 2

Note 3

3,4,4. 1(e)

Output

Calculations

Reference

= 942.5 mm2

Assuming that the steel llas yielded, T = (0.87)fy.~ = (0.87)(460)(942.5) = 377189 N Hence, balancing compressive force = 377189 N (0,45)feu .b(O.9)x = 377189 . (0,45)(25)(225)(O.9x) = 377189 x = 166 mm Since x/d = 166/375 = 0.44 < = 0.64, steel has yielded and original assumption is correct. z = d - (0,45)x = 375 - (0,45)(166) = 300 mm Note :- z/d = 300/375 = 0.8 < 0.95, Hence O.K. Moment of resistance

= (377189)(300) = 113.16 x106 Nmm = 113 kNm

T

= 377189 N

x

=

z

= 300 mm

M

=

166 mm

113 kNm

Notes on Calculations 2.

Most singly reinforced sections will be under-reinforced in practice. Hence, assuming that the steel has yielded is the most convenient way of starting. (This assumption 5

should be checked later on, of course, using the xJd value.) 3.

The condition that tensile reinforcement has ~ielded when the concrete strain is 0.0035, is x/d < = 0.64 (for fy = 460 N/mm ) and x/d < = 0.76 (for fy = 250 N/mm 2). This can be shown by assuming a linear strain distribution. However the code recommends that x/d < = 0.50, in order to accommodate redistribution up to 10% (Clause 3.4.4.4).

Concluding Notes 4.

The lever arm is the distance between the centroids of the tensile and compressive forces. This separation between two opposite forces is what creates the moment of resistance in a flexural element.

5.

Because this distance has to be accomodated within the depth of the section, flexural elements tend to have larger cross'sections than compressive elements.

EXAMPLE 2 - ANALYSIS OF OVER-REINFORCED SECTION Determine the moment of resistance of the section shown. (

150

)

2-25

o

Id=300

f cu = 25 N/mm2 f y = 460 N/mm 2

0

(All dimensions in mm)

Introductory Notes 1.

This section is different from that in Example 1, in that it is over-reinforced. The calculation procedure is more complicated here.

6

Reference

Output

Calculations Area of steel

= 981.7 mm2

Assuming that the steel has yielded, T = (0.87)fy ' " = (0.87)(460)(981.7) = 392876 N Hence, C = (0.45)fcu .b(0.9)x = 392876 (0.45)(25)(150)(0.9)x = 392876 x = 259 mm Note 2

But, x/d = 259/300 = 0.86 :>0.64 Hence, steel has nQ1 yielded. We shall try to find a value for x, by trial and error, such that T and C are approximately equal. Try x = 200 mm C = (0.45)fcu .b(0.9)x

Note 3

= (0.45)(25)(150)(0.9)(200) = 303750N

= (0.0035)(300-200)/200 = 1.75 xlO-3 Hence, fs = (1.75 xlO-3)(200 xloJ) = 350 N/mm 2 , and T = (350)(981.7) = 343595 N Es

For a better approximation, try x = 205 mm. Then C = 311344 Nand T = 318454 N.

Note 4

For a still better approximation, try x = 206 mm. Then C = 312863 Nand T = 313572 N. This approximation is sufficient. Note:- x/d = 206/300 = 0.69 (> 0.64)

x

z = d - (0.45)x = 300 - (0.45)(206) = 207 mm M = C.z = (312863)(207) = 64.763 x106 Nmm = 64.8 kNm

M

Note:- Alternative method of finding x. Once it is established that the steel has not reached yield point, for any given value of x, Es = (0.0035)(300-x)/x fs = [(0.0035)(300-x)/x](200 xloJ) N/mm 2 T = ((0.0035)(300-x)/x](200 xloJ)(981.7) N C = (0.45)(25)(150)(0.9,qN Putting T = C, we have the quadratic equation x2 + (452.47)x - 135741 = 0, giving x = 206 or -659 mm

x

7

= 206 mm = 64.8 kNm

= 206 mm

Notes on Calculations

2.

In some rare cases, as in this one, a beam may be over-reinforced, meaning that the yielding of steel will not take place before the crushing of concrete. If such a beam fails, it will do so suddenly, without warning, and hence over-reinforced beams are discouraged in practice.

3.

Since the steel has not yielded, the stress can no longer be assumed to be 0.87fy. Rather, the stress is the steel is obtained by (i) determining the strain in the steel, assuming a linear strain distribution across the section and (ii) using the stress-strain curve in Figure 2.2 of the code to arrive at the stress. ~

'"8

8

........

0.87x460=400 N/mm 2

z

I

, , I

I

200 ,

kNAnm 2

Strain Strain diagram

4.

Stress-Strain diagram

It is possible to use this method because the stress-strain curve for steel below the

yield point is a single straight line.

Concluding Notes

5.

One way of ensuring that the beam failure is ductile is to introduce some compression steel, so that x/d will be reduced to 0.5 (See Example 3).

8

EXAMPLE 3 - ANALYSIS OF DOUBLY REINFORCED SECTION

1

Detennine the amount of compression steel required, in order to make Example 2. Find also the moment of resistance of the resulting beam.

~ .150 .) ~ A'd'

s

d=300

f eu fy

x/d

=

0.5 in

= 25 N/mm2 = 460 N/mm2

2-25

o

0

(All dimensions in mm)

Introductory Notes 1.

If it is found that a singly reinforced beam is over-reinforced and it is desired to make it under-reinforced or balanced, this may be achieved by (i) increasing the depth of the section, (ii) increasing the breadth of the section or (iii) introducing compression steel.

2.

Increasing the breadth of the section will generally be uneconomical. Therefore, if the depth of the section cannot be increased due to non-structural reasons, option (iii) above is used.

Reference Note 3

Calculations

Output

Assume a suitable value for d', say 50 mm.

d' = 50 mm

For equilibrium of the section, the compression in the top steel plus the concrete must equal the tension in the bottom steel. Setting x = (0.5)d = 150 mm (which automatically ensures the yielding of tension steel), we have d'/x = 50/150 = 0.33 < = 0.43, which means that the compression steel will yield as well. 3.4.4.4

Note 4

(0. 87)fy.As ' + (0.45)feu ·b(0.9)x = (0.87)f As (0. 87)(460)As ' + (0.45)(25)(150)(0.9)(150 = (0.87)(460)(981.7) Hence, As' = 412 mm 2 , Use 4Tl2 (As' = 452.4 mm 2).

r

9

A s '= 412 mm 2

Use 4T12 (452.4 mm2)

Reference

Output

Calculations

Table 3.27 Note 5

Note:- lOOA s ' / Ac = (100)(452.4) / (150)(350) = 0.86 (> 0.2), Hence O.K. Lever arm for balanced section = d - (0.45~1.) = (0.775)d = (0.775)(300) = 232.5 mm Distance between top and bottom steel = 250 mm Hence, taking moments about level of tension steel, moment. of resistance = (0.45)(25)(150)(0.9)( 150)(232.5) + (0.87)(460)(412)(250) = 94187006 Nmm = 94.2 kNm

Note 6

M = 94.2 kNm

Notes on Calculations 3.

The value of d' will depend on the cover, and other requirements (See Example 8).

4.

If the compression steel provided is greater than that required, the neutral axis depth will be reduced slightly; this is desirable, as it will increase the ductility of the section. When providing four bars within a width of 150 mm, it may be necessary to use the bars as two pairs of bars.

5.

When compression steel is provided, a minimum percentage is required. The area of concrete is based on the gross section, and the overall depth is taken as (300 + 50) = 350 mm.

6.

In general, the most convenient way of fmding the moment of resistance for a doubly reinforced section, is to take moments about the level of tension steel. The amount of compression steel to be used in the calculation is the amount required (412 mm 2), and not the amount provided (452.4 mm2).

Concluding Notes

7.

The moment of resistance of a doubly reinforced section can be considered to be the sum of the moments of resistance of (i) a balanced section and (ii) a steel section consisting of equal amounts of tension and compression steel, separated by (d-d'). It

~

150

t - d'=50 I 41~m2 _

d=300

_

1o

150 ""

982 mm2

o:~~~

1

i

~:.-G.....&-'l 232.5 570mm2

0

-

10

I

~

+

It ,

EXAMPLE 4 - ANALYSIS OF NON-RECTANGULAR SECTION Determine the moment carrying capacity of the trapezoidal beam section· shown below.

f

300

I

d=400

h=450

1

1

fcu = 25 N/mm 22fy = 460 N/mm

156

(All dimensions in mm)

Introductory Notes 1.

As in previous examples, the moment carrying capacity has to be found by working from first principles. The additional complication in this example is that the section is non-rectangular.

Reference

Output

Calculations Assume values for the neutral axis, x until the compression in concrete is equal to the tension in steel. The area of the section under compression = (0.5)(0.9)x[600 - {(3OO-150)/450}(O.9)x] ~300~ Area of steel = 981.7 mm 2 Assume also that the steel bas yielded. \10.9><

II

W:

Try x = 100 mm 0 Area in compression, Ac ~ = (O.5)('JO){600 - (O.33)(O.9)(IOO)) = 25650 mm2 C = (0.45)fcu .Ac = (0.45)(25)(25650) = 288563 N T = (0.87)(460)(981.7) = 392876 N Try x = 139 mm Then, C = 392868 Nand T = 392876 N. This approximation is satisfactory. Note also that x/d = 139/400 = 0.35 < 0.5; hence assumption that steel has yielded is O.K.

11

x

= 139 mm

Note 2

The centroid of the compression zone from the top of the section will be given by y = {(150)(139)(139/2) + (O.5)(150)(139)(139/3)} 1 {(150)(139) + (0.5)(150)(139)} = 61.8 mm Hence, lever arm = 400 - 61.8 = 338.2 mm M

Note 3

Output

Calculations

Reference

= C.z = (392868)(338) = 132.8 x106 Nmm = 133 kNm

Note:- Alternative method of finding x. Assuming that steel has yielded, T = (0.87)(460)(981.7) = 392876 N For any x, the area under compression is Ac = (O.5)(O.9)x[600 - {(300-150)/450}(O.9)x] C = (O.45)(25)A c Putting T = C, we have the quadratic equation, x2 - (2000)x + 258684 = 0, giving x = 139 or 1861 mm Since x/d = 139/400 = 0.35 0.13; hence O.K.

3.4.4.4

r

13

.dmin = 425 mm d =475 mm h = 525 mm b = 225 mm 0(

225 •

~I~I4Th As = 935 mm 2 Use 21'25 (981.7 mm 2)

Calculations

Reference (b)

Note 5

Note 6

Output

Overall depth restricted

If the overall depth is restricted to 400 mm, h = 400 mm, d = 400 - 50 = 350 mm, b = 225 mm (assuming the same breadth as before)

d = 350 mm b = 225 mm

Now K = M 1 {b.d2.fcu> = (150 x106 ) 1 {(225)(350)2(25)} = 0.218 > 0.156 (Le. K') Hence, compression reinforcement is required. Let us assume that d' = 50 mm. As'

= (K-K')fcu .b.d2 1 {(O.87)f (d-d'))

= {(O.218-0.156)(25)(225)(350f} 1

{(O. 87)(460)(350-50)}

= 356 mm2 Use 2Tl6 (~' = 402.1 mm 2) looAs '/Ac = (100)(402.1)1 (400)(225) = 0.45

Table 3.27

z 3.4.4. 1(e)

Use2Tl6 (402.1 mm 2)

< (0.95)(350)

= 333 mm; hence O.K.

Table 3.27 Note 8

Hence, use 3T25 (bottom) and 2Tl6 (top).

10:: Li~ (225 )

= d[O.5 + {0.25 - K'/(O.9)}O.s] = (350)[0.5 + {0.25 - (0. 156)/(O.9)}O.s] = 272 mm

= 356 mm 2

> 0.2; hence O.K.

As = {(K'.fcu .b.d2) 1 (O.87)fy'z} + As' ={(O.156)(25)(225)(350)21 (O.87)(460)(272)) = 1344 mm2 Use 3T25 (As = 1473 mm2) looA/Ac = (100)(1473) 1 (400)(225) = 1.64 > 0.13; hence O.K.

Note 7

A' s

400

000

+ 356

As = 1344 mm 2 Use 3T25 (1473 mm 2)

Notes on Calculations

2.

In practice, the ratio of depth to breadth for a beam will have a value between 1.5 and 2.5.

3.

Many designers still choose dimensions for beams and columns in steps of 25 mm, because 1 inch is approximately 25 mm. Furthermore, depths considerably in excess of the minimum depth for a singly reinforced section may be chosen, in order to reduce the steel requirement.

4.

The check for minimum reinforcement is almost always satisfied for tension steel in 14

beams. A little care should be excercised, however, for compression steel. 5.

The overall depth of the beam may have to be restricted, due to architectural requirements. On the other hand, there may be some economy in designing beams with a marginal amount of compression steel, because longitudinal steel on the compression face will be required anyway, in order to support the shear links.

6.

This is keeping with the idea that the difference between overall and effective depths is 50 mm.

7.

When calculating the are of tension steel, it is sufficient to use the value of compression steel required (as opposed to that provided), in this equation.

8.

When providing reinforcement, a combination of bar sizes should be adopted, such that the maximum and minimum spacing between bars is kept within specified limits (see Example 12).

Concluding Notes 9.

Design charts (in Part 3 of the code) could also have been used to design the steel required for these sections. The relevant charts are Chart No. 2 for the singly reinforced section and Chart No.4 for the doubly reinforced section, since d'/d = 50/350 = 0.143.

10.

The design charts are given for· ,.. ' d'/d values ranging from 0.10 to 0.20, in steps of 0.05. The chart with d'/d value closest to the actual value should be used for design. If the actual d'/d value lies exactly between the chart values, the chart with the higher d'/d value should be used in the design, as this is more conservative.

EXAMPLE 6 - DESIGN OF SECTION WITH REDISTRIBUTION If the beam section in part (a) of Example 5 (Le. h = 525 mm, d = 475 mm and b = 225 mm) was carrying an ultimate moment of 150 kNm after a 30% downward redistribution of moment, design the steel reinforcement required. Assume that d' = 50 mm, feu = 25 N/mm 2 and fy = 460 N/mm 2. Use the methods of formulae and design charts.

Introductory Notes 1.

If the moment at a section has been reduced by downward redistribution, that section must have adequate rotational capacity at ultimate limit state, in oder for plastic hinge action to take place. This capacity is ensured by restricting the x/d ratio to a specified value.

15

Output

Calculations

Reference Cal Using formulae 3.2.2.1(b) 3.4.4.4

I3b

= (1-0.3) 1 1 = 0.7 K' = (0.402)(l3b-0.4) - (0.18)(~-0.4)2 = (0.402)(0.7-0.4) - (0,18)(0.7-0.4)2

= 0.104

Now, K = M 1 (b.d 2.feu> = (150 x106) 1 {(225)(475)2(25)} = 0.118 > 0.104 Hence, compression steel is required. z

= d[O.5 + {0.25 - K'/(0.9)}0.5]

= (475)[0.5 + {0.25 3.4.4. 1(e)

Table 3.27

= 412 mm

- (0.104)/(0.9)}O.5]

< (0.95)(475) = 451 mm; hence O.K.

As' = (K -K')feu ·b.d2 1 {(0.87)fy 45° and ~ is restricted to l.S(d..(i'), this assumed value of 67.5° for fJ is reasonable and easy for calculation purposes.

4.

This is the same approach described in Note 3 of Example 10, The links designed can be used from the support upto the point where the main bars are cranked up.

S.

Although 2 bars are bent up, they also continue for at least distance "d" from any point in this section of the beam. Hence, the value of vc will be the same as in the middle area.

6.

If the link spacing is less than around ISO mm, it will be difficult for concreting to be carried out. Hence, as in this case, 2 links can be placed together, spaced wider apart. An alternative would have been to use 12 mm dia. links; however fabrication will be easier if links of the same diameter are used throughout the beam.

a

Concludina Notes 7.

It is not very common practice to use bent up bars as described in this example, although it was in the past.

30

EXAMPLE 12 - SERVICEABnJTY CHECKS AND DETAll.JNG

Carry out serviceability checks on the beam analysed in Example 7 and designed in Example 8. Also carry out detailing of reinforeement, including curtailment and lapping. Assume that type 2 defonned bars are used as reinforcement.

Introductory Notes 1.

The serviceability checks consist of spanldepth ratio calcu1ations for deflection and bar spacing rule checks for cracking. If these simplified checks are satisfied, the beam is "deemed to satisfy" the serviceability limit state requirements.

Refereace

Calculations

Output

Check for deflection fSRanIde,pth rules) Note 2 3.4.1.3 Table 3.10 Example 8

Table 3.11 Notes 3&4

3.4.1.4 Table 3.10 Example 8

Table 3.11 Notes 3&4

Consider the man BC; effective span = 6000 mm bwlb = 0.22 < 0.3 .Hence, basic span/depth = 20.8 for continuous, flanged beam. Mlbd2 = 1.00 and f. = (S/8)(460){(1224)/1295)} = 272 N/mm2 Hence, P l -= I.4S (for tension reinforcement) F 2 = 1.0 (as there is no compression reinforcement) Hence, allowablespanldepth ratio = (20.8)(1.45) = 30.16 Actual spanldepth= (6000)/(397.5) = 15.09 < 30.16; hence O.K. Consider fP8D AD: effective span = 2000 mm Basic spanldepth = 7 for cantilever with rectangular beamaetion. Mlbd2 = 2.32 and f. = (5/8)(460){(799)/829)} = 277 N/mm2 Hence F l == 1.07 (for tension reinforcement) F2 = 1.0 (as there is no compression reinforcement) Hence, allowable spanldepth ratio = (7)(1.07) = 7.5 Actual spanldepth = (2000)/(397.5) = 5.03 < 7.5; hence O.K.

Curtailment of reinforcement The bending moment diagram envelope must first be dmwn

31

All. span! depth = 30.2

Act spanI

depth = 15.1 Hence O.K.

All. span! depth = 7.5 Act. span/ depth = 5.03 . Hence O.K.

Cakulatiolls

Reference NoteS

Por 111M Be. the controlling 1oa4.c:ase is when AS has the minimum designultimate1Qed 'aDd Be has the maximum designultimateIoad~ This case bas already been considered in Example 7 .

7.0 /6.34 kN/m

/45.28 kN/m

~•• nn~ 1.95m

Example 7

For span BC,

6.Om

x

Mx = (128.1)x - (45.28)x212

Mx = Oatx - O. Mx is max. at x = 2.83 and equal to 18l.2 kNm Mx .. 0 apiA (192 mm) or -d- (397.5 mm). The difference between the larger y values is (3.10-2.30) = 0.80 m or 800 mm,which is alsogIeater than (12)41 or -d-. Hence, the practical cut-off points are y = 1.22 m and y= 3.10 m. Length of 16 mm bar required = 3.10 - 1.22 = 1.88 m Distances to B are (2.0 - 1.22) =O.78m (span AD) and (3.10 - 2.0) = 1.10 m (span Be) Since the distances to either side of B > = (40)41 {Le. (40)(16)= 640 mm}, anchorage is satisfied.

Curtail I T16 top bar 0.78 m (left) and 1.10 MCright)

ofB. I..eIlgth of bar is 1.88 m.

J 'Imine of bars

Note 12

3. 12.9. 1(c) Table 3.29 Note 13

The continuing 21'20 top bars at B can be curtailed at the point of contraflexure closer to B in span BC.and lapped with 2T12 bars (which will anchor the shear links). Similarly, the continuing 2T25 bottom bats in span Be can be curtailed attbe point of contraflexure closer to B in span BC and lapped with 2T12 bus. For top bars, distance of point of contraflexure from A is 4.23 m. This would be the theoreticalcut-off point To find the practical cut-off point, continue the bars for an effective depth {Le. 397.5 mm (>

12cP)}. Heace, cut-off paint is 4.23

+ 0.4= 4.63 m

3.12.8.13

from A, Le. 4.63 - 2.0 = 2..63 m.to the right of B. The lapped 2T12 bars will start (40)(12) =4S0mm before the curtailment of the 21'20 bars, Le. 2.63 0.48 =2.15 m to the right of B. ~:- Min. lap length = gtQterof (15~ (::: ISO mm) or 300 mm is satisfied; also distance between laps will be greater than 75 mm and (6)4> (=72 mm).

Note 15

For bottom bars, distance of point of contraflexure from C is 5.66 m, Le. 6.00 - 5.66 = O.34m to the right of B. As before, the practical cut-off point would be 397.5 mm beyond this. Hence, it would be 0.4 - 0.34 = 0.06 m to the left of B. The nl2 bars will start 0.48 - 0.06 = 0.42 m to the right of B.

3.12.8.13 Note 14 3.12.8.11

34

Curtail 2T20 top bars 2.63 m to .right of B. Start 2T12 top bars 2.1~ m to right of B.

Curtail 2T25 bottom bars 0.06 m to left ofB. Start 2Tl2 bottom bars 0.42 m to right ofB.

Relerenee

Output

Calculations lT16 ZIal

mo

I A

>

Zf12

214> <

•

<

t

>'

Zf12

,

),

21Q

2I25 l'IID

B

21"25t c

Crack width check 0.13); hence min. steel O.K. As = (0.26)(1000)(150) 1 (100) = 390 mm2/m Use TI0@175 mm (As = 448 mm2/m) Half theSe be Q1rtailed at (0.2)1 - i.e (0;2)(5) == 1.0 mfrom the centre-line of support. Thenr/f will be TI0 @ 350 mm {«3)(150) "" 450} lOOA/Ac =i (100)(44812) / (1000)(175) = 0.13 Hence crack cootroland minimum steel O.K.

bars_

deflection

Note 9

Check for

Table 3.11 Table 3.10

MJbd2 = 0.96 and f s = (5/8)(460){(390)/(448)} = 250 N/mm2 Hence F 1 = 157 Allowable spanldepth = (26)(1.57) = 40.8 Actualspanldepth =(5000)/(150) = 33.3 < 40.8; ~ O.K. ~n for

Fig. 3.25

Span steel TIO @ 175 mm

Deflection O.K.

bendin& at sUllJlOrt

Since the moment is identical to that in the span, steel provided also can be identical. Half these bars can be curtailed at (0.15)1 = (0.15)(5) = 0.75 m from the face of support (Note: 45 tP = 450 mm< 750mm) and all the steel curtailed at (0.3)1 = (0.3)(5) = 1.5 m from the face of support.

39

Support steel TIO @ 175 mm

Output

calculations

Reference

,

Check for shear

Note 10 Table 3.9 Table 3.17 Note 11

v = (34.3 xl 0.23N/mm2 Hence, no shear reinforcement required.

v

= 0.23

N/mm 2

vc - 0.54 N/mm.2.

Seconda[y reinforcement Table 3.27

3.12.11.2.7

l00As/A c = 0.13 As = (0.13)(1000)(175) I (100) = 227.5 mm 2/m Use TlO @ 350 mm (As = 224 mm2/m) Max. spacing = (3)(150)- 450 mm > 350 mm.

~ o. 7fm J>=7~

~7EmMO.7~ Tl~175

Notel2

cr

nOl3OO~75

TlQfJ350

.. ....

~~Tl~

·--n~·---TlOOI75

, I

0(

I.On

(

'>

SeCondary steel TI0@350mm

,

"

.. ~

Tl............I

'f f

( I.On )

5.0m

)

Notes on Calculatioas 3.

Although the bending moment is the controlling factor in the choice of depth for beams, where slabs are concerned, the controlling factor is the spanldepth ratio, representing the check for deflection. Atrial'V8lue has to be used initially; a value of around 34 is a reasonable estimate for lightly loaded one way continuous slabs; this should be reduced to around 30 for heavily loaded .s1abs. A lightly loaded slab would have an imposed load of around 4 kN/m2 , while a heavily loaded slab would bave one of around 10 kN/m2 •

4.

Slabs are generally designed such that shear links are not required; hence, no allowance need be made for link diameter.

5.

One way and two way slabs are generally designed - Le. loads evaluated and reinforcementcalculated - on the basis of a strip of unit width (e.g. 1 m wide).

6.

The minimum steel requirement is in fact based on looA/ Ac. However. since the 40

lOOA/bd is obtained from the design charts, it provides an approximate check on the minimum steel requirement. 7.

Although we can use the sllghtlylarger spacing·of 200 mm (giving As = 392.5 mm 2/m), we adopt this smaller spacing, as it results in the minimum steel requirement being satisfied even after half the steel is curtailed.

8.

Although 60% of the steel can be curtailed, in practical slabs, curtailing 50% is easier, because every other bar can be curtailed.

9.

The assumption regarding spanldepth ratio must becbecked as early as possible in the design. Hence span moments should be designed for first and the deflection check made soon after.

10.

The area of steel used here is that of the top (tension) steel at the support.

11.

In general, apart from .some cases in flat slabs, it is sought to avoid shear reinforcement in flat stabs. Hence, if v is greater than vC' the slab thickness is increased. This should always be borne in mind, and perhaps an approximate check for shear made early in the design,especially if the slab is heavily loaded (e.g. with a water load).

12.

Where the curtailment of steel is'toncemed, the distances corresponding to top steel are given from the face of the support and those corresponding to bottom steel from the centre-line of support.

Concluding Notes 13.

Although it is quite easy to satisfy minimum steel requirements and maximum bar spacing rules at critical sections (such as midspan and support), care should be taken to ensure that the above checks are not violated after curtailment of reinforcement.

14.

The simplified approach to the design of slabs, using Table 3. 13·can be used in most practical situations. Such an approach is given for the design of continuous beams as well, in Table 3.6. The coefficients in this latter table are higher than those for slabs, because the slab coefficients are based on the less stringent single load case of all spans loaded, with support moments redistributed downwards by 20%.

41

EXAMPLE 14 - ONE WAY SLAB A garage roof in a domestic building is to function as an accessible platform, surrounded by a parapet wall; the slab is supported on two parallel 225 mm brick walls, the clear distance between walls being 3.5 m. Design the slab,taking fcu .·2S'N/mm2, f y = 460 N/mm2 and density of reinforced concrete 24 kN/m3 •

=

Introductory Notes 1.

This example has more unknowns than the previous one. It describes a "real" situation, where design assumptions will have to be made. The imposed load and load from finishes and parapet wall have to be assumed and a decision taken regarding the end fixity of the slab.

...i..

The imposed load could be taken as 1.5 kN/m 2, since it is a domestic building. The finishes (onboth top surface and soffit) can be assumed to be a uniformly distributed load of 1 kN/m2 .

3.

The parapet wall which is constructed on the slab perpendicular to its span will give a degree of fixity to the slab. However, the most conservative approach is to idealize this slab as a one way simply supported slab. Any fiXing moments caused by the above partial fixity can be accomodated by taking SO% of the midspan steel into the top face of the slab at the support.

4.

The parapet wall parallel to the span will have to be carried by the slab. It can be assumed that the wall is 1.0 m high and 120 mm thick and that the density of the (brick) wall is 23 kN/m2• The load from this wall will be distributed only over a limited width of the slab (Clause 3.5.2.2).

Calculations

Reference

Output

Slab thickness Note 5 Note 6

TABLE 1 Example 8 Note 7 3.4.1.2

Approximate span = 3500 mm Assuming Spanldepth ratio of 28 (for a simply supported 1 way slab), effective depth = (3500)/(28) = 125 mm If we take cover = 30 mm (moderate exposure conditions and TABLE 1 values modified by Notes 5 and 6), and bar diameter = 10 mm, we can choose h = 160 mm and d = 160 -30 - 10/2 = 125 mm. Hence, effective· span = lesser of (3500+225) = 3725 mm or (3500+ 125) = 3625 mm

42

h d

= =

160 mm 125 mm

eff. span 3.625 m

=

Reference

Calculations

Output

Loadine (for 1 m wide strip) Self load = (0.16)(1)(24) Finishes (1.0)(1) Total dead load Imposed load = (1.5)(1)

= 3.84 kN/m

=

=

1.00 kN/m 4.84 kN/rn = 1.50 kN/rn

=

Design load = (1.4)(4.84) + (1.6)(1.5) = 9.2 kN/m 3.5.2.2 Note 8

Strip carrying parapet wall = (0.3)(3.615) + 0.12 = 1.21 m Additional dead load in that area = (1.0)(0.12)(23) 1 (1.21) = 2.28 kN/m

design udl = 9.2 kN/rn

Ultimate bending moment and shear force Note 3

Since we assume the slab to be simply supported, Mid span moment = w.l 2/8 = (9.2)(3.625)21 8 = 15.1kNmlm Shear force at support = w.l/2 = (9.2)(3.625) 12 = 16.7 kN/m

Mspan = 15.1 kNmlm V = 16.7 kN/m

Desim for bending Chart 2 (Part 3)

Mlbd2 = (15.1 xl 0.13); hence min. steel O.K. = (0.26)(1000)(125) I (100) = 325 mm2/m Use TI0 @ 225 mm (As = 349 mm2/m) Max. spacing allowed = (3)(125) = 375 mm > 225 mm; hence crack width O.K.

A. 3.12.11.2.7

Note 9

Fig. 3.25

However, bar spacing as well as minimum steel requirement will be violated if bars are curtailed. Hence, use TlO @ 187.5 mm (As =·419 mm2/m) Spacing after curtailment = 375 mm. l00A/ Ac after curtailment = (100)(419/2) 1 (1000)(160) = 0.131 > 0.13 Hence, min. steel andbar spacing are O.K. after curtailment. The steel should be curtailed at (0.1)1 = (0.1)(3625) = 362.5 mm from the point of support, Le. 362.5 (225/2) = 250 mm from the face of support.

43

span steel TIO @ 187.5 mm

RefeNilee Note 10

3.12.10.3.2

Output

Calculations The rest of the steel could be taken into the support and bent back into the span as top steel to extend a distance from support face of (0.15)1 = (0.15)(3625) = 544 mm {> (45)cP = (45)(10) = 450 mm}, say 0.55 m

support steel TI0 @ 375 mm

Check for deflection

Note II Table 3.11 Table 3.10

Mlbd2 = 0.93 f s = (5/8)(460)(325/419) = 223 N/mm 2 Hence, F 1 = 1. 71 (for tension steel)

Allowable span/depth = (20)(1.71) = 34.2 Actual span/depth = 3625/125 = 29 < 34.2; hence O.K.

Deflection O.K.

Check for shear

v = (16.7 xlol) 1(1000)(125) = 0.13 N/mm 2

looA/bvd Table 3.9 Table 3.17

Hence,

Vc

=

0~131

= 0.45 N/mm2 > 0.13 N/mm2 ; hence shear r/f is not required.

Shear O.K.

SecondflO' reinforcement

3.12.11.2.7

Note 12

looA/Ac = 0.13 As = (0.13)(1000)(160) I (100) = 208 mm2/m Use TIO @ 375 mm (i.e. max. spacing allowed - 3d) (As = 209 mm2/m) Note:- It can be shown that the spacing of the reinforcement in the edge strips of 1.21 m should be T10 @ 175 mm at midspan (and hence TIO @ 350 mm at supports). rt1 110075 TIotm5 ~o.~ II~

r

I :

t

•

I

•

1100187.5

o.ti: ~.

0.:rAJn

44

secondary steel TI0@375 rom

under parapets TI0@ 175 mm (span) TI0@350 (support)

Notes on Calculations 5~

6.

In order to use Clause 3.4.1.2 tofind the effective span, the clear distance between supports is taken as a first approxl.mation of the span. For a lightly loaded one-way simply supported slabs, a span/depth ratio of around

26-28 may be assumed. Tbisshould be .reduced to around 24 for a heavily loaded slab. 7.

In this instance, we have taken a value for h, such that slab thicknesses are assumed to vary in steps of lO mill. To use steps of 25 mm (corresponding to 1 inch) would be too Conservative for slabs. Hence either 10 mm steps· or 12.5 mm steps (corresponding to 0.5 inches) should be adopted.

8.

The edge areas of the slab, Le. the 1.21 m strips carrying the parapet loads, will be more heavily reinforced than the rest of the slab. However, only the central part of the slab is actually designed in this example.

9.

There may be other alternatives to increasing the mid-Span steel, bot this approach makes the detailing for curtailment very simple and also helps to satisfy the deflection check, which is very critical in slabs. This approach also facilitates the detailing of steel for support restraint, as shown in the figure. One possible alternative is to use smaller diameter bars, but bars smaller than 10 mm, if used as main steel, will not be very stiff and may deflect significantly during concreting, thus losing their cover.

lO.

As shown in the fIgure,this is a very neat method of providing top steel at partially restrained ends of slabs and beams.

11.

Since we have provided more steel than required at mid-Span (see Note 9), advantage should be.taken of this by generally calculating the service stress, which will be lower than (5/8)fy and lead to a greater allowable span/depth ratio.

12.

It may be convenient to reinforee the entire slab with TlO @ 175 mm at mid span and TlO@ 350 mm at support, since the central part of the slab already has TlO @ 187.5 mm and TlO @ 375 mm at span and support respectively. The small penalty in cost will probably be worth the simpler detailing arrangement.

Concluding Notes 13.

It is important to keep in mind curtailment, bar spacing rules and minimum 'steel requirements while designing the reinforcement, because these detailing considerations may lead to the design being altered, as was the case here.

45

EXAMPLE 15 - TWO WAY SLAB

A .two way spanning slab which has several bays in each direction.ha$ a panel.size of S m x 6 m. The imposed load on the slab is 3 kN/m2• The loading fronl finisheaand light partitions can each be taken as 1 kN/m2 • Design a typical interior panel, using feu == 2S N/mm2 , fy == 460 N/mm2 and density of reinforced concrete == 24kN1m3. Introductory Notes

1.

The short span length and loading for this example have been anade ideAtica1to those in Example 13 for a one-way spanning slab. Hence, results can be compared.

2.

It will be assumed that the comers of this slab are prevented from lifting and that adequate provision is made for torsion. .

Calculatiops

ReferellCe

Note 3 3.5.7 TABLE I

Note 4 Note 5

Assultle.a spanldepth ratio of 40 (for a continuous 2 way slab) effective depth == (5000)/(40) == 125 mm If we take cover == 20 mm (mild exposure conditions and concrete protected by 10 mm·l:3 cement:sand rendering) and bar diameter as 10mm, then we can choose h == 150 mm and dmort == 150·20 - 10/2 == 125 mm and ~oog == 125 . 10 == 115 mm

h == 150 mm dlbort

==

125 mm ~== l1Smrn

Loadine (udl) == 3.6 kN/m2 Self load == (0.15)\:, 1(24) Finishes == (1,0>, == 1.0 kNlm2 Total dead load == 4.6 tN/mt Imposed load == (3.0). == 3.0kN/m2 Partitions = (1,0)' = 1,0 kNlm 2 Total imposed load == 4.0 kN/m 2 Design load= (1,4)(4.6) + (1,6)(4.0) == 12.8 IcN/m 2 Bendine moments This interior panel- has lyIlx == 615 == 1,2

46

n == 12.8 kN/m2

Reference Table 3.15

Chart 2 (Part 3)

3.12.11.2.7 Table 3.27

Note 6

Table 3.11 Table 3.10

Chart 2 (part 3) :l.12.11.2.7 Table 3.27

Calculations

Short way, Short way, Long way, Long way,

edge span edge span

Output

=(0.042)(12.8)(5)2= 13.44 kNm/m =(O.0~)(l2.8)(5)2= 10.24 kNmlm =(0.032)(12.8)(5)2= 10.24 kNmlm =(0.02~)(12.8)(5)2= 7.68 kNm/m

I Desien of reinforcement / Short way, mid-man:Mlbd 2 = (10.24 xld') 1 (lOOO)(l25t = 0.66 100A/bd = 0.17 As = (0.17)(1000)(125) 1 100 = 213 mm2/m Use TlO @ 350 mm (As :: 224 mm2/m) Max. spacing = (3)(125) = 375 mm > 350 mm l00A/Ac = (100)(224) 1 (1()()())(150) = 0.15 >0,13 Hence, bar spacing and min. steel are O.K. but if steel is curtailed, they will be violated.

Short way, span TlO@ 350 mm

(Check for, deflection):Mlbd2 = 0.66 f s = (5/8)(460)(213/224) = 273 N/mm 2 F 1 = 1.65 (for tension steel) Allowable span/depth = (26)(1.65) = 42.9 Actual spanJdepth = (5000)/(125) = 40 < 42.9; hence O.K.

Deflection O.K.

Short way. cts. edge:MJbd2 = 0.86, l00AJbd = 0.23, As '7' 288 mm2/m Use TlO @ 250 mm (A. = 314 mm2/m) Bar spacing and min. steel areIO.K.

Short way, edge TlO@250 mm

Long way. cts. edge:Mlbd2 = (10.24 xlW) 1 (1000)(115)2 = 0.77 l00Aibd = 0.21 As = (0.21)(1000)(115) 1 (100) = 242 mm2 Use TlO @ 325 mm (A. = 242 mm 2/m) Max. spacing = (3)(115) = 345 mm > 325 mm l00AJAc = (100)(242) 1 (1000)(150) = 0.16 >0.13 Hence, bar spacing and min. steel are O.K., but steel cannot be curtailed. Long way. mid-s.pan:= 0.58, 100Aibd =' 0.15, As = 173 mm 2/m Use TlO @ 350 mm (As = 224 mm2/m), since max. clear spacing (345 mm) governs.

"

Long way, edge TlO @ 325 ~m

Mlbd 2

47

Long way, span TlO@ 350 mm

CalcuJatiOllS

Reference

Output

Edge strips:3.5.3.5

Note 7

Table 3.16

Table 3.9

l00A/Ac =0.13 As = (0.13)(150)(1000) I (100) = 195 mm2/m Use TlO @ 375 mm (governed by max. spacing rule in short way direction) Use only in short way cts. edge; at other locations, use middle strip steel for edge steel.

Edge strip nO@375 mm (only for short way, cts. edge)

Check for shear Short way sypport:V = (0.39)(12.8)(5) = 25.0 kN/m v = (25.0 xloJ) I (1000)(125)= 0.2 N/mm 2 l00A/bd = (loo~(314) I (1000)~125) = 0.25 vc = 0.53 N/mm > 0.2 N/mm ; hence O.K.

LoDa l)!&Y R>J!Ort:-

Table 3.16

Table 3.9

V == (0.33)(12.8)(5) = 21.1 kN/m v = (21.1 xloJ) I (1000)(115) = 0.18 N/mm2 l00A/bd = (100)(242) I (1000)(115) = 0.21 vc = 0.50 N/mm 2 > 0.18 N/mm2; hence O.K.

o

Fig. 3.25

~

No shear r/f required

< 1800

~ 16T106325'11 ~

600

, "" E-o ...

1.0

l:"-

M

@

0

E-o

N

~

-~lo

1.0 N

, ~l-

0

1.0 M

--@

@

0

0

E-o

E-o

O'l

00

48

-

...1-

E-o

1.0

r-

M

@

0

E-o

N

-

Notes on CalcuIatiODS 3.

A trial value for span/depth ratio of 40 is reasonable for a lightly loaded, continuous square2-way slab; a ratio of 38 would be appropriate for heavily loaded slabs. This will of course reduce with the ratio of long to short span, ~hing the value for 1way slabS when the latter ratio becomes 2. The span/depth ratio is calculated with respect to the shorter span, as it is this that controls d~flection.

4.

It should be noted that the slab thickness required for a two-way slab is less than that required for a one-way slab of similar span and loading - cf. 175 mm required for the slab in Example 13.

5.

In arranging the reinforcement in the slab, the short way reinforcement should be placed outermost, in order to have the greatest effective depth, since the shorter span controls deflection and since the bending moments and shear forces are greater in the short way direction as well.

6.

Two way spanning slabs are, in general, very lightly reinforced, so that curtailing is often not possible because of the minimum steel requirement or the maximum spacing requirement, or both.

7..

Since the main steel requirements are also fairlysmaIi, for practical detailing it may be it' may be convenient to use the same reinforcement as the middle strip for the edge strips, except in the case of the short way continuous edge.

Concluding Notes 8.

Where an edge or comer panel is concerned, in addition to the main and edge steel, the requirements of torsional steel reinforcement have to bernet at the top and bottom of the slab according to Clause 3.5.3.5; in many cases, the main and edge steel provided would meet those requirements.

9.

Although the loads on a beam supporting a two-way slab will be either triangular or trapezoidal, the code gives coefficients for an equivalent uniformly distributed load over three quarters of its span.

10.

In the calculation of moment coefficients from Table 3.15, if there are significantly differing coefficients on either side of a common edge, the code suggests a method of moment distribution to rectify the situation, in Clause 3.5.3.6.

49

EXAMPLE 16 - FLAT SLAB A. flat~, which _several bays in each direction, has a panel size 0,{5m x 6 m. The ~gn imposed ontbesJab is 3 kN/m2 • Tbe ioa4ingfrom finishes and liaht PlU'titions can each be considered to be 1 kN/m2. Design a typical interior panel, using feu == 2S N/mm2 , fy 460 N/mm 2 and density of reinforced concrete 24 kNlmJ • It may be assUmed that the columns supporting the slab are braced.

=

=

Introductory Notes 1.

This example, too, can be compared with Examples 13 and 15.

2.

As the columns are braced, and as the I1ab has several bays in each. direction, the be simplified method of analysis described in Clause 3.7:2.7 and Table 3.19 employed.

3.

It will be assumed that the slab is without drops, and the maximum value of effective diameter will be employed for column .beads.

will

Calculations

Reference Slab thickness 3.7.1.4 Note 4 3.7.8 TABLE 1

NoteS

=

Max. value of he =.(114)(5.0) 1.25 m Assuming a trial span/depth of 32, (6000)/(32) 187.5 mm effective depth If we take cover = 20 mm (mild exposui-e condi\ions and concrete protected by 10 mm 1:3 et:sand render) and bar diameter = 10 mm, we can choose h = 212.5 mm, d long = 212.5-20-10/2 == 187.5 mm, dshort == 187.5-10 == 177.5 mm, dave == 182.5 mm

=

Loadin~ (for

entire panel)

he

= 1.25 m

=

h == 212.5 mm d y == 187.5 mm dx == 177.5 mm davg==182~5 mm

>"

Note 6 Panel area == (5)(6) == 30 m2 Self load = (0.2125)(30)(24) == 153lcN Finishes == (1.0)(30) = ..1U:.kN Total dead load = 183 leN . Imposed load = (3.0)(30) = 90 leN Partitions = (1.0)(30) = 30 leN Total imposed load = 120 leN Design load = (1.4)(183) + (1.6)(120) = 448 leN

50

.il"

F = 448 leN

Reference Table 3.19 Note 7

Fig. 3.12 Table 3.20

Fig. 3.12

Note 8 Table 3.11 3.7.8

Chart 2 (part 3) Note 9 3.12.11.2.7 Table 3.27 Note 10

Chart 2 (Part 3)

Calculations

Output

Bending mOments Long way:1 = 6.0 - (2/3)(1.25) = 5.17 m Span moment = (0.071)(448)(5.17) = 164 kNm Col. strip (2.5 m) = (0.55)(164) = 90.2 kNm Mid. strip (2.5 m) = (0.45)(164) = 73.8 kNm Support moment = (0.055)(448)(5.17) = 127 kNm COL strip (2.5 m)= (0.75)(127) = 95 kNm MId. strip·(2.5 m) = (0.25)(127) = 32 kNm Short way:1 = 5.0 - (2/3)(1.25) = 4.17 m Span moment = (0.071)(441)(4.17) = 133 kNm Col. strip (2.5 m) = (0.55)(133) ::: 73 kNm Mid. strip (3.5 m) = (0.45)(133) = 60 kNm Support moment = (0.055)(448)(4.17) ::: 103 kNm Col. strip (2.5 m) = (0.75)(103) = 77 kNm Mid. strip (3.5 m) = (0.25)(103) = 26 kNm

Design of reinfOrcement Long way. an:(Check for deflection) Total ~ moment = 164 kNm Mlbd2 = (164 xlo6) 1 (5000)(187.5)2 = 0.93 2 If AS,reqd = Aa.(Jrov. fs = 288 N/mm and F 1 = 1.41 for tension steel) Allowable spanldepth = (26)(1.41)(0.9) = 33.0 Actual spanldepth = (6000)/(187.5) = 32 < 33.0; hence O.K.

Deflection O.K.

(Column strip - 2.5 m wide) Mlbd 2 = (90.2 xl 0.J.tN/mm2 ; hence shear r/f not required.

2

Top steel 3.6.2

Output

Cakulations

shear r/f not required

over sup,port

This is to control cracking and should be 25 % of midspan steel. As = (114)(246) = 61.5 mm2 Use 1 TIO bar (As = 78.6 mm2), extending (0.15)1 = (0.15){5.00) = 0.75 m into span on each side.

over support 1 TIO per rib

Notes on CalculatioR§ 3.

This trial ratio is reasonable for simply supported one~way slabs - see Note 6 in Example 14.

4.

It should be noted that the effective thickness of this slab (reflecting the volume of concrete that will be used) is much lower than the one-way solid slab of similar span and loading in Example 13.

5.

It is convenient to calculate the loading for a strip of width equal to a repeating cross sectional unit.

6.

Since support details are not given, the shear force is calculated at a distance "d" from the centre-line of support (and not from the face of the support - Clause 3.4.5.10). The approach here is conservative.

7.

The average width of web below the flange is used for shear stress calculations.

Concluding Notes 8.

Fire resistance considerations will, to a large extent, govern the choice of form of ribbed slabs.

9.

The design of these slabs is essentially the same as the design of flanged beams. Generally the neutral axis will lie within the flange.

10.

Although the code suggests a single layer mesh reinforcement for the topping, it does not demand it (Clause 3.6.6.2). It will be quite difficult to place such a mesh in a 50 mm topping while maintaining the top and bottom cover requirements.

11.

These ribbed slabs probably have a lower material cost than solid slabs, but their construction costs would be greater, because of non-planar formwork requirements. 57

EXAMPLE 18 - COLUMN CLASSIFICATION

PI. four storey building has columns on a grid 9f 5.0 m x 5.0m, supporting beams of dimension 525 mm x 300 mm in one direetiononly and a one-way $lab of 175 mm thickness. The roof also has a beam-slab ammgement identical to other. floors. The oolumns are of dimension 300 mm x 300 mm and the soffit to soffit heigbtoffloors is.3.5m; the height from the top of the pad foundation (designed to resist moment) to the soffit of U1e first floor beams is 5.0 m. If the frame is braced. classify a typical internal column for different storeys as short or slender. Introductory Notes 1.

Columns are classified as unbraced or braced on the one hand (depending on whether they' take lateral loads or not) and .as slender or short on the other (depending on whether they should be designed to carry additional moments due to deflection or not).

2.

The effective length of a column will depend on the degree of fixity at its ends.

Caknlations

Reference

Clear height between end restraints, (for ground floor columns) lox = 5.0 m. loy = 5.0 + (0.525-0.175) (for other floor columns) lox = 3.5-0.525 = 2.975 m, loy = 3.5-0.175 ::: 3.~25 m 3.8.1.6.2

Table 3.21 Note 3

3.8.1.3 3.8.1.3

Output

= 5.35 m .

The end conditions for the columns in the direction of beams are all conditionl. H(mce, fJ ::: 0.75 In the other direction. fJ = 0.80 (ground floor columns) fJ = 0.85 (other columns) lex = (0.75)(5000) = 3750 mm (ground floor) ::: (0.15)(2975) ::: 2231 mm (other floors) ley::: (0.80)(5350) ::: 4280 mm (ground floor) = (0~85)(3325) ::: 2826 mm (other floors) Hence. for ground floor columns, le/h = (3750)/(300)= 12.5 < 15, leylb ::= (4280)/(300) = 14.3 < 15; hence short. for other columns. le/h = (2231)/(300) = 7.44 < 15, l~jb (2826)/(300) 9.42. < 15; hence short.

=

=

58

All columns are short.

Notes on Calculations 3.

The values of {3 in Tables 3.21 and 3.22 have been obtained from the more rigorous method for calculating effective column lengths in framed structures, given in (i.e. sum of equations 3 to 6 in section 2.5 of Part 2 of the code. The ratios column stiffnessesl sum of beam stiffnesses) have been assumed to be 0.5, 1.5, 3.0, and 7.0 for conditions 1, 2, 3 and 4 in Clause 3.8.1.6.2 (Part I) respectively.

"c

Concluding Notes

4.

Where edge columns are concerned, they will not have beams "on either side" as specified in the provisions of Clause 3.8.1.6.2. In this case, an approximate value for {3 can be interpolated, based on the actual C¥c value and the values used in Tables 3.21 and 3.22 (see Note 3 above); otherwise the method in Section 2.5 of Part 2 can be adopted.

5.

For a column to be considered short, both le/h and le/b have to be less than 15 (for braced columns) and less than 10 (for unbraced columns), as specified in Clause 3.8.1.3. The ratiO"c in section 2.5 of Part 2 or the value {3 in Clause 3.8.1.6 has to be obtained for beams in one plane at a time.

EXAMPLE 19 - SYMMETRICALLY LOADED SHORT COLUMN Assuming that the density of reinforced concrete is 24 kN/of, fcu == 2S N/mm 2, fy = 460 N/mm 2 , and that the imposed loads on the roof and the floors are 1.0 kN/m2 and 2.5 kN/m 2 respectively and that the allowance for partitions and finishes are 1.0 kN/m 2 each, design the ground floor part of an internal column of the framea structure described in Example 18. : . " . ".

Introductory Notes 1.

Since the ground floor .part of an internal column has been found to be short, and since the arrangement ,of loads symmetrical, this design can be carried out according to the provisions of Clause 3.8.4.3, using equation 38.

2.

The major part of this exercise consists of a load evaluation, taking into account the appropriate reduction factors for imposed loads specified in "BS 6399: Part 1 (1984): Design loading for buildings: Code of practice for dead and imposed loads". The partition loads are taken as imposed loads, since their positions are not fixed. 59

Reference

Calculations Column grid dilDClDsions are 5.0 x 5.0 m.

Hence, area conesponding to column" (5)2 .. 25 m2

Dead loads

Note 3

From 4 slabs = (4)(24)(0.175)(25) = 420 kN From beams = (4)(0. 525-D. 175)(0.3)(24)(5) =50.4 kN From columns={(3)(2.975)+5}(0.3Y(24) =30.1 kN From finishes = (4)(1.0)(25) = 100 kN Total dead load = 600.5 kN

Ok = 601 kN

Imposed loads From roof = (1.0)(25) = 25 kNFrom 3 floors = (3)(2.5)(25) = 187.5 kN From partitions- (3)(1.0)(25) =~ Total imposed load = 287.5.kN I.L. reduction due to floor area = (0.05)(25/50)(287.5)(4) = 28.75 kN I.L. reduction due to 4·floors = (0.3)(287.5) = 86.25 kN Hence, imposed load = 287.5 - 86.25 == 201.25 kN

o.c .. 201 kN

N = (1.4)(601) + (1.6)(201) = 1163 kN

N .. 1163 kN

Desim of main steel 3.8.4.3 equation 38

Note 4 Note 5 Table 3.27

For short columns resisting axial load = (O.4)fcu 'Ac -t (0.75)A,.,.fy Assuming we use 4 T16, Awe =8Q4 mm2 Ac .. (300)2 - 804 III 89196 mm2 (1163 xl03) .. (0.4)(25)(89196) + (0.75)Awe(460) Awe .. 785 mm2 < 804 mm2;hence O.K. ' .' . Hence, use 4 T16 (A.c = 804 mm 2) . Note:- lOOAwc/Ac = (lOO){804) I (300)2 == 0.89 > 0.4; hence min. steel O.K.

N

Use 4 T16 (at column comers)

Notes on .Calculations 3.

The total imposed load can be reduced either on the basis of the area supported by the column or the number of floors - supported by the column. In this case, the reduction allowed as a result of the latter is greater and is hence applied - see as 6399: Part 1 (1984), referred to in Note 2 above.

4.

The term Ac is the net area of concrete. A trial value of Awe C3ft be obtained from equation 38 assuming the gross area of concrete for Ac as a flrst approximation; this 60

area of A~ can then be deducted from the gross area to obtain Ac' The value of Asc obtained from the formula should be less than the original trial value of A.c. 5.

In some cases, a negative value may be obtained for A~; this indicates that nominal steel will be sufficient. In any case, bar diameters under 12 mm are generally not used for columns, because they will not be stiff enough for the erection of the reinforcement cage.

Concluding Notes 6.

This method of design is applicable for short braced columns, where moments. are negligible, due to a symmetrical arrangement of loads. Even if this symmetry is only approximate, provided the columns are short and braced, equation 39 can be used in place of equation 38.

7.

In addition to the main reinforcement, columns should be reinforced by links which

surround the main reinforcement as well. Ttiis will be shown in the next example.

EXAMPLE 20 - SHORT COLUMN WITH AXIAL LOAD AND MOMENT A short column of 300 mm x 400 mm cross section carries an ultimate axial load of 800 leN. If an ultimate moment of 80 kNm is applied (a) about the major axis, (b) about the minor axis"

(c) about both axes determine the column reinforcement required. Note that fcu = 25 N/mm 2 and fy d 460

N/mm2 •

Introductory Notes

1.

This column canies a substantial moment as well as an. axial load. Hence, we shall have to use the design charts, which will give us a symmetrically reinforced section. 61

,.

Output

Calculations

Reference

N = 800 kN, M == 80 kNm Nlbh = (800 xI(3) 1 (400)(300) = 6.67 (for all cases)

Note 2 Chart 23 (part 3)

Chart 23 (Part 3)

(al Major axis bendine b = 300 mm, h = 400 mm Mlbh2 = (80 xl()6) 1 (300)(400)2 = 1.67 lOOA.Jbh = 0.4 A~ = (0.4)(300)(400) I (100) = 480 mm 2 Use 4 T16 (Asc = 804 mm2) (b) Minor axis bendine b '''';' 400 mm,h = 300 mm Mlbh2 = (80 xl()6) 1(400)(300)2 = 2.22 lOOAaclbh = 0.8 A~ = (0.8)(300)(400) 1 (lOOi = 960 mm 2 Use 4 TIO (Ase = 1256 mm ) ( (16/4) = 4 mm} @ 175 mm{ < (12)(16) = 192 mm}. For minor axis bending, use R6{ > (20/4) = 5 mm} @ 225 mm {< (12)(20) = 240 mm}. For biaxial bending, use R 8 {> (25/4) :;: 6.25 mm}! @ 175 mm {< (12)(16) = 192 mm}.

0 • • OeO

biaxial

4 T25 + 4 T16

JiDb

major axis R6@ 175 mm. minor axisR6@225 mm.

biaxial R8@ 175 mm.

Check for shear

= (126)/(800) = 0.158 « (0.75)(0.3) = 0.225 m);

3.8.4.6

MIN

Note 7

hence, shear is not critical.

62

Shear check oot required.

Calculations

Reference 3.8.6

Crack control (0.2)fcu .Ac = (0.2)(25){(3OO)(400) - 804} = 596 kN Axial load = 800 kN > 596 kN. Hence, no check is required.

Output Crack width check not required.

Notes on Calculations 2.

If we assume a cover of around 30 mm (moderate exposure conditions and TABLE 1 values modified by Notes 5 and 6), links of 8 mm and a bar diameter of 25 mm, then d/h will be (400 - 50.5) 1 (400) = 0.87 for major axis bending and (300 - 50.5)1 (300) = 0.83 for minor axis bending. Hence, Chart 23 (Part 3) - which corresponds to a d/h value of 0.85 - can be used. If there is a doubt, the lower d/h value should be used, as this is more conservative. It should be noted that the column design charts have a lower limit of looAsclbh = 0.4, thus ensuring that the minimum steel requirement of Table 3.27 is met.

3.

In this case too, the difference between hand h' and b and b' is taken as 50 mm, by a similar argument as in Note 2 above.. '

4.

If the steel requirement for bi-axial bending is greater than that which can be provided as corner steel, the additional amount required has to be provided in each of the, two mutually perpendicular directions, distributed along the faces of the section. Other approaches, which are less conservative and more accurate, perhaps, are given in "Allen, A.H., Reinforced concrete design to BS 8110 simply explained, E. & F.N. Spon, London, 1988" and in "Rowe, R.E. et al., Handbook to British Standard BS 8ll0: 1985 - Structural use of concrete, Palladian, London, 1987".

5.

Although smaller diameter bars (e.g. TIO) could have been used, the Tl6 bars'are used, so that the link spacing would no~ be too small; furthermore, bars smaller than T12 are not used as column reinforcement, as they would not be stiff enough during erection.

6.

Generally plain mild steel is used for links as it is easier to bend into shape. Furthermore, where bars other than corner bars are used, multiple links may have to be used if (i) there is more than one intermediate bar or (ii) the intermediate bar is greater than 150 mm ~way from a restrained bar (see Clause 3.12.7.2)

7.

Strictly speaking, howe~er, the shear stress should be found in order to check for the limits on v max '

Concluding Notes 8.

In general,sh.ear and crack control are not very critical for columns. 63

EXAMPLE 21 - SLENDER COLUMN

A braced slender column of 300 mm x 400 mm cross section carries an ultimate axial load of 800 kN. It is bent in double curvature about the major axis, carrying ultimate moments of 80 kNm and 40 kNm at its ends. The effective length of the column corresponding to the major axis is 7200 mm. Determine the column reinforcement if feu = 25 N/mm 2 and fy =

460 N/mm 2 •

Introductory Notes 1.

This example can be compared with Example 20, where the short column was of the same dimensions and carried similar loads.

Reference 3.8.1.3 3.8.3.3 Note2

3.8.3.2 equation 36 3.8.2.4 equation 34 equation 32 equation 35

Output

Calculations Type of column Vh = (7200)/(400) = 18 Also h/b = (400)/(300) = hence, single axis bending.

> IS; hence < 20, 1.33 < 3;

s~nder.

Design moments M 1 = -40 kNm; M 2 = +80 kNm M i = (0.4)(-40) + (0.6)(80) = 32 kNm {= (0.4)(80) = 32 kNm} emin = (0.05)(400) = 20 mm N.emin = (800)(0.020) = 16 kNm (3a = (1I2000)(IJb')2= (112000)('200/300Y= 0.288 au = #a· K.h = (0.288)(1)(0.4) = 0.115 M add == N.au = (800)(0.115) = 92 kNm Hence, critical moment is M j + Madd = 32 + 92 = 124 kNm. However, as K is reduced, if M j + Madd becomes < 80 kNm, M2 will become critical. Design of reinforcement

TABLE 1

Chart 23 (Part 3)

Assuming cover = 30 mol (moderate exposure conditions and TABLE 1 values modified by Notes 5 & 6), link diameter of 8 mm and main bar size of 25 mm, d/h = (400-50.5)/(400) = 0.87. N/bh = (800 x103) I (300)(400) = 6.67 M/bh 2 7 (124 xI06) I (300)(400)2 = 2.58 K = 0.9

64

slender column bent about major axis.

M j = 32 kNm

Madd = 92 kNm

Reference

Output

Calculations

Madd = (0.9)(92) = 83 kNm M = 32 + 83 = 115 kNm (>

Chart 23 (Part 3)

Note 3 Chart 23 (part 3)

80 kNm) Mlbh 2 = (115 xldi) 1 (300)(400)2 = 2..40 K = 0.85 Madd ~ (0.85)(92) = 78 kNm M = 32 + 78 = 110 kNm ·Mlbh2 = 2.29; K = 0.85 (again). Hence, lOOAsc /bh 0.8 Asc (0.8)(300)(400) 1 (100) :;: 960 mm2 Use 4 1'20 (Asc 1256 mm2)

=

=

=

aJnksl 3.12.7.1

Use R 6 {> (20/4) :;: 5mm} @ 225 mm {< (12)(20) :;: 240 mm}

K

= 0.85

main steel -

::I R6@ 225 mm

Notes on Calculations 2.

When major axis bending takes place, if either the Vh value is greater than 20 or the bib value is not less than 3, in order to account for the deflection due to,'Slendemess about the minor axis, the column has to be designed asbiaXiaUy Dent, with zero initial moments about the minor axis (see Clauses 3.8.3.4 and 3.8.3.5).

3.

In general; around 2 iterations are sufficient to arrive at a value of K that is virtually constant. It should be noted that the factor K should be applied to the original value of M add alone. "

Concluding Notes 4.

The reinforcement required for this column is the same as for minor axis bending of the short column in Example 20.

65

~LE

22· PAD FOOTING

Design a square pad footing for a 300 mm x 300 mm internal column, which carries an ultimate load of 1100 leN (service load of 760 kN), if the allowable bearing pressure of the soil is 150 kN/m 2 • Use feu = 25 N/mm2 , fy = 460 N/mm2 (deformed type 2) and density of reinforced concrete = 24 kN/m3 •

Introductory Notes 1.

Square pad footings are the most common foundation type for columns of framed structures. Pad footings are essentially inverted cantilever flat slab elements.

2.

The design of pad footings involves the choice of (i) footing area (which is based on soil bearing pressure), (ii) footing depth (which is based on shear resistance) and (iii) reinforcement to resist bending moment.

"

Calculations

Reference

Output

Dimensions of base Note 3 Note 4

=

=

=

=

= = =

NoteS

=

=

=

= =

=

Ultimate bearing pressure

= (1100) 1 10(N)°'~ 10(1100)°·5 332 mm; hence overall depth of 400 mm is O.K.

=

Assume a cover of 40 mm, for moderate exposure conditions. H bar size of 16 mm is assumed, dmin = 400-40-16-16/2 336 mm and daV2 = 344 mm.

=

66

M

~

= 253 leNm

= 336 mm dan = 344 mm d min

Reference

Calculations

Chart 2 (Part 3)

Mlbd 2 = (253 xIW) I (2400)(336)~ = 0.93

3.11.3.2

= (3/4)(300) + (9/4)(336) = 981 mm < Ie = 1200 mm; hence reinforcement should be banded. Use 7 Tl6 @ 200 mm in band of 1200 mm {< (3)(336) + 300 = l308}; Use (3+3) T12 @ 200 mm in two outer bands. (As = 1407 + 678 = 2085 mm2; 1407/2085 > 2/3) Max. spacing = 750 mm; hence O.K. 100AiAe = (100)(2085) I (2400)(400) = 0.22 > 0.13; hence O.K. Anc~ge length :; (40)(16) = 640 mm < (2400-300)/2 = 1050 mm; hence O.K.

3.12.11.2.7

Note 8

l00A/bd = 0.25 As = (0.25)(2400)(336) I (100) = 2016 mm2 (3/4)c

+ (9/4)d

Check for yerticalline 3.4.5.10

Table 3.9

shear

Consider a section at "d" from the column face, .and assume no enhancement to ve. V = (191)(2.4){(2.4-o.3)/2 - 0.336} = 327 kN v = (327 xl 0.41 N/mm2; hence O.K. Check for punchine shear

3.7.7.2

= (1100 xl (11.5)(Nf·'5

6.

As the weight of the footing is considered to be a uniformly distributed load which is taken directly by the soil reaction, It should not be considered when designing for the ultimate limit states of flexure and shear.

7.

If the values of TABLE 1 are modified by Note 5, a cover of 35 mm will suffice for moderate exposure conditions. However, the cover is increased by a further 5 mm, in case the footing comes into contact with .any contaminated ground water. The minimum value of "d" should be used in the design for flexure and vertical line shear, while the average value of lid" can be used in checking for punching shear.

8.

If the distance between the column face and the end of the footing is smaller than the anchorage length, the bars will have to be bent up near the end. of the footing; otherwise, they can be straight. .

9.

In most cases, punching shear is more critical than vertical line shear. Furthermore, if a distance "d" is not available from the critical perimeter to the end of the footing, the value of Vc should correspond to 100A/bvd < = 0.15 in Table 3.9.

Concluding Notes

10.

I.f the footing carries a bending moment in addition to the axial load, the maximum and minimum pressures under the footing will be given by (l/BL}(N ± 6M1L), with symbols having usual meanings. The maximum pressure should be kept below the allowable bearing pressure.

11.

If the difference between maximum and minimum pressures is small (say upto 20%

of the maximum pressure) it may be convenient to design for bending and vertical shear by assuming that the pressure distribution is uniform and equal to the maximum pressure. 12.

Where the design for punching shear is concerned, the average pressure can be taken for calculations, but a factor of 1.15 applied to the shear force, according to the provisions of Clause 3.7.6.2. 68

EXAMPLE 23 • COMBINED FOOTING Let us assume that an external columns is flush with the property line and that the footings for the external and first internal columns have to be combined, as shown. While the internal column carries an ultimate axial load of 1100 kN, the external column carries an ultimate moment of 60 kNm in addition to an ultimate axial load of 600 leN. The allowable bearing pressure of the soil is 150 kN/m2 • Use feu = 25 N/mm 2 and fy = 460 N/mm2 •

o~_""

4_._7m_~_--7>1~~'r

\lOkN C

~D

E

Introductory Notes 1.

The situation described above is often found in crowded urban areas where buildings are constructed on very small plots of land.

2.

It is difficult to provide an isolated pad footing for the external column, because of eccentric loading on the footing. Hence, it can be combined with the first internal column footing as shown above.

Reference

Calculations Service loads

Note 3 External column load External column moment ,Internal column load

= (600)/(1.45) = 414 leN = (60)/(1.45) = 41.4 kNm = (1100)/(1.45) = 759 leN

Dimensions of footine

Note 4

Distance of C.of G. from A is given by x, where (759+414)x = (414)(0.15) + 41.4 + (759)(5.15) Hence, x = 3.4 m; therefore, for uniform pressure under base, use a base of length (2)(3.4) = 6.8 m

69

Output

Note 5

Note 6

Outppt

CalcuJatloDS

Reference

If we assume a thickness of 0.8 m for the base, the allowable bearing pressure is 150 - (0.8)(24) • 131 kN/rr Width of base required = (759+414) I (6.8)(131) = 1.32 m Use base of 6.8 III x 2.0 m x 0.8 m

footing size 6.8 m·x 2.0 m xO.8 m

Analysis of footing Assuming that the C.of G. for ultimate loads is the same as that for service loads, the footing can be

idealised as follows:udl = (1100+600) / (6.8) Note 7

:1T A-&;::':_= ,6E = 250 leN/m

~~

w•

250 kN/m/

x~

Max. moment at C = (250)(1.65Y I 2 = 340 leNm To find max~ moment in AC. Mx == -(250)x212 + llOO(x-1.65) = -(125)x2 + (lloo)x - 1815 putting dMx/dx == -(250}x + 1100 = 0, we have x = 4.4 m Mmax = -(125)(4.4)2 + (1100)(4.4) - 1815 = 605 kNm Max. shear force at C == 1100 - (250)(1.65) = 688 leN Shear force at A = 600 leN Design for bending TABLE 1 Note 8 Chart 2 (part 3)

Assume cover of 40 mm, bar size (longitudinal) of 25 mm; hence, d == 800 - 40 - 25/2 = 747.5 mm (Section AC) Mlbd2 == (605 xl06) I (2000)(747.5)2 = 0.54 looAJbd == 0.15 . A. -= (0. 15)(2000}(747.5) I (100) = 2243 mm2 Use 5 ns (As = 2455 mm2) - on top surface; these can be curtailed if required.

70

longitudinal top steel- S n5

Reference

Table 3.27

3.11.3.2

Table 3.27

Table 3.29

Output

Calculations

~fb~~ = (340 x106) 1 (2000)(747.5)2 = 0.30

100A/bd = 0.08; hence use lOOA/Ae = 0.13 As = (0.13)(800)(2000) 1 (100) = 2080 mm2 Use 5 TI5 (As = 2455 mm2) - on bottom surface; these .~ c.ould be curtailed as well. Note:- (3/4)c + (9/4)d = (3/4)(300) + (9/4)(747.5) = 1907 > Ie = 1000 mm; hence, longituninal steel can be evenly distributed.

longitudinal rlf to be evenly spaced

crransverse direction) M = (25012){(2.0-0.3)/2}2(112) = 45 kNm/m d = 747.5 - 25/2 - 20/2 = 725 mm (assuming bar size of 20 mm) M/bd 2 = (45 x106) 1 (1000)(725)2 = 0.09 Use 100A/Ae = 0.13 As = (0.13)(1000)(800) 1 (100) = 1040 mm 2/m Use TIO @ 300 mm (Ag = 1047 mm 2/m) Anchorage length = (20)(40) = 800 mm < (2000-300)/2 - 40(cover) = 810 mm; hence O.K. This steel too can be evenly distributed, as it is nominal reinforcement; the same nominal steel can also be used as distribution steel for the top longitudinal bars.

t

4.7m 5T25 \. , ·~T20@300 ........

s

..

,

1 '5T25

k

6.&n

,

>I

In the longitudinal direction, check at a distance "d" from the internal column face. V = (688) - {(688+600)/(5.0)}(0.15+0.7475) = 457 kN v = (457 x1 0.31 N/mm2; hence O.K. In transverse direction, a distance "d" from column face is virtually at edge of footing; hence O.K.

71

transverse bottom steel T20@3oo mm, to be evenly spaced.

J

Check for vertical line shear 3.4.5.10

longitudinal bottom steel 5TI5

Reference

Calculations

Output

Check for punCbio& shear 3.7.7.2

Notes

00

vmax (internal column) = (1100 xlol) I (4)(300)(735) = 1.25 N/mm 22 S < (0.8)(25f· = 4 N/mm 2 < 5 N/mm ; hence O.K. As the critical perimeter, (1.5)d from column face, is outside the footing, only vertical line shear need be checked.

Shear r/f is not required

Calculations

3.

Where service loads are not specified or known, they can be estimated by dividing ultimate loads by 1.45 for.reinforced concrete structures. When converting service to ultimate" loads, it is safer to multiply the former by 1.5.

4.

If the footing dimensions are given, as opposed to being designed, the pressure distribution under the base may riot be uniform.

5.

This base thickness is fairly high, and is governed primarily by shear considerations. If the distance between columns is large, bending moment considerations will also require a, fairly deep base.

6.

This fairly large width has been chosen to reduce the pressure under the footing and satisfy the shear criteria. Although increasing the depth is generally more efficient than increasing the width, having a large depth may also cause excessive build up of h5t of hydration temperatures, leading to thermal cracking.

7.

Assulning the column loads to be point loads is conservative. In reality, the load will be spread over a (mite and the resulting bending moments and shear fol'CeS' at the column faces will ~, somewhat~maller than those obtained from this analysis.

8.

The argument used to choose the cover is the same as that in Note 7 of Example 22.

area

Coocludina Notes 9.

If the perimeter or section at which shear should be checked falls outside the footing,

the ,footing can be considered safe for shear. 10.

The analysis of the above footing has been performed assuming that both footing and subgrade are rigid. If elastic foundation assumptions had been used, the soil pressure near the columns (i.e. loaded areas) would increase, but the midspan bending moment would decrease.

72

EXAMPLE 24 - PILE CAP.

A two-pile group of pile diameter 500 mm and spacing 1250 mm centres supports a 450 mm square column carrying an ultimate load of 2500 leN. Design the file cap, using concrete of grade 25 and type 2 deformed reinforcement of fy = 460 Nlmm . Introductory Notes

1.

The minimum centre-to-centre distance for piles is twice the least width of piles for end bearing piles and thrice the least width of piles for friction piles.

2.

A pile cap can be considered as a deep beam, and the most appropriate way to analyse forces is to consider truss action in the pile cap.

Reference

Calculations

Output

Pile cap dimensions

Note 3

Use an outstand beyond the piles of half the pile diameter. Hence, pile cap di~ensions are: length = 1250 + 500 + 500 = 2250 mm width = 500 + 500 = 1000 mm Try overall depth of 700 mm; hence, effective depth = 700 - 40 - 25/2 = 647.5 mm > (1250)/2; hence O.K.

dimensions 2250 mm x 1000 mm x 700mm

N

61

Truss action Note 4

The force T is given by T N.l/(2)d = (2500)(625) 1 (2)(647.5) T = 1207 leN N/2 N/2 As = (1207 x1oJ) 1 (0.87)(460) , 21 ~ = 3016 mm2 Use 7 1'25 (As = 3437 mm2) Banding is not required, as pile spacing < (3) (4)(180)(1.4) == 1008 kN; hence O.K.}

84

l1w is O.K.

Caleulatio..

Refereace

Minimum reinfOfCQDCOt

4"

Min. r/f Note 8

= (0.25)(1000)(175) I (100) = 437.5 mm2/m (both directions)

Use vertical steel Tl2 @ 300 mm in both faces

(A. = 753 mm2/m)

Horizontal steel 1'8 @ 225 mm in both faces (As = 446 mm2/m)

- -

Output

vertfcal steel 2 x Tl2 @ 300 horizontal steel 2xT8@225

Notes on Calculations 4.

A column is considered braced in a given plane if it is not required to carry the lateral forces in that plane. A wall bowever is considered braced if lateral stability is given to it by other structural elements, when it is carrying in-plane loads. If the wall alone has to resist transverse loads, it is unbraced.

5.

Since the end conditions in the given wall are "midway· between those specified in Clause 3.9.4.3., the effective length factor is also midway between the factors given.

6.

The imposed load here is factored by 0.8, according to BS 6399: Part 1 (1984): Design loading for buildings: Dead and imposed loads, since loads from 3 floors are involved. Later on, when cbecldng the Ow value for the wall .panel, a factor of 0.7 is used, since loads from 4 floors are involved.

7.

Equations 43 and 44 for braCed walls correspond to the top (maximum initial eccentricity) and midway (maximum eccentricity due to deflection) sections. However OW is calculated at .the bottom of the wall,. taking into account the self weight of the wall and maximum inoment due to wind. This is slightly inconsistent but conservative. A similar approach is used in column design.

8.

Since reinforcement to control thermal and hydration sbrinkage should be fairly closely spaced, a spacing of 300 mm is not exceeded. 12 mm dia. bars are used for vertical steel, in order to give stiffness to the reinforcement cage prior to concreting. The horizontal reinforcement sbould be placed outside the vertical steel on both faces, to ensure better crack control, as thenna! and shrinkage movements will generally be in the horizontal direction; furthermore it is easier to fix the horizontal steel on the outside.

Concluding Notes

9.

The wall reinforcement should also be checked for satisfying tie reinforcements. This is dealt with in Example 33.

85

EXAMPLE 28 - CORBEL Design a corbel that will carry a vertic:alloed of 350 kN into a 300 mm x 300 nun column. assuming the line of action of the load to be 150 mm from the face of the column. Take ~u = 30 N/mm 2 and f y = 460 N/mm2 (defcmned type 2). Introductory Notes 1.

A cos:bcl can be considered to bea -deep cantilever"• where truss Don. as opposed beam action, predominates and where shearing action is critical..

2.

Compatibility of strains between the strut-and-tie system of the truss must be ensured at the root of the corbel (Clause 5.2.7.2.1 (b».

Calc:ulatioos

Output

Corbel dimensions

5.2.3.4

Note 3

The width of the corbel can be the same as that of the column, Le. 300 mm. The length of the bearing plate can also be taken as 300 mm, and if dry bearing on concrete is assumed, the width of thebeariag plate, b, will be given by (350 xloJ) I (3OO)b = (350 xloJ) I (300)(0.4)(30)= 97 nun Hence, choose bearing width of 100 mm. Since the corbel has to project out from the bearing area a distance that wouJd accommodate a. stressed bend radius choose corbel projection as 400 mm. Corbel depth bas lObe ·suclltbat max. aIL shear is not exceeded - i.e. (O;8)(3Of.5 = 4.38 N/mm2 Hence, d > (350 xl(3) I (300)(4.38) = 266mm Choose h = 375 nun and assuming cover of 20 mm (mild exposure conditions, concrete protected by lQ mm 1:3 et: sand rendel) and bar dia. of 20 mm, d = 375 - 20 - 2012 = 345 mm. Let the depth. vary from 375 mm to 250 mm. cAy

~l~n:

350 kN (ultimate)

'l5o L 1:- t

T~C'

T

125 Oo9xl[J 0.45f

-A

v

86

cu

bearing width 100 mm

total proj~tion 400nun

h =375 mm d::;: 345 mm

Reference

5.2.7.1

Output

Calculations

Now avid == (150)/(345) == 0.43 < 1

Also, depth at outer edge of bearing area > 375/2 mm; hence, definition of corbel is satisfied. Main reinforcement From strain compatibility and stress block, C == (0.45)fcu(0.9)b.x.CosP .......... (1) Since the line of action of C must pass thro' the centroid of stress block, (J == tan- 1(zlI50), Le. {J == tan-I {(d - 0.45x)IlSO} Furthermore, from the triangle of forces for P, T andC, C == P 1 Sin{J ........................(2) We need to find a value of x, and hence {J, that will satisfy (1) and (2) simultaneously. x == 216 mm will give {J == SS.SO and C == 409 kN T == 350 1 Tan{J == 212 kN

Note 4 5.2.7.2.1

.

NoteS

Since x == 216 mm, by strain compatibility, strain in steel is {(d-x)/x}(0.0035) == 2.090 xlO-] Hence, steel has just ~ded and fs == (0.87)fy Hence, As == (212 d()3) 1 (0.87)(460) == 530 mm2 Use 3 T16 (As == 603 mm2) . Min. area required == (112)(350 xloJ) / (0.87)(460) == 437 mm2 < 603 mm 2 ; hence O.K. Also l00A/bd == (100)(603) 1 (300)(345) == 0.58 > 0.4 and < 1.3; hence O.K.

main steel 3 T16

Detailing O.K.

Shear reinforeeroent

Table 3.9 3.4.5.8 Table 3.8

5.2.7.2.3

v == (350 xloJ) 1 (300)(345) == 3.38 N/mm2 l00AJbd == 0.58 Vc == (0.546)(30/25)°.J3(2d1ay) == (0.58)(2/0.43) == 2.69 N/mm1 < 3.38 N/mm 2 Provide An > == bv.Sy(v-vJ 1 (0.87)£ v Aav!Sy > == (300)(3.38-2.69) 1 (0.87J(460) == 0.517 Use lOT @ 300 mm. Since this has to be provided over (213)(375) == 250 mm, 2 bars will suffIce. Min. requirement is 603/2 == 302 mm2 Use 2 TI0 links @ 175 mm (As == 314 mm2 > 302 mm2; hence O.K.)

87

links

2TlO

Reference

Bendine main reinforcement Note 6

3.12.8.25.2

equation 50

The bend in the main reinforcement should start a cover distance (20 mm) from the bearing plate. It should end a cover + bar di.a. (20+ 10+ 16 = 46 mm) from the end of the corbel. Hence. distance available for bend radius == 200 - 20 - 46 = 134 mm Critical value of 3tJ= 20 + 10 (link) + 16 = 46 mm Stress in bars = (0.87)(460)(5301603) = 352 N/mm 2 F bt I (vI» < = (2)~ I {I + (2)(qJ~} (352)(201) I (16)r < = (2)(30) I {I + (2)(l6l46)}

r > = 125 m m b e n d radius Choose r 130 mm < 134 mm; hence O.K. 130 mm

=

=

100· 100.. 200

.1

r-l30,., 250

2TIOfH75(

Notes on Calculations 3.

Varying the depth from a full depth at the root to 2/3 of the depth at the end ensures that one of the conditions for a corbel in Clause 5.2.7.1 is automatically met - i.e. that the depth at the outer edge of bearing is greater than half the depth at the root. Furthermore, it facilitates the placing of horizontal shear links in the upper two-thirds of the effective depth of corbel as specified in Clause 5.2.7.2.3.

4.

Using Figure 2.2, the strain at yield is (0.87)(460) 1(200 xloJ) = 2.0 x 10-3 for steel of fy '= 460 N/mm 2 , since the Young's Modulus specified is 200 kN/mm2 .

5.

Although these limits on l00A/bd, where d is the effective depth at th~ root of the corbel, are not given in BS 8110, they are specified in "Rowe, R.E. et aI., Handbook to British Standard BS 8110: 1985 : Structural use of concrete, Palladian, London, 1987".

88

6.

Although the code allows the bend to start at the edge of the bearing plate itself, the allowance of a cover distance from the outer edge of the bearing plate will ensure the spreading of load from the bearing plate to the level of tie steel before the bend commences.

Concluding Notes 7.

Since a fairly large distance is involved in accomodating the bend radius, an alternative way of anchoring tie bars is to weld a transverse bar of equal strength, subject to the detailing rules in Clause 5.2.7.2.2. In any ease, the actual projection of the corbel beyond the bearing plate can be adjusted right at the end of the design, and will not affect preceding calculations.

89

EXAMPLE 19 - DESIGN FOR TORSION A cantilever slab of clear span 2.0 m functions as a hood over a porch. Its thickness varies from 200 mm at the support to 100 mm at the free end, while it carries finishes amounting to 0.5 kN/m2 and an imposed load of 0.5 kN/m 2 • It is supported by a beam 600 mm x 300 mm, which spans 4.0 m between columns, which are considered to provide full bending and torsional restraint. Design the beam for bending and torsion, assuming feu == 30 N/mm 2 , f y = 460 N/mm2 (deformed type 2), fyv = 250 N/mm2 and density of reinforced concrete = 24 kN/m3 •

IDtroductory Notes

1.

It is instructive to classify torsion into two types. Compatibility torsion, which may arise in statically indeterminate situations, is generally not significant; torsional moments will be shed back into the elements carrying bending moments (at right angles to the element carrying torsion), because torsional stiffnesses are lower than bending stiffnesses. Any torsional cracking will be controlled by shear links. However, equilibrium torsion in statically determinate situations, where torsional resistance is required for static equilibrium, will have significant magnitudes, and has to be designed for. The example above is such a case (see Clause 2.4.1, Part 2).

2.

Assuming that the columns provide full bending restraint implies that they have infinite stiffness. In practice, of course this will not be the case and the deformation of the columns will reduce the beam fixed end moments. However, full torsional restraint has to be provided by the columns, in order to preserve static equilibrium, where equilibrium torsion is invol"lled.

CaleuIatiODS

Reference

~oILJ

2000

Output

~$OO )

~.

Loadin 2 on beam

= {(0.2+0.1)/2}(2.0)(24) = 7.2 kN/m HoOd = (0.5)(2.3) Finishes = 1.15 kN/m Self weight = (0.6)(0.3)(24) = 4.32 kN/m Total dead load = 12.67 kN/m Imposed load = (0.5)(2.3) = 1.15 kN/m Design load ={(1.4)(l2. 7)+(1.6)(1.15)} = 19.6 kN/m

90

bending udl 19.6 kN/m

=

Calculations

Reference

Output

Torsional loadio& (assume shear centre is at centroid of beam section) Hood = (7.2)(2/3)(0.15+ 1.0) + (7.2)(1/3){0.15+(2.0/3)} = 7.48 kNmlm = 1,15 kNm/m Finishes == (0.5)(2.0)(0.15+ 1.0) Total dead load torsion == 8,63 kNm/m Imposed load torsion = (0.5)(2.0)(0,15+ 1.0) == 1,15 kNm/m Design load=={(1.4)(8,6)+(1.6)(1.15)} = 13.9 kNmlm

torsional udl =

13.9 kNm/m

Desim for bendin& TABLE 1 Example 8

Chart 2 (Part 3)

Table 3.27

Table 3.9

Assume cover == 30 mm (moderate exposure conditions, TABLE 1· values modified by Notes 5 & 6), link dia. ,.; 10 mm and main bar dia, == 20 Mm. hence, d = 600 - 30 - 10 - 20/2 = 550 mm Take M == (1I12)w.l2 (for built in beam) == (1112)(19.6)(4)2 == 26.1 IeNm Mlbd 2 = (26.1 xlo6) I (300)(550f = 0,29 lOOA,lbd == 0.08 Use lOOAJAc == 0,13 As = (0.13)(300)(600) I (loo) = 234 mm2 Same nominal steel r/f can be used at span. Shear Force = (19.6)(4) I 2 = 39.2 leN (max.) v = (39.2 xloJ) I (500)(300) = 0,24 N/mm 2 < Vc

d = 550 mm

Desi&D for torsion y I = 600 - (2)(30 + 10/2) = 530 mm

YI

= 530 mm

= 300 - (2)(30 + 10/2) = 230 mm

Xl

== 230 mm

Xl

Note 3

Total torsional moment = (13.9)(4) = 55,6 kNm Torsional restraint at each end= 55,6/2 = 27,8 kNm The torsional moment will vary as follows:27 .8 ~33

I

1.2m

--..~

"

~

-27.8

91

Refereace

CaJaJlatioos

equation 2 (Part 2)

Max. value of vt = (2)T I (bmiJi~ - huuJ3) = (2)(27.8 xld') I (300) (600 - 300/3) = 1.24 N/mm2 < 4.38 N/mm 2 (vtu ) > 0.37 N/mm2 (vt,min) Thus, beam section is O.K. but requires torsional r/f.

Table 2.3 (Part 2)

Output

Proyision of reinforcement Table 2.4 (part 2)

Since v < Vc for the entire beam, the area where vt < = vt,min bas to be provided with nominal shear r/f and the area where vt > vt.min with designed torsion r/f.

equation 2 (Part 2) Table 2.3 (Part 2)

Torque corresponding to edge of nominal shear r/f is given by T = V t ,min.hmm29tmax - ohznm/3) I 2 . = (0.37)(300f(600 - 30013) I 2 Nmm = 8.33 kNm Distance from beam elL = (8.33/27.8)(2.0)= 0.6 m Hence, length of beam for nominal shear links = (2)(0.6) = 1.2 m

Table 3.8

A,;.)Sv > = (0.4)(300) I (0.87)(250) = 0.55 For 10 mm links, Asv = 157 mm2 ; Sv < = 285 mm

Nominal shear links given by

Use RIO links @ 250 mm {< (0.75)d = 413 mm}

Nominal links RIO@250mm (middle 1.2 m)

Designed torsional links given by 2.4.7 (Part 2)

2.4.8 (Part 2)

A,;.)Sv > = T I

(0.8)~'Yl(0.87)f

= (27.8 xl

0

) I (0.8)(130)(530)(0.87)(250)

= 1.31 For 10 mm links, Asv = 157 mm2; Sv < = 120 mm Use 2RI0 links @ 200 mm {< = 200 mm, Xl' yl /2} Length of beam at each end for torsional links = (4.0 - 1.2) I 2 = 1.4 m Designed additional longitudinal steel given by

2.4.7 (Part 2)

2.4.9 (Part 2)

Note 4

As! > (A,;.)Sv)(fyvlf )(xl +Yl) = (157/120)(2501460)(230 +

530) = 540 mm 2 If this is divided between 8 bars, each requires 67.5 mm2 (3 at top and bottom, 2 in middle). Since beam length is small, assume bending reinforcement is not curtailed; longitudinal reinforcement for torsion also cannot be curtailed.

92

Torsion links 2RIO@2oo mm (1.4 m from both ends)

Reference

Table 3.30

2.4.9 (part 2)

Output

Calculations Total steel requirement at top and bottom levels = (67.5)(3) + 234 = 436.5 mm 2 Use 2Y16 + YIO at top and bottom levels (As = 481 mm2) and 2 Y 10 at intennediate level (As = 157 mm 2) This arrangement will satisfy (a) max. spacing for tension rlf < = 160 mm (b) max. spacing for torsional rlf < = 300 mm (c) torsional rlf provided in 4 comers

top & bottom 2Y.l6+YIO middle 2YlO

Notes on Calculations

3.

The torsional moment variation in beams, whether for a distributed moment such as this or for a point moment, is geometrically identical to the shear force variation corresponding to distributed or point loads respectively.

4.

Longitudinal torsion reinforcement has to be extended at least a distance equal to the largest dimension of the section beyond the point where it is theoretically not required. In this example, that would extend the reinforcement by 600 mm, exactly to the mid point of the beam. Hence, curtailment is not possible

Concluding Notes 5.

The links provided for torsion have to be of the closed type as specified in Clause 2.4.8 (Part 2), whereas even open links are permissible for shear links.

6.

If the section carrymg torsion is a flanged beam, it has to be divided into component (non-intersecting) rectangles, such that hmin3.hmax is maximized. This can generally be achieved by making the widest rectangle as long as possible (see Clause 2.4.4.2Part 2). The torque is divided up amoung the rectangles in the ratios of their (hmin3 .h max) values and each rectangle designed for torsion. The torsional links should be placed such that they do intersect.

division into 2 rectangles

IC

l-

J ~intersecting torsional II

93

links

EXAMPLE 30 - FRAME ANALYSIS FOR VERTICAL LOADS A typical internal braced transverse· frame for a multi-storey office building is shown below. The frames are located at 5 m centres and the length of the building is 40 m. The cross sectional dimensions of members are as follows. (i) Slab. thickness (roof and floors) - 150 mm (il) Beams (roof and floors) - 600 mm x 300 mm (iii) Columns (for all floors) - 300 mm x 300 mm The vertical loading is as follows:. (i) Load corresponding to finishes = 0.5 kN/m 2 (for roof and floors) (il) Load corresponding to light partitions = 1.0 kN/m 2 (for floors only) (iii) Imposed load on roof = 1.5 kN/m 2 (iv) Imposed load on floors = 2.5 kN/m2 (v) Density of reinforced concrete = 24 kN/m3 Obtain the design ultimate moments and shear forces from vertical loads for the beam ABC at the first floor level. Roof j 4.0m 1\~

2nd FIoar

4. Om B

A

C

\

1st FIoar

4.2f.m

, T75m. ~". ~

6.0m

7- ~

6.0m

-.

?!~ - - - -

Ground Level Footing Level

Introductory Notes '. L

The next 4 examples (including this one) deal with the entire structure, as opposed to structural elements.

2.

The loading for partitions and imposed loads is the minimum permissible under"as 6399: Part I (1984): Design loading on buildings: Dead and imposed loads~

3.

In general, most frames are braced, the lateral load being taken by masonry infill or

lift!stair wells. 4.

Since the frame is braced, it is possible to use either a beam level sub-frame analysis or a continuous beam analysis. Since the latter over-estimates moments considerably, the fanner will be performed. 94

Output

Calculations

Refereoce Stiffnesses

(IlL) of columns above 1st floor

=

(1112)(300)4/ (4000) = 0.169 xl()6 mm3 (Ill,tof columns below 1st floor = (1/12)(300)4/ (5000) = 0.135 x106 mm3 Since T-beam action will prevail in the beam , eff. flange width = 300 + (0.7)(6000)/5 = 1140 mm « 5000 mm). I of beam section = 1140 9.388 xl09 mm4 , (I/L)ofbeams = = (9.388 x109) I (6000) 1.565 xl~ mm3

b f = 1140 mm

°1¥

15 4 so

)

~ Distribution factors Only the beam factors will be considered. DAB = D CB =(1.565)/(1.565 +0. 169+0. 135) = 0.84 DBA =DBC = 1.565/{(1.565)(2)+0.169+0.135} =0.46

Loading on beam = (5)(0.15)(24) = 18 kN/m = (0.45)(0.3)(24) = 3.24 kN/m Finishes = (0.5)(5) = 2.5 kN/m Total dead load = 23.74 kN/m Imposed load (floor) = (5)(2.5) = 12.5 kN/m Partitions = (5)(1.0) = 5.0 kN/m Total imposed load = 17.5 kN/m Since a beam span carries 30 m2 of floor area, reduced imposed load = (0.97)(17.5) = 17.0 kN/m Slab

Beam

BS 6399: Part 1

Load arrangements 3.2.1.2.2

95

ik= 23.7 kN/m

Qk= 17.0 kN/m

Output Arrangement 3 will be the mirror image, about B, of Arrangement 2.

Moment distribution (kNm) Note 5 (Arrangement 1)

0.84

Note 6

0.46

AB -181.2 + 152.2.....,.

- 29,0

0.46

0.84

BA Be CB + 181.2 -181.2 + 181.2 + 76,1 - 76,1 f cu = 25 N/mm2

1

f y ;"'460 N/mm 2

690

750

,

3T25 000

Es

= 200 kN/mm 2

(All dimensions in mm)

Introductory Notes 1.

This crack width calculation can be performed when the bar spacing rules are not satisfied, to see whether this more accurate method will satisfy the crack width requirements in Clause 3.2.4 of Part 2. It can also be used to estimate the actual crack width in a flexural element.

Reference

Output

Calculations Sectional data

I

= 24S kNm

Note 2

Ms = (20+20)(7) 2 18

equation 17

E c = 20 + (0.2)(25) = 25 kN/mm 2 Eeff = (0.5)Ee · = 12.5 kN/mm 2 CX e = E/Eeff =' (200 xloJ) 1 (12.5 xHf) = 16 P = (3)(491) 1 (690)(450) = 0.00474 x/d = -cxe.p + {cxe.p (2 + cxe.p)}o.S = -(16)(0.00474) + [(16)(0.00474){2+(16)(0.00474)}]O.S = 0.321 x = 221 mm ~/bd3 = (1/3)(x1d)3 + a.,..p {I - (x/d)}2 = (1/3)(0.321)3 + (16)(0.00474)(1-Q.321Y = 0.046 Ie = 6.80 x10 9 mm4

3.8.3 (Part 2) Note 3 Note 4

Note 5

107

M s = 245 kNm

x = 221 mm

Ie

= 6.80 xl09 mm~

Calculations

Reference

Output

Calculation of strains Strain in steel .

equation 13

(part 2) equation 13

(12.5 xl