Design Example of Reinforce Concrete Beams

Chapter 2

Design of Reinforced Concrete Beams

2.0 NOTATION

D' Compression face to point on surface of concrete where crack width is calculated

Db Centre-to-centre distance between bats or groups of bars ~. Po int on surface of concrete to nearest face of a bar Ac Gross area of concrete in a section As Area of steel in tension A.; Area of steel in compression A,v Area of steel in vert ical links b Width of reinforced conCrete section be Breadth of compression face of beam mid-way between restraints hi Width of section at centroid of tensile steel bw Average web width c Coefficie nt of torsional " jffness emin Minimum cover to tensile re inforceme nt C Torsional stiffness d Effective depth of tensile reinforcement d' Effective depth of compressive reinforcement d l From tension face of concrete section to centre of tensile

reinforcement Ec Modulus of elasticity of concrete £, Modulus of elasticity of stcel I, Service stress in steel reinforcement Iy Characteristic yield strength of steel I; Revised compressive stress in steel taking into account depth of neutral

axis laJ Characteristic cube strength of concrete at 28 days Iyv Characteristic yield strength of reinforcement used as links F Coefficient for calculation of cracked section moment of inenia F bI Tensile force in a bar at start of a bend G Shear modulus h Overall depth of a concrete section hr Thickness of flange in a T ·beam hmu. Maximum overall dimensio n of a rectangular concrete beam hmin Minimum overall dimensio n of a rectangular concrete beam I Moment of inertia I Clear span or span face·to--face of support

41

42 Reinforced Concrete

M M, M, N

P p' , S. S. T T,

V' .min

v V. V, V,

V"'~" Voom

Vro~ Wmax

X

X,

y , ,

Effective span Centrc-to-centre distance between supports modular ratio = Est Ec Applied bending moment Design bending moment modified to account for axial load Moment of resistance of concrete in flanged beams Axial load Percentage of tensile reinforcement Percentage of compressive reinforcement Internal radius of a bend in a bar Spacings of bent bars used as shear reinforcement Spacing of vertical links Applied torsion Proportion of total torsion carried by each rectangle of an 1-, T- or L-scction Shear stress in concrete (N/mm2) Design concrete shear stress (Ntmm2)

Shear stress in concrete due to torsion (N/mm2) Ultimate permissible torsional shear stress (N/mm2) Design concrete torsional shear stress (N/mm2) Shear force in concrete section Shear force carried by bent bars Shear force capacity of concrete section Shear force carried by vertical lin ks Uhimate maximum shear forces allowed on sectio n

Shear force capacity of concrete section with minimum vertical links Design shear resistance of concrete Maximum crack width (mm) Depth of neutral axis from compression face Centre-to-centre of two external venical legs of a link Centre-to-centre of two external horizontal legs of a link Depth of leve r arm

Angle of inclination to horizontal of shear reinforcement Angle of inclination to horizontal of concrete strut in truss analogy Ratio of redistributed moment over clastic analysis moment Factor governing moment of resistance of concrete T-section Material factor Calculated strain in concrete at depth h Strain with stiffening effect corrected Strain at centre of steel reinforcement Yield strain in steel reinforcement Strain at centre of compressive reinforcement Strain at depth h corrected for stiffening effect Calculated strain in concrete ignoring stiffening effect Poisson's ratio Diameter of a reinforcing bar or equivalent diameter of a group of bars

Design of Reinforced Concrete Beams 43

2. 1 ANALVSIS OF BEAMS

2.1.1 Eftedive spans

l

l

\

SK 2/2 Continuous beam. SK 213 Cantilever beam. SK 211 Simply supponed beam.

Simply supported or cncastre le:E smaller of (/ + d) or 10

Continuous

Cantilever d

le= I + -2

where to = centre-to-ccnlre distance between supports Ie ::: effective span I = clear span or span to face of support d "" effective depth of tension reinforcement.

2.1.2 Effective width or compression flange

r. SK 2/4 Effective wid th of ~ k.

~ ~ ~ ZI

l h. compression fta nge. SECTION A - A

Simply supported T-beam

Simply supported L-beam

Continuous or encastrt T ·beams

Continuous or encastre L-beams

I. b=-+b 5 •

I. b = 10 + by;

I. b -- +b 7.14 w

~


where b = effective width of compression flange hw = average width of web.

Notl!: Use actual b if it is less than the calculated b using the above formulae. A typical example may be a precast T-beam.

2.1.3 Moment of inertia

... ~

~.

Mdhod I Gross concrek section only Find moment of inenia of gross concrete section - see Table 11 .2.

Method 2 Uncracked transformed concrete If reinforcement quantities arc known. find moment of inertia of transformed concrete section using Table 11.2.

Method 3 Average of gross concrete section and cracked section

A' • y

~

b A.

SK ZlS Doubly reinforced beam.

where I = moment of inertia of rectangular concrete section b = width of rectangular concrete section h .: overall depth of rectangular concrete section F = facto r - see Fig. ] 1. 1 for values of F.

A, P = 100 bd

where A. = area of tensile reinforcement d = effective depth to tensile reinforcement.

A' p' = 100 --.!

bd

where A; = area of compressive reinforcement.


. E. m = modular ratiO = -

E<

The graphs in Fig. IL l have been drawn for p' = 0 and p = p'. Intermediate values may be interpolated .

Note: The preferred method is Method 3 for rectangular sections. Where reinforce· ment quantities are not known . an assumption may be made of the percentage of reinforcement.

T-beams and L-beams in a frame or continuous beam structure should be treated as rectangular beams for the purpose of determining moment of inertia. The width of the beam will be taken equal to bw •

2.1.4 Modulus of elasticity

Modulus of elasticity of reinforcement stee l

E. = 200 kN /mm2

Modulus of elasticity of concrete, Ec. for short-term and long-term loadings is given in Table 2. 1.

Table 2.1 Modulus of elasticity of concrete: shon·term and long-term loading.

i<" Short-term loading. E~ Long-term loading, Ec (N/mm2) (kN/mm2) (kN/mm2)

2() 2' 12 25 25 12.5 30 26 13 40 28

" '0 30 15 60 32 16

Note: Wind load is short-term loading and dead load is long-term loading.

2.1.5 Torsional stiffness

For a rectangular section the torsional stiffness. C, is given by

C = ch!'mhmax

where c = coefficient from Table 2.2 hm .... = maximum overall dimension of rectangular section hmin = minimum overall dimension of rectangular section.


r i ~

:J=:Ih ••

hlllirl

h~.

I' SK 216 Component rectangles of a beam to find torsional sti ffness.

The torsional stiffness of a non-rectangula r section may be obtained by dividing the section into a se ries of rectangles and summing the torsional stiffness of these rectangles.

TaMe 2.2 Values of coefficient c.

h",,,,, 1.5 2 3 hm;n

, 0 . 14 0 .20 0 .23 0.26

The coeffi cient c is given by the following formula:

1 ['6 (k')] c = 16 3" - 3.36k 1 - 12

where k = hmin.

lima"

5 JO

0. 29 0. 31

2.1.6 Shear modulus

Shear modulus, G. is given by

E G = '2 (t + !1) = 0 .42E~ fo r concrete

where !1 "" Poisson's ratio.

Note: In normal slab and beam or framed construction . torsional rigidity of RC beams may be ignored in the analysis and the torsional stiffness may be give n a very small value in the computer analysis. Torsional rigidity becomes important only whcre torsion is relied on to carry the load . as in curved beams.


2.1.7 Poisson's ratio

Poisson·s ralio for concrete::: 0.2

2.1.8 Shear area

Shear a rca of concrete = O.8A c

where Ac::: gross cross-sectional area of concrete.

Not~: The sbear area of concre le is entered as input to some computer programs when the analysis is required 10 lake into account the deformations due to shear.

2.1.9 Thermal slrain

The coefficients of thennal expansion are given in Table 2.3 for different types o f aggregate used.

Table Z.3 Coefficient of thermal expansion.

Aggregate type Flint. Quar1zile

Granite. Basalt Limestone

Coefficient (x IO- nr c) 12 10 8

Note: Normally for ultimate limit slate no specific calculations are necessary fo r Ihennal loads. Thermal calculations should be produced for st ructures in contact with hOi gases o r liquid .

2.2 LOAD COMBINATIONS

2.2.1 General rules

The following load combi na tions and partial load factors should be used in carrying out the analysis of beams:

Le i: 1.4 DL + 1.6 LL + 1.4 EP + 1.4 WP L e 2: 1.0 DL + 1.4 EP + 1.4 WP Le l : 1.4 DL + 1.4 WL + 1.4 EP + 1.4 WP Le.: 1.0 DL + 1.4 WL + 1.4 £ P + 1.4 WP LCs: 1.2 DL + 1.2 LL + 1.2 WL + 1.2 EP + 1.2 WP

lVote: Load combinations LC2 and LC. should be considered when the effects of dead load and live load are beneficial.


where DL = dead load LL = live load or imposed load WL = wind load WP = water pressure EP = earth pressure.

The general principle of load combination is to leave out the loads which have beneficial effeci. If the load is of a permanent nature. like dead load , earth load or water load. use the partial load factor of 1 for that load which produces a beneficial rather than adverse effect. This rule of combination will be used for design as well as for the check of stability of a structure.

2.2.2 Rules of load combination for continuous beams

SK 217 Continuous beam loading sequences.

(1) Load all spans with Lei'

liT'!''!' '!''!' I! lei ON ALL SPANS

LC, LC, LCI

~ 1&. ALTERNATE WITH LC2 FOR MAXIMUM MIDSPAN MOMENT

(2) Load alternate spans with Lei and other spans with LC2•

(3) Load beam in the repeated sequence of two adjacent spans loaded with LCI and one span loaded with LC2. This sequence givcs the maximum support moment between adjacent spans. This is not a normal requirement . as per clause 3.2.1.2.2 of BS 8110: Part 1: 1985.111


2.2.3 Redistribution of moments

2.2.3.1 Continuous beams

SK 2/8 Continuous beam - typical moment redistribution.

• •

Usually 10% redistribution of moments may be allowed from those obtained by elastic analysis . Redraw be nding moment diagram with redis· tributed moments. Calculate revised shear. Reduction of support moment means a corrcsponding increase in span moment. For structu ral frames over four stories high providing lateral stability , the redistribution of moments should not exceed 10%. Resistance moment at any section must be at least 70% of moment at that section obtained by elastic analysis.

1.1.3.1 Frame structures No reduction or redistribution of moments is allowed from the columns.

2.1.3.3 Continuous one· way spanning slab panels Usually 10% redistribution of moments may be allowed from those obtained by elastic analysis.

2.2.4 Exceptional loads

Exceptional loads may be any of the following.

( I) Accidental loads of very low probability properly quantified . The definition of low probability may vary from project to project and will be agreed with the client.

(2) Probable misuse and its effect accurately quantified. (3) Once in a lifetime very short-lerm loads which are accurately quantified.

Note: With exceptional loads some rectification of local damage after the incident may be necessary.

Load combination to be considered:

LC6 "" 1.05DL + 1.05LL. + 1.05£L + l.05WL j

where DL = full expected dead load LL. = full expected live load if th is is a storage building, other·

wise, one-third of expected maximum live load EL = exceptional load

WL l = one· third o f expected maximum wind load .


2.3 STEP·BY·STEP DESIGN PROCEDURE FOR BEAMS

Step J Analysis Carry out analysis - follow Section 2.1.

Step 1 Moment I!n"l!wpe

SK 219 Typical moment envelope of a continuous beam.

Draw maximum-minimum ultimate load bending moment envelope after rcdistribution.

SK 2110 Typical shear envelope of a continuous beam.

11111111111111 1 Vmax

Draw maltimum-minimum ultimate load shear forcc envelope after redistribution .

Stl!P 4 Axial loads Determine coacting axial loads with maximum and minimum bending moments respectively. Ignore axial load if less than O.lfcubh.

Step 5 Torsions Determine coacling torsions with maximum and minimum shear forces respectively.

SK 2/ 11 Dimensions to compute axial load in beam.

St~p 6 Cov~r to reinforc~menl

Design of Reinforced Concrete Beams Beams 51

Determine cover required to rei nforcement as per Tables 11.6 and 11.7. Find effective depth d. assuming re inforcement diameter.

SK Zl 12 Rectangular beam -effective depth .

Sl~p 7 EJf~cti,,~ span Dete rmine effective span - see Section 2. 1. 1.

Slep 8 E/J~cti,,~ width of compression flilnge Determine effective width of compression flange - see Section 2. 1.2.

SI~P 9 Sletukrn~ss rati&

r 1

L

SK 2/ 13 Simply supported and continuous beams.

SK 2114 Cantilever beam.

Check slenderness of beam as per clause 3.4. 1.6 of BS 8UO: Part I : 1985.111

For simply supported and continuous beams.

I < 60bc or 250 b~ d

For cantilever beams.

100 b~ d


where d = J(0. 15~fcub ) or actual d, whichever is lesser

b = effecTive width of the compression fl ange

M = design ultimate moment.

Step /0 Design for moment - rectangular beam

I

I +--

I

!

Select critical sections on beam for bending moment. Find the following parameters al all critical sections, for rectangular beams and flanged beams.

A' j'

~ , <

---- I ~ I-~ , <

~1 M

N

SK 2116 Rectangular beam -moment and axial load.

II; d'

~]~~.~ T

~ •

" , ~

SK 2117 Rectangular beam -stress diagram.

SK 2115 Rectangular beam -doubly reinforced.

(or N :s: O.l fcubd

Note: For N > O. l fcubd, design as column (sec Chapter 4). Md may also be taken equal to M where NS O.l fcuhd and N may be totally ignored. (Sign convention: N is +vc for compression .)

M, K - -

- fm bd'

, = +5 + J(025 - OK9) 1 '" 0.95d d - ,

x = --0.45

M, N A - -----

• - 0.87 fyz 0.87 fy

K' := 0.156 when redistribution does not exceed 10%

K' = 0 .402(~, - 0.4) - O . 1 8(~, - OA)' exceeds 10%

M' " = - < 0.9 >'b M'

when redistribution

Design of Reinforced Concrete Bcams S3

where M2 == moment after redistribution; M' = moment before redistribution

When K > K',

,- d[05 + J(025 - :~) l d -,

x - -- ::5 O.5d 0.45

A' _ (K - K')f~bd2 • O.f':7f , (d - d')

A. = K" bd2 N

Jeu +A'- --O·f':7f" • O.f':7f,

SK 2fl8 Strain dillgram.

If d'lx> 0.43x,

A ' _ (K - K' )f, .bd' • f; (d - d' )

K'/cu bdz , N A= + A- --

$ 0 .87 /y1. I 0.87 /y

M ~

X X ~ ~

0

r---l.'-(X - d') ,

I - 0.57x Ym because steel strain e,; =

Ii I E, I

E', - -- -

(X - d') 0.S7x Ey

where Ey corresponds to steel stress /ylYm, as in Section 1.4.2.

" 0 ,


u

Note: The flanged beam becomes a rectangular beam if the bending moment produces tension in the fl ange .

Design charts in 8S 811O: Part 3: 1985111 may be used if design parameters fall within the scope of the charts.

Step II Design jor moment - jlJJnged beam

b

'/////~/' A

• /'

w SK 2119 Flanged beam - singly reinforced.

A fl anged beam should be designed as a rectangular beam with width equal to the effective width of fl ange in compression if x s lo1h,. If x > Uhf. find blb w and dlh,. Obtain fJr from Fig. 2.1-

Calculate M. = i3r/cubd1

IfM, 2:: Md•

Md + O.lfcub .. d(0.45d - hd N A = - - -

• O."'[,(d O.5h,) 0.87[,

If M, < Md. follow Section 1.5.4 using the second approach to design of Hanged beams.

SUp 12 Maximum allowable shear stress Find maximum shear in beam from shear envelope at the face of support or under a concentrated load. Find v = Vlbd.

Check. v s O.8 V/c ... s 5N /mm2

Change beam geometry if this condition is not satisfied .

A._ lo-J

SK 2120 Rectangular beam. SK 2121 Flanged beam.

Design of Rei.nforced Concrete Beams S5

Step 13 De$igtl for shear

>3d i1: , ,

SK 2/22 Critical section for shear. Shear check based on bottom reinforcement adequately anchored .

SK 2/23 Critical section for shear. Shear check based on top reinforcemenl adequately anchored (A ; to continue at least 3d from face of support).

Select critical sections on beam which are a l a distance of 2d from the face of support or concentrated load. Find the following parameters for rectangular and flanged beams.

V = design ultimate shear force

v V"" bd s 0.8 V/eu S 5 N/mm2

lOOA. p =-

bd

Find Ve from Figs 11.2 to 11.5. When axial load in compression. N. is present,

. (NVh)., , Ve "" Ve + 0.6 -- S 0.8 v/eu :S 5N/ mm

A<M

Note: N1A e is average stress in the concrete section. VhIM:s 1 and moment and shear at the section under consideration must be for the same load combination . N is +ve for compression and - vc for tension. To avoid shear cracks at ultimate load, limit shear stress to

Replace Vc by v~ where axial load is present.

F' d O.4bS.,. In A sv = ---

0 .87/ ., ...

I.,,, :s 460 N /mm2 for links.

Provide minimum area of links, A • .,. , at a spacing of S" for the zone where shear is less or equal to Vnom . From the shear force envelope determine zones where V exceeds Vnom= ( V" + O.4)bd


SK 2/24 Elevation of beam showing shear reinforcement. 5

SK 2125 Single system of bentup shear reinforcement.

SK 2126 Multiple system of bent-up shear reinforcement.

F· d A "" bS .. (v - v,) In .v .

0.87["

Replace l ie by v~ where ax ial load is present. Provide area of links. Am at a spaci ng of Sv at the section of the beam in consideralion. For a mixture of links and benl-UP bars.

Vb + V. ~ ( v - ve)bd

where . (d-d' )

Vb "" A ~b(O.87fy ... )(COSO: + sma cot ~) Sb

d V. = A s,,(O.87fyv)- ~ 0.5(v - vdbd

S,

a and ~ i!!: 4SO

Replace Vc by v; where axial load i n compression is prese nt .

Note: Slep 14 be low may be omitted if at Step 13 the crit ical section is selected at a distance of d from the face of support or from Ihe concenlrated load . No further checks wi ll be necessary at the face of sup pori o r a t the concentrated load .

Step 14 Afternatit'e design for shear Find Vma• "" 0.8 V feu bd, o r = Sbd, whichever is less . Complete the table below:

Distance from face of v ....... = V ..... _+O.4bd support or conce ntrated load

2.00d (v~ + U.4)bd 1.75d ( 1.1431'" + 0.4 )bd !.SOd ( 1.3331'" + O.4)bd 1.25d ( 1.61'" + O.4 )bd !.OOJ (2.0v" + O.4)bd O.75d (2.67v" + O.4)bd 0.50£1 (4v,, +0.4) bd O.25d (8 v" + O.4) bd

v......, < Vmn

I'chd 1.1 43v~bd

1.333vebd 1.611,.bd 2.0v.,bd 2.67v .. btl 4.0v.,bd 8 .0v.,bd

Design of Reinforced Concre te Beams 57 '

v Distance from face of $Upport or concentrated (act ual $hear force)

''''. 2.00d 1.75d 1.50d 1.25tl 1.00d O.75d 0.5Od O.25d

Satisfy the following conditions: when V :s Vnom ,

A = O.4bS~ I~ O.ff7[p

when V > Vnom• calculate

d V, = A.v (O.ff7[y,,)- ~ V - Ven"" s.

V - V.,.,...,

The shear resistance may be provided by a combination of links and bcntup bars.

Step 15 Minimum tension reinforcement

b Effective' width

A, A,

SK 2/1.7 Flanged beam . SK 2128 Rectangular beam.

For flanged beam web in te nsion. find b .• /b. If b..,lb < 0.4

A . 2: O.OOlSb ... h

If b""lb2: 0A

A s 2: O.OOt3b ... h

for [ y ;: 460N/mm2

for [ y = 460N/mm2

· 58 Reinforced Concrete

For flanged beam fl ange in tension.

for T·beam As i'! O.OO26b",h

for L-beam A~ 2:: O.OO2Obwh

For rectangular beams.

for fy""' 460N/mm2

for fy = 460N/mm1

A. ~ O.OO l3bh for f 'J = 460 N/mm2

Step /6 Minimum compresswn reinforcenunt - when designed as doubly reinforced For fl anged beam fla nge in compression.

A; O!: O.OO4b11,

For flanged beam web in compression.

A; 2:: O.OO2bwh

Fo r r<X:tangular beam ,

A; O!: O.OO2bh

Note: Minimum compression reinforcement in beams will be used o nly when compression rei nforcement is required.

Step 17 Minimum tranSllerse reinfon:ement in jUlnge

SK 2129 Minimum transverse reinforcement in flange of flanged beam.

Fo r flanged beams over fult effeclive fl ange width near top surface. use 1.5h, mm1/m reinforcement for the whole length of the beam. Normally this amount of reinforcement is providL-d in the slab at the top surface over the beam as part of slab reinforcement when the flanged beam forms part of a beam- slab construction.

Step 18 Minimum reinforcement in side face of ~ams

d ... dia. of bar 2:: J(StP) /,

Sb S 250 mm b"" actual. or 5OOmm. whichever is the lesser.

Design of Reinforced Concrete Beams S9

SK 2/30 Minimum reinforcement side face of beam .

\1 .G

.G

/' L ~ , -

./ L ~ , N

AS~O.0012Sbh

Note: To control cracking on the side faces of beams use small diameter bars at close spacings. The distribution of these bars should be over two-thirds of beam's overall depth measured from tension face.

A. :2: O.OOI25bh on each side face as shown.

Step 19 De./kclion

, Ellec:t iw width

A~

t_ .. lJ

SK lI3t Doubly reinforced Hanged beam .

•

SK 2/32 Simply supported or continuous beam. M = moment at midspan.

Find b .... /b for Hanged beams. Find Ie/d.

" /.

SK 2133 Cantilever beam. At= moment at suppon.

Find basic span/effective depth ratio from Table 11.3.

Note: If b",lb is greater than 0.3, then interpolate between values in Table 11.3 assuming bw/b equal to I for rcctangular beams and 0.3 for Hanged beams.

60 Reinforced Concre te

_. . ,,(5)(~) Fmd service stress ,. = Iy R R .... b AI prov

where ~ ::::: M /M ' M = moment after redistribution

M' = moment before redistribution A s .eqd = area of steel required from calculations AI p ..... = area of stee l actually provided.

Find M lbd2•

Find modification factor for tension reinforcement from Chart 11 .5.

Find lOOA ~/bd.

Find modification factor for compression reinforcement from Chart 11.4.

Find modified span/depth ratio by multiplying the basic span/depth ratio by the modification factor of tension and compression reinforcement.

Check ielli < modified spanfdeplh ratio

Nou: Table 11 .3 can be used for up to a 10m span. Beyond a 10m span multiply these values by lO/span except for cantilevers where deflection should be calculated (see Section 1.8 for calcula tion of deflection .)

Stf!p 10 Maximum arf!as of reinforcement For all beams,

A~ :5 O.04bwh A; :S O.04bw h

Step 11 Containment of compression reinforcement Designed compression reinforceme nt in a beam should be contained by links . Minimum diameter of links = 0.25 times diameter of largest compression bar, or 6 mm. whichever is greate r . Maximum spacing of links= 12 times diameter of smallest bar in compression.

SK 2134 Containment of compression reinforcement.

Design of Reinforced Concre te Beams 61

Step 22 Bearing stress inside bend

Ix

x o

SK 2135 Bearing stres.~ inside bend. ~_"",_w _...st:CT I ON )( - X

Cheek bearing stress inside bend where it is required to extend the bar for more than 4 x diamete r beyond the bend because the anchorage requirement is not o therwise satisfied. Satisfy that

F~ _-,2,,1~7-:-' bearing stress = - " -'$ 1 + 2(!)

where Fbc = tension in bar at the sta rt of bend a b = centre 10 centre of bar, o r, cover plus diameter of bar.

Step 23 Curtl1iJnuml of bars Follow simplified de tailing rules for beams where the load is predominantly uniformly distributed and spans in a cont inuous beam are approximately equaL Follow bending moment diagram fo r other cases.

Step 24 Spacing of ban Minimum clear spacing horizontall y = MSA + 5?; diameter of bar

where MSA = maximum size of aggregate.

. . I 2MSA Mimmum clear spacing vertically between ayers = --3-

47 000 Maximum clear spacing of bars in tension :5 - - " 300

f .

.. (5 )(~) ServICe stress In bar f. = f y 8f\, A , pru~


(Sec Step 19 for explanation of ~b')

Thc distance between the corner of the beam and the nearest longitudinal bar in tension should not be greate r than half the maximum d ear spacing.

Nole: In normal internal or external condition of exposure where the limita tion of crack widths to O.3 mm is appropriate, Step 24 will deem to sa tisfy the crack width criteria.

Step 2S Torsional shear slTess

• I =

SK 2/36 Rectangular section torsional shear stress.

Check torsional shcar stress.

hili •• .!

1 "I

I L

SK 2/37 Division intn rectangles of composite section_

Find ultimate torsion T from analysis. For a rectangular section, torsional shear stress, v" is given by

2T Y, = --;-=--;-~

h' (h hm

;") .... in max -""3

For I,T or L-scetion. divide each section into component rectangles.

Proportion of tolal torsion carried by each rectangle ::: Ts

Th!..lnh..."x ) = Ts

1: (hm;n hrnu)

Torsional shear stress for cach section

;


For hollow and other box sections. follow the method in Chapter 8. If wall thickness in a rectangular hollow section exceeds one quarter of the dimension in that direction . treat the hollow section as a solid rectangle. Calculate

v""" n = 0.067 yfeu < O.4N/mm2

Vtu = 0.8 y/eu :S 5 N/mm2

H v, < V,.min, no torsional reinforcement is required .

I-I-I--I---l-++++ - Spacing of links,. Sy

SySII't Vd2 and 200 A.y (Gl"fl of two outllckle9,1

A.IICll'l9/tUdInaI I"I lnforc.r.mt I SK 2139 Elevation of torsional reinforce ment in beam.

SK 2138 Torsional reinforcement in beam.

If VI.min < VI < V1\1 ' provide torsional shear reinforcement by closed links and longitudinal bars.

Check (v + v,) < VIII

where v = flexural shear s tress.

Cheek V , < VIU (:~)

Nok: Add torsional reinforcement to already calculated shear reinforcement.

Sy < XI> or ~ , or 2OO mm, whichever is the least

Step 26 Crack widJh in jkxurt!

Serv~eabiJity limil stale

Load combination LC1 = 1.0DL + 1.0LL + 1.0EP + 1.0WP + 1.0WL

Note: Omit loadings from LC1 which produce beneficial rather than adverse effect.


b, (h-.I{,'-. ) 3E,A.(d-.)

h , - :r~ ___ ___ _ b.

SECTION Of BEAM

"1 ,-1----

0

AI - -

Wmu = __ --::3a?'''~'~m'___c_

+ =2(~a~,~, _~'m~;",") (h - x)

b (h ) (a' - x) - , - x 3E,A,(d _ x)

SK 2/40 Critical dimensions for crack widlh calculalions.

SK 2141 Slr.!in diagram for crack width calcu lativlL~ .

SK 2/42 Doubly reinforced rectangular beam .

Nole: 10 1 is the strain due to load combination LC1 .

Design of Reinforced Concrete Beams 6S

where b. is the width of th e section at the centroid of tensile reinforcement .

For a rectangular section .

Note: A Hanged beam is a rectangular sectio n if x s 1. lh,.

E, m =-

E,

x = d{[ (mp + (m - 1)p')2 + 2(mp + (m -

- (mp + (m - I)P'») (See Section 1.13.1.)

A, p =-

bd

, A~ p=

bd

k' : (~) (1- ;) k,~ (m - I) ( I - ~)

/, ~m/, (; - I) /.

E."" -E,

(h - x)

Eh= -- " d-x

Note: In normal internal or external condition of exposure where the limitation of crack widths to 0.3 mm is appropriate. Step 24 will deem to satisfy the crack width criteria.

Step 27 Design of connections to other components Follow Chapter 10.

2.4 WORKED EXAMPLES

Example 2.1 Simply supported rectangular beam Clear span = 6.0 m


1I11 11 11UIII'il l lllllll l llll

:!o21_L-___ 6"OO"O'--__ J~~

Overall depth = 500 mm Width = 300 mm Width of supporting walls"" 200 mm

SK 2/43 Simply supported beam.

All rei nforcement to be used is high yield steel with / y= 460N/mm2.

Note: Steps 1-5 form part of the analysis and a re excluded from the worked example. For a Iypical ana lysis see Example 2.3.

Step 6 Determinillion of cOlier Maximum size of aggregate"" 20 mm Maximum bar size assumed = 32 mm Maximum size of link assumed = lOmm Exposure condition = severe Fin;! resistancc rcquircd = 2 ho urs.

Refer to thc following tables in Chapter 11:

Table 11 .6 grade of concre te"" C40 for severe exposure Table 11 .6 minimum cement content"" 325 kg/m3 Table 11.6 maximum free wate r/cement ratio = 0.55 Table 11.6 nominal oover = 40mm Table 11.7 nominal cover to beams for 2 hours fire resistancc = 40mm

For 2 hours fire resistance, minimum width of beam "" 200 mm, from Figure 3.2 of 8S81lO: Part 1: 1985.1l1

I"' •

SK 2/44 Section of rectangular beam.


Effective depth. d. is given by:

d = overall depth - nominal cover - dia . of link - half dia. of bar :::500-40-10- 16 = 434mm

Sup 7 Determination of effective span 1+ d = 6.0 + 0.434 = 6.434 m

J" = 6.2 m

Therefore Ie == 10 = 6.2 m

Sup 8 Determin4lion of effectil'e width Not required.

Step 9 Checlc slenderness of beam 1= 6.0m

60bc = 60 x 300 mm = IS.Om

250 b ~ 250 x 301'Y - -= =Sl.84m d 434

Satisfied

Step 10 Design for moment - rectanguUu btam

SK 2/4.5 Stress diagram of rectangular beam.

Maximum ultimate bending moment = 216 kNm Maximum shear at face of support = 140kN Shear a t 2d from face of support = 96 kN Shear a t d from face of support = 116 kN Direct load, N=OkN Md = M == 216kNm f cu= 40N/mm2

r U·4,·r ... ~18Nfmm2


Md 216 x lW K : -- "" = 0.0956 < 0.156

fcybd2 40 x 300 X 4342

z : +5 + J(025 - OK9)]

. +5 + J(025 - 0:':6)] = 0.8&1 = 382 mm

d -z 434-382 x :-- = "" 116mm

0.45 0.45

M, A,=---

0 .87fyz

216 x I(f = = 1413mm2

0.87 x 460 x 382

Use 3 no. 25 mm dia. Grade 460 = 1472 mmz

Step II Dtsign for moment - flanged bellm Not required.

Step 12 Check maximum allo .... able shear V

v "" - at face of support bd

140 x 10' 300 x 434

= 1.075N/ mm2 < 0.8 vfr;u = 5 N/ mm l

Step 13 Dtsign for shear 2d "" 870mm

v = 96kN al 2d from suppan face

IJ6 x lcP I ' =

300 x 434

= 0.74N/mm2 < 0.8 Vfeu = 5 N / mmz

100 A. p : - -

bd

100 x 1472 300 x 434

= 1.13%

v( "" 0.65 x 1.17N/mm2 for Grade 40 concrete = 0.76N/mmz from Fig. 1l .5


VIOOm = (v( + 0.4) bd "" 151kN > 140kN at face of support

V < Vnoon at all poinls in the beam.

.. O.4bsv Nommal lmks A $~::: --' -0,87[,.

Assume Sv ::: 3(X) mm

A sv ::: 0.4 x 300 x 300

0.87 x 460

= 9Omm2

Use 8mm dia. single closed link = A 5V::: lOOmm2 (f~ "" 460N/ mm2) at

300 mm centre to centre.

Step 14 AlJernQ/ille design for sheQr Vnom > Vat face of support so Step 14 is superfluous - use nominal links everywhere on the beam.

Step J 5 Minimum tensile reinforcement Minimum tensile reinforccmCnI "" O.OO13bh

=0.0013 x 300 x 5OOmm2

= 195 mm2 < 1472 mm2 provided

2 no. 12 diameter (::: 226mm2) providcd at top of beam.

Step 16 Minimum compression reinforcement Not required.

Step J 7 Minimum franSllerse reinforcement in jiQnge Not required.

Step 18 Minimum reinforcement in side fQCe of beams

u , u

0 0 M

• ~ ~ , ~

~ • 0 0 N

SK U46 Sectmn through beam.

'--'--'·3-;25

ALL REINFORCEMENT HIGH-Y IELD

0 0 ~

70 Reinforced Concre te

b = actual . o r 500 mm . whichever is the lesser.

Minimum dia . of bar in side face of beam = J(S"b) f,

(assume Sb = 2oo mm)

= Jeoo ;'1

300) = 11.4 mm

Use 12 dia . Grade 460 bars at approximalely 200 centres on the side face of beam.

Reinforcement on each side face of beam = 2 no. 12 dia. + I no. 25dia. = 7l6mm2

A, = O.OO125bll = 0.00125 x 300 x 500 = 188 mm 2 < 716 mm2 O K

Note: Strictly speaking these bars on tile side face arc not required for beams less than 750 mm ove rall depth but it is good practice to use them in order to

avoid shrinkage cracks.

Step /9 Clleck dejkction

&. = 6200 "" 14 .3 d 434

Basic span/depth ratio = 20 from Table 11 .3

~b = I ,

f, : f,(~)(A, ,",oj : 460 (~)( 1413) : 275N/mm' 8 A, pro, 8 1472

M 216 x Ht -: : 3.8 bd2 300 X 4342

Modification factor fo r !ension reinforcement = 0.90

Modified span/dept h ratio = 20 x 0.90 =

Hence deHection is OK .

Step 10 Maximum areas of reinforcement As is less than 4%.

Step 11 Containment of compression reinforcement Not required.

Step 11 Check bearing stress inside bend Not required.

10 18> - =

d

from Table 11.5

14.3


Step 23 Cuf14ilmenl of bors 0.081 = 0.08 x 6000 "" 480 mm

The central 25 mm dia . bar will be stopped 250 mm from the face of the support.

Step 14 Spacing of bars

2-,15

300 300 300

500

SK l/48 Arrangement of bar.; li t the bottom of beam.

SK l/47 Elevation of beam near suppon.

Oear spacing between bars in tension = 64.5 rnm

Minimum required spacing = 20 + 5 - 25 mm

. . 47(0) MaXimum spacmg "" -

f. 47000

= -- = 170mm 275

where I. = 275 N/mm1

Spacing of bar.; is OK.

Step 25 Check torsional shear stress Not req uired.

Step 26 Crack width cakulalions

(sec Step 19)

Service maximum momenl = 144 kNm A,= 1472mm2 d'=54mm A; = 226mm2 d = 439mm

m=~=200""l0 E, 20

Ec assumed halfway between long and short·term.


l_b J

~ • • " ~

SK 2149 Doubly reinforced rectllngular beam . SK 2/50 Strain diagram.

As 1472 p =~= = 0.0112

bd 30J X 439

SK 2151 Crack width calculations.

A ' p' = --.! =' 0.0017

bd

x =' d{[(mp + (m - 1)p')2 + 2(mp + (m -

(mp + (m - I)P'l =- 160mm

k , : (;)(1 - ;) : C ~~39)( 1 160 )

3 x 439

= 0.16

k3 -= (m - 1)(1 - :) = (10 - \)( 1 - 15~) = 5.96

M

f. : k bJ' + k A ' (d - d') 2 3 ,

: _____ -,-:-I44:..:..cX-'-"IO"'---,-,---_--:-:_-,-,-0.16 x 300 x 4392 + 5.96 x 226 X (439 - 54)

= 14.74N/mm2


:: to x 14.74 x (~: - I) = 257N/mm2

E.. = /5 = 257 = 1.285 x 10-3 Es 200 x 103

(" - x) "'= --,,= d - x (:)es = 1.566 x 10- 3

b(1l - x)2

Emh = Et! - 3E,A.(d x)

300 X 3402 = 1.566 X 10- 3 - :----:=-'-'-'-:;~-=,=-=

3 x 200 X 103 x 1472 x 279

= 1.425 X 10- 3

a~ 1 :: 1.414 x 60.5 - 12.5 = 73.0mm

ad = \1(60.52 + 452) - 12.5 :: 62 .9 mm

w = :lacr Em

"' I + [ 2(<,<, - em;")] II - x

3 x 73.0 x 1.425 x 10- 3

[2(73 - 48)]

1 + 340

:: 0.27 mm < 0.3 mm OK

Exampu 2.1 Three span continuous /Ham

SK 2152 Three-span continuous beam.

G o .1.10000 .f. l0000 J A 8 C 0

Three equal spans of 10 m centre-ta-centre of columns.

Width of column ::: 0.4 m clear span = 9.6m slab depth = 150 mm beam spacing = 4.0m beam overall depth = 550 mm beam width = 300 mm

Redistribution of moments = 10%

Nole: Steps 1-5 form part of the analysis and have been excluded . For a typical analysis see Example 2.3.


I

All reinforcement to be used will be high yield steel with fy = 460N/mm1.

It is expected that the analysis will be carried out using a computer program with the load combination shown in Section 2.2. From moment and shear envelope ,

MA=O V"a=300kN VA'B=250kN MAB = +600kNm VAH = negligible Ma =-650 kNm VnA =370kN Vn'A=320kN VBC = 325 kN Va'c = 275kN M(lc=+370kNm or -150kNm where V A'B = shear at a distance of d from face of support.

Step 6 Deurminalion of cover Maximum size of aggregate = 20 mm maximum bar size = 32 mm maximum size of link = 8 mm exposure condition = severe fire resistance required = 2 hours grade of concrete = C40 maximum cement content = 325 kg/m3 max imum free water cement ratio = 0.55 nominal cove r = 40 mm from Tables 11.6 and 11 .7 effective depth, d = 550 - 40 - 8 - 16 = 486mm

Step 7 Effective span Ie = 10 = 1O.Om

Sup 8 Effective I4IUllh of compression folnge Actual b = 4.0 m (centre-to-centre of beams)

1700

I. Calculated b = -- + b.

7.14

"I

10000 =--+300

7.14

= 1700 mm

SK 2/53 Effective width of compression flange.

Design or Reinrorced Concrete Beams 7S

Step 9 Slenderness check May be ignored.

Sup JO Design for moment M A8 =- 6CO kNm

Flanged beam Md =- MA8 = 6(X)kNm

K =~ fnbd2

600 x ul' 40 x 1700 X 4862

~ 0.0373 ,~ +5 + J(025 - OK9)]

-+.5+ J(025-0~3~)] = O.95d = 462 mm

d - z x ~--0.45

486 - 462 0.45

= 53 mm < II, = ISOmm

Neuual axis jn the slab

M, As =---

0.87/" 600 x Hf' 2

= 0.87 x 460 x 462 = 3245mm

Use 3 no. 32dia . bars in bottom layer plus 2 no. 25 dia. bars in top layer.

SK 1154 Anangemenl of reinforcement at bottom of beam al mKlspan .

2-2S¢

3-3211'

8

40 cover 10 -1-+ links


Total area of steel provided = 3394 mm2

Check effeclive depth . Centre of gravity of group of 5 bars

3 x 804 x 64 + 2 x 491 x 124.5 x ~

3394

"'" 81.5 mm

d "'" 550 - 81.5 = 468.5 mm

Rechec~ reinforce ment requirement with revised effective depth:

K ~ 0.040

1.:= 0.95 x 468.5 = 445mm

(required)

A . provided "" 3394 mm2 OK

Mn "'" -650 kNm

Reclongu/ar beam

M<J = 650kNm

Effective depth , d = 550 - 40 - 32 - 16 - 8 = 454 mm

(assuming two layers of 32dia. bars)

K = 650 x 1<1' 40 x 300 X 4542

= 0.263 > 0.1 56

Compression reinforcement required . Redistribution is 10%

A ' ~ (K - O.l56)f"bd' , 0.S7 f,( d - d')

(0.263 - 0.156) x 40 x 300 X 4542

0.87 x 460 x (454 - 64)

= 1696mm2

Use 3 no. 32dia. bars (2412mm2) - bottom of beam.

,~ -[o,J(0 25 - 0~1~6)l = O.775d = 352 mm

x = O.5d = 227mm and

_ O. 156/cubd2 •

A. - 0.87 /yZ + As

d' 64 ~ ~ - = 0.28 < 0.43 x 227

Design of Reinforced Concrete Beams n

= 0. 156 x 40 x 300 X 4542 + 1696 "" 4435 mm2

0.87 X %0 X 352

Use 6 no. 32 dia . bars (4824mm2) - top of beam in two layers.

300 I,(),o wr to linkS

6-32¢

U~ ~ -:- g -~-

~8 +';nk 5

SK 2155 Arrangement of reinfon:emenl al top of beam over support.

Mac= +370 kNm

Fhzng~d beam

b = 1700mm

d = 550 - 40 - 8 - 16 = 486 mm

M. K ~ -- ~ 0.023

f~bd2

z = 0.95d = 426mm

M. A. = --

O.87fyz

370 x lW _ 2 = 0.87 x 460 x 462 - 2001 mm

3 32¢ -

Use 3 no. 32dia . bar (2412mm2) - bottom of beam

Moe "" - 150kNm

Rectangular Inam

b = 300 mm

d = 486 mm

K = 150 x llt = 0.053 300 x 4862 X 40

J 4O <:0 WI' to links

!


z ~ + 5 + J (025 - OK9) 1 "" 0.94d "" 456mm

ISO x lIf A "" "" 822 mm2

• 0.87 x 460 x 456

Use 2 no. 32 dia. bar (1608 mm2) - top of beam,

SK 2/56 Section through 3-~3'2 midspan Be.

Step 11 Design for moment - flanged warn Not required ,

Step 12 Maximum shear stress

Vm~~ 1 2 V "" bd "" 2.716N/mm < 5Nfmm

Step /J Design for shear Maximum shear"" 370kN "" VBA

V 370xHP V"" - =

bd 300 x 454

"" 2.716N/mm 2 < 0.8 v/cu = 5N/mm2 OK

Check shear stress al d from face of column,

V A 'B = 250kN

d = 468.5mm

250 x lW , ~

300 x 468.5

= 1.78N/mm2

for span AB

lOOAs 100 x 3394 P = /;;f = 300 x 468.5

"" 2.41


V, ::: 0.85 x 1.17 from Fig. 11 .5 ::: 0.99 N /mm2

V""", = (v,+O.4)bd = 195 kN

v > Ve + 0.4 :::: 1.39N/mm2

bSv(v - ve) A.v = 0.87/y

(assumc 5v = 15O mm)

300 x ISO x (1.18 - 0.99) =

0.87 x 460

= 89 mm2

Use 8 mm dia. links= lOOmm2 (two legs) at ISO centre-to-centre up to the point where shear falls 10 195 kN. High yield reinforcement (fy = 460N/mm2),

. O.4bS. Nominal A.v "" --,

O.87J y

= 0.4 x 300 x J<X) = 90 mm1 0.87 x 460

Use Smm dia. links = lOO mm2 (two legs) at 300 centre-to-centre as nominal links (fy = 460N/mm2).

VB' A = 320 kN

320 x 10' v = J<X) x 454 (d = 454 mm at B)

= 2.35N/mm2

= lOOA. =- 100 x 4435 = 3.25 p bd 300 x 454

II, == 0.91 x 1. 17 = I.065N/mm2 from Fig. 11 .5

bSv( v - ve) A.v = 0.S7/

y

= 300 x ISO x (2.35 - 1.063)

0.S7 x 460

= 144.5 mm2

Use S mm dia . links = 150 mm2 (3 legs) at 150 cenlre to centre up to the point where shear falls to 195 kN .

Sup 14 Aiternali)le ihsign for shear Omitted.

Step IS Minimum tension reinforument


Flo.nged beam

bw 300 - = - = 0.176 < 0.4 b 1700

For web in tension

As > O.OOIBbwh "" 297 mm2

For flange in tension

A, > O.OO26bwh = 429 mm2

Both conditions satisfied.

Step 16 Minimum compression reinforcement A~ > O.OO2bwh "" 330mm2

Provided A; = 2412 mm2

Condition satisfied .

Step 17 Transverse reinforcement in flonge Minimum transverse reinforcement in flange = 1.5ll f mm2/m

= 1.5 x 150 mm2/m = 225 mm2lm

Reinforcement in the slab over the beam will be a lot morc than this quantity.

Step 18 Reinforcement in sitk face of beam For a 550 mm overall depth of beam with 150 mm slab, side reinforcement will not be required.

Step 19 Check deflection Ie 10(0) - = -- = 21.3 d 468.5

d = 468.5 mm

b. - = 0.176 < 0.3 b

for span AB

Basic span/effective depth ratio from Table 11.3 = 20.8

Since the ultimate moment at midspan is greater after redistribution than the ultimate clastic momcnt , the service e lastic stress may be laken as (518)1,.

Service stress, f. = ~fy (assumed)

5 =-x 460 8

= 288N/ mm2

Design of Rei nforced Concrete Beams 81

M -= bd'

600 x to<' = 1.6

1700 x 468.52

Modification factor = 1. 19 from Chart 11 .5

Modified span/effective depth ratio = 20.8 x 1.19

Step 20 Maximum areas of reinforcement A. :5 O.04bwh _ 66OO mm2

=24.75>21.3 OK

Maximum tensile reinforcement used = 4824 mm2 OK

Step 21 COnlainment of compression reinforc~mtmt Minimum dia . or links = 0.25 x max . dia. of bar

=O.25x 32 =8 mm OK

Max imum spacing of links = 12 x dia. of bar = 12x32 mm =384 mm OK

N()I~: At least one link at the centre of columns Band C will be required for containment.

Step 21 Ch~ck bearing stress jnsUle bend Not required.

Step 13 Curtoilm~nt of ban

0.151 = 1500 mm 0.101 = lOOOmm 0.25/ = 2500 mm

Span AB Continue 3 no. 32dia. + 2 no. 32 dia. up to lOOOmm from A (end support).

Stop I no. 32 dia. and 2 no . 32dia. at IOOOmm from A.

(See Step 26: reinrorcement in span AB increased .)

Over support B (lop bars)

Continue 6 no. 32 dia. bar top up to lSOOmm on either side of B.

Stop 2 no. 32dia. bar at 1500 from B.

Stop 2 no. 32dia. bar at 2500 from B.

Continue 2 no. 32dia. thro ugh span.

Step 24 Spacing of bon Minimum clear spacing = MSA + 5 =25 mm


Oear spacing of bars in tcnsion = 54 mm > 25 mm

47(0) 47(0) Maximum clear spacing:: -- = -- = 163 mm

f. 288

(See Step 19 for Il) At span BC lop tension reinforcemen t

Clear spacing of bars (2 no. 32dia.) = 140 mm OK

Nou: Under normal circumstances this step will deem to satisfy the 0.3 crack width limitation criteria. but, as Step 26 will prove, when crack width calculations are actually carried out this may not be the casco In span AB the maximum clear spacing criterion is satisfied but the calculations show that the crack widths may be exceeded.

Step 25 Check wrsionaJ shear stress Not required .

Step 26 Crack widJh calcuJations

Span AB Maximum service moment = 400 kNm

e min = 48 mm d = 468.5mm

b = 1700mm

A. = 3394mm2

A~ "" 1608mm2 (ignored in the computation)

E. m=-=lO

E,

x = J[ ('n:.)' + 2m:.d] _ mA,

= 118 mm < hr(= 150mm)

x z = d - - = 429mm

3

400 x 111 h = = 274N/mm2

s 429 x 3394

b

Eo = II = 274 = 1.37 x 10- 3 E, 200 x I!I'

Eh = (h - d) ... = ( 550 - 118) x 1.37 X 10- 3 d - :x 468.5 - 118

= 1.69 x 10- 3

Design of Reinforced Concrete Beams 8J

b(h - X ) 2

Emh = Eh - 3EsA.(d _ x )

"" 1.69 X 10- 3 _ 300 X 4322

3 x 200 x 103 x 350.5 x 3394

"" 1.61 X 10- 3

a.:! :::I V(642 + 642) - 16 = 74 .5 mm

ael = y(642 + 432) - 16 = 61.1 mm

a.:r = 74 .5 mm at the corner of the beam

__ ..,::3a~,~,~e~m,----,- : 3 x 74.5 x t .61 X 10- 3

Wcr = 1 + 2("" - ' m;. ) 2(74.5 - 48)

(h x) 1 + (550 118)

"" O.32 mm > O.3 mm

The calculated crack width is greate r than allowable. Increase reinforcement to 5 no. 32dia . bar instead of 3 no. 32 dia. plus 2 no. 25dia. No more checks are necessary .

SK US7 Arrangement of bars over support.

At face of column ,

maximum service moment = 390 kNm

d = 454 mm

b = 3(X) mm

As = 4824mm1

A; = 2412 mm2

d' = 64 mm

_ . • .. r-

• • • 1.80.1. '" ·1-,oJ


Cm in = 48mm

m=1O

See Step 26 of Example 2.l for explanation of symbols and the equations.

x = 225mm

K2 = 0.2068

KJ = 6.44

M {. = = 20.69 N/mm 2

c K2bd2 + KJA;(d - d')

f. = 2ll.6N/mm1

E,. "" 1.058 X 10- 3

Eh "" l .S02 X 10- 3

aCt = 74.5 mm at the top corner

Wcr ;; O.297 mm < O.3mm OK

SUp 17 Design of connections 10 olher elemenls See Chapter 10.

A

'so. T.nsioo

I ""'.,. C-· 12

"'" - I

I .- l l- .32 .,.

I !81ink$ (1800)

2-+12(sc

.1 150e/C SK 21SSA Detail of beam at A.

22SO <! 3- +32 2250 l- t 32ts«ono:! ~)

2- +321 1 [2 -.~

,-+" 1 I I I .. ,-

_t8to,ksfJlcQsll

l rH12 ,

_ f .B links 50 _, l_tBlinks I3t.QS) tl Unks ~ 300 de ;1\ I50clC ilt ISO cit .. JOOelc

SK 2fS88 Detail of beam al B.

Design of Reinforced Concrele Beams 8S

Example 2.3 Design of beam with torsion

SK 2/59 Two-span edge beam with nib.

Edge beam to carry precast floor slabs on nibs.

Clear gap between beams = 4.5 m Effective span of beam = 9.0 m

,

9. .!. 9.

See Example 5.2 for details of precast floor slabs and nib geometry computations.

Two-span beam is fully restrained at the rigid supports.

SUp 1 AnlJ/Y$;$ of beam

Properties of section

SK 2160 Section of beam with nib5.

Area of section = 500 x 290 + 2 x 110 x 105 = 168 I00 mm2

110

Self-weight of beam = 0. 1681 x 24 kNfm3 = 4 kNfm

x= 500 x 290 x 145 + 2 x 110 x 105 x (290 + 55)

168100

= 172.5 mm

1 1 J "" - x 400 X 5003 - - x 110 x 29(Y xx 12 12

= 3.943 X 109 mm" (gross section)

~ 0 0 ~

29'


x

p' Assume - :: 0

p

Assume p :: 1 %

E, Assume m :: - :: 10

Eo

From Fig. 11.1 .

F"" 6 X 10- 2

Cracked moment of inertia:: Fbd3

:= 6 X 10- 1 x (400 x 5OC.P - 110 x 29(f) == 2.839 X lOll mm4

Average moment of inertia. Ixx = 0.5(3.943 + 2.839) x 109 mm4

= 3.391 x 1OII mm4

- '-- X .r:

b = 400 - 145 = 255

SK 2161 Beam geometry to fi nd shcllr centre e.

h = 500 - 105 = 395 t = lOS

2552 X 3952 x 105 e = 4 x 3.391 X 109 = 78.5 mm

Loading Dead load from slab = 5 kNfm2 x 2.25 m = 11.25 kN/m.

Self-weight of beam "" 0.1681 x 24 kN/m) = 4.OkN/m

Total dead load on beam including self-weight ::: 15.25 kN/m

Live load from slab @ 5 kN /m2 = 5 x 2.25 = 11.25 kN/m

Sleps 1 and 3

Boundary condition

A and C fully restrained

Plastic hinge at A, C fully restrained


Ultimate limit state,

LC I = 1.4DL + 1.6LL"", 1.4 x 15.25 + 1.6 x 11.25 = 22kN/m + 18kN/m LC2 "'" 1.0DL = 15.25kN/m

load both spans with Lei 10 get maximum support moment at B.

load span AB with LCI and spall BC with LC2 10 get maximum support moment A and maximum span moment at AB.

Draw mOfTU!nl and shear envelope Non-linear analysis with 10% redistribution.

Loading Fo= Support Span Support Span Support A AB B BC C

Lei on both BM - 270 +135 +270 +135 -270 kNm spans Shear 180 180 180 180 kN

I.4DL on AB BM -159.8 +79.9 - 125.9 +46.0 -92.0 kNm l.ODL on BC Shear 102.8 95.2 72.6 65.1 kN

Lei on AB BM -311.7 +155.9 - 186.7 +28.0 - 70.6 kNm Le2 on BC Shear 193.9 166.1 kN

1.0kN/m LL BM -8.44 +4.22 -3.38 - 1.0 +1.19 kNm on AB Shear 5.06 3.94 kN

1.0kN/m LL BM 0 +7.23 -5.78 - 1.45 +2.89 kNm on AB Shear 3.86 5.14 kN

Assume 10% redistribution. Support moment at A is fixed at 0.9 x 311.7 = 280kNm. The support moment at A reaches 280kNm elastically with live load on span AB equal to (280 - 159.8)/8.44 = 14.24 kNm. At that point a plastic hinge forms at A and the boundary condition of the structure changes. The remaining live load to go on the span with changed boundary condition is (18kN/m - 14.24kNfm) = 3.76kN/m.

Design bending moment al support A = 280 kNm

Design bending moment at midspan AB at centre of span = 79.9 + 14.24 x 4.22 +3.76 x 7.23 = 167.2 kNm

A conservative design span moment = 175 kNm

allowing for the maximum span moment to occur away from the centre of span.

Design bending moment at support B=270kNm (Lei on both spans)

from elastic analysis

Design shear at support A = 102.8 + 5.06 x 14.24 + 3.86 x 3.76 = 189.5kN say 190kN


Design shear at support B "" ISO kN ( L eI on both spans)

Step 4 Determine axial loadJ Nol required.

Step 5 Iktermine torsion Ultimate load from slab = 1.4 x 11.25 + 1.6 x 11.25

= 34kN/m

Load assumed 10 act on edge of nib.

Eccentricity of load from shear centre of beam"" 110 - 15(chamfer)

Torsion per unit lenglh

Ultimate self-weight of beam

Eccentricity of se lf-weight from sllear centre

Torsion per unit length

Total ultimate torsion in beam

Step 6 Corer 10 reinforcement Maximum size of aggregatc = 20mm Maximum size of bar"" 25 mm assumed Maximum size of link = 10 mm Exposure condition = mild Fire resistance requircd ::;: I hour Grade of concrete = C40 Minimum cement content ::: 325 kg/m3

Maximum free water/cement ratio = 0.55 Nominal cover = 20 mm

290 + 2' + 78.5

(e "" 78.5 "" shear centre)

"" 318.5 mm

"" 34 x 0.3185 "" 10.83 kNm/m

"" 5.6kN/m

290 "" i - - + ,

2 "" 172.5 - 145 + 78.5 "" l06 mm

"" 5.6 x 0.106 "" 0.59kNm/m

: (10.83 + 0.59) x 4.5 = 51.4kNm at the

supports restrai ning rOlalion

Effective depth, d = 500 - 20 - 10 - 12.5 = 457.5 mm

Step 7 Effective span Effeclive span = 9.0 m

Sup 8 Effective widJh of flange Not required.


Step 9 Slendeml!lS ratio 1= 8.5 m = clear span

be = 400 mm 60be = 60 X 400 = 24 000 mm > 8500 mm

d = ( M )1 = ( 280 X 10' )1 = 335 mm O.l56fcub 0.156 X 40 X 400

250b~ 250 X 4(Kf -- = :::: 119402 mm > 8500 mm

d 335

Slenderness check is satisfied.

Step 10 Design for flexure

L

"\ SK 2162 Calculation of tensile steel al support.

Support bending moments at A or C=280 kNm

M K=--

fcubd2

280 X It1' = -40-'X:0400::"':'-X=45-7-.5~2

= 0.0836 < 0.156

No eompres. .. ive reinforcement required.

2 = d[05 + J(O.25 - O~) 1 = 457.5[0.5 + J (0.25 - 0.::6)]

= 4IOmm

d - 2 x =--

0.45

457.5 - 410 =

0.45

I.

As "

• • ~ 0 -.. ~

~

- --, m

b " 400mm .1

~

~ ~ v .. u


= 105mm = hr

:. Neutral axis is in the ftange .

M As = - - -

o.m!" 280 x 10'

;:-;=,':'-::;;~--,,:;; "" 1706 mm2

0.87 x 460 x 410

Use 4 no. 25 mm dia . bars (1964 mm' ) .

Midspan bending moment = 175 kNm

M K =-

f cybd2

175 x Iff' 40 x 400 x 457.52

= 0.052

z = 0.94d = 430mm

d - , x = - -

0.45 = 61mm < 105mm = hr

M A . = - --

0.87/yz. 175 x 1<1'

= = =-:-:::=--= 0.87 x 460 x 430

= 1017mm2

Use 2 no. 25 mm dia. bars (982 mm2) + I no . 12 mm dia . bar (113 mm' ).

SUp 11 Flanged beam Not required.

Step 12 Check. maximum shear strelS at support V

v =-bd

190 x HP 290 x 457 .5

= 1.43 N/ mm2

O.8V/cu = 0.8 x '140 = 5 N/mm2

Step 13 Check. flexural shear stress d = 457.5 mm

VA := 190 - 40 x 0.457 := InkN at effective depth away from support


v v =-

bd

In x llY ~

290 x 457 .5

= 1.30N/mm2

lOOA$ p ~--

bd

100 x 1964 :

290 x 457.5

= 1.48

vc = 0.72 x 1.7 = O.B4 N /mm2

Vnom = (vc + O.4)bd

From Fig. 11.5

= (0.84 + 0.4) x 290 x 457.5 x 10-3

= 164.5kN

v > Vc + 0.4

bS~(v - v,) As .. = O.87{y

290 x 200 x (1.30 - 0.84) :

0.87 x 460 at 200 mm dc (2 legs)

A.v 66.7 -5-

v = 200 x 2 = 0.17 for each leg

. As.. 0.4b Nommal s: = O.87/y

0.4 x 290 :

0.87 x 460

: 0.29 (2 Jegs) = 0.145 (for each leg)

Area of tension reinforcement required 10 carry weight of slab on the nib

34kN/m 0.87 x 460

= 85 mm21m

As = ~ = 0.085 for each leg 5 1000

St~p 14 AlI~rtr4ti1'e design for slr~ar

Not required since design shear is calculated at d from suppon.

St~p 15 Minimum tension reinforumenJ Assume channel section as L-beam.


A. > O.OO2Oh ... h == 0.0020 x 290 x 500 = 21JO mm2 < 1964 mm2 provided

Step 16 Minimum compression reinforcement Not required.

Step 17 Transverse reinforcement in fklnge As = 1.5hrmmP m

= 1.5 x 105 ::: 158 mm2/m minimum

(See Example 5.2.) Reinforcement in nib = 201 mm2/m provided.

Step 18 Minimum reinforcement in side face of lHams Nol required .

Step 19 Check deflection

bw = 290 = 0.725 > 0.3 b 400 From Table 11.3, Basic span/effective depth ratio

for rectangular section ::: 26 for fl anged beams = 20.8

(or b",lb::: 1.0 for b..,/b = 0.3

Interpolated basic ratio::: 20.8 + ( 26 - ZO.S) x (0.725 _ 0.3) = 24 0 .7

II>, = M = 167.2 M' 155.9

Midspan service stress = (~)f (~) 8~ y A~ p<OV

= G) x G!~:~) x 460 x (1017) 1095

= 249N/mm2

M 175 x 1If' - = =2.09 bd2 400 x 457.52

Modification factor for tcnsion reinfo rcemen t from Table Il .5"" 1.20

Modified span/depth ratio = 24 x 1.20'" 28.8

I, 9000 -=--d 457.5

::: 19.67 < 28 .80 OK

Step 20 Maximum areas of reinforcement A . < O.04b",11 ",. 5800 mm2

Satisfied.

(i)


Step 21 Containment of compression reinforcement Not required.

Step 22 Check bearing stress inside bend Not required.

Step 23 Curtailment of bars 45xbar dia. = 45x25 = 1125mm

0.15/ = 0.15 X 9000 = 1350mm 0.25/=0.25 X 9000 = 2250mm

2 no. 25 mm dia. top and bottom throughout . 2 no. 25 m dia. extra top at A, Band C - 5000 long at B, 2500 mm into

span at A and C and properly anchored at A and C. 1 no. 12mm dia. bottom in spans AS and Be. Follow simplified detailing rules for beams as in Fig. 2.2.

Step 24 Spacing of bars Minimum clear spacing = MSA + 5 = 20 + 5 = 25 mm

Actual minimum clear spacing used = 43 mm (support)

Actual maximum clear spacing used = 84 mm (midspan)

4700:1 4700:1 Maximum clear spacing allowed = Is "" 249 =

where f. = 249N/mm2

Step 25 Check torsional shear stress Ultimate torsion = 51.4 kNm

(see Step 19.)

(see Step 5)

189mm > 84mm

Divide section imo 3 rectangles of maximum total torsional stiffness. First choice

IL---,_ (i)_' __ _

'1--'----o FIRST CHOICE SECOND CHOICE

SK 2163 Calculation of torsional shear stress.

SK 2/64 Calculation of torsional shear stress.


500 x 290 - stiffness = h~i"hn'u = 29IY x S{X) = 1.22 x 10m

2 X 110 x 105 - stiffness = 2 x 1053 X 110 = 0.025 x lOw TOTAL = 1.245 x lOw

Second choice

290 x 290 - stiffness = 29W x 290 ;: 0.707 x tOm 2 X 400 x lOS - stiffness = 2 x 1053 X 400 = 0.0926 X lOll)

TOTAL;: 0.7996 x lOw

Hence the first choice is critical.

Proportion or torsional moment carried by the web ::: T x 1.22 X lO"!

1.24S x 10141

S1.4 x 1.22 1.245

= 50.4kNm

Torsion carried by flanges = 0.S(SI.4 - SO.4) = O.5kNm

2T Torsional shear stress V, in web = ( h . ) ," " .. Imin 'mu -""3

2 x 50.4 x 1(1) :

290'(500 _ 2~) = 2.97N/mm2

2x0.5x Uf Torsional shear stress VI in flange = ( 105)

1052 110 --3

VI.min = O.067V/cu : 0.061\140 = 0.4N/mm2

Vcu = O.8v/cu "" 5N/mm2

Vlu Yl 5 x 450 --: 550 550

= 4.1 N/m m2

= L.21 N /mm2

VI > VI. min. torsional reinforcement required .

Torsional shear stress + Hexu ral shear stress = 2.97 + 1.30 (sec Step 13) = 4.27N/mm2

< 5N/mm2 OK

Torsional reinforcement in web (vertical)


A ,y T

Sy O.8x1Yl0.87 fy

~ =-C-="50,,.4~X--,l:.:O'==--~ 0.8 x 238 x 448 x 0.87 x 460

:: 1.48 (for 2 legs)

== 0.74 (for each leg)

Longitudinal reinforcement for (orsion-

A , = (~:')(f)(X' + y,)

= 1.48 x I x (238 + 448)

= 1012mm2

Use 10 no. bars at 101 mm2 each in the longitudinal direction placed on the pcrimeter of web cross~section (fy = 460N/mm2).

evenly

SK 2f6S A"IS. diagram for Example 2.3.

0-0

0-"

'" ~ ~ -~

0 .,. Q

12.tll0

:1·0

660

"'-svlsv" 0·145 (fLEXURAl SHEAR)

AsvAv"0 ·085 INIB OIReCT TENSION)

!TORSION) A'iY/5V -, 0·10

{COMBINED) AS'I/'iY 0-3-;10 ·23

12 at 130 12 at 175

0·81 0·61;. 12 a~ .:~

1300 17""

Torsional reinforcement in flange

A sv T s: = 0.8xLY10.87fy

I ,

,

,

,


0.5 x 1<1' = ~O.~8~X~(I~05~~~~)~X~(~~~7.48~)~X~O~.~~~X~~~ = 0.078

Maximum spacing = XI' or ~. or 2IXlmm

= 57mm

Use 8 mm dia. li~ks at 50 mm ce ntres (1006 mm2/ m) (ly = 460 N/mm2) Could also use 6mm dia. mild steel links at SOmm centres (566mm2/m) (fy = 250N/mm~.

See Example 5.2, Step 4.

As ~ = 131 mm2/ m (ror fle xu re) = 241 mm21m

(460 grade steel) (mild steel Grade 250)

~:v = 0.078 for torsion (Grade 460)

A.v = 0.078 x ~

1000 x-250

(Grade 250)

= 144 mm2/ m (for 2 legs of mild steel)

= 72 mm2/m (for each leg - hori7.0ntal)

Total requirement = 241 + 72::::: 313mm2/m < S66 mm2/m

Longitudinal reinforcement for t()~ion in fI~nge

(A •• ) (I,,)

A~ :: s: fr (XI + y,)

460 = 0.078 x 250 x (57 + 352)

= 59 mm2

See Step 10.

(4 no. 6mm dia. mild steel: If = 250N/mm2)

At support. A. required = 1706mm2

Torsional A. required at corners (2 bars) = 202 mm2 (SlOp 25)

Total top rei nforcement required ::::: 1706 + 202 ::::: 1908 mm2

Provided = 4 no. 25 mm dia . = 1964 mm2 OK

Step 26 Flexural t:rat:k width t:alt:ulations By elastic analysis: no redistribution.

Maximum support moment at A or C = 201 kNm (serviceability limit state)

d = 500 - 20 - 12 - 12.5 = 455.5 b = 400mm

AI = 1964mm2 p = 0.0108

, . Design of Reinforced Concre te Beams Ir1

SK lI66 Typical section .. t support. . SOmnlc/c

A; = 982 mm! m o;; l0 hf "", 105 mm h = SOOmm bl = 290mm d' = 42.5 mm

2- 111 25

p ' = 0.0054

x = dU (mp + (m - I)p')' + 2 (mp + (m - I) (~)p,)y - (mp + (m - I )p' ) }

= 156.3 mm > ht = 105 mm

Using Reference 10, Table 117.

mdA5 + O.5b11 r x =

mA s + bhr

10 X 455 .5 x 1964 + 0.5 X 400 X 1052

to X 1964 + 400 X 105

= 181 mm

1. = d _ ~hf,,(::3x~--=2h=f) 3(2x - h,)

= 455.5 _ 105(3 x 181 - 2 X 105) 3(2 x 181 - 105)

= 410mm

98 Reinforced Concrele

M [, =

Asz

201 x Iff' = 1964 x 410

= 250N/ mm2

lOs = f. = 250 = 1.25 X 10- 3

E. 200 x JlY

(h - x)

Eh = -- " d -x

( 500 - 181 )

= 455.5 _ 181 x 1.25 x 10-3

=- 1.45 X 10- 3

bl(h - xf E - E -

mh - h 3EsAs(d _ x)

=- 1.45 X 10-3 _ 290 x (500 - 18 1)2 3 x 200 x 16' x 1964 x (455.5 - 181)

= 1.36 X 10- 3

a~. = V(44.51 + 44.52) - 12.5 = 50.4 mm

__ -;;3",a""-,',,m,,' _.,. ~V~< =

:::2(",a,,"_-'" m",'"") 1+-(h - x)

3 x 50.4 x 1.36 x 10- 3

2(50.4 - 32) I + (500 _ 181)

= O.18 mm < O.3mm

Slep 27 Design of connections to other components Follow ChapleT to.

Design of Reinforced Concre te Beams 99

2.S FlGURES FOR CHAPTER 2

\ \ \ \ 1 \ . 1 1 \ \ \ bIb . • i bit • 1

\ .\ \ \ , bib • • 3 "-.

11 ,\ \ \ e-- / bI' •• --

1\\\ \ \- '" V ,lb • • , ,,\:

~ ,/' ~ :'- blb" 6

'( ' ''' ,-/ / b/b • ?

~ ~y '\. V Y ~Ib • •

~>< 10

iIiiiij; .-

b/b ~

1

0." 0.0' !b O.Db O. . 0 0.09 0 . 10 O. o. " 1] 0.1. 0 , 1::>

" F'tg. 2.1 Values of ~.

100 Reinforced Concrele

0·101. 0·151 0-151 .~

I, ,1-'" 'n ""') I ". ~" I 100"1, '1'1 [ 20"1,

i ~

lOO'I,J

10.151

I

''''1 l 1OO'I,

O·15l O·lOl for mOsupporl

i (ff«li .... SD ..... l EIf«IM: S , l

CQNTlNUO!JS 8fAM . CURTAilMENT Of RE INfORCEMENT

""",

SIMPlY SI,JPPQRT(O BUM: CURTAILMENT Of REINFORCEMENT

1

0·51.;!: 45+

r 1OO'I. r ~'I,

I

I ,

< ~,

CANTILEVER BEAM : ClfiTAI\J'!ENT !,F REINfORCEMENT

Fig. 2.2 Simplified dctailing rulcs for beams.

Documents

Design Example of Reinforce Concrete Beams