LOADING & ANALYSIS Influence Lines
A bending moment influence line for any given point X on a structure is a line whose ordinate at any point B qives the bending moment at X when a load is placed at B ( keep reading it and look at the example below - it will make sense eventually !). Influence lines can be obtained for bending
moments shear forces and reactions. A bending moment influence line is drawn for one point only which may not necessarily be the point of maimum bending. A bending moment diagram by comparison with a influence line will give the bending moment at all points for one position of load.
Example - Simply Supported Single Span
Influence Line utorial for !A "DL and !# Loading #ending Effects Example$ "hree span dec# with continuity over pier supports.
Step%$ $etermine the position of the point of maimum bending moment in each element for a
single point load.
%oint A & maimum sagging moment in span ' %oint B & maimum hogging moment over pier ' %oint & maimum sagging moment in span *ote+ as end spans are equal then critical points over pier and in span , can be obtained from point A and B by symmetry.
Step $ $etermine influence line diagram for point A+
Step '$ $etermine influence line diagram for point B+
Influence Line utorial for !A "DL and !# Loading #ending Effects Example$ "hree span dec# with continuity over pier supports.
Step%$ $etermine the position of the point of maimum bending moment in each element for a
single point load.
%oint A & maimum sagging moment in span ' %oint B & maimum hogging moment over pier ' %oint & maimum sagging moment in span *ote+ as end spans are equal then critical points over pier and in span , can be obtained from point A and B by symmetry.
Step $ $etermine influence line diagram for point A+
Step '$ $etermine influence line diagram for point B+
Step ($ $etermine influence line diagram for point +
Step )$ $etermine loading for critical cases+ *oint A
"he maimum sagging moment is achieved by loading spans ' and , however we also need to chec# -A $/ for loading in span ' only.
-A 0pan ' only+ loaded length 1 '2m hence udl 1 3'.4 #*5m (B$,36table ',) -A 0pan ' and ,+ loaded length 1 2m hence udl 1 78.' #*5m (B$,36table ',) 9:/+ 1 '2 #* (B$,36 lause ;..)
-B loading will produce worst sagging moment with an ale at the maimum ordinate (.'8). Any one of the 7 ales can be located at this position< the vehicle is however positioned with the other , ales to achieve the maimum total ordinates+
*ote+ "he -B vehicle has a range of spacings between the centre ales in this case the ;m spacing gives the worst effect.
*oint #
"he maimum hogging moment is achieved by loading spans ' and however we also need to chec# -A $/ for loading in span only.
-A 0pan only+ loaded length 1 2m hence udl 1 78.' #*5m (B$,36table ',) -A 0pan ' and + loaded length 1 ,2m hence udl 1 ,7.7 #*5m (B$,36table ',)
9:/+ 1 '2 #* (B$,36 lause ;..)
sually -B loading will produce the worst hogging moment with an ale at the maimum ordinate (.28'). Any one of the 7 ales can be located at this position< the vehicle is however positioned with the other , ales to achieve the maimum total ordinates. In the case below the sum of the ordinates is 2.;33 = 2.3' = .28' = '.>;; 1 8.727
?ther cofigurations of -B loading need be chec#ed and in this case the ;m vehicle will produce a greater value with the vehicle in the position shown below. "he sum of the ordinates for this configuration 1 '.;27 = '.>7, = '.;8' = '.,28 1 ;.82,
*oint +
"he maimum sagging moment is achieved by loading span only.
-A 0pan only+ loaded length 1 2m hence udl 1 78.' #*5m (B$,36table ',) 9:/+ 1 '2 #* (B$,36 lause ;..)
-B loading will produce worst sagging moment with an ale at the maimum ordinate (,.'8). Any one of the 7 ales can be located at this position< the vehicle is however positioned with the other , ales to achieve the maimum total ordinates+
*ote+ "he -B vehicle has a range of spacings between the centre ales in this case the ;m spacing gives the worst effect.
Step ,$ $etermine load effects on dec#.
"he following assumptions will be made to demonstrate principles of influence lines+ @ Assume loads applied to ' notional lane width of dec# (,.;8m wide). @ Assume ultimate limit state hence use load factor f/ of '.8 for -A loading and '., for -B loading. @ Assume ,2 units of -B. Span %
aimum sagging moment due to -B loading+ 1 '., ,2 '2 (.'3 = '.,78 = 2.244 = 2.'24) 1 '774 #*m
aimum sagging moment due to -A loading at point A+ ase ' & 0pan' loaded 1 '.8 (3'.4 '2.48 = '2 .'3) 1 %( .Nm (-A critical) ase & 0pan' and , loaded 1 '.8 C78.' ('2.48 = 2.;>8) = '2 .'3D 1 '',7 #*m
*ier %
aimum hogging moment due to -A loading at point B+ ase ' & 0pan loaded 1 '.8 (78.' 7.33; ='2 .28') 1 278 #*m ase & 0pan' and loaded 1 '.8 C,7.7 (7.33; = 7.;7') = '2 .28'D 1 '443 #*m
aimum hogging moment due to -B loading+
1 '., ,2 '2 ('.;27 = '.>7, = '.;8' = '.,28) 1 )', .Nm (-B critical)
Span
aimum sagging moment due to -A loading at point + ase ' & 0pan loaded 1 '.8 (78.' 8.8 = '2 ,.'8) 1 3' #*m
aimum sagging moment due to -B loading+ 1 '., ,2 '2 (,.'8 = .4; = 2.4 = 2.,;;) 1 ),) .Nm (-B critical)
*ote+ -B loading is shown to be critical for two of the cases however if the loads are distributed using a computer analysis such as a grillage analysis then the -B moments will be reduced considerably.
!A and !# ype Loading !istory
"he first standard vehicle load for highway bridges in the 9 was introduced in '>. British 0tandards introduced a traffic live load requirement in B0 '8, %art , in '>, which was later revised in '>8 and '>,3. "he "ype - A uniformly distibuted loading was introduced in '>78 and the concept of a "ype -A and -B load was included in the '>87 edition of B0 '8,+ %art ,A. In '>;' the -B load was specified in terms of units and varied depending on the class of road with 78 units required for otorways and "run# Eoads and ,3.8 units for class i and class ii roads. A requirement for all public roads to be designed for at least ,2 units of -B was introduced in '>3,.
!A Loading
B$ ,352' Appendi A of the $esign anual for Eoads and Bridges says that "ype -A loading is the normal design loading for Freat Britain and adequately covers the effects of all permitted normal vehicles other than those used for abnormal indivisible loads. *ormal vehicles are governed by the Eoad Gehicles (Authorised Height) Eegulations '>>4 referred to as the AH Gehicles and cover vehicles up to 77 tonne gross vehicle weight. /oads from these AH vehicles are represented by a uniformly distributed load and a #nife edge load. "he loading has been enhanced t o cover+ i) impact load (caused when wheels bounce i.e. when stri#ing potholes or uneven epansion Joints). ii) ?verloading iii) /ateral bunching (more than one vehicle occupying the width of a lane). "he magnitude of the uniformly distributed load is dependent on the loaded length as determined from the influence line for the member under consideration. Kor simply supported dec#s this usually relates to the span of the dec#.
!# Loading
B$ ,352' Appendi A of the $esign anual for Eoads and Bridges says that "ype -B loading requirements derive from the nature of e ceptional industrial loads (e.g. electrical transformers generators pressure vessels machine presses etc.) li#ely to use the roads in the area. "he vehicle load is represented by a four aled vehicle with four wheels equally spaced on each ale. "he load on each ale is defined by a number of units which is dependant on the class of road and is specified in B$ ,352' hapter 7 as follows+ otorways and trun# roads require 78 units %rincipal roads require ,3.8 units and other public
roads require ,2 units. ?ne unit of -B is equal to '2#* per ale. "here are five -B vehicles to chec# although most vehicles can be discounted by inspection.
Design
"he design procedure is to analyse the bridge for -A and -B load effects applying the appropriate load factors. "he member is then deisgned for the worst effects of -A or -B loading.
#S)(//$*t Loading Idealisation for Grillage Analysis of #ridge Dec.s Index
'.-A $/=9:/ .-B Gehicles ,.%edestrian /oad 7.Accidental Hheel /oad 8.%arapet ollision /oad
sing a prestressed L7 beam with reinforced concrete dec# slab as the dec# eample as shown in Kig.'< the dec# having a '2M s#ew a span of 2m and carrying a 3.,m carriageway with two m footpaths.
B0 8722 %t.+22; l.,..>.,.' 3.,m carriageway has notional lanes hence lane width 1 ,.;8m. l.;., "he dec# shall carry 78 units of -B l.;.3 Assume bridge requires high containment parapets hence collision loading needs to be considered. %0!A "DL 1 2EL
-A $/ can be applied to each longitudinal member as a uniformly distributed load the intensity of the load is proportional to the width of the lane directly above the longitudinal member for eample+
-A $/ for a 2m span 1 78.'#*5m of notional lane. *otional lane width 1 ,.;8m -A $/5m width 1 78.' 5 ,.;8 1 '.,;#*5m -A $/ on member 1 2.'8 '.,; 1 '.48#*5m -A $/ on members ,7 N 8 1 '.2 '.,; 1 '.,;#*5m -A $/ on member ; 1 2.8 '.,; 1 ;.'4#*5m
$/ applied to each longitudinal grillage member to represent -A $/ in lane '. Alternatively if the program has the facility of applying patch loads then a patch width equal to the lane width and length equal to the loaded length may be applied. "he patch load is usually positioned by the centroid of the patch area in relation to the grid co6ordinates.
-A 9:/ can also be applied as a uniformly distributed load to the transverse members. As loads are initially proportioned to the adJacent members and Joints then the worst effects will always be achieved by positioning the 9:/ directly above a transverse member. If the dec# is s#ewed then the postion of the 9:/ to give the worst effect will be different to a square dec# and two or three
positions may need to be chec#ed to find the critical case.
It is therefore useful to separate the -A $/ and -A 9:/ into different load cases to avoid repeating the calculation for the effects of the $/. "he $/ and the various positions of the 9:/ can be added together in different combination cases. 0imilar load cases are produced for the -A $/ and 9:/ in the second lane. Kull -A live load will have the -A $/ and 9:/ in both lanes whilst -B live load has the -B vehicle in one lane and the -A $/ and 9:/ in the second lane. All these variations in load cases can be developed in the combination cases.
0 !# 3E!I+LES
"he -B vehicle consists of four ales with four wheels on each ale and is applied to the grillage as a series of point loads. lause ;.,. and ;.,., allow t he wheel loads to be applied as patch loads however there is little to be gained in a global analysis by applying this refinement and point loads will be a suitable representation for t he wheel loads.
"here are five variations of the inner ale spacing for the -B vehicle that can be applied to the dec#. A line beam analysis incorporating moving point loads will indicate the positions of the critical -B vehicle to achieve the design moments and shears. An :cel spread sheet using moment distribution to carry out a line beam analysis of standard moving vehicles can be downloaded by clic#ing here.
"he result of the line beam analysis shows that the maimum sagging moment occurs at 4.8m from the end of the dec# with the leading ale at ';.,m from the end.
All critical load cases are produced from the vehicle with the ;m inner ale spacing. As the loading is symmetrical and both ends of the single span dec# are simply supported then the position of maimum moment can be measured from either end of the dec#. "he transverse position of the -B vehicle will depend on which member is being considered however it is usual to design all internal beams for the critical loading condition for vehicles on the carriageway. "he edge beams will require special consideration to support the additional loading from the cantilever.
"he diagram shows one of the inner wheels on the critical ale positioned over the nearest transverse member at 4.8m from the support. "his would produce the critical loading condition for the bending moment on the internal beam for an orthogonal dec# however other positions need to be considered to ta#e account of the s#ew effects. As a chec# on the data the total of the reactions should equal the total load of the vehicle 1 7 782 1 '422#*. Also the line beam analysis gives a total moment of 8;>.8#*m< so as there are four longitudinal members supporting the vehicle then the moment from t he grillage should be in the order of (but less than) 8;>.8 5 7 O '722#*m in t he longitudinal member.
'0 *edestrian Load
lause ;.8.' states that the pedestrian live load shall be ta#en as 8.2 #*5m but reduced to 2.4 8.2 1 7.2#*5m for members supporting both footway and carriageway loading. onsequently the edge beam should be designed for 8.2#*5m and the net6to6edge beam designed for 7.2#*5m . "he $/s can be applied to these two members in a similar manner to the -A $/ described in 0ection '. above however as there is no barrier between the carriageway and footway lause ;.; requires that the footway members are designed for Accidental Hheel /oad which is generally more onerous than the pedestrian live load.
(0 Accidental 45eel Load
Accidental Hheel /oading consists of a 22#* ale and a '82#* ale with two wheels on each ale and is applied to the grillage as four point loads. lause ;.;. and ;.;., allow the wheel loads to be applied as patch loads however there is little to be gained in a global analysis by applying this refinement and point loads will be a suitable representation for the wheel loads.
0imilarly as with the -B vehicle a line beam analysis incorporating moving point loads will indicate the critical positions of the vehicle to achieve the design moments and shears. An :cel spread sheet using moment distribution to carry out a line beam analysis of standard moving vehicles can be downloaded by clic#ing here. "he Abnormal /oad facility is used in the line beam proforma to input the accidental wheel vehicle.
"he result of the line beam analysis shows that the maimum sagging moment occurs at '2.;m from the end of the dec# under the leading ale. "he vehicle will be positioned over the parapet beam as shown to obtain the critical loading condition for bending in this member. "his may also be the critical position for the design moment in the main edge beam however the '22#* wheel should be positioned at Joint B to confirm the critical case.
?ther positions on adJacent transverse members need to be considered to ta#e account of the s#ew effects. As a chec# on the data the total of the reactions should equal the total load of the vehicle 1 22 = '82 1 ,82#*. Also the line beam analysis gives a total moment of ';83.8#*m< so as there are
two longitudinal members supporting the vehicle then the moment from the grillage should be in the order of (but less than) ';83.8 5 O 422#*m in the longitudinal member.
)0*arapet +ollision Load
/oads due to collision with parapets need only be considered in a grillage analysis if high level containment parapets (-7a) are required. ollision loads on other types of parapet need only be considered for local effects (how the load is transferred to the main members).
lause ;.3..' describes the three loads that are to be applied to the top of the parapet over a ,.2m length. "he point loads need to be transferred down to the datum level of the grillage which is at the centroid of the dec# slab and distributed over a ,.2m length. "he high containment parapet is '.8m high above the bac# of footpath level. "he centroid of the dec# slab is about 2.,m below the bac# of footpath level consequently the two horiPontal loads will induce moments on the grillage with a lever arm of '.4m. "he 822#* horiPontal load will produce a moment of >22#*m at the centre6line of the dec#. "his moment is distributed along a ,.2m length giving ,22#*m5m moment to be applied to the grillage.
"he horiPontal load of ';3#*5m will be ta#en into the dec# which as it is very stiff aially compared to bending will distribute evenly between all longitudinal members and therefore have negligible effect in the grillage. "he load is however considered in determining local effects in accordance with lause ;.3.'.
"he '38#* vertical load can be idealised as a uniformly distributed load 84#*5m along a ,.2m length of the parapet beam. "he '22#* horiPontal load acts in the plane of the parapet and there is an argument that the load will be resisted by the framing effect of the parapet rails with the posts and will therefore be transferred to the dec# as a series of horiPontal and vertical loads at the base of the posts.
As the loads are to be applied over a ,.2m length then the moment of '22#* ('.8 = 2.,) 1 '42#*m can be represented by a vertical couple of ;2#* ,.2m. "he horiPontal load of '22#* will be ta#en into the dec# which as it is very stiff aially compared to bending will distribute evenly between all longitudinal members and therefore have negligible effect in the grillage. "he load is however considered in determining local effects in accordance with lause ;.3.'.
"he three loads can be combined in one load case. "he ,.2m length can be positioned anywhere along the parapet beam and positions are generally chosen to coincide with the critical positions for the accidental wheel load.
!A & !# Loading Example BS 5400 Part 2 : 2006 Clause 6.2 Type HA Loading Problem: How do you work out the HA loading and bending moment for a bridge deck?
Example:
Carriageway = 6m wide Deck span = 34m centre to centre of bearings for a simply supported single span! Design for a metre width of deck " Cl 3#$#%#3#&#
'umber of notional lanes = $ 'otional lane width = 6#()$ = 3#(m Cl 6#$#&#
*oaded length = 34m + = 336&)*!(#6, k')m per notional lane! + = 3 k')m per notional lane! Cl 6#$#$#
-nife .dge *oad = &$( k' per notional lane! Cl 6#4#&#&# /able &4#
0$ = (#(&3,1b*4(2*!3#6*2$(!5 0$ = (#(&3,13#(4(234#(!3#634#(2$(!5 = (#%4, 'ote" or loaded lengths less than $(m the load is proportioned to a standard lane width of 3#6m7 i#e# (#$,4b* = b*)3#6# or a metre width of deck" + = 3 8 (#%4,!)3#( = &(#( k')m -.* = &$( 8 (#%4,!)3#( = 3,#99 k' :a8imum mid span ;ending :oment with -.* at mid span" : = &(#( 8 34$!)9 3,#99 8 34!)4 = &,6, k'm Cl 6#$#,#
iceability limit state 2 combination &!
; 4(( @t#4 2
Assume the road o>er the bridge is not a @rincipal Boad then we need to check for 3( units type H; loading see ;D 3,)(& Chapter 4!# Cl 6#3#&
'ominal load per a8le = 3(units 8 &(k' = 3((k' /he ma8imum bending moment will be achie>ed by using the shortest H; >ehicle i#e# with 6m spacing see ; 4((2$"$((6 ig &$!# /he ma8imum moment for a simply supported span occurs under the inner a8le when the >ehicle is positioned such that the mid span bisects the distance between the centroid of the load and the nearest a8le# +ith a 34m span and the 6m H; >ehicle with eual a8le loads7 the inner a8le is placed at &#m from the mid span#
B* = 3((&(#,&$#&9#$(#3!)34 = 4, k' BB = 483((24, = 63k' :oment at = 4,8&# 2 3((8 = ,%3%k'm Cl 6#4#$
/he H; >ehicle occupies one lane with HA load in the adEacent lane# Assume for the e8ample that the H; load is carried by a standard lane width of 3#6m# Hence the moment per metre width of deck = ,%3%)3#6 = $&,k'm Cl 6#3#4#
iceability limit state 2 combination &! ern although a grillage or finite element type distribution would reduce the H; moment considerably#
6einforced +oncrete to #S )(// *art (
Index
'.Introduction .0erviceability /imit 0tate ,.ltimate /imit 0tate 7.0hear
%0 Introduction
Both the 0erviceability and ltimate /imit 0tates need to be considered. 0erviceability /imit 0tate ensures that crac# widths do not eceed values specified for different environmental conditions and also ensures that concrete and reinforcement stresses are maintained below a safe limit. ltimate /imit 0tate ensures that the structure will not collapse.
0 Ser7icea8ility Limit State
i) rac# ontrol rac#s in concrete can be caused by+
•
corrosion of the reinforcement which causes the concrete to spall
•
thermal movements particularly cooling from heat of hydration (called early thermal crac#ing)
•
structural actions such as bending shear or torsion
orrosion of reinforcement is controlled by use of suitable concrete grades and providing adequate cover to the reinforcement. rac#s due to thermal movements are controlled by providing minimum nominal steel area and restricting the maimum bar spacing. B$4543 is used to calculate the minimum steel area and bar spacing to control early thermal crac#ing. "he width of shear crac#s is controlled by ultimate strength calculations. ?nly the crac# widths caused by bending and tension need to be calculated and clause 8.4.4. is used.
Kor calculating the crac# width only combination ' loading is used and a maimum of ,2 units of -B is applied. rac#s caused by higher loadings will not eist for long enough to affect corrosion. "he maimum design stress limits will ensure that the crac#s close up when the loads are removed. A crac#ed elastic section analysis is used to determine the strain Q m at the level where crac#ing is being considered. sing a rectangular section as an eample+
He first need to determine the position of the neutral ais. /et+ Re 1 :s 5 :c :s 1 22#*5mm (clause 7.,..) "he short term value of : c is obtained from "able , which is appropriate to the live load portion of the moment (q). "he permanent load portion of the moment ( g) has an :c value equal to half the short term value. "he modified value of : c used for the crac# width calculation is an intermediate value between the short and long term values (clause 7.,..'(b)). odified :c 1 :c(short6term)S'62.8Cg 5(g=q)DT U 1 As 5 btd and U 1 As 5 btd "hen d c5d 1 VSCReU = (Re 6 ')UD = CReU = (Re 6 ')Ud5dDT 6 CReU = (Re 6 ')UD 0econd oment of Area of rac#ed 0ection I c 1 btdc, 5, = (Re6')As(dc6d) = ReAs(d6dc) 0train at level of tension reinforcement Q s 1 Re(g=q)(d6dc) 5 (Ic:s) $epth to notional surface where nominal cover cnom is provided 1 a 0train at notional surface Q' 1 Qs(a6dc) 5 (d6dc) "here is a correction formula (equation 8 in clause 8.4.4.) which allows for the stiffening effect of concrete in the tension Pone. $ue to fatigue the stiffening effect does not wor# under fluctuating load. "his is allowed for in equation 8 by providing no advantage if more than half the moment is due to live load.
"he corrected strain at notional surface Q m is thus obtained using the value of Q ' in equation 8. acr 1 VS(cnom=W5) = (s5)T 6 W5 "he $esign rac# Hidth can now be calculated using equation 7 in clause 8.4.4..
ii) 0tress /imits "he methods given in the code for chec#ing crac# widths assumes a linear elastic behaviour. It is therefore necessary to chec# the stress limits to ensure this assumption is correct. in particular that the strains caused by transient loads will not become permanent. "he stresses are calculated using a crac#ed elastic section analysis similar to that used to determine the crac# width described above. All load combinations need to be chec#ed to ensure that the stress limits are not eceeded. "he stress limits are given in "able of the code which are+ 2.38f y for reinforcement in tension or compression. "his is critical for most members in bending. 2.8f cu for concrete in compression with triangular stress distribution. "his is critical for members in bending with significant aial load or with heavy reinforcement. 2.,4f cu for concrete in compression with uniform stress. "his is critical for members in aial compression.
'0 "ltimate Limit State
"o determine the moment of resistance of a member at failure by limit state analysis the following assumptions are made+ a. "he distribution of strain across any section is linear. "his means that plane sections before bending remain plane after bending and the strain at any point is proportional to its distance from the neutral ais. b. "he resistance of concrete in tension is ignored. c.
"he relationship between the stress and strain in the reinforcement is as shown in Kigure of the code with m 1 '.'8.
d. "he relationship between the stress and strain in the concrete is as shown in Kigure ' of the code with m 1 '.8. "he strain in the concrete at the outermost fibre is ta#en as 2.22,8. Alternatively the distribution of stress in the concrete at failure may be represented by a uniform stress of 2.7f cu acting over the whole of the compression Pone.
"he deisgn formulae given in clause 8.,.., of the code are based on a uniform compressive stress of 2.7f cu for concrete and stresses of 2.43f y in tension and 2.3f y in compression for steel. "he steel stresses are the maimum values provided by the stress6strain curves where 2.3f y is a simplification of the epression for steel in compresssion. "he design formulae are also based on a maimum depth of concrete in compression of 2.8d< this ensures a strain 2.22,8 in the tension reinforcement. Kor values of Y 2.8d the use of the design stress of 2.43f y in tension is invalid the design becomes inefficient and the failure less ductile. Kor the singly reinforced rectangular section+
"a#ing moments about the centre of compression for the tensile force 9u : ;/0<=f y>As?
"a#ing moments about the centre of tension for the compressive force 1 (2.7f cu)bP 1 (2.7f cu)b(d 6 2.8) "he maimum moment of resistance is obtained when 1 2.8d so substituting for we get+ 9u : /0%)f cu8d
"he depth to the neutral ais depends upon the reinforcement provided and is obtained by equating the forces+ (2.43f y)As 1 (2.7f cu)b Eearranging and dividing both sides by d we get+ (5d) 1 (.f yAs) 5 (f cubd) /ever arm P 1 (d 6 2.8) so substuting for we get+ ? : ;% - @%0%f yAs B @f cu8d>d
"he value of P is not to be ta#en greater than 2.>8d.
Kor the doubly reinforced rectangular section+
"a#ing moments about the centre of tension for the compressive forces 9u : /0%)f cu8d 1 ;/0=f y>ACs;d - dC>
:quating the tensile and compressive forces ;/0<=f y>As : /0f cu8d 1 ;/0=f y>ACs
"hese two equations are based on a value of d5d Z 2. which ensures a strain 2.22,8 2.; 1 2.22' in the compression reinforcement. Kor values of d5d Y 2. the use of a design stress of 2.3f y in compression becomes invalid.
(0 S5ear
"he design rules for shear in beams are based on the results of tests carried out on beams with and without shear reinforcement. "he results from the beams tested without shear reinforcement showed that for a constant concrete strength and longitudinal steel percentage the relationship between the ratio of the bending moment at collapse ( c) to the calculated ultimate fleural moment ( u) and the ratio of shear span (av) to effective depth (d) is as shown below+
"he diagram has four distinct regions each of which has a different mode of failure.
Eegion (i) fails by crushing of a compression strut running from the load to the support. Eegion (ii) fails by diagonal tension causing splitting along the line from the load to the support. Eegion (iii) fails when a fleural crac# develops into a shear crac#. Eegion (iv) fails in fleure. "he fleural failure in region (iv) is prevented by designing the beam for u in accordance with lause 8.,..,. "he test results show that this is unsafe for regions (ii) and (iii) and c needs to be controlled by considering the shear force. /et the shear force at failure 1 G c and the nominal shear stress v c 1 Gc 5 bd then+ c 1 Gcav 1 vcbdav "he test results relationship with a v 5d can be reproduced by dividing both sides of the above equation by bd which gives+ c5bd 1 vc av 5d
"he dashed line is u 5bd which assumes fleural failure. "he chain dotted line is constructed to cut off the unsafe side of the graph< the slope of this line is v cand is the allowable shear stress. Galues of vc are given in "able 4 of the code< these values will ensure that the moment to cause collapse will fall below the test values. It can be seen from "able 4 that the steel area A s has a more significant effect on the value of v c than does the concrete strength.
Hhen the shear stress v is greater than [ svc as given in "able 3 of the code then lin#s need to be designed. "hese are designed on the basis that the beam and lin#s act as a pin6Jointed truss.
"he lin#s are designed to carry the shear in ecess of that which can be carried by the concrete. "he horiPontal tie force in the truss analogy has to be provided by the tension reinforcement this is in addition to that required to resist any bending effects and is determined from A sa G5S(2.43f y)T in clause 8.,.,.. "he beam may fail by crushing of the compression struts regardless of the amount of shear reinforcement. "he maimum allowable shear stress is therefore limited to a value of 2.>V(f cu 5'.8) as given in clause 8.,.,.'
6einforced +oncrete Dec. Example to #ritis5 Standards ein!or"ed Con"rete #e"$ #esign to BS 5400 Part 4 Problem: Design a simply supported reinforced concrete deck slab using a unit strip method# /he deck carries a &((mm depth of surfacing7 together with a nominal HA li>e load udl of &,# k')m$ and knife edge load of 33k')m # /he deck should also be designed to carry 3( units of H; load# /he span of the deck is &$#(m centre to centre of bearings#
.8posure Class D& used for bridge deck soffits# cl# A#3
i8ing tolerence for reinforcement Gc = &mm for insitu concrete# /able A#
'ominal co>er for C3$)4( concrete = 4 Gc = 60%% with ma8imum water2cement ratio = (# and minimum cement content of 3$( kg)m 3 Loading per metre width of deck unit strip! 'ote" /he loading has been simplified to demonstrate the method of designing the slab ee ; 4(( @t$7 or ;D 3,)(& for full design loading!#
'ominal Dead *oads " deck slab = $ (#6 &#( = &6#3 k')m surfacing = $4 (#& &#( = $#4 k')m 'ominal *i>e *oad " HA = &,# &#( 33#( = &,# k')mudl! 33k'kel! 3( units H; = 3( &( ) 4 per wheel = , k' per wheel
-ey" I Jt is usually assumed that there is local plasticity at the critical sections at ltimate *imit tate and the self2euilibrating stresses due to nonlinear temperature distribution can be ignored in combination loadings# Bef"KConcrete bridge engineering" performance and ad>ancesK by B#L#Cope!# Te%perature #i!!eren"e &!!e"ts Apply temperature differences gi>en in ; 4(( @t$ ig#% Froup 4! to a &m wide deck section# Cl# #4#6 2 Coefficient of thermal e8pansion = &$ &(26 per MC# rom ; 4(( @t4 /able 3 " .c = 3& k')mm$ for fcu = 4(')mm$ Hence restrained temperature stresses per MC = 3& &(3 &$ &(26 = (#3,$ ')mm$
ection @roperties Area = &((( 6( = (#6 &(6 mm$ econd :oment of Area = &((( 6(3 ) &$ = $$#% &(% mm4 a! @ositi>e temperature difference orce to restrain temperature strain" (#3,$ &(3 1 &( 3#( #( ! &% &#! &% &#(!5 &(23 = 634#$ k' /aking moments about centroid of section to determine reuired moment : to restrain cur>ature due to temperature strain " (#3,$ &(3 1&( 3#( $( #( $,! &, (#3 9,# &&6#,! 2 $( (#& 6#,! 2 &% &#( $6(!5 &(26 = &&&#k'm
b! Be>erse temperature difference orce to restrain temperature strain " 2 (#3,$ &(3 1 &3( $# &# &#% ! &63 (#% (#, !5 &(23 = 2 4,$#4k' /aking moments about centroid of section to determine reuired moment : to restrain cur>ature due to temperature strain " 2 (#3,$ &(3 1&3( $6( $# $9$ 2 &# $6( 2 &#% $9$ ! &63 (#% &4& 2 (#, &4& !5 &(26 = 2&3#34k'm
'ote" ign con>ention is compressi>e stresses are positi>e# #ead ' Superi%posed #ead Loading (per %etre )idt* o! de"$+ * = er>iceability *imit tate * = ltimate *imit tate
Design * moment = N
'ominal HA mid span moment = &,# &$#( $ ) 9 33#( &$#( ) 4 = 44$,% /he ma8imum moment for the H; >ehicle occurs at point in the diagram below with the >ehicle positioned as shown#
CF = position of the centre of gra>ity of the three ,k' wheel loads! 'ominal H; moment at = %%#4 #3 2 , = 3/2$,%
Co%ination Loading Design HA * moment = e! = 34 4%, = 142 $,%
Design HA * moment = e! = 44 693 = 3 $,% Co%ination 3 Loading Design HA * moment = e! = 34 4&4 = 5/ $,%
Design HA * moment = e! = 44 6% = 023 $,% lti%ate Capa"ity o! #e"$ Sla ltimate Design :oment = 3 $,%
; 4(( @t 4 cl# #&#$#&
Jt is usual to design reinforced concrete for the ultimate limit state and check for ser>iceability conditions# cl# #4#$
se clause #3#$ for the resistance moments in slabs# cl# #3#$#3
Try 32%% dia. rein!or"e%ent at 25%% "entres: 'ominal co>er to reinforcement in deck soffit = 6(mm d = 6( 2 6( 23$)$ = ,4 As = R&6$ &((( ) &$ = 6434mm$)m f y = ((')mm$ fcu = 4(')mm$ S = 1& 2 O&#&f yAsP)OfcubdP!5d S = 1& 2 O&#& (( 6434P)O4( &((( ,4P!5d = (#9d Q (#% d ∴ S = (#9 ,4 = 411%% :uteel = (#9,f yAsS = (#9, (( 6434 499 &(26 = &366 k'm)m :uConcrete = (#&fcubd$ = (#& 4( &((( ,4$ &(26 = &%,, k'm)m T &366 ∴ u 366 $,%% T &&3,k'm)m ∴ U-#
C*e"$ Ser-i"eaility Li%it State Co%ination * Design :oment = 142 $,% 34D* 4%,**!
Determine depth V7V to neutral a8is of cracked section" cl# 4#3#$#& /able 3
Woungs :odulus for concrete for short term loading = .c = 3& k')mm
$
cl# 4#3#$#$
Woungs :odulus for steel reinforcement = .s = $(( k')mm$ Case ! +hen the bridge has Eust opened when only a small amount of creep has occurred!" :odular Batio = .s ) .c = $(( ) 3& = 6#4 /aking first moments of area about the neutral a8is" &((( 7$ ) $ = 6#4 6434 ,4 2 7! ((7$ 4&&(7 2 $3#93&(6 = ( 7 = &,, mm econd :oment of Area of cracked section" J88 = &(((&,,3 ) 3 6#46434,42&,,!$ = 9#3%&(% mm4 Cl 4#&#
:a8 compressi>e stress in concrete = 94$&(6 &,, ) 9#3%&(% = &,#9 ')mm$ /able $ $
Allowable compressi>e stress = (#fcu = $( ')mm T &,#9 ∴ UCase 2! +hen creep and shrinkage in the bridge are substantially complete" cl# 4#3#$#&b! $
Woungs :odulus for concrete for long term loading = .c)$ = &# k')mm Hence :odified .c for 34D* 4%,**! = 34 &# 4%, 3&! ) 94$ = $4#6 k')mm$ cl# 4#3#$#$ $
Woungs :odulus for steel reinforcement = .s = $(( k')mm :odular Batio = .s ) .c = $(( ) $4#6 = 9#& /aking first moments of area about the neutral a8is" &((( 7$ ) $ = 9#& 6434 ,4 2 7! ((7$ $&&7 2 3(&(6 = ( 7 = &%9 mm econd :oment of Area of cracked section" J88 = &(((&%93 ) 3 9#&6434,42&%9!$ = %#%6&(% mm4
Cl 4#&# 6
%
$
:a8 compressi>e stress in concrete = 94$&( &%9 ) %#%6&( = &6#, ')mm
/able $
Allowable compressi>e stress = (#fcu = $( ')mm$ T &6#, ∴ U/ensile stress in reinforcement = 94$&(6 ,4 2 &%9! 9#& ) %#%6&(% = $,# ')mm$ /able $ $
Allowable tensile stress = (#,f y = 3, ')mm T $,# ∴ U-
Crack Control" train in reinforcement = ∈& = $,# ) $((((( = (#((&$% /able &3
'otional surface for crack calculation = 3mm co>er to reinforcement
#9#9#$ $
$
acr = X1&$)$! &63! 5 2 &6 = 6 Allow for stiffening effect of concrete" en $
= ∈& 2 1O3#9bthaV2dc!P ) O∈sAsh2dc!P5 1&2:):g!&(2%5 ∈m = ∈& 2 1O3#9&(((6(6$ 2 &%9!P ) O(#((&$%64346( 2 &%9!P5 1& 2 4%,)34!&(2%5 ∈m = ∈& 2 12(#(((&$5 but not greater than ∈& Hence no stiffening effect ∈m = ∈& = (#((&$% 6$ 2 &%9! ) ,4 2 &%9! = (#((&46 Design crack width = 3 6 (#((&46 ) 1& $ 6 2 3! ) 6( 2 &%9!5 = (#$ mm ∈m
en $4
Design crack width = 3acr∈m ) 1&$acr2cnom!)h2dc!5 /able &
:a8imum allowable crack width = (#$ mm ∴ UCo%ination 3 * Design :oment = 5/ $,% 34D* 4&4**!
Determine depth V7V to neutral a8is of cracked section" cl# 4#3#$#& /able 3
Woungs :odulus for concrete for short term loading = .c = 3& k')mm
$
cl# 4#3#$#&b! $
Woungs :odulus for concrete for long term loading = .c)$ = &# k')mm Hence :odified .c for 34D* 4&4**! = 34 &# 4&4 3&! ) ,% = $3#% k')mm$ cl# 4#3#$#$ $
Woungs :odulus for steel reinforcement = .s = $(( k')mm :odular Batio = .s ) .c = $(( ) $3#% = 9#3 /aking first moments of area about the neutral a8is" &((( 7$ ) $ = 9#3 6434 ,4 2 7! ((7$ 3,$47 2 3(#94&(6 7 = $(( mm
econd :oment of Area of cracked section" J88 = &((($((3 ) 3 9#36434,42$((!$ = &(#&9&(% mm4T Cl 4#&#
:a8 compressi>e bending stress in concrete = ,%&(6 $(( ) &(#&9&(% = &4#% ')mm$ :a8 compressi>e stress due to positi>e temperature difference = < f* $#3& = (#9 $#3& = ')mm$ /otal compressi>e stress in concrete = &4#% = &6#, ')mm$ /able $ $
Allowable compressi>e stress = (#fcu = $( ')mm T &6#, ∴ U/ensile stress in reinforcement = ,%&(6 ,4 2 $((! 9#3 ) &(#&9&(% = $3$#9 ')mm$ /ensile stress due to re>erse temperature difference =
Allowable tensile stress = (#,f y = 3, ')mm T $3, ∴ UHen"e B32 ars at 25 "entres are ade8uate !or t*e %id span. S*ear #esign
hear is designed for ultimate limit state# cl# #4#4
Y = shear force due to ultimate loads# :a8imum Dead *oad Y =
:a8imum HA Y = < f3 &# &$ &,# ) $ 33 &$6 ) &$ 2 &,# (#,4! :a8imum HA Y = &#& &%( k' = $(% k'
:a8imum H; Y = < f3 , &$6 %#6$6 3#6$6 $6! ) &$ :a8imum H; Y = &#& $& k' = $3, k' T $(% ∴ H; loading critical :a8imum Y = && $3, = 399 k' hear stress = Y ) bd = 399&(3 ) &((( ,4! = (#69 ')mm$ Design for no shear reinforcement condition then Z s>c T (#69 ')mm$ /able %
Zs = (()d!&)4 = ((),4!&)4 = (#%, /able 9 &)3
&)3
>c = (#$,)c = (#$, ) &#$! 1&(( 6434 ) &((( ,4!5&)3 4(!&)3 = (#,, ')mm$ Zs>X = (#%, (#,, = (#, ')mm$ T (#69 ∴ Ucl #3#3#&
Check that the ma8imum allowable shear stress is not e8ceeded" :a8imum allowable shear stress = (#,Xfcu or 4#, ')mm$ (#,Xfcu = (#,X4( = 4#,4 T (#69 ∴ UHen"e B32 ars at 25 "entres are ade8uate !or s*ear at t*e ends o! t*e de"$.
'ote" Jntermediate sections between mid span and the ends of the deck will ha>e a smaller moment than at mid span and a small shear than at the ends of the deck# /hese sections need to be checked to determine where the reinforcement may be reduced to ;$ at &$c)c# cl# #9#4#&
:inimum area of reinforcement = (#&[ of b ad = (#& &((( ,4 ) &(( = 96& mm$)m ∴ use ;&$ bars at &$ centres As = %( mm$)m! for distribution reinforcement#
*restressed +oncrete
0ince concrete is wea# in tension in normal reinforced concrete construction crac#s develop in the tension Pone at wor#ing loads and therefore all concrete in tension is ignored in design. %restressing involves inducing compressive stresses in the Pone which will tend to become tensile under eternal loads. "his compressive stress neutraliPes the tensile stress so that no resultant tension eists (or only very small values within the tensile strength of the concrete). rac#ing is therefore eliminated under wor#ing load and all of the concrete may be assumed effective in carrying load. "herefore lighter sections may be used to carry a given bending moment and prestressed concrete may be used over much longer spans than reinforced concrete. "he prestressing force also reduces the magnitude of the principal tensile stress in the web so that thin6webbed I 6 sections may be used without the ris# of diagonal tension failures and with further savings in self6weight. "he prestressing force has to be produced by a high tensile steel and it is necessary to use high quality concrete to resist the higher compressive stresses that are developed. "here are two methods of prestressing concrete+ ') %re6cast %re6tensioned ) %re6cast %ost6tensioned Both methods involve tensioning cables inside a concrete beam and t hen anchoring the stressed cables to the concrete.
%> *re-tensioned #eam +onstruction
Stage %
"endons and reinforcement are positioned in the beam mould.
Stage
"endons are stressed to about 32\ of their ultimate strength.
Stage '
oncrete is cast into the beam mould and allowed to cure to the required initial strength.
Stage (
Hhen the concrete has cured the stressing force is released and the tendons anchor themselves in the concrete.
Stage )
/oss of prestress due to elastic deformation of the concrete and relaation of the steel need to be considered. Kurther loss of prestress will also occur due to shrin#age and creep of the concrete< as these are time related then the effects will need to be considered both at short term and long term.
> *ost-tensioned #eam +onstruction
Stage %
able ducts and reinforcement are positioned in the beam mould. "he ducts are usually raised towards the neutral ais at the ends to reduce the eccentricity of the stressing force
Stage
oncrete is cast into the beam mould and allowed to cure to the required initial strength.
Stage '
"endons are threaded through the cable ducts and tensioned to about 32\ of their ultimate strength. "he diagram above indicates Jac#ing from both ends of the beam.
Stage (
a) Hedges are inserted into the end anchorages
b) the tensioning force on the tendons is released. Frout is then pumped into the ducts to protect the tendons.
Stage )
/oss of prestress due to elastic deformation of the concrete and relaation of the steel need to be considered. Kurther loss of prestress will also occur due to shrin#age and creep of the concrete< as these are time related then the effects will need to be considered both at short term and long term.
Loss of *restress
"otal losses in prestress can amount to about ,2\ of the initial tensioning stress.
*restressed +oncrete #eam Example to #ritis5 Standards Prestressed Con"rete Bea% #esign to BS 5400 Part 4
Problem: Design a simply supported prestressed concrete W beam which carries a &(mm thick concrete slab and &((mm of surfacing7 together with a nominal li>e load udl of &(#( k')m$ and kel of 33k')m # /he span of the beam is $4#(m centre to centre of bearings and the beams are spaced at &#(m inter>als#
e *oad " HA = &( &#( 33#( = &( k')m 33k' $ units H; = $ &( ) 4 per wheel = 6$# k' per wheel
BS 5400 Pt. 4 Se"tion Properties cl#,#4#&
:odular ratio effect for different concrete strengths between beam and slab may be ignored#
Te%perature #i!!eren"e &!!e"ts Apply temperature differences gi>en in ; 4(( @t$ ig#% Froup 4!to a simplified beam section# Cl# #4#6 2 Coefficient of thermal e8pansion = &$ &(26 per MC# rom ; 4(( @t4 /able 3 " .c = 34 k')mm$ for fcu = (')mm$ Hence restrained temperature stresses per MC = 34 &(3 &$ &(26 = (#4(9 ')mm$
a! @ositi>e temperature difference orce to restrain temperature strain " (#4(9 &((( 1 &( 3#( #$ ! 5 &(23 (#4(9 3(( $( &# ,( $(( &#$ ! &(23 = (4#% &$$#4 = 6$,#3 k' :oment : about centroid of section to restrain cur>ature due to temperature strain " (#4(9 &((( 1 &( 3#( ($ #$ $, ! 5 &(26 (#4(9 3(( $( &# 344 2 ,( $(( &#$ 6 ! &(26 = $6&# 2 $6#, = $34#9 k'm
b! Be>erse temperature difference orce to restrain temperature strain " 2 (#4(9 1 &((( &( 3#6 $#3 ! 3(( %( (#% ! 5 &(23 2 (#4(9 3(( $(( (#4 &( (#4 ! &(23 2 (#4(9 ,( 1 ( (#% (#& ! $4( &#$ $#6 ! 5 &(23 = 2 39#% 2 &%#3 2 $%#& = 2 ,((#3 k' :oment : about centroid of section to restrain cur>ature due to temperature strain " 2 (#4(9 1 &(((( 3#6 ($ $#3 $, ! $,((( (#% 39$ 3%, ! 5 &(2 6
2 (#4(9 3(( $(( (#4 $,( 2 &( (#4 $93 ! &(26 (#4(9 ,( 1 ( (#% 39 (#& 366 ! $4( &#$ (3 $#6 43 ! 5 &(26 = 2 &%4# 2 (#6 &3#9 = 2 4 k'm
#i!!erential S*rin$age &!!e"ts ; 4(( @t#4 cl#,#4#3#4
se cl#6#,#$#4 /able $% " /otal shrinkage of insitu concrete = 3(( &(26 Assume that $)3 of the total shrinkage of the precast concrete takes place before the deck slab is cast and that the residual shrinkage is &(( &(26 7 hence the differential shrinkage is $(( &(26 ; 4(( @t#4 cl#,#4#3#
orce to restrain differential shrinkage " = 2 \ diff .cf Acf ] = 2$(( &(26 34 &((( &( (#43 = 243% k' .ccentricity acent = ($mm Bestraint moment :cs = 243% (#($ = 2$$(#4 k'm
elf2weight of beam and weight of deck slab is supported by the beam# +hen the deck slab concrete has cured then any further loading superimposed and li>e loads! is supported by the composite section of the beam and slab# #ead Loading (ea% and sla+ /otal load for ser>iceability limit state = &#( 3#6!&#( &(#,9! = &4#4k')m Design ser>iceability moment = &4#4 $4 $ ) 9 = 03 $,% Co%ination Loading uper# HA li>e load for *" = 1&#$ $#4!&#$ &(!5udl 1&#$ 33!5kel = $#99 &$#(!udl 3%#6kel = 4./ $,% 9 3/.6$,
uper# H; li>e load for *" = $#99 4 wheels ^ &#& 6$# = 2./ $,% 4 wheels ^ 69#, k' /otal load for ultimate limit state" = 1&#& 3#6!&#& &(#,9!&#, $#4!&# &(!5udl 1&# 33!5kel = 4#&4 &$#4( 4#$( &#(!udl 4%#kel = 35. $,% 9 4/.5$,
HA Design ser>iceability moment" = &4#% $4#($ ) 9 3%#6 $4 ) 4 = 30 $,% $ units H; Design * moment" = $#% $4#($ ) 9 %9$#3from grillage analysis! = /. $,% Design ultimate moment" = 3#, $4#($ ) 9 4%# $4 ) 4 = 216 $,% Co%ination 3 Loading uper# HA li>e load for *" = 1&#$ $#4!&#( &(!5udl 1&#( 33!5kel = $#99 &(#(!udl 33kel = 2./ $,% 9 33$,
/otal load for ultimate limit state" = 1&#& 3#6!&#& &(#,9!&#, $#4!&#$ &(!5udl 1&#$ 33!5kel = 4#&4 &$#4( 4#$( &$#!udl 4kel = 33.2 $,% 9 4.3$, Design ser>iceability moment" = &$#% $4#($ ) 9 33 $4 ) 4 = 2 $,% Allo)ale stresses in pre"ast "on"rete At transfer " cl#6#3#$#$ b!
Compression /able $3 ! (#fci _ (#4fcu! = $( ')mm$ ma8# cl#6#3#$#4 b!
/ension = &#( ')mm$ At ser>iceability limit state " cl#,#4#3#$
Compression &#$ /able $$! &#$ (#4fcu = $ ')mm$ /ension = ( ')mm$ class &! 3#$ ')mm$ class $ 2 /able $4!
Prestressing or"e and &""entri"ity sing straight7 fully bonded tendons constant force and eccentricity!# Allow for $([ loss of prestress after transfer# Jnitial prestress at *e>el & to satisfy class $ reuirement for * Comb# 3!# tress at transfer = &,#6, 2 3#$ ! ) (#9 = &9#& ')mm$ use allowable stress of $( ')mm$!
/he critical section at transfer occurs at the end of the transmission Sone# /he moment due to the self weight at this section is near Sero and initial stress conditions are" @)A @e)`le>el & = $( ##################### en# &! @)A 2 @e)`le>el $ T= 2 &#( ##################### en# $! en# &! `le>el & en# $! ` le>el $ gi>es " @ T= A $( ` le>el & 2 &#( `le>el $! ) `le>el & `le>el $! @ = 44%#$$ &(3 $( &&6#($ 2 9%#(66! ) &&6#($ 9%#(66! &(23 = 4999 k' Allow &([ for loss of force before and during transfer7 then the initial force @ o = 4999 ) (#% = 43&k' sing &#$mm class $ rela8ation standard strand at ma8imum initial force of &,4k' (#, @u! Area of tendon = &3%mm$ 'ominal tensile strength = f pu =&6,( ')mm$
Hence 3$ tendons reuired# Jnitial force @o = 3$ &,4 = 5561 $, @ = (#% 69 = 50 $, ubstituting @ = (&& k' in en# $! e Q= `le>el $ ) A `le>el $ ) @ = 9%#(66 &(6 ) 44%#$$ &(3! 9%#(66 &(6 ) (&& &(3! e = &%9 &9 = 26 %% Arrange 3$ tendons symmetrically about the W2W a8is to achie>e an eccentricity of about $&6mm#
cl# 6#,#$#3 $
@ = (#%% @o ) 1 & .s Aps ) A! & A e ) I! ) .ci 5 @ = (#%% @o ) 1 & &%6 3$ &3% ) 44%$$(! & 44%$$( $&%$ ) $#%( &(%! ) 3& 5 @ = (#%& @o = (#%& 69 = 506 $, Jnitial stresses due to prestress at end of transmission Sone " *e>el & " @ ) A & A e ) `le>el & ! = & & $&% ) $9 ! = $(#9% ')mm$ (20.89 N/mm2 is slightly greater than the allowable of 20 N/mm2 so a number of tendons will need to be debonded near the ends of the beam) # *e>el $ " @ ) A & 2 A e ) `le>el $ ! = & & 2 $&% ) &%9 ! = 2 &#$( ')mm$ :oment due to self weight of beam at mid span = &(#,9 $4$ ) 9 = ,,6#$ k'm
tress due to self weight of beam at mid span " ^ *e>el & = 2 ,,6#$ ) &&6#($ = 2 6#6% ')mm$ ^ *e>el $ = ,,6#$ ) 9%#(66 = 9#,& ')mm$ Jnitial stresses at mid span "
cl# 6#,#$#
Allowing for $[ rela8ation loss in steel after transfer7 concrete shrinkage \cs = 3(( &(26 and concrete specific creep ct = &#(3 49 &(26 per ')mm$ *oss of force after transfer due to " cl# 6#,#$#$
teel rela8ation = (#($ 69 = &&& cl# 6#,#$#4 26
Concrete shrinkage = \cs .s Aps ! = 3(( &( &%6 3$ &3% = $6$ cl# 6#,#$#
Concrete creep = ct fco .s Aps ! = &#(3 49 &(26 &$#,6 &%6 3$ &3% = ( /otal *oss = &&& $6$ ( = %$3 k' inal force after all loss of prestress = @e = (6, 2 %$3 = 444 $, @e)@ = (#9$! inal stresses due to prestress after all loss of prestress at " *e>el & f&7(#9$@ = (#9$ $(#9% = &,#(9 ')mm$ *e>el $ f$7(#9$@ = (#9$ 2 &#$( = 2 (#%9 ')mm$ Combined stresses in final condition for worst effects of design loads7 differential shrinkage and temperature difference " *e>el &7 combination & H; " f = &,#(9 2 &6#,& = (#3, ')mm$ T ( hence U#-#! *e>el &7 combination 3 " f = &,#(9 2 &,#6, = 2 (#% ')mm$ T 2 3#$ hence U#-#! *e>el $7 combination & " f = 2 (#%9 &(3, ) 9%#(66 &3&( ) $4$#4$4 @ = &,#,& Q $ U#-#! *e>el 3# combination 3 " f = &&$, ) &,%#4($! (#9 3#&! = 9#9 ')mm$ Q $ U#-#!
lti%ate Capa"ity o! Bea% and #e"$ Sla (Co%posite Se"tion+
ltimate Design :oment = < f3 : = &#& $96, = 354 $,% cl# 6#3#3
Unly steel in the tension Sone is to be considered " Centroid of tendons in tension Sone = 66( &(&&( 9&6( 4$6(! ) $9 = &3mm .ffecti>e depth from *e>el 3 = &$(( 2 &3 = &(6mm Assume that the ma8imum design stress is de>eloped in the tendons7 then " /ensile force in tendons p = (#9, $9 &3% &6,( &(23 = 6 k' Compressi>e force in concrete flange " f = (#4 4( &((( &( &(23 = $4(( k' *et = depth to neutral a8is# Compressi>e force in concrete web " w = (#4 ( 13%3 2 3%3 2 $((! 2 &(! ) 6,& $!5 2 &(! &(23 w = 2$#9,6$ 9,$$#94 2 &$43,&,! &(23 .uating forces to obtain " 6 = $4(( 2$#9,6$ 9,$$#94 2 &$43,&,! &(23 = 6% mm tress in tendon after losses = f pe = 4&44 &(3 ) 3$ &3%! = %3$ ')mm$ @restrain \pe = fpe ) .s = %3$ ) $(( &(3 = (#((4,
Determine depth to neutral a8is by an iterati>e strain compatibility analysis /ry = 6% mm as an initial estimate +idth of web at this depth = $4,mm \pb6 = \6 \pe = 24% (#((3 ) 6% (#((4, = (#(($$ \pb = \ \pe = 23% (#((3 ) 6% (#((4, = (#(($9 \pb4 = \4 \pe = $9& (#((3 ) 6% (#((4, = (#((6$
\pb3 = \3 \pe = 39& (#((3 ) 6% (#((4, = (#((6, \pb$ = \$ \pe = 43& (#((3 ) 6% (#((4, = (#((6% \pb& = \& \pe = 49& (#((3 ) 6% (#((4, = (#((,$ fpb6 = (#(($$ $(( &(3 = 444 ')mm$ fpb = (#(($9 $(( &(3 = & ')mm$ fpb4 = &&6$ $%( (#((6$ 2 (#((9! ) (#((6 = &&,9 ')mm$ fpb3 = &&6$ $%( (#((6, 2 (#((9! ) (#((6 = &$(& ')mm$ fpb$ = &&6$ $%( (#((6% 2 (#((9! ) (#((6 = &$&3 ')mm$ fpb& = &&6$ $%( (#((,$ 2 (#((9! ) (#((6 = &$$ ')mm$ /ensile force in tendons " p6 = $ &3% 444 &(23 = &$4 p = $ &3% & &(23 = &3 23 p4 = 4 &3% &&,9 &( = 6 23 p3 = 9 &3% &$(& &( = &336 p$ = &( &3% &$&3 &(23 = &696 23 p& = 6 &3% &$$ &( = &($$ t = N p& to 6 = 4%,6 k' Compressi>e force in concrete " f = (#4 4( &((( &( &(23 = $4(( w = (#4 ( (# 3%3 $4,! 6% 2 &(! &(23 = 3$9 c = f w = 69 k' c T t therefore reduce depth to neutral a8is and repeat the calculations# sing a depth of 6mm will achie>e euilibrium# /he following forces are obtained " p6 = &34 f = $4(( p = &69 w = $,6 p4 = 6, c = &6 p3 = &39$ p$ = &,46 p& = &(6( t = &6
/aking :oments about the neutral a8is " :p6 = &34 2(#36 = 24% :p = &69 2(#$6 = 24 :p4 = 6, (#3, = $3 :p3 = &39$ (#4, = 66 :p$ = &,46 (#$ = %&, :p& = &(6( (#, = 6&( :f = $4(( (#4% = &&,6 :w = 3$9 (#$(, = 6,4 :u = N :p& to 6 :f :w = 4/2 $,% T 3&4 k'm hence U#-# cl# 6#3#3#&
:u ) : = 4&%$ ) 3&4 = ! T &#& ! hence strain in outermost tendon U#-#
cl# 6#3#4
/he hear Besistance of the beam needs to be determined in accordance with clause 6#3#4# and compared with the ultimate shear load at critical sections#
