Design of Simply Supported Deep Beams using IS 456: 2000 and Strut and Tie Method P Nagarajan, Non–member Dr T M M Pillai, Member Dr N Ganesan, Fellow Structural members can be broadly divided into two regions, namely, B (or Bernoulli) regions where the strain distributions are linear, and D (or Disturbed) regions where the strain distributions are non– linear. While well defined theories are available for designing B regions, thumb rule or empirical equations are still being used to design D regions, though B and D regions are equally important. It has been recently understood that the strut and tie method (STM) is an effective tool for the design of both B and D regions. Since the STM is a realistic approach, this has found place in many codes like Euro code, American code, Canadian code, Australian code and New Zealand code. The STM approach is not described in detail in the current IS 456:2000 code of practice for plain and reinforced concrete. The main objective of this paper is to highlight the STM concept as a powerful design concept for the analysis and design of concrete structural elements. For this purpose a simply supported deep beam subjected to a uniformly distributed loading (UDL) is considered and it is designed using both IS 456 codal recommendations and with STM. It is shown that deep beams can be economically designed using STM. Keywords : Strut and tie method; Deep beam; Concrete
W ,Wb ,WT : widths of strut, beam and tie, respectively
NOT ATION NOTA
Asi
: area of crack control reinforcement
WU
: factored load
Ast
: area of tension steel reinforcement
Z
: lever arm
C
: compressive force
βs , βn
D
: depth of beam
: stress reduction factors for strut and node, respectively
F
: force in the truss member
φi
: angle made by crack control reinforcement with the axis of the strut
fck
: characteristic compressive strength of concrete
θ
: strut inclination
fcn, fcs
: permissible stresses in nodes and struts, respectively
fy
: yield stress of steel
h
: height of truss
L
: effective span
LB
: length of bearing plate
MU
: factored moment
Si
: spacing of crack control reinforcement
T
: tensile force
P Nagarajan, Dr T M M Pillai and Dr N Ganesan are with the Department of Civil Engineering, National Institute of T echnology Technology echnology,, Calicut 673 601. This paper (modified) was received on August 24, 2006. Written discussion on this paper will be entertained till July 31, 2007.
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INTRODUCTION Concrete structural members having depth comparable to the span are generally termed as deep beams. In these members, the distribution of strain across the depth of the cross section is nonlinear and a significant amount of load is carried to the supports by a compression strut joining the load and the reaction. These structural elements belong to D (disturbed) regions, which have traditionally been designed using empirical formulae or using past experience. The design procedure for deep beams recommended by IS 456 : 20001 is empirical in nature and based on the experimental investigations conducted by Leonhardt and Walter at University of Stuttgart. These provisions are valid for deep beams subjected to uniformly distributed loads (UDL)2. The strut and tie model (STM) offers an alternative to such empirical methods. Also, the STM provides design engineers with a more flexible and intuitive option for designing structural elements. In this method, complex stress flows in a cracked concrete structure is approximated with simple truss elements that
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can be analysed and designed using basic structural mechanics. The concept of using uniaxially stressed truss members to model the complex stress flows in cracked reinforced concrete was used by the pioneers of reinforced concrete design like Ritter and Morsch. Schliach, Schafer and Jennewein3,4 have made significant contributions to this approach for the design of D region. Following this work, STM began appearing in many international codes of practice5. In order to popularise the use of STM for routine design work, design examples using STM are necessary.
Table 1 Side face reinforcement as given in IS 456 : 2000 T ype of reinforcement
Percentage of side face reinforcement of gross area of concrete Horizontal Ve r t i c a l
Bars of diameter ≤ 16 mm and yield stress (fy ) ≥ 415 N/mm2 Bars of other types than the above Welded wire fabric made with bars of diameter ≤ 16 mm
0.12 0.15
0.20 0.25
0.12
0.20
In this paper, a simply supported deep beam subject to UDL is designed using the recommendations given in IS 456 : 2000. The same is designed using STM and the results are compared. The various design steps that have to be followed in STM is also highlighted.
members and join them by nodes. These models are generally used for the analysis, design and detailing of D regions, such as, vicinities of point loads, frame corners, corbels and also where sudden changes in cross section occurs.
IS 456: 2000 CODAL PROVISIONS FOR THE DESIGN OF SIMPL Y SUPPORTED DEEP SIMPLY BEAMS
The STM is based on the lower bound theorem of plasticity. Therefore, the actual capacity of the structure is considered to be equal to or greater than that of the idealised truss. Hence designs done by using this method will be always on the safer side.
As per the code, when the ratio of the effective span (L ) to the overall depth (D ) of a simply supported beam is less than or equal to 2.0, then the beam can be treated as a deep beam. The lever arm (Z ) is given as (1)
(2)
Struts Struts are compression members in a strut and tie model. These represent concrete stress fields whose principal compressive stresses are predominantly acting along the centreline of the strut. Struts are often idealised as prismatic or as bottle shaped elements. The different types of struts are shown in Figure 1. Ties
where L is the effective span taken as centre to centre distance between the supports, or 1.15 times the clear span, which ever is smaller; and D is the overall depth. The tensile reinforcement required to resist the positive bending moment shall
Ties are tension members and represent reinforcing steel. Nodes
§ be embedded to a length not less than 0.8 times the development length beyond the face of the support; and
Nodes in STM are the intersection points of three or more straight struts and ties. These are analogous to joints in a real truss. Depending on the nature of forces, nodes can be classified as CCC, CCT, CTT and TTT nodes (Figure 2). C is used to denote the compression force and T is used to denote the tension force.
§ be placed within a zone of depth equal to 0.25 D – 0.05 L from the extreme tension fibre at mid–span.
The design procedure using STM generally involves five major steps as
§ extend without curtailment between the supports;
The side face reinforcement is provided both in the vertical and horizontal directions at a spacing limited to, smaller of three times the width of the beam or 450 mm. The minimum areas of vertical and horizontal side face reinforcement are given in Table 1.
§ Identify the D region; § Sketch the truss and determine the equivalent boundary loads and analyse the truss to get the member forces;
REVIEW OF STRUT AND TIE METHOD (STM) In the strut and tie method, a reinforced concrete member is replaced by a system of truss members that can resist the applied loads. For analytical purposes, the strut and tie models condense all stresses in compression and tension
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Prismatic strut
Bottle shaped strut
Figure 1 Different types of struts
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C
T
T
Table 2 β s for different types of struts
C C
C C
C
CCC node
T
T
CCT node
T
T
CTT node
TTT node
T ype of strut Prismatic Bottle shaped (with crack control reinforcement) Bottle shaped (with no crack control reinforcement)
βs 1.00 0.75 0.60
Figure 2 Different types of nodes Table 3 β n for different types of nodes T ype of node CCC CCT CTT, TTT
§ Evaluate the dimensions of the struts and nodes, such that, the capacities of the struts and nodes are sufficient to carry the member forces; § Provide sufficient steel reinforcement for the required tie capacity and ensure that this reinforcement is properly anchored; and § Provide distributed reinforcement to ensure ductile behaviour of the D region. In the case of a real truss, the identification of member areas and joint details and their design is fairly straight forward. However, in the case of an implicit truss embedded in concrete, the determination of appropriate member cross sectional areas and node dimensions is not so simple, especially for the determination of the concrete strut and node dimensions. Although IS 456 : 2000 recommends the use of the strut and tie method (for corbel design), no guidelines are given for the determination of the dimensions of the struts and nodes and for the permissible stresses in these elements. Hence, the design recommendations given in ACI 318–026 are used in this paper and the salient details are given here. The recommendations are slightly modified by incorporating the safety factors and notations followed in IS 456 : 2000.
βn 1.0 0.8 0.6
where Asi refers to the crack control reinforcement provided at a spacing of Si in a layer of reinforcement with bars at an angle of φi to the axis of the strut. Inclination of Compression Strut The inclination of the compression strut with the horizontal (θ) should lie between 25° and 65°. DESIGN PROBLEM A typical beam is considered for the analysis and design. The details of the beam are shown in Figure 4. The grade of concrete is M20 and steel is Fe 415. The size of bearing plate is 0.45 m × 0.5 m. The beam is designed using IS 456 : 2000 and STM and the results are critically compared. Design using IS 456 : 2000 Codal Recommendations The L /D ratio of the beam is 2.0 and the beam is classified as a deep beam. The total factored load (including dead load)
Permissible Stresses in Struts and Nodes The permissible stresses in different types of struts (fcs ) is given as
WU = 1.5(200 + (25 × 3 × 0.5)) = 356.25 kN/m S1
(3)
Axis of strut
where the values of βs are given in Table 2. Strut boundary
The permissible stresses in different types of nodes (fcn ) is given as (4) where the values of βn are given in Table 3.
S2
φ1 φ2 AS2
Crack Control Reinforcement ACI 318–026 recommends an orthogonal grid of bars (Figure 3) on each face to control the crack width of bottle shaped struts and it should satisfy the criteria (5) 40
A S1
Figure 3 Crack control reinforcement crossing a bottle shaped strut
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WU L/2
WU L/2
200 kN/m
L/4
B
C
h
3m
L/4
L/2
0.5 m
6m
Bearing plate
D
A
Figure 4 Simply supported deep beam Figure 5 Strut and tie model for simply supported deep beam subjected to UDL
The maximum mid–span moment (factored) Support reaction
The lever arm (equation (1)) Bearing pressure at node A and D = Area of steel required to resist MU
=
4.75 MPa (< fcn, hence safe).
Determination of Strut Inclination ( θ)
The main reinforcement should be anchored properly. The side face reinforcement according to Table 1 must be provided. Design using S TM (based on ACI 318–02 6 Codal STM Recommendations) The strut and tie model for the deep beam subjected to a UDL is shown in Figure 5. The UDL is replaced by two point loads. In the Figure 5, struts are shown by dotted lines (since they are not real members) and ties are shown by solid lines. Strut BC is prismatic and struts AB and CD can be considered as bottle shaped struts. From the figure it can be seen that nodes B and C are CCC nodes and nodes A and D are CCT nodes. Member AC is included so that the truss is stable. Since the truss is symmetrically loaded, the force in member AC is zero. The step by step procedure for the design using STM is given here.
Check for Bearing Pressure at Support Points (A or D) Since, A and D are CCT nodes, the bearing pressure must not exceed the permissible stress for a CCT node. Permissible stress (equation (4))
Since, the position of the nodes A, B, C and D are not known (in the vertical direction), it is not possible to find the strut inclination θ and hence the member force cannot be evaluated. Hence, it is assumed that the capacity of the prismatic strut BC is fully utilised. If WBC is the depth of the strut (in mm), the force in the member BC (FBC ) must be equal to
where fcs is determined using equation (3). Assuming WAD and FAD to be the depth and force of tie AD, FAD must be equal to FBC (in magnitude). Since, the permissible stress of node A or D is less than the permissible stress of tie AD ( ie, 0.87 fy ), the nodal stress should be used to determine WAD . Equating FAD and FBC
Hence
WAD = 1.25 WBC . Then, the height of the truss ( h ) will be equal to .
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Hence
h = D – 1.125 WBC = 3000 – 1.125 WBC C
By taking summation of moments about joint D W
From this equation WBC is determined. For the given problem, WBC = 124.57 mm and hence, WAD =155.7 mm. Try WBC = 130 mm and WAD = 165 mm
.
θ
T
WT
LB
Hence
Figure 7 Strut dimension at node A
and
Check for Stresses in Concrete Struts
θ = 62.26° (< 65°, ok)
Strut BC
Member Forces Using truss analysis, the member forces are determined as
Since the depth provided for strut BC is more than that is required, the stress in the member will be less than the permissible value. Strut AB or CD
FBC = FAD = 562.06 kN
The depth of strut WAB for the member AB at node A can be determined from the geometry.
FAB = FDC = 1207.53 kN FAC = 0.0 kN
WAB at node A From Figure 7
W = WT cosθ + LB sinθ .
WU L/2
where W is depth of the strut (W = WAB ); WT , the depth of tie (WT = WAD); LB , the length of bearing plate (LB = 0.45 m) and θ is the strut inclination (θ = 62.26°). FBC
C
From the formula, WAB is 475.1 mm.
h
Permissible stress in strut AB , (equation (3), f cs ) = 0.45 f ck (0.75)=6.75 MPa (assuming that crack control reinforcement according to ACI 318–02 is provided).
θ
FAD
D
L/4
Stress in strut AB
Area of Tensile Reinforcement (Ast) Now the area of tensile reinforcement, Ast is calculated as
WU L/2 Figure 6 Section of the truss
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This area of steel is less than that required as per IS 456: 2000 codal recommendations. Steel reinforcement satisfying the above criteria should be provided over a depth equal to WAB , such that, the centroid of the tension steel
must be extended upto the support. Thus, the strut and tie model helps to understand the behaviour of the structural elements and will be extremely useful for detailing reinforced concrete members.
from the soffit of the beam. The main
Hence, concrete structural elements can be economically designed using the strut and tie method and recommendations for designs using STM should be incorporated in the next revision of IS: 456.
lies at
reinforcement should be anchored sufficiently into the support and side face reinforcement (crack control reinforcement) should be provided as per ACI 318–02 recommendations (equation( 5)). CONCLUSIONS § Though STM is a conservative method, the area of steel calculated using the STM is less than that required as per IS 456 : 2000 recommendations. § STM can be used to design deep beams subject to any type of loading, but IS 456 : 2000 recommendations are applicable only to deep beams subject to UDL. § From the strut and tie model, it is seen that the force in the bottom member is constant through out its length. This means that arch action is predominant in a deep beam. Further this implies that the reinforcement provided at the mid–span should not be curtailed and
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REFERENCES 1. IS : 456–2000. ‘Plain and Reinforced Concrete — Code of Practice’. Bureau of Indian Standards, Manak Bhavan, New Delhi, India. 2. R Park and T Paulay. ‘Reinforced Concrete Structures’. John Wiley and Sons, New York, 1975. 3. J Schliach, K Schafer and K Jennewein. ‘Towards a Consistent Design of Structural Concrete’. Journal of the Prestressed Concrete Institute, vol 32, 1987, p 75. 4. J Schliach and K Schafer. ‘Design and Detailing of Structural Concrete using Strut–and–Tie Models’. The Structural Engineer, vol 69, 1991, p 113. 5. CSA Standard A23.3–94. ‘Design of Concrete Structures’. Canadian Standards Association, Rexdale, Ontario, Canada. 6. ACI 318–02. ‘Building Code Requirements For Structural Concrete and Commentary’. American Concrete Institute, Detroit, USA.
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