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Is code Cracked | Reinforced Concrete | Bending
this refers to various design methods of concrete
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CE 2306-DESIGN OF RC ELEMENTS
Prepared by M.UMAMAGUESVARI M.Tech Sr. Lecturer Department of Civil Engineering RAJALAKSHMI ENGINEERING COLLEGE
Reinforced concrete, as a composite material, has occupied a special place in the modern construction of different types of structures due to its several advantages. Italian architect Ponti once remarked that concrete liberated us from the rectangle. Due to its flexibility in form and superiority in performance, it has replaced, to a large extent, the earlier materials like stone, timber and steel. Further, architect's scope and imaginations have widened to a great extent due to its mouldability and monolithicity. Thus, it has helped the architects and engineers to build several attractive shell forms and other curved structures. However, its role in several straight line structural forms like multistoried frames, bridges, foundations etc. is enormous.
Design of reinforced concrete structures started in the beginning of last century following purely empirical approach. Thereafter came the so called rigorous elastic theory where the levels of stresses in concrete and steel are limited so that stress-deformations are taken to be linear. However, the limit state method, though semi-empirical approach, has been found to be the best for the design of reinforced concrete structures (see sec. 1.1.3.1 also).
Every structure has got its form, function and aesthetics. Normally, we consider that the architects will take care of them and the structural engineers will be solely responsible for the strength and safety of the structure. However, the roles of architects and structural engineers are very much interactive and a unified approach of both will only result in an "Integrated" structure, where every material of the total structure takes part effectively for form, function, aesthetics, strength as well as safety and durability. This is possible when architects have some basic understanding of structural design and the structural engineers also have the basic knowledge of architectural requirements. Both the engineer and the architect should realize that the skeletal structure without architecture is barren and mere architecture without the structural strength and safety is disastrous. Safety, here, includes consideration of reserve strength, limited deformation and durability. However, some basic knowledge of architectural and structural requirements would facilitate to appreciate the possibilities and limitations of exploiting the reinforced concrete material for the design of innovative structures.
Before proceeding to the design, one should know the objectives of the design of concrete structures. The objectives of the design are as follows:
1 .The structures so designed should have an acceptable probability of performing satisfactorily during their intended life.
This objective does not include a guarantee that every structure must perform satisfactorily during its intended life. There are uncertainties in the design process both in the estimation of the loads likely to be applied on the structure and in the strength of the material. Moreover, full guarantee would only involve more cost. Thus, there is an acceptable probability of performance of structures as given in standard codes of practices of different countries.
2. The designed structure should sustain all loads and deform within limits
for construction and use.
Adequate strengths and limited deformations are the two requirements of the designed structure. The structure should have sufficient strength and the deformations must be within prescribed limits due to all loads during construction. The structure having insufficient strength of concrete which fails in bending compression with the increase of load, though the deformation of the structure is not alarming. On the other hand another situation where the structure, having sufficient strength, deforms excessively. Both are undesirable during normal $construction and use. However, sometimes structures are heavily loaded beyond control. The structural engineer is not responsible to ensure the strength and deformation within limit under such situation. The staircases in residential buildings during festival like marriage etc., roof of the structures during flood in the adjoining area or for buildings near some stadium during cricket or football matches are some of the examples when structures get overloaded. Though, the structural designer is not responsible for the strength and deformations under these situations, he, however, has to ensure that the failure of the structures should give sufficient time for the occupants to vacate. The structures, thus, should give sufficient warning to the occupants and must not fail suddenly.
3. The designed structures should be durable.
The materials of reinforced concrete structures get affected by the environmental
conditions. Thus, structures having sufficient strength and permissible deformations may have lower strength and exhibit excessive deformations in the long run. The designed structures, therefore, must be checked for durability. Separate checks for durability are needed for the steel reinforcement and concrete. This will avoid problems of frequent repairing of the structure.
4. The designed structures should adequately resist to the effects of misuse and fire. Structures may be misused to prepare fire works, store fire works, gas and other highly inflammable and/or explosive chemicals. Fire may also take place as accidents or as secondary effects during earthquake by overturning kerosene stoves or lantern, electrical short circuiting etc. Properly designed structures
should allow sufficient time and safe route for the persons inside to vacate the structures before they actually collapse.
All the above objectives can be fulfilled by understanding the strength and deformation characteristics of the materials used in the design as also their deterioration under hostile exposure. Out of the two basic materials concrete and steel, the steel is produced in industries. Further, it is available in form of standard bars and rods of specific diameters. However, sample testing and checking are important to ensure the quality of these steel bars or rods. The concrete, on the other hand, is prepared from several materials (cement, sand, coarse aggregate, water and admixtures, if any). Therefore, it is important to know the characteristic properties of each of the materials used to prepare concrete. These materials and the concrete after its preparation are also to be tested and checked to ensure the quality. The necessary information regarding the properties and characteristic strength of these materials are available in the standard codes of practices of different countries. It is necessary to follow these clearly defined standards for materials, production, workmanship and maintenance, and the performance of structures in service.
Three methods of design are accepted in cl. 18.2 of IS 456:2000 (Indian Standard Plain and Reinforced Concrete - Code of Practice, published by the Bureau of Indian Standards, New Delhi). They are as follows:
The term “Limit states” is of continental origin where there are three limit states - serviceability / crack opening / collapse. For reasons not very clear, in English literature limit state of collapse is termed as limit state.
As mentioned in sec. 1.1.1, the semi-empirical limit state method of design has been found to be the best for the design of reinforced concrete members. More details of this method are explained in Module 3 (Lesson 4). However, because of its superiority to other two methods (see sections 2.3.2 and 2.3.3 of Lesson 3), IS 456:2000 has been thoroughly updated in its fourth revision in 2000 takinginto consideration the rapid development in the field of concrete technology and incorporating important aspects like durability etc. This standard has put greater emphasis to limit state method of design by presenting it in a full section (section 5), while the working stress method has been given in Annex B of the same standard. Accordingly, structures or structural elements shall normally be designed by limit state method.
This method of design, considered as the method of earlier times, has several limitations. However, in situations where limit state method cannot be conveniently applied, working stress method can be employed as an alternative.
It is expected that in the near future the working stress method will be completely replaced by the limit state method.
Method based on experimental approach
The designer may perform experimental investigations on models or full size structures or elements and accordingly design the structures or elements. However, the four objectives of the structural design (sec. 1.1.2) must be satisfied when designed by employing this approach. Moreover, the engineer in charge has to approve the experimental details and the analysis connected therewith.
Though the choice of the method of design is still left to the designer as per cl. 18.2 of IS 456:2000, the superiority of the limit state method is evident from the emphasis given to this method by presenting it in a full section (Section 5), while accommodating the working stress method in Annex B of IS 456:2000, from its earlier place of section 6 in IS 456:1978. It is expected that a gradual change over to the limit state method of design will take place in the near future after overcoming the inconveniences of adopting this method in some situations.
Structures when subjected to external loads (actions) have internal reactions in the form of bending moment, shear force, axial thrust and torsion in individual members. As a result, the structures develop internal stresses and undergo deformations. Essentially, we analyse a structure elastically replacing each member by a line (with EI values) and then design the section using concepts of
limit state of collapse. Figure 1.1.1 explains the internal and external reactions of
a simply supported beam under external loads. The external loads to be applied
on the structures are the design loads and the analyses of structures are based on linear elastic theory (vide cl. 22 of IS 456:2000).
The design loads are determined separately for the two methods of design as mentioned below after determining the combination of different loads.
1.In the limit state method, the design load is the characteristic load with appropriate partial safety factor (vide sec. 2.3.2.3 for partial safety factors).
In the working stress method, the design load is the characteristic load only.
What is meant by characteristic load? Characteristic load (cl. 36.2 of IS 456:2000) is that load which has a ninety-five per cent probability of not being exceeded during the life of the structure.
The various loads acting on structures consist of dead loads, live loads, wind or earthquake loads etc. These are discussed in sec. 1.1.6. However, the researches made so far fail to estimate the actual loads on the structure. Accordingly, the loads are predicted based on statistical approach, where it is assumed that the variation of the loads acting on structures follows the normal distribution (Fig. 1.1.2). Characteristic load should be more than the average/mean load. Accordingly, Characteristic load = Average/mean load + K (standard deviation for load)
The value of K is assumed such that the actual load does not exceed the characteristic load during the life of the structure in 95 per cent of the cases.
The following are the different types of loads and forces acting on the structure. As mentioned in sec. 1.1.5, their values have been assumed based on earlier data and experiences. It is worth mentioning that their assumed values as stipulated in IS 875 have been used successfully.
These are the self weight of the structure to be designed . Needless to mention that the dimensions of the cross section are to be assumed initially which enable to estimate the dead loads from the known unit weights of the materials of the structure. The accuracy of the estimation thus depends on the assumed values of the initial dimensions of the cross section. The values of unit weights of the materials are specified in Part 1 of IS 875.
They are also known as live loads and consist of all loads other than the dead loads of the structure. The values of the imposed loads depend on the functional requirement of the structure. Residential buildings will have comparatively lower
values of the imposed loads than those of school or office buildings. The standard values are stipulated in Part 2 of IS 875.
3 .Wind loads
These loads depend on the velocity of the wind at the location of the structure, permeability of the structure, height of the structure etc. They may be horizontal or inclined forces depending on the angle of inclination of the roof for pitched roof structures. They can even be suction type of forces depending on the angle of inclination of the roof or geometry of the buildings Wind loads are specified in Part 3 of IS 875.
4 .Snow loads
These are important loads for structures located in areas having snow fall, which gets accumulated in different parts of the structure depending on projections, height, slope etc. of the structure. The standard values of snow loads are specified in Part 4 of IS 875.
5 .Earthquake forces
Earthquake generates waves which move from the origin of its location (epicenter) with velocities depending on the intensity and magnitude of the earthquake. The impact of earthquake on structures depends on the stiffness of the structure, stiffness of the soil media, height and location of the structure etc. Accordingly, the country has been divided into several zones depending on the magnitude of the earthquake. The earthquake forces are prescribed in IS 1893. Designers have adopted equivalent static load approach or spectral method.
6 .Shrinkage, creep and temperature effects
Shrinkage, creep and temperature (high or low) may produce stresses and cause deformations like other loads and forces. Hence, these are also considered as loads which are time dependent. The safety and serviceability of structures are to be checked following the stipulations of cls. 6.2.4, 5 and 6 of IS 456:2000 and Part 5 of IS 875.
7. Other forces and effects
It is difficult to prepare an exhaustive list of loads, forces and effects coming onto the structures and affecting the safety and serviceability of them. However, IS 456:2000 stipulates the following forces and effects to be taken into account in case they are liable to affect materially the safety and serviceability of the structures. The relevant codes as mentioned therein are also indicated below:
• Foundation movement (IS 1904) (Fig. 1.1.3)
• Elastic axial shortening
• Soil and fluid pressures (vide IS 875 - Part 5)
• Impact (vide IS 875 - Part 5)
• Erection loads (Please refer to IS 875 - Part 2) (Fig. 1.1.4)
• Stress concentration effect due to point of application of load and the like.
8. Combination of loads
Design of structures would have become highly expensive in order to maintain their serviceability and safety if all types of forces would have acted on all structures at all times. Accordingly, the concept of characteristic loads has been accepted to ensure that in at least 95 per cent of the cases, the characteristic loads considered will be higher than the actual loads on the structure. However, the characteristic loads are to be calculated on the basis of average/mean load of some logical combinations of all the loads mentioned in sec. 1.1.6.1 to 7. These logical combinations are based on (i) the natural phenomena like wind and earthquake do not occur simultaneously, (ii) live loads on roof should not be present when wind loads are considered; to name a few. IS 875 Part 5 stipulates the combination of loads to be considered in the design of structures.
Slabs, used in floors and roofs of buildings mostly integrated with the supporting beams, carry the distributed loads primarily by bending. the integrated slab is considered as flange of T- or L-beams because of monolithic construction. However, the remaining part of the slab needs design considerations. These slabs are either single span or continuous having different support conditions like fixed, hinged or free along the edges . Though normally these slabs are horizontal, inclined slabs are also used in ramps, stair cases and inclined roofs . While square or rectangular plan forms are normally used, triangular, circular and other plan forms are also needed for different functional requirements. This lesson takes up horizontal and rectangular /square slabs of buildings supported by beams in one or both directions and subjected to uniformly distributed vertical loadings.
The other types of slabs, are given below. All these slabs have additional requirements depending on the nature and magnitude of loadings in respective cases.
(a) horizontal or inclined bridge and fly over deck slabs carrying
heavy concentrated loads,
(b) horizontal slabs of different plan forms like triangular, polygonal
flat slabs having no beams and supported by columns only,
inverted slabs in footings with or without beams,
slabs with large voids or openings,
(f) grid floor and ribbed slabs.
the share of loads on beams supporting solid slabs along four edges when vertical loads are uniformly distributed. It is evident from the figures that the share of loads on beams in two perpendicular directions depends upon the aspect ratio ly /lx of the slab, lx being the shorter span. For large values of ly, the triangular area is much less than the trapezoidal area (Fig.8.18.4a). Hence, the share of loads on beams along shorter span will gradually reduce with increasing ratio of ly /lx. In such cases, it may be said that the loads are primarily taken by beams along longer span. The deflection profiles of the slab along both directions are also shown in the figure. The deflection profile is found to be constant along the longer span except near the edges for the slab panel of Fig.8.18.4a. These slabs are designated as one-way slabs as
they span in one direction (shorter one) only for a large part of the slab when ly /lx > 2.
On the other hand, for square slabs of l y /l x = 1 and rectangular slabs of l y /l x up to 2, the deflection profiles in the two directions are parabolic (Fig.8.18.4b). Thus,
they are spanning in two directions and these slabs with l y /l x up to 2 are
designated as two-way slabs, when supported on all edges. It would be noted that an entirely one-way slab would need lack of support on short edges. Also, even for l y /l x < 2, absence of supports in two parallel edges will
render the slab one-way. In Fig. 8.18.4b, the separating line at 45 degree is tentative serving purpose of design. Actually, this angle is a function of l y /l x .
Design Shear Strength of Concrete in Slabs
Experimental tests confirmed that the shear strength of solid slabs up to a depth of 300 mm is comparatively more than those of depth greater than 300 mm. Accordingly, cl.40.2.1.1 of IS 456 stipulates the values of a factor k to be multiplied with cτ given in Table 19 of IS 456 for different overall depths of slab. Table 1 presents the values of k as a ready reference below:
Table1 Values of the multiplying factor k
Thin slabs, therefore, have more shear strength than that of thicker slabs. It is the normal practice to choose the depth of the slabs so that the concrete can resist the shear without any stirrups for slab subjected to uniformly distributed loads.
However, for deck slabs, culverts, bridges and fly over, shear reinforcement should be provided as the loads are heavily concentrated in those slabs. Though, the selection of depth should be made for normal floor and roof slabs to avoid stirrups, it is essential that the depth is checked for the shear for these slabs taking due consideration of enhanced shear strength as discussed above depending on the overall depth of the slabs.
The primary design considerations of both one and two-way slabs are strength and deflection. The depth of the slab and areas of steel reinforcement are to be determined from these two aspects. The detailed procedure of design of one-way slab is taken up in the next section. However, the following aspects are to be decided first. (a) Effective span (cl.22.2 of IS 456) The effective span of a slab depends on the boundary condition. Table 8.2 gives the guidelines stipulated in cl.22.2 of IS 456 to determine the effective span of a slab. Table 8.2 Effective span of slab (cl.22.2 of IS 456)
(b) Effective span to effective depth ratio (cls.23.2.1a-e of IS 456)
The deflection of the slab can be kept under control if the ratios of effective span to effective depth of one-way slabs are taken up from the provisions in cl.23.2.1a- e of IS 456. These stipulations are for the beams and are also applicable for one- way slabs as they are designed considering them as beam of unit width. These provisions are explained in sec.3.6.2.2 of Lesson 6.
(c) Nominal cover (cl.26.4 of IS 456)
The nominal cover to be provided depends upon durability and fire resistance requirements. Table 16 and 16A of IS 456 provide the respective values.
Appropriate value of the nominal cover is to be provided from these tables for the particular requirement of the structure.
(d) Minimum reinforcement (cl.26.5.2.1 of IS 456)
Both for one and two-way slabs, the amount of minimum reinforcement in either direction shall not be less than 0.15 and 0.12 per cents of the total cross- sectional area for mild steel (Fe 250) and high strength deformed bars (Fe 415 and Fe 500)/welded wire fabric, respectively.
(e) Maximum diameter of reinforcing bars (cl.26.5.2.2)
The maximum diameter of reinforcing bars of one and two-way slabs shall not exceed one-eighth of the total depth of the slab.
(f) Maximum distance between bars (cl.26.3.3 of IS 456)
The maximum horizontal distance between parallel main reinforcing bars shall be the lesser of (i) three times the effective depth, or (ii) 300 mm. However, the same for secondary/distribution bars for temperature, shrinkage etc. shall be the lesser of (i) five times the effective depth, or (ii) 450 mm.
The procedure of the design of one-way slab is the same as that of beams. However, the amounts of reinforcing bars are for one metre width of the slab as to be determined from either the governing design moments (positive or negative) or from the requirement of minimum reinforcement. The different steps of the design are explained below.
Step 1: Selection of preliminary depth of slab
The depth of the slab shall be assumed from the span to effective depth ratios as given in section 3.6.2.2 of Lesson 6 and mentioned here in sec.8.18.5b.
Step 2: Design loads, bending moments and shear forces
The total factored (design) loads are to be determined adding the estimated dead load of the slab, load of the floor finish, given or assumed live loads etc. after multiplying each of them with the respective partial safety factors. Thereafter, the design positive and negative bending moments and shear forces are to be determined using the respective coefficients given in Tables 12 and 13 of IS 456 and explained in sec.8.18.4 earlier.
Step 3: Determination/checking of the effective and total depths of slabs
The effective depth of the slab shall be determined is given below
M u,lim = R ,lim bd 2 …. (3.25)
where the values of R ,lim for three different grades of concrete and three different
grades of steel .The value of b shall be taken as one metre. The total depth of the slab shall then be determined adding appropriate nominal cover (Table 16 and 16A of cl.26.4 of IS 456) and half of the diameter of the larger bar if the bars are of different sizes. Normally, the computed depth of the slab comes out to be much less than the assumed depth in Step 1. However, final selection of the depth shall be done after checking the depth for shear force.
Step 4: Depth of the slab for shear force
Theoretically, the depth of the slab can be checked for shear force if the design shear strength of concrete is known. Since this depends upon the percentage of tensile reinforcement, the design shear strength shall be assumed considering the lowest percentage of steel. The value of cτ shall be modified after knowing the multiplying factor k from the depth tentatively selected for the slab in Step 3. If necessary, the depth of the slab shall be modified.
Step 5: Determination of areas of steel
Area of steel reinforcement along the direction of one-way slab should be determined employing Eq.3.23 of sec.3.5.5 of Lesson 5 and given below as a ready reference. M u = 0.87 f y A st d {1 – (A st )(f y )/(f ck )(bd)} …. (3.23)
The above equation is applicable as the slab in most of the cases is under- reinforced due to the selection of depth larger than the computed value in Step 3. The area of steel so determined should be checked whether it is at least the minimum area of steel as mentioned in cl.26.5.2.1 of IS 456 and explained in
sec.8.18.5d.
Alternatively, tables and charts of SP-16 may be used to determine the depth of the slab and the corresponding area of steel. Tables 5 to 44 of SP-16 covering a wide range of grades of concrete and Chart 90 shall be used for determining the depth and reinforcement of slabs. Tables of SP-16 take into consideration of maximum diameter of bars not exceeding one-eighth the depth of the slab. Zeros at the top right hand corner of these tables indicate the region where the percentage of reinforcement would exceed p t,lim . Similarly, zeros at the lower left
and corner indicate the region where the reinforcement is less than the minimum stipulated in the code. Therefore, no separate checking is needed for the allowable maximum diameter of the bars or the computed area of steel exceeding the minimum area of steel while using tables and charts of SP-16. The amount of steel reinforcement along the large span shall be the minimum amount of steel as per cl.26.5.2.1 of IS 456 and mentioned in sec.8.18.5d earlier. Step 6: Selection of diameters and spacings of reinforcing bars The diameter and spacing of bars are to be determined as per cls.26.5.2.2 and 26.3.3 of IS 456.
continuous slab showing the different reinforcing bars in the discontinuous and continuous ends (DEP and CEP, respectively) of end panel and continuous end of adjacent panel (CAP). The end panel has three bottom bars B1, B2 and B3 and four top bars T1, T2, T3 and T4. Only three bottom bars B4, B5 and B6 are shown in the adjacent panel. Table 8.3 presents these bars mentioning the respective zone of their placement (DEP/CEP/CAP), direction of the bars (along x or y) and the resisting moment for which they shall be designed or if to be provided on the basis of minimum reinforcement. These bars are explained below for the three types of ends of the two panels. Table 8.3 Steel bars of one-way slab (Figs.8.18.5a and b)
0.5 M x for each,
- 0.5 M x for each,
Notes: (i) DEP = Discontinuous End Panel (ii) CEP = Continuous End Panel
(iii) CAP = Continuous Adjacent Panel
(i) Discontinuous End Panel (DEP)
• Bottom steel bars B1 and B2 are alternately placed such that B1 bars are
curtailed at a distance of 0.25 lx 1 from the adjacent support and B2 bars
are started from a distance of 0.15lx 1 from the end support. Thus, both B1
and B2 bars are present in the middle zone covering 0.6lx 1 , each of which
is designed to resist positive moment 0.5M x . These bars are along the
direction of x and are present from one end to the other end of l y .
• Bottom steel bars B3 are along the direction of y and cover the entire
span lx 1 having the minimum area of steel. The first bar shall be placed at
distance not exceeding s/2 from the left discontinuous support, where s
the spacing of these bars in y direction.
Top bars T3 are along the direction of x for resisting the negative
moment which is numerically equal to fifty per cent of positive M x . These
bars are continuous up to a distance of 0.1lx 1 from the centre of support at
the discontinuous end.
• Top bars T4 are along the direction of y and provided up to a distance of
0.1lx 1 from the centre of support at discontinuous end. These are to satisfy
the requirement of minimum steel.
(ii) Continuous End Panel (CEP)
• Top bars T1 and T2 are along the direction of x and cover the entire l y .
They are designed for the maximum negative moment M x and each has a
capacity of -0.5M x . Top bars T1 are continued up to a distance of 0.3lx 1 ,
while T2 bars are only up to a distance of 0.15lx 1 .
• Top bars T4 are along y and provided up to a distance of 0.3lx 1 from the
support. They are on the basis of minimum steel requirement.
(iii) Continuous Adjacent Panel (CAP)
• Bottom bars B4 and B5 are similar to B1 and B2 bars of (i) above.
• Bottom bars B6 are similar to B3 bars of (i) above.
(a) Problem1
Design the one-way continuous slab of Fig.8.18.6 subjected to uniformly
distributed imposed loads of 5 kN/m 2 using M 20 and Fe 415. The load of floor
finish is 1 kN/m 2 . The span dimensions shown in the figure are effective spans. The width of beams at the support = 300 mm.
The basic value of span to effective depth ratio for the slab having simple support at the end and continuous at the intermediate is (20+26)/2 = 23 (cl.23.2.1 of IS
Modification factor with assumed p = 0.5 and f s = 240 N/mm 2 is obtained as 1.18
from Fig.4 of IS 456. Therefore, the minimum effective depth = 3000/23(1.18) = 110.54 mm. Let us take the effective depth d = 115 mm and with 25 mm cover, the total depth D = 140 mm.
Step 2: Design loads, bending moment and shear force
Dead loads of slab of 1 m width = 0.14(25) = 3.5 kN/m Dead load of floor finish =1.0 kN/m Factored dead load = 1.5(4.5) = 6.75 kN/m Factored live load = 1.5(5.0) = 7.50 kN/m
Total factored load = 14.25 kN/m Maximum moments and shear are determined from the coefficients given in Tables 12 and 13 of IS 456. Maximum positive moment = 14.25(3)(3)/12 = 10.6875 kNm/m Maximum negative moment = 14.25(3)(3)/10 = 12.825 kNm/m Maximum shear V u = 14.25(3)(0.4) = 17.1 kN
Step 3: Determination of effective and total depths of slab
From Eq.3.25 of sec. 3.5.6 of Lesson 5, we have
bd 2 where R
is 2.76 N/mm 2 from Table 3.3 of sec. 3.5.6 of Lesson
5. So, d = {12.825(10 6 )/(2.76)(1000)} 0.5 = 68.17 mm
Step 4: Depth of slab for shear force
Table 19 of IS 456 gives cτ = 0.28 N/mm 2 for the lowest percentage of steel in the slab. Further for the total depth of 140 mm, let us use the coefficient k of cl.
40.2.1.1 of IS 456 as 1.3 to get c ττkc= = 1.3(0.28) = 0.364 N/mm 2 .
Table 20 of IS 456 gives maxcτ = 2.8 N/mm 2 . For this problem bdVuv/ =τ = 17.1/115
= 0.148 N/mm 2 . Since, maxc cvτττ<<, the effective depth d = 115 mm is acceptable.
From Eq.3.23 of sec. 3.5.5 of Lesson 5, we have
M u = 0.87 f y A st d {1 – (A st )(f y )/(f ck )(bd)}
(i) For the maximum negative bending moment
12825000 = 0.87(415)(A st )(115){1 – (A st )(415)/(1000)(115)(20)}
or - 5542.16 A2stA st + 1711871.646 = 0
Solving the quadratic equation, we have the negative A st = 328.34 mm
(ii) For the maximum positive bending moment
= 0.87(415) A st (115) {1 – (A st )(415)/(1000)(115)(20)}
or - 5542.16 A2stA st + 1426559.705 = 0
Solving the quadratic equation, we have the positive A st = 270.615 mm
Alternative approach: Use of Table 2 of SP-16
(i) For negative bending moment
M /bd 2 = 0.9697
Table 2 of SP-16 gives: p s = 0.2859 (by linear interpolation). So, the area of
negative steel = 0.2859(1000)(115)/100 = 328.785 mm 2 . (ii) For positive bending moment
M /bd 2 = 0.8081
Table 2 of SP-16 gives: p s = 0.23543 (by linear interpolation). So, the area of
positive steel = 0.23543(1000)(115)/100 = 270.7445 mm 2 . These areas of steel are comparable with those obtained by direct computation using Eq.3.23.
Distribution steel bars along longer span l
Distribution steel area = Minimum steel area = 0.12(1000)(140)/100 = 168 mm 2 . Since, both positive and negative areas of steel are higher than the minimum area, we provide:
(a) For negative steel: 10 mm diameter bars @ 230 mm c/c for which A st =
mm 2 giving p s = 0.2965.
For positive steel: 8 mm diameter bars @ 180 mm c/c for which A st =
mm 2 giving p s = 0.2426
(c) For distribution steel: Provide 8 mm diameter bars @ 250 mm c/c for
which A st (minimum) = 201 mm 2 .
Step 6: Selection of diameter and spacing of reinforcing bars
The diameter and spacing already selected in step 5 for main and distribution bars are checked below:
For main bars (cl. 26.3.3.b.1 of IS 456), the maximum spacing is the lesser of 3d and 300 mm i.e., 300 mm. For distribution bars (cl. 26.3.3.b.2 of IS 456), the maximum spacing is the lesser of 5d or 450 mm i.e., 450 mm. Provided spacings, therefore, satisfy the requirements. Maximum diameter of the bars (cl. 26.5.2.2 of IS 456) shall not exceed 140/8 = 17 mm is also satisfied with the bar diameters selected here.
2.Design the cantilever panel of the one-way slab shown in Fig.8.18.8 subjected
to uniformly distributed imposed loads 5 kN/m 2 using M 20 and Fe 415.
The load of floor finish is 0.75 kN/m 2 . The span dimensions shown in the figure are effective spans. The width of the support is 300 mm.
Step 1: Selection of preliminary depth of slab Basic value of span to depth ratio (cl. 23.2.1 of IS 456) = 7 Modification factor = 1.18 (see Problem 8.1) Minimum effective depth = 1850/7(1.18) = 223.97 mm Assume d = 225 mm and D = 250 mm. Step 2: Design loads, bending moment and shear force
Factored dead loads = (1.5)(0.25)(25) = 9.375 kN/m Factored load of floor finish = (1.5)(0.75) = 1.125 kN/m Factored live loads = (1.5)(5) = 7.5 kN/m Total factored loads = 18.0 kN/m Maximum negative moment = 18.0(1.85)(1.85)(0.5) = 30.8025 kNm/m Maximum shear force = 18.0(1.85) = 33.3 kN/m Step 3: Determination of effective and total depths of slab From Eq.3.25, Step 3 of sec. 8.18.6, we have
d = {30.8025(10 6 )/2.76(10 3 )} 0.5 = 105.64 mm, considering the value of R = 2.76
N/mm 2 from Table 3.3 of sec. 3.5.5 of Lesson 5. This depth is less than assumed depth of slab in Step 1. Hence, assume d = 225 mm and D = 250 mm. Step 4: Depth of slab for shear force Using the value of k = 1.1 (cl. 40.2.1.1 of IS 456) for the slab of 250 mm depth,
we have cτ (from Table 19 of IS 456) = 1.1(0.28) = 0.308 N/mm 2 . Table 20 of IS
456 gives maxcτ= 2.8 N/mm 2 . Here, The depth of the slab is safe in shear as .N/mm
0.148 33.3/225 / 2===bdVuvτ. maxccvτττ<<
Step 5: Determination of areas of steel (using table of SP-16) Table 44 gives 10 mm diameter bars @ 200 mm c/c can resist 31.43 kNm/m > 30.8025 kNm/m. Fifty per cent of the bars should be curtailed at a distance of larger of L d or 0.5l x . Table 65 of SP-16 gives L d of 10 mm bars = 470
mm and 0.5l x = 0.5(1850) = 925 mm from the face of the column. The curtailment
distance from the centre line of beam = 925 + 150 = 1075, say 1100 mm. The above, however, is not admissible as the spacing of bars after the curtailment exceeds 300 mm. So, we provide 10 mm @ 300 c/c and 8 mm @
300 c/c. The moment of resistance of this set is 34.3 kNm/m > 30.8025 kNm/m
(see Table 44 of SP-16).
Two-way slabs subjected mostly to uniformly distributed loads resist them primarily by bending about both the axis. However, as in the one-way slab, the depth of the two-way slabs should also be checked for the shear stresses to avoid any reinforcement for shear. Moreover, these slabs should have sufficient
depth for the control deflection. Thus, strength and deflection are the requirements of design of two-way slabs.
Design shear strength of concrete in two-way slabs is to be determined incorporating the multiplying factor k from Table 1 Computation of shear force
Shear forces are computed following the procedure stated below The two-way slab of Fig. 8.19.1 is divided into two trapezoidal and two triangular
zones by drawing lines from each corner at an angle of 45 o . The loads of triangular segment A will be transferred to beam 1-2 and the same of trapezoidal segment B will be beam 2-3. The shear forces per unit width of the strips aa and bb are highest at the ends of strips. Moreover, the length of half the strip bb is
equal to the length of the strip aa. Thus, the shear forces in both strips are equal and we can write, V u = W (l x /2) (8.1)
where W = intensity of the uniformly distributed loads. The nominal shear stress acting on the slab is then determined from bdVuv/ =τ (8.2) Computation of bending moments Two-way slabs spanning in two directions at right angles and carrying uniformly distributed loads may be analysed using any acceptable theory. Pigeoud’s or Wester-guard’s theories are the suggested elastic methods and Johansen’s yield line theory is the most commonly used in the limit state of collapse method and suggested by IS 456 in the note of cl. 24.4. Alternatively, Annex D of IS 456 can be employed to determine the bending moments in the two directions for two types of slabs: (i) restrained slabs, and (ii) simply supported slabs. The two methods are explained below:
(i) Restrained slabs
Restrained slabs are those whose corners are prevented from lifting due to effects of torsional moments. These torsional moments, however, are not computed as the amounts of reinforcement are determined from the computed areas of steel due to positive bending moments depending upon the intensity of torsional moments of different corners. This aspect has been explained in Step 7 of sec. 8.19.6. Thus, it is essential to determine the positive and negative bending moments in the two directions of restrained slabs depending on the various types of panels and the aspect ratio l y /l x .
Restrained slabs are considered as divided into two types of strips in each direction: (i) one middle strip of width equal to three-quarters of the respective length of span in either directions, and (ii) two edge strips, each of width equal to one-eighth of the respective length of span in either directions. Figures 8.19.2a and b present the two types of strips for spans l x and l y separately.
The maximum positive and negative moments per unit width in a slab are determined from
where yxαα and are coefficients given in Table 26 of IS 456, Annex D, cl. D-1.1. Total design load per unit area is w and lengths of shorter and longer spans are represented by l x and l y , respectively. The values of,
given in Table 26 of IS 456, are for nine types of panels having eight aspect ratios of l y /l x from one to two at an interval of 0.1. The above maximum bending
moments are applicable only to the middle strips and no redistribution shall be made. Tension reinforcing bars for the positive and negative maximum moments are to be provided in the respective middle strips in each direction. Figure 8.19.2 shows the positive and negative coefficients yxαα and .
The edge strips will have reinforcing bars parallel to that edge following the minimum amount as stipulated in IS 456. The detailing of all the reinforcing bars for the respective moments and for the minimum amounts as well as torsional requirements are discussed in sec.
8.19.7(i).
(ii) Simply supported slabs The maximum moments per unit width of simply supported slabs, not having adequate provision to resist torsion at corners and to prevent the corners from lifting, are determined from Eqs.8.3 and 8.4, where yxαα and are the respective coefficients of moments as given in Table 27 of IS 456, cl. D-2. The notations M x ,
M y , w, l x and l y are the same as mentioned below Eqs.8.3 and 8.4 in (i) above.
8.19.7(ii).
The coefficients yxαα and of simply supported two-way slabs are derived from the Grashoff-Rankine formula which is based on the consideration of the same deflection at any point P (Fig.8.19.3) of two perpendicular interconnected strips
containing the common point P of the two-way slab subjected to uniformly distributed loads.
The design considerations mentioned in sec. 8.18.5 of Lesson 18 in (a), (c), (d), (e) and (f) are applicable for the two-way slabs also. However, the effective span to effective depth ratio is different from those of one-way slabs. Accordingly, this item for the two-way slabs is explained below. Effective span to effective depth ratio (cl. 24.1 of IS 456) The following are the relevant provisions given in Notes 1 and 2 of cl. 24.1.
• The shorter of the two spans should be used to determine the span to effective depth ratio.
• For spans up to 3.5 m and with mild steel reinforcement, the span to
overall depth ratios satisfying the limits of vertical deflection for loads up to
3 kN/m 2 are as follows:
Simply supported slabs 35 Continuous slabs 40 • The same ratios should be multiplied by 0.8 when high strength deformed bars (Fe 415) are used in the slabs.
The procedure of the design of two-way slabs will have all the six steps mentioned in sec. 8.18.6 for the design of one-way slabs except that the bending moments and shear forces are determined by different methods for the two types of slab. While the bending moments and shear forces are computed from the coefficients given in Tables 12 and 13 (cl. 22.5) of IS 456 for the one-way slabs, the same are obtained from Tables 26 or 27 for the bending moment in the two types of two-way slabs and the shear forces are computed from Eq.8.1 for the two-way slabs. Further, the restrained two-way slabs need adequate torsional reinforcing bars at the corners to prevent them from lifting. There are three types of corners having three different requirements. Accordingly, the determination of torsional reinforcement is discussed in Step 7, as all the other six steps are common for the one and two-way slabs. Step 7: Determination of torsional reinforcement Three types of corners, C1, C2 and C3, shown in Fig.8.19.4, have three different requirements of torsion steel as mentioned below.
At corner C1 where the slab is discontinuous on both sides, the torsion reinforcement shall consist of top and bottom bars each with layers of bar placed parallel to the sides of the slab and extending a minimum distance of one-fifth of the shorter span from the edges. The amount of reinforcement in each of the four
layers shall be 75 per cent of the area required for the maximum mid-span moment in the slab. This provision is given in cl. D-1.8 of IS 456.
(b) At corner C2 contained by edges over one of which is continuous, the
torsional reinforcement shall be half of the amount of (a) above. This provision is given in cl. D-1.9 of IS 456. (c) At corner C3 contained by edges over both of which the slab is continuous, torsional reinforcing bars need not be provided, as stipulated in cl. D-1.10 of IS
As mentioned in Step 5 explains the two methods of determining the required areas of steel required for the maximum positive and negative moments. The two methods are (i) employing Eq.3.23 as given in Step 5 of sec. 8.18.6 or (ii) using tables and charts of SP-16. Thereafter, Step 7 of sec. 8.19.6 explains the method of determining the areas steel for corners of restrained slab depending on the type of corner. The detailing of torsional reinforcing bars is explained in Step 7 of sec. 8.19.6. In the following, the detailings of reinforcing bars for (i) restrained slabs and (ii) simply supported slabs are discussed separately for the bars either for the maximum positive or negative bending moments or to satisfy the requirement of minimum amount of steel. (i) Restrained slabs The maximum positive and negative moments per unit width of the slab calculated by employing Eqs.8.3 and 8.4 as explained in sec. 8.19.4.2(i) are applicable only to the respective middle strips (Fig.8.19.2). There shall be no redistribution of these moments. The reinforcing bars so calculated from the maximum moments are to be placed satisfying the following stipulations of IS
• Bottom tension reinforcement bars of mid-span in the middle strip shall extent in the lower part of the slab to within 0.25l of a continuous edge, or 0.15l of a discontinuous edge (cl. D-1.4 of IS 456). Bars marked as B1, B2, B5 and B6 in Figs.8.19.5 a and b are these bars.
• Top tension reinforcement bars over the continuous edges of middle strip
shall extend in the upper part of the slab for a distance of 0.15l from the support, and at least fifty per cent of these bars shall extend a distance of 0.3l (cl. D-1.5 of IS 456). Bars marked as T2, T3, T5 and T6 in Figs.8.19.5 a and b are these bars.
• To resist the negative moment at a discontinuous edge depending on the
degree of fixity at the edge of the slab, top tension reinforcement bars equal to fifty per cent of that provided at mid-span shall extend 0.1l into the span (cl. D-1.6 of IS 456). Bars marked as T1 and T4 in Figs.8.19.5 a and b are these bars. • The edge strip of each panel shall have reinforcing bars parallel to that edge satisfying the requirement of minimum amount as specified in sec. 8.18.15d of Lesson 18 (cl. 26.5.2.1 of IS 456) and the requirements for torsion, explained in Step 7 of sec. 8.19.6 (cls. D-1.7 to D-1.10 of IS 456). The bottom and top bars of the edge strips are explained below.
• Bottom bars B3 and B4 (Fig.8.19.5 a) are parallel to the edge along l x for
the edge strip for span l y , satisfying the requirement of minimum amount of
steel (cl. D-1.7 of IS 456).
• Bottom bars B7 and B8 (Fig.8.19.5 b) are parallel to the edge along l y for
the edge strip for span l x , satisfying the requirement of minimum amount of
• Top bars T7 and T8 (Fig.8.19.5 a) are parallel to the edge along l x for the
edge strip for span l y , satisfying the requirement of minimum amount of
• Top bars T9 and T10 (Fig.8.19.5 b) are parallel to the edge along l y for
The detailing of torsion bars at corners C1 and C2 is explained in Fig.8.19.7 of Problem 8.2 in sec. 8.19.8. The above explanation reveals that there are eighteen bars altogether comprising eight bottom bars (B1 to B8) and ten top bars (T1 to T10). Tables 8.4 and 8.5 present them separately for the bottom and top bars, respectively, mentioning the respective zone of their placement (MS/LDES/ACES/BDES to designate Middle Strip/Left Discontinuous Edge Strip/Adjacent Continuous Edge Strip/Bottom Discontinuous Edge Strip), direction of the bars (along x or y), the resisting moment for which they shall be determined or if to be provided on the basis of minimum reinforcement clause number of IS 456 and Fig. No. For easy understanding, plan views in (a) and (b) of Fig.8.19.5 show all the bars separately along x and y directions, respectively. Two sections (1-1 and 2-2), however, present the bars shown in the two plans. Torsional reinforcements are not included in Tables 8.4 and 8.5 and Figs.8.19.5 a and b. Table 8.4 Details of eight bottom bars
Max. + M
8.19.5a,
Min. Steel
4 B5,
8.19.5b,
Notes: (i) MS = Middle Strip
LDES = Left Discontinuous Edge Strip
ACES = Adjacent Continuous Edge Strip
BDES = Bottom Discontinuous Edge Strip
Table 8.5 Details of eight top bars
-0.5 M y for
(ii) Simply supported slabs Figures 8.19.6 a, b and c present the detailing of reinforcing bars of simply supported slabs not having adequate provision to resist torsion at corners and to prevent corners from lifting. Clause D-2.1 stipulates that fifty per cent of the tension reinforcement provided at mid-span should extend to the supports. The remaining fifty per cent should extend to within 0.1l x or 0.1l y of the support, as
Design the slab panel 1 of Fig.8.19.7 subjected to factored live load of 8 kN/m 2 in addition to its dead load using M 20 and Fe 415. The load of floor finish is 1
kN/m 2 . The spans shown in figure are effective spans. The corners of the slab are prevented from lifting. Solution Step 1: Selection of preliminary depth of slab The span to depth ratio with Fe 415 is taken from cl. 24.1, Note 2 of IS 456 as 0.8 (35 + 40) / 2 = 30. This gives the minimum effective depth d = 4000/30 = 133.33 mm, say 135 mm. The total depth D is thus 160 mm. Step 2: Design loads, bending moments and shear forces
Dead load of slab (1 m width) = 0.16(25) = 4.0 kN/m
Dead load of floor finish (given) = 1.0 kN/m
Factored dead load = 1.5(5) = 7.5 kN/m
Factored live load (given) = 8.0 kN/m
Total factored load = 15.5 kN/m The coefficients of bending moments and the bending moments M x and M y per
unit width (positive and negative) are determined as per cl. D-1.1 and Table 26 of IS 456 for the case 4, “Two adjacent edges discontinuous” and presented in Table 8.6. The l y / l x for this problem is 6/4 = 1.5.
Table 8.6 Maximum bending moments of Problem 8.2
Maximum shear force in either direction is determined from Eq.8.1 (Fig.8.19.1) as V u = w(l x /2) = 15.5 (4/2) = 31 kN/m
Step 3: Determination/checking of the effective depth and total depth of slab Using the higher value of the maximum bending moments in x and y directions from Table 8.6, we get from Eq.3.25 of Lesson 5 (sec. 3.5.5):
or d = [(18.6)(10 6 )/{2.76(10 3 )}] 1/2 = 82.09 mm,
where 2.76 N/mm 2 is the value of R ,lim taken from Table 3.3 of Lesson 5
(sec. 3.5.5). Since, this effective depth is less than 135 mm assumed in Step 1, we retain d = 135 mm and D = 160 mm. Step 4: Depth of slab for shear force
Table 19 of IS 456 gives the value of cτ = 0.28 N/mm 2 when the lowest percentage of steel is provided in the slab. However, this value needs to be modified by multiplying with k of cl. 40.2.1.1 of IS 456. The value of k for the total depth of slab as 160 mm is 1.28. So, the value of cτ is 1.28(0.28) = 0.3584
N/mm 2 .
Table 20 of IS 456 gives maxcτ = 2.8 N/mm 2 . The computed shear stress vτ = V /bd
= 31/135 = 0.229 N/mm 2 . Since, vτ < cτ < maxcτ, the effective depth of the slab as 135 mm and the total depth as 160 mm are safe. Step 5: Determination of areas of steel The respective areas of steel in middle and edge strips are to be determined employing Eq.3.23 of Step 5 of sec. 8.18.6 of Lesson 18. However, in Problem 8.1 of Lesson 18, it has been shown that the areas of steel computed from Eq.3.23 and those obtained from the tables of SP-16 are in good agreement. Accordingly, the areas of steel for this problem are computed from the respective Tables 40 and 41 of SP-16 and presented in Table 8.7. Table 40 of SP-16 is for the effective depth of 150 mm, while Table 41 of SP-16 is for the effective depth
of 175 mm. The following results are, therefore, interpolated values obtained from the two tables of SP-16. Table 8.7 Reinforcing bars of Problem 8.2
Short span l
(kNm/
(kNm/m
The minimum steel is determined from the stipulation of cl. 26.5.2.1 of IS 456 and is
A s = (0.12/100)(1000)(160) = 192 mm
and 8 mm bars @ 250 mm c/c (= 201 mm 2 ) is acceptable. It is worth mentioning that the areas of steel as shown in Table 8.7 are more than the minimum amount of steel. Step 6: Selection of diameters and spacings of reinforcing bars The advantages of using the tables of SP-16 are that the obtained values satisfy the requirements of diameters of bars and spacings. However, they are checked as ready reference here. Needless to mention that this step may be omitted in such a situation. Maximum diameter allowed, as given in cl. 26.5.2.2 of IS 456, is 160/8 = 20 mm, which is more that the diameters used here. The maximum spacing of main bars, as given in cl. 26.3.3(1) of IS 456, is the lesser of 3(135) and 300 mm. This is also satisfied for all the bars. The maximum spacing of minimum steel (distribution bars) is the lesser of 5(135) and 450 mm. This is also satisfied.
Figures 8.19.8 and 9 present the detailing of reinforcing bars.
Torsional reinforcing bars are determined for the three different types of corners as explained in sec. 8.19.6 (Fig.8.19.4). The length of torsional strip is 4000/5 = 800 mm and the bars are to be provided in four layers. Each layer will have 0.75
times the steel used for the maximum positive moment. The C1 type of corners will have the full amount of torsional steel while C2 type of corners will have half of the amount provided in C1 type. The C3 type of corners do not need any torsional steel. The results are presented in Table 8.8 and Figs.8.19.10 a, b and
Table 8.8 Torsional reinforcement bars of Problem 8.2
Dimensions along
Cl. no. of IS
Singly , doubly and flanged Beams
Types of Problems Two types of problems are possible: (i) design type and (ii) analysis type. In the design type of problems, the designer has to determine the dimensions b, d, D, A st (Fig. 3.6.1) and other detailing of reinforcement, grades of concrete and steel
from the given design moment of the beam. In the analysis type of the problems, all the above data will be known and the designer has to find out the moment of resistance of the beam. Both the types of problems are taken up for illustration in the following two lessons.
Design Type of Problems
The designer has to make preliminary plan lay out including location of the beam, its span and spacing, estimate the imposed and other loads from the given functional requirement of the structure. The dead loads of the beam are estimated assuming the dimensions b and d initially. The bending moment, shear force and axial thrust are determined after estimating the different loads. In this illustrative problem, let us assume that the imposed and other loads are given. Therefore, the problem is such that the designer has to start with some initial dimensions and subsequently revise them, if needed. The following guidelines are helpful to assume the design parameters initially.
1 Selection of breadth of the beam b
Normally, the breadth of the beam b is governed by: (i) proper housing of reinforcing bars and (ii) architectural considerations. It is desirable that the width
of the beam should be less than or equal to the width of its supporting structure like column width, or width of the wall etc. Practical aspects should also be kept in mind. It has been found that most of the requirements are satisfied with b as 150, 200, 230, 250 and 300 mm. Again, width to overall depth ratio is normally kept between 0.5 and 0.67.
2 Selection of depths of the beam d and D
The effective depth has the major role to play in satisfying (i) the strength requirements of bending moment and shear force, and (ii) deflection of the beam. The initial effective depth of the beam, however, is assumed to satisfy the deflection requirement depending on the span and type of the reinforcement. IS 456 stipulates the basic ratios of span to effective depth of beams for span up to 10 m as (Clause 23.2.1) Cantilever 7 Simply supported 20 Continuous 26 For spans above 10 m, the above values may be multiplied with 10/span in metres, except for cantilevers where the deflection calculations should be made. Further, these ratios are to be multiplied with the modification factor depending
on reinforcement percentage and type. Figures 4 and 5 of IS 456 give the different values of modification factors. The total depth D can be determined by adding 40 to 80 mm to the effective depth.
3 Selection of the amount of steel reinforcement A st
The amount of steel reinforcement should provide the required tensile force T to resist the factored moment M u of the beam. Further, it should satisfy
the minimum and maximum percentages of reinforcement requirements also. The minimum reinforcement A s is provided for creep, shrinkage, thermal and
other environmental requirements irrespective of the strength requirement. The minimum reinforcement A s to be provided in a beam depends on the f y of steel
and it follows the relation: (cl. 26.5.1.1a of IS 456)
The maximum tension reinforcement should not exceed 0.04 bD (cl. 26.5.1.1b of IS 456), where D is the total depth. Besides satisfying the minimum and maximum reinforcement, the amount of reinforcement of the singly reinforced beam should normally be 75 to 80% of p t,
lim . This will ensure that strain in steel will be more than
as the design stress in steel will be 0.87 f y . Moreover, in many cases, the depth
required for deflection becomes more than the limiting depth required to resist M u, lim . Thus, it is almost obligatory to provide more depth. Providing more depth
also helps in the amount of the steel which is less than that required for M u, lim .
This helps to ensure ductile failure. Such beams are designated as under- reinforced beams.
4 Selection of diameters of bar of tension reinforcement
Reinforcement bars are available in different diameters such as 6, 8, 10, 12, 14,
16, 18, 20, 22, 25, 28, 30, 32, 36 and 40 mm. Some of these bars are less available. The selection of the diameter of bars depends on its availability, minimum stiffness to resist while persons walk over them during construction, bond requirement etc. Normally, the diameters of main tensile bars are chosen from 12, 16, 20, 22, 25 and 32 mm.
5 Selection of grade of concrete
Besides strength and deflection, durability is a major factor to decide on the grade of concrete. Table 5 of IS 456 recommends M 20 as the minimum grade under mild environmental exposure and other grades of concrete under different environmental exposures also.
6 Selection of grade of steel
Normally, Fe 250, 415 and 500 are in used in reinforced concrete work. Mild
steel (Fe 250) is more ductile and is preferred for structures in earthquake zones
or where there are possibilities of vibration, impact, blast etc.
It may be required to estimate the moment of resistance and hence the service
load of a beam already designed earlier with specific dimensions of b, d and D and amount of steel reinforcement A st . The grades of concrete and steel are also
known. In such a situation, the designer has to find out first if the beam is under- reinforced or over-reinforced. The following are the steps to be followed for such problems.
3.7.2.1 x
u, max
The maximum depth of the neutral axis x u, max is determined from Table 3.2 of
Lesson 5 using the known value of f y .
3.7.2.2 x
The depth of the neutral axis for the particular beam is determined from Eq. 3.16
of Lesson 5 employing the known values of f y , f ck , b and A st .
3.7.2.3 M u and service imposed loads
The moment of resistance M u is calculated for the three different cases as
(a) If x u < x u, max , the beam is under-reinforced
(b) If x u = x u, max , the M u is obtained from Eq. 3.24
(c) If x u > x u, max , the beam is over-reinforced for which x u is taken as x u, max and
then M u is calculated using x u = x u, max .
With the known value of M u , which is the factored moment, the total factored load
can be obtained from the boundary condition of the beam. The total service imposed loads is then determined dividing the total factored load by partial safety factor for loads (= 1.5). The service imposed loads are then obtained by subtracting the dead load of the beam from the total service loads.
1.Determine the service imposed loads of two simply supported beam of same
effective span of 8 m (Figs. 3.7.1 and 2) and same cross-sectional dimensions, but having two different amounts of reinforcement. Both the beams are made of
M 20 and Fe 415.
Given data: b = 300 mm, d = 550 mm, D = 600 mm, A st = 1256 mm 2 (4-20 T), L eff
= 8 m and boundary condition = simply supported (Fig. 3.7.1). 1 x
From Table 3.2 of Lesson 5, we get x u, max = 0.479 d = 263.45 mm
Concrete has very good compressive strength and almost negligible tensile strength. Hence, steel reinforcement is used on the tensile side of concrete. Thus, singly reinforced beams reinforced on the tensile face are good both in compression and tension. However, these beams have their respective limiting moments of resistance with specified width, depth and grades of concrete and steel. The amount of steel reinforcement needed is known as A st,lim . Problem will
arise, therefore, if such a section is subjected to bending moment greater than its limiting moment of resistance as a singly reinforced section. There are two ways to solve the problem. First, we may increase the depth of the beam, which may not be feasible in many situations. In those cases, it is possible to increase both the compressive and tensile forces of the beam by providing steel reinforcement in compression face and additional reinforcement in tension face of the beam without increasing the depth (Fig. 4.8.1). The total compressive force of such beams comprises (i) force due to concrete in compression and (ii) force due to steel in compression. The tensile force also has two components: (i) the first provided by A st , lim which is equal to the compressive
force of concrete in compression. The second part is due to the additional steel in tension - its force will be equal to the compressive force of steel in compression. Such reinforced concrete beams having steel reinforcement both on tensile and compressive faces are known as doubly reinforced beams. Doubly reinforced beams, therefore, have moment of resistance more than the singly reinforced beams of the same depth for particular grades of steel and concrete. In many practical situations, architectural or functional requirements may restrict the overall depth of the beams. However, other than in doubly reinforced beams compression steel reinforcement is provided when:
(i) some sections of a continuous beam with moving loads undergo
change of sign of the bending moment which makes compression zone as tension zone or vice versa.
(ii) the ductility requirement has to be followed.
(iii) the reduction of long term deflection is needed.
It may be noted that even in so called singly reinforced beams there would be longitudinal hanger bars in compression zone for locating and fixing stirrups.
(i) The assumptions of sec. 3.4.2 of Lesson 4 are also applicable
(ii) Provision of compression steel ensures ductile failure and
hence, the limitations of x/d ratios need not be strictly followed
(iii) The stress-strain relationship of steel in compression is the
same as that in tension. So, the yield stress of steel in compression is 0.87 fy. Design type of problems In the design type of problems, the given data are b, d, D, grades of concrete and
steel. The designer has to determine A sc and A st of the beam from the given
factored moment. These problems can be solved by two ways: (i) use of the equations developed for the doubly reinforced beams, named here as direct computation method, (ii) use of charts and tables of SP-16.
(a) Direct computation method
Step 1: To determine M u, lim and A st, lim from Eqs. 4.2 and 4.8, respectively.
Step 2: To determine M u2 , A sc , A st2 and A st from Eqs. 4.1, 4.4, 4.6 and 4.7,
respectively. Step 3: To check for minimum and maximum reinforcement in compression and tension as explained in sec. 4.8.5. Step 4: To select the number and diameter of bars from known values of A sc and
4.8.6.2 Analysis type of problems In the analysis type of problems, the data given are b, d, d', D, f ck , f y , A sc and A st .
It is required to determine the moment of resistance M u of such beams. These
problems can be solved: (i) by direct computation method and (ii) by using tables of SP-16.
Step 1: To check if the beam is under-reinforced or over-reinforced.
The beam is under-reinforced or over-reinforced if ε st is less than or more than
the yield strain.
Step 2: To determine M u,lim from Eq. 4.2 and A st,lim from the p t, lim given in Table
3.1 of Lesson 5. Step 3: To determine A st2 and A sc from Eqs. 4.7 and 4.6, respectively.
Step 4: To determine M u2 and M u from Eqs. 4.4 and 4.1, respectively.
Problem Design a simply supported beam of effective span 8 m subjected to imposed loads of 35 kN/m. The beam dimensions and other data are: b = 300 mm, D = 700 mm, M 20 concrete, Fe 415 steel (Fig. 4.9.1). Determine f sc from
It may be noted that A st is provided in two layers in order to provide adequate
space for concreting around reinforcement. Also the centroid of the tensile bars is at 70 mm from bottom
This lesson explains the three failure modes due to shear force in beams and defines different shear stresses needed to design the beams for shear. The critical sections for shear and the minimum shear reinforcement to be provided in beams are mentioned as per IS 456.
Failure Modes due to Shear
Bending in reinforced concrete beams is usually accompanied by shear, the exact analysis of which is very complex. However, experimental studies confirmed the following three different modes of failure due to possible combinations of shear force and bending moment at a given section (Figs. 6.13.1a to c):
(i) Web shear (Fig. 6.13.1a) (ii) Flexural tension shear (Fig. 6.13.1b) (iii) Flexural compression shear (Fig. 6.13.1c) Web shear causes cracks which progress along the dotted line shown in Fig. 6.13.1a. Steel yields in flexural tension shear as shown in Fig. 6.13.1b, while concrete crushes in compression due to flexural compression shear as shown in Fig. 6.13.1c. An in-depth presentation of the three types of failure modes is beyond the scope here.