The three common types of reinforced concrete beam section are
a. rectangular sections with tension steel only (this generally occurs when designing a given
width of slab as a beam.)
b. rectangular sections with tension and compression steel
c. flanged sections of either T or L shape with tension steel and with or without compression steel
Beam sections are shown in Fig.4.1. It will be established later that all beams of structural
importance must have steel top and bottom to carry links to resist shear.Fig.4.1
(a) Rectangular beam and slab, tension steel only;
(b) rectangular beam, tension and compression steel;
(c) flanged beams.
4.2 REINFORCEMENT AND BAR SPACING
Before beginning section design, reinforcement data and code requirements with regard to
minimum and maximum areas of bars in beams and bar spacing are set out. This is to enable sections to be designed with practical amounts and layout of steel. Requirements for cover were discussed in section 2.9.4.2.1 Reinforcement DataIn accordance with BS8110: Part 1, clause 3.12.4.1, bars may be placed singly or in pairs or in bundles of three or four bars in contact. For design purposes the pair or bundle is treated as a single bar of equivalent area. Bars are available with diameters of 6, 8, 10, 12, 16, 20, 25, 32 and 40 mm and in two grades with characteristic strengths fy: Hot rolled mild steel fy=250 N/mm2High yield steel fy=460 N/mm2For convenience in design calculations, areas of groups of bars are given in Table 4.1.
Table 4.2 gives equivalent diameter of bundles of bars of same diameter.
| Table 4.1 Areas of groups of barsDiameter of bar in mm Numbers of bars in group 1 2 3 4 5 6 7 8 | ||||||||
| 6 | 28 | 57 | 85 | 113 | 141 | 170 | 198 | 226 |
| 8 | 50 | 101 | 151 | 201 | 251 | 302 | 352 | 402 |
| 10 | 79 | 157 | 236 | 314 | 393 | 471 | 550 | 628 |
| 12 | 113 | 226 | 339 | 452 | 566 | 679 | 792 | 905 |
| 16 | 201 | 402 | 603 | 804 | 1005 | 1206 | 1407 | 1609 |
| 20 | 314 | 628 | 943 | 1257 | 1571 | 1885 | 2109 | 2513 |
| 25 | 491 | 982 | 1473 | 1964 | 2454 | 2945 | 3436 | 3927 |
| 32 | 804 | 1609 | 2413 | 3217 | 4021 | 4826 | 5630 | 6434 |
| Table 4.2 Equivalent diameters of bars in groupsDiameter in mm of bars in group Number of bars in group 1 2 3 4 | ||||
| 6 | 6 | 8.5 | 10.4 | 12 |
| 8 | 8 | 11.3 | 13.9 | 16 |
| 10 | 10 | 14.1 | 17.3 | 20 |
| 12 | 12 | 17.0 | 20.8 | 24 |
| 16 | 16 | 22.6 | 27.7 | 32 |
| 20 | 20 | 28.3 | 34.6 | 40 |
| 25 | 25 | 35.4 | 43.3 | 50 |
| 32 | 32 | 45.3 | 55.4 | 64 |
Detailed drawings should be prepared according to Standard Method of Detailing Structural Concrete. Institution of Structural Engineers,London, 1989.Bar types are specified by letters:
R mild steel bars
T high yield bars
Bars are designated on drawings as, for example, 4T25, i.e. four 25 mm diameter bars of
grade 460. This system will be used to specify bars in figures.4.2.2 Minimum and Maximum Areas of Reinforcement in BeamsThe minimum areas of reinforcement in a beam section to control cracking as well as to resist tension or compression due to bending in different types of beam section are given in BS 8110: Part 1, clause 3.12.5.3 and Table 3.25. Some commonly used values are shown inFig.4.2 and Table 4.3. Other values will be discussed in appropriate parts of the book.
| Table 4.3: Minimum steel areasPercentage fy=250 N/mm2 fy=460 N/mm2 | |||
| Tension reinforcement | |||
| Rectangular beam | 100As/Ac | 0.24 | 0.13 |
| Flanged beam—Web in tension: bw/b<0.4 | 100As/bw h | 0.32 | 0.18 |
| Flanged beam—Web in tension: bw/b≥0.4 | 100As/bw h | 0.24 | 0.13 |
| Compression reinforcement | |||
| Rectangular beam | 100Asc/Ac | 0.2 | 0.2 |
| Flanged beam—flange in compression: | 100Asc/bwhf | 0.2 | 0.2 |
| Ac=total area of concrete, As=minimum area of reinforcement, Asc=area of steel in compression, b, bw, hf=dimensions. |
The maximum area of both tension and compression reinforcement in beams is specified in
BS8110: Part 1, clause 3.12.6.1. Neither should exceed 4% of the gross cross-sectional area of
the concrete.4.2.3 Minimum Spacing of BarsThe minimum spacing of bars is given in BS8110: Part 1, clause 3.12.11.1. This clause states the following:
1. The horizontal distance between bars should not be less than hagg+5 mm
2. Where there are two or more rows
(a) the gap between corresponding bars in each row should be vertically in line and
(b) the vertical distance between bars should not be less than 2hagg/3 where hagg is the maximum size of coarse aggregate. The clause also states that if the bar size exceeds hagg+5 mm the spacing should not be less than the bar size. Note that pairs or bundles are treated as a single bar of equivalent area. The above spacing ensures that the concrete can be properly compacted around the reinforcement. Spacing of top bars in beams should also permit the insertion of a vibrator. The information is summarized in Fig.4.3.Fig.4.3 (a) Flanged beam; (b) minimum spacing.
4.3 BEHAVIOUR OF BEAMS IN BENDING
The behaviour of a cross section subjected to pure bending is studied by loading a beam at
third points as shown in Fig.4.4(a). Under this system of loading, sections between the loads
are subjected to pure bending. Initially the beam behaves as a monolithic elastic beam till the
stresses at the bottom fibre reach the tensile strength of concrete. Because of the very low
tensile strength of concrete (about 10% of its compression strength), vertical cracks appear at a fairly low load. As the load is increased, cracks lengthen and penetrate deeper towards the compression face. Simultaneously, the strain in steel also increases. The final failure depends on the amount and yield stress of steel. The three possible modes of failure are:
1. Steel yields first: If the tensile force capacity of steel is ‘low’, then steel yields before
the strain in the concrete at the compression face reaches the maximum permissible value of
0.0035. Because steel is a ductile material, steel elongates while maintaining its strength. The beam continues to deform at constant load and the neutral axis moves up. The beam finally fails when the depth of the compression zone is too small to balance the tensile force in steel. This type of failure is the desired type because there is ample warning before failure. All beams, if overloaded, should be designed to fail in this manner. Fig.4.4(b) shows the
qualitative load versus deflection curve and Fig.4.4(c) shows the stress distribution at elastic
and ultimate stages.
2. Simultaneous ‘yielding’ of steel and concrete: If the tensile force capacity of steel is
‘moderate’, yielding of steel is simultaneously accompanied by the crushing of concrete.
Unlike the failure mode where the steel yields first, there is little warning before failure. This
is not a desirable mode of failure.
3. Concrete crushes first: If the tensile force capacity of steel is ‘high’, then steel does not
yield at all before concrete crushes. Because concrete is a fairly brittle material, it fails in an
explosive manner without any significant residual load bearing capacity. This form of failure
is to be avoided at all costs!
4.4 SINGLY REINFORCED RECTANGULAR BEAMS
4.4.1 Assumptions and Stress-Strain Diagrams .The ultimate moment of resistance of a section is based on the assumptions set out in BS8110:Part 1, clause 3.4.4.1. These are as follows:
1. The strains in the concrete and reinforcement are derived assuming that plane sections
remain plane;Fig.4.4
(a) Flexural cracks at collapse;
(b) load-deflection curve;
(c) effective section and stress distribution.
2. The stresses in the concrete in compression are derived using either:
(a) the design stress-strain curve given in Fig.4.5(a) with γm=1.5 or
(b) the simplified stress block shown in Fig.4.6(d) where the depth of the stress block is 0.9 of the depth to the neutral axis denoted by x.
Note that in both cases the maximum strain in the concrete at failure is 0.0035;Fig.4.5 Stress-strain diagrams (a) Concrete; (b) Steel.3. The tensile strength of the concrete is ignored;
4. The stresses in the reinforcement are derived from the stress-strain curve shown inFig.4.5(b) where γm=1.05;
5. Where the section is designed to resist flexure only, the lever arm should not be assumed
to be greater than 0.95 of the effective depth, d. This is because of the fact that at the top face during compaction water tends to move to the top and causes a higher water cement ratio than the rest of the beam. In addition weathering also affects the strength. Because of that a layer of concrete at the top is likely to be weak and by limiting the value of the lever arm z, one avoids the possibility of expecting a weak layer of concrete to resist the compressive stress due to bending.
On the basis of these assumptions the strain and stress diagrams for the two alternative
stress distributions for the concrete in compression are as shown in Fig.4.6, where the
following symbols are used:Fig.4.6 (a) Section; (b) strain; (c) rectangular parabolic stress diagram; (d) simplified stress diagram. h overall depth of the section
d effective depth, i.e. depth from the compression face to the centroid of tension steel
b breadth of the section, x depth to the neutral axis, fsstress in steel ,As area of tension reinforcementεc maximum strain in the concrete (0.0035)εsstrain in steelThe alternative stress distributions for the compressive stress in the concrete, the rectangular parabolic stress diagram and the simplified stress block, are shown in Figs 4.6(c) and 4.6(d) respectively.
The maximum strain in the concrete is 0.0035 and the strain εs in the steel depends on the
depth of the neutral axis. Stress-strain curves for concrete and for steel are shown in Figs4.5(a) and 4.5(b) respectively.4.4.2 Moment of Resistance: Rectangular Stress BlockFig.4.6 (d) shows the assumed stress distribution.
The concrete stress is0.67 fcu/γm=0.67 fcu/(γm=1.5)=0.447fcuwhich is generally rounded off to 0.45fcu.
The total compressive force C is given by
C=0.447 fcu×b×0.9x=0.402 b×x×fcuThe lever arm z is
z=d-0.9x/2=d-0.45x
If M is the applied moment, thenM=C×z=0.402 b×x×fcu×(d-0.45x)Setting
k=M/(b d2 fcu)
k=0.402(x/d) (1–0.45 (x/d))
Rearranging,
0.1809(x/d)2–0.402 (x/d)+k=0
Solving for x/d
x/d={0.402-√(0.1616–0.7236 k}/ 0.3618
=1.11-√(1.2345-5.5279k)
z/d=1–0.45(x/d)
z.d=0.5+√(0.25-k/0.9)
Total tensile force T in steel is T=As×fsFor internal equilibrium, total tension T must be equal to total compression C. The forces T and C form a couple ata lever arm of z.
M=T z=As fs z
As=M/(fs z)
The stress fs in steel depends on the strain εs in steel. As remarked in section 4.3, it is highly
desirable that final failure is due to yielding of steel rather than due to crushing of concrete. It is useful therefore to calculate the maximum neutral axis depth in order to achieve this.
Assuming that plane sections remain plane before and after bending, an assumption validated by experimental observations, if as shown in Fig.4.6(b), the maximum permitted strain in concrete at the compression face is 0.0035, then the strain εs in steel is calculated from the strain diagram by Strain εs in steel at a stress of fy/γm is given by
where fy=yield stress, γm=material safety factor and Es is Young’s modules for steel.
Taking fy=460 N/mm2, γm=1.05, fy/γm=438 N/mm2, Es=200 kN/mm2, εs= 0.0022
For εs=0.0022, the depth of neutral axis x is given by x/d=0.6140
However in order to ensure that failure is preceded by steel yielding well before the strain in
concrete reaches 0.0035 resulting in the desirable ductile form of failure, in clause 3.4.4.4, the code limits the ratio x/d to a maximum of 0.5. If x=0.5 d, then
C=0.447 fcu×b×0.9x=0.402×b×0.5d×fcuC=0.201×fcu×b×d
z=d-0.45x=d-0.45×0.5d=0.775 d
M=C×z=0.156×b×d2×fcuk=M/(bd2fcu)=0.156
This is the maximum value of the applied moment that the section can resist because it utilises fully the compression capacity of the cross section. This formula can be used to calculate theminimum effective depth required in a singly reinforced rectangular concrete section.
In practice the effective depth d is made larger than the required minimum consistent with the required headroom.
The reason for this is that with a larger depth, the neutral axis depth is smaller and hence the lever arm is larger leading for a given moment M, to a smaller amount of reinforcement. It has the additional advantage that in the event of unexpected overload, the beams will show large ductility before failure. If x/d≤0.5, steel will always yield,
fs=fy/1.05=0.95fyM=T z=As 0.95 fy z,
As=M/(0.95 fy z)4.4.3 Procedure for the Design of Singly Reinforced Rectangular BeamThe steps to be followed in the design of singly reinforced rectangular beams can be
summarised as follows.
| • From the minimum requirements of span/depth ratio to control deflection (see Chapter 6),estimate a suitable effective depth d. • Assuming the bar diameter for the main steel and links and the required cover as determined by exposure conditions, estimate an overall depth h. |
| h=d+bar diameter+Link diameter+Cover |
| • Assume breadth as about half the overall depth. • Calculate the self-weight. • Calculate the design live load and dead load moment using appropriate load factors. The load factors are normally 1.4 for dead loads and 1.6 for live loads. • In the case of singly reinforced sections, calculate the minimum effective depth using the formula |
| • Adopt an effective depth greater than the minimum depth in order to reduce the total tension reinforcement. • Check that the new depth due to increased self-weight does not drastically affect the calculated design moment. If it does, calculate the revised ultimate moment required. • Calculate k=M/(b d2 fcu) • Calculate the lever arm z |
| z=d{0.5+√(0.25–k/0.9)}≤0.95d |
| Note that z/d≤0.95 if k≥0.0428 • Calculate the required steel As |
| As=M/{0.95 fy z} |
| • Check that the steel provided satisfies the minimum and maximum steel percentages specified in the code. |
Examples of Design of Singly Reinforced Rectangular SectionsExample 1: A simply supported reinforced rectangular beam of 8 m span carries uniformly distributed characteristic dead load, which includes an allowance for self-weight of 7 kN/m and characteristic imposed load of 5 kN/m.
The breadth b=250 mm. Design the beam at mid-span section. Use grade 30 concrete and
high yield steel reinforcement, fy=460 N/mm2.
Design load=(1.4×7)+(1.6×5)=17.8 kN/m
Design ultimate moment M at mid-span:
M=17.8×82/8=142.4 kNm
Minimum effective depth to avoid any compression steel is given by
Using this value of d,
x=0.5d
z=d-0.45x=0.775 d.
The area of steel required is
However, if a value of d equal to say 400 mm, which is larger than the minimum value is used, then one can reduce the area of steel required.Fig.4.7: Mid-span section of the beam.Assuming d=400 mm
Provide four 20 mm diameter bars in two layers as shown in Fig.4.7. From Table 4.1, A=1257 mm2. Assuming cover of 30 mm and link diameter of 8 mm, the overall depth h of the beam is h=400+30+8+20=458, say 460 mm. Check that the percentage steel provided is greater than the minimum of 0.13. 100 As/(bh)=100×1257/(250×450)=1.12>0.13.
Note that this is only one of several possible satisfactory solutions.Example 2: Determination of tension steel cut-off.
Fig 4.8 (a) Section at mid-span;
(b) section at support;
(c) loading and bending moment diagram.
In simply supported beams bending moment decreases towards the supports. Therefore the
amount of steel required towards the support region is much less than at mid-span. For the
beam in Example 4.1, determine the position along the beam where theoretically two of the
four 20 mm diameter bars may be cut off.
The section at cut-off has two 20 mm diameter bars continuing: As=628 mm2. The effective
depth here is 410 mm (Fig.4.8(b)). The neutral axis depth can be determined by equating total compression in concrete to total tension in the beam.
T=0.95 fy As=0.95×460×628×10-3=274.44 kN
C=(0.445 fcu b 0.9x)10-3C=(0.445×30×250×0.9x)×10-3=3x kN
Equating C=T,
x=91 mm
z=d-0.45x=369 mm
z/d=369/410=0.90<0.95
Moment of resistance MR
MR=T z=264.44×369×10-30=97.6 kNm
Determine the position of p along the beam such that M=97.6 kN m (Fig.4.8c).Left hand reaction V is V=17.8×8/2=71.2 kN 97.6=71.2a-0.5×17.8a2The solutions to this equation are a=1.76 m and a=6.24 m from end A.
| Table 4.4 Table to be used for calculating steel areas in slabs, walls, etc. TOTAL REINFORCEMENT AREA (mm2/m) | |||||||
| Bar diameter mm | |||||||
| Bar spacing mm | 6 | 8 | 10 | 12 | 16 | 20 | 25 |
| 50 | 566 | 1010 | 1570 | 2260 | 4020 | 6280 | 9820 |
| 75 | 378 | 670 | 1050 | 1510 | 2680 | 4190 | 6550 |
| 100 | 283 | 503 | 785 | 1130 | 2010 | 3140 | 4910 |
| 125 | 226 | 402 | 628 | 904 | 1610 | 2510 | 3930 |
| 150 | 189 | 335 | 523 | 753 | 1340 | 2090 | 3270 |
| 175 | 162 | 288 | 448 | 646 | 1150 | 1790 | 2810 |
| 200 | 141 | 251 | 392 | 565 | 1010 | 1570 | 2460 |
| 250 | 113 | 201 | 314 | 452 | 804 | 1260 | 1960 |
| 300 | 94 | 167 | 261 | 376 | 670 | 1050 | 1640 |
| 350 | 81 | 144 | 224 | 323 | 574 | 897 | 1400 |
| 400 | 70 | 126 | 196 | 282 | 502 | 785 | 1230 |
| 450 | 63 | 112 | 174 | 251 | 447 | 697 | 1090 |
| 500 | 57 | 101 | 157 | 226 | 402 | 628 | 982 |
| Note: A=(πd /4) {1000/(c to c spacing in mm} |
Example 3:
Singly reinforced one-way slab section
A slab section 1 m wide and 130 mm deep with an effective depth of 100 mm is subjected
to a design ultimate moment of 10.5 kNm. Find the area of reinforcement required. The
concrete is grade C30 and the reinforcement grade Constrain z to, z=0.95×100=95 mm
In the case of slabs, reinforcement is not usually specified as a fixed number of bars but in
terms of the diameter of the bar and its spacing. Using Table 4.4, provide 8 mm diameter bars at 175 mm centres. As=288 mm2.
The percentage steel=100 As/(bh)=100×288/(1000×130)=0.22>0.13. The reinforcement for the slab is shown in Fig.4.9.Fig.4.9 Reinforcement in slab4.4.5 Design ChartUsing the equations developed in section 4.4.2, a chart for the design of singly reinforced rectangular beams can be constructed as follows.
• Choose a value of (x/d)≤0.5
• (z/d)={1-0.45 (x/d)}≤0.95
• C=0.45 fcubd 0.9 (x/d)
• M=C z=0.401 b d2 fcu (x/d) (z/d)
• As=M/(0.95 fy z)=0.4221 b d (fcu/fy)
• shows the calculations and the design chart is shown in Fig.4.10.
| Table 4.5: Calculations for the design chartx/d 1–0.45 x/d z/d M/(b d2 fcu) 100(As/bd)(fv/fcu) | ||||
| 0.001 | 1.00 | 0.95 | 0 | 0 |
| 0.025 | 0.99 | 0.95 | 0.0095 | 1.055 |
| 0.05 | 0.98 | 0.95 | 0.0191 | 2.111 |
| 0.10 | 0.96 | 0.95 | 0.0381 | 4.221 |
| 0.15 | 0.93 | 0.93 | 0.0561 | 6.332 |
| 0.20 | 0.91 | 0.91 | 0.0730 | 8.442 |
| 0.25 | 0.89 | 0.89 | 0.0890 | 10.553 |
| 0.30 | 0.87 | 0.87 | 0.1041 | 12.663 |
| 0.35 | 0.84 | 0.84 | 0.1183 | 14.774 |
| 0.40 | 0.82 | 0.82 | 0.1315 | 16.884 |
| 0.45 | 0.80 | 0.80 | 0.1439 | 18.995 |
| 0.50 0.78 0.78 0.1554 21.105 |
Example 1: Use the design chart to calculate the area of steel for the beam in Example 1,section 4.4.4.From Fig.4.9, As=978 mm compared with 967 mm previously calculated.Example 2: Calculate the moment of resistance for the beam in Example 2, section 4.4.4, for the section where steel is curtailed to 2T20.
From Fig.4.10, M=100.9 kNm (Exact answer M=97.6 kNm)Fig.4.10 Design chart for singly reinforced rectangular concrete beams. Fig.4.11 (a) section; (b) Strain diagram; (c) stress diagram.4.4.6 Moment of Resistance Using Rectangular Parabolic Stress BlockIn the previous sections, the simplified rectangular stress block was used to derive design
equations. In this section, the stress-strain curve for concrete shown in Fig.4.5a will be used to derive the corresponding design equations. As shown in Fig.4.11(b), the maximum strain at the top is 0.0035. The strain ε0 is where the parabolic part of the stress strain-strain curve ends. If the neutral axis depth is x, the distance ‘a’ from the neutral axis to where the strain is ε0 is given by a=x (ε0/0.0035) where ε0 from Fig.4.5(a) is given by
The compressive force C1 in the rectangular portion of depth (x–a) of the stress block is given by
The lever arm z1 for C1 from the centroid of steel area is z1=d-0.5(x-a) Using the ‘well-known’ result that the area of a parabola is equal to two-thirds the area of the enclosing rectangle, the compressive force in the parabolic portion of depth ‘a’ of the stress block is given by The centroid of C2 is at a distance of 5a/8 from the neutral axis. The lever arm z2 for C2 from the centroid of steel area is z2= d-x+5a/8
Therefore taking moments about the centroid of steel area,M=C2z2+C1z1For fcu=30 N/mm2 and γm=1.5,ε0/0.0035=0.306
a=x(ε0/0.0035)=0.306x
C1=0.31 fcubx
C2=0.091 fcubx
z1=(d-0.347x)
z2=d-0.8088x
The corresponding equation for rectangular stress block assumption as derived in section
4.4.2 is
The two equations differ by very little from each other. The Rectangular stress block
assumption is therefore accurate for all practical calculations.
4.5 DOUBLY REINFORCED BEAMS
The normal design practice is to use singly reinforced sections. However if for any reason, for
example headroom considerations, it is necessary to restrict the overall depth of a beam, then it becomes necessary to use steel in the compression zone as well because concrete alone cannot provide the necessary compression resistance.4.5.1 Design Formulae Using the Simplified Stress BlockThe formulae for the design of a doubly reinforced beam are derived using the rectangular stress block.
Let M be the design ultimate moment. As shown in section 4.4.2, a rectangular section as asingly reinforced section can resist a maximum value of the moment equal to
Msr=0.156bd2fcuThe corresponding neutral axis depth x=0.5 d. The compressive force Cc in concrete is Cc=0.45fcub0.45 d=0.2fcubd
The lever arm zczc=0.775d
If M>Msr, then compression steel is required. The compressive force Cs due to compression
steel of area As ̀is Cs=As ̀fs ̀where fs’ is the stress in compression steel. As shown in Fig.4.12, the lever arm zs for compression steel is zs=(d-d ̀)
The stress in the tensile steel is 0.95 fy because the neutral axis depth is limited to 0.5d.
However the stress fs ̀in the compressive steel depends on the corresponding strain strain εsc in concrete at steel level. εsc is given by
If the strain εsc is equal to or greater than the yield strain in steel, then steel yields and the
stress fs ̀in compression steel is equal to 0.95 fy. Otherwise, the stress in compression steel is given by
fs̀=Es εsc.
If fy=460 N/mm2 and Es=200 kN/mm2, then the yield strain in steel is equal to
Therefore, steel will yield if
If mild steel is used, then fy=250 N/mm2. The above equations then become
Therefore, steel will yield if
Taking moments about the tension steel,
M=Cczc+CszsM=0.2 fcu bd(0.775 d)+As ̀fs ̀(d-d ̀)
M=0.156 bd2 fcu+A ̀fs ̀(d-d’)
As ̀=(M-0.156 bd2 fcu)/{fs ̀(d-d ̀)}
From equilibrium, the tensile force T is
T=As0.95fy=Cc+Cs
One important point to remember is that to prevent steel bars in compression from buckling, it is necessary to restrain them using links. Clause 3.12.7 of the code says that links or ties at
least one quarter of the size of the largest compression bar or 6 mm whichever is greater
should be provided at a maximum spacing of 12 times the size of the smallest compression
bar.Fig.4.12 Doubly reinforced beam.4.5.2 Examples of Rectangular Doubly Reinforced Concrete BeamsThe use of the formulae developed in the previous section is illustrated by a few examples.Example 1: A rectangular beam is simply supported over a span of 6 m and carries characteristic dead load including self-weight of 12.7 kN/m and characteristic imposed load of 6.0 kN/m. The beam is 200 mm wide by 300 mm effective depth and the inset of the compression steel is 40 mm. Design the steel for mid-span of the beam for grade C30 concrete and grade 460 reinforcement.
design load=(12.7×1.4)+(6×1.6)=27.4 kN/m
Required ultimate moment M:
M=27.4×62/8=123.3 kN m
Maximum moment that the beam section can resist as a singly reinforced section is
Msr=0.156×30×200×3002×10-6=84.24 kNm
M>Msr, Compression steel is required.
d’/x=40/150=0.27<0.37
The compression steel yields. The stress fs’ in the compression steel is 0.95fy.
As’={M-0.156 b d2 fcu}/[0.95 fy (d-d)]
As̀={123.3–84.24}× 106/[0.95×460×(300–40)]=344 mm2From equilibrium:
As 0.95 fy=0.2bdfcu+As ̀fs ̀
As 0.95×460=0.2×200×300×30+344×0.95×460
As= 1168 mm2
For the tension steel (2T25+2T16) give As=1383 mm2. For the compression steel 2T16 give
As=402 mm2. The beam section and flexural reinforcement steel are shown in Fig.4.13.Fig.4.13 Doubly reinforced beam.Example 2: Design the beam in Example 4.6 but with d’=60 mm. d ̀/x=60/150=0.40>0.37
Compression steel does not yield. Strain in compression steel Stress in compression steel is
fs̀=Es εsc=200×103×0.0021=420 N/mm2As ̀={M-0.156 b d2 fcu}/[420 (d-d’)]
As̀= {123.3-84.24}×106/[420×(300-40)]=358 mm2From equilibrium:
As 0.95 fy=0.2bdfcu+As ̀fs ̀
As 0.95×460=0.2×200×300×30+358×420, As=1168 mm24.6 FLANGED BEAMS4.6.1 General ConsiderationsIn simple slab-beam system shown in Fig.4.14, the slab is designed to span between the beams. The beams span between external supports such as columns, walls, etc. The reactions from the slabs act as load on the beam. When a series of beams are used to support a concrete slab, because of the monolithic nature of concrete construction, the slab acts as the flange of the beams. The end beams become L-beams while the intermediate beams become T-beams.
In designing the intermediate beams, it is assumed that the loads acting on half the slab on the two sides of the beam are carried by the beam. Because of the Fig.4.14 Beam-slab system.comparatively small contact area at the junction of the flange and the rib of the beam, the distribution of the compressive stress in the flange is not uniform. It is higher at the junction and decreases away from the junction. This phenomenon is known as shear lag. For simplicity in design, it is assumed that only part of full physical flange width is considered to sustain compressive stress of uniform magnitude. This smaller width is known as effective breadth of the flange. Although the effective width actually varies even along the span as well, it is common to assume that the effective width remains constant over the entire span.
The effective breadth b of flanged beams (Fig.4.15) as given in BS 8110: Part 1, clause
3.4.1.5:Fig.4.15: Cross section of flanged beams. 1. T-beams: b={web width bw+ℓz/5} or the actual flange width if less;
2. L-beams: b={web width bw+ℓz /10} or the actual flange width if less
where bw is the breadth of the web of the beam and ℓz is the distance between points of zero
moment in the beam. In simply supported beams it is the effective span where as in
continuous beams ℓz may be taken as 0.7 times the effective span. The design procedure for flanged beams depends on the depth of the stress block. Two possibilities need to be considered.4.6.2 Stress Block within the FlangeIf 0.9x≤hf, the depth of the flange (same as the total depth of the slab) then all the concrete below the flange is cracked and the beam may be treated as a rectangular beam of breadth b and effective depth d and the methods set out in sections 4.4.6 and 4.4.7 above apply. The maximum moment of resistance when 0.9x=hf is equal to Mflange=0.45 fcubhf(d-hf/2)
Thus if the design moment M≤Mflange, then design the beam as singly reinforced rectangular
section b×d.4.6.3 Stress Block Extends into the WebAs shown in Fig.4.16, the compression forces are: In the flange of width (b-bw), the compression force C1 is
C1=0.45 fcu(b-bw)hfIn the web, the compression force C2 is
C2=0.45 fcubw0.9x
The corresponding lever arms about the tension steel are
z1=d-hf/2
z2=(d-0.9x/2)Fig.4.16: T-beam with the stress block extending into the web.The moment of resistance MR is given by MR=C1z1+C2z2MR=0.45 fcu(b-bw)hf(d-hf/2)+0.45fcubw0.9x (d-0.9x/2)From equilibrium,
T=Asfs=C1+C2If the amount of steel provided is sufficient to cause yielding of the steel, then fs= 0.95fy. The maximum moment of resistance without any compression steel is when x=0.5d. Substituting x=0.5d in the expression for MR, the maximum moment of resistance is
Mmax=0.45 fcu(b-bw)hf(d-hf/2)+0.156 fcubwd2or Mmax=ßfbd2fcuwhere
Thus if the design moment Mflange <M≤Mmax, then determine the value of x from
where x≤0.5d and the reinforcement required is obtained from the equilibrium condition,
As0.95fy=C1+C24.6.3.1 Code formulaAs an alternative, a slightly conservative formula for calculating the steel area is given in clause 3.4.4.5 of the code. The equation in the code is derived using the simplified stress block with x=0.5d (Fig.4.16). depth of stress block=0.9x=0.45d
The concrete forces in compression are
C1=0.45fcuhf(b-bw)
C2=0.45fcu×0.45dbw=0.2fcubwd
The values of the lever arms for C1 and C2 from the steel force T are:
z1=d-0.5hfz2=d-0.5x 0.45d=0.775d
The steel force in tension is T=0.95fyAsThe moment of resistance of the section is found by taking moments about force C1:
M=Tz1-C2(z1-z2)
M=0.95fyAs(d-0.5hf)-0.2fcubwd (0.225d-0.5hf)
M=0.95fy As(d-0.5hf)-0.1fcubwd(0.45d-hf)
from which
This is the expression given in the code. It gives conservative results for cases where x is less
than 0.5d. The equation only applies when hf is less than 0.45d, as otherwise the second term in the numerator becomes negative.4.6.4 Steps in Reinforcement Calculation of a T-or an L-Beam
| • Calculate the total design load (including self-weight) and the corresponding design moment M using appropriate load factors. • Calculate the maximum moment Mflange that can be resisted, when the entire flange is incompression. |
| Mflange=0.45fcubhf(d-hf/2) |
| • Calculate the maximum moment that the section can withstand without requiring compression reinforcement. |
| Mmax=0.45fcu(b-bw)hf (d-hf/2)+0.156fcubwd2 |
| • If M≤Mflange, then design as a rectangular beam of dimensions, b×d. • If Mflange <M≤Mmax, then the required steel area can be determined to sufficient accuracy from the code formula |
| • If M>Mmax, then compression steel is required or the section has to be revised. Compression steel is rarely required in the case of flanged beams. |
4.6.5 Examples of Design of Flanged Beams
Example 1: A continuous slab 100 mm thick is carried on T-beams at 2 m centres. The
overall depth of the beam is 350 mm and the breadth bw of the web is 250 mm. The beams are 6 m span and are simply supported. The characteristic dead load including self-weight and finishes is 7.4kN/m2 and the characteristic imposed load is 5 kN/m2. Design the beam using the simplified stress block. The materials are grade C30 concrete and grade 460 reinforcement. Since the beams are spaced at 2 m centres, the loads a the beam are:
Dead load=7.4×2=14.8 kN/m
Live load=5×2=10 kN/m
design load=(1.4×14.8)+(1.6×10)=36.7 kN/m
ultimate moment at mid-span=36.7×62/8=165 kN m
effective width b of flange: b=250+6000/5=1450 mm
The beam section is shown in Fig.4.17. From BS8110: Part 1, Table 3.4, the nominal cover on
the links is 25 mm for grade 30 concrete. If the links are 8 mm in diameter and the main bars are 25 mm in diameter, thend=350–25–8-12.5=304.5 mm, say 300 mm.
First of all check if the beam can be designed as a rectangular beam by calculating Mflange.
Mflange=0.45fcubhf(d-hf/2)
Mflange=0.45×30×1450×100 (300–0.5×100)×10-6=489.3 kNm
The design moment of 165 kNm is less than Mflange. The beam can be designed as a
rectangular beam of size 1450×300. Using the code expressions in clause 3.4.4.4
k=M/(bd2fcu)=165×106/(1450×3002×30)=0.042z/d={0.5 +√(0.25–0.042/0.9)}= 0.95
z=0.95d=285 mm
As=165×106/(0.95×460×285)=1325mm2.
Provide 3T25; As=1472 mm2.Fig.4.17 Cross section of T-beam.Example 2: Determine the area of reinforcement required for the T-beam section shown inFig.4.18 which is subjected to an ultimate moment of 260 kNm. The materials are grade C30
concrete and grade 460 reinforcement. Calculate Mflange to check if the stress block is inside the flange or not.
Mflange=0.45×30×600×100(340–0.5 x 100)×10-6=234.9 kNm
The design moment of 260 kNm is greater than Mflange. Therefore the stress block extends into the web.
Check if compression steel is required.
Mmax=0.45fcu(b-bw)hf(d-hf/2)+0.156 fcubwd2Mmax={0.45×30×(600–250)×100×(340–100/2) +0.156×30×250×3402}×10-6
Mmax=(137.0+135.3)=272.3 kNm
Mmax>(M=260 kNm)
The beam can be designed without any need for compression steel. Two approaches can be
used for determining the area of tension steel required.Fig.4.18 Cross section of T-beam.(a) Exact approachDetermine the depth of the neutral axis from setting x/d=a
0.1250=0.0659+0.1667 a-0.075 α2Simplifying
α2-2.22 α+ 0.788=0
Solving the quadratic in a,
a=x/d=(2.22–1.3328)/2=0.444< 0.5
×=0.444×340=151 mm
T=0.95fyAs=C1+C2T=0.45fcu(b-bw)hf+0.45fcubw0.9 x
T=(0.45×30×(600–250)×100+0.45×30×250×0.9×151)×10-3T=0.95 As=(472.5+458.7)=931.2 kN
As=931.2×103/(0.95×460)=2131 mm2
(b) Code formulaCalculation of As using simplified code formula which assumes x/d=0.5
This is only 1% more than that calculated using the exact neutral axis depth! Provide 5T25,
A=2454 mm2
4.7 CHECKING EXISTING SECTIONS
In the previous sections methods have been described for designing rectangular and flanged
sections for a given moment. In practice it may be necessary to calculate the ultimate moment capacity of a given section. This situation often occurs when there is change of use in a building and the owner wants to see if the structure will be suitable for the new use. Often
moment capacity can be increased either by
• Increasing the effective depth. This can be done by adding a well bonded layer of concrete at the top of the beam/slab
• Increasing the area of tension steel by bonding steel plates to the bottom of the beam.
4.7.1 Examples of Checking for Moment Capacity
Example 1: Calculate the moment of resistance of the singly reinforced beam section shown
in Fig.4.19(a). The materials are grade C30 concrete and grade 460 reinforcement. The
tension reinforcement is 4T20 giving As=1256 mm2 Assuming that tension steel yields, total
tensile force T is given by
T=0.95fyAs=0.95×460×1256×10-3=548.7 kN
If the neutral axis depth is x, then the compression force C is
C=0.45fcu (0.9x×b)=0.45×30×0.9x×250×10-3=3.0375x kN
For equilibrium, T=C. Solving for x
x=181 mm<(0.5 d=200 mm)
Check the strain in steel
Steel yields. Therefore the initial assumption is valid.
z=d-0.45x=400–0.45×181=310 mm
z/d=310/400=0.775<0.95
Moment of resistance M
M=T z=548.7×310×10-3=169.9 kNm
One can also use the Design Chart shown in Fig.4.10 to solve the problem.
From design chart Fig.4.10,
M=174.0 kNmFig 4.19 Cross section of rectangular beam.Example 2: Determine the ultimate moment capacity of the beam in Fig.4.19, except,
As=6T20=1885 mm2Proceeding as in Example 1, assume that steel yields and calculate
T=0.95fyAs=0.95×460×1885=8.24×105 N
C=0.45×fcu×0.9×x×b=3037.5x N
For equilibrium, T=C.
x=271 mm, x/d=0.68>0.5
Check strain in steel to check the validity of the initial assumption.
Since the strain in steel is less than yield strain, tension steel does not yield indicating that the initial assumption is wrong. Assume that the tension steel does not yield. For an assumed
value of neutral axis depth x, strain εs in tension steel is
Since the steel is assumed not to yield, if Young’s modulus for steel is Es=200 kN/mm2, then
stress fs in tension steel is given by
T=As×fs={1885×700×(d-x)/x}×10-3=1319.5(400-x)/x kN
C=0.45 fcub 0.9x={0.45×30×250×0.9x}×10-3=3.0375x kN
For equilibrium, T=C1319.5×(400-x)/x=3.0375x
Simplifying
x2+434.40x-173761.3=0
Solving the quadratic equation in x,
x=253 mm, x/d=253/400=0.63>0.5
Calculate the strain in steel.
fs=Es×εs=200×103×0.00204=408 N/mm2z=d-0.45x=286 mm
M=T×z=769×286×10-3=220 kNm
Since x/d>0.5, it is sensible to limit the permissible ultimate moment to a value less than 220
kNm. Assuming that x=0.5d=200 mm,
C=0.45 fcub 0.9x={0.45×30×250×0.9×200}×10-3=607.5 kN
Lever arm z=0.775 d=0.775×400=310 mm
Taking moments about the steel centroid, M=Cz=188.3 kNmExample 3: Calculate the moment of resistance of the beam section shown in Fig.4.20. The materials are grade C30 concrete and grade 460 reinforcement.
As=4T25=1963 mm2, As̀=2T20+T16–829 mm2Assume that both tension and compression steels yield and calculate the tension force T and compression force Cs in the steels.
T=0.95fyAs=0.95×460×1963 ×10-3=857.8 kN
Cs=0.95 fyAs ̀=0.95×460×829 ×10-3=362.3 kN
The compression force in concrete is
Cc=0.45fcu(0.9x×b)=0.45×30×0.9x×250×10-3=3.0375x kN
For equilibrium,
Cc+Cs=T
3.0375x +362.3=857.8.
Solving x=163 mm, x/d=0.47<0.5.
Fig.4.20 Cross section of doubly reinforced beam.
Calculate strain in tension and compression steels to verify the assumption.
Both strains are larger than yield strain of 0.0022. Therefore both steels yield and the initial
assumption is correct.Cc=0.45fcu b0.9x=0.45×30×250×0.9×163 ×10-3=495.1kN
Taking moments about the tension steel,M=Cc(d-0.45x)+Cs(d-d ̀)M=495.1(350–0.45×163)×10-3+362.3×(350–50)×10-3=245.8 kNm4.7.2 Strain Compatibility MethodIn the previous section, examples were given for calculating the moment of resistance of a given section. It required making initial assumptions about whether the steel yields or not.
After calculating the neutral axis depth from equilibrium considerations, strains in tension and compression steels are calculated to validate the assumptions. The problem can become
complicated if one steel yields while the other steel does not. A general approach in this case
is the method of Strain Compatibility which has the advantage of avoiding the algebraic
approach. The basic idea is to assume a neutral axis depth. From the assumed value of neutral axis depth, strains in steel in compression and tension are calculated. Thus
From the stresses, calculate the forces
T=Asfs,Cs=As’fs’,Cc=0.45fcub0.9x, C=Cs+CcFor equilibrium, T=C. If equilibrium is not satisfied, then adjust the value of x and repeat until equilibrium is established. Normally only two sets of calculations for neutral axis depth are required. Linear interpolation can be used to find the appropriate value of x to satisfy equilibrium. The following example illustrates the method.4.7.2.1 Example of Strain-Compatibility MethodExample 1: Calculate the moment capacity of the section with b=250 mm, d = 350 mm,
d’=50 mm,
As ̀=3T20=942.5 mm2, As=6T25=2945.2 mm2Trial 1: Assume x=220 mm
Strain εs in compression steel is given byεs’=0.0035(x–d ̀)/x=0.0035×(220–50)/220=0.0027>0.0022
Therefore compression steel yields and the stress fs’ is equal to 0.95 fySimilarly, strain εs in tension steel is given by
εs=0.0035(d-x)/x=0.0035×(350–220)/220=0.00207<0.0022
Therefore tension steel does not yield and the stress fs is equal to
fs=εsEs= 0.00207×200 ×103=413.6 N/mm2T=As×fs=2945.2×413.6×10-3=1218.1 kN
C=0.45 fcu×b×0.9x+As̀×fs ̀
C={0.45×30×250×0.9×220+942.5×0.95×460}×10-3C=(668.25+411.87)=1080.1 kN
T-C= 138.0 kN
Total tensile force T is greater than the total compressive force C. Therefore increase the
value of x in order to increase the compression area of concrete and also reduce the strain in
tension steel but increase the strain in compression steel.Trial 2: Assume x=240 mm say
Strain εs in compression steel is given byεs ̀=0.0035(x–d’)/x=0.00277>0.0022
Therefore compression steel yields and the stress fs ̀is equal to 0.95 fySimilarly, strain εs in tension steel is given byεs=0.0035(d–x)/x=0.0016<0.0022
Therefore tension steel does not yield and the stress fs is equal to
fs=εs E=0.001604×200×103=320.8 N/mm2T=As×fs=2945.2×320.8×10-3=944.8 kN
C=0.45fcu×b×0.9x+As̀×fs ̀
C={0.45×30×250×0.9×240+942.5×0.95×460}×10-3C=(729.0+411.87)=1140.9 kN
T-C=-196.04 kN
As shown in Fig.4.21, linearly interpolate between x=220 and 240 to obtain the value of x
giving T–C=0.
x=220+(240–220)×(138.0)/(138.0+196.04)=228 mm
x/d=228/350=0.65>0.5
Fig.4.21 Linear interpolation
As a check calculate T and C for x=228 mm
Strain εs’ in compression steel is given byεs̀=0.0035(x-(d ̀)/x=0.0027>0.0022
Therefore compression steel yields and the stress fs’ is equal to 0.95 fySimilarly, strain εs in tension steel is given by
εs=0.0035(d-x)/x=0.00187<0.0022
Therefore tension steel does not yield and the stress fs is equal to
fs=εs E=0.00187×200×103=374.6 N/mm2T=As×fs=2945.2×374.6×10-3=1103.2 kN
C=0.45fcu×b× 0.9x+As’×fs’ As’
C={0.45×30×250×0.9×228+942.5×0.95×460}×10-3C=(692.6+411.87)=1104.4 kN
T-C=-1.22kN
This is close enough to be zero.
Taking moments about the tension steel, the lever arm for compression force in concrete is
(d–0.45x) and for the compression force in steel it is (d–d’).
M=(692.6×(350–0.45×228)+411.87×(350–50)}×10-3=294.9 kNm
Since x/d>0.5, it is sensible to limit the permissible ultimate moment to a value less than
294.9 kNm. Assuming that x=0.5d=175 mm,
Cc=0.45fcub0.9x={0.45×30×250×0.9x 175}×10-3=531.6 kN
Lever arm zc=0.775 d=0.775×350=271 mm
Strain εs ̀in compression steel is given by εs ̀=0.0035(x–d’)/x=0.0025>0.0022 Therefore
compression steel yields and the stress fs ̀is equal to 0.95 fyCs={942.5×0.95×460}×10-3=411.9 kN, Lever arm zs=d-d ̀=300 mm
Taking moments about the steel centroid, M=Cczc+Cszs=267.6 kNm
