The document summarizes the analysis of reinforced concrete beam cross sections to determine their moment of resistance at the ultimate limit state. It outlines the key assumptions of the strength design method and describes the behavior of beams under small, moderate and ultimate loads. It also discusses balanced, under-reinforced and over-reinforced beam sections, and introduces the concept of the equivalent stress block to simplify calculations. Worked examples are provided to demonstrate how to determine the depth of the neutral axis and moment of resistance for various beam cross sections.

1. 1

2. Analysis of the Section
1.1 Strength design method assumption
The analysis of a cross section to determine its moments of resistance at
ultimate limit state is based on the following assumptions:
1- A plane section (perpendicular to the axis of bending) before bending
remains a plane after bending, so that the strains distribution in the
concrete and the reinforcement (tension or compression) must be
linear distribution of strains.
2- The tensile strength of concrete is zero and the reinforcement carries
all the tensile forces (in fact the tensile strength of concrete is only
about 10 per cent of the compressive strength)
3- The bond between the concrete and the reinforcement is perfect, so no
slip occurs to ensure that the tensile forces to be transferred by bond to
the reinforcement
4- A compressive concrete stress is approximately proportional to strain
only up to moderate loading, with an increase in load, the compressive
stress-strain diagram takes shape similar to the concrete compressive
stress-strain curve shown in figure 1.1;
Figure 1.1: Short term stress strain curve for normal weight concrete
design curve of the
BS8110
characteristic curve
0.67 fcu
/γm
fck
fcu
0.0035
Stress
Strain
Notes:
As assumed by the
BS8110, the ultimate
compressive strain is
0.0035,
0.67 is the size factor,
γm
equal to 1.5 (Table 2.2
of the BS8110)
2

3. 5- The maximum useable concrete compressive strain at the extreme
fiber is assumed to be equal to 0.0035
6- For stress in steel reinforcement less than yield stress (fy ), the steel
stress is proportional to strain (εs) and equal to Es εs. Stress-strain
diagram of steel reinforcement is shown in Figure 1.2.
Figure 1.2: Stress strain diagram of steel reinforcement
1.2 Section Analysis
Based on the assumption previously stated in section 1.1, we can now
examine the strains, stresses and forces that exist in reinforced concrete
beam as the load increases from zero to the magnitude that would cause
failure:
A) At very small load before cracks occurring (at cracking load); both concrete
and steel reinforcement resist the tensile forces at the bottom of the beam
(tension side, below the natural exist). In this case, the stresses are vary
linearly from the neutral axis (zero position) of the section and proportioned to
strains as shown in figure 1.3.
Characteristic curve
Design curve
E = 200 kN/mm2
fy
0.95 fy
Stress
Strain
3

4. 4
Figure 1.3: Flexural behaviour of RC beam at
very small load (below cracking load)
b) Section A - A c) Strains
distribution
d) Stresses
distribution
fcu
(compression)
fcu
(tension)
fs (tension)
εc
(compression)
εS
(tension)
h d
εc
(tension)
a) Reinforced concrete beam
A B
A
A
N.A.

5. B) At moderate load which starts from cracking load and up to before yielding
load. When the tensile stresses at the bottom of the beam exceed the tensile
strength of the concrete, the cracks in concrete (hair crack) occur. Since the
concrete cannot transmit any tension forces across a crack, the steel
reinforcement then resist the entire tension forces. In this case; concrete
compression on stresses are still assumed to be proportional to strains in
concrete as shown in Figure 1.4.
5
A
A
A B
a) Reinforced concrete
beam
N.A.
fcu
(allowable stress)
fs (allowable stress)
εc
(tension)
εc
(compression)
εs
(tension)
b) Section A-A
beam
c) Strains distribution
d) Stresses
distribution
Figure 1.4: Flexural behaviour of RC beam at moderate load

6. C) At yielding load, this is started from the yield point of the steel
reinforcement up to the ultimate load (before failure). In this case the trend of
stress strain distribution in concrete section of the beam (above the neutral
axis) is similar to that of concrete stress-strain curve (shown in Figure 1.1).
the stresses and strains distribution in the section of the beam are shown in
Figure 1.5.
6
concrete crush
A
A
A B
a) Reinforced concrete
beam
N.A.
fs = fy (yield stress)
0.67 fcu (ultimate stress)
εs
εs
= 0.0035 (as a limit)
b) Section A-A c) Strains
distribution
d) Stresses
distribution
Figure 1.5 : Flexural behaviour of RC beam at ultimate load

7. As shown in Figure 1.5 (c), the shape of stresses distribution is regtangular
parabolic shape. Some problems are encountered with the rectangular
parabolic shape; such as calculation of the area under the curve and
determine the centroid location.
1.3: Equivalent rectangular stress block
Equivalent rectangular stress block was adopted by BS8110 for purposes of
simplification and practical application. With respect to equivalent stress
distribution block shown in figure 1.6(c); the average stress intensity is taken
as 0.45 fcu and is assumed to be acted on the upper edge of the RC beam
cross section. The area is defined by the width b and a depth of s.
7
N.A.
fy
x
fs
= fy
(yield stress)
0.67 fcu
/γm
0.67 fcu
/ 1.5
= 0.45 fcu
s =0.9 x
d
b
a) Section b) Rectangular parabolic
stress block
c) Equivalent rectangular stress
block
Figure 1.6: Rectangular parabolic stress block and equivalent rectangular stress
block.

8. The equivalent rectangular stress block does not extend to the neutral axis of
the section but has a depth of s, which is equal to 0.9x. This will result the
centroid of the stress block being s/2, which is equal to 0.45x from the top
edge of the section. The area and the centroid of the two stresses block
(rectangular parabolic and equivalent stress block) are approximately equal.
Thus the moment of resistance of the section will be similar using calculation
of either type of stress block.
1.4: Types of failure and beam section classification.
Three failure types could be happened to RC beam loaded in flexure and that
depends on the amount of steel provided in the section. The three types of
failure are as following.
A – Balanced reinforced beam: In this type of failure the concrete crush at the
compression area (above the neutral axis) and the steel reinforcement yield
simultaneously at ultimate load. The amount of steel reinforcement required
can be calculated by equating the compression force in the concrete and the
tension force in the steel (equilibrium principle).
Example 2.1:
For the section shown in the Figure 1.7, determine the depth of the neutral
axis for the balanced design case. fy = 460 N/mm2
and Es = 200 KN/mm2
8

9. Solution:
Section Strains
xd
y
x
c
−
∑
=
∑
9
b
h d
h d
b
N.A.
εc
= 0.0035
εy
x
d-x
Figure 1.7

10. 0.615dexceedshouldnotxcrush;concrete
beforestatelimintultimatetheatareatensioninentreinforcemsteeltheofyieldingensuretoHence,
limitstateultimatetheat
1equation
d.x
x.x)(d.
.c
.y
.
y
mmkNE sand.γ m
E s
γ mf y
y
.........x .......yx)(dc
6150
00219000350
1equationinngsubstitutithatso
00350
and
002190
000200
051460
200051 2
=
∗=−∗
∑ =
∑ =
∑ =
==
∑ =
∑ ∑ ∗=−∗












B – Under reinforced beam: if the amount of steel reinforcement provided is
less than that of case balanced reinforced beam, the steel will reach at
ultimate yielding stress before the concrete fails at the ultimate load. This type
of failure is gradual and giving warning of failure. In this case the depth of the
neutral axis must be less than 0.615d. The BS8110 was adopted for the
ultimate state design, (where the moment redistribution not greater than 10%);
x ≤ 0.5d
C – Over reinforced beam: If the amount of the steel reinforcement provided is
greater than that of case of balanced reinforcement beam; then the concrete
fails before steel reaches yielding stress. This type of failure is sudden and
without warning of failure. The depth of the neutral axis in this case is greater
than 0.615d.
Example 2
10

11. The beam shown in the Figure 1.8 is made of concrete with compressive
strength (fcu) of 35N/mm2
and steel reinforcement with strength (fy) of
460N/mm2
. Determine the ultimate moment of resistance of the cross section.
Solution:
Fcc = Stress ∗ Area
Fcc = 0.45fcu ∗S ∗ b
S = 0.9∗x
Fcc = 0.45 ∗35∗0.9∗300
Fcc = 4252.5∗x
Assume the steel has been yielded; fst = 0.95fy
11
b = 300mm
As
= 1450mm2
d = 550 mm
0.45 fcu
x s
N.A.
Fcc
Fst
z
b = 300mm
As = 1450mm2
d = 550 mm
Figure 1.8
cross section stress block

12. Fst = 0.95 fy ∗ As
Fst = 0.95 ∗ 460 ∗ 1450
Fst = 633650 N
For equilibrium;
Fcc = Fst
4252.5 ∗ x = 633650
4252.5
633650
x =
x = 149 mm
Since, x (=149) < 0.615d (=338.25), therefore the steel has been yielded and
fst = 0.95fy as assumed.
Moment of resistance of the section
M = Fst ∗ Z
= 0.95fy As (d -0.9x/2)
= 0.95 fy As (d – 0.9x / 2)
= 0.95∗460 ∗1450 (550 – (0.9 ∗149) /2) ∗ 10 -6
= 306.02 KN.m
Example 3
12

13. As = 1450mm2
305 mm305 mm
h =610mm d =590mm
0.9x
0.45fcu
Fcc
N.A.
Fst
Compression zone
area = (0.9x)2
/ 2
cross - section
305 mm 305 mm
Determine the ultimate moment of resistance of the cross section shown in
Figure 1.9, given that the characteristic strength of the steel reinforcement is
460N/mm2
and the characteristic strength of concrete is 30 N/mm2
.
13
Solution:
Z
0.9x
stresses block
(2/3) (0.9x)
Figure 1.9

14. Fcc = 0.45fcu ∗ area of compression zone area
= 0.45 ∗ 30 ∗ (0.9x)2
/ 2
= 5.467 x2
Assume the steel reinforcement has been yielded,
Fst = 0.95fy
Fst = 0.95 fy As
= 0.95 ∗ 460 ∗1450
= 633650 N/mm2
From equilibrium
Fcc = Fst
5.467∗ x2
= 63350
x (=340.44 mm) < 0.615d (=362.85), therefore, the steel reinforcement has
been yield and
Fst = 0.95fy as assumed.
Moment of resistance of the section
Z = d – 2/3 (0.9x)
= 590 -2/3 (0.9 x 340.44)
= 385.736 mm
M = Fst x Z
= 0.95 fy As + Z
= 0.95 x 460 x1450 x 385.736 x 10-6
= 244.421 KN.m
14

15. 1.5 Analysis of flanged section
In the analysis of flanged section (T-section or L- section) is important to
determine the location of the neutral axis whether within the flange or within
the web.
Example 4
A flanged beam section (T- section) is shown in the Figure 1.10. Determine
the maximum moment which can be applied to the section. The characteristic
strength of the materials are fy = 460 N/mm2
and fcu = 35 N/mm2
.
Solution
Assume that the neutral axis lies within the flange and the steel reinforcement
is yielded ( Fst = 0.95fy )
15
bf = 850mm
hf = 150mm
d = 450mm
As = 1450 mm2
Figure 1.10

16. Fcc = 0.45 fcu ∗bf ∗ 0.9x
= 0.45 ∗ 35 ∗850 ∗0.9x
= 12048.7 x
Fst = 0.95fy As
= 0.95 x 460 x 1450
= 633650 N
From equilibrium
Fcc = Fst
12048.7 x = 633650
x = 633650 / 12048.7
= 52.6 mm < hf (=150)
The neutral axis lies within the flange as assumed.
X ( = 52.6mm ) < 0.615d (= 276.75mm)
The steel will have yielded as assumed thus, fs = 0.95fy
The resistance moment of the section
M = Fst ∗z
16
Fcc
d = 450mm
x
0.45fcu
S=0.9x
0.9x /2
Z
Fst
N.A.
bf
= 850mm
hf
= 150 mm
cross section stress block

17. = 633650 ∗ (d – 0.9x / 2) ∗10-6
= 270.144 KN.m
Example 5
Determine the moment of resistance of the T –section shown in the Figure
1.11. fy = 460 N/mm2
and fcu = 20N/mm2
Solution
Assume that the neutral axis lies within the flange and the steel reinforcement
is yielded so that fs = 0.95fy
17
bf = 850mm 0.45fcu
0.9xx Fcc
As= 2200 mm2
300mm
d = 450mm
bf
= 850mm
hf
= 100mm
Figure 1.11

18. Fcc = 0.45 fcu ∗bf ∗0.9x
= 0.45 ∗20 ∗850 ∗0.9x
= 6885 x
Fst = 0.95fy As
= 0.95 ∗ 460 ∗2200
= 96.1400
Fcc = Fst
6885 x = 961400
x = 961400 / 6885
= 139.6 mm > hf ( = 100mm)
The neutral axis lies within the web
18
Fst
N.A.
bf = 850mm 0.567fcu
Fst
Z2 = d-100/2
Z1 = d-100-(0.9x-100)/2
Fcx
Fcw
100
mmS=0.8x
N.A
Cross section Stress block
d=450mm

19. Fcf = 0.45 fcu bf hf
= 0.45 ∗ 20∗850 ∗100
= 765000N
Fcw = 0.45 fcu bw (0.9x – 100)
= 0.45 ∗20 ∗300 (0.9x – 100)
= 2430 x – 270000
Fst = 0.95 fy As
= 961400
Fst = Fcf + Fcw
961400 = 765000 + 2430x – 270000
x = 191.9 mm < 0.615d (= 276.75)
Therefore, the steel will have yielded as assumed thus,
Fs = 0.95fy
Taking moment about the centroid of the steel reinforcement.
M = Fcw ∗Z1 + Fcf ∗Z2
= [(2430 x – 27000) ∗ [ d – (0.9x – 100) / ] + 765000 ( d – (100/2)] ∗10-6
= [(2430 x 191.9 – 27000)x[450–100-(0.9x-100)/2 ]+765000 (450 – (100/2)] ∗
10-6
= [ 196317 x 313.645 + 306000000] 10-6
= 367.57 KN.m
1.6 The ultimate moment of resistance of singly reinforced section and
analysis of doubly reinforced section:
19
bw = 300mm

20. When the applied design moment exceeds the concrete capacity;
compression steel reinforcement is required to supplement the load - carrying
capacity of the concrete. As specified by BS 8110 that the upper limit of the
lever arm (Z) is 95 d and the lower limit of Z is 0.755d, while the depth of the
neutral axis (x) should not exceeds 0.5d to ensure that the steel reinforcement
will have yielded before the concrete in compression area is failed, therefore
the RC beam will give signs of warning of the failure. To determine the
moment capacity of the single reinforced section shown in Figure 1.12; the
depth of the neutral axis assumed to be 0.5d.
a) section b) strain distribution c) stress block
Figure 1.12: strain distribution and stress block of single reinforced section
Fc = 0.45 fcu b ∗0.45 d
= 0.202 b d fcu
And
Z = d – 0.9x /2
= d – 0.45d / 2
= 0.775d
Taking moment about the centroid area of the steel reinforcement
20
h d
b
x =0.5d
0.0035 0.45fcu
0.9x = 0.45d Fcc
Fst
z
N.A

21. M = Fc ∗ Z
= 0.202 bd fcu ∗0.775d
= 0.156 bd2
fcu (capacity of the singly reinforced section based on concrete
strength)
Taking moment about the centroid of the compression area
M = Fs ∗ Z
= 0.95 fy As Z
= 0.95 fy As (d – 0.45d / 2)
= 0.736 fy As d (capacity of the singly reinforced based on steel
reinforcement strength)
Therefore, the capacity of the singly reinforced section (the ultimate moment
of resistance) is given by the lesser of
M = 0.156 fcu bd2
fcu (based on concrete strength)
or
M = 0.736 fy As d (based on steel reinforcement strength)
Example 6
Determine the maximum ultimate moment which can be carried by the section
shown in Figure 1.13. The characteristic strengths are fy = 460 N/mm2 for
steel reinforcement and fcu = 30 N/mm for concrete.
21
b = 300mm
h = 450mm d=400mm

22. Solution
Section capacity based on concrete strength
M = 0.156 bd2
fcu
= 0.156 ∗300∗4002
∗ 300∗10-6
= 224.64 KN.m
Section capacity based on steel strength
M = 0.736 fy As d
= 0.736 ∗460 ∗1200 ∗400 ∗ 10-6
= 162.5 KN.m
The maximum ultimate moment which can be carried by the section is 162.5
kN.m
In the analysis of doubly reinforced section, to ensure that the compression
steel will have yielded; d'/x should be less than 0.37 (for fy = 460 N/mm2
) as
shown in the analysis performed on the doublly reinforced section shown in
Figure 1.14.
22
As = 1200mm2
As′
x
0.0035
εsc
d'
x-d'
Figure 1.13

23. Assume steel reinforcement in compression area are yielded
εsc= 0.95 fy / Es
= 0.95 ∗460 ∗200000
= 0.00219
0.0035 / x = 0.00219 /( x-d')
(x – d') / x = 0.00219 / 0.0035
1 – (d' / x) = 0.62
d' / x = 0.37
for other grades of steel, with x = 0.5d
d' / d = 0.185
If d'/ x > 0.37 (or d'/d > 0.185), then it is necessary to determine
the value of εsc from strains distribution, then and then to compute fsc from
fsc = Es xεsc
where Es = 200 000 N/mm2
Example 7
23
As
d-x
N.A
Est
Figure 1.14

24. Compute the design strength (the ultimate moment of resistance) of the
doubly reinforced cross-section as shown in Figure 1.15; fy = 460 N/mm2
and
fcu = 30 N/mm2
Solution
Form equilibrium.
Fst = Fcc + Fsc (assume that the steel reinforcement in the tension and
compression are yielded values ( fst = fsc = 0.95 fy)
0.95 fy As = 0.45 fcu b ∗ 0.9 x + 0.95 fy As'
x = 0.95 fy (As – As') / 0.405 fcu b
x = 0.95 ∗ 460 (2325 – 403) / 0.405 ∗30 ∗250
x = 276.51 mm < 0.615d ( = 319.8mm)
Therefor, steel reinforcement in tension will have yielded as assumed
24
b=250mm
As' = 403mm2
As = 1325mm2
d = 520mm
d' =50mm
As′
As
0.45fcu
Fst
N.A
Z2=d-50
Z1=d-0.9x/2
Fsc
Fcc0.9xx
Figure 1.15

25. d' / x = 50 / 276.51 = 0.18 < 0.37
So the steel reinforcement in the compression will have yielded as assumed.
Taking moment about the centroid of the steel in tension
M = Fcc x Z1 + Fsc ∗Z2
M = 0.45 fcu b ∗0.9 x (d – 0.9x / 2) + 0.95 fy As' (d – 50)
= [0.45 x 30 x 250 x 0.9 276.51 (520 – 0.9x276.51/ 2) + 0.95 x 460 x 403
(520-50) ∗10-6
= 415 KN. m
1.7 Derivation of Design Formulae for Singly and Doubly Rectangular
Reinforced Concrete Beams:
To derive the equations stated in clause 3.4.4.4 of the BS8110:1995; Figure
1.16 is referred:
Force= stress ∗ area
Fs = 0.95 fy As
Fc = 0.45 fcu ∗b ∗0.9x
= 0.405 fcubx
b
dh
As
Z=d-0.45x
Fs
Fc
0.45fcu
N.A
25
Figure 1.16

26. For equilibrium
Fc= Fs
[0.405 fcu b x = 0.95 fy As] ∗ d
1
d
x
= 2.345 ∗
fcu
fy
∗
bd
As
Note that
d
x
increases with increasing of As and decreases with increasing of
the section area (bd).
Taking moment about the centroid of the steel area:
M = Fc ∗z
= (0.405 fcu b x) (d - 0.45 x)
Rearrange the above formulae
= (0.405
d
x
) (1-0.45 x) fcu b d2
Note that moment increases with increasing the value of
d
x
Assume; k= (0.405
d
x
) (1-0.45 x)
∴ M=kfcubd2
( the applied bending moment on the section)
Where the simplified stress block is used; BS8110:1995 limits
d
x
should not
exceed 0.5 and
x
d′
(for doubly reinforced section) should not exceed 0.37
(for fy=460 N/mm2
). This limit in
d
x
and
x
d′
is to ensure that the steel
reinforcement will have yielded (fs=0.95fy) and the design will be under
26

27. reinforced section. Therefore, the maximum moment capacity of a singly
reinforced beam based on concrete strength is calculating form (limiting
x=0.5d)
Mu= 0.202 fcu bd2
z
Where; z= d-0.45 x
= d-0.45 (0.5 d)
= 0.775 d (minimum value of z)
∴Mu = (0.202 fcu bd) (0.775d)
= 0.156 fcu bd2
Where k′ = 0.156
∴ Mu= k′ fcu bd2
(the maximum moment capacity of singly reinforced beam
based on concrete stregth)
Where the applied moment exceeds the maximum moment capacity of the
section then the excess (M-Mu) to be resisted by using steel reinforcement (As′
) in the concrete compression area to supplement the load-carrying capacity
of the concrete. The neutral axis depth (x) to be maintained at the maximum
permitted value i.e 0.5d. By referring to the Figure 1.17:
b
dh
As
Z=d-0.45x
Fs
Fc
0.45fcu
N.A
As′
d′ Fs′
z1
=d-d′
27
Figure 1.17

28. F s′ = 0.95 fy As′
M-Mu=Fs′ ∗ z1
= 0.95 fy As′ ∗ (d-d′)
)d(df0.95
MuM
As
y
′−
−
=′
Where
Mu= k′ fcu bd2
∴
)d(dfy0.95
d2bfcukM
As ′−
′−
=′
An area of steel in tension must then provided to balance the total
compressive forces in concrete and the compression reinforcement:
Fs=Fs′ + Fc
0.95 fy As = 0.95 fy As′ + 0.202 fcu bd
∴ As
fy0.95
bdfcu0.202
As ′+=′
Where; Mu=0.202fcubd∗z
∴ As
zfy0.95
Mu
As ′+=
Note: to prove that z= d {0.5+ 9.025.0 k− }; refer to Figure 1.16
M = Fc ∗ z
= 0.45 fcu b∗ 0.9x ∗z
Where; z = d -
2
0.9x
and 0.9x = 2(d-z)
∴M= 0.45 fcu b∗ 2(d-z) ∗ z
28

29. = 0.9 fcu b (d-z) ∗ z
Where; k =
dbfcu
M
2
∗∗
d
z
d
z
0.9
k
2
2
−=
And (z/d)2
– (z/d) + k/0.9 = 0
Solve the above quadratic equation
z= d {0.5+ 9.025.0 k− }
note: the above equations are also applicable to flanged beams where the
neutral axis lies within the flange.
Example 8
The ultimate design moment to be resisted by the section in the Figure 1.18 is
150 kN.m. Determine the area of tension reinforcement (As) required. The
characteristic strength of steel fy=460 N/mm2
and the characteristic strength of
concrete fcu = 40 N/mm2
:
29
b=200mm
d=
387mm
h= 450
mm
As
Figure 1.18

30. Solution
k =
dbfcu
M
2
∗∗
=
40387200
10150
2
6
∗∗
∗
= 0.125 < 0.156
Therefore, the section is singly reinforced:
z= d {0.5+ 9.025.0 k− }
= d {0.5+ 9.0125.025.0 − }
= 0.83d < 0.95 d
As= zfy0.95
M
= mm
3870.834600.95
10150 2
6
=
∗∗∗
∗
Example 9
Design the steel reinforcement for the section of dimension; 200 mm wide and
300 mm effective depth, while the inset of compression steel is 40 mm. the
section to resist an applied moment of 123.3 kN.m.
Solution
k =
dbfcu
M
2
∗∗
=
300200
123.3
2
6
∗∗
∗
30
10
= 0.22 > k′ (=0.156)
Therefore, compression steel reinforcement is required
30

31. z= d {0.5+ 9.0156.025.0 − }
0.775 d (zmin) < 0.776 d < 0.95 d (Z max)
∴z = 0.775 d
= 0.776 ∗300
= 232.8 mm
x =
0.45
z)(d −
=
45.0
)8.232300( −
= 149.33 mm
37.0267.0
33.149
40
x
d
<==
′
∴The compression steel will have yielded
)d-d(f y0.95
2bdf cu)k-(k
′
′
=′As
=
)40300(46095.0
300200300.156)-(0.22 2
−∗∗
∗∗∗
= 304.17 mm2
As = sA
0.95f
bdfk
y
2
cu
′+
′
z
= 17.304
232.84600.95
300200300.156 2
+
∗∗
∗∗∗
= 1132.215 mm2
1.8 Derivation of design formulae for flanged reinforced concrete beams:
The design procedure of flanged beam depends on the location of the neutral
axis. The neutral axis may lie in the flange or in the web as shown in the
Figure 1.19
31

32. A) Neutral axis in the flange
Z= d – (0.9/2)x
Or 0.9x = 2(d-z)
If 0.9 x ≤ hf
xhf
d
bf
N.A
N.A
x
a) Neutral axis in the flange b) Neutral axis in the web
xhf
d
bf
N.A
0.9x Fc
0.45x
Fs
Z=d-0.45x
32
Figure 1.19

33. ∴the neutral axis does lie within the flange as assumed and As=
zfy0.95
M
B) The neutral axis in the wed
Fcf = 0.45fcu ∗ hf ∗ (b-bw)
Fcw = 0.45fcu ∗0.9x∗bw
Where;
d
x
= 0.5
x = 0.5∗d
0.9 x = 0.9∗0.5d
Therefore, Fcw=0.202 fcu bw d
Z1 = d – 0.5 hf
Z2 = d – 0.5 ∗0.9x
= d – 0.5 ∗0.45d
= 0.775 d
Fs = 0.95 fy As
Taking moment about the centroid of the flange
M = Fs ∗z1 - Fcw ∗(z1-z2)
= 0.95fyAs (d - 0.5hf)-0.202 fcu bw d (d-0.5hf - 0.775d)
x
hf
d
bf
N.A
0.9x
Fcf
0.45fcu
Fs
Z1
Fcw
Z2
bw
33

34. = 0.95 fy As (d – 0.5hf) – 0.202 fcu bw d (0.225 d – 0.5 hf)
∴As =
( )
( )h0.5df0.95
h0.45ddbf0.1M
f
fw
y
cu
−
−+
As stated in the BS8110; the above equation to calculate As for singly
reinforced flanged section only when hf< 0.45d. For section with tension
reinforced only; the applied moment must not exceed the moment of
resistance of the concrete which can be calculated as following:
M = Fcf.z1+Fcw.z2
= 0.45 fcu hf (b-bw) (d-0.5hf)+ 0.155 fcu bw d2
Thus;
b
bw0.15
2d
h1
b
bw1
d
h
0.45
dbfcu
M ff
2
+







−







−=
or M = βf fcubd2
Where βf can be obtained from Table 3.6 of the BS8110
The equation for the steel area As only applies when the ultimate moment to
be resisted by the section is less than βf fcubd2
. If the applied moment is
greater than the resistance moment of the concrete section then the excess
(M-Mu) to be resisted by steel reinforcement in the compression area:
x
hf
d
bf
N.A
0.9x Fcf
0.45fcu
Fs
dFcw
bw
Fs′
d′
34

35. ∴ M – Mc = Fs′ ∗z
Where z = (d-d′) and Fs′ = 0.95 fy As′
∴ M – Mc = 0.95 fy As′ (d - d′)
As′ =
)d(df0.95
M
y
′−
−
c
M
For equilibrium;
Fs = Fcf + Fcw +Fs′
So that: As =
( )
As
fy0.95
bbhff0.45dbf0.2 wfcuwcu
′+
−+
Take note that
x
d′
should be less than o.37 to ensure that the steel in
compression area will have yielded.
Example 10
A ‘T’ beam has effective flange width 900 mm, flange thickness 150 mm, web
width 250 mm, and overall depth 500 mm. characteristic strength for concrete
30 N/mm2
and for steel 460 N/mm2
. Assume that the centroid of reinforcement
is placed 50 mm from the top or bottom of the beam, design suitable
reinforcement for ultimate design moments:
a) 300 kN.m,
b) 700 kN.m,
c) 800 kN.m
Solution
a) Determine the location of the neutral axis:
35

36. the moment resistance of the flange (Mf):
Mf= 0.45 fcu bf hf (d-hf/2)
= [ ] 6
10)2/150(450(1509003045.0 −
∗−∗∗∗∗
= 683.437 kN.m > the applied moment (=300 kN.m)
∴the neutral axis lies within the flange
cu
2
fbd
M
k =
=
30450900
10300
2
6
∗∗
∗
= 0.054 < /
k (=0.156)
∴ the section is singly reinforced section
}k/0.9-0.25d{0.5z +=
= }0.054/0.9-0.25d{0.5 +
= 0.935d < zmax (=0.95d)
4500.935z ∗=
= 420.75 mm
zf95.0
M
As
y
=
=
75.42046095.0
10300 6
∗∗
∗
= 1631.6 mm2
b) Mf (=683.437 kN.m) < 700 kN.m
∴the neutral axis lies within the web
2
wcufwffcu dbf155.0)0.45h-d)(b(bh0.45fMc +−=
36
1
2 2
x

37. = [ ] 62
1045025030155.0)15045.0450)(250900(1503045.0 −
∗∗∗∗+∗−−∗∗∗
= 738.87 > 700 kN.m
∴the section is singly reinforced section
hf (=150 mm) < 0.5d (0.5∗450=225)
)0.5h-d(f95.0
)h-d(0.45db0.1fM
As
fy
fwcu+
=
=
)1505.0450(46095.0
)15045045.0(450250301.010700 6
∗−∗∗
−∗∗∗∗∗+∗
= 4379 mm2
c) Mc (=738.87 kN.m) < 800 kN.m
∴the section is doubly reinforced section
x = 0.5d
= 0.5∗450
= 225 mm
37.0222.0
225
50
x
d/
<==
∴the compression reinforcement will have yielded
)d-(d0.95f
Mc-M
As /
y
/
=
=
)50450(46095.0
10)87.738800( 6
−∗
∗−
= 349.7 mm2
/
y
wffcuwcu
As
f95.0
)b-b(h0.45fdb0.2f
As +
+
=
= 7.349
46095.0
)250900(1503045.0450250302.0
+
∗
−∗∗∗+∗∗∗
= 4906 mm2
37

38. 1.10 Minimum and maximum percentages of reinforcement in
beams and minimum clear spacing between bars:
As stated in cl. 3.12.5 of the BS 8110:Part 1:1997, the minimum percentages
of tension and compression reinforcement of rectangular and flanged beams
are given in Table 3.25 of the BS 8110. While the area of tension or
compression reinforcement should not exceed 4% of the cross sectional are
of the beam as stated in cl. 3.12.6.1 of the BS 8110.
As stated in cl.3.12.11.1 of the BS 8110, the horizontal distance between bars
should be not less than the maximum size of coarse aggregate (hagg.) + 5 mm,
while the vertical distance between bars should be not less than 2hagg./3.
Example 11
A simply supported beam of 7 m span carries a characteristic dead load of 10
kN/m and live load of 5 kN/m. The beam dimensions are breadth 250 mm,
effective depth (d) 450 mm and overall height (h) 500mm. The concrete grade
is 35 N/mm2
and the steel grade is 460 N/mm2
, while the nominal cover is
37.5 mm. The maximum aggregate size is 20 mm, design the beam to resist
the design flexural moment
Solution
Design load = 1.4∗10+1.6∗5
= 22 kN/m
38

39. 134.75 kN.m
Ultimate moment =
8
2wl
=
8
722 2
∗
= 134.75 kN.m
cu
2
fbd
M
k =
=
35450250
10134.75
2
6
∗∗
∗
= 0.076 0.156)(k =′<
∴singly reinforced section
}k/0.9-0.25d{0.5z +=
= }0.076/0.9-0.25d{0.5 +
= 0.906d < zmax (=0.95d)
450906.0z ∗=
= 407.7 mm
0.45
z)-(d
x =
39
7m
22 kN/m

40. =
0.45
407.7)-(450
= 94 mm < 0.5d (0.5 ∗450=225mm), ∴tension steel reinforcement will have
yielded
zf95.0
M
As
y
=
=
7.40746095.0
1075.134 6
∗∗
∗
= 756.32 mm2
100
50025013.0
Asmin
∗∗
=
= 162.5 mm2
500250
100
4
Asmax ∗∗=
= 5000 mm2
)5000(As756.32)As()5.162(As maxmin =<=<=
use steel bar size 25 mm; Area of one steel bar = =∗µ2
5.12 490.87 mm2
No. of steel bars =
87.490
32.756
= 1.54 2≅
As provided = 2∗490.87 = 981.74 mm2
spacing between bars =
1
2522.372250 ∗−∗−
= 125 > hagg max. +5 (=20+5=25mm)
40
2Y25