good

1. Design of RCC Structures to
BS8110
Part E: Design of Rectangular Beams

2. Choice of Section
Considerations
-avoid compression steel
-reduce tension steel (large d values)
Depth –d
-Choose such that the section is singly reinforced
-for initial beam sizing we can use trial span/depth ratios
cantilever---------------------------6
simply supported -----------------12
continuous-------------------------15
If span >10m multiply above by 10/span (m)
Width -b
in general choose ‘b’ such that
𝑑
1.5
≥ b ≥
𝑑
2.5
(say b=d/2)

3. Singly reinforced section
C=0.45fcu.0.9x.b=0.405fcubx
T=0.87fyAs (assuming steel has yielded, i.e x/d ≤ 0.5)`
For equilibrium 0.405fcubx=0.87fyAs x = 2.15 (fy/fcu)(As/b)
hence x/d = 2.15 (fy/fcu)(As/bd)
z=d-0.45x and M=C.z=0.405fcu.bx(d-0.45x)= 0.405 x/d (1-0.45x/d)fcubd2
M=Kfcubd2 and M=f(x/d)=f(a)
And we have
𝑑𝑀
𝑑𝑎
= 0 at a=x/d=1.11; where
𝑑𝑀2
𝑑𝑎2 <0 therefore from x/d= 0 to x/d= 1.11 M will increase
Maximum allowable x/d=0.5 (for under reinforced section)
Then M-=k’fcubd2 if applied moment is M and assuming b=b/2, the minimum depth required
M=0.156fcu. (d/2)d2
dmin =
3
12.82𝑀/𝑓𝑐𝑢 choose d>dmin
d
b
C
0.45x
0.45fcu
0.9x
T
Z=d-0.45x
K
v

4. Design Formulae
Eqn V -- K=(0.405
𝑥
𝑑
)(1-0.45
𝑥
𝑑
) --------------------------- VII
also z=d-0.45x
hence x =
𝑑−𝑧
0.45
, then above become
K=(
0.405
0.45
)(1-
𝑧
𝑑
)(
𝑧
𝑑
)
(
𝑧
𝑑
)2 – (
𝑧
𝑑
) + 1.111𝐾 = 0
Then z/d =
1+ 1−4.444𝐾)
2
= 0.5 + 0.25 −
𝑘
0.9
------- z = d 0.5 + 0.25 −
𝑘
0.9
T.z = M, 0.87fyAs.z=M , Then As = 𝑀
0.87𝑓𝑦𝑧
As BS8110 suggested for the yielding of tension steel x/d≤0.5
i.e permissible maximum x=0.5d
Eqn vii------- Kmax=K’=0.405x0.5(1-0.45x0.5)=0.156
As =
𝑀
0.87𝑓𝑦𝑧
K=
𝑀𝑢
𝑓𝑐𝑢𝑏𝑑2
provided that
K ≤ 0.156

5. Design Formulae
Notes
i. The above formula can be used only if K≤K’.
ii. K’=0.156 if % redistribution > 10%. Otherwise K’=0.402 (βb-0.4)-
0.18(βb-0.4)2 ; where βb =
100−(% 𝑟𝑒𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛)
100
,
iii. K=fcubd2= applied ultimate moment
iv. K′fcubd2 =
maximum moment that can be carried as a singly reinforced section.
v. Z should not be taken as greater than 0.95d. (only good concrete
considered).

6. Design Formulae (Doubly reinforced section)
Cs=0.87fyAs’
Cc
Ts=0.87fyAs
As’
As
b
h
d’
d
0.45fcu
Provided d’/d > 0.21
0.9x=0.45d
taking
𝑥
𝑑
= 0.5
Cc = 0.45fcu(o.9x)b
= 0.45fcu(0.9)(0.5d)b
= 0.2fcubd
As1
As2
As’
≡ +
As = As1+As2
M = K’fcubd2 + 0.87fyAs’ (d-d’); where K’=0.156
As’=
𝑘−𝑘′ 𝑓𝑐𝑢𝑏𝑑2
0.87𝑓𝑦(𝑑−𝑑′)
As=
𝐾′𝑓𝑐𝑢𝑏𝑑2
0.87𝑓𝑦𝑧
+ As’

7. Design Formulae
Notes
i. It can be also shown that in general z = d 0.5 + 0.25 −
𝐾′
0.9
for
section requiring compression steel.
ii. Kfcubd2 = applied moment
K’fcubd2 = maximum moment for a singly reinforced section (where
K’=0.156)
iii. If K ≤ K’ singly reinforced section
K > K’ doubly reinforced section

8. Design Formulae (Flanged beams)
 Monolithic beam-slab construction facilitate flanged beam action.
 Flanged beam action is effective only if slab helps to take compression
(eg. In sagging moment conditions for down stand beams).
Bf (L –beam) Bf (T –beam)
Flanged beam
action
Rectangular
beam action

9. Design Formulae (Flanged beams)
Design Procedure
1. Choose d assuming that we are designing rectangular beam (take b ≈
d/2, d > dmin ).
2. Calculate x/d, using b=bf
z/d = 0.5 + 0.25 −
𝐾
0.9
; where K=
𝑀
𝑓𝑐𝑢𝑏𝑓𝑑2
also Z=d-0.45 x
It can be shown x/d=1.11- 2.22 (0.25 −
𝐾
0.9
)

10. Design Formulae (Flanged beams)
Design Procedure
Choose
‘d’
assuming
rectangular
beam
Calculate
x/d
using
b=bf
Is
x/d
≤
h
f
/0.9d
Is
x/d
≤
0.5
Singly
reinforced
Doubly
reinforced
Design
as
a
rectangular
beam
with
b=b
f
YES
NO
YES
NO
Is
compression
within flange
Calculate
indirectly
Compression
is in web too

11. Design Formulae (flanged beam-compression
in web too)
d
b b
hf
0.9x
bw bw
As As
Fig 1 Fig 2
As =
𝑀+0.1𝑓𝑐𝑢𝑏𝑤𝑑(0.45𝑑−ℎ𝑓)
0.87𝑓𝑦(𝑑−0.5ℎ𝑓)
If βffcubd2 is the largest moment that can be taken by above
section (i.e. at x/d=0.5), then using fig 2 we can show
Βf = 0.45
ℎ𝑓
𝑑
(1 −
ℎ𝑓
2𝑑
)(1-
𝑏𝑤
𝑏
)+0.15
𝑏𝑤
𝑏

12. Design Formulae (flanged beam-compression
in web too)
Notes
1. If applied moment M is less than maximum capacity (βffcubd2 ), then x/d < 0.5 and
then above equation can be used to find As.
2. βf values are given in table 3.6.
3. If applied moment M> βffcubd2 then x/d >0.5 for a singly reinforced section and
then section shall be designed as doubly reinforced section to keep x/d=0.5
then;
A’s =
𝑀+βffcubd2
0.87𝑓𝑦(𝑑−𝑑′)
As = A’s +
{ 𝑏−𝑏𝑤 ℎ𝑓+0.45𝑏𝑤𝑑}(0.45𝑓𝑐𝑢)
0.87𝑓𝑦
Cl 3.4.4.4:
BS8110, prt 2

13. The End