UNIT 11. DESIGN AND DETAILING
BEAMS & LINTELS
OF
Structure
1 1.1
Introduction
Objectives
1 1.2
1 1.3
1 1.4
1 1.5
Q p e s of Beams and their Design
Design of Lintels
,
11.3.1
General
11.3.2
Evaluation o f Loads on a Lintel
Scmmary
Answers to SAQs
11.1 INTRODUCTION
Design of a beam or a lintel means fixing the size of concrete section, determining the
areas of tensile as well as compressive (if necessary) reinforcements, provision of shear
and torsion reinforcements and curtailment of tensile reinforcements. Analysis for design
is based on design (actual) load. The safety and serviceability is ensured by keeping
stresses deflection and cracking below the permissible limits. The detailing iules
enunciated in the code ensures complex, unevalulated or semi empirical stresses below
the allowable liinits.
Objectives
i
After going through the text and illustrations of this unit, you should be able to design a
simply supported beam, a cantilever beam, a continuous beam, and a lintel.
11.2 TYPES OF BEAMS AND THEIR DESIGN
I
1
From analysis and detailing point of view, a beam may be of one of the following types:
(a)
Simply supported beam
(b)
Cantilever beam
(c)
Continuous beam, and
(d)
Overhanging beam
I
i
I
I
1
For the design of an overhanging beam, the detailings of overhang portion can be done as
a cantilever beam and that of supported.side as at intermediate support of a continuous
beam. Hence, only first three types of beams have been illustrated here.
Example 1111.11
Design a simply supported rectangular beam of clear span 6m and supported on
375 thick walls. The beam is loaded with a dead load of 16 kNm and a live load
of 12 kNm. Use M 15 concrete and Fe 250 steel.
Working Stress Method
Solution
Oeprh ( D )
(i)
From Thumb Rule
lef and 1
D lying between -
10
20
lef
Assuming =10 and taking le, 6 + 0.375= 6.375m
D
Accordingly D = 700
1
2
and b (taken between - rd to - rd of D)= 400
3
3
(ii)
From control of deflection criteria
d>
K~K1K2K3
ler= 6 + 0.375 = 6.375m at the outset as effective depth is not known
K,= 20
K,for 0.714% balanced steel = 1.68
Kz= K 3 =
1
Calculation of Loads
Self = 0.70x 0.40x 1 x 25
Total DL
= 23.00 kNm
=
Total Loads
12.00kNm
= 35.00 kNm
wlef2 35 x 6 . 3 7 ~ ~
Maximum B.M.,
M = -- =
= 170.80 kNm
9
8
Taking steel in two layers and keeping a clear vertical spacing of 25 mm in
between the layers,
D = 716.103+ 25 + 20 + 12.5 = 773.53 775
Hence, provided D = 775 and d = 775- 25-20- 12.5 = 717.5
:.Ier = 6 + 0.375m= 6.375m
Design and Detailing of
Beams & Lintels
Figure 11.1 :Designed Section of the Beam
Check for b
lC=6000<60b
or
b > 100
Again
or
b > 131.23
Hence provlded b = 400
Provision of Reinforcement
Provided 8@20in two layers (A, = 25 12 mm2)
Evaluation of Moment of Resistance of the Section
Taking moment of area about n.a.
Working Stress Method
Hence the section is over reinforced.
= 191.07 kNm > 177.80 kN. Hence, O.K.
Check
Loads
Total D L
LL
= 12.00 kN/m
Total D L
+ LL
= 35.75 kN/m
Provision of Shear Reinforcement
Figure 11.2 : S. E D. for Design
SF at face of support
Taking 50% of tensile steel being extended to the support,
Design and Detailing of
Beams & Lintels
or7, = 0.273 MPa
tc,max
= 1.6 MPa for M 15 concrete
since 2,<2, < 2,. max
Vs = V-7,bd = 107.25- 0.273 x 400 x 717.5 x10s3= 28.90 kN
Providing @ 8 two-legged stirrups
Spacing s,, shall also be less than the followings :
(ii)
s V = 0.75, d = 0.75 x 717.5
(iii)
sv = 450
= 538.13
Hence provided 08 two-legged vertical stirrups (3135
Curtailment of Reinforcement
MRfor @20
Let upper 4 bars be curtailed. Then the effective depth for the remaining bars,
d=775-25-10=740
Taking moment of area about n.a. (Figure 11.3)
Hence the section is over-reinforced,
Working Stress Method
Figure 11.3 :Section of the Beam after Curtailment
Let x' be the distance from support where be above M, occur, (Figure 11.4)
6.375
i.e. theoretical cut-off section from mid span = -1.27 = 1.918m
2
Actual cut off point from mid span = 1.918 + ( lesser of 12@or d)
= 1.918 + 0.740 = 2.658m
Length available from the actual cut-off point to the centre line of the support
Therefore, portion of L, to be provided beyond centre line of support
Available straight portion of proposed extension beyond centre line of support,
X
=
Support width
- (clear cover + dia of bar of bend
2
+ radius of bend)
Length of straight portion to the provided beyond the bend,
e=
-
Total extension reqd. beyond centre line of support - equivalent length of
bend-available straight portion beyond centre line of support ,
Check
(i)
Extension of positive reinforcement into the support
O.K.
M
(ii)
At support Ld 5 1.3-
v
+
L,
= 1.167 m
MI
= 115.853 kNm
V
= S.E at simple support due to design load
where
and
= 35.75 x 6.37512 = 113.95 kN
and
= 0.637 m
Lo
Substituting these values in the above equation
< 1.96 m Hence O.K.
Provision of stirrups in excess at cut-off point
As per Code ( cl 25.2.3. 2b) distance beyond cut-off point upto which shear
3
stirrups in excess to be provided = - d
4
3
= - x 7 4 0 = 555 > 342.5 (distance of
4
face of support)
0.4bs
Excess area,
4ve
=
fy
(for 08 - two-legged vertical stirrups, AsV+ Asve= 100)
Hence provided 08 two-legged vertical stirrups @ 105
The details of reinforcement is shown in Figure 11.4
Design and Detailing of
Beams & Lintels
Working Stress Method
Figure 11.4 : Curtailment and Detailing of the Reinforcement
Example 11.2
Design a cantilever beam shown in Figure 11.5 below. Use M 15 concrete and
Fe 250 steel.
Figure 11.5 :A Cantilever Beam Projecting from a RCC Column
Design and Detailing of
Beams & Lintels
Solution
Effective span
Le,= 4m
Depth (D)
(i)
From Deflection Criteria
where K, = 7
K, = 1.68for 0.714% of balanced steel
K, = K, = 1
Taking D = 850 ; b = 400
and d = 850- 50 = 800
Check forb
le, < 25 b
100b2
Again le, < d
(ii)
From Moment of Resistance Consideration
Loads
Self = 0.85x 0.4x 1 x 25
-
8.50kN/m
Super imposed load
-
40.00 kN/m
-
388.00 kNm
Total load
wle: = 485 x 4
'
Maxm. B.M = M =2
2
Taking D = 1150
d = 1150 .- 25 -20- 12.5 = 1092.5
48.30 kN/m
--
Checking for Lateral Stability
Working Stress M e t h o d
Loads
Self = 1.15 x 0.4 x 1 x 25
--
11.50 kN/m
Super imposed load
-
40.00 kN/m
Total load
-
5 1.50 kN/m
Taking 0 = 1150
d = 1150- 25-20-
12.5 = 1092.5
Hence provided 12@20in two layers ( A , = 3770 mm2)
Curtailment of Steel
Curtailing all the six @20bars of the lower layer (Figure 11.6)
d = 1150-50-10= 1115
and taking moment of area about N.A.
k=
-35796k J3.321138562~10"
2 x 223000
k = 0.328 < 0.404 (k,)
Hence the section is under-reinforced.
Design and Detailing of
Beams & Lintels
Figure 11.6 :Section of the Beam after Curtailment
Let x be the distance from free end where 6 lower bars can be curtailed
(Figure 1 1.5)
Actual cut-off point from free end
= 3.189-(12$or d)
= 3.189- 1.1 15 = 2.074 > 1.167m (L,)
i.e. distance of actual curtailment fromjxed end = 4-2.074 = 1.925 > 1.167
(L,) .....(Figure 1 1.7)
Inside the support, length of the bar to be provided *= L, = 1167 = 5 15 + 8 0 + x
or
x = 492 = length of straight bar to be extended beyond bend
Provision for Shear
(i)
Ar support
Maximum Shear Force. V = wl,, = 51.5 x 4= 206 kN
*
As the width of support = 600, straight portion of bar inside the column ( Figure 11. 7)
= 600 - (clear cover + dia of bar bend + radius of bend)
~ 6 0 0 - ( 2 5 + * + 2p) = 600- ( 2 5 + 3 x 20) = 5 1 5
Working Stress Method
Figure 11.7 : Extension of Main Reinforcement in the
Support for Development Length
Adopting @8 two-legged verical stirrups
From minimum shear force reinforcement consideration, maximum
spacing of stirrups are given by
(b)
sV= 0.75 d = 0.75 ~ ' 1 0 9 2 . 5= 819.375
(c)
sV=450mm
.'. Provided (i8 two-legged vertical stirrups @ 135
(ii)
At cut off point
Total shear reinforcement
(Adopting Q8 two-legged verical stirrups, AsV+ Asv, = 100.53 mm2)
sV= 103.48
Hence provided 48 two-legged vertical stirrups @ 100 for a distance of
3
X 11 15 = 836.25(=900) from actual cut-off point section
4
As D > 750, side reinforcement reqd., AS = 0.1 % of Total Web Area
0.1
= -x
100
400x 1150 = 460 mm2 i.e. 46012 = 230 mm2 on each face at
spacing not exceeding 300.
The detailing of reinforcement has been shown in Figure 11.8.
Figure 11.8 :Detalling of Reinforcement
Example 11.3
Design a continuous beam for the following data Figure 11.9 :
Super imposed dead load= 10 kNm
I
Super imposed live load = 12 kNm
Width of support
= 350 m m
Use M 15 concrete and Fe 250
Design and Detalling of
Beams & Untels
Working Stress Method
Figure 11.9 :A Continuous Beam
Solution
Depth (D)
(i)
From Thumb Rule
lef
-lying
d
between 10 to 20
ref
Let = 10
d
(ii)
From deflection criteria
Corresponding to p,% = 0.714 , K, = 1.68
K2= KJ = 1
(iii)
From Moment of Resistance Consideration
Taking D = 760 mm
Loads
Self load = 0.76 x 0.3 x 25
-
5.70 kN/m
--
12.00 kN/m
DL
Total DL
LL
Total Load (DL + LL)
*
Here K,, is the average basic value for -for
lef
d
27.70 kN/m
simple and continuous supports
Evaluation of BMs and SFs at Critical ~ e c t i o n s p i ~ u 11.10)
re
(vide. Table A1 & A2 of Appenidx A of Unit 6)
Due to DL only
Ma= 0.125~15.75~6
(a) Span and Support EMS.
@) Maximum S.Fs. at the Supports for DL
Due to LL only
l2Wm
M, = 0 . 1 2 5 ~ 1 2 ~ 6 '
= 54 LNm
(c) Span and Sopport B.Ms. for LL
(d) Muimum S.F. at the Supports for LL
(e) Span and Support B.Ms. for DL+LL
(fJ Maximum S.F. at the Supports for DL+LL
Figure 11.10 : Evaluation of R. Ms and S. Fs at Critical Sections
Design and Detailing of
Beams & Lintels
Working Stress Method
Maximum B.M.(M) occurs at intermediate support,
D = 692.27 + 25 + 10 = 727.27
Provided D = 760; d = 760- 25- 10 = 725
M
a,,jBd
Ast = --
124.65x lo6
= 1416.47 mm2
140 x 0.867 x 725
Hence provided 5420 (AS, = 1570 mm2) at intermediate support
Moment of Resistance of the Section for Intermediate Support Section
(Figure 11.1 1).
Figure 11.11 :Section at Support
Taking moment of area about N.A.
kd
b x k d x -=mA,(d-kd)
2
300
-xk2x725=19x314x5(1-k)
2
108750 k2 + 29830 k- 29830 = 0
k = 0.404 = k, (=0.404)
#.'
M,,= ostA,, jd
= 140 x 5 x 314 x 0.867 x 725
= 138.16 kNm
For +ve B.M
Hence provided 5Q16as positive reinforcement at mid span
Moment of resistance of the section at mid span
Taking moment of area about N.A. (Figure 1 1.12).
Figure 11.12 :Section at Mid Span
b x kd x
kd
2
- = tnA, ( d - kd)
300
2
-x kZx727 = 19 x 1005.3 x 5 ( 1 - k )
109050 kZ + 19100.7 k- 19100.7 = 0
k = 0.34
k < k, (=0.404)
The section is under-reinforced
.:
M =u,,
A, jd
= 90.72 kNm>
The detailing of the reinforcement has been shown in the Figure 11.13.
Design and Detailing of
Beams & Linlels
Working Stress Method
0.6m
ff
=O.ll,
0.151,4.%
1
.
0.15 I@.%
I-+
3
%
fiiG3
5--1
Figure 11.13 : Reinforcement Detailing for the Designed Beam
11.3 DESIGN OF LINTEL
11.3.1 General
Linrel is a beam provided over an opening to support louds from masonry (allowing for
*
arching and dispersion, where applicable) and from any other part of the structure.
The design of a lintel is done exactly in the same way as that of a rectangular beam except
that the minimum shear reinforcement requirements as per code may not be complied with
where the maximum shear stress calculated is less than half the permissible value.
1
Length of bearing of lintel on each side shall neither be less 90 rnrn nor less than -- th of
10
the span.
Area of bearing on each side should be sufficient to transfer the reaction of lintel and laod
due to arching action without exceeding the permissible stresses on masonry.
11.3.2 Evaluation of Design Loads on a Lintel
The selfweight of a wall or any other load from structural component upon it over an
opening are transferred to the sides of the opening through a lintel and by arching action.
The proportion of the above mentioned loads carried by lintel depends upon support
condition, height of masonary above the opening, location of concentrated or distributed
gravity loads etc.
For arching action to take place, masonary must have good shearing strength, good bond
and enough spread of wall on both sides of an opening.
Arching action transfers the loads of floor and rnasonary above the equilateral triangle to
the sides of wall. Any other opening or load below the horizontal plane, 250 above the
apex of the dbc,effects the loading on the lintel (Figure 11.14)
Arch~ngi s a phenomenon by which part o f the load over an opening in the wall gets
transferred to the s ~ d e sof the opening.
Design and Detailing of
Beams & Lintels
Figure 11.14 : Diagrammatic Representation of Arching Action and Design Load
when I, & I, 2 I,! 2 and h 2 0.867 1, + 250
Keeping above facts in view design loads on a lintel is evaluated as under :
Condition I
When wall above the opening, h L (0.867 le, + 250) high, and supporting
lef
walls are spread for more than - on both sides of the opening
2
Figure 11.14
Due to arching action shown by thick dotted line Figure 11.14, the load of masonry on
the lintel (W) is only that of an Equilateral Triangular Portion of wall formed on effective
span (I,) as base regardless of any other opening or load outside the dotted arch.
In other words,
Design Load,
W = Weight of Equilateral Aarportion of masonary wall on lintel.
where t = thickness of wall and p = density of masonry
Wl,f
1 lef W 1 -1 = Maximum B.M., M m = - W--2 2 2'2'3
6
W
and maximum S.F., Vmax= 2
+
*
lc = clear span of the opening
leris the same as that for simply supported beam given in the Code.
Working Stress Method
Condition I1
lef
When 1, and /or l2 < 2
In case if wall on any one or both sides of the opening spreads less than half the effective
span of lintel, the design load is equal to the sum of the masonry and floor load falling
above the effective span of the lintel- regardless of the height of the floor slab
(Figure 11.15)
In other words.
W = Ie, x h x p x r + le, x Floor LoadIUnit length
rFloor
Load on Lintel
Level
Figure 11.15 : Design Load on Lintel when 1, and I or I, < Ie,/ 2
lef andfloor or roof slab falls within a part of the triangle
Condition 111 When I , & 1, 2 2
In this case design load consists of trapezoidal portion (ABCD) of the triagnular load
below the floor slab and floor load falling within the triangle. If the wall continuous on
the upper floor or there is a parepet wall over the wall, the triangular portion (EFG) of the
wall will be included in the design load (Figure 11.16).
In other words,
Wt. of masonry wall (ABCD) (or wt. of masonary wall ABCD +
wt. of rnasonary wall EFG if the wall continuous above floor) +
length CD x wt. of floortunit length
Design load,
W=
Condition IV
lef
When I, & l 2 2 -and
h < 0.867 and another opening comes within
2
the horizonal plane 250 above the A"( Figure 1 1.17)
Design and Detailing of
Beams & Lintels
The lintel of floor A may be designed for
Design laod W, =
Masonry Load of Storey A on lintel (Area ABGH)
-I-
Load of floor A on CD
Load of floor B plus masonry load of storey B on lintel
i-
Masonry Load on lintel of floor B on lee,
Figure 11.16 : Design Load when I, & I, 2 I,/ 2 and h < 0.867 Ier
C.G. ofother Losd within the Horizontal rime at 250
c a b o v e l e bkor Lows
LodofMarony on
Account of lnflumce
the a h e r L o d
Lod from Otkr Lmd
within rhe Influence
Assumed Udl and Di
a1 30' hom rhc Vcrti
Floor
Level
Figure 11.17 :Load when l1 & 12 2 l e f / 2 and the A''~ on Lintel
is within the Influence of any other Load
Working Stress hfc khod
ConditionV
lef
When 1, & 1, 2 -and
any other load lying between the lintel and
2
hurizuntal plane 250 above the triunglr ( Figure 1 1.16)
Floor B
ad of floor B Plvr
Msronry Loed of Lin
Figure 11.18 :Load on Lintel when I, & I,> lA*,I 2 and another opening on Upper
Floor falling within Horizontal Plane 250 above the Equilateral
Design Load, W = Masonry Load of equilateral triangle over the lintel
Masonry load on account of influence of the other load
Load from the other load within the influence of equilateral
triangle assuming udl and dispersed at an angle of 30' from the
vertical
Example 11.4
Design a lintel for 2rn wide central opening in a brick wall of total length of 3.5m
of a room of total inside dimension 3.5m x 3m for the following data :
(1)
thickness of wall
= 40
(ii)
thickness of R.C. roof slab
= 120
(iii)
Lime terrace thickness
= 150
(iv)
parapet wall over the roof of Im height and 100 thick
(v)
Height of wall above lintel
= l .OO m
Use M 15 concrete and Fe 250 steel
Solution
4,
As the spread of the wall on both sides of the opening are less than --and
height
2
of wall above lintel is less than 0.867 le,, the load on the lintel will be that a s
shown in condition 11. Assuming total thickness of lintel 270 and effective cover
35 and bearing on each side = 250.
I , = 2 + 0.235 = 2.235 m
Design and Detailing of
Beams & Lintels
or
I , = 2 + 0.25 = 2.25 m
.. let= 2.25 m
Loads
self weight of lintel
= 2.25 x 0.4 x 0.27 x 25
wt. of wall above lintel = 2.25 x 0.4 x 1 x 18.85
wt. of slab
= 2.25 x 1.5 x 0.12 x 25
wt. of lime concrete
= 2.25 x 1.5 x 0.15 x 19
live load
= 2.25 x 1.5 x 1.5
Total Load W
:
, Maximum B.M.
Design coefficients
For balanced section
:.
D = 196.2 + 25 + 6 = 227.2 = 270 (adopted)
.'. d = 270- 25*st
6 = 239 (using I$ 12 bars)
=
-
6.075
kN
16.965
kN
10.125
kN
9.619
kN
5.063
kN
47.847
kN
Working Stress Method
Hence provided 5412 bars ( A , = 565.49 mm2> 460.16 mm3
Bent up 2)12 at
7
=
2.25
= 300 frbm inner edge of support
7
Provision for shear
S.E at d from inner edge of support
n
At support, A , = 3 x - x 12'= 339mm2
4
From Table 8.3
7,
Since zv a,,but zv > - only nominal reinforcement will be provided.
2
Adopting f6 two-legged vertical stirrups
or
0.87fy4
0.46
s, I
v
- 0.87 x 250 x 56 = 76.125
0.4 x 400
The spacing of stirrups shall not also exceed the following
(i)
0.75 d = 0.5 x 233 = 179.25
(ii)
450
Hence provided 46 two-legged retical stirrups @ 75 througout
Check<for Development Length of Support
with 3Q12 at support the section is under-reinforced, and therefore first of
all n.a. depth be calculated to know M,
Taking moment of equivalent concrete areas about n.a.
Design .and Detailing of
Beams & Lintels
or
L
Lo = S-x'
2
+ 13$ for standard hook
where Ls = support width = 250
x'= side cover = 25
1.3-Ml
V
+ Lo = 1.3 10.187~lo6 + 256 = 809.4 mm > 699.6 Hence O.K.
2 3 . 9 3 ~lo3
SAQ 1
(i)
Design and detail a simply supported K.C. beam of clear span 7m
supported on 400 mm wall on both sides. The beam is loaded with a dead
load of 20 kN/m and a live load of 12 kN/m. Use M 20 concrete and
Fe 415 steel.
(ii)
Design and draw a R.C. cantilever beam of 3.5 m span loaded with
45 kN/m of design laod inclusive of self load. The beam is projecting
from a R.C. column. Use M 20 concrete and Fe 415 steel.
(iii)
Design and detail a lintel for the specifications of Example 11.4 except
that the wall horizontal length is 5m instead of 3.5m.
Working Stress Method
11.4 SUMMARY
Design of three types of recatangular R.C. beams have been done for design (actual or
charactenistic) loads keeping in view their safety by ensuring that permissible stresses
(allowabie stresses in working stress design) are no where exceeded. Secondly, while
fixing the size of concrete and steel and detailing the reinforcements. control of deflection
and cracking (serviceability limit) were the main concern. The design of a lintel is the
same as that of a simply supported beam except that the loading over it is different due to
arching action. Variation in loading patterns on lintel due to different parameters and
evaluation of corresponding design load have been explained.
-
-
11.5 ANSWERS TO SAQs
SAQ
1
(i)
Refer Example 11.1
(ii)
Refer Example 11.2
(iii)
Refer Example 11.4 using condition 111of 11.3.2
-
-
-