1. DESIGN OF RCC LINTEL
-ARUN KURALI

2. Lintel
A lintel is a horizontal member which is placed across the openings like doors, windows etc.
It takes the load coming from the structure above it and gives support. It is also a type beam, the width of which is equal to the width of wall, and the ends of which are built into the wall. These are very easy to construct as compared to arches.
Bearing of lintel: The bearing provided should be the minimum of following 3 cases.
i) 10 cm.
ii) Height of lintel beam
iii) 1/10th to 1/12th of span of the lintel.

3. Types of Lintels Timber lintel Stone lintel Brick lintel Steel lintel
Reinforced concrete lintel
Reinforced brick lintel

4. Timber Lintels
In olden days of construction, Timber lintels were mostly used. But now a days they are replaced by several modern techniques, however in hilly areas these are using. The main disadvantages with timber are more cost and less durable and vulnerable to fire.
If the length of opening is more, then lintel is provided by jointing multiple number of wooden pieces with the help of steel bolts which was shown in fig (a). In case of wider walls, lintel is composed of two wooden pieces kept at a distance with the help of packing pieces made of wood. Sometimes, timber lintels are strengthened by the provision of mild steel plates at their top and bottom, called as flitched lintels.

5. Stone Lintels
These are the most common types of lintels especially where stone is abundantly available. The thickness of these are most important factor of its design. These are also provided over the openings in brick walls. Stone lintels are provided in the form of either one single piece or more than one piece.
The depth of this type is kept equal to 10 cm / meter of span, with a minimum value of 15 cm. They are used up to spans of 2 meters. In the structure is subjected to vibratory loads, cracks are formed in the stone lintel because of its weak tensile nature. Hence caution is needed.

6. Brick Lintels
When the opening is less than 1m and lesser loads are acting, brick lintels are used. The depth of brick lintel varies from 10 cm to 20 cm, depending up on the span. Bricks with frogs are more suitable than normal bricks because frogs when filled with mortar gives more shear resistance of end joints. Such lintel is known as joggled brick lintel.

7. Reinforced Brick Lintels
If loads are heavy and span is greater than 1m, then reinforced brick lintels are useful.
The depth of reinforced brick lintel should be equal to 10 cm or 15 cm or multiple of 10 cm. the bricks are so arranged that 2 to 3 cm wide space is left length wise between adjacent bricks for the insertion of mild steel bars as reinforcement. 1:3 cement mortar is used to fill up the gaps. Vertical stirrups of 6 mm diameter are provided in every 3rdvertical joint. Main reinforcement is provided at the bottom consists 8 to 10 mm diameter bars, which are cranked up at the ends.

8. Steel Lintels
If the superimposed loads are heavy and openings are large then we can go for steel lintels.  These lintels consist of channel sections or rolled steel joists. We can use one single section or in combinations depending up on the requirement. When used singly, the steel joist is either embedded in concrete or cladded with stone facing to keep the width same as width of wall. When more than one units are placed side by side, they are kept in position by tube separators.

9. Shear stirrups are provided to resist transverse shear
Reinforced Cement Concrete Lintels
At present, the lintels of R.C.C are widely used to the openings for doors, windows, etc. in a structure because of their strength, rigidity, fire resistance, economy and ease in construction. R.C.C lintels are suitable for all the loads and for any span.
The width of lintel is equal to width of wall. main reinforcement is provided at the bottom and half of these bars are cranked at the ends.
Shear stirrups are provided to resist transverse shear

10. Types of Design Loads on lintel
Self-weight or dead load of the lintel
Dead load of the wall above the opening
Dead load and live load transferred from the roof or the floor supported by the wall over the opening
Self-Weight or Dead Load of Masonry Lintel
All solid masonry and concrete lintels must be adequately grouted and the dead load of the lintel can be computed if the enough information about the dimension of the lintel is available. For the design purposes, initial or preliminary cross sectional dimensions for the lintel are considered. Since, lintels are an important part of the wall so the width of the lintel is the same as the wall and the only unknown that is required to estimate the self-weight of the lintel is its depth. Depth of approximately 20 mm per linear of 300 mm of the span can be considered for preliminary design. Not only does the computation of masonry self-weight depend on masonry unit types for example light weight, medium weight, or normal weight but also on the unit weight of the grout that is employed for the wall and it may be taken as Kg/m3 or Kg/m3. As an alternative for the method of dead load calculation mentioned in the above section, the dead load of the lintels that have specific height and breadth which are offered by NCMA could be utilized and considerable errors can be avoided.

11. Dead Load of the Wall Above the Lintel
The dead load masonry above the lintel is the weight of masonry contained in forty-five-degree triangular area above the lintel if arching action is considered to occur. Consequently, the dead load for which the masonry lintel must be design for consists of masonry dead load in triangular area plus self-weight of masonry lintel. It may be claimed that, the degree which made the triangular area above masonry lintel vary from 45 to 60 degree. The dead load for case where triangle is formed due to 60o over lintel effective span is greater than dead load of masonry contained in triangle created by 45o over the effective span of the lintel. It is advised to employ the triangular formed by 45o for the calculation of wall dead load over the lintel

12. Design steps Assume effective depth of lintel d= 𝑙 10
overall depth D = effective depth + effective cover
D = d + Effective cover=d+50
2. Width of lintel, b = Thickness of masonry (eg: 230 to 300mm)
3. Effective span (l)
Must be least of the following two
1= clear span (L)+ bearing
l = clear span(L) + effective depth
4. Load and moment calculation
transfer of loads to the sides by arch action height of wall above Lintel
H≥1.25h
h=l sin 60°
W = 1 2 𝑥 𝑙x l sin600 x t x 𝜌m (Triangular load)

13. t= Thickness of masonry Moment due to masonry load, M1, = 𝑤𝑙 6
l=effective span
t= Thickness of masonry
𝜌m = density of masonry (Assume 19.2kN/m3 )
Moment due to masonry load, M1, = 𝑤𝑙 6
Self weight of lintel = b x D x density of RCC (𝜌), w=_ _ _kN/m
Moment due self weight. M2 = 𝑤𝑙2 8
Total bending moment M =M1 + M2
ultimate moment Mu = 1 .5 X M
5. Check for depth
Equating ultimate moment(Mu) = moment of resistance
Mu = fck bd2 – Fe 250
Mu = fck bd2 – Fe 415
drequired = 𝑀 𝑢 𝑜.138 𝑓𝑐𝑘 𝑏 𝑓𝑜𝑟 𝐹𝑒 415 𝑠𝑡𝑒𝑒𝑙
If drequired < d -----design is safe
If drequired > d -----design is not safe Revise the depth

14. 6. Design of Tension reinforcement
Mu= 0.87fyAstd(1- Astfy bdf𝑐𝑘 )
Solve for A st
Assume suitable diameter of the bars (8mm,10mm,12mm…)
Area of one bar No, of Bars = 𝐴 𝑠𝑡 𝑎 𝑠𝑡
7. Design of Shear reinforcement
Shear force v = 𝑤𝑙 𝑊 2 , Vu = v x 1.5
Shear stress 𝜏v = 𝑉 𝑢 𝑏𝑑
Percentage steel =100 𝐴 𝑠𝑡 𝑏𝑑
Refer table No.19, 1S for 𝜏c value
If 𝜏c is less than 𝜏v then shear reinforcement is to be designed.
If 𝜏c is less than 𝜏v then minimum shear reinforcement is to be provided (For design steps refer design of singly reinforced beam)

15. Problems:
Design a lintel using following Data: width of opening = 2.4m, Height of brick wall above lintel = 3m, thickness of walls = 230mm, bearing width of wall = 230mm, concrete M20 and steel Fe415, Weight of concrete = 25kN/m3, Weight of brick wall = 19.2kN/m3. Sketch reinforcement details.

16. Design steps Assume effective depth of lintel
d= 𝑙 10 = = = 240mm
overall depth D = effective depth + effective cover
D = d + 35mm Effective cover= = 275mm
2. Width of lintel, b = Thickness of masonry = 230mm
3. Effective span (l)
Must be least of the following two
1= clear span (L)+ bearing = = 2.63m
ii) l = clear span(L) + effective depth = = 2.64m
l=2.63m
4. Load and moment calculation
For transfer of loads to the sides by arch action,
Height of wall above Lintel
H≥1.25h , H=3m, h = l sin600 = 2.63 x sin600 = 2.27m, 1.25h = 1.25 x 2.27 = 2.83m
H>1.25h

17. Masonry load(W)-Triangular load W = 1 2 𝑥 𝑙x l sin600 x t x 𝜌m W = 1 2 𝑥 2.63 x 2.63 sin600 x 0.23 x 19.2 = 13.22kN Moment due to masonry load, M1 = Wl 6 = 13.22x = 5.79kN-m Self weight of lintel, w = b x D x Density of RCC(𝜌) = 0.23 x x 25 = 1.58kN/m Moment due to self weight , M2 = wl2 8 = 1.58x = 1.36kN-m Total bending moment, M= M1+M2 = = 7.15kN-m Ultimate moment Mu=1.5 x M = 1.5 x 7.15 = 10.73kN-m
t= width of wall

18. 5) Check for depth Equating, M=Mu Mu = 0. 138xfckbd2 drequired = 𝑀𝑢 0
5) Check for depth Equating, M=Mu Mu = 0.138xfckbd2 drequired = 𝑀𝑢 0.138𝑥 𝑓𝑐𝑘𝑥 𝑏 = 𝑥 𝑥 20 𝑥 230 = 130mm <240mm(dprovided) Design is safe. 6. Design of tension reinforcement Mu= 0.87fyAstd(1- Astfy bdf𝑐𝑘 )

19. 10.73 x106 = 0.87 x 415 x Astx240[1- 𝐴𝑠𝑡x 𝑥 240 𝑥 20 ] x106 = 86.65x103Ast-32.57Ast Ast x103Ast x106 = 0 Ast=130.24mm2 Providing 10mm dia bars Area of one bar, ast= 𝜋𝑥 =78.5m2 No. of bars = 𝐴𝑠𝑡 𝑎𝑠𝑡 = = 1.65 say 2 Nos

20. 7. Design of Shear reinforcement
Shear force v = 𝑤𝑙 𝑊 2 = 𝑥 =8.68kN
Vu = v x 1.5 = 8.68 x 1.5 = 13.02kN
ii) Shear stress 𝜏v = 𝑉 𝑢 𝑏𝑑 = 𝑥 𝑋 240 = 0.235N/mm2
Ast = = 2𝑥 𝜋𝑥 = 157mm2
iii) Percentage steel =100 𝐴 𝑠𝑡 𝑏𝑑 = 100 x 𝑥240 = 0.28
Refer table No.19, 1S 𝜏c value after the interpolating available values.
𝜏c=0.37 N/mm2 > 0.235N/mm2
Entire shear is taken by the concrete alone hence providing nominal shear reinforcement, Provide 6mm dia bars

21. Asv = 2 x 𝜋 𝑥 62 4 = 56.55mm2 (Area of 2-Legged stirrups)
Spacing of stirrups should be least of the following three
Sv= 0.87𝑥𝑓𝑦 𝑥𝐴𝑠𝑣 0.4𝑥 𝑏 = 0.87𝑥415 𝑥 𝑥 230 = mm say 220mm
Sv should not be greater than 0.75d = 0.75x 240 = 180mm
Sv should not be greater than 300mm
Therefore provide 6mm dia 180mm c/c

22. Reinforcement details:

23. PROBLEM 2
A lintel beam 230 x 150 mm is reinforced with 3 — 10 mm bars at an effective cover of 35 mm. If lintel is simply supported over an opening of effective span 1.35m, determine the total uniformly distributed load the lintel can carry including its self weight by assuming that load of the wall on the lintel is uniform. Take fck = 15 Mpa and fe 250 Mpa
Given: b = 230mm, D = 150mm, effective cover = 35mm l=35m , Ast = 3 – 10 Ø fck =15N/mm2 ,fy =250N/mm2
Ast = 3x𝐴=3x 𝜋 𝑑2 4 = mm2
depth of neutral axis
xu = 𝑓 𝑦 𝐴𝑠𝑡 0.36 𝑓𝑐𝑘 𝑏 = 𝑥 250 𝑥 𝑥 15 𝑥 230 =41.26𝑚𝑚
d = =115mm

24. Xu max =0.53d = 0.53 x 115 =60.95mm
Xu < Xumax
Section is under reinforced ,
moment of resistance is given by
Mu = 0.87 fy A st d x (1− 𝐴 𝑠𝑡 𝑓𝑦 𝑏𝑑 𝑓𝑐𝑘 )=0.87 𝑥 250 𝑥 𝑥 115 ( 1− 𝑥 𝑥 115 𝑥 15 )=5.01x106 N-mm
M = Mu / 1.5 = =3.34𝑘𝑁−𝑚
The Maximum BM for simply supported beam carrying udl
Equating maximum BM = Moment of resistance
𝑤𝑙2 8 = 𝑤 𝑥 =0.228𝑤
3.34 = w
w= 14.64kN/m