Design of cantilever beam, point load on beam and doubly reinforced beam

Cantilever beam

If the one end is fixed and another end is free in a beam, then the beam is called a cantilever beam.

The top surface or the top extreme fiber experience a tension, while bottom fiber experience compression. 

That's why we provide tension reinforcement at the top.

Cantilever beam design example
Cantilever beam

We see the cantilever beam or slab mostly in the balcony, chajja, etc. but, in some of the case, we need to provide the cantilever portion in the house or bungalow also.

To provide the cantilever beam, we must aware of the calculation for the cantilever beam and the design process.

In our plan, there are two balconies on the first floor and three cantilever beam which are similar in length. 

We will calculate the cantilever beam, point load on simply supported and cantilever beam, and the doubly reinforced section in this article.

Let's calculate and design the cantilever beam BC-14 first.

Design calculation for cantilever beam

The design process of a cantilever beam is the same as simply supported beam except for the moment calculation and tensile reinforcement detail.

cantilever beam example
Cantilever beam BC- 14

We will calculate the cantilever beam BC-14, as shown in the plan.

There are no slab loads that will transfer in BC-14 and also the length is only 0.91 m. so, from the previous calculation experience, we can reduce the depth from 450 mm to 300 mm.  

You can see that the Beam B-1 is creating the point load in BC-14. So, we need to calculate the point load and UDL load on BC-14.

B-1 will transfer the half load to BC-14 and half load to BC-18. So, we need to simply divide the previously calculated load of B-1 by 2 and multiply by the total length of the beam.

load on BC-14 form left of B-1 =  UDL X length/2

                                                       =  5.79 X 5.1/2

                                                        = 14.76 kN

Now, we need to calculate the total UDL acting on BC-14.

Slab load = 0

Wall load = 12.64 kN/m ( 230mm thick wall )

Self-weight of beam = 0.23 X 0.300 X 25

                                     = 1.72 kN/m

Total UDL = 12.64 + 1.72

                   = 14.36 kN/m  

The moment on BC-14 will be,

= moment due to UDL + moment due to point load

`=\frac{w l^{2}}{2} + wl `

`=\frac{5.79 X 0.91^{2}}{2} + 14.36 X 0.91`

= 2.42 + 13.43

= 15.85 kN⋅m

Factored moment = 15.85 X 1.5

                           Mu = 23.77 kN⋅m

Mulim = 0.138 fck ⋅bd2

           
           = 0.138 X 20 X 230 X 2702

           = 46.27 kN⋅m

MMulim  ...... Under reinforced section )

To find p,

`p_{t} = 50 \frac{f_{ck}}{f_{y}} [1 - \sqrt{1-\frac{4.6M_{u}}{f_{ck} b d^{2}}} ]`

      `= 50 X \frac{20}{415} [1 - \sqrt{1 - \frac{4.6 X 27.77 X 10^{6}}{20 X 230X 270^{2}}} ]`

pt = 0.513 %

To find the area of steel, 

`A_{st} = \frac{p_{t}}{100}bd`

       `= \frac{0.513}{100} X 230 X 270 `

`A_{st}`  = 318.57 mm2

Provide, 318.57/ 113.04 = 2.81

Provide, 3 nos of 12 mm bars.

So, the detailing of the cantilever beam BC-14 is 3 nos 12 mm dia. bars at the top and 2 nos. of 12 mm dia. bars at the bottom.

Point load on the simply supported beam

point load on simply supported beam
Point load on B-4


From the above plan, we can see the point load B-4 from B-22.

Let's simplify this plan.

UDL with point load
Point load on beam free body diagram

As shown in the figure, 

The total length of the beam is 18.5 feet, point load is acting at 13.125 feet distance from the left side, supported on C-2 and C-3 and the condition is UDL with point load.

To calculate the total moment, we need to calculate the UDL plus point load on B-4.  

First, we will calculate the UDL on B-4.

On the upper side of B-4, two loads are coming S-4 trapezoidal and S-5 triangular. we will take the bigger slab only. so, we will consider only the S-4 trapezoidal load. 

The total slab loads are coming from S-2 rectangular plus S-4 trapezoidal load.

S-2 is a balcony and S-4 is a master bedroom. hence, the live load will be 3 kN/m2 and 2 kN/m2 respectively and the floor finish will be 1 kN/m2 for both the slab.  

S-2 total slab weight = ( 0.125 X 25 ) + 3 +1

                                      = 7.125 kN/m

S-4 total slab weight = (0.125 X 25 ) + 2+ 1

                                      = 6.125 kN/m

S-2 load = Shortar span X slab weight / 2

                = 1.14 X 7.125 / 2

                = 4.06 kN/m

To calculate the S-4 trapezoidal load,

we have,

Lx = shorter span = 2.63 m 

Ly = longer span  = 4.001 m

= `\frac{l_{y} + (l_{y} - l_{x})}{2} X \frac{l_{x}}{2} X \frac{load}{l_{y}}`

= `\frac{4 + (4 - 2.63)}{2} X \frac{2.63}{2} X \frac{6.125}{4}`

= 5.54 kN/m

Total slab weight on B-4 = S-2 weight + S- 4 weight

                                             = 4.06 + 5.54

                                             = 9.60 kN/m

B-4 Wall weight = 0.230 X (3.04 - 0.45) X 20

                             = 11.95 kN/m

B-4 self-weight = 0.230 X 0.450 X 25

                            = 2.58 kN/m

Total UDL on B-4 = Slab weight + wall weight + s/w of beam 

                                = 9.60 + 11.95 + 2.58

                                = 24.13 kN/m

Factored moment due to UDL `=\frac{w l^{2}}{8}` 

                                                      `=\frac{24.13 X 1.5 X 5.64^{2}}{8}`

                                                       = 143.91 kN⋅m

For point load, we previously calculated the UDL on B-22.
 
Load on B-4 from the left side of B-22 = UDL x length of B-22/ 2

                                                                     = 16.525 X 2.62/ 2  

                                                                     = 21.64 kN

we have, a = 4.01 (13.125 feet) and b = 1.64 (5.375 feet)

Moment  `= \frac{wab}{l}`

                `= \frac{21.64 X 4 X 1.64}{5.64}`

                   = 25.17 kN⋅m   

Factored moment due to point load = 25.17 X 1.5

                                                                 = 37.76 kN⋅m

Total factored moment = moment due to UDL + moment due to point load  

                                           = 143.91 + 37.76  

                                            = 181.67 kN⋅m

To calculate the required depth, we assume the balanced section.

xu = xumax

pt = ptlim

Mu = Mulim

we have, d = 420

(IS: 456-2000, P. - 96) 

For Fe 415, 

`\frac{x_{umax}}{d} = 0.479`

`M_{ulim} = 0.36 X \frac{x_{umax}}{d} (1 - 0.42 \frac{x_{umax}}{d}) bd^{2} X f_{ck}`

`181.67 X 10^{6} = 0.36 X 0.479 (1 - 0.42 X 0.479) 230 X d^{2} X 20`

d = 535.45 mm 

we can provide D = 600 mm if we design as a singly reinforced beam.

Mu = 181.67 kN/m

Mulim = 0.138fckb⋅d2  (SP. - 16, P- 10, table - C )

Mulim for 9 X 18 = 111.97 kN/m < 181.67 kN/m (d = 420)

            for 9 x 21 = 158.7 kN/m 181.67 kN/m ( d = 500)

             for 9 x 24 = 206.24 kN/m > 181.67 kN/m ( d = 570)

To use a 9 X 18 or 9 X 21-inch beam, we must design the beam as a doubly reinforced section. because the moment is higher than the limiting moment.

To design as a singly reinforced section, we can use the 9 X 24-inch section. as the moment is less than the limiting moment.

For a two-story building, the use of a 230 X 600 mm section could be heavier.

So, the question is should we use a 9 X 21-inch doubly reinforced beam section or should we use a 9 X 24-inch singly reinforced section?

To get the answer, let's calculate both the section.

The beam size of 9 X 24-inch

Total slab weight = 9.60 kN/m 

 

Wall weight = 0.230 X (3.04 - .600) X 20


                      = 11.22 kN/m


Self-weigth of beam = 0.230 X 0.600 X 25


                                      = 3.45 kN/m  


Total UDL = 9.60 + 11.22 + 3.45

                   = 24.27 kN/m

Factored moment = 1.5 X 24.27 X 5.642 / 8

                                 = 144.75 kN⋅m

Total factored moment = 144.75 + 37.755

                                           = 182.50 kN⋅m

(which is less than the limiting moment 206.24 kN⋅m. hence, design as a singly reinforced beam)

 `p_{t} = 50 \frac{f_{ck}}{f_{y}} [1 - \sqrt{1-\frac{4.6M_{u}}{f_{ck} b d^{2}}} ]`

pt = 0.814 %

`A_{st} = \frac{p_{t}}{100}bd`

Ast = 1067.15 mm2   


Provide, 1067.15 / 201

5 nos 16 mm bars.


If we design as a singly reinforced beam area of steel is 1067.15  mm2. 


what if the size is limited to 9 X 21 for B-4?


For that, we can design a doubly reinforced beam section.


Doubly reinforced beam

Self-weight of beam = 0.230 X 0.530 X 25

                                     = 3.04 kN/m

Wall weight of beam = 0.230 X (3.04 - 0.530) X 20

                                      = 11.54 kN/m

Slab weight = 9.60 + 3.04 + 11.54 


                       = 24.18 kN/m

Factored moment due to Udl = 144.22 kN⋅m


Total factored moment      Mu = 144.22 + 37.755 


                                                     = 181.98 kN⋅m

Mulim = 158.7 kN⋅m


Mu2 Mu - Mulim


Mu2 = 181.98 - 158.7

        = 23.28 kN⋅m


fcc = 0.446 fck    ( SP. 16, P-13)


      = 0.446 X 20


      = 8.92


 d'/d = 30/500  ( SP. 16, P-13, Table - F)


          = 0.06 ≈ 0.10


fsc = 353  N/mm2


To Find Asc (IS: 456-2000, P- 96 , CL. G-1.2)  


$$M_{u2} = (f_{sc} - f_{cc}) A_{sc} (d-d')$$


`23.28 X 10^{6} = (353 - 8.92) A_{sc} (500-30)`


`A_{sc} = 143.95 mm^{2}`


Find Ast1, 

`p_{tlim} = 19.82 X (\frac{f_{ck}}{f_{y}})`


`p_{tlim}` = 0.96 %


`\therefore A_{st1} = \frac{p_{tlim}}{100} X bd`


       `A_{st1}` = 1104 mm2


Find Ast2 (IS: 456-2000, p-96),


$$A_{st2} = \frac{(f_{sc} - f_{cc}) A_{sc}}{0.87 f_{y}}$$


               $$= \frac{(353- 8.92) X 143.95}{0.87 X 415} = 137.18 mm^{2}$$


`A_{st} = A_{st1} + A_{st2}`


`A_{st}` = 1241.18 mm2


Provide reinforcements,

Top = 2 nos. 12 mm dia. bars

Bottom = 2 nos. 20 mm dia. and 3 nos. 16 mm dia. bars.


Summary

We have seen the different size calculation on B-4. The size of 9 X 24 requires 1067.15 mmand 9 X 21 requires 1241.18 mm2 area of steel. 

If we use a 9 X 24-inch beam, then the minimum size of supporting columns should be 9 X 24 inches. 

As we discussed earlier that the size of the column should be higher or equal to the size of a beam.

I recommend using the 9 X 21-inch beam.

The use of 20 mm steel for a two-story building could be costly. so, how we can reduce the required area and minimize the numbers of bars?

There are some ways to reduce the area of steel.

Revise the beam layout plan and try to reduce the length of the beam and avoid point load if possible.

The wall resting in the beam is 230 mm thick which creates heavy UDL. the partition wall is possible on the first-floor balcony. 

Do the discussion with the architect to revise the plan and make the wall a partition wall which has a size of 115 mm thick and which will reduce the half the wall weight.

We use the factor of safety for concrete = 1.5 which raises the moment on the beam. Instead of 1.5, we can even use 1.4,1.3 or 1.2 according to some other standard code.

The manual design gives higher design value compare to structure design software. try to use the software such as ETABS, Staad pro, struds, SAP 2000, etc. 

Conclusion

I have seen that so many people ask that for 15 feet beam span what will be the size of the beam and what are the reinforcement detailing? 

Or what is the standard size for the beam for a 5-meter span?

If you have ever designed the beam then the answer for all that type of question is let me calculate that beam. 

Because after calculating the beams we know that the size and the area of steel of the beam depend on the moment.

The moment depends upon the loads ( UDL and point loads) and the length of the beam.

UDL depends upon the slab weight, beam weight, and wall weight. 

Slab weight depends on the thickness and the function of a room.
for example, L.L for bedroom is 2 kN/㎡ and L.L for balcony is 3 kN/㎡.

The beams B-2 and B-4 have the same length of 18.5 feet.

B-2 is resting on cantilever beams and supports the one-way slab only and not even the wall is resting on it. After the calculation, we provide 3 nos of 12 mm bars at the bottom and 2 number of 12 mm bars at the top with the 9 X 12-inch beam size. 

While B-4 has the same length but different reinforcement detail and the size. 

You can see the difference between 12 inches and 21 inches with the same length.

So, the point is the beam size and bars detailing is not depends only on the length of the beam.

We have calculated some of the beams with different conditions on the first floor. you can practice the beam design calculation of the second-floor plan as I have provided in the previous article.

After calculating the beams, we will move ahead to the design of the slabs.





       





    


  



                 




















 



                       




















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