A beam is a structural element that resists loads applied laterally to its axis. The structural capabilities of beams can be exploited in supporting walls, floors, ceilings, decks and garages. Beams support vertical loads and are commonly built to resist bending moment and shear force. Beams are also quite useful to a structural system as carry out a number of functions such as:

- Providing a uniform distribution of forces from the loads into the supports.
- Connecting the structural frame together
- Bearing loads above them.

Beams can be made from different materials such as timber, steel and reinforced concrete. Reinforced concrete beams which are the subject of this article are made by combining concrete together with steel bars which improve the performance of the material in tension.

## Types and Classification of beams

Based on the various capabilities of beams, they can be grouped into a few types depending on their purpose such as:

### a) Based on purpose

**Ring beams**: These are a type of support used in construction to connect walls together and increase the load capacity for the wall. They usually span the entire top of the wall of the building. the wall.**Lintels**: These are short span beams used to support the wall above an opening such a window or door.**Primary beams:**These are beams directly connected to the columns and transfer the loads from slabs or floors or any other beams connected to them.**Secondary beams:**These are beams which transfer their loads onto the supporting primary beams.

### b) Based on material

**Steel beams**: Steel beams are usually I beams (wide flange) as this kind of cross section resists the highest bending moment. They are mainly used in structures such as, ware houses, high rise buildings and bridges. This is due to their light weight as compared to concrete beams. They possess a high strength to weight ratio.**Concrete beams**: Concrete beams are normally made in the rectangular cross section due to the ease associated with making such a cross section. Concrete beams are very strong in compression but very weak in tension and this justifies the relevance for reinforcement with steel so as to resist high bending moments that beams are subjected to.

### c) Based on nature of supports

**Simply supported beams:**Such beams are supported to freely rotate at both ends. They are free to rotate at the supports. They can be used in short span foot bridges.-
**Cantilever beams:**Cantilever beam are fixed at one end and free at the other end. The fixed end is restricted from rotation, horizontal and vertical deflection. The upper portion of a cantilever beam is always in tension while the lower part is in compression due to the applied forces. Cantilever beams may be used to support balconies and also in cranes. **Continuous beams:**Continuous beams span on more than two supports. Although similar to simply supported beams they are more economical than other beams since the supports reduce the length of the spans. They are majorly used in bridges and complex roof structures. A ring beam is a continuous beam.**Simply supported beams with a hangover:**In such a beam, the beam extends beyond one or both supports. Some continuous beams may also possess an over hung.

## Principles in design of reinforced concrete beams

Beams are mainly designed for bending (flexural) moments, shear effects and deflection. With regard to bending moment, the design consider the maximum moment within the bend for each of the possible directions (sagging and hogging). Maximum sagging occurs within the span of the beam while the hogging moment occurs at the support of the beam. Once the design moment is established, the required area of the longitudinal (main) steel reinforcement can be calculated, and this helps to establish the number of bars to place in the beam at both the top and bottom.

In the case of the shear effects, the maximum shear stress can be estimated , and this normally occurs at the supports of the beam. The concrete beam is then first checked to confirm whether it is sufficient under shear without needing shear reinforcement (vertical links). If the beam is inadequate, the required shear reinforcement can then be computed.

The deflection of the beam also needs to be limited so that both functional and safety requirements of its design are meet. The deflection of the beam is mainly controlled by its size, and deeper beams have more stiffness which limits their deflections. With regard to design, the deflection of a beam can be checked using the span to depth ratio.

**Example on design of RC beams **

Consider the design of a 6m span simply supported beam with a dead load of 30kN/m and live load of 20kN/m. Let’s assume the concrete strength of the beam is 30 MPa, and the reinforcement has a strength of 500MPa.

**Solution**

Lets first determine the depth of the beam using the table below.

Structural system | K | Highly stressed concrete | Lightly stressed concrete |
---|---|---|---|

Simply supported beams and slabs | 1 | 14 | 20 |

End span of continuous beams and slabs | 1.3 | 18 | 26 |

Interior spans of continuous beams and slabs | 1.5 | 20 | 30 |

Cantilever beams and slabs | 0.4 | 6 | 8 |

L/d = 6000/d = 14

d=429 mm

The depth of the beam can be taken as 450 mm.

If the depth of the beam is 450, the width can be assumed to be about half of its depth. The width is about 225 mm. However, to be safe we can take the width as 300mm.

The cover to reinforcement can also determined on exposure conditions.

Assuming XC1 condition as per Eurocode is considered for corrosion risk and inside structure, then minimum cover is 25mm.

C_{nominal} = C_{min }+ ∆C_{dev}

C_{nominal }= 25 + 10

C_{nominal} = 35mm

Assuming the diameter of links is 8mm and main reinforcement is 25mm. The effective cover, C_{eff } can be determined:

C_{eff} = 35 + 8 + (25/2)

C_{eff }can be taken 50 mm

Therefore, overall depth = 450 + 50 = 500mm.

*Checking fire resistance*

Standard fire resistance (min) | Minimum Dimension of a & b (mm) | |||
---|---|---|---|---|

Possible combination of "a" and "b" | ||||

1 | 2 | 3 | 4 | |

R60 | b=120 a=40 | b=160 a=35 | b=200 a=30 | b=300 a=25 |

R90 | b=150 a=55 | b=200 a=45 | b=300 a=40 | b=400 a=35 |

R120 | b=200 a=65 | b=240 a=60 | b=300 a=55 | b=500 a=50 |

From the above table, the beam is sufficient to resist fire an hour.

*Loading on the beam*

Total ultimate load, w_{Ed} = 1.35(30) + 1.5(20) = 70.5 kN/m

The span of the beam , L = 6m

Design bending moment, M = (w_{Ed} L^{2})/8 = 318 kNm

##### Calculating the K value

f_{ck} = 30 N/mm^{2}

b = 300 mm

K = M/(b d^{2} f_{ck}) = 0.175

Since K is bigger than 0.167, compression reinforcement is required.

##### Determining area of reinforcement

Compression reinforcement must have yielded:

d’/d =35/450 = 0.07 < 0.171

*Area of reinforcement for compression:*

Strength of the steel, f_{yk} = 500 N/mm^{2}

As^{1} = (k-k_{bal}) (f_{ck} b d^{2})/(0.87f_{yk} (d-d’))

As^{1} = (0.175-0.167) (30x300x450^{2})/(0.87x500x(450-35))

As^{1} = 81mm^{2}

^{ }*Area of reinforcement for tension:*

As = [(k_{bal }f_{ck} b d^{2})/(0.87 x f_{yk } x z_{bl})] + As^{1}

As = [(0.167x30x300x450^{2})/(0.87x500x0.82×450) ] + 81

As = 1980 mm^{2}

The areas are the required areas of reinforcement, but to account for durability and control of cracks, the following should be meet:

**As**should be >=^{1}**As**= 0.2% gross area (b x h)^{1}_{min }**As**should be >=**As**= 0.26 f_{min}_{ctm}(bd)/ [ (0.26 f_{ctm}b d )/(f_{yk})] > 0.0013(bd)

*Checking for the minimum area requiremets:*

0.2%x (bxh) = (0.2 x 500 x 300)/100 =300mm^{2}

Hence for area of compression reinforcement can be taken as 300mm^{2}

*For tension reinforcment:*

f_{ctm} = 0.3 f_{ck} ^{(2/3)} = 2.9

As_{min} > 0.26x 2.9bd/500 = 0.15% bd > 0.13%bd

Hence As_{min} = 0.15bd/100 = 202.5 which is less than the required area of steel.

Maximum area of steel, A_{max} = 4% bh = 4 x300x 500/100 = 6000 mm^{2}

*Determination of the actual number of steel bars*

Number of bars | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|

Bar size (mm) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

6 | 28.3 | 56.6 | 84.9 | 113 | 142 | 170 | 198 | 226 | 255 | 283 |

8 | 50.3 | 101 | 151 | 201 | 252 | 302 | 352 | 402 | 453 | 503 |

10 | 78.5 | 157 | 236 | 314 | 393 | 471 | 550 | 628 | 707 | 785 |

12 | 113 | 226 | 339 | 452 | 566 | 679 | 792 | 905 | 1020 | 1130 |

16 | 201 | 402 | 603 | 804 | 1010 | 1210 | 1410 | 1610 | 1810 | 2010 |

20 | 314 | 628 | 943 | 1260 | 1570 | 1890 | 2200 | 2510 | 2830 | 3140 |

25 | 491 | 982 | 1470 | 1960 | 2450 | 2950 | 3440 | 3930 | 4420 | 4910 |

32 | 804 | 1610 | 2410 | 3220 | 4020 | 4830 | 5630 | 6430 | 7240 | 8040 |

Take As^{1 }=339, hence 3 bars of 12mm diameter

Take As = 2200, hence 7 bars of 20mm diameter.

The beam will be provided with 3H12 at the top for compression steel and 7H20 for tension reinforcement at the bottom.

However, since the space between rebars must be greater than the nominal aggregate size (20mm), the bars at the bottom need to be put separate layers. This is means putting 5 bars in the lowest layer and 2 bars in the upper bottom layer as shown below.

##### Determining the Shear reinforcements

Due to great shear force near the supports, the shear links are closely packed.

V_{max} = (w_{Ed} L)/2 =(70.5 x 6)/2 =211.5kN

V_{eff} =V_{max }– w_{Ed} (support width)/2

V_{eff} = 211.5 -70.5 x 0.3 x 0.5 = 201kN

V_{Ed} = V_{eff} -w_{Ed}xd = 201 – 70.5 x 0.45 = 170kN

Now V_{ED} has to be compared with the shear resistance, V_{Rd,c}

V_{Rd,c} = [C_{Rd,c} k (100 ρ f_{ck})^{(1/3)}+ 0.15σ_{cp}] bd

The constants can be calculated as follows:

C_{Rd,c} = 0.18/γ_{c} = 0.18/1.5 = 0.12

k = 1 + √200/ √d = 1 + √200/ √450 = 1.67

Assuming the axial force in the beam is zero, σ_{cp} = 0

when half the reinforcement bars are extended, ρ = (2200×0.5) /(300×450) =0.008 ,

Therefore V_{Rd, c }= [0.12×1.67x(100×0.008×30)^{(1/3)}+ 0] x 300 x 450 = 78kN

Since V_{Rd, c} < V_{Ed, }there is need to provide shear links.

For shear steel:

A_{sw} / s = v_{Ed} /(z f_{ywd} cot Ɵ)

The angle Ɵ ranges from 22^{0 }to 45^{0}_{,}

But z = 0.9d= 405mm

V= 0.6(1 – f_{ck}/250)= 0.528

Also, f_{cd } = 0.85xf_{ck}/1.5 =17 MPa

V_{Rd,max} = (300x 405 x 0.528 x 17)/(cot 22 + tan 22 ) = 375.5 kN

Since V_{Rd, max} > V_{Ed} , Lets Ɵ as 22^{o}

Let f_{ywd }be equal to 500MPa.

Then A_{sw}/s = (170000)/(0.9×450 x (500/1.15)x 2.5)= 0.386

Stirrup diameter (mm) | Stirrup spacing (mm) | ||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

85 | 90 | 100 | 125 | 150 | 175 | 200 | 225 | 250 | 275 | 300 | |

8 | 1.183 | 1.118 | 1.006 | 0.805 | 0.671 | 0.575 | 0.503 | 0.447 | 0.402 | 0.366 | 0.335 |

10 | 1.847 | 1.744 | 1.57 | 1.256 | 1.047 | 0.897 | 0.785 | 0.698 | 0.628 | 0.571 | 0.523 |

12 | 2.659 | 2.511 | 2.26 | 1.808 | 1.507 | 1.291 | 1.13 | 1.004 | 0.904 | 0.822 | 0.753 |

16 | 4.729 | 4.467 | 4.02 | 3.216 | 2.68 | 2.297 | 2.01 | 1.787 | 1.608 | 1.462 | 1.34 |

From the table above , the shea r link can be taken 8 mm at spacing 250 mm.

ρ_{w,min} = (0.08√f_{ck})/(f_{yk}) = (0.08√30)/(500) = 8.76 x 10^{-4}

Since the stirrups are vertical, then α = 90^{o}

8.76 x 10^{-4} = A_{sw} / (s x b x sin α )

A_{sw} / s = 0.26

Maximum spacing = 0.75(450)(1+0) = 338mm, hence approximately 300mm

Since 250 < 300 mm, the stirrup spacing is ok.

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