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Reinforced concrete beams are structural elements that designed to carry transverse external loads. The loads cause bending moment, shear forces and in some cases torsion across their length. Moreover, concrete is strong in compression and very weak in tension. Thus, Steel reinforcement used to take up tensile stresses in reinforced concrete beams. Furthermore, beams support the loads from slabs, other beams, walls, and columns. They transfer the loads to the columns supporting them. Additionally, beams can be simply supported, continuous, or cantilevered. they can be designed as rectangular, square, T-shaped, and L-shaped sections. Beams can be singly reinforced or doubly reinforced. The latter are used if the depth of the beam is restricted. Finally, in this article, the design of rectangular reinforced concrete beam will be presented.Contents:
Design guidelines
Prior to the design of reinforced concrete beam begin, there are certain assumption that need to be made. these guidelines are provided by certain codes and researchers. It should be known that, experience of designer plays significant role in making these assumptions.Beam depth (h)
There is no unique procedure for computing overall beam depth(h) for design. However, certain guidelines can be followed to compute beam depth such that the deflection requirements can be satisfied.- ACI 318-11 provides suggested minimum thickness for non-prestressed beams unless deflections are calculated. t
- Canadian Standard Association (CSA) provides similar table except for one end continuous which is l/18.
|
Minimum thickness, h |
|||
| Simply supported | One end continuous | Both end continuous | Cantilever |
| Members not supporting or attached to partitions or other construction likely to be damaged by large deflections | |||
| l/16 | l/18.5 | l/21 | l/8 |
| Notes: Values given shall be used directly for members with normal weight concrete and Grade 420 reinforcement. For other conditions, the values modified as follows: a) For lightweight concrete having equilibrium density (wc) in the range of 1440 to 1840 kg/m3, the values shall be multiplied by (1.65 – 0.0003wc) but not less than 1.09. b) For fy other than 420 MPa, the values shall be multiplied by (0.4 + fy/700). | |||
- The depth of beam can also be estimated based on span/depth ratio. IS 456 2000 provides span to depth ratio to control deflection of beam as provided in Table 2.
| Beam span | Beam type | Span/depth ratio |
| Up to 10m | Simply supported | 20 |
| Cantilever | 7 | |
| Continuous | 26 | |
| Greater than 10m | Simply supported | 20*10/span |
| Cantilever | - | |
| Continuous | 26*10/span |
Beam width (b)
The ratio of beam depth to its width is recommended to be between 1.5 to 2 with upper bound 2 being the most common used. The reinforcement arrangement is one of major factors that specify beam width. So, minimum bar spacing shall be considered while beam width is estimated. The width of the beam shall be equal or less than the dimension of the column supporting the beam.
Steel reinforcement
ACI 318-11 provided minimum and maximum reinforcement ratio. The reinforcement ratio is an indicator of the amount of steel in a cross section. So, any values between this range can be used for beam design. Nonetheless, the choice is influenced by ductility requirement, construction and economic consideration. lastly, it is recommended to use 0.6*maximum reinforcement ratio.
Reinforcing bar sizes
Generally, it is advised to avoid the use of large bar sizes for beams. This is because such bars cause flexural cracking and required greater length to develop their strength.
However, large bar size placement cost is smaller than the installation cost of large number of small bar sizes.
Moreover, common bar sizes for beams range from NO.10 to NO.36 (SI unit) or NO.3 to NO.10 (US customary unit), and the two larger diameter bars NO.43 (NO.14) and NO.57 (NO.18) are used for columns.
Furthermore, it is possible to mix different bar diameter to meet steel area requirements more closely.
Finally, the maximum number of bars which can be installed in a beam of given width is controlled by the bar diameter, minimum spacing, maximum aggregate size, stirrup diameter, and concrete cover requirements.
Spacing between bars
ACI 318-11 specify minimum spacing between bars equal to bar diameter or 25mm. This minimum spacing shall be maintained to guarantee proper placement of concrete around steel bars. Additionally, to prevent air pockets below reinforcements, and ensure good contact between concrete and bars so as to achieve satisfactory bond. If two layer of steel bars are placed in a beam, then the distance between them shall not be less than 25 mm.
Concrete protection for reinforcement
the designer must maintain minimum thickness or concrete cover outside of the outermost steel to provide the steel with adequate concrete protection against fire and corrosion. According to ACI Code 7.7, concrete cover of 40 mm for cast in place beams, not exposed directly to the ground or weather. At least 50mm cover, if the concrete surface is to be exposed to the weather or in contact. To simplify construction and thereby to reduce costs, the overall dimensions of beams, b and h are almost rounded up to nearest 25 mm.
Design of rectangular reinforced concrete beam procedure
The design of concrete beam includes the estimation of cross section dimension and reinforcement area to resist applied loads. There are two approaches for the design of beams. Firstly, begin the design by selecting depth and width of the beam then compute reinforcement area. Secondly, assume reinforcement area, then calculate cross section sizes.The First approach will be presented below
The following procedure used for the design of rectangular reinforced concrete beam:- Initially, select beam effective depth (d) and width (b). Effective depth can be computed using beam depth (h).
- Then, calculate the required flexural resistance factor assume ?=0.9
- After that, find reinforcement ratio corresponding to the computed flexural resistance compute above,
- The reinforcement ratio shall be less than maximum reinforcement ratio and greater than minimum reinforcement ratio.
- Minimum reinforcement ratio,
- Maximum reinforcement ratio
- Either reinforcement ratio can be employed but the latter will ensure that the strain in steel is at least 0.005.
- Thereafter, compute reinforcement area,
- Then, find number of bar by dividing reinforcement area over the area of a single bar.
- Lastly, check whether the bar can be placed within selected width of the cross section,
- The value of S shall not be less than 25 mm which the minimum required spacing between adjacent bars.
: strength reduction factor
b: cross section width
d: effective depth of beam cross section from the top of the beam to the center of reinforcement layer
fc': compressive strength of concrete
fy: yield strength of steel bars
p_u: ultimate strain in concrete which is equal to 0.003 according to ACI Code and 0.0035 according to EC
p_0.004: reinforcement ratio at steel strain equal to 0.004
p_0.005: reinforcement ratio at steel strain equal to 0.005
As: reinforcement area
S: spacing between adjacent bars
n: number of bars in a single layer
Shear design of rectangular beam
Shear design include estimation of stirrup spacing to support ultimate shear. Commonly, part of concrete will resist shear force but the portion of which is not supported by concrete will be carried by shear reinforcement.- Firstly, compute ultimate shear force at distance d which is the depth the cross section. There are exceptions in which shear at the face of the support shall be used for shear design. For example, when loads are applied at the bottom of the beam.
- Secondly, estimate design concrete shear strength,
- No shear reinforcement is needed if Vu< 0.5
Vc . - If 0.5Vc>Vu< Vc , then provide minimum shear reinforcement only.
- Provide shear reinforcement when Vu> Vc.
- Thirdly, select trial web-steel area based on standard stirrup sizes ranges from NO.10 to NO.16.
- Multiply shear reinforcement area by number of stirrup legs to calculate shear reinforcement area.
- Forthly, find spacing for the stirrup for vertical and inclined stirrups respectively using equation 12 and 13.
- Do not space vertical stirrups closer than 100mm. Therefore, the size of stirrups should be chosen to prevent closer spacing.
- Distribute stirrups uniformly over short span beams.
- However, over long spans, it is more economical to compute the spacing required at several sections. And, place stirrups accordingly in groups of varying spacing.
- The required spacing shall be equal or less than maximum spacing which is equal to the smallest of 600, d/2, and equation 14.
- If Vs is greater than 0.33fc’^0.5bwd, then these maximum spacing shall be halved.
- Finally, draw the design beam with longitudinal and shear reinforcement.
