### Design of Rectangular Beams

Design of RC beams involves determining beam dimensions and required reinforcement to resist factored loads. there is a desirable beam proportion shall be followed to produce economic beam cross-section.

- beam proportion: if there are no architectural or other restrictions that control the dimension of the beam, then the most economical beam will be obtained for short beam span of 20ft(6m) to 25ft(7.5m) by using d to b ratio of 1.5 to 2. for longer spans the depth of beam shall be increased and the width should be reduced to obtain economical sections. despite the economic benefits of using deep beams, designer tends to use shallow and wide beam due to height restrictions of the floor.
- Deflection: excessive deflection of a beam will cause damage to the surrounding structure. ACI table 9.5(a) (Table 1) providing a minimum thickness of beam and one-way slab for which deflection calculation is not required. if deflection calculation for a lesser thickness is calculated and found satisfactory then no need to abide by the minimum thickness.

Table 1

- Estimation of beam weight: at the beginning of the designing process the beam dimension is unknown and the same for the beam weight. therefore we should propose the beam dimension then check if our section is satisfactory. as a guide, we can use ACI table 9.5(a), Table 1 to calculate the minimum thickness h, the beam width can be assumed to be 0.5*h. then we calculate ultimate moment for external loads and estimated beam weight, after that we can select beam dimension, then we calculate ultimate moment (Mu) for the new beam weight. if it is similar to the estimated one, then we can stop and no need for further checking, but if there is a big difference then we need to recheck the beam capacity. if it is not sufficient then we need to redesign the beam again.
- selection of bars: after calculating the required area of steel. we will select the required number of bars that provide area equal to the calculated area.
- cover: reinforcing steel shall not be placed at the edge of the beam, enough space and cover shall be provided to protect steel from fire and environments. ACI code section 7.1 to 7.3 providing values for concrete covers for different applications such as cast-in-place concrete, prestressed cast-in-place concrete, and precast concrete

**Cast-in-place concrete (nonprestressed)**

**Concrete**

**cover, mm**

**(a) Concrete cast against and**

**permanently exposed to earth ..............................75**

**(b) Concrete exposed to earth or weather:**

**No. 19 through No. 57 bars ................................50**

**No. 16 bar, MW200 or MD200 wire, and smaller.....40**

**(c) Concrete not exposed to weather**

**or in contact with ground:**

**Slabs, walls, joists:**

**No. 43 and No. 57 bars ....................................40**

**No. 36 bar and smaller.................................... 20**

**Beams, columns:**

**Primary reinforcement, ties, stirrups,**

**spirals ...............................................................40**

**Shells, folded plate members:**

**No. 19 bar and larger ...................................... 20**

**No. 16 bar, MW200 or MD200 wire, and smaller.. 13**

Figure 1

- minimum spacing of bars: spacing between reinforcing bars shall not be less than the minimum requirement to allow placing concrete readily between reinforcing bars and to prevent concrete honeycombs. ACI section 7.6 states that the minimum clear distance between parallel bars shall be 25 mm (1in) or more, for two layers of steel placed vertically above each other, the clear distance between this layer shall be at least 25 mm(1in). ACI3.3.2 code limiting the maximum size aggregate to be the lowest of:

- 1/5 of the narrowest dimension between formwork
- 1/3 of slab depth
- 3/4 clear distance between steel bars.

minimum beam width shall be checked to confirm it is the ability to host longitudinal bars, shear stirrups while providing the required cover.

**steps of beam design:**

1.assuming beam dimensions using table ACI 9.5a(table 1), for example for simply supported beam (beam depth)h=L/20.

2.assumuing beam width equal half of beam depth b=0.5h

3. after assuming the beam dimension we can calculate beam dead weight

4.now we will multiply all expected loads with factors to calculate ultimate loads, then we calculate ultimate bending moment (Mu), Mu for simply supported beam=(wu.(L^2))/8

5. we will assume reduction factor=0.90 and we can calculate ρ

**ρ=(0.85*fc'/fy)*(1-√(1-(2*Rn/0.85*fc'))**

6. area of steel can be calculated now As=ρ*b*d, now we will select the number and diameter of reinforcing bars. beam width shall be checked again to verify if it is sufficient or no.

7. we will check if the provided reinforcement is more than the minimum required reinforcement if it is less we will use the minimum reinforcement. also, we should check section type if it is tension or compression control to confirm if the used reduction factor is correct or no.

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