Eng. Hasan Shehada Eng. Ruba Awad

Eng. Hasan Shehada Eng. Ruba Awad

Introduction:

It can be noticed that one of the important factors that affect the design of the structure is

economy factor. For that reason, we must use less materials in order to minimize the cost of this

structure, and one of the materials that significantly affects the cost of the project is steel

reinforcement.

ACI code provides reinforcement detailing that help saving a lot of reinforcement while

designing the structure. The code gives two choices for this purpose:

1- Development length requirement, which will be time wasting for large projects.

2- ACI 315-02 provides some values as portions with respect to lengths and spans of beam.

This method will be discussed in details in this lecture.

Of course, the values given will vary between negative moment, positive moment, and

cantilever reinforcement. Each one will be discussed in details below.

Negative Moment Reinforcement:

The reinforcement in both positive and negative moment sections, there will be two types of

reinforcement, one is continuous, and the other is cut and found only in the sections where

maximum moment exists.

- For continuous reinforcement:

For continuous reinforcement at the top of the beam, there must be at least:

Two bars for beams with a width less than 60 cm.

Four bars for beams with a width between 60 cm and 100 cm.

Six bars for beams with a width more than 100 cm.

Note:

Since we have bars with limited lengths (practically), then the continuous bars must be cut

and then tied with the following bars with a shared length that equal 60*diameter of bar. This

cut must be at the middle of the span where negative moment is at its minimum.

These continuous bars must go through the support (column) with at least 0.5*width of

support or 15 cm. The figure below shows the details of continuous bars.

Eng. Hasan Shehada Eng. Ruba Awad

- For cut reinforcement:

The reinforcement that is cut will be at the support area where negative moment is at its

maximum. The remaining bars of required reinforcement will be extended a length equals to

width of support plus (L/3) from both sides, where L is the largest of the two spans. The

figure below shows these values.

Eng. Hasan Shehada Eng. Ruba Awad

Positive Moment Reinforcement:

As we saw for negative moment sections, we have two types of bars, continuous bars and cut

bars.

- For continuous reinforcement:

The minimum for continuous bars must be the greatest of:

Two bars for less than 60 cm width beam, four bars for beams with a width of 60 cm to 100

cm, and six bars for beams with a width greater than 100 cm.

One third of the required reinforcement.

Minimum required reinforcement.

And knowing that we dont have enough length of bars to use along the beams, it must be cut

and linked with the following bar. The cut must be at the supports where the positive bending

moment is at its minimum. The figure below shows the details of continuous reinforcement

in the positive moment section.

- For Cut reinforcement:

The remaining reinforcement will be put in the middle of the span where maximum positive

moment exists. The length of bar will be equal to length of span subtracted from it 0.15Lc

from middle of support, and 0.10Lc for end support. The figure below shows the details of

these bars.

Eng. Hasan Shehada Eng. Ruba Awad

Cantilever Reinforcement:

For cantilever reinforcement, it must be extended along the length of the span plus 1.5* length of

the span and this extra length must be extended in the following span. The figure below shows

the details of cantilever beam reinforcement.

Eng. Hasan Shehada Eng. Ruba Awad

Example:

The beam shown above was analyzed by ROBOT structural analysis software. The results of

analysis are:

The dimensions of columns are as follows:

C1width is 25 cm, the remaining columns width is 40 cm. Design the beam for bending

moment.

Solution:

First, we find the percentage of steel needed at each critical section in both positive and negative

moments using the following excel sheet:

Eng. Hasan Shehada Eng. Ruba Awad

Now we will take an example for negative, positive, and cantilever moments.

- For continuous reinforcement in negative moment sections:

No. of bars that must be extended all over the beams is four bars because the width of beam

is 60 cm, and this reinforcement must be cut at the middle of spans where negative moment

is at its minimum.

Take D to C as an example:

Length of bar = [0.5W1 > 15 cm] + L1 + W2 + L2/2 - 30

= 0.15 + 3.775 + 0.4 + 3.1/2 + 30*0.014 = 6.3 m (Use 6.5 m)

- For Cut Reinforcement in negative moment sections:

The remaining reinforcement will be put at each negative moment section. Take section C as

an example:

No. of bars remaining = 4 bars

Length of bars = 2*(L/3) [where L is the larger of L1 and L2]

= 2*(4/3) = 2.7 m (use 3 m length)

- For continuous reinforcement in positive moment sections:

No. of bars that will be extended must be the largest of:

4 bars (due to width of beam)

(1/3) of maximum No. of bars [ (1/3)*7 = 2.33 ] which means 3 bars

Minimum reinforcement (4 bars)

So we choose to extend four bars.

Length of bars = span length + (W1/2 > 15 cm) + W2 + 30

= 4 + 0.15 + 0.2 + 30*0.014 = 4.1 m [use 4 m bars]

Position A Mid AB B Mid BC C Mid CD D f'c (kg/cm2) 280

Moment (t.m) -7.02 4.94 -10.26 2.28 -10.26 8.31 0 fy (kg/cm2) 4200

0.005279 0.00366 0.007907 0.001659 0.007907 0.006309 0 b (cm) 60

(chosen) 0.005279 0.00366 0.007907 0.00333 0.007907 0.006309 0.00333 d (cm) 24.8

As (cm2) 7.854493 5.445651 11.76587 4.95504 11.76587 9.387334 4.95504 0.9

Diameter of bar (mm) 14 14 14 14 14 14 14 (min) 0.00333

No. of bars 6 4 8 4 8 7 4 (=0.005) 0.018

No. of Layers 1 1 1 1 1 1 1 Cover (cm) 2

Spacing 9.12 16.13333 6.114286 16.13333 6.114286 7.366667 16.13333

Eng. Hasan Shehada Eng. Ruba Awad

- For Cut Reinforcement in Positive moment sections:

The remaining reinforcement will be put in the middle of the span at maximum positive

moment section.

Length of bar = (Span length center to center - width of columns)*(1-0.1-0.15)

= (4 0.325)(0.75) = 2.75 m (Use 3.0 m)

- For cantilever reinforcement:

For cantilever reinforcement, no cut of bars. Al required bars will be extended the required

distance.

Length of bar = Lc + 1.5*Lc = 1.2 + 1.5*1.2 = 3.5 m

The final drawing will be as follows