Eng. Hasan Shehada Eng. Ruba Awad

Eng. Hasan Shehada Eng. Ruba Awad


Introduction:

Direct design method is the first method of two way slab analysis in ACI 318. The

method relies on some factors at which the bending moment is distributed among the

different spans of the slab. This method has some limitations that must be considered

in order to use it. If one of the limitations is not applicable, then the second method

must be used which is the Equivalent Frame Method.

Limitations:

1- There must be at least three or more spans in each direction (three or more

slabs). The minimum number of slabs is shown in the figure below.

2- The ratio of length to width must not be more than 2 (all slabs must be two

way slabs).

3- Successive span must be different by not more than one third.

4- Offset of columns must be not more than 10% of the total span.

5- All loads applied on the slab must be gravity loads (no lateral loads).

6- The live load must not exceed double the dead load (LL < 2DL).

7- Relative stiffness between beam and slab must be more than 0.2 but not more

than 5 (0.2 < α < 5). Calculations for this ratio is discussed in the examples

below.

General Steps:

Direct design method is consisted of five main steps as follows:

1- Divide the slab into strips. Each strip will be analyzed separately.

2- Find the static moment on each span of the strip. In this step, it is assumed that

each span is a simply supported span.

3- Distribute the static moment into positive and negative moments, assuming

that the spans are continuous (not simply supported).

4- Distribute the bending moment between column strip and middle strip.

5- Distribute the bending moment in the column strip between the beam and the

slab. Details are discussed in the examples below.


Example 1:

Design the slab above using the Direct Design Method using the following data:

Wu = 1.5 t/m2 Beams are 20x40 cm Thickness of slab is 12 cm

Solution:

1- Divide the slab into strips. The strips are divided as shown in the figure below:

Width of strip = 0.5* L2' + 0.5*L2"


Another four strips in the perpendicular direction must be taken also. The hatched

strip will be analyzed and the same calculations are applied for all interior strips.

2- Find the static moment on the strip. To do so, use the equation:

𝑀𝑜 =𝑤𝑢 ∗ 𝐿2 ∗ 𝐿𝑛2

8

Where Wu is the load on slab in t/m2, L2 is the width of strip, and Ln is the clear span.

Before finding Mo, strip with its clear spans is drawn as follows:

𝑀𝑜(1) =1.5 ∗ 5 ∗ 3.72

8= 12.8 𝑡. 𝑚

𝑀𝑜(2) =1.5 ∗ 5 ∗ 3.82

8= 13.5 𝑡. 𝑚

𝑀𝑜(3) =1.5 ∗ 5 ∗ 2.72

8= 6.8 𝑡. 𝑚


3- Distribute the static moment into negative and positive moments using table (6)

which is shown below:

Since there are beams between all supports, no need to use cases (1) and (3). Cases

(2) and (4) are used. Case (2) is used for exterior spans and case (4) is used for

interior span. The table below summarizes the results.

Spans 1 2 3

Exterior Negative

Moment (0.16)(12.8) (0.65)(13.5) (0.16)(6.8)

Positive Moment (0.57)(12.8) (0.35)(13.5) (0.57)(6.8)

Interior Negative

Moment (0.7)(12.8) (0.65)(13.5) (0.7)(6.8)


4- Distribute the bending moment resulted from step 3 between column strip and

middle strip using table (7).

Before using the table, values of α and β must be found. By the following equations:

𝛼 =𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑏𝑒𝑎𝑚

𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏=

𝐸𝑐𝑏 ∗ 𝐼𝑏

𝐸𝑐𝑠 ∗ 𝐼𝑠

𝛽 =𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑒𝑑𝑔𝑒 𝑏𝑒𝑎𝑚𝑠

2 ∗ 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏=

𝐶

2 ∗ 𝐼𝑠

𝐼𝑠 =𝑏ℎ3

12=

500 ∗ 123

12= 72000 𝑐𝑚4

Since the beam has a T-shape cross section, the figure below is used to find the

stiffness of it.


𝐼𝑏 =𝑏ℎ3

12∗ 𝑓 =

20 ∗ 403

12∗ 1.74 = 185600

𝛼 =185600

72000= 2.58 < 5 𝑎𝑛𝑑 > 0.2 → 𝑜𝑘

Now find the torsional stiffness using the following equation:

𝐶 = ∑(1 − 0.63 (𝑥

𝑦))(

𝑥3 ∗ 𝑦

3)

Where x is the smallest dimension of rectangle.

The edge beam, which has an L-shape cross section, is divided into two rectangles.

There are two ways to divide the section into two rectangles as shown below.

Calculate the torsional stiffness for both of them and choose the highest one.

𝐶1 = (1 − 0.63 ∗ (12

28)) ∗ (

123 ∗ 28

3) + (1 − 0.63 ∗ (

20

40)) ∗ (

203 ∗ 40

3) = 84840.1

𝐶2 = (1 − 0.63 ∗ (12

48)) ∗ (

123 ∗ 48

3) + (1 − 0.63 ∗ (

20

28)) ∗ (

203 ∗ 28

3) = 64360.1

So choose C1 which is the largest. Now find β.

𝛽 =84840.1

2 ∗ 72000= 0.59

Now table (7) can be used.


Span (1) Span (2) Span (3)

L2 / L1 = 5/2.9 = 1.72 L2 / L1 = 5/4 = 1.25 L2 / L1 = 5/3.9 = 1.3

α (L2 / L1 )= 2.58*1.72 α (L2 / L1 )= 2.58*1.25 α (L2 / L1 )= 2.58*1.3

β = 0.59 β = 0.59 β = 0.59

Exterior Negative = 0.91 No Exterior Moment Exterior Negative = 0.92

Interior negative = 0.56 Interior negative = 0.69 Interior negative = 0.66

Positive = 0.56 Positive = 0.69 Positive = 0.66

The portions shown are for column strip. The portion of middle strip is the remaining of

this percentage.

Column Strip Moment

5- Distribute the moment on the column strip into beam moment and slab moment.

Since α (L2 / L1 ) > 1 then the beam will take 0.85 of the moment on the column

strip.


Example 2:

Design the slab above using the Direct Design Method using the following data:

Wu = 1.5 t/m2 Beams are 20x40 cm Thickness of slab is 12 cm

Solution:

1- Divide the slab into strips. The strips are divided as shown in the figure below:

Width of strip = 0.5* L2' + 0.5*L2"


Another four strips in the perpendicular direction must be taken also. The hatched

strip will be analyzed and the same calculations are applied for all exterior strips.

2- Find the static moment on the strip. To do so, use the equation:

𝑀𝑜 =𝑤𝑢 ∗ 𝐿2 ∗ 𝐿𝑛2

8

Where Wu is the load on slab in t/m2, L2 is the width of strip, and Ln is the clear span.

Before finding Mo, strip with its clear spans is drawn as follows:

𝑀𝑜(1) =1.5 ∗ 2.5 ∗ 3.72

8= 6.4 𝑡. 𝑚

𝑀𝑜(2) =1.5 ∗ 2.5 ∗ 3.82

8= 6.75 𝑡. 𝑚

𝑀𝑜(3) =1.5 ∗ 5 ∗ 2.72

8= 3.4 𝑡. 𝑚


3- Distribute the static moment into negative and positive moments using table (6)

which is shown below:

Since there are beams between all supports, no need to use cases (1) and (3). Cases

(2) and (4) are used. Case (2) is used for exterior spans and case (4) is used for

interior span. The table below summarizes the results.

Spans 1 2 3

Exterior Negative

Moment (0.16)(6.4) (0.65)(6.75) (0.16)(3.4)

Positive Moment (0.57)(6.4) (0.35)(6.75) (0.57)(3.4)

Interior Negative

Moment (0.7)(6.4) (0.65)(6.75) (0.7)(3.4)


4- Distribute the bending moment resulted from step 3 between column strip and

middle strip using table (7).

Before using the table, values of α and β must be found. By the following equations:

𝛼 =𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑏𝑒𝑎𝑚

𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏=

𝐸𝑐𝑏 ∗ 𝐼𝑏

𝐸𝑐𝑠 ∗ 𝐼𝑠

𝛽 =𝑇𝑜𝑟𝑠𝑖𝑜𝑛𝑎𝑙 𝑠𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑒𝑑𝑔𝑒 𝑏𝑒𝑎𝑚𝑠

2 ∗ 𝑆𝑡𝑖𝑓𝑓𝑛𝑒𝑠𝑠 𝑜𝑓 𝑠𝑙𝑎𝑏=

𝐶

2 ∗ 𝐼𝑠

𝐼𝑠 =𝑏ℎ3

12=

250 ∗ 123

12= 36000 𝑐𝑚4

Since the beam has an L-shape cross section, the figure below is used to find the

stiffness of it.


𝐼𝑏 =𝑏ℎ3

12∗ 𝑓 =

20 ∗ 403

12∗ 1.45 = 154667 𝑐𝑚4

𝛼 =154667

36000= 4.29 < 5 𝑎𝑛𝑑 > 0.2 → 𝑜𝑘

Now find the torsional stiffness using the following equation:

𝐶 = ∑(1 − 0.63 (𝑥

𝑦))(

𝑥3 ∗ 𝑦

3)

Where x is the smallest dimension of rectangle.

The edge beam, which has an L-shape cross section, is divided into two rectangles.

There are two ways to divide the section into two rectangles as shown below.

Calculate the torsional stiffness for both of them and choose the highest one.

𝐶1 = (1 − 0.63 ∗ (12

28)) ∗ (

123 ∗ 28

3) + (1 − 0.63 ∗ (

20

40)) ∗ (

203 ∗ 40

3) = 84840.1

𝐶2 = (1 − 0.63 ∗ (12

48)) ∗ (

123 ∗ 48

3) + (1 − 0.63 ∗ (

20

28)) ∗ (

203 ∗ 28

3) = 64360.1

So choose C1 which is the largest. Now find β.

𝛽 =84840.1

2 ∗ 360001.17

Now table (7) can be used.


Span (1) Span (2) Span (3)

L2 / L1 = 2.5/2.9 = 0.86 L2 / L1 = 2.5/4 = 0.625 L2 / L1 = 2.5/3.9 = 0.64

α (L2 / L1 )= 4.29*0.86 α (L2 / L1 )= 4.29*0.625 α (L2 / L1 )= 4.29*0.64

β = 1.17 β = 1.17 β = 1.17

Exterior Negative = 0.88 No Exterior Moment Exterior Negative = 0.93

Interior negative = 0.81 Interior negative = 0.87 Interior negative = 0.87

Positive = 0.81 Positive = 0.87 Positive = 0.87

The portions shown are for column strip. The portion of middle strip is the remaining of

this percentage.

5- Distribute the moment on the column strip into beam moment and slab moment.

Since α (L2 / L1 ) > 1 then the beam will take 0.85 of the moment on the column

strip.

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Eng. Hasan Shehada Eng. Ruba Awad