chapter 8 mat foundation - uoqasim.edu.iq

CHAPTER EIGHT

Shallow Foundation – Mat Foundation

8.1 Introduction Under normal conditions, square and rectangular footings such as those described in

previously are economical for supporting columns and walls. However, under certain

circumstances, it may be desirable to construct a footing that supports a line of two or

more columns. These footings are referred to as combined footings. When more than

one line of columns is supported by a concrete slab, it is called a mat foundation.

Common Types of Mat Foundations

The mat foundation, which is sometimes referred to as a raft foundation. Mat

foundations are sometimes preferred for soils that have low load-bearing capacities, but

that will have to support high column or wall loads. Under some conditions, spread

footings would have to cover more than half the building area, and mat foundations

might be more economical. Several types of mat foundations are used currently. Some

of the common ones are shown schematically in Figure 8.1 and include the following:

1. Flat plate (Figure 8.1a). The mat is of uniform thickness.

2. Flat plate thickened under columns (Figure 8.1b).

3. Beams and slab (Figure 8.1c). The beams run both ways, and the columns are located

at the intersection of the beams.

4. Flat plates with pedestals (Figure 8.1d).


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AL-Qasim Green University College Of Water Resources Engineering

Foundation Engineering Mr. Wissam Nadir

5. Slab with basement walls as a part of the mat (Figure 8.1e). The walls act as stiffeners

for the mat.

Mats may be supported by piles, which help reduce the settlement of a structure built

over highly compressible soil. Where the water table is high, mats are often placed over

piles to control buoyancy.


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8.2 Structural Design of Mat Foundations The structural design of mat foundations can be carried out by two conventional

methods:

The conventional rigid method

The approximate flexible method

Finite-difference and finite-element methods can also be used, but this section

covers only the basic concepts of the conventional rigid method are discussed in this

chapter.

8.3 Design Procedure for the Conventional (Rigid) Method The procedure for design a mat foundation, using the conventional method, consists of

the following steps:

Step 1. All the columns and walls are numbered and their total unfactored (working)

axial loads, moments and any overturning moment due to wind or other causes, are

calculated separately. It may be useful if a suitable table is used for this purpose.

Step 2. The line of action of the resultant R of all the axial loads and moments is

determined using statics; summing moments about two of the adjacent mat edges, and

computing the x and y moment arms of the resultant force. Then, the eccentricities ex

and ey are computed (Figure 8.2a).

Step 3- Determine the maximum contact pressure (unfactored), qmax, below one of the

mat corners using Equation below, repeated here for convenience.

𝑞 =𝑅

𝐵𝐿1 ∓

6𝑒

𝐵∓

6𝑒

𝐿

In this equation, the B and L dimensions are in x and y directions, respectively.

Step 4. Compare the computed qmax with the allowable soil pressure qa furnished by the

geotechnical engineer to check if qmax ≤ qa.


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Step 5. Compute the contact pressure (unfactored) at selected points beneath the mat.

These selected points are corners of continuous beam strips (or combined footings with

multiple columns) to which the mat is divided in both x and y directions, as shown in

Figure 8.2b. The contact pressure q at any point below the mat is computed using

Equation:

𝑞 , =𝑅

𝐴∓

𝑀

𝐼𝑥 ∓

𝑀

𝐼𝑦

Where:

A = B × L

𝐼 = 𝐵𝐿 12⁄ , 𝐼 = 𝐿𝐵 12⁄

𝑀 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑 𝑎𝑏𝑜𝑢𝑙 𝑥 − 𝑎𝑥𝑖𝑠 = 𝑅𝑒

𝑀 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑 𝑎𝑏𝑜𝑢𝑙 𝑦 − 𝑎𝑥𝑖𝑠 = 𝑅𝑒

Step 6. Divide the mat into several strips in the x and y directions. (See Figure 8.2). Let

the width of any strip be B1

Step 7. Draw the shear, V, and the moment, M, diagrams for each individual strip (in the

x and y directions). For example, the average soil pressure of the bottom strip in the x

direction of Figure 8.2 a is

𝑞 =𝑞 + 𝑞

2

Where qI and qF = soil pressures at points I and F, as determined from Step 5.

The total soil reaction is equal to q B1B.

Now obtain the total column load on the strip as Q1+ Q2 + Q3 + Q4. The sum of the

column loads on the strip will not equal q B1B, because the shear between the

adjacent strips has not been taken into account. For this reason, the soil reaction and


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the column loads need to be adjusted, or

𝑎𝑣𝑎𝑟𝑒𝑎𝑔𝑒 𝑙𝑜𝑎𝑑 =q B B + (Q1 + Q2 + Q3 + Q4)

2

Now, the modified average soil reaction becomes

𝑞 ( ) = 𝑞𝑎𝑣𝑎𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑

q B B

In addition, the column load modification factor is

𝐹 =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑

(Q1 + Q2 + Q3 + Q4)

So the modified column loads are FQ1, FQ2, FQ3 , and FQ4. This modified loading on the

strip under consideration is shown in Figure 8.2b. The shear and the moment diagram

for this strip can now be drawn, and the procedure is repeated in the x and y directions

for all strips.


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Figure 8.2


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Step 8. Determine the effective depth (d) of the mat by checking for diagonal tension

shear near various columns. For the critical section,

∅𝑉 ≥ 𝑉

According to ACI Code 318-14 (Section 22.6.5.2) for nonprestressed slabs and footings,

Vc shall be the smallest of the equations

ACI Table 22.6.5.2 Tow Way Shear

𝑣

Least of (a),(b) and (c)

0.33 𝜆 𝑓 𝑏°𝑑 a

0.17 1 +2

𝛽 𝜆 𝑓 𝑏°𝑑 b

0.083 2 +𝛼 𝑑

𝑏° 𝜆 𝑓 𝑏°𝑑 c

Step 9. From the moment diagrams of all strips in one direction (x or y), obtain the

maximum positive and negative moments per unit width (i.e.,Mu= M/B1).

Step 10. Determine the area of steel per unit width for positive and negative

reinforcement in the x and y directions.


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Example: The plan of a simple mat foundation and the tabulated column working loads

are shown in the figure below. All the columns are 0.4 × 0.4 m and carry no moment of

significant value. The design soil pressure or net qa and average modulus of subgrade

reaction Ks,av, recommended by the geotechnical engineer, are 60 kPa and 7200 kN/m3,

respectively. Design the mat by the conventional rigid method, and then check if this

design method could be considered appropriate. Use:𝑓 = 24𝑀𝑃𝑎 , 𝑓𝑦 = 420𝑀𝑃𝑎

Col. No.

Axial col. loads D, kN L, kN

1 80 160 2 120 240 3 100 200 4 320 640 5 320 640 6 270 540 7 320 640 8 320 640 9 270 540 10 80 160 11 120 240 12 90 180


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Solution:

Col. No.

Axial col. loads D, kN L, kN D+L

1 80 160 240 2 120 240 360 3 100 200 300 4 320 640 960 5 320 640 960 6 270 540 810 7 320 640 960 8 320 640 960 9 270 540 810 10 80 160 240 11 120 240 360 12 90 180 270 sum 2410 4820 7230

Step 1. Compute the resultant of working load and its location.

a. 𝑅 = 2410 + 4820 = 7230 𝑘𝑁 b. Find the location of resultant and calculate the eccentricity in X and Y direction.

Take moments about left edge to find �̅�

�̅� =0.2 × (2 × 240 + 2 × 960) + 6(2 × 360 + 2 × 960) + 11.8(300 + 2 × 810 + 270)

7230

= 5.831𝑚

𝑒 = 𝐵 2⁄ − �̅� = 6 − 5.83 = 0.17 <𝐵

6= 2 𝑜𝑘

Take moments about bottom edge to find 𝑦 𝑦

=0.2(240 + 360 + 270) + 7.4(2 × 960 + 810) + 14.6(960 × 2 + 810) + 21.8(240 + 360 + 300)

7230

= 11.044𝑚

𝑒 = 𝐿 2⁄ − 𝑦 = 11 − 11.044 = −0.04 <𝐿

6= 3.66 𝑜𝑘


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Step 2. Determine the maximum contact pressure 𝑞 .and compare with net bearing capacity.

𝑞 =𝑅

𝐵𝐿1 +

6𝑒

𝐵+

6𝑒

𝐿

𝑞 . =7230

12 × 221 +

6 × 0.17

12+

6 × 0.04

22= 30.01 𝑘𝑁 𝑚⁄

𝑞 . = 60 𝑘𝑁 𝑚⁄ > 𝑞 . = 30.01 𝑘𝑁 𝑚⁄

Step 3. Divide the the mat as continuous beam strip in both x and Y direction, then compute the unfactored contact pressure in each corner of strip.

point X (m) Y (m) q kN/m2 A 6 11 30.00 E 2.9 11 28.80 F -2.9 11 26.55 B -6 11 25.35 I 6 7.2 29.9 J -6 7.2 25.24 K 6 0 29.7 L -6 0 25.05 M 6 -7.2 29.51 N -6 -7.2 24.85 C 6 -12 29.38 G 2.9 -12 28.18 H -2.9 -12 25.93 D -6 -12 24.72


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𝑞 , =𝑅

𝐴∓

𝑀

𝐼𝑥 ∓

𝑀

𝐼𝑦

𝑀 = 𝑅 × 𝑒 = 7230 × 0.17 = 1229.1 𝑘𝑁. 𝑚

𝑀 = 𝑅 × 𝑒 = 7230 × 0.04 = 289.2 𝑘𝑁. 𝑚

𝐼 =𝐿 𝐵

12= 3168 𝑚 , 𝐼 =

𝐵 𝐿

12= 10648 𝑚

𝑞 , =7230

12 × 22∓

1229.1

3168𝑥 ∓

289.2

10648𝑦 = 27.38 ∓ 0.388 𝑥 ∓ 0.027 𝑦

Step 4: check the equilibrium for each strip and modify the column load and contact pressure accordingly.

Y-Direction a. Beam strip (ACGE) (3.1m X22 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 240 + 960 + 960 + 240 = 2400 𝑘𝑁

𝑎𝑣𝑎𝑟𝑎𝑔𝑒 𝑠𝑜𝑖𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 22 × 3.130 + 29.38 + 28.18 + 28.8)

4= 1983.94 𝑘𝑁

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 ≠ 𝑠𝑜𝑖𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

𝑎𝑣𝑎𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑 =𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 + 𝑠𝑜𝑖𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛

2=

2400 + 1983.94

2= 2191.97 𝑘𝑁

𝑡ℎ𝑒 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 𝐴𝐶𝐺𝐸 =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑=

2191.97

2400= 0.913

Find the modification factored column load and ultimate bearing capacity

Column No. Modified ultimate load kN 1 0.913 × (1.2 × 80 + 1.6 × 160) 321.376 4 0.913 × (1.2 × 320 + 1.6 × 640) 1285.5 7 0.913 × (1.2 × 320 + 1.6 × 640) 1285.5


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10 0.913 × (1.2 × 80 + 1.6 × 160) 321.376 sum 3213.75 kN Since the strip is symmetrically loaded, the factored R falls at the center and the factored contact pressure qult is assumed uniformly distributed.

𝑞 . =𝑃

𝐴=

3213.75

22 × 3.1= 47.122 𝑘𝑁 𝑚⁄

b. Beam strip (EGHF) (5.8m X22 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 360 + 960 + 960 + 360 = 2640 𝑘𝑁

𝑎𝑣𝑎𝑟𝑎𝑔𝑒 𝑠𝑜𝑖𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 22 × 5.828.8 + 28.18 + 25.93 + 26.55)

4= 3491.77 𝑘𝑁



2=

2640 + 3491.77

2= 3065.88 𝑘𝑁

𝑡ℎ𝑒 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 𝐸𝐺𝐻𝐹 =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑


3491.77

3065.88= 1.139


Column No. Modified ultimate load kN 2 1.139 × (1.2 × 120 + 1.6 × 240) 601.39 5 1.139 × (1.2 × 320 + 1.6 × 640) 1603.71 8 1.139 × (1.2 × 320 + 1.6 × 640) 1603.71 11 1.139 × (1.2 × 120 + 1.6 × 240) 601.39 sum 4410.2kN

Since the strip is symmetrically loaded, the factored R falls at the center and the factored contact pressure qult is assumed uniformly distributed.


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𝑞 . =𝑃

𝐴=

4410.2

22 × 5.8= 34.563 𝑘𝑁 𝑚⁄

c. Beam strip (FHDB) (3.1m X22 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 300 + 810 + 810 + 270 = 2190 𝑘𝑁


4= 1740.8 𝑘𝑁



2=

2190 + 1740.8

2= 1965.4 𝑘𝑁

𝑡ℎ𝑒 𝑚𝑜𝑑𝑖𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 FHDB =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑


1965.4

2190= 0.897


Column No. Modified ultimate load kN 3 0.897 × (1.2 × 100 + 1.6 × 200) 394.68 6 0.897 × (1.2 × 270 + 1.6 × 540) 1065.64 9 0.897 × (1.2 × 270 + 1.6 × 540) 1065.64 12 0.897 × (1.2 × 90 + 1.6 × 180) 355.212 sum 2881.17 Since the strip is unsymmetrically loaded, the factored R does not fall at the center; it has an eccentricity eL or ey, calculated as follows:

𝑦 =394.68 × 0.2 + 1065.64 × 7.4 + 1065.64 × 14.6 + 355.21 × 21.8

2881.17= 10.85 𝑚

𝑒 = 𝐿 2⁄ − 𝑦 = 11 − 10.85 = 0.15 < 𝐿 6⁄

𝑞 . ., =𝑅

𝐴∓

𝑒 𝑅

𝐼𝑙


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𝑞 . . =2881.17

3.1 × 22+

0.15 × 2881.17. ×

× 11 = 44 𝑘𝑁 𝑚⁄

𝑞 . . =2881.17

3.1 × 22−

0.15 × 2881.17. ×

× 11 = 40.51 𝑘𝑁 𝑚⁄

X- Direction d. Beam strip (AIJB) (3.8m X12 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 240 + 360 + 300 = 900 𝑘𝑁

𝑎𝑣𝑎𝑟𝑎𝑔𝑒 𝑠𝑜𝑖𝑙 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛 = 12 × 3.830 + 29.9 + 25.4 + 25.35)

4= 1261.41 𝑘𝑁



2=

900 + 1261.41

2= 1080.7 𝑘𝑁

𝑚𝑜𝑑𝑒𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 AIJB =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑


1080.7

900= 1.20


Column No. Modified ultimate load kN 1 1.2 × (1.2 × 80 + 1.6 × 160) 422.6 2 1.2 × (1.2 × 120 + 1.6 × 240) 633.6 3 1.2 × (1.2 × 100 + 1.6 × 200) 528 sum 1584.2kN Since the strip is unsymmetrically loaded, the factored R does not fall at the center; it has an eccentricity eB or eX, calculated as follows:

�̅� =422.6 × 0.2 + 633.6 × 6 + 528 × 11.8

1584.2= 6.386 𝑚

𝑒 = 𝐵 2⁄ − �̅� = 6 − 6.386 = −0.386 < 𝐿 6⁄

𝑞 . ., =𝑅

𝐴∓

𝑒 𝑅

𝐼𝑙


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𝑞 . . =1584.2

3.8 × 12+

0.386 × 1584.2. ×

× 6 = 41.13 𝑘𝑁 𝑚⁄

𝑞 . . =1584.2

3.8 × 12−

0.386 × 1584.2. ×

× 6 = 28.036 𝑘𝑁 𝑚⁄

e. Beam strip (IKLJ) (7.2m X12 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 960 + 960 + 810 = 2730 𝑘𝑁


4= 2373.62 𝑘𝑁



2=

2730 + 2373.62

2= 2551.81 𝑘𝑁

𝑚𝑜𝑑𝑒𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 IKLJ =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑


2551.81

2730= 0.935


Column No. Modified ultimate load kN 4 0.935 × (1.2 × 320 + 1.6 × 640) 1316.48 5 0.935 × (1.2 × 320 + 1.6 × 640) 1316.48 6 0.935 × (1.2 × 270 + 1.6 × 540) 1110.78 sum 3743.74kN

Since the strip is unsymmetrically loaded, the factored R does not fall at the center; it has an eccentricity eB or eX, calculated as follows:

�̅� =1316.48 × 0.2 + 1316.48 × 6 + 1110.78 × 11.8

3743.74= 5.68 𝑚

𝑒 = 𝐵 2⁄ − �̅� = 6 − 5.68 = 0.319 < 𝐿 6⁄


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𝑞 . ., =𝑅

𝐴∓

𝑒 𝑅

𝐼𝑙

𝑞 . . =3743.74

7.2 × 12+

0.319 × 3743.74. ×

× 6 = 50.178 𝑘𝑁 𝑚⁄

𝑞 . . =3743.74

7.2 × 12−

0.319 × 3743.74. ×

× 6 = 36.42 𝑘𝑁 𝑚⁄

f. Beam strip (KMNL) (7.2m X12 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 960 + 960 + 810 = 2730 𝑘𝑁


4= 2356.77 𝑘𝑁

We can assume the beam strip (KMNL) it’s similar to Beam strip (IKLJ)

g. Beam strip (MCDN) (3.8m X12 m)

𝑐𝑜𝑙𝑢𝑚𝑛 𝑙𝑜𝑎𝑑𝑠 = 240 + 360 + 270 = 870 𝑘𝑁


4= 1236.44 𝑘𝑁



2=

870 + 1236.44

2= 1053.22 𝑘𝑁

𝑚𝑜𝑑𝑒𝑓𝑖𝑐𝑎𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑐𝑜𝑙𝑢𝑚𝑛𝑠 𝑜𝑓 𝑠𝑡𝑟𝑖𝑝 MCDN =𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑜𝑎𝑑


1053.22

870= 1.21


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Column No. Modified ultimate load kN 10 1.21 × (1.2 × 80 + 1.6 × 160) 425.92 11 1.21 × (1.2 × 120 + 1.6 × 240) 638.88 12 1.21 × (1.2 × 90 + 1.6 × 180) 479.16 sum 1543.96kN

Since the strip is unsymmetrically loaded, the factored R does not fall at the center; it has an eccentricity eB or eX, calculated as follows:

�̅� =425.92 × 0.2 + 638.88 × 6 + 479.16 × 11.8

1543.96= 6.2 𝑚

𝑒 = 𝐵 2⁄ − �̅� = 6 − 6.2 = 0.2 < 𝐿 6⁄

𝑞 . ., =𝑅

𝐴∓

𝑒 𝑅

𝐼𝑙

𝑞 . . =1543.96

3.8 × 12+

0.2 × 1543.96. ×

× 6 = 37.24 𝑘𝑁 𝑚⁄

𝑞 . . =3743.74

7.2 × 12−

0.319 × 3743.74. ×

× 6 = 30.47 𝑘𝑁 𝑚⁄

Step 6. Draw factored load, shear and moment diagrams for the continuous beam strips in both directions.


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Moment kN.m/m

y- Strips x- Strips ACGE EGHF FHDB AIJB IKLJ MCDN

+ M 543.8 199.77 289.57 36 -M 93.3 134.74 160.89 212.53 306.26 179.18

Step 7. Determine the minimum mat thickness considering punching shear (two-way or

diagonal tension shear) at critical columns.

Considering the factors which contribute to two-way shear (column load, soil reaction,

column size and length of shear perimeter at each column), by inspection, it appears

that the maximum shear occurs at column No. 4 (or column No. 7) with a three-side

shear perimeter.

Factored load = 1.2 × 320 + 1.6 × 640 = 1048kN 𝑉 = 1048 𝑘𝑁 Note that the maximum value of d is selected as the design value and it corresponds to the minimum value of Vc obtained from equations

ACI Table 22.6.5.2 Tow Way Shear

𝑣

Least of (a),(b) and (c)

0.33 𝜆 𝑓 𝑏°𝑑 a

0.17 1 +2

𝛽 𝜆 𝑓 𝑏°𝑑 b

0.083 2 +𝛼 𝑑

𝑏° 𝜆 𝑓 𝑏°𝑑 c


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𝑏 = 2 400 + 𝑑2 + (400 + 𝑑) = 1200 + 2𝑑

𝛼 = 30

𝑉 ≤ ∅𝑉

From equation (a)

1408 × 10 = 0.33 𝜆 𝑓 𝑏°𝑑

1408 × 10 = 0.33 × 1 × √24 × (1200 + 2𝑑)𝑑 → 𝑑 = 424.9 𝑚𝑚 𝒄𝒐𝒏𝒕𝒓𝒐𝒍 From equation (b)

1408 × 10 = 0.17 1 +2

𝛽 𝜆 𝑓 𝑏°𝑑

1408 × 10 = 0.17 × 3 𝜆√24(1200 + 2𝑑)𝑑 → 𝑑 = 309.8 𝑚𝑚 From equation (c)

1408 × 10 = 0.083 2 +𝛼 𝑑

𝑏° 𝜆 𝑓 𝑏°𝑑

1408 × 10 = 0.083 2 +30 × 𝑑

(1200 + 2𝑑) 𝜆 𝑓 (1200 + 2𝑑)𝑑 → 𝑑 = 285.8 𝑚𝑚

𝑈𝑠𝑒 𝑑 = 450 𝑚𝑚

𝐻 = 𝑑 + 𝑐𝑜𝑣𝑒𝑟 +𝑑

2= 450 + 75 + 12.5 = 537.5𝑚𝑚 ≈ 550𝑚𝑚


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Step 9. Obtain the design factored positive and negative moments per metre width using the factored moment diagrams in Step 6.

1. Positive moment for long direction 543.8 kN.m /m

∅𝑀 ≥ 𝑀 → ∅𝑀 = ∅𝜌𝑏𝑑 𝑓𝑦 1 − 0.59 ×𝑓𝑐

′ × 𝜌

543.8 × 10 = 0.9 × 𝜌 × 1000 × 450 × 420 1 − 0.59 ×420

24× 𝜌

→ 𝜌 = 0.00772 → 𝐴 = 3474 𝑚𝑚 𝑚⁄

𝐴 = 0.0018 × 𝐵 × 𝑡 = 990 𝑚𝑚 𝑚⁄

𝑛 =3474

491= 7.075 𝑏𝑎𝑟 𝑠 =

1000

7.075= 140𝑚𝑚

2. Negative moment for long direction 160.89 kN.m /m

∅𝑀 ≥ 𝑀 → ∅𝑀 = ∅𝜌𝑏𝑑 𝑓𝑦 1 − 0.59 ×𝑓𝑐

′ × 𝜌

160.89 × 10 = 0.9 × 𝜌 × 1000 × 425 × 420 1 − 0.59 ×420

24× 𝜌

→ 𝜌 = 0.00242 → 𝐴 = 1028.5 𝑚𝑚 𝑚⁄

𝐴 = 0.0018 × 𝐵 × 𝑡 = 990 𝑚𝑚 𝑚⁄

𝑛 =1028.5

201= 5.11 𝑏𝑎𝑟 𝑠 =

1000

5.11= 195 𝑚𝑚

3. Positive moment for short direction 36 kN.m /m

∅𝑀 ≥ 𝑀 → ∅𝑀 = ∅𝜌𝑏𝑑 𝑓𝑦 1 − 0.59 ×𝑓𝑐

′ × 𝜌

36 × 10 = 0.9 × 𝜌 × 1000 × 450 × 420 1 − 0.59 ×420

24× 𝜌

→ 𝜌 = 0.00047 → 𝐴 = 211 𝑚𝑚 𝑚⁄

𝐴 = 0.0018 × 𝐵 × 𝑡 = 990 𝑚𝑚 𝑚⁄


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𝑛 =900

201= 4.477 𝑏𝑎𝑟 𝑠 =

1000

4.477= 223𝑚𝑚

4. Negative moment for short direction 306.26 kN.m /m

∅𝑀 ≥ 𝑀 → ∅𝑀 = ∅𝜌𝑏𝑑 𝑓𝑦 1 − 0.59 ×𝑓𝑐

′ × 𝜌

306.26 × 10 = 0.9 × 𝜌 × 1000 × 425 × 420 1 − 0.59 ×420

24× 𝜌

→ 𝜌 = 0.004715 → 𝐴 = 2003.87 𝑚𝑚 𝑚⁄

𝐴 = 0.0018 × 𝐵 × 𝑡 = 990 𝑚𝑚 𝑚⁄

𝑛 =2003.87

491= 4.08 𝑏𝑎𝑟 𝑠 =

1000

4.08= 245 𝑚𝑚

chapter 8 mat foundation - uoqasim.edu.iq

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