chapter 8 mat foundation - uoqasim.edu.iq
Transcript of chapter 8 mat foundation - uoqasim.edu.iq
CHAPTER EIGHT
Shallow Foundation โ 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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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.
๐ =๐
๐ต๐ฟ1 โ
6๐
๐ตโ
6๐
๐ฟ
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 โค 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:
๐ , =๐
๐ดโ
๐
๐ผ๐ฅ โ
๐
๐ผ๐ฆ
Where:
A = B ร L
๐ผ = ๐ต๐ฟ 12โ , ๐ผ = ๐ฟ๐ต 12โ
๐ = ๐๐๐๐๐๐ก ๐๐ ๐๐๐๐ข๐๐ ๐๐๐๐ ๐๐๐๐ข๐ ๐ฅ โ ๐๐ฅ๐๐ = ๐ ๐
๐ = ๐๐๐๐๐๐ก ๐๐ ๐๐๐๐ข๐๐ ๐๐๐๐ ๐๐๐๐ข๐ ๐ฆ โ ๐๐ฅ๐๐ = ๐ ๐
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
๐ =๐ + ๐
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
๐๐ฃ๐๐๐๐๐๐ ๐๐๐๐ =q B B + (Q1 + Q2 + Q3 + Q4)
2
Now, the modified average soil reaction becomes
๐ ( ) = ๐๐๐ฃ๐๐๐๐๐ ๐๐๐๐
q B B
In addition, the column load modification factor is
๐น =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
(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,
โ ๐ โฅ ๐
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
๐ฃ
Least of (a),(b) and (c)
0.33 ๐ ๐ ๐ยฐ๐ a
0.17 1 +2
๐ฝ ๐ ๐ ๐ยฐ๐ b
0.083 2 +๐ผ ๐
๐ยฐ ๐ ๐ ๐ยฐ๐ 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 ร 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:๐ = 24๐๐๐ , ๐๐ฆ = 420๐๐๐
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. ๐ = 2410 + 4820 = 7230 ๐๐ b. Find the location of resultant and calculate the eccentricity in X and Y direction.
Take moments about left edge to find ๏ฟฝฬ ๏ฟฝ
๏ฟฝฬ ๏ฟฝ =0.2 ร (2 ร 240 + 2 ร 960) + 6(2 ร 360 + 2 ร 960) + 11.8(300 + 2 ร 810 + 270)
7230
= 5.831๐
๐ = ๐ต 2โ โ ๏ฟฝฬ ๏ฟฝ = 6 โ 5.83 = 0.17 <๐ต
6= 2 ๐๐
Take moments about bottom edge to find ๐ฆ ๐ฆ
=0.2(240 + 360 + 270) + 7.4(2 ร 960 + 810) + 14.6(960 ร 2 + 810) + 21.8(240 + 360 + 300)
7230
= 11.044๐
๐ = ๐ฟ 2โ โ ๐ฆ = 11 โ 11.044 = โ0.04 <๐ฟ
6= 3.66 ๐๐
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Step 2. Determine the maximum contact pressure ๐ .and compare with net bearing capacity.
๐ =๐
๐ต๐ฟ1 +
6๐
๐ต+
6๐
๐ฟ
๐ . =7230
12 ร 221 +
6 ร 0.17
12+
6 ร 0.04
22= 30.01 ๐๐ ๐โ
๐ . = 60 ๐๐ ๐โ > ๐ . = 30.01 ๐๐ ๐โ
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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๐ , =๐
๐ดโ
๐
๐ผ๐ฅ โ
๐
๐ผ๐ฆ
๐ = ๐ ร ๐ = 7230 ร 0.17 = 1229.1 ๐๐. ๐
๐ = ๐ ร ๐ = 7230 ร 0.04 = 289.2 ๐๐. ๐
๐ผ =๐ฟ ๐ต
12= 3168 ๐ , ๐ผ =
๐ต ๐ฟ
12= 10648 ๐
๐ , =7230
12 ร 22โ
1229.1
3168๐ฅ โ
289.2
10648๐ฆ = 27.38 โ 0.388 ๐ฅ โ 0.027 ๐ฆ
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)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 240 + 960 + 960 + 240 = 2400 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 22 ร 3.130 + 29.38 + 28.18 + 28.8)
4= 1983.94 ๐๐
๐๐๐๐ข๐๐ ๐๐๐๐๐ โ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
๐๐ฃ๐๐๐๐๐ ๐๐๐๐ =๐๐๐๐ข๐๐ ๐๐๐๐๐ + ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
2=
2400 + 1983.94
2= 2191.97 ๐๐
๐กโ๐ ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐ข๐๐๐ ๐๐ ๐ ๐ก๐๐๐ ๐ด๐ถ๐บ๐ธ =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
๐๐๐๐ข๐๐ ๐๐๐๐=
2191.97
2400= 0.913
Find the modification factored column load and ultimate bearing capacity
Column No. Modified ultimate load kN 1 0.913 ร (1.2 ร 80 + 1.6 ร 160) 321.376 4 0.913 ร (1.2 ร 320 + 1.6 ร 640) 1285.5 7 0.913 ร (1.2 ร 320 + 1.6 ร 640) 1285.5
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10 0.913 ร (1.2 ร 80 + 1.6 ร 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.
๐ . =๐
๐ด=
3213.75
22 ร 3.1= 47.122 ๐๐ ๐โ
b. Beam strip (EGHF) (5.8m X22 m)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 360 + 960 + 960 + 360 = 2640 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 22 ร 5.828.8 + 28.18 + 25.93 + 26.55)
4= 3491.77 ๐๐
๐๐๐๐ข๐๐ ๐๐๐๐๐ โ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
๐๐ฃ๐๐๐๐๐ ๐๐๐๐ =๐๐๐๐ข๐๐ ๐๐๐๐๐ + ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
2=
2640 + 3491.77
2= 3065.88 ๐๐
๐กโ๐ ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐ข๐๐๐ ๐๐ ๐ ๐ก๐๐๐ ๐ธ๐บ๐ป๐น =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
๐๐๐๐ข๐๐ ๐๐๐๐=
3491.77
3065.88= 1.139
Find the modification factored column load and ultimate bearing capacity
Column No. Modified ultimate load kN 2 1.139 ร (1.2 ร 120 + 1.6 ร 240) 601.39 5 1.139 ร (1.2 ร 320 + 1.6 ร 640) 1603.71 8 1.139 ร (1.2 ร 320 + 1.6 ร 640) 1603.71 11 1.139 ร (1.2 ร 120 + 1.6 ร 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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๐ . =๐
๐ด=
4410.2
22 ร 5.8= 34.563 ๐๐ ๐โ
c. Beam strip (FHDB) (3.1m X22 m)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 300 + 810 + 810 + 270 = 2190 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 22 ร 3.126.55 + 25.93 + 24.27 + 25.35)
4= 1740.8 ๐๐
๐๐๐๐ข๐๐ ๐๐๐๐๐ โ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
๐๐ฃ๐๐๐๐๐ ๐๐๐๐ =๐๐๐๐ข๐๐ ๐๐๐๐๐ + ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
2=
2190 + 1740.8
2= 1965.4 ๐๐
๐กโ๐ ๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐ข๐๐๐ ๐๐ ๐ ๐ก๐๐๐ FHDB =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
๐๐๐๐ข๐๐ ๐๐๐๐=
1965.4
2190= 0.897
Find the modification factored column load and ultimate bearing capacity
Column No. Modified ultimate load kN 3 0.897 ร (1.2 ร 100 + 1.6 ร 200) 394.68 6 0.897 ร (1.2 ร 270 + 1.6 ร 540) 1065.64 9 0.897 ร (1.2 ร 270 + 1.6 ร 540) 1065.64 12 0.897 ร (1.2 ร 90 + 1.6 ร 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:
๐ฆ =394.68 ร 0.2 + 1065.64 ร 7.4 + 1065.64 ร 14.6 + 355.21 ร 21.8
2881.17= 10.85 ๐
๐ = ๐ฟ 2โ โ ๐ฆ = 11 โ 10.85 = 0.15 < ๐ฟ 6โ
๐ . ., =๐
๐ดโ
๐ ๐
๐ผ๐
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๐ . . =2881.17
3.1 ร 22+
0.15 ร 2881.17. ร
ร 11 = 44 ๐๐ ๐โ
๐ . . =2881.17
3.1 ร 22โ
0.15 ร 2881.17. ร
ร 11 = 40.51 ๐๐ ๐โ
X- Direction d. Beam strip (AIJB) (3.8m X12 m)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 240 + 360 + 300 = 900 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 12 ร 3.830 + 29.9 + 25.4 + 25.35)
4= 1261.41 ๐๐
๐๐๐๐ข๐๐ ๐๐๐๐๐ โ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
๐๐ฃ๐๐๐๐๐ ๐๐๐๐ =๐๐๐๐ข๐๐ ๐๐๐๐๐ + ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
2=
900 + 1261.41
2= 1080.7 ๐๐
๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐ข๐๐๐ ๐๐ ๐ ๐ก๐๐๐ AIJB =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
๐๐๐๐ข๐๐ ๐๐๐๐=
1080.7
900= 1.20
Find the modification factored column load and ultimate bearing capacity
Column No. Modified ultimate load kN 1 1.2 ร (1.2 ร 80 + 1.6 ร 160) 422.6 2 1.2 ร (1.2 ร 120 + 1.6 ร 240) 633.6 3 1.2 ร (1.2 ร 100 + 1.6 ร 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:
๏ฟฝฬ ๏ฟฝ =422.6 ร 0.2 + 633.6 ร 6 + 528 ร 11.8
1584.2= 6.386 ๐
๐ = ๐ต 2โ โ ๏ฟฝฬ ๏ฟฝ = 6 โ 6.386 = โ0.386 < ๐ฟ 6โ
๐ . ., =๐
๐ดโ
๐ ๐
๐ผ๐
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๐ . . =1584.2
3.8 ร 12+
0.386 ร 1584.2. ร
ร 6 = 41.13 ๐๐ ๐โ
๐ . . =1584.2
3.8 ร 12โ
0.386 ร 1584.2. ร
ร 6 = 28.036 ๐๐ ๐โ
e. Beam strip (IKLJ) (7.2m X12 m)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 960 + 960 + 810 = 2730 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 12 ร 7.229.9 + 29.7 + 25.05 + 25.24)
4= 2373.62 ๐๐
๐๐๐๐ข๐๐ ๐๐๐๐๐ โ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
๐๐ฃ๐๐๐๐๐ ๐๐๐๐ =๐๐๐๐ข๐๐ ๐๐๐๐๐ + ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
2=
2730 + 2373.62
2= 2551.81 ๐๐
๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐ข๐๐๐ ๐๐ ๐ ๐ก๐๐๐ IKLJ =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
๐๐๐๐ข๐๐ ๐๐๐๐=
2551.81
2730= 0.935
Find the modification factored column load and ultimate bearing capacity
Column No. Modified ultimate load kN 4 0.935 ร (1.2 ร 320 + 1.6 ร 640) 1316.48 5 0.935 ร (1.2 ร 320 + 1.6 ร 640) 1316.48 6 0.935 ร (1.2 ร 270 + 1.6 ร 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:
๏ฟฝฬ ๏ฟฝ =1316.48 ร 0.2 + 1316.48 ร 6 + 1110.78 ร 11.8
3743.74= 5.68 ๐
๐ = ๐ต 2โ โ ๏ฟฝฬ ๏ฟฝ = 6 โ 5.68 = 0.319 < ๐ฟ 6โ
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๐ . ., =๐
๐ดโ
๐ ๐
๐ผ๐
๐ . . =3743.74
7.2 ร 12+
0.319 ร 3743.74. ร
ร 6 = 50.178 ๐๐ ๐โ
๐ . . =3743.74
7.2 ร 12โ
0.319 ร 3743.74. ร
ร 6 = 36.42 ๐๐ ๐โ
f. Beam strip (KMNL) (7.2m X12 m)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 960 + 960 + 810 = 2730 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 12 ร 7.229.7 + 29.51 + 24.85 + 25.05)
4= 2356.77 ๐๐
We can assume the beam strip (KMNL) itโs similar to Beam strip (IKLJ)
g. Beam strip (MCDN) (3.8m X12 m)
๐๐๐๐ข๐๐ ๐๐๐๐๐ = 240 + 360 + 270 = 870 ๐๐
๐๐ฃ๐๐๐๐๐ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐ = 12 ร 3.829.51 + 29.38 + 24.72 + 24.85)
4= 1236.44 ๐๐
๐๐๐๐ข๐๐ ๐๐๐๐๐ โ ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
๐๐ฃ๐๐๐๐๐ ๐๐๐๐ =๐๐๐๐ข๐๐ ๐๐๐๐๐ + ๐ ๐๐๐ ๐๐๐๐๐ก๐๐๐
2=
870 + 1236.44
2= 1053.22 ๐๐
๐๐๐๐๐๐๐๐๐ก๐๐๐ ๐๐๐๐ก๐๐ ๐๐๐ ๐๐๐๐ข๐๐๐ ๐๐ ๐ ๐ก๐๐๐ MCDN =๐๐ฃ๐๐๐๐๐ ๐๐๐๐
๐๐๐๐ข๐๐ ๐๐๐๐=
1053.22
870= 1.21
Shallow Foundation โ Mat Foundation
Page 17 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
Find the modification factored column load and ultimate bearing capacity
Column No. Modified ultimate load kN 10 1.21 ร (1.2 ร 80 + 1.6 ร 160) 425.92 11 1.21 ร (1.2 ร 120 + 1.6 ร 240) 638.88 12 1.21 ร (1.2 ร 90 + 1.6 ร 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:
๏ฟฝฬ ๏ฟฝ =425.92 ร 0.2 + 638.88 ร 6 + 479.16 ร 11.8
1543.96= 6.2 ๐
๐ = ๐ต 2โ โ ๏ฟฝฬ ๏ฟฝ = 6 โ 6.2 = 0.2 < ๐ฟ 6โ
๐ . ., =๐
๐ดโ
๐ ๐
๐ผ๐
๐ . . =1543.96
3.8 ร 12+
0.2 ร 1543.96. ร
ร 6 = 37.24 ๐๐ ๐โ
๐ . . =3743.74
7.2 ร 12โ
0.319 ร 3743.74. ร
ร 6 = 30.47 ๐๐ ๐โ
Step 6. Draw factored load, shear and moment diagrams for the continuous beam strips in both directions.
Shallow Foundation โ Mat Foundation
Page 18 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
Shallow Foundation โ Mat Foundation
Page 19 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
Shallow Foundation โ Mat Foundation
Page 20 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
Shallow Foundation โ Mat Foundation
Page 21 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
Shallow Foundation โ Mat Foundation
Page 22 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
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 ร 320 + 1.6 ร 640 = 1048kN ๐ = 1048 ๐๐ 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
๐ฃ
Least of (a),(b) and (c)
0.33 ๐ ๐ ๐ยฐ๐ a
0.17 1 +2
๐ฝ ๐ ๐ ๐ยฐ๐ b
0.083 2 +๐ผ ๐
๐ยฐ ๐ ๐ ๐ยฐ๐ c
Shallow Foundation โ Mat Foundation
Page 23 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
๐ = 2 400 + ๐2 + (400 + ๐) = 1200 + 2๐
๐ผ = 30
๐ โค โ ๐
From equation (a)
1408 ร 10 = 0.33 ๐ ๐ ๐ยฐ๐
1408 ร 10 = 0.33 ร 1 ร โ24 ร (1200 + 2๐)๐ โ ๐ = 424.9 ๐๐ ๐๐๐๐๐๐๐ From equation (b)
1408 ร 10 = 0.17 1 +2
๐ฝ ๐ ๐ ๐ยฐ๐
1408 ร 10 = 0.17 ร 3 ๐โ24(1200 + 2๐)๐ โ ๐ = 309.8 ๐๐ From equation (c)
1408 ร 10 = 0.083 2 +๐ผ ๐
๐ยฐ ๐ ๐ ๐ยฐ๐
1408 ร 10 = 0.083 2 +30 ร ๐
(1200 + 2๐) ๐ ๐ (1200 + 2๐)๐ โ ๐ = 285.8 ๐๐
๐๐ ๐ ๐ = 450 ๐๐
๐ป = ๐ + ๐๐๐ฃ๐๐ +๐
2= 450 + 75 + 12.5 = 537.5๐๐ โ 550๐๐
Shallow Foundation โ Mat Foundation
Page 24 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
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
โ ๐ โฅ ๐ โ โ ๐ = โ ๐๐๐ ๐๐ฆ 1 โ 0.59 ร๐๐
โฒ ร ๐
543.8 ร 10 = 0.9 ร ๐ ร 1000 ร 450 ร 420 1 โ 0.59 ร420
24ร ๐
โ ๐ = 0.00772 โ ๐ด = 3474 ๐๐ ๐โ
๐ด = 0.0018 ร ๐ต ร ๐ก = 990 ๐๐ ๐โ
๐ =3474
491= 7.075 ๐๐๐ ๐ =
1000
7.075= 140๐๐
2. Negative moment for long direction 160.89 kN.m /m
โ ๐ โฅ ๐ โ โ ๐ = โ ๐๐๐ ๐๐ฆ 1 โ 0.59 ร๐๐
โฒ ร ๐
160.89 ร 10 = 0.9 ร ๐ ร 1000 ร 425 ร 420 1 โ 0.59 ร420
24ร ๐
โ ๐ = 0.00242 โ ๐ด = 1028.5 ๐๐ ๐โ
๐ด = 0.0018 ร ๐ต ร ๐ก = 990 ๐๐ ๐โ
๐ =1028.5
201= 5.11 ๐๐๐ ๐ =
1000
5.11= 195 ๐๐
3. Positive moment for short direction 36 kN.m /m
โ ๐ โฅ ๐ โ โ ๐ = โ ๐๐๐ ๐๐ฆ 1 โ 0.59 ร๐๐
โฒ ร ๐
36 ร 10 = 0.9 ร ๐ ร 1000 ร 450 ร 420 1 โ 0.59 ร420
24ร ๐
โ ๐ = 0.00047 โ ๐ด = 211 ๐๐ ๐โ
๐ด = 0.0018 ร ๐ต ร ๐ก = 990 ๐๐ ๐โ
Shallow Foundation โ Mat Foundation
Page 25 of 25
AL-Qasim Green University College Of Water Resources Engineering
Foundation Engineering Mr. Wissam Nadir
๐ =900
201= 4.477 ๐๐๐ ๐ =
1000
4.477= 223๐๐
4. Negative moment for short direction 306.26 kN.m /m
โ ๐ โฅ ๐ โ โ ๐ = โ ๐๐๐ ๐๐ฆ 1 โ 0.59 ร๐๐
โฒ ร ๐
306.26 ร 10 = 0.9 ร ๐ ร 1000 ร 425 ร 420 1 โ 0.59 ร420
24ร ๐
โ ๐ = 0.004715 โ ๐ด = 2003.87 ๐๐ ๐โ
๐ด = 0.0018 ร ๐ต ร ๐ก = 990 ๐๐ ๐โ
๐ =2003.87
491= 4.08 ๐๐๐ ๐ =
1000
4.08= 245 ๐๐