Design of Reinforced Concrete Elements 275

5.16.14 Example 5.33: Inverted T-Beam Combined Foundation An inverted T-beam combined foundation is required to support two square columns transferring axial loads as shown in Figure 5.125. Using the data provided, design suitable reinforcement for the base.

Design Data: Characteristic dead load on column A 450 kN Characteristic imposed load on column A 450 kN Characteristic dead load on column B 750 kN Characteristic imposed load on column B 750 kN Characteristic concrete strength fcu = 40 N/mm2

Characteristic of reinforcement fy = 460 N/mm2 Net permissible ground bearing pressure pg = 175 kN/m2

Nominal cover to centre of main reinforcement 40 mm Column A dimensions 350 mm × 350 mm Column B dimensions 350 mm × 350 mm

Figure 5.125

5.16.15 Solution to Example 5.33

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure : Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 1 of 7 Date :

References Calculations Output

The foundation should be designed as a T-beam between the columns where the rib is in tension and as a rectangular beam at the column positions where the bottom slab is in tension.

Using the service loads it is first necessary to determine a suitable base area and dimensions X and Y such that the centre-line of the base coincides with the centre-of-gravity of the column loads. This ensures a uniform earth pressure under the base.

Gk = 450 kN Qk = 450 kN

Gk = 750 kN Qk = 750 kN

A B

Columns 350 mm × 350 mm

0.6 m 5.0 m X m

400 mm

800 mm

460 mm Y m

Design of Structural Elements 276

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure :Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 2 of 7 Date :

References Calculations Output

Base Area: Total service load = (450 + 450 + 750 + 750) = 2400 kN

Minimum area required = pressure bearing epermissibl

load service

= 175

2400 = 13.71 m2

Actual Load System

Equivalent Load System

Equate the moments of the force systems about the centre-line of column B: Mcentre-line column B = (900 × 5.0) = (2400 × x) ∴x = 1.875 m

The centre-line of the base should coincide with this position.

5.0 m

A B

900 kN 1500 kN

xB

Total load = 2400 kN

5.0 m

900 kN 2400 kN

0.6 m 3.125 m 1.875 m X m

1.875 m

B

Design of Reinforced Concrete Elements 277

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure :Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 3 of 7 Date :

References Calculations Output

Length of the base = [2 × (3.125 + 0.6)] = 7.45 m ∴ X = (7.45 − 5.6) = 1.85 m

Width of the base Y = (13.71 / 7.45) = 1.84 m

Ultimate bearing pressure: Ultimate load on column A = (1.4 × 450) + (1.6 × 450) = 1350 kN Ultimate load on column B = (1.4 × 750) + (1.6 × 750) = 2250 kN

Ultimate design pressure under the base = ( )( )841457

22501350

.. ×+

= 262.6 kN/m2

The combined base can be regarded as a beam 1.84 m wide. Longitudinal load/m = (262.6 × 1.84) = 483.2 kN/m

Shear force at Q = (483.2 × 0.6) = + 289.9 kN and = (289.9 − 1350) = − 1060 kN

Shear force at R = −1060 + (483.2 × 5.0) = +1356 kN and = (1356 − 2250) = −894.1 kN

Position of zero shear = 2483

1060

. = 2.19 m

Shear force diagram

Bending moment at Q = shaded area

=

× 9289

2

60.

. = 86.8 kNm

1350 kN 2250 kN

483.2 kN/m

0.6 m 5.0 m 1.85 m

P Q R S

894.1 kN

1356 kN

289.9 kN

1060 kN

Design of Structural Elements 278

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure :Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 4 of 7 Date :

References Calculations Output

Bending moment at R = shaded area

=

×894

2

851. = 827 kNm

Maximum bending moment at point of zero shear:

= resultant shaded area = 87 −

×1060

2

192.= − 1074 kNm

Design the base as a T-beam between A and B and as a rectangular beam at the column locations.

Assume 32 mm diameter bars for the main steel.

Effective depth: d = (1260 – 40 – 16 ) = 1204 mm (Note: This is the mean effective depth)

Bending between the columns A and B (T-beam):

Design bending moment between the columns = 1074 kNm

Clause 3.4.4.4cufbd

MK

2= =

4012041840

1010742

6

×××

= 0.01 < K´ ( = 0.156)

Section is singly reinforced

827 kNm

1074 kNm

87 kNm

460 mm

1840 mm

400 mm

d =

120

4 m

m

Design of Reinforced Concrete Elements 279

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure :Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 5 of 7 Date :

References Calculations Output

x = (d − z)/0.45 = (0.05 × 1204)/0.45 = 133.8 mm < 460 mm The neutral axis lies in the flange

Z =

−+

9025050

.

K..d =

−+

90

01025050

.

...d

= 0.98d The lever arm is limited to 0.95d

As = Zf.

M

y950 =

1204950460950

101074 6

××××

..= 2149 mm2

Design bending moment at column B (Rectangular section):

= ( )

−×2

17508512483

2... = 678 kNm

Clause 3.4.4.4cufbd

MK

2= =

401204400

106782

6

×××

= 0.03 < K´ ( = 0.156)

Section is singly reinforced

Z =

−+

90

03025050

.

...d = 0.96d

The lever arm is limited to 0.95d

As = Zf.

M

y950 =

1204950460950

10678 6

××××

..= 1356 mm2

Table 3.25 Minimum area of steel = hb

A

w

s100= 0.26%

As ≥100

1260400260 ××. = 1310 mm2 < 1356 mm2

The bottom reinforcement required at column A should be determined by the reader.

460 mm

1840 mm

400 mm

d =

120

4 m

m

TopReinforcement

Select:3/32 mm diameter bars providing2410 mm2

BottomReinforcement

Select:3/25 mm diameter bars providing 1470 mm2

Design of Structural Elements 280

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure :Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 6 of 7 Date :

References Calculations Output

The reinforcement required to resist transverse bending can be determined in a similar manner. Consider a 1m width strip:

Assume 25 mm diameter bars:

d = (460 − 40 − 25 − 13) = 382 mm

Transverse bending moment =

×

2

7202483

2..

= 125.2 kNm/m width

Clause 3.4.4.4cufbd

MK

2= =

403821000

102.1252

6

×××

= 0.02 < K´ ( = 0.156)

Section is singly reinforced

Z =

−+

90

02025050

.

...d = 0.98d

The lever arm is limited to 0.95d

As = Zf

M

y95.0 =

38295.046095.0

102.125 6

××××

= 789 mm2 /m

Table 3.25 Minimum area of steel = 0.15%

As ≥100

4601000150 ××. = 690 mm2/m < 789 mm2/m

Clause 3.4.5 Shear resistance of the T-beam: Consider shear at the face of Column B Table 3.8 Critical shear stress

db

A

v

s100 =

1204400

2410100

××

= 0.5 ∴ vc = 0.5 N/mm2

Modify to allow for using C40 concrete vc = (0.5 × 1.17) = 0.59 N/mm2

Table 3.7 (vc + 0.4) = 0.99 N/mm2

1.0 m

BottomReinforcement

Select:20 mm diameter bars@ 400 mm centres providing 786 mm2/m

720 mm

483.2 kN/m2

Design of Reinforced Concrete Elements 281

Contract : Foundations Job Ref. No. : Example 5.33 Calcs. by : W.McK. Part of Structure :Inverted T-Beam, Combined Foundation Checked by : Calc. Sheet No. : 7 of 7 Date :

References Calculations Output

Shear force at the face of the column: V = (1356 − (483.2 × 0.175) = 1271.4 kN

v = db

V

v =

1204400

1041271 3

××.

= 2.64 N/mm2

Since (vc + 0.4) < v < 0.8 cuf

< 5.0 N/mm2

Design links are required

Asv ≥yv

cvsv

950

)(

f.

sb υυ −

Assume two-legged 10 mm diameter links ∴ Asv = 157 mm2

Sv ≤ ( )

−×××

590642400

250950157

..

. = 45.5 mm

A similar calculation should be carried out by the reader to determine the links required over the full length of the T-beam.

Clause 3.4.5 Shear resistance of the flange: Consider shear at the face of rib

Table 3.8 Critical shear stress

db

A

v

s100 =

3821000

786100

××

= 0.2 ; d = 382 mm

vc = 0.37 N/mm2 ; modify to allow for using C40 concrete vc = (0.37 × 1.17) = 0.43 N/mm2 (vc + 0.4) = 0.83 N/mm2

Shear force V = (483.2 × 1.0 × 0.72) = 348 kN

v = db

V

v =

3821000

10348 3

××

= 0.91 N/mm2 < 0.8 cuf

< 5.0 N/mm2

Table 3.16 Since the shear stress v is greater than vc shear reinforcement is required. Shear links are not normally provided in slabs.

Alternatives to providing links are to increase the % of reinforcement As and hence increase the value of vc sufficiently, or to increase the thickness of the slab, which will reduce the value of v.

Adopt double 10 mm diameter two-legged links @ 100 mm centres for the 1st

metre from column B

720 mm

483.2 kN/m2