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Structural Engineering Design Calculations

Lusaka Residential Perimeter Walls 55 Kabulonga Road Kabulonga Lusaka – Zambia

Brian C Chutu Bsc (Hons) Civil Engineering (UK), REng, MEIZ, MConsEZ Checked By:

Prepared for:

Bricch Consultants Limited

Subject Drg. Ref

Sheet No.

Project Job No. Made by Rev

Date Checked by

1. DESIGN STATEMENT

Introduction

Design Proposal

a)

b)

Design Codes

a) - BS 8110: Structural use of concrete

b) - BS 5628: Code of practice for use of masonryStructural Masonry

Wall

The wall shall be constructed using 7N Concrete Hollow Blocks filled with Grade 10

(10Mpa) mass concrete. The wall strength shall be enhanced by inserting vertical

reinforcing bars through block cells. Horizontal reinforcing bars shall be provided in every

mortar joint and tied to vertical bars. The horizontal bars shall thus act as distribution

bars to vertical main bars.

1.3

The following Standard Codes of Practice shall be employed for the design of structural

elements:

Structural Concrete

GENERAL LOADING

CALCULATION SHEET 1.1

1720 BCC

C

It is required to provide detailed structural engineering design for a perimeter wall on plot

no. 55 Kabulonga Road, Kabulonga, Lusaka Zambia property belonging to Embassy of The

United States of America.

Client proposed a solid masonry wall reinforced with steel bars for additional strength.

The wall height to be not less than 2.8m high according to Client's standards.

A free standing wall shall be considered and designed to resist wind pressure and gravity

load from its self weight.

The wall shall be founded on traditional reinforced concrete strip footing.

The foundation shall be designed using soil investigation parameters provided in the

Geotechnical Report.

1.1

1.2

Foundation

12.02.2018

LUSAKA RESIDENTIAL PERIMETER WALLS

55 KABULONGA ROAD KABULONGA

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Date Checked by

DESIGN DATA

Characteristic Dead Load

Concrete Density =

200mm thick Solid Concrete Blockwork =

Characteristic Imposed Load

Wind Load (to Client's Standard) =

Soil Properties (obtained from Geotechnical Investigation Report)

Ground Allowable Bearing Capacity =

Angle of Internal Friction ( Ø) =

Soil Density (assumed) =

CALCULATION SHEET

1720

12.02.2018

DESIGN DATA



2.1

C

1.20

4.31

BCC

kN/m²

kN/m²

35 °

18 kN/m ³

24 kN/m ³

100 kN/m²

Subject Drg. Ref

Sheet No.


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CASE 1: WIND PRESSURE OUTSIDE OF PROPERTY

CHECK WALL STABILITY

Sliding

Consider wind pressure acting on 1m wall length

= kN

= kN

Consider gravity loads acting on 1m wall length

= kN

= kN

= kN

= kN

= kN

3.60

16.38

10.08

11.52

2.88

40.86

12.02.2018 C

3.60

WALL DESIGN


1720 BCCLUSAKA RESIDENTIAL PERIMETER WALLS


𝑊𝑖𝑛𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑊𝑘 = 1.2 x 3

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑊𝑎𝑙𝑙,𝑊𝑤 = 3.8 x 4.31

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝐵𝑎𝑠𝑒,𝑊𝑏 = 1.4 x 0.3 x 24

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑆𝑜𝑖𝑙 1,𝑊𝑠1 = 0.8 x 0.8 x 18

𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑆𝑜𝑖𝑙 2,𝑊𝑠2 = 0.8 x 0.2 x 18

𝑻𝒐𝒕𝒂𝒍 𝑮𝒓𝒂𝒗𝒊𝒕𝒚 𝑳𝒐𝒂𝒅,𝑾𝒕

𝑻𝒐𝒕𝒂𝒍 𝑾𝒊𝒏𝒅 𝑳𝒐𝒂𝒅,𝑾𝒌

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Sheet No.


Date Checked by

= 0.4 x 40.86 = kN

=

x

> OK

Overturning

Take moments about Point P1

∑ of overturning moment,

= x

= x = kNm

∑ of restoring moment,

x =

x =

x =

x =

kNm

Factor of Safety against Overturning,

> OK Wall is satisfactory

against overturning

=

2.00

= 3.16

1.20 3.00

9.36

16.34

12.02.2018

= 4.54

2.6

3.6 2.6 9.36

1.50

16.34

2.88 1.10 3.17

29.57

29.57

C


1720 BCC

WALL DESIGN



Wall is satisfactory

against sliding

16.38 0.90 14.74

10.08 0.70 7.06

11.52 0.40 4.61

𝑇𝑎𝑘𝑖𝑛𝑔 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛, 𝐹𝑓 = 𝜇𝑊𝑡

𝐴𝑠𝑠𝑢𝑚𝑒 𝑝𝑎𝑠𝑠𝑖𝑣𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑜𝑟𝑐𝑒, 𝐹𝑝 = 0

Therefore factor of safety against sliding = 𝐹𝑓

𝑊𝑘

𝑀𝑜𝑣𝑒𝑟

𝑊𝑘

𝑀𝑟𝑒𝑠𝑡

Subject Drg. Ref

Sheet No.


Date Checked by

GROUND BEARING PRESSURE

Consider Moment about centreline of base

M = x =

x =

x =

x =

kNm

N = =

M

N

Therefore, maximum ground pressure occurs at toe, which is given by:

=

=

<

Ground pressure at the heel:

=

=

= 5.93 kN/m ²1.4 1.96

-

N 6M

D D ²

40.86 45.58

-

52.44 kN/m ²

Ground Bearing Pressure

OK

Allowable

(200kN/m ²)

1.96

45.58

1.4

40.86=+

D ²D

6MN+

7.38

m

12.02.2018 C

3.60 2.05

2.30

40.86

11.52 0.20

2.88 0.50

=

-3.53-10.08 0.35

= 0.186

D

6

7.60

40.86

7.60

1.44

<6

=1.4

0.23=


1720 BCC

WALL DESIGN



𝑊𝑡

𝑃𝑡𝑜𝑒

𝑃ℎ𝑒𝑒𝑙

Subject Drg. Ref

Sheet No.


Date Checked by

REINFORCEMENT

Wall

Stem Height = m

Horrizontal pressure on stem due to wind:

= x = kNm

Design Moment at Wall Base:

x x

Effective Depth

Assume Bar Ø = mm

Thus d = - - 6 = mm

= x x

x x = kNm

Steel Area

x

x x

Z = [ - / )]

= mm

x 10 ⁶

x xA =

Y10@200mm c/c

(393mm ²) is adequatemm ²/m

76

0.87 500 75.94

8.64= 262

94 0.50 0.25 0.140 0.9

= 0.140=7 10 ᶟ 94 ²

8.64 10 ⁶

94

0.156 15 10 ᶟ

94 ² 10 ⁻⁶ 20.68

200 100

= 8.64 kNm

12

=1.60 3.60 3.00

2

M =

12.02.2018 C


1720 BCC

WALL DESIGN

3.80

1.20 3.00 3.60



𝐹𝑤𝑖𝑛𝑑

𝛾𝑡𝐹𝑤𝑖𝑛𝑑ℎ𝑠2

𝑀𝑢

𝑆𝑖𝑛𝑐𝑒 𝑀𝑢 > 𝑀, 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑖𝑠 𝑛𝑜𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑

𝐾 =𝑀

𝑓𝑐𝑢𝑏𝑑²

+ √(

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Sheet No.


Date Checked by

REINFORCEMENT Cont'd

Provide nominal reinforcement for horizontal or distribution bars

0.13%bh = x x = mm ²/m

Base

Toe

Soil Pressure:

= x = kN/m ²

= x = kN/m ²

( - )

= kN/m ²

Loading:

= x = kN/m ²

= x = kN/m ²2.88 1.4 4.032

= 8.3 +8.30

1.4

50.16

11.52 1.4 16.13

1.4 52.44 73.41

1.4 5.93 8.304

0.90 73.41

12.02.2018 C

0.13% 10 ᶟ 200 260 Y10@200mm c/c

(393mm ²) is adequate


1720 BCC

WALL DESIGN



𝑃1

𝑃2

𝑃3

𝑊1

𝑊2

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Sheet No.


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REINFORCEMENT Cont'd

Design Moment @ Point P2

x x x x

x x x

2 x 3

= kN/m ²

Steel Area

x 10 ⁶

x x

Y10@200mm c/c

(393mm ²) bottom is

adequate0.87 500 89.30A =

16.31= 420 mm ²/m

-16.3

10.08 0.4

42+=M

-50.16 0.9 ²

2-

65.11 0.90 0.90


1720 BCC

12.02.2018 C

1.4016.13 0.90 0.90

WALL DESIGN



Subject Drg. Ref

Sheet No.


Date Checked by

CASE 2: CONSIDER REVERSE WIND PRESSURE

CHECK WALL STABILITY

Sliding

Consider wind pressure acting on 1m wall length

= kN

= kN

Consider gravity loads acting on 1m wall length

Weight of Wall, = x = kN

Weight of Base, = x x = kN

Weight of Soil 1, = x x = kN

Weight of Soil 2, = x x = kN

= kN

WALL DESIGN


1720 BCC

12.02.2018 C

3.60

3.60

16.38

14.40

2.88

43.74

0.8 0.2 18



3.8 4.31

1.4 0.3 10.0824

1.0 0.8 18

𝑊𝑖𝑛𝑑 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒, 𝑊𝑘 = 1.2 x 3

𝑊𝑤

𝑻𝒐𝒕𝒂𝒍 𝑮𝒓𝒂𝒗𝒊𝒕𝒚 𝑳𝒐𝒂𝒅,𝑾𝒕

𝑻𝒐𝒕𝒂𝒍 𝑾𝒊𝒏𝒅 𝑳𝒐𝒂𝒅,𝑾𝒌

𝑊𝑏

𝑊𝑠1

𝑊𝑠2

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Sheet No.


Date Checked by

Stability of wall against sliding similar to CASE 1, therefore check not required

Overturning

Take moments about Point P1

∑ of overturning moment,

= x

= x = kNm

∑ of restoring moment,

x =

x =

x =

x =

kNm

Factor of Safety against Overturning,

> OK

GROUND BEARING PRESSURE

Consider Moment about centreline of base

M = x =

x =

x =

x =

kNm


1720 BCC

12.02.2018 C

WALL DESIGN

2.6

3.6 2.6 9.36

16.38 0.30 4.91

2.439.36

2.00 Wall is satisfactory

against overturning

10.08 0.60 6.05

14.40 0.80 11.52

2.88 0.10 0.29

3.60 2.60 9.36

16.38 0.30 4.91

-14.40 0.20 -2.88



2.88 0.50 1.44

12.83

22.77

=22.77

=

𝑀𝑜𝑣𝑒𝑟

𝑊𝑘

𝑀𝑟𝑒𝑠𝑡

Subject Drg. Ref

Sheet No.


Date Checked by

GROUND BEARING PRESSURE Cont'd

N = = kN

M

N

Therefore, maximum ground pressure occurs at toe, which is given by:

=

=

<

Ground pressure at the heel:

REINFORCEMENT

C

WALL DESIGN


43.74

=12.83

= 0.343.74

1720 BCC

12.02.2018

>D

6

=1.4

= 0.2 m6

Ground Bearing Pressure

OK

=N

-6M

D D ²

N+

6M

D D ²

43.74+

77=

Moments in CASE 2 would be similar but in reverse of CASE 1, therefore provide nominal

reinforcement Y10@200mm c/c in Top, main and distribution bars

=43.74

-77

= -8.04 kN/m ²1.4 1.96

70.53 kN/m ²1.4 1.96

Allowable

(100kN/m ²)



𝑊𝑡

𝑃𝑡𝑜𝑒

𝑃ℎ𝑒𝑒𝑙

structural engineering design calculations - … · structural engineering design calculations ......

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