Structure One One

1

Prof. Dr. Raddi M. Al Zubaidi

PhD Civil Engineering

United Kingdom 1984

2

Subjects to be Covered

* Structure of building

* Code of Practice

*Reactions and Supports

*Stability and determinacy

* Trusses

*Shear and moment concepts

* Shear and moment diagrams

3

Assessment

Exam 1 & Assignments 30%

Midterm Exam 20%

Final Exam 50%

4

Dependent Lecture

-Course Text book

- J.B. Gauld, Bryan. Structures for

Architects.Longman.1995

- Reference Books

- Ambrose,James, Simplified Mechanics and Strenght of

Materials, John Wiley 2000

- -Schodek,L.Daniel ,Structures, Pearon,2004

.

5

Road base PCCFoundation

G.L

Tie beam Column

Beam

Slab

Structure of Building

6

Types of Footing

1- Wall Footing

2- Speard Footing

3- Combined Footing

4- Mat or Raft Foundation

5- Piles Foundation

7

Wall Footing

8

Load from column

Spread Footing

L

B

9

P1 P2COMBINED FOOTING

10

Mat or Raft Foundation

11

Piles Foundation

12

One Way Slab

13

Ribs Slab

14

Flat Slab

15

Flat Slab

16

Structure

____________________

The World structure describes much of what is seen in nature . Living plant from the frailest of fern to the most

rugged of trees ,posses a structure ,other than aircraft ,ships and floating structures is to transfer applied

loads to the ground

17

Types of Structure

1- Concrete

2- Steel

3- Wood

18

Structure

19

Structure

20

Structure

21

Type of Structure

22

Type of Structure

23

Type of Structure

24

Type of Structure

25

Stress-Strain Curve

26

Stress-Strain Curve

When external force is applied to abeam as shown in the previous slide the beam will exert stress and strain

27

Stress-Strain Curve

28

Stress- Strain Curve

When a load is applied the beam will deflect ,but if the beam is behaving elastically it will return to its original position when the load is removed

When the load is increased then the beam enter the plastic zone and the plastic deformation increases until the beam will fail. The point of breaking is called ultimate strength of the material

29

Stress-Strain Curve

30

Stress – Strain Curve

Yield point =The point at which the permanent deformation occur and the material begins to yield

Permissible safety=Yield point of material/safety factor

Permissible safety=(Ultimate strength of material/safety factor) +(Applied loads x Safety factor)

31

Approximate Sizes of Structural Members

During the planning stages of any project ,the architect requires to understand how the project is to be structured .Where at the

detailed planning stage it is needed to know the sizes of the structural members.

32

Approximate Sizes of Structural Members

50

Example No.1

Apiece of wire hanging from a stairwell with a weight on the end. The weight will cause the wire to stretch as

shown in Fig.1 .If the length and the diameter of the wire and the magnitude of the weight are known. Determine the extension of the wire ?If the following information are available? .

Fig 1

51

Length of wire = 12 m

Diameter of wire = 2 mm

Weight = 31.5 kg

Assume 1 kg = 10 N

E = 200 KN /mm2

54

Example 2

For the same information in Fig.1 but the weight is increased to 72 kg .Determine the extension

55

Force = 72 x 10 = 720 N

Area = 3.14 x 12 = 3.14 mm2

Stress = Force / Area = 720/3.14 =229.3 N/mm2

Strain = Extension / Original length =Extension/12000

Young s modulus for mild steel=200 KN/mm2

Stress =E x Strain

229.3 =200000x strain

Strain =229.3 /200000 =0.00114

Extension =12000 x 0.00114 = 13.758 mm

56

Example 3For the hard steel road shown in Fig 2 .There is a tension force equal to 250000 N acting on both sides . Determine the extension if the following information's are available ?

Diameter = 50 mm ,E= 300 KN/ mm2

250000 N250000N

Fig.2

L = 3 m

57

SolutionLength = 3m

Diameter = 50 mm

E =300 KN /mm2

Area of rod = 3.14 x(25)2 = 1962.5 mm2

Stress = Force /Area =250000/1962.5

=127.38 N/mm2 Stress = E x Strain 127.38 =300000 x strain Strain =127.38 /300000 =0.0004246

58

Solution Example 3 Strain = Extension /Original

Extension= 0.0004246 x 3000 =1.272mm

59

Example 4

A reinforcement bar shown in Fig.3 .The bar undergoes a tension force of 300000 N at both ends .Determine the extension with the following information's available :

E = 300 KN/mm2 ,Diameter= 24 mm

300000 N 300000 N

Fig.3

L = 15 m

60

Solution Example 4Length = 15 m

Diameter =24 mm

E = 300 KN/mm2

61

Solution Area = 3.14 (12)2 = 452.16 mm2

Stress =300000 / 452.12 = 663.48 N /mm2

Strain =Stress / E =663.48 /300000 =0.00221

Strain =Extension / Original length

Extension = 0.00221 x15000 = 33 mm

62

Effect of Loads on Members

h

h

h

h

h

A

B

C

D

E

F

G

H

I

J

K

L

ι

q

Girder G,H,I,J,K,LGirder A,B,C,D,E,F

63

Beam AL

qιh /2

A Lι

64

Beam BK

B K

q ι h

ι

65

Beam EH

ιE H

q ι h

66

Beam FG

ι

q ι h/2

67

(qh/4)ι(qh/2)ι(qh/2)ι(qh/2)ι(qh/2)ι(qh/4)ι

5(qh/4)ι 5(qh/4)ι

Girder GHIJKL

G H I J K L

column

5h

column

68

(qh/4)ι(qh/4)ι (qh/2)ι (qh/2)ι (qh/2)ι (qh/2)ι

5(qh/4)ι 5(qh/4)ι

Girder FEDCBA

5hF E D C B A

column column

69

Example 2For the concrete slab shown in Fig.1 .The

following information's are available : Dead load = 4 KN /m2 Live load =1.5 KN/m2 L=20 m h= 4 m The slab rests on a number of beams and the

beams are rests on two main girders. Determine the resulting effect on beams: AL, BK ,EH,FG

The Girders GHIJKL and FEDCBA and the columns at G and F

70

Solution Factored load q= 1.2 D + 1.6 L

q = 1.2 x 4 +1.6 x 1.5 = 7.2 KN /m2

Beams AL and FG Tributary area =(h/2) l= 4/2 x 20 =40 m2

Total load = q x 40 =7.2 x40 = 288 KN Uniform load = w = Total load/l = 288/20 =14.4

KN /m End support load =288/2 =144 KNBeams BK,CG ,DI and Eh Tributary area = 2(h/2) x20 =2x4/2x20 =80 m2

71

Total load = 80 x7.2 = 576 KN

Uniform load = 576/ 20=28.8 KN/m

End load = 576 / 2= 288 KN

72

Beams AL and FG

144 kN144 kN

20 m

14.4 KN/m

73

Beams BK and CG

288 KN 288 KN

28.8 KN/m

74

Girders FEDCBA and GHIJKL

144 KN 288 KN 288 KN 288 KN 288 KN 144 KN

720 KN 720 KN

75

Example 2For the floor system shown in Fig.1.25.Assum

that the surface load q is applied on panel

”abef ” if q = 100 lb/ft2 determine the resulting effect on beams ab, cd, ef, girder ae and the columns at a and e

Assume ι = 20 ft and h= 10 ft

84

Example 3

For the slab shown in Fig.3 .The following information's are available:

Dead load = 4.5 KN /m2

Live load= 1.5 KN/m2

Find the total and uniform load on beam CD and BE and the loads on girder DEF and columns D

and F ?

85

Example 3

6 m

6 m

30 m

A

B

CD

E

F

Fig.3

86

Solution

q = 1.2 x 4.5 + 1.6 x 1.5 = 7.8 KN/ m2

Load on beam CD

Total load = 3 x30 x 7.8 = 702 KN

Uniform load = 702 / 30 = 23.4 KN /m

Beam CDTotal load = 702 KN

uniform load = 3.4 KN/m

351 KN 351 KN

87

Beam BF

Total load =6 x30 x 7.8 = 1404 KN

Uniform load =1404 /30= 46.8 KN/m

Total load = 1404 KNUniform load = 46.8 KN/m

702 KN 702 KN

88

Girder FED

702 KN 702 KN

351 KN 702 KN 351 KN

F DE

89

Example 4For the reinforced concrete slab shown 4 the

following information are available

Dead load = 5 KN/ m2

Live load = 1.5 KN /m2

Find the total and uniform load on beams :

EF, DG, CH, BI and AJ and the loads on girder

FGHIJ and the column F . Determine also the size of the column ?

90

4 m

6 m

8 m

3 mA

B

C

D

E F

G

H

I

J20 m

Fig.4

91

Solutionq =Factored load= 1.2 x5 + 1.6 x 1.5 = 8.4

KN/ m2

Beam EF

Total load = 2 x 20 x 8.4 = 336 KN

Uniform load =336 /20 = 16.8 KN/m

92

Beam EF

168 KN 168 KN

Total load = 336 KN

Uniform load = 16.8 KN/M

93

Beam DGTotal load =5 x 20 x 8.4 =840 KN Uniform load = 840 /20 =42 KN /m

420 KN420 KN

94

Beam CHTotal load = 7 x 20 x 8.4 = 1176 KN

Uniform load = 1176 /20 =58.8 KN/ m

588 KN 588 KN

95

Beam BI

Total load = 5.5 x 20 x 8.4 = 924 KN Uniform load =924 /20 = 46.2 KN/m

462 KN 462 KN

96

Beam AGTotal load =1.5 x 20 x 8.4 = 252 KNUniform load = 252 / 20 = 12.6 KN

126 KN 126 KN

97

4 m6 m8 m3 m

126 KN

462 KN 588 KN 420 KN

168 KN

J I H G F

GIRDER FGHIJ

98

Load on column F can be found from bending moment as

168x 21 + 420 x 17 + 588 x 11 + 462 x3 – F x 21 = 0

F =

= 882 KN

The size of the column should be 30 x 30 cm

1852521

99

Approximate size of column

100

Example 5For the reinforced concrete structure shown in

Fig.5 The following information are available

Dead load =5.5 KN/m2

Live load =1.8 KN/m2

Find the total and uniform load on beams ABCD,EFGH,IJKL and MOQR .The total load on girder CGKQ and the columns C and Q are required. Find also the depth of the slab, beam BC, the girder CGKQ and the cantilever CD ,also find the size of the columns?

101

6 m

8 m

7 m1 m

1 m

8 m

AB C D

E

FG H

IJ K

L

MO

QR

Fig.5

102

SolutionQ = 1.2 x 5.5 + 1.6 x 1.8 = 9.48 KN / m2

Beam ABCD

Total load = 9.48 x 3x 10 =284.4 KN

Uniform load = 284.4 / 10 = 28.44 KN /m

142.2 KN 142 KN

103

Beam EFGH

Total load = 9.48 x 7 x10 = 663.6 KN

Uniform load = 663.6 / 10 = 66.36 KN /m

331.8 KN 331.8 KN

104

Beam IJKL

Total load = 9.48 x 7.5 x 10 = 711 KN

Uniform load = 711 /10 = 71.1 KN /m

355.5 KN 355.5 KN

105

Beam MOQR

Total load = 9.48 x 3.5 x 10 =331.8 KN

Uniform load = 331.8 / 10 = KN /m

165.9 KN 165.9 KN

106

Girder CGKQ

6 m8 m7 m

Q K G C

165.9 KN 355.5 KN331.8 KN

142.2 KN

107

Load on columns C & Q

Take the bending moment at C

Q x 21 = 165.9 x 21 + 355.5 x 14 + 331.8 x 6

Q = 497.7 KN

C = 497.7 KN

108

Depth of the slab

Slab depth = 8/ 30 = 27 cm

109

Depth of beam BC

Depth = 8 /18 = 45 cm

110

Depth of girder CGKQ

Depth = 21 /18 = 117 cm

111

Depth of cantilever

Depth = 1 /7 = 15 cm

112

Size of column c

Size of column = 24 x 24 cm

113

Example 6For the reinforced concrete structure building shown in

Fig.6 .The following information are available:-

Dead Load = 5.6 KN/ m2

Live Load = 1.6 KN / m2

Find the total and uniform loads on beams AD, EH, IL, MN, OP, QR, ST, the load on girder NT and the load on column N and C . Find also the depth of :-

the first and ground slabs

Beam BC and QR

Girder NT

Cantilever CD

The size of column at N and C

114

8 m2 m 2 m

Fig.6

115

10 m

8 m

AB

C

DE

F

G

H

I

J

K

L

2 m

8 m

2mGround Floor

116

M N

O P

Q R

S T

5 m

7 m

6 m

First Floor

8 m

117

Solution

q = 1.2 x5.6 +1.6 x1.6 = 9.28 KN /m2

First Floor

Load on beam MN

Total load = 9.28 x8 x 2.5 = 185.6 KN

Uniform load = 23.2 KN /m

M N

92.8 KN 92.8 KN

118

Beam OPTotal load = 9.28 x8 x6 = 445.44 KN

Uniform load = 55.68 KN/m

O P

227.7 KN227.7 KN

119

Beam QRTotal load = 60.32 KN /m

Uniform load= 9.28 x 8 x 6.5 = 482.56 KN

Q R

241.3 KN 241.3 KN

120

Beam STTotal load = 9.28 x 8 x 3 = 222.7 KN

Uniform load = 27.84 KN/m

S T

111.36 KN 111.36 KN

121

Load on Girder NT ,total load on column N =(111.3+241.3+222.7+92.8)/2= 334 KN

T NPR

92.8222.7241.3111.36 KN KN KN KN

5 m7 m6 m

122

Ground Floor

Beam AD

Total load = 9.28 x 12 x5 = 556.8 KN

Uniform =46.4 KN/m

A B C D

278.4 KN 278.4 KN

123

Beam EHTotal load = 9.28 x12 x9 = 1002 KN


EF

G H

501 KN 501 KN

124

Beam IL

Total load = 9.28 x 12 x4 = 445.44 KN


IJ K

L

222.7 KN 222.7 KN

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Girder CK

Load on column C

C x 18 = 278.4 x 18 + 501 x 8

C = 501 KN

8 m 10 m222.7 KN 501 KN 278.4 KN

K G C

126

First Floor

Depth of Slab = 8/ 30 = 27 cm

Depth of beams = 8 / 18 = 45 cm

Depth of girder = 18 /18 = 100 cm

Total load on column N =334 KN

Size of column at N =23 x23 cm

127

Ground floorDepth of slab =1000/30 = 34 cm

Depth of beams = 800 / 18 = 45 cm

Depth of girders = 1800 / 18 =100 c m

Depth of cantilever = 200 / 7 = 29 cm

Total load on column C = 501 + 334 = 835 KN

Size of column = 30 x30 cm

128

Reactions Forces

Reactions forces develop at support points of structure to equilibrate the effects of applied forces. It

is recalled that a force is completely defined by its magnitude ,direction and point of application since

applied forces are fully specified ,these three characteristics are necessarily known for each.

Reaction forces are not specified as priori however but develop as a consequences of applied forces and in conformance with the support mechanisms.

138

RbY = Rb cos Φ = Rb = RbY √5 /2 Rbx = Rb sin Φ Rbx = Rb / √5 = Rb= Rbx √5 Rbx= RbY / 2

Φ2

1

Rbx

Rb

RbY

√5

Or tanΦ = 12

= RbxRby

Rbx = Rby2

140

Determine the reactions at points A&B ?

3 m 4 m 4 m 5 m

12

2

1

50 KN60 KN

75 KN

30 KN

A B

141

Free Body diagram

Rax

Ray

30 KN 53.6 KN

26.83 KN50 KN 67 KN

33.54 KN

Rby

A B

3 m 4 m 4 m 5 m

142

Solution

∑Fx =0

Rax + 53.6 + 30 – 33.54 = 0

Rax = - 50 KN

∑MA =0

Taking the moment at point A

26.83x 3 + 50 x7 + 67x 11 –Rby x 16 = 0

Rby = 73 KN

Ray – 26.83 – 50 – 67 + 73 =0

Ray = 71 KN

149

tanΦ =Rby

Rbx=

1

√3

Rby =Rbx

√3

301

√3

2=Rby

=Rbx

155

1

√3

8 m

4 m

6 KN/M

2 KN/m30 KN

15 KN

10 KN

4 m 6 m 8 m

a

b c

d

EXAMPLE 6

156

15 KN 15 KN 16 KN

10KN 26 KN

Rdx

Rdy

Ray

a

d

2.67 m

5.33 m

4 m 10 m 4 m

8 m

4 m

24 KNb c

157

30 x √3 /2

30 x 1/2

158

Solution Taking the moment at point d a clock wise positive

Ray x 18 + 10 x 8 - 15 x 18 - 15 x 14 - 16 x 4 - 24 x 5.33 -26 x 8 = 0

Ray = 44.4 KN

Summation of Y axis forces equal to zero up positive

Ray + Rdy - 15 -15 -16 = 0

44.4 + Rdy - 46 = 0

Rdy = 1.6 KN

Summation of X axis equal to zero right is positive

10 - 26 -24 – Rdx = 0

Rdx = 40 KN

159

5 m 5 m 5 m

4 KN/m

6 KN/m2

1

12 KN

a bc

Example 7

160

Free body diagram

Rax

5 m

7.5 m

10 m

a

4 KN4 KN

10.73 KN 4 x 5=20 KN(2x15)/2=15 KN

5.36 KN

b c

Ray Rby

161

SolutionSummation of forces in the X- axis equal to zero

right is positiveRax – 5.36 = 0Summation of moment at a is equal to zero clock

wise is positive 10.73 x 5 + 60 x 7.5 + 15 x 10 –Rby x 10 = 0 Rby = 65.36 KN Summation of forces in the Y –axis is equal to

zero up is positiveRay + Rby – 10.73 – 60 – 15 = 0Ray = 20.4 KN

167

Example 1

175

Question3

1-What are the reactions at the supports A and C?2-Draw the shear force diagram?

3-Draw the bending force diagram ?

176

Solution

177

Shear force diagram

178

Bending Moment Diagram

179

Question 4 1- What are the reactions at the supports A and C ?

2-Draw the shear force diagram? 3-Draw the bending moment diagram ?

180

Solution to Q 4 Part 1

181

Part 2

182

Part 3

183

Question 51-What are the reactions at A&C2-Draw the shear force diagram

3-Draw the bending moment diagram

184

Solution Q 5Part 1

185

Part 2

186

Part 3

187

QUESTION 61-What are the reaction at A&C ?2- Draw the shear force diagram ?

3- Draw the bending moment diagram ?

188

Solution Q 6Part 1

189

Part 2

190

Part 3

375337

191

QUESTION 7 1- What are the reaction at A&D ?2-Draw the shear force diagram ?

3-Draw the bending moment diagram ?

192

Solution Q 7 Part 1

193

Part 2

194

Part 3

196

Question 81-What are the reaction at A & D ?2- Draw the shear force diagram ?


197

Solution Q8Part 1

198

Part 2

199

Part 3

200

Question 91-What are the reaction at A & C ?2- Draw the shear force diagram ?


201

Solution Q 9Part 1

202

Part 2

203

Part 3

204

Question 10

1-What are the reaction at A& C2- Draw the shear force diagram3-Draw the bending moment diagram

205

Part 1

206

X /4-x =33.33/166.67X = 0.67 m

207

277.7266.6

209

Question 111-what are the reactions at A& D2-Draw the shear force diagram

3-Draw the bending moment diagram

210

Part 1

211

Part 2

212

Part 3

214

QUESTION NO 12

1- What are the reactions at A &D

2- Draw the shear force & bending moment diagrams ??

ΛΛΛΛΛΛ ΛΛΛΛ

200 KN 350 KN

50 KN/m

3 m 4 m 2 mA B C D

215

Taking the bending moment at point A

9 RD = 200 x 3 + 50 x4x5 +350 x7

RD = 450 KN

RA + RD = 200 + 4 x50 + 350

RA = 300 KN

216

Shear Force Diagram

300 200

350450

3 m 2 mA B C DF

217

The bending moment at A& D are zero

The maximum bending moment at F, zero shear force

MF = 300 x5 – 200 x2 – 50x 2 x 1 = 1000 KN-m

MB = 300 x3 = 900 KN-m

MC = 450 x 2 = 900 KN-m

218

100o

900 900••

•BF

C

219

Question No.13

1- Find the reactions at A&D

2- Draw the shear force and the bending moment diagrams ?

ΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛΛ

4 m 10 m 6 m

180 KN 130 KN

A C

20 KN/m

B D

220

Taking the moment at point A

180x 4 +130 x14 + 20x20x10 – RD x 20 = 0

RD =327 KN

Taking the resultant of the vertical forces equal

To zero

RA – 180 – 130 – 400 + 327 = 0

RA = 383 KN

221

383

80

180

123

6.153.85

77

130

120

327

Shear force diagram

222

The max. bending moment at zero shear force

383x10.15 – 180x6.15 – 20x10.15x10.15/2=

= 1750 KN-m

223

Bending Moment Diagram

4 m 6.15 m 3.85 m 6 m

1372

1750 1602

224

Question 14

For the beam shown below find the shear force diagram and bending diagram

a b

2 KN

3 m 6 m 6 m 3 m

6 KN2 KN

225

The reaction at point a and b could be found as follows Taking the bending moment at point a

6 x 6 – Rb x 12 + 2 x15 – 2 x3 = 0

Rb = 5 Kn

Ra = 5 Kn

227

Question 15

For the uniform load acting on the shown beam it s required to draw the shear force and bending diagrams

⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂⌂ 1 KN/m

3 m 12 m 3 mA B

228

Reactions at supports-Rb x 12 – 3 x 1 x 1.5 + 12 x 1 x 6 + 3x 1 x 13.5 =

0

Rb = 9 KN

Ra = 9 KN

Ma= 3 x 1.5 = 4.5 KN – m

M at mid span = - 3 x 7.5 + 9 x 6 – 6 x1x 3

M = 13.5 KN - m

232

Show house contains 15 apartments

233

The building is of a high standard and has large balconies

235

State-of-the-art casting molds for wall production

236

Walls being cast with high-performance concrete

237

Wall in the middle of the manufacturing process

238

Wall being transported on air cushions in the factory

239

Windows and doors being assembled in the factory

240

Wall under production in the factory.

241

Complete wall ready for delivery to the assembly site

242

Exterior wall ready for delivery to the assembly site

243

Complete wall being transported on air cushions by two people

244

Complete exterior wall ready for delivery to the assembly site

245

Kitchens are built entirely in the factory

246

Kitchens are built in the factory

247

Kitchens are manufactured in the factory & delivered complete to the assembly site

248

Kitchen cabinets being installed in an installation shaft

249

Kitchens are of a high standard & are delivered in a fully finished state from the factory

250

Kitchens have smart storage compartments

251

Floor production in the factory

252

Waterjet cutting of board materials

253

Production of ceilings in the factory

254

Strip-forming machine for manufacturing edgings for floors and ceilings

255

Manufacturing of edgings for floors and ceilings

256

Waterjet cutting of board materials such as plasterboard and façade insulation

257

All installations are completed in the factory

258

Ceiling being papered

259

Floor being laid in the factory

260

Floors being completely finished in the factory

261

Wall-mounted toilets and sinks. Completely tiled, clinkers and floor heating as standard features

262

The Factory

264

Complete wall being positioned with the help of the crane

265

Connecting device keeping the wall frames together

266

Kitchen being assembled using the overhead crane

267

Assembly hall: building assembled inside

Structure One One

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