Post on 13-Nov-2014
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Prof. Dr. Raddi M. Al Zubaidi
PhD Civil Engineering
United Kingdom 1984
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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
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Assessment
Exam 1 & Assignments 30%
Midterm Exam 20%
Final Exam 50%
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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
.
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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
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Wall Footing
8
Load from column
Spread Footing
L
B
9
P1 P2COMBINED FOOTING
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Mat or Raft Foundation
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Piles Foundation
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One Way Slab
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Ribs Slab
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Flat Slab
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Flat Slab
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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
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Types of Structure
1- Concrete
2- Steel
3- Wood
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Structure
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Structure
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Structure
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Type of Structure
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Type of Structure
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Type of Structure
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Type of Structure
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Stress-Strain Curve
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Stress-Strain Curve
When external force is applied to abeam as shown in the previous slide the beam will exert stress and strain
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Stress-Strain Curve
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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
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Stress-Strain Curve
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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)
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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.
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Approximate Sizes of Structural Members
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40
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45
46
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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
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53
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Example 2
For the same information in Fig.1 but the weight is increased to 72 kg .Determine the extension
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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
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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
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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
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Solution Example 3 Strain = Extension /Original
Extension= 0.0004246 x 3000 =1.272mm
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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
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Solution Example 4Length = 15 m
Diameter =24 mm
E = 300 KN/mm2
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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
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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
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Beam AL
qιh /2
A Lι
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Beam BK
B K
q ι h
ι
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Beam EH
ιE H
q ι h
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Beam FG
ι
q ι h/2
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(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
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(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
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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
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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
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Total load = 80 x7.2 = 576 KN
Uniform load = 576/ 20=28.8 KN/m
End load = 576 / 2= 288 KN
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Beams AL and FG
144 kN144 kN
20 m
14.4 KN/m
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Beams BK and CG
288 KN 288 KN
28.8 KN/m
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Girders FEDCBA and GHIJKL
144 KN 288 KN 288 KN 288 KN 288 KN 144 KN
720 KN 720 KN
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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
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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 ?
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Example 3
6 m
6 m
30 m
A
B
CD
E
F
Fig.3
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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
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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
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Girder FED
702 KN 702 KN
351 KN 702 KN 351 KN
F DE
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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
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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
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Beam EF
168 KN 168 KN
Total load = 336 KN
Uniform load = 16.8 KN/M
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Beam DGTotal load =5 x 20 x 8.4 =840 KN Uniform load = 840 /20 =42 KN /m
420 KN420 KN
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Beam CHTotal load = 7 x 20 x 8.4 = 1176 KN
Uniform load = 1176 /20 =58.8 KN/ m
588 KN 588 KN
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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
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Beam AGTotal load =1.5 x 20 x 8.4 = 252 KNUniform load = 252 / 20 = 12.6 KN
126 KN 126 KN
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4 m6 m8 m3 m
126 KN
462 KN 588 KN 420 KN
168 KN
J I H G F
GIRDER FGHIJ
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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
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Approximate size of column
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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
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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
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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
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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
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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
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Girder CGKQ
6 m8 m7 m
Q K G C
165.9 KN 355.5 KN331.8 KN
142.2 KN
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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
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Depth of the slab
Slab depth = 8/ 30 = 27 cm
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Depth of beam BC
Depth = 8 /18 = 45 cm
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Depth of girder CGKQ
Depth = 21 /18 = 117 cm
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Depth of cantilever
Depth = 1 /7 = 15 cm
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Size of column c
Size of column = 24 x 24 cm
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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
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8 m2 m 2 m
Fig.6
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10 m
8 m
AB
C
DE
F
G
H
I
J
K
L
2 m
8 m
2mGround Floor
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M N
O P
Q R
S T
5 m
7 m
6 m
First Floor
8 m
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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
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Beam OPTotal load = 9.28 x8 x6 = 445.44 KN
Uniform load = 55.68 KN/m
O P
227.7 KN227.7 KN
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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
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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
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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
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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
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Beam EHTotal load = 9.28 x12 x9 = 1002 KN
Uniform load = 83.52 KN /m
EF
G H
501 KN 501 KN
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Beam IL
Total load = 9.28 x 12 x4 = 445.44 KN
Uniform load = 37.12 KN /m
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
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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
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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
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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.
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134
135
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137
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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
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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
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144
145
146
147
148
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tanΦ =Rby
Rbx=
1
√3
Rby =Rbx
√3
301
√3
2=Rby
=Rbx
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152
153
154
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
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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
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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
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Example 1
168
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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
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Part 2
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Part 3
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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
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Part 3
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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
195
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Question 81-What are the reaction at A & D ?2- Draw the shear force diagram ?
3- Draw the bending moment 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 ?
3- Draw the bending moment 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
208
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
213
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
226
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
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Show house contains 15 apartments
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The building is of a high standard and has large balconies
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State-of-the-art casting molds for wall production
236
Walls being cast with high-performance concrete
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Wall in the middle of the manufacturing process
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Wall being transported on air cushions in the factory
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Windows and doors being assembled in the factory
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Wall under production in the factory.
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Complete wall ready for delivery to the assembly site
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Exterior wall ready for delivery to the assembly site
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Complete wall being transported on air cushions by two people
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Complete exterior wall ready for delivery to the assembly site
245
Kitchens are built entirely in the factory
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Kitchens are built in the factory
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Kitchens are manufactured in the factory & delivered complete to the assembly site
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Kitchen cabinets being installed in an installation shaft
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Kitchens are of a high standard & are delivered in a fully finished state from the factory
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Kitchens have smart storage compartments
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Floor production in the factory
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Waterjet cutting of board materials
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Production of ceilings in the factory
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Strip-forming machine for manufacturing edgings for floors and ceilings
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Manufacturing of edgings for floors and ceilings
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Waterjet cutting of board materials such as plasterboard and façade insulation
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All installations are completed in the factory
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Ceiling being papered
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Floor being laid in the factory
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Floors being completely finished in the factory
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Wall-mounted toilets and sinks. Completely tiled, clinkers and floor heating as standard features
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The Factory
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Complete wall being positioned with the help of the crane
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Connecting device keeping the wall frames together
266
Kitchen being assembled using the overhead crane
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Assembly hall: building assembled inside
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