Building Structures (ARC2522)
BEAMS (Part 1) By Mohd Adib Ramli Taylor’s School of Architecture, Building, Design
Introduction to beam 
Usually a horizontal structural member that is subjected to a load that tends to bend it

Designed for moment, shearing forces and deflections brought about by the loads and spans of a building structure
Bending Moment - the bending force induced into the beam which is caused by:
External loads i.e.: floor, wall, machineries, occupants, chandelier etc. Own weight i.e.: weight of beam’s material, reinforcement, etc.
External reactions
55kg
Different forms of beams
Types of Beam Continuous beam  Has more than two points of support along its length, usually in the same horizontal plane  It extends
without break in itself over two or more spans
Continuous beam  Provides resistance to bending when a load or force is applied  Commonly used in bridges and buildings
Fixed beam  Fixed at both ends
 Can resist vertical and horizontal forces as well as moment
Fixed beam  i.e. steel beams which are welded together & a cast-in-place concrete structure  Most common in building structures
Simply Supported Beam Made up of beams that span between only
two supports Both ends are free to rotate
Simply Supported Beam Often found in zones of frequent seismic activity Not often found in normal building structures Commonly used in bridges
Cantilever Beam  A fixed beam projecting from a vertical support  The upper part is tension & the lower part in compression
Cantilever Beam  Examples of one-way and two-way span cantilever beams
Cantilever Beam
Shapes of Beam Rectangular  Commonly in concrete and timber beam  If the direction of loading can be predicted then the beam can be designed so that most of the material are placed in that plane
Rectangular
Minimum material in the direction of the force
Maximum material in the direction of the force
Steel beams Different shape would have different stiffness and resistance to torsion Most stiff
Torsion ‘H’ channel square angle
Least stiff
‘tee’
Box girders  High resistance towards torsion
 Stresses tend to concentrate at the corners
Concrete box girder
stress
Steel box girder
Cylindrical tube  Best if the direction of the load could not be predicted  Resistance towards bending and torsion
Characteristics of Beam A beam should satisfy the following requirements: Adequate strength to resist bending
moments Least excessive amount of deflections under pressure
Able to resist lateral buckling Able to resist shear forces
Bending The beam deflects under a heavy load The amount of deflections are greater when the beam is flat The beam deflects less under uniformly distributed load
 When a beam deflects under a load it, results in the formation of stresses such as tension and compression Sponge Test  Unbent sponge - the grid lines are spaced equally along the top and bottom edges of the sponge. Unbent sponge  Bent sponge - the grid lines are closer along the top edge and farther along the bottom edge. Bent sponge
Compression
Tension
 Cantilever beam the stresses are reversed with the top of the beam being in tension and the bottom in compression
Tension
Compression
Deflections Deflection is the degree to which a structural element is displaced under a load The amount of deflection depends on: • how the load is distributed - point load or UDL
• how the ends of the beam are supported – fixed or freely supported
Simply Supported Beam  Deflection becomes more of a problem as the span increases
Fixed Beam  Deflects less than a simply supported beam
 Assuming that the beams are the same size and material, then the deflection of the simply supported beam will be 8 times that of the fixed beam
Lateral or Sideways Buckling  A beam may buckle if its depth is much greater than its width when subjected to compressive stresses on top of the beam Force
Failure Due to Shear There are three different types of shear
Vertical  Imagine a beam made up of blocks
 When a force acts down on the beam, it deflects and causes the blocks to slide down one another
Horizontal  usually occurs when a beam is made up of layers  when a force acts down on the beam, it deflects and causes the layers to slide over each other
At an angle
Forces/Load on Beams Point Load  In point load, the load is focused in one point; depicted by a single arrow in diagrams  Examples: column sitting on a beam, connection points between secondary beam & primary beam (girder), a chandelier hanging down from a beam 50N
50N
Forces/Load on Beams Point Load In point load, the load is focused in one point – depicted by a single arrow in diagrams
Column
Point load
50N
Primary beam
Secondary beam
Uniformly Distributed Load (UDL) In UDL, the load is spread evenly along the structure – depicted by a series of arrows or curves in diagrams Examples of UDL on beams in building structures: beam self weight, wall on beam, slab on beam, etc. 10N/m
10N/m
Example of UDL on beams
Flat roof Brick wall Balcony wall
Floor slab
Self weight
Important Calculation 1) Beam self weight Assume the initial beam size, i.e. 200mm x 300mm = beam size x density of reinforced concrete = 0.2m x 0.3m x 24kN/m3 = 1.44kN/m (depends on initial beam size) 2) Brick wall weight = wall height x wall thickness x density of brick = 3m x 0.15m x 19kN/m3 = 8.55kN/m * The density of different materials can be obtained from schedule 4 of UBBL
Combination In reality most times the total load acting on a structure is a combination between point load and UDL i.e.: UDL from wall & beam’s self weight plus point load from beams carrying slab and columns carrying load from roof
20N 10N/m
Point load UDL
Load Transfer in Beams
beam UDL
Beams generally carries
vertical forces Can also be used to carry horizontal loads i.e. due to wind or earthquake The load are transferred to columns, walls or girders which then transfer the force to adjacent structural
compression members
wall floor
PL
tension beam
compression
ground
Load transferred to columns
Load transferred to girders
Load transferred to wall
E
F
A
B
G
void C
D
H
A C
E
F
Point Load UDL Combination E
B
D
Identify the type of load for each beam without considering the beam’s self weight
H G
F
G
A
B
C
D H
Free body diagram (FBD) A beam is converted into a free body diagram before the load is quantified A
B
C
D
A
B
C
D
5m
3m
1.5m
3m
Section
Plan 10kN/m
A
5kN/m
10kN/m
B B
D 1.5m C
3m
3m
25kN/m
A
D B
7.5m
C
√ X
Case study Calculate the load on Beam 2/B-D
Architectural plan
Steps: • Draw the structural plan & indicate all beams and columns • Determine the type of each slab whether one-way or two-way & draw the load distribution (refer to lecture notes on “Slabs”) • Identify which slab(s) contributes its load to the beam • Identify which part(s) of the beam carries a brick wall
Case study
Structural plan
Load distribution
Case study Beam 2/B-D carries: • Dead load from beam self weight: gridline B-D • Dead load from brick wall weight: gridline B-C only • Dead load from slab: gridline B-D/1-2, B-C/2-3 & C-D/2-3 • Live load from slab: gridline B-D/1-2, B-C/2-3 & C-D/2-3 • Point load at point C/2 from beam C/2-3 Steps: • Find the total amount of dead load according to grid B-C, C-D • Find the total amount of live load according to grid B-C, C-D • Include free body diagrams for reference • Find the ultimate load and draw the ultimate load diagram • Analise Beam C/2-3 using the same steps to know the reaction force at C/2 which would become the point load
Case study Dead Load • Beam self weight = 1.44kN/m (beam size is 200mm x 300mm)
• Brick wall weight = 8.55kN/m (wall height is 3000mm) • Dead load on slab B-D/1-2 = 7.2kN/m • Dead load on slab B-C/2-3 = 3.6kN/m • Dead load on slab C-D/2-3 = 3.6kN/m • Total dead load For B-C = 1.4 + 8.55 + 7.2 + 3.6 = 20.75kN/m For C-D = 1.4 + 7.2 + 3.6 = 12.2kN/m
Case study Live Load • Live load on slab B-D/1-2 = 4kN/m
• Live load on slab B-C/2-3 = 2kN/m • Live load on slab C-D/2-3 = 2kN/m • Total live load For B-C =4+2 = 6kN/m For C-D =4+2 = 6kN/m
Case study Ultimate Load
• • • •
Ultimate dead load at B-C = 20.75kN/m x 1.4 Ultimate dead load at C-D = 12.2kN/m x 1.4 Ultimate live load at B-C = 6kN/m x 1.6 Ultimate live load at C-D = 6kN/m x 1.6
= 29.05kN/m = 17.08 kN/m = 9.6kN/m = 9.6kN/m
• Total ultimate load at B-C = 29.05 + 9.6 = 38.65kN/m • Total ultimate load at C-D = 17.08 + 9.6 = 26.68kN/m • Include reaction force for beam C/2-3 at point C/2 in the diagram; in this case say 32.5kN
Quantifying the Internal Forces  The loads (and reactions) bend the beam and try to shear through it Bending
Shear
 When a beam is loaded by forces, stresses and strains are created throughout the interior of the beam
Quantifying the Internal Forces  To determine these stresses and strains, the internal forces that act on the beam must be identified: 1. Find all the forces (PL & UDL) 2. Make the beam into a free body diagram 3. Find the reactions using the conditions of equilibrium
Moment Moment M = Force F x Perpendicular distance d  It tends to rotate a structure (i.e.: beam or truss) about a point Distance d
Force F
+ve A
M=Fxd
B -ve
 However, it does not always cause rotation
Moment i.e.:
A
MA = F x dA
dA
F
M B = F x dB
dB
The force tends to rotate the structure clockwise about the support A but also tends to rotate the beam counter clockwise about the support B. Thus, there is no rotation, only bending Rotation would occur if either support is removed
B
Calculate moment about each point 1) 50kN A
2)
4m
2m
B
4m 2.5kN/m
A
B 10kN
The total load is 2.5kN x 4m = 10kN
This load is equivalent to a single point load of 10kN in the middle of the beam
Building Structures (BLD61003/ARC2523)
BEAMS (Part 2) By Mohd Adib Ramli Taylor’s School of Architecture, Building, Design
Quantifying the Internal Force Reaction Force
Applied to a structure when it rests against something
Reaction forces can be solved by applying two simple rules in conditions of equilibrium: • A body in equilibrium has no resultant force in any direction • A body in equilibrium has no resultant turning moment in any direction
In a state of equilibrium the sum of forces acting on a structure should be 0 Total load on structure + reaction forces = 0 ∑F = 0 2m
y
10kN/m
+ A 10kN
B 10kN
_
x
Reaction force on beam
A uniform load on a beam is shown below with a value of 5kN per meter (kN/m) 6m 5kN/m
A Ra 15kN
B 30kN
Rb 15kN
The total load is 5kN x 6m = 30kN This load is equivalent to a single point load of 30kN in the middle of the beam In this case the reactions must be equal to half the total load; 30kN ÷ 2 = 15kN both acting up
30kN _
M B1 3m
A Ra + 15kN
B
6m
Rb 15kN
M B2
1) Choose ONE reference point, in this example, point B 2) Calculate ALL moments from point B to get Ra 3) Balance ALL forces to get Rb ∑M = 0 = MB2 – MB1 = (Ra x 6) - (30 x 3) = 6Ra - 90 6Ra = 90 Ra = 15kN
∑Fy
=0 = Ra + Rb - 30 = 15 + Rb - 30 Rb = 30 - 15 Rb = 15kN
Now consider a uniform load of 2.5kN/m over part of the beam 4m 1.5m
2.5kN/m
A
B
3.75kN
Ra 0.75m
Rb
The total load is 2.5kN x 1.5m = 3.75kN & acts in the middle of the UDL length (0.75m) from A
_
MB1
4m
0.75m
3.75kN A Ra
B +
MB2
∑M = MB2 – MB1 = 0 = (Ra x 4) - (3.75 x 3.25) = 0 = 4Ra - 12.19 = 0 4Ra = 12.19 Ra = 3.05kN
Rb ∑Fy = 0 Ra + Rb - 3.75 = 0 3.05 + Rb - 3.75 = 0 Rb = 3.75 – 3.05 Rb = 0.7kN
Exercise 1 Calculate the reaction forces Ra & Rb 6m 2m
2.5kN/m
2m
1m
4kN/m
10kN
A
B
Ra ∑M = 0 = Ra(6) – 5(5) – 8(3) – 10(1) = 6Ra – 25 – 24 – 10 6Ra = 59 Ra = 9.8kN
Rb ∑Fy Rb
=0 = 9.8 + Rb – 5 – 8 – 10 = 13.2kN
Exercise 2 Calculate the reaction forces Ra & Rb 8m
3m
2kN/m
2m
5kN
3m
8kN
3kN/m
A Ra ∑M
=0 = Ra(8) – 6(6.5) – 5(5) – 8(3) – 9(1.5) = 8Ra – 39 – 25 – 24 – 13.5 8Ra = 101.5 Ra = 12.7kN
B Rb ∑Fy = 0 = 12.7 + Rb – 6 – 5 – 8 – 9 Rb = 15.3kN
Shear & Bending Moment
Shear and moment occur internally in a beam These forces can be identified in any section of a beam in the condition of equilibrium For shears and moments calculations, uniformly distributed loads cannot be simply replaced with point load
Sign conventions for shear force (V): +ve shear R.H. down L.H. up
-ve shear L.H. down R.H. up
Positive shear: denoted by an internal shear force that causes clockwise rotation of the member on which it acts Negative shear: opposite to the positive shear
 Sign conventions for bending moment (M): +ve (sagging/happy)
-ve (hogging/sad)
 Positive moment: denoted by an internal moment that causes compression, or pushing on the upper part of the member  Negative moment: opposite to the positive moment
Shear & Moment Diagrams  The magnitude of these internal shears & moments will vary depending on the location at which the beam is cut
 A shear & moment diagram shows the values of these forces at different locations along the beam
Example of shear & moment diagrams in simply supported beam (point load) L x
L x
A 0
F
A
F B
C
Load Diagram
B C
• Identify and quantify all forces acting on the beam using the conditions of equilibrium • Use the information to produce shear force and bending moment diagrams
A
+
C
B
_
Shear Force Diagram C + A
B
Bending Moment Diagram
 Example of shear & moment diagrams in simply supported beam (uniformly distributed load) F/m A
F/m
B
Load Diagram A
B (+)
B
A
(-)
Shear Force Diagram A
B (+)
Bending Moment Diagram
 Example of shear & moment diagrams in cantilever beam F/m A
F/m
L
Load Diagram
B A
L
B
B
A (-)
Shear Force Diagram A
B (-)
Bending Moment Diagram
 Example of shear & moment diagrams in continuous beam (uniformly distributed load) F/m A
C
B
Load Diagram (+)
B
A
(+)
C (-)
(-)
Shear Force Diagram A
(+)
B (-)
Bending Moment Diagram
(+)
C
Easy steps in drawing load, shear force and bending moment diagrams: 3m
18kN
6m
A
B C
1. Identify and quantify all forces acting on the beam then draw the load diagram ∑MA = RB (9) – 18(3) = 9 RB – 54 RB = 6kN ∑Fx = 0 3m
∑Fy = 0 = RA – 18 + 6 = RA – 12 RA = 12kN
18kN
A
6m B
C 12kN
6kN
2. Plot the shear force diagram based on forces in the load diagram 3m
18kN
6m
A
B C
12kN
6kN
12kN
V
12kN (+)
0kN
A
C -6kN
B (-)
0kN -6kN
•
At point A there is a 12kN force acting upwards (+ve)
•
No other force placed between point A and C, thus the line remain constant at 12kN mark
•
At point C there is an 18kN force acting downwards (-ve), thus 12kN – 18kN = -6kN
•
No other force placed between point C and B thus the line remain constant at -6kN mark
•
At point B there is a 6kN force acting upwards (+ve), thus -6kN + 6kN = 0kN (the diagram is correct only if it ends with 0)
3. Plot the bending moment diagram based on the areas below the shear force diagram 3m
6m
12kN
V
12kN (+)
0kN
A
C
B (-)
-6kN
0kN -6kN
36kNm
V
(+)
M 0kNA
Relationship between lines in shear & moment diagrams
C
B
0kN
of the bending moment diagram isorequivalent to the M area underneiagram
• At point A there is only a line so no area = 0kN • At point C, M = area of rectangular between A and C = 12 x 3 = 36 kNm • At point B, M = area of rectangular AC (+ve) + area of rectangular CB (-ve) = 36 + (-6 x 6) =0
or
4. Redraw load, shear force and bending moment diagrams 3m
18kN
6m
A
B C
12kN
6kN
12kN
V
12kN (+)
0kN
A
C
B (-)
-6kN
0kN -6kN
36kNm
(+)
M
0
A
C
B
0
Example 25N 2N/m B A
C 10m
10m 25N
2N/m A
B 10m
RAy = 27.5N Reactions: ∑Fy = 0; ∑MA = 0 = RCy (20m) – 25N(10m) – [2N(10m)(5m)] = 20RCy – 350 RCy = 17.5N
10m
C RCy = 17.5N
∑Fy = 0; = RAy + RCy – 25 – 2(10) = RAy + 17.5 – 45 = RAy – 27.5 RAy = 27.5
25N 2N/m B
A
10m
C
10m
17.5N
27.5N
27.5N
(+) 0N
7.5N (27.5 – 20) B
A
C
√
0N (-17.5 + 17.5)
(-) -17.5N (7.5 – 25)
-17.5N
Shear Force Diagram (V) Note: • UDL was converted to PL only for calculation of reaction forces • When drawing V diagram, it should be treated as UDL
27.5N 20kN (+) 0N
7.5N B
A 10m
-17.5N
C
0N
(-) 10m
-17.5N
175Nm
0 A
B
C
Bending Moment Diagram (M)
25N 2N/m B
A
C 17.5N
27.5N
Load Diagram 27.5kN
(+) 0kN
A
7.5kN B
C
0kN
(-)
-17.5kN
-17.5kN
Shear Force Diagram 175kNm
0 A
B
Bending Moment Diagram
C
Example 8kN
16kN B
C D
A 2m
3m
3.5m 8kN
16kN B
A
C 3.5m
2m
D 3m
15.06kN
8.94kN
15.06 + (15.06 – 16 = –0.94)
0 A
B –0.94
– C
–
(–0.94 – 8 = –8.94)
–8.94
Shear Force Diagram (V)
0 D
Reactions: ∑MD = 0 = RA (8.5) – 16(6.5) – 8(3) = 8.5RA – 104 – 24 = 8.5RA – 128 RA = 15.06kN ∑Fy = 0 = RD + RA – 16 – 8 = RD + 15.06 – 24 = RD – 8.94 RD = 8.94kN
√
(–8.94 + 8.94 = 0)
–8.94
15.06 + 0 A
(15.06 x 2 = 30.12)
2m B –0.94
– 3.5m
D C
–
(–0.94 x 3.5 = –3.29) (–8.94 x 3 = –26.82)
–8.94
3m
Shear Force Diagram 30.12
0
A
B
[30.12 +(– 3.29) = 26.83]
26.83
C
D
√
0
[26.83 +(– 26.82) = 0.01 ≈ 0]
Bending Moment Diagram
Case study (continued from previous lecture on beams)
Steps • Find the reaction forces at B & D • Draw the shear force diagram • Draw the bending moment diagram
Case study
Reactions forces ∑MA = 0 = RB (6) – 115.95(4.5) – 32.5(3) – 80.04(1.5) = 6RB – 521.78 – 97.5 – 120.06 RB = 739.34/6 = 123.22kN ∑Fy = 0 = RD + 123.22 – 115.95 – 32.5 – 80.04 RD = 105.27kN
Case study
32.5kN 38.65kN/m 26.68kN A 105.27kN
123.22kN Load diagram 123.22kN
7.27kN 0kN -25.23kN
Shear force diagram
-105.27kN
Case study 123.22kN
7.27kN 0kN -25.23kN
Shear force diagram
-105.27kN
195.74kNm
0 Bending moment diagram
How to calculate the highlighted area?
2
3
Using triangular proportion
4
17kN/m 10.34
36.3kN/m
34kN/m
x 2.9m
2m
94.93 95.34kN/m
Load Diagram
94.93kN/m
95.34 2.9
+ V
x/2.9 = 10.34/(10.34+94.93) x = 0.28
95.34 – 68 = 27.34 27.34 – 17 = 10.34
-94.93 + 94.93 =0
_ 124
√
10.34 – 105.27 = -94.93
M
√
124 - 124= 0
Bending Moment Diagram Positive area – negative area in V +ve = [0.5(95.34 + 27.34)(2) + [0.5(10.34)(0.28)] = 122.68 + 1.45 = 124.13 ≈ 124 -ve
= 0.5(-94.93)(2.9-0.28) = -124.35 ≈ -124
Building Structures (ARC2522)
COLUMNS
By Norita Johar Taylor’s School of Architecture, Building, Design
Columns A structural member usually inclined at 90º to the horizontal Carrying an axial* compressive load Found in buildings supporting beams floor & roofs
Columns Axial load is a load applied through the center of gravity of a column’s cross section along an axis Eccentric load is a load applied at any other point on a column’s cross section
Axial load
Eccentric load
Columns for Structure
Columns supporting roofs Columns supporting slab
Columns supporting beams
Columns supporting roofs
Columns for Structure  Strut - a structural member in any position & carrying an axial compressive load
Types of Columns ( Shape ) The main types of columns are: 1) Simple member with uniform crosssection 2) Simple member with non-uniform cross-section 3) Built-up columns
Types of Columns ( Shape ) 1)Simple member with uniform cross section
Types of Columns ( Shape ) 1)Simple member with uniform cross section  The connections to the circular section is expensive and difficult to design
Types of Columns ( Shape ) 1)Simple member with uniform cross section Possible to use square or rectangular hollow sections whose geometrical properties are good The connections are easier to design but rather expensive Hot-rolled sections are the most common cross-sections used for compression members
Types of Columns ( Shape ) 2) Simple members with non- uniform cross-sections In tapered members the cross- section geometry changes continuously along the length Usually concrete columns In steel it can be either open or box shapes formed by welding together i.e. tapered webs or flanges or both
Types of Columns ( Shape ) 2) Simple members with nonuniform cross-sections  Steel tapered members can be formed by welding together two reversed halves of a hot-rolled H or I-section whose web is cut
Types of Columns ( Shape ) 2) Simple members with non- uniform cross-sections In stepped columns, the cross-section varies in steps Usually found in industrial buildings with overhead travelling cranes Can also be used in multistorey buildings to resist the loads in the columns at the lower levels
Types of Columns ( Shape )
Compression Members ( Shape ) 3)Built-up Columns Fabricated from various different elements Consist of two or more main components, connected together at intervals to form a single compound member i.e.: I-section laced with small U-section or Usection battened with flat bars
Types of Columns ( Shape ) 3)Built-up Columns
Types of Columns ( Shape ) 3)Built-up Columns
Compression Members  Examples of built-up columns in buildings Stepped built-up columns
Elasticity vs Plasticity
Force (F)
shortening of the column
 An ideal column is one that is perfectly straight, homogeneous, and free from initial stress.  When an ideal column is subjected to a compressive axial force, the only deformation that takes place is a shortening of the column
Force (F)
Elasticity vs Plasticity Force (F)  For low values of F, if the column were to be deflected laterally by a perpendicular force and the force is then removed, the column would return to its straight position.  This indicates a condition of stability or elasticity
Force (F)
Elasticity vs Plasticity
Force (F)
 If the load F were increased, there is a value of F for which the column would remain in the deformed shape (plasticity)  Yield strength, or the yield point, is defined in engineering as the amount of stress that a material can undergo before moving from elastic deformation into plastic deformation.
Reaction Force (F)
Column Test  To measure the compressive strength of any material  Measured by plotting the applied force against deformation in a testing machine  Usually the tests are made on small specimens in the shape of cubes or circular cylinder
Column Failure 1) Crushing If the force or load is gradually increased, the column will reach a stage when it will be subjected to the ultimate crushing stress Beyond this stage the column will fail by crushing The load corresponding to the crushing stress is called crushing or critical load/force
Column Failure 1) Crushing Main bars
Horizontal circular ties
 Short columns normally fail due to crushing  Failure of a column may be attributed to the insufficient number and spacing of horizontal circular ties.  i.e. during earthquake shaking the ties failed to confine the concrete and the vertical reinforcing main bars
Column Failure 1) Crushing  Stable if loaded within design limit  Failure occurs through crushing if heavily overloaded Unstable load Stable load
Column Failure 2) Buckling Sometimes a compression member fails by bending, i.e. buckling Buckling can also be described as bending or bowing of a column due to compressive load Happens in the case of long columns. For a long (slender) column, buckling occurs way before the normal stress reaches the strength of the column’s material
Column Failure
Force (F)
2) Buckling Buckling
 Buckling is a failure mode; failure to react to the bending moment generated by compressive load  The load at which the column just buckles is called buckling load, critical load or crippling load.
Force (F)
Long compression member
Column Failure 2) Buckling  Stable when provided with lateral restraints  Without lateral restraints it would fails through buckling Stable load
Unstable load
Lateral restraints i.e. from beams & floors
Deformation
Critical Load The critical load depends on: Material properties (i.e. yield strength, modulus of elasticity; can be found in reference books) Length of member (can be determined by slenderness ratio) Section properties
Critical Load Slenderness Ratio (SR or λ) The Slenderness Ratio is the (effective) length Le of the column divided by radius of gyration Is a means of classifying columns whether it is short, intermediate or long column Sometimes expressed with lambda (λ) A generally accepted relationship between the slenderness ratio (SR) and the type of column is as follows: 0 ≤ SR ≤ 60 = short column 60 ≤ SR ≤ 120 = intermediate column 120 ≤ SR ≤ 300 = long column
Critical Load Slenderness Ratio (SR or λ) Formula for slenderness ratio:
SR = KL r Where
or
SR = Le r
SR = slenderness ratio K = column effective length factor KL Le L ==unsupported length of column Le = effective length of column r = radius of gyration
3000mm
SR = 3000 = 10 300 0 ≤ SR ≤ 60= short column
300mm
Critical Load Slenderness Ratio (SR or λ)
SR = 3000 = 100 30 60 ≤ SR ≤ 120= intermediate column
SR = 6000 = 200 30
3000mm
120 ≤ SR ≤ 300= long column
30mm
Critical Load Slenderness Ratio (SR or λ) The end conditions of a column significantly influence the buckled shape Identifying the end conditions helps in computing the buckling load Force (F)
Examples of pinjointed columns
Slenderness Ratio (SR or Ν)  Buckled shape of columns with different end conditions
Critical Load: Moment of Inertia Is a property which measures the efficiency of a shape in respect of its resistance to bending In general a shape is more efficient when the greater part of its mass is as far as possible from its center of gravity y
Center of gravity
y
y x
+
x
x
+
x
+
y
y
y
Column A (400cm2)
Column B (400cm2)
Column C (400cm2)
i.e.: Column B & C have larger moment of inertia than column A thus they will be stronger/stiffer (because their area is spread out away from the center of gravity
Critical Load: Moment of Inertia  The strength of a column may be increased by distributing the material as far from the center of gravity as possible to increase the moment of inertia  At the same time the material needs to be thick enough to prevent local buckling  Therefore an I-section is much better in resisting bending compared to a rectangular section with similar crosssectional area
Column Load Calculation Basically there are 2 types of column load calculation methods: 1) Beams reaction method 2) Tributary area method
Column Load Calculation: Beams reaction method B3
RB3b
B1 RB1a
C3
RB3a
RB1b
C4
C2
C1
B4
B2 RB2a
RB2b
RB4a
RB4b
Which beams contribute to the column load for column C1? RB1b RB3a RB2b RB4a
50kN
50kN
5m
100kN 2m
B3
B1 RB3a
RB1b
150kN
B4
B2 RB4a RB2b
C1 Calculate the column load for column C1 if beam self-weight is 10kN/m
∑FB1 = 10kN(2m) ∑FB1 = 20kN RB1b = 10kN# ∑FB3 = 50kN + 10kN(5m) ∑FB3 = 50kN + 50kN ∑FB3 = 100kN RB3a = 50kN# ∑FB2 = 10kN(2m) ∑FB2 = 20kN RB2b = 10kN# ∑FB4 = 75kN + 10kN(5m) ∑FB4 = 75kN + 50kN ∑FB4 = 125kN RB4a = 62.5kN# Column Load C1 = RB1b+RB3a+RB2b+RB4a = 10+50+10+62.5
= 132.5kN
100kN
100kN
8m
200kN 3m
B3
B1 RB3a
RB1b
400kN
B4
B2 RB4a RB2b
C1
Example 1
Calculate the column load for column C1 if beam self-weight is 10kN/m
100kN
100kN
8m
200kN 3m
B3
B1 RB3a
RB1b
400kN
B4
B2 RB4a RB2b
C1 Calculate the column load for column C1 if beam self-weight is 10kN/m
∑FB1 = 10kN(3m) ∑FB1 = 30kN RB1b = 15kN# ∑FB3 = 100kN + 10kN(8m) ∑FB3 = 100kN + 80kN ∑FB3 = 180kN RB3a = 90kN# ∑FB2 = 10kN(3m) ∑FB2 = 30kN RB2b = 15kN# ∑FB4 = 200kN + 10kN(8m) ∑FB4 = 200kN + 80kN ∑FB4 = 280kN RB4a = 140kN# Column Load C1 = RB1b+RB3a+RB2b+RB4a = 15+90+15+140
= 260kN
Column Load Calculation: Tributary Area Method A column does not support the entire weight of the whole floor. It only supports the load in its tributary area at all available levels. a) Live load= Tributary area x live load factor b) Dead Load= TA x weight of the material
Column Load Calculation: Tributary Area Method Calculate the tributary area labelled2A to I. 1 6000
3
8000
A 4000
B
A
B
C
D
E
F
G
H
I
3500
C
Column Load Calculation
In analyzing column load, few components need to be considered: Dead Load on roof Roof frame weight or Slab ( Flat Roof) Roof Beam Dead Load on 1st Floor Walls Slabs SW Beams SW Column SW Live Load on 1st Floor Dead Load on Grd Floor Walls Slabs SW Beams SW Column SW Live Load on Grd Floor
Column Load Calculation Tributary Area Method How to determine RC Column Self Weight? Width x length x height of column (m) x weight of RC concrete ( 24 kN/m3) Example: Calculate the dead load of the RC columns provided that: 1) Column size = 400mm x 450mm Building height=4.75m Column = 0.45m x 0.4m x 4.75m x 24 kN/m3 = 20.52 kN 2) Column size = 300mm x 400mm Building height=3m Column = 0.3m x 0.4m x 3m x 24 kN/m3 = 8.64 kN
Tributary area method ( Live Load Only) • Half distance of slabs to adjacent columns • Also known as column area method • Column load = area of slab (xy) X load factor 4m
4.5m
3m
4m
Calculate the shaded area of slab x = 2m + 2m = 4m y = 2.25m + 1.5m = 3.75m Area of slab = xy = 4m x 3.75m = 15m2 Load factor: Office = 1.5kN/m2 Residence = 2.0kN/m2 Hospital/Factory = 4.0kN/m2
Exercise1: Find column load for column B2 of an office A
5m
B
5m
C
1
5.5m
x = 5m y = 6.25m
2 3.5m
Area of slab = 5 x 6.25 = 31.25m2
Column Load= 31.25m2 x 1.5kN/m2 =46.87kN
1500 1500
Load factor: Office = 1.5kN/m2 Residence = 2.0kN/m2 Hospital/Factory = 4.0kN/m2
Calculate column load for column A if the building is a residence Column load A =(2m x 6m) x 2.0kN/m2 = 24kN Calculate column load for column B if the building is a factory Column load B =[(2m x 3m) + (3m x 3.5m)] x 4kN/m2 = 66kN
• Column load in multi-storey building • Start calculation from the highest to the lowest level 5m
5m
5m
5m
5.5m
B
A
First Floor (gymnasium) Load factor: Office = 1.5kN/m2 Gymnasium = 4.0kN/m2
4m
4m
5.5m
B
A
Ground Floor (office)
5m
5m
5m
5m B
A First Floor – gymnasium (4.0kN/m2)
Column load for column 1A = (5m x 4.75m) x 4.0kN/m2 = 95kN Column load for column GA = (5m x 4.75m) x 1.5kN/m2 = 35.63kN Total Column load for column A = 95kN + 35.63kN = 130.63kN#
4m
4m
5.5m
5.5m
B
A Ground Floor – office (1.5kN/m2)
Column load for column 1B = (2.5m x 2.75m) x 4.0kN/m2 = 27.5kN Column load for column GB = (2.5m x 2.75m) x 1.5kN/m2 = 10.31kN Total Column load for column B = 27.5kN + 10.31kN = 37.81kN#
Exercise • Find the column load for each column • The load factor is 10kN/m2
Level 2
Level 1
• Identify the tributary area for each column on level 2 • Name all columns according to grid lines
A1
B1
C1
D2 A2
B2
B3 A4
Level 2
C3
• Identify the tributary area for each column on level 1 • Name all columns according to grid lines
A1
B1
A2
B2
C1
D2 B3 C3 A4
Level 1
• Calculate the column load based on the tributary area
D
D
D D
Column Area Load 2C A/1- 6 -60 2C A/2- 16.5 - 165 2C A/3- 10.5 - 105 2C B/1- 13.14 - 131.4 2C B/2- 25.5 - 255 2C B/3- 18.75 - 187.5 2C C/1- 7.14 -71.4 2C C/3- 8.75 - 87.5 2C D/2- 9 -90
Load Column Area 1c A/1 - 9 - 90 + 60 - 150 1C A/2- 24.75 - 247.5 + 165 -412.5 1C A/3- 15.75 - 157.5 + 105 - 262.5 1C B/1 - 10 - 100 + 131.4 - 231.4 1C B/2- 25.5 - 255 + 255 - 510 1C B/3- 23.75 -237.5 +187.5- 425 1C C/1- 7.14 - 71.4 + 71.4 - 142.8 1C C/3- 8.25 -82.5 + 87.5 - 170 1C D/2 - 9.5 - 95 + 90 - 185
Tributary area method ( Both Live & Dead Load ) The example given is the layout plan for a house. Analyze the columns based on the specifications given. Height of the floors=3m Brick wall 150mm thick Dead Load= 3.0m x 0.15m x 19 kN/m3= 8.55kN/m Beam size is 150 x 300mm Dead Load= 0.15 m x 0.3 m x 24 kN/m3= 1.08kN/m Slab thickness is 150mm, Dead Load= 0.15m x 24 kN/m3= 3.6 kN/m2 Column size=300 x 400mm Dead Load= 0.3m x 0.4m x 3m x 24 kN/m3 = 8.64 kN
The example begins analyzing column B1. 1
6000
2
8000
A 2000 4000
Living
Dining
7. 5m2
B
3750 3500
C GROUND FLOOR PLAN
Kitchen
3
1
6000
2
3
8000
A 2000 4000
Bedroom 1
Bedroom 2
7. 5m2
B
3750 3500
C FIRST FLOOR PLAN
Gymnasium
Toilet
1 A 4000
B 3500
C ROOF PLAN
6000
2
8000
3
1
6000
2
3
8000
A 2000 4000
Bedroom 1
Bedroom 2
7. 5m2
B
3750 3500
C STRUCTURAL PLAN
Gymnasium
Toilet
Determine the load acting on column B1. Roof Level Dead Load Slab ( Flat Roof) 7.5m2 (total slab area) x 3.6 kN/m2 =27 kN Roof Beam 6.75 m (total beam length) x 1.08 kN/m = 7.29kN 1st Floor Dead Load Live Load Walls 6.75 m x 8.55kN/m= 57.7 kN Bedroom Slabs 7.5m2 x 3.6 kN/m2= 27 kN (3 X 2m) x 1.5 kN/m2 = 9 kN Beams 6.75m x 1.08 kN/m= 7.29 kN Gymnasium Column = 8.64 kN (3 X 1.75m) x 5 kN/m2= 26.25kN Total 100.63 kN Grd Floor Dead Load Live Load Walls 3.75 m x 8.55kN/m= 32 kN Livingroom Slabs 7.5m2 x 3.6 kN/m2= 27 kN (3 X 2m) x 1.5 kN/m2=9 kN Beams 6.75m x 1.08 kN/m= 7.29 kN Column = 8.64 kN Total 74.93 kN TOTAL Dead Load 209.85 kN TOTAL Live Load =44.25kN Apply 1.4 factor 293.79 kN Apply 1.6 factor 70.8kN So, ultimate load acting on column B1= 364.59 kN
Column for Residential Units
In order to know whether the designed column is able to hold the load, we may use the formula below in accordance to BS 8110;
N = 0.4fcuAc + 0.8 fyAsc N = capacity of concrete Fcu = concrete strength (N/mm2) Ac = cross section of concrete column fy = yield strength of steel (N/mm2) Asc = steel content in a column
Example :
Given that fcu = 30N/mm2 and fy = N = 0.4fcuAc + 0.8 fyAsc N = capacity of concrete 2 460N/mm . Assuming percentage of Fcu = concrete strength (N/mm2) Ac = cross section of concrete column steel reinforcement in a rectangular fcu = yield strength of steel (N/mm2) concrete column is 2%, determine Asc = steel content in a column the capacity of the column. Ac = 300 x 400 = 120,000 Asc = 2 % x 120,000 = 2,400 N = 0.4fcuAc + 0.8 fyAsc = 0.4(30)(120000) + 0.8(460)(2400) = 1440,000 + 883,200 = 2323200 N = 2323.2kN
300mm
400mm
Thus, this column can sustain any ULTIMATE load below 26323kN
Forces
Introduction • The main function of a structure is to transfer load to the earth • What are load that act upon structure???
Accidental Load
100kg Force incurred to portal frame = mass x gravitational acceleration = 100 kg x 9.81 (ď ž 10) = 1000 N = 1 kN
Force System • In any structure, forces can be applied in a variety of directions • Coplanar forces – line of action of forces that acting on the same plane (2D arrangement) • Non coplanar forces – forces that do not act on the same plane (3D arrangement)
Coplanar Force (2D arrangement) Line of action lie in the same plane
F1
F3
F2
Non - Coplanar Force (3D arrangement) Line of action do not lie in the same plane
F1 F3
F2
When it comes to structural analysis, coplanar force or non coplanar force analysis is commonly adopted?
Although structures are three dimensional, but most assemblies are two dimensional.
F5
F4 F1
F2 F3
F6
F7
F8 Coplanar forces analysis is easier than non coplanar forces analysis
Non concurrent force – Forces that do not intersect
Collinear Force – Forces that have the same line of action
F4
F5
F2
Joint A F3
F6
F7
F8 Concurrent Force – forces intersect at a common point
A force is a vector quantity. It must have magnitude, direction and point of action.
To indicate that it is a vector, it is usually indicated as F or F or F
F
x
There can be many forces of different direction and magnitude acting on a particle. These forces can be replaced by ONE single force that produce the same effect.
100kN
50kN
150kN
Both are equivalent
There can be many forces of different direction and magnitude acting on a particle. These forces can be replaced by ONE single force (named as resultant force) that produce the same effect.
100kN
50kN
150kN
Both are equivalent
20kN + y -
50kN
30kN x +
Cartesian system - To simplify the direction of forces
250kN + y -
x +
20kN
Resultant force, Fr = -270kN in y direction
270kN
Fb = 100kN
45⁰ Triangle Rule Polygon Rule Resolution of Force
Fa = 200kN
35⁰ x
A single force can be replaced by two or more forces that produce the same effect. These forces are called as components of a forces There could be infinite pairs of components. In order to simplify structural analysis, force is usually resolved into components X and Y (similar to Cartesian system)
Similarly, two or more forces can be replaced by a single force that produces the same effect.
This process is named as resolution of force
Components can be in forces are called as components of a forces y
x
FA
FAX
FAY
y
x
Basic understanding of trigonometry is important!!! In a right angled triangle, đ?‘œđ?‘?đ?‘?đ?‘ đ?‘–đ?‘Ąđ?‘’ đ??šđ??´đ?‘Œ sin đ?œƒ = = â„Žđ?‘Śđ?‘?đ?‘œđ?‘Ąđ?‘’đ?‘›đ?‘˘đ?‘ đ?‘’ đ??šđ??´ đ?‘Žđ?‘‘đ?‘—đ?‘Žđ?‘?đ?‘’đ?‘›đ?‘Ą đ??šđ??´đ?‘‹ cos đ?œƒ = = â„Žđ?‘Śđ?‘?đ?‘œđ?‘Ąđ?‘’đ?‘›đ?‘˘đ?‘ đ?‘’ đ??šđ??´ đ?‘œđ?‘?đ?‘?đ?‘œđ?‘ đ?‘–đ?‘Ąđ?‘’ đ??šđ??´đ?‘Œ tan đ?œƒ = = đ?‘Žđ?‘‘đ?‘—đ?‘Žđ?‘?đ?‘’đ?‘›đ?‘Ą đ??šđ??´đ?‘‹
FA
Also Pythagorean theorem, FR = đ??šđ?‘Ś 2 + đ??šđ?‘Ľ 2
FAX
FAY
Resolve the following forces into x and y components.
Fb = 100kN
Fa = 200kN
y
45â °
x
35â °
Fa = 200kN
y
35⁰ x
Fax
Fay
sin 𝜃 =
𝐹𝑎𝑦 𝐹𝑎 𝐹𝑎𝑦 200𝑘𝑁
sin 35 = 𝐹𝑎𝑦 = 200𝑘𝑁 × 𝑠𝑖𝑛35
=114.7kN
𝐹𝑎𝑥 cos 𝜃 = 𝐹𝑎 𝐹𝑎𝑥 cos 35= 200𝑘𝑁
𝐹𝑎𝑥 = 200𝑘𝑁 × 𝑐𝑜𝑠35
=163.8kN
sin 𝜃 =
Fb = 100kN
Fby
sin 45 = 𝐹𝑏𝑦 = 100𝑘𝑁 × 𝑠𝑖𝑛45
=70.7kN
y
45⁰ Fbx
𝐹𝑏𝑦 𝐹𝑏 𝐹𝑏𝑦 100𝑘𝑁
x
𝐹𝑏𝑥 cos 𝜃 = 𝐹𝑏 𝐹𝑏𝑥 cos 45= 100𝑘𝑁
𝐹𝑏𝑥 = 100𝑘𝑁 × 𝑐𝑜𝑠45
=-70.7kN
Fb = 100kN
Fby
Fa = 200kN y
y
45⁰ Fbx
x
35⁰
Fax
Fay
Fy = 185.4 kN x
Fx = 93.1 kN
𝐹𝑥 = Fax + Fbx = 163.8 + - 70.7kN = 93.1 kN 𝐹𝑦 = Fay + Fby = 114.7 + 70.7kN = 185.4 kN FR = 𝐹𝑦 2 + 𝐹𝑥 2 = 207.6 kN
tan đ?œƒ =
đ??šđ?‘Ś đ??šđ?‘Ľ 185.4 93.1
tan đ?œƒ = đ?œƒ = 63.33° y
63.33ď‚° x
Resolve the following forces into x and y components.
Fb = 250kN y
Fc = 150kN
40â °
x
25â °
Fa = 150kN
Fbx y
𝐹𝑏𝑦 = 250𝑘𝑁 × 𝑠𝑖𝑛40
Fby
= 160.7kN
𝐹𝑏𝑥 = −250𝑘𝑁 × 𝑐𝑜𝑠40
= -191.5kN
Fay
Fc = 150kN
40⁰
x
25⁰
Fa = 150kN Fax 𝐹𝑎𝑦 = −150𝑘𝑁 × 𝑐𝑜𝑠25
= -135.9kN
𝐹𝑎𝑥 = −150𝑘𝑁 × 𝑠𝑖𝑛25
= -63.3kN
FR = 107.7kN y
13.3⁰
x
24.8𝑘𝑁 −104.8
tan 𝜃 = 𝜃 = 13.3°
𝐹𝑦 = 160.7 − 135.9
= 24.8kN
𝐹𝑥 = −191.5 − 63.3 + 150 =−104.8kN 𝐹𝑅 = 24.8 2 + −104.8 2 = 107.7𝑘𝑁
An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force. An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.
Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object). F = ma = mass x acceleration For every action there is an equal and opposite re-action.
Building is a static structure, remain at rest all the time. Thus resultant force acting on a structure must always be ZERO! đ??šđ?‘Ľ = 0 đ??šđ?‘Ś = 0
đ?‘€ =0
Determine if the object shown is in equilibrium. đ??šđ?‘Ľ = 0 đ??šđ?‘Ś = - 100kN + 100kN = 0
đ?‘€đ?‘œ = -100kN(2m) + 100kN(3.5m) = +150kNm ď‚š 0 Moment is also a vector!
Clockwise moment = +ve
Anticlockwise moment = -ve
Determine resultant force for each of the following case.
To illustrate the idea of axial force, consider a simple column subject to a vertical load as shown. Compression
Axial Force Internal External
Internal
Tension External
Restraint either direction
Restraint both directions
Unable to take moment!!
Restraint both directions & able to take moment
The End Thank you!
INTRODUCTION TO STRUCTURAL DESIGN By Mohd Adib Ramli Taylor’s School of Architecture, Building, Design
Lecture outcomes At the end of this lecture students should be able to : Know the basic issues that affect the design of building structures Recall previous lesson on forces in Building Construction II Understands how load is transferred within a building
Why learn building structures? 1. So that architects can solve structural problems 2. So that they can talk more intelligently to their engineering consultants
3. So that architectural design, becomes more rational and therefore, perhaps, more beautiful
Basic Concerns • All physical objects have structures • The design of structures is part of the general design problems
Basic Concerns Safety • Life safety is the major concern in structural design. Two critical considerations are: a) Fire resistance
b) Low likelihood of collapse under load
Basic Concerns a) Major elements of fire resistance: i) Combustibility of the structure ii) Loss of strength at high temperature iii) Containment of the fire
Basic Concerns b) Sustaining loads i) a building should have some margin of structural capacity beyond what is required for the actual task • The margin of safety is defined by the safety factor and calculated as follows actual capacity of the structure Safety Factor, SF =
required capacity of the structure
Basic Concerns Example: • If a structure is required to carry 20,000kg and is actually able to carry 35,000kg before collapsing, the safety factor is expressed as follows: SF = 35,000/20,000 = 1.75 • Building structures are generally designed with an average safety factor of about 2
Basic Concerns Other concerns of structural design apart from fire resistance & sustaining loads are: c) Feasibility d) Economy e) Optimization f) Integration Feasibility Structures must use materials & products which are available & can be handled by the production organization
Basic Concerns Economy • Structures are usually less appreciated unless it constitutes a major design feature • Expensive structures do not often add value in the way that expensive finishes or hardware may.
Basic Concerns Optimization • Building designers are subjected to conflicting needs • Often changes that tend to improve one factor would degrade others
Basic Concerns Integration • Good structural design requires integration of the structure into the whole physical system of the building
Building Construction II Revision Structural Fundamentals
Building Construction II Revision Forces
Building Construction II Revision Internal Forces
Building Construction II Revision External Forces
Structural Requirements To perform its function of supporting a building in response to whatever loads that being applied to it, a structure must : 1. be capable of achieving a state of equilibrium. 2. be stable. 3. have adequate strength. 4. have adequate rigidity.
Structural Requirements Equilibrium
• It is the state of rest under the action of applied load. • All applied loads are balanced exactly by the reaction forces generated at its foundations.
Structural Requirements Stable
• It is the property which preserves the geometry of a structure and allows its elements to act together to resist loads. • It is still possible for a structure to fall due to its being geometrically unstable even if it has sufficient strength & rigidity
Structural Requirements Strength
• It depends upon material properties. • The strength of a material depends on its capacity to withstand axial stress, shear stress, bending, and torsion.
Structural Requirements Rigidity
• It is a property that makes a structure does not bend or flex under an applied force. • Some structures are formed by collections of objects that are rigid bodies, with pairs of objects connected by flexible hinges.
Structural Failures Structure fails due to the following: 1. Stability failures • Relates to the overall structural system 2. Strength failures
• Relates to the strength of the individual structural members
Structural Failures
Structural Members Generally two types:
1. Horizontal members • Floors • Roofs • Beams • Girders 2. Vertical members • Columns • Loadbearing walls.
Load Path • The path that a load travels through the structural system
• “Tracing” or “chasing” the loads
• Each structural element must be designed for all loads that pass through it
HVAC
Load Path • Every load applied to the building will travel through the structural system until it is transferred to the supporting soil. APPLIED LOAD
Load Path
Load Path
Load Path
beam
5kN
B 10kN
6m
A
1m
girder
1m
Tributary Area Is the area of a slab that contributes to the load on a specific beam The yellow area constitute the total load on Beam A The green area constitute the total load on Beam B
Load Path 1
2 8m
C
1.5 1.5 1
2m
B
1
3m
A
Force on slab AB 1-2 = 100kN BC 1-2 = 80kN Beam A1-2 = 50kN or 6.25kN/m Beam B1-2 = 90kN or 11.25kN/m Beam C1-2 = 40kN or 5kN/m
FLOOR SLAB Mdm Norita Johar
Definition A flat piece of concrete, supported by the walls or beams and columns of a structure. It serves as a walking surface and technically it serves as a load bearing member.
Functions of Flooring • To provide a uniform, flat surface to place building facilities • To act as sound, heat and fire insulator • Act as a vertical divider for the occupants • Upper slab became the ceiling for the storey below • To support dead & live load
Design Types • Flat slab • Ribbed / waffle • Hollow core Hollow core slab
Flat Slab
Waffle slab
Design Types
Flat slab with capital
Two-way slab with beams
Load Distribution & Units 20kN
(point) 40kN 5m
(point) 20kN/m
20kN/m2 (area) UDL 2m
(line) UDL
COLUMN
2m
BEAM & WALL
(point) 200kN
SLAB
Load Distribution & Units Slabs can be supported by load bearing walls, beams or straight to column. Loads are uniformly distributed to these supporting elements.
Single Path Slab on Columns
Load Distribution & Units 20kN (point)
COLUMN
Load Distribution & Units
Single Path Slab On Walls / Beams
Load Distribution & Units (point) 40kN
20kN/m (line) UDL
2m
BEAM & WALL
Load Distribution & Units
Dual Path Slab On Beams, Beams on Columns
Slab Analysis Factors • Types of Load – dead load + floor finishes – live load
• Types of system - one way -two way- trapezoid - triangle
Slab Analysis Factors -Types of Load- Dead Load The calculation for dead load from the slab transferred to the beam needs to consider the: • Thickness of the slab • Self weight of the slab • Weight of walls sitting on the slab • Self weight of the beam
Slab Analysis Factors -Types of Load- Dead Load The standard weight of the materials can be referred in UBBL.
Slab Analysis Factors -Types of Load- Live Load The calculation and analysis on a slab for live load is determined according to the functions of the rooms as stipulated in UBBL . Generally , a room with heavy equipments or larger user like a hospital or gymnasium has a larger amount of standard live load.
Slab Analysis Factors -Types of Load- Live Load
Slab Analysis Factors -Types of Slab System Slab systems can be divided in 2 types: • One way • Two way Ly Ly=Longer side of slab Lx= Shorter side of slab When Ly / Lx > 2, it is a one way slab When Ly /Lx < 2 or =2,it is a two way slab
Lx
Slab Analysis Factors -Types of Slab System One Way System
Lx Ly
Ly / Lx > 2
One-way slabs carry load in one direction. Bending primarily about the long axis. Thus,main reinforcement is required in the one direction.
Slab Analysis Factors -Types of Slab System Two Way System Lx
Ly
Ly / Lx <= 2
Two-way slabs carry load in two directions. Bending will take place in the two directions in a dish-like form. Accordingly, main reinforcement is required in the two directions.
Slab Analysis Factors -Types of Slab System Examples: one way slab Ly / Lx > 2 two way slab Ly / Lx <= 2 Ly=6m
Lx=2m
Ly=3m
Lx=2m
Slab Analysis Factors -Types of Slab System Exercise: Determine the system of the slabs below:
Ly=6m
Lx=5m
Ly=8m
Lx=3m
Ly=10m
Lx=5m
Slab Analysis Factors -Types of Slab System Exercise-Determine the system of the slabs below:
Ly=6m
Lx=5m
Ly / Lx < 2
two way slab
Ly=8m
Lx=3m
Ly / Lx > 2
one way slab
Ly=10m
Lx=5m
Ly / Lx = 2
two way slab
Load Distribution-Tributary Area Loads exerted on slabs are calculated per unit area kN/m2. Loads are uniformly distributed on the slabs. Many floor systems consist of a reinforced concrete slab supported on a rectangular grid of beams. The distribution of floor loads on floor beams is based on the geometric configuration of the beams forming the grid. When the loads are transferred to a beam, they are calculated per unit meter kN/m.
Load Distribution-One Way Slab When the length of slab is two times its width, almost all of the floor load goes to the longer beams AB,CD & EF B
Slab ABDC
D
Slab CDFE
F
one way slab
Ly / Lx > 2
50kN
20kN
30m
A
10m
C
8m
E
Load Distribution-One Way Slab The load transferred to these beams will depend on the area of the slabs. Beam CD will receive more slab load than beam AB and EF. B D Slab CDFE F Slab ABDC
1/2
1/2
1/2
1/2
30m
A
5m
5m
4m
4m
25kN
25kN
10kN
10kN
10m
C
8m
E
One Way Dead/Live Load (UBBL) x (Lx/2) Std weight of materials Sch Std Live Load Sch
Area of slab affected
Load Distribution-Two Way Slab Triangular form
When the ratio of the long span Ly to the short span Lx as Trapezoidal shown is less or more than 2, form 45° the floor load is carried in both directions to the four supporting beams around the two way slab panel.
Ly / Lx <= 2
Load Distribution-Two Way Slab Loads from slabs transferred to beams has 2 different patterns: • trapezoidal Trapezoidal • triangular pattern form
Triangular form
45°
two way slab
Ly / Lx <= 2
Two Way Dead/Live Load (UBBL) x (Lx/2) Std weight of materials Sch
Area of slab affected
Std Live Load Sch
[Dead/Live Load (UBBL) x (Lx/2)] x 2/3
Slab Analysis-Dead Load Area of the slab affected
Slab self weight (thickness x concrete density ) Eg, Assume slab thickness is 150mm, 0.15m x 24 kN/m3= 3.6 kN/m2
One Way Dead Load (UBBL) x (Lx/2)
Two Way ( Trapezoidal ) Dead Load (UBBL) x (Lx /2) Two Way ( Triangular ) [Dead Load (UBBL) x (Lx /2) ] 2/3 (need to times with 2/3)
Slab Analysis-Dead Load Exercise 1:
C 6m
A
5m 9m slab 2
4m slab 1
D
B
Calculate total dead loads transferred from slabs 1 & 2 onto beam CD. Beam CD receives load from: slab 1 (one way) Dead Load (UBBL) x (Lx/2) from slab 2 and the triangular pattern (2 way). [Dead Load (UBBL) x (Lx /2) ] 2/3
Slab Analysis-Dead Load Dead load from slab 1 self weight Dead Load (UBBL) x (Lx/2) =3.6 kN/m2 x (4m/2) =7.2 kN/m
C 6m
A
5m 9m slab 2
4m slab 1
B
Dead load from slab 2 self weight
[Dead Load (UBBL) x (Lx /2) ] 2/3
=[3.6kN/m2 x (5m/2) ] 2/3 =[9 kN/m ] 2/3 =6 kN/m
D
Total dead load from both slabs: = 7.2 kN/m + 6 kN/m =11.2 kN/m
Slab Analysis-Dead Load Slab self weight is not the only dead load on the beam. Slab self weight (thickness x concrete density ) Eg, Assume slab thickness is 150mm, 0.15m x 24 kN/m3= 3.6 kN/m2 There are other factors to be included that sit on the beam; Wall Self weight (height x thickness x brick density ) Eg, 3.0m x 0.15m x 19 kN/m3= 8.55kN/m
Beam Self weight (beam size x concrete density ) Eg, Assume beam size is 150mm x 300mm 0.15 m x 0.3 m x 24 kN/m3= 1.08kN/m
Slab Analysis-Dead Load Total dead load from both slabs: =11.2 kN/m
Beam Self weight =1.08kN/m Wall Self weight = 8.55kN/m Total dead load =
20.83 kN /m
Eg, Assume beam size is 150 x 300mm 0.15 m x 0.3 m x 24 kN/m3= 1.08kN/m Eg, 3.0m x 0.15m x 19 kN/m3= 8.55kN/m
Slab Analysis-Live Load One Way Live Load (UBBL) x (Lx /2) Two Way ( Trapezoidal ) Live Load (UBBL) x (Lx /2) Two Way ( Triangular ) [Live Load (UBBL) x (Lx /2) ] 2/3 (need to times with 2/3)
Slab Analysis-Live Load-One way Slab Example 1: The diagram is showing a one way slab in a house. Find the live load transferred from the slab to beam AB. A
C
12m
D
5m
B
From the UBBL, a house has 1.5 UDL standard live load amount. So, live load from slab transferred to beam AB is: Live Load Factor (UBBL) x (Lx /2) = 1.5 kN/m2 x ( 5m / 2) =2.75 kN/m
Slab Analysis-Live Load-Two way Slab Example 2: The diagram is showing a two way slab in a house. Find the live load transferred from the slab to beam AB and AC. A
C
8m
D
5m
B
To beam AB and CD - trapezoidal form to beam AC and DB - triangular form
Slab Analysis-Live Load-Two way Slab Live Load Factor (UBBL) x (Lx /2)
To beam AB and CD - trapezoidal form So, live load = 1.5kN/m2 x ( 5m / 2) =2.75 kN/m [Live Load Factor (UBBL) x (Lx/2) ] 2/3
To beam AC and DB - triangular form (need to times with 2/3) So, live load = [1.5 kN/m2 x ( 5m / 2) ] x 2/3 = [ 2.75 kN/m ] x 2/3 =1.8 kN/m
Slab Analysis-The Ultimate Load The ultimate load is generally the amount of all loads after the consideration of safety load factor. The purpose of load factor is to allow extra safety than the intended use. The load factor needs to be applied of 1.4 and 1.6 for dead load and live load respectively. Example; Total Dead Load= 20.3 x 1.4= 29.1 kN/m Total Live load= 3.0 x 1.6 = 4.8 kN/m Ultimate Load= 33.9 kN/m
Truss
Truss â&#x20AC;˘ A truss is a structure built up of three or more members which are normally considered as being pinned or hinged at the various joints.
â&#x20AC;˘ Any loads which are applied to the truss are usually transmitted to the joints, so that individual members are in pure tension or compression.
Why Truss? • Quick Installation – Can be installed quickly – Usually fabricated at factory and delivered to site for installation only. • Longer Span – truss is able to span longer with minimal deflection • Accessibility – Triangular spaces allow installation of utilities
A Simple Truss â&#x20AC;˘ Members in compression - strut â&#x20AC;˘ Members in tension tie
Types of Truss • Perfect Frame, Imperfect Frame and Redundant Frame • Perfect frame – consists of member just sufficient for being stable • Imperfect frame – consists too few members to prevent collapse • Redundant frame – consist more than sufficient members for being stable
Number of members in Perfect Frame â&#x20AC;˘ The simplest perfect frame â&#x20AC;&#x201C; three (3) members in form of triangle â&#x20AC;˘ As long as triangles are added, the frame remains as perfect frame. 2đ??˝ = đ?&#x2018;&#x161; + 3 J = number of joints m = number of members
Which of the following is perfect frame?
Analysis of Perfect Frame • Resolution of forces at joints • Recall equilibrium – forces in any direction should be = 0 (fx=0 and fy=0 and M=0) • Thus, equilibrium of external and internal forces should be achieved at every joint of frame
Example 1 â&#x20AC;˘ Determine the force in each member. State that it is compression or tension.
Analysis Steps • Determine the reaction force • Analyze joint by joint – start with joint with least members – Assume direction of force, F in member – Resolve F into x and y components – Equilibrium @ every joint, Fx=0, Fx=0, M=0
Determine reaction force @ A & C
MA = +100kN (1.25/2) – RC(1.25) 0 = +62.5 – 1.25RC RC = 50kN Fy Ray
= Ray + Rc -100kN = 50 kN
At Joint â&#x20AC;&#x2DC;Aâ&#x20AC;&#x2122; FABx
0.75
tan đ?&#x153;&#x192; = 0.625 đ?&#x153;&#x192; = 50.2°
Resolve FAB into component x and y, FABy FABy = FAB sinď ą = -FAB sin50.2 FABx = FAB cosď ą = -FAB cos50.2
50kN
The structure must be in equilibrium, hence ď &#x201C;Fx = 0 and ď &#x201C;Fy = 0
ď &#x201C;Fx = 0 = -FABcos50.2 + RAX - FAC ď &#x201C;Fy = 0 = -FAB sin50.2 + 50kN
At Joint ‘C’ FCBx
FCBy
50kN
Forces in member
Example 2:
Space Frame â&#x20AC;˘ 3 Dimensional Trusses
BMW Welt in Munich, Germany