ACR 3413 BASIC STRUCTURAL ENGINEERING 3 Lecture 3 Univers rsit - - PowerPoint PPT Presentation

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ACR 3413 BASIC STRUCTURAL ENGINEERING 3 Lecture 3 Univers rsit - - PowerPoint PPT Presentation

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ACR 3413 BASIC STRUCTURAL ENGINEERING 3 Lecture 3 Univers rsit ity y Putra a Malaysia ysia - Communication - Talk to Architect, M&E Engineer and Other Consultants of their Requirements Item Verti tical Load {V} Horizon zonta tal

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Univers rsit ity y Putra a Malaysia ysia

ACR 3413 BASIC STRUCTURAL ENGINEERING 3 Lecture 3

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  • Communication - Talk to Architect, M&E Engineer and Other

Consultants of their Requirements

  • Quality Control (QA) (V & H) - Do It All Again and Again

Item Verti tical Load {V} Horizon zonta tal Load {H} Conceptual Design  X Loading  X Scheme Design TODAY’S LECTURE X Analysis TODAY’S LECTURE X Design X X

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 Mecha

chanis nism: Unstable structure, as < 0.

 Sta

Staticall tically Deter etermina nate te:

as = 0. The equilibrium equations provide both the necessary and sufficient conditions for equilibrium. When all the forces in a structure can be determined strictly from these equations.

 Sta

Staticall tically Indet ndeter ermina nate te : as > 0. Structures having

more unknown forces than available equilibrium equations aka Redundant.

3

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Statica cally lly Deter ermina inate e Struc uctures ures

  • Simply Supported
  • Cantilever
  • 3 Pinned Arch
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Lesso sons ns From History ry

  • Provide redundancy (statical indeterminacy) i.e. it

has alternate load path

  • If a structure is statically determinate, ensure that

the design is not at its code limit, i.e. not on a knife-edge, this is ENGINEERING JUDGEMENT

  • These 2 principles will avoid disproportionate

collapse

  • Finally, and further, ensure safe method of failure

(ductile bending failure, not brittle shear failure)

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1) 1) Mass (kg) (3-D) D) The mass of an object is a measure how heavy the

  • bject

is. It is measured in units of grams (g) or kilograms (kg). 2) Pressure (kN kN/m /m2) (2-D) D) Pressure is the force applied perpendicular to the surface of an

  • bject per unit area (kN/m2) over

which that force is distributed. 3) Unifor

  • rml

mly y Distr tribu bute ted Load (UDL) (kN kN/m) m) (1-D) D) UDL is a load that is evenly spread along a length such as brick wall on

  • slab. It is

measured in units of (kN/m). 4) Point t Load (kN kN) (0-D) D) A point load is a load applied to a single, specific point on a structural

  • member. It is

measured in units of (kN).

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1) 1) Mass (kg) (3-D) D) Made e up of the e followi

  • wing

g types es of loading  DL DL, SDL, LL, NHL, WL, EQ 2) Pressure (kN kN/m /m2) (2-D) D) Made up of the followi

  • wing

g types of loading  DL, SDL, LL, NHL, WL, EQ 3) Unifor

  • rml

mly y Distr tribu bute ted Load (UDL) (kN kN/m) m) (1-D) D) Made up of the followi

  • wing

g types of loading  DL, SDL, LL, NHL, WL, EQ 4) Point t Load (kN kN) (0-D) D) Made up of the followi

  • wing

g types of loading  DL, SDL, LL, NHL, WL, EQ

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Mercedes One car = 2500 kg = 2.5 tonnes = 25kN Proton Saga One car = 1000 kg = 1.0 tonnes = 10kN Normal Person Average one person mass = 80 kg = 0.8kN Heavy Person Average one person mass = 100 kg = 1.0kN African Bush Elephant

[LL]

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Building Functi tion

  • n

Load (kPa)

Classrooms, lecture rooms, tutorial rooms, computer rooms 3.0 Domestic uses & residential activities 2.0 Wards, bedrooms and toilet rooms in hospitals, nursing homes and residential care homes. 2.0 Kitchens 2.0 Floors for offices 3.0 Conference rooms 5.0 Stair Case 4.0 Department stores, supermarkets, markets, shops for display and sale

  • f merchandise.

5.0 Cold storage 5.0 for each meter height

[LL]

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[DL+SDL]

These figures do not include vertical elements but is the DL+SDL for the floor !!!

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[DL+SDL]

These figures do not include vertical elements but is the DL+SDL for the floor !!!

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13 Burj Khalifa 171 Storeys (659m High)  Concrete Building Load = 9,460,000 kN Area = 280,000 m2 Pressure = 33.8 kN/m2 Sears Tower 113 Storeys (454.8m High)  Steel Building Load = 3,800,000 kN Area = 408,922 m2 Pressure = 9.3 kN/m2 Taipei 101 94 Storeys (405.8m High) Load = 3,650,000 kN Area = 187,110 m2 Pressure = 19.5 kN/m2 Petronas Twin Tower 93 Storeys (403.8m High) Load = 3,300,000 kN Area = 213,750 m2 Pressure = 15.5 kN/m2 Skyview Penang 43 Storeys (147.3m High) Load = 1,140,000 kN Area = 66,365 m2 Pressure = 17.2 kN/m2 Ampang Condo 18 Storeys (63.4m High) Load = 690,000 kN Area = 51,621 m2 Pressure = 13.3 kN/m2 Ipoh Hospital New Block 10 Storeys (46.1m High) Load = 1,830,000 kN Area = 103,717 m2 Pressure = 17.7 kN/m2

Note All Figures Are Indicative and Not Exact and Should Not Be Relied Upon for Detailed Structural Analysis.

These figures now include vertical elements as well in DL+SDL+LL !!!

[SLS]=[DL+SDL+LL]

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[DL+SDL+LL+NHL+WL+EQ]

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[DL+SDL+LL+NHL+WL+EQ]

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  • Communication - Talk to Architect, M&E Engineer and Other

Consultants of their Requirements

  • Quality Control (QA) (V & H) - Do It All Again and Again

Item Verti tical Load {V} Horizon zonta tal Load {H} Conceptual Design  X Loading  X Scheme Design TODAY’S LECTURE X Analysis TODAY’S LECTURE X Design X X

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Load / Eleme ment Verti tical Load {V} Horizon zonta tal Load {H} Effect Column mns Beams Column mns Beams Action Axial Force Primary y Effect None Primary y Effect Primary y Effect Shear Force Secondary Effect Primary y Effect Primary y Effect Primary y Effect Bending Moment Secondary Effect Primary y Effect Primary y Effect Primary y Effect Torsion Moment None Special Cases None None Kine mati c Deflection Primary y Effect Primary y Effect Primary y Effect Primary y Effect

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Axial Force: is a force that tend to elongate or shorten a member and normally measured in kN.

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Shear ar Force ce

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Because of loading we apply on a member, the member will experience

  • bending. The bending will lead to compression and tension in member.

Since

  • ur

design will be in concrete and concrete is strong in compression, we interested to know how much tension the member is experiencing. This is because concrete is weak in tension, steel will be required at that zone. Concrete is cracking at bottom part of beam due to tension force in member.

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Concrete te is stron

  • ng

g in compr press ssion

  • n.

Concrete te is Weak in Tensi sion

  • n.
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Analysis Methods Available Statically y Determi minate te Structu tures

  • 1. Use Statics – Practical to do by Hand
  • 2. Use Tabulated Coefficients – Practical to do by hand
  • 3. Use Stiffness Method – Not practical to do by hand, must

use computers Statically y Indete etermin minate te Structu tures

  • 1. Cannot Use Statics but Instead Use Moment Distribution

Method / Moment Area Method / Flexibility Method – Practical to do by hand but superceded in practice by the stiffness method !!

  • 2. Use Tabulated Coefficients – Practical to do by hand
  • 3. Use Stiffness Method – Not practical to do by hand, must

use computers

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We need to find moment at any point i.e. x along the beam span. x = 3m say x w w = 20 kN/m A M R = 60kN V First, we need to find the reactions at the supports. Step 2: Force Equilibrium RA + RB – w.L = 0 RB = w.L – RA RB = 20kN/m x 6m – 60kN RB = 60kN A B Span, L = 6 m Step 1: Moment Equilibrium RA.L - wL.L/2 = 0 RA = wL/2 RA = 20kN/m x 6m / 2 = 60kN For Statically Determinate Structures

Use 2 Moment Equilibriums SMABOUT A = 0 SMABOUT B = 0 Use 1 Moment Equilibrium and 1 Force Equilibrium SMABOUT A = 0 SF = 0

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x = 3m say x w w = 20 kN/m A Mx R = 60kN Vx Step 4: Draw Shear Force Diagram RA + Vx – w.x = 0 Vx = w.x – RA Vx = 20kN/m x 3m – 60kN Vx = 0kN A B Span, L = 6 m Step 3: Draw Bending Moment Diagram Mx - RA.x + wx.x/2 = 0 Mx = RA.x - wx.x/2 Mx = 60kN x 3m – 20kN/m x 3m x 3m/2 Mx = 90kNm

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Beam and External Load Effects Bending Deflection Simply supported beam uniformly loaded w Mmid = wL2/8  = 5wL4 / (384EI) Simply supported beam mid-span point load P Mmid = PL/4  = PL3 / (48EI) Cantilever uniformly loaded w Mfixed-end = wL2/2  = wL4 / (8EI) Cantilever free-end point load P Mfree-end = PL  = PL3 / (3EI) Fixed-ended beam (both sides) uniformly loaded w Mfixed-end = wL2/12 Mmid = wL2/24  = wL4 / (384EI) Fixed-ended beam (both sides) mid-span point load P Mfixed-end = PL/8 Mmid = PL/8  = PL3 / (192EI) Propped cantilever uniformly loaded w Mfixed-end = wL2/8 Msag = 9wL2/128 sag = wL4 / (185EI) Propped cantilever mid-span point load P Mfixed-end = 3PL/16 Msag = 5PL/32 sag = 0.00932PL3/EI i.

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  • GSA is a general purpose structural analysis software; it can

analysis any type and shape of building.

  • It can show bending moment, torsion moment, shear force,

axial force diagram for any member or frame. Also can use any type or form of loading.

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Nodes Element nt (Inc. Section n and Material) Loading ng Boundar ary Cond ndition Analysis

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Select nodes and insert the element coordinates.

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After we define the nodes, now we need to connect the nodes with an element.

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Connect nodes by using the Add Element function. Also add the section and material properties.

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1) Need to define load cases titles 2) Load combination 3) Assign load on desire member

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Define your load type, then close the window.

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Define your load combination.

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Assign Loading by press right click on member and select “Create Element Loading”.

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Again same thing for live load.

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Select nodes, then right click and select modify nodes.

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Bending Moment Diagram

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Shear Force Diagram

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Definitions

  • 1. Slab – Horizontal flat member supporting loads
  • 2. Beam - Horizontal member supporting slabs
  • 3. Column / Wall – Vertical member supporting

beams and/or slabs

  • 4. Foundations – Vertical member supporting

columns Conc nceptua ual Design

  • 1. Discretization of Physical Model - Mechanism /

Determinate / Indeterminate Structures Loading ng

  • 1. Load – externally applied load
  • mass - kg / tonnes
  • load – kN
  • pressure - kPa
  • 2. Dead load - externally applied v. DL (self-

weight)

  • 3. Superimposed dead load - externally applied v.

SDL

  • 4. Live load - externally applied v. LL
  • 5. NHL load - externally applied h. NHL
  • 6. Wind load - externally applied h. WL
  • 7. EQ load - externally applied h. EQ

Scheme Design

  • 1. RC Two-Way Slab With RC Beams
  • 2. RC One-Way Slab With RC Beams
  • 3. RC Flat Slab
  • 4. PT Flat Slab
  • 5. ST Composite Slab With ST Beams

Analys ysis

  • 1. ULS and SLS loading combinations
  • 2. Structural analysis - mathematics
  • 3. Force – internal distribution of effects
  • bending moment (kNm)
  • axial (kN)
  • shear (kN)
  • torsion (kNm)
  • 4. Deflections – externally displacements

Design

  • 1. ULS Capacity
  • Stress
  • normal (direct) stress
  • shear stress
  • 2. SLS Capacity