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 Load {H} Conceptual Design X Loading X Scheme Design TODAY’S LECTURE X Analysis TODAY’S LECTURE X Design X X - Quality Control (QA) (V & H) - Do It All Again and Again 2
Mecha chanis nism: Unstable structure, a s < 0. Sta Staticall tically Deter etermina nate te: = 0. The equilibrium a s 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 : a s > 0. Structures having more unknown forces than available equilibrium equations aka Redundant. 3
Statica cally lly Deter ermina inate e Struc uctures ures - Simply Supported - Cantilever - 3 Pinned Arch 4
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) 5
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1) Mass (kg) (3-D) 1) D) 3) Unifor orml mly y Distr tribu bute ted Load (UDL) (kN kN/m) m) (1-D) D) The mass of an object is a measure UDL is a load that is evenly spread how heavy the object is. It is along a length such as brick wall on measured in units of grams (g) or slab. It is measured in units of kilograms (kg). (kN/m). 2) Pressure (kN kN/m /m 2 ) (2-D) D) 4) Point t Load (kN kN) (0-D) D) Pressure is the force applied A point load is a load applied to a perpendicular to the surface of an single, specific point on a structural object per unit area (kN/m 2 ) over member. It is measured in units of which that force is distributed. (kN). 7
1) Mass (kg) (3-D) 1) D) 3) Unifor orml mly y Distr tribu bute ted Load (UDL) (kN kN/m) m) (1-D) D) Made e up of the e followi owing g types es of loading DL DL, SDL, LL, NHL, WL, EQ Made up of the followi owing g types of loading DL, SDL, LL, NHL, WL, EQ 2) Pressure (kN kN/m /m 2 ) (2-D) D) 4) Point t Load (kN kN) (0-D) D) Made up of the followi owing g types of Made up of the followi owing g types of loading DL, SDL, LL, NHL, WL, EQ loading DL, SDL, LL, NHL, WL, EQ 8
Mercedes One car = 2500 kg = 2.5 tonnes = 25kN Proton Saga [LL] 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 9
Building Functi tion on Load (kPa) Classrooms, lecture rooms, tutorial 3.0 rooms, computer rooms Domestic uses & residential 2.0 activities Wards, bedrooms and toilet rooms in hospitals, nursing homes and 2.0 residential care homes. Kitchens 2.0 Floors for offices 3.0 Conference rooms 5.0 Stair Case 4.0 Department stores, supermarkets, 5.0 markets, shops for display and sale [LL] of merchandise. Cold storage 5.0 for each meter height 10
These figures do not include [DL+SDL] vertical elements but is the DL+SDL for the floor !!! 11
These figures do not include vertical elements but is the DL+SDL for the floor !!! [DL+SDL] 12
Burj Khalifa 171 Storeys (659m High) Concrete Building Load = 9,460,000 kN Area = 280,000 m 2 Pressure = 33.8 kN/m 2 Sears Tower 113 Storeys (454.8m High) Steel Building Load = 3,800,000 kN Area = 408,922 m 2 Pressure = 9.3 kN/m 2 Taipei 101 94 Storeys (405.8m High) Load = 3,650,000 kN Area = 187,110 m 2 Pressure = 19.5 kN/m 2 Petronas Twin Tower 93 Storeys (403.8m High) Load = 3,300,000 kN Area = 213,750 m 2 These figures Pressure = 15.5 kN/m 2 now include Skyview Penang 43 Storeys (147.3m High) Load = 1,140,000 kN vertical Area = 66,365 m 2 Pressure = 17.2 kN/m 2 elements as Ampang Condo 18 Storeys (63.4m High) well in Load = 690,000 kN Area = 51,621 m 2 DL+SDL+LL !!! Pressure = 13.3 kN/m 2 Ipoh Hospital New Block 10 Storeys (46.1m High) Load = 1,830,000 kN [SLS]=[DL+SDL+LL] Area = 103,717 m 2 Pressure = 17.7 kN/m 2 Note All Figures Are Indicative and Not Exact and Should Not Be Relied Upon for Detailed Structural Analysis. 13
[DL+SDL+LL+NHL+WL+EQ] 14
[DL+SDL+LL+NHL+WL+EQ] 15
- Communication - Talk to Architect, M&E Engineer and Other Consultants of their Requirements 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 - Quality Control (QA) (V & H) - Do It All Again and Again 16
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Load / Verti tical Load {V} Horizon zonta tal Load {H} Eleme ment Effect Column mns Beams Column mns Beams Axial Primary y Primary y Primary y None Force Effect Effect Effect Shear Secondary Primary y Primary y Primary y Force Effect Effect Effect Effect Action Bending Secondary Primary y Primary y Primary y Moment Effect Effect Effect Effect Torsion Special None None None Moment Cases Primary y Primary y Primary y Primary y Kine mati Deflection Effect Effect Effect Effect c 22
Axial Force: is a force that tend to elongate or shorten a member and normally measured in kN. 23
Shear ar Force ce 24
Because of loading we apply on a member, the member will experience bending. The bending will lead to compression and tension in member. Since our 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. 25
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Concrete te is Weak in Tensi sion on. Concrete te is stron ong g in compr press ssion on. 27
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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 30
Span, L = 6 m w = 20 kN/m We need to find moment at any point i.e. x along the beam span. A x = 3m say B First, we need to find the reactions at the supports. w M A Step 1: Moment Equilibrium x R A .L - w L.L/2 = 0 R A = w L/2 R A = 20kN/m x 6m / 2 = 60kN R = 60kN V Step 2: Force Equilibrium For Statically Determinate Structures R A + R B – w .L = 0 Use 2 Moment Equilibriums S M ABOUT A = 0 R B = w .L – R A S M ABOUT B = 0 R B = 20kN/m x 6m – 60kN Use 1 Moment Equilibrium and 1 Force Equilibrium R B = 60kN S M ABOUT A = 0 S F = 0 31
Step 3: Draw Bending Moment Diagram M x - R A .x + w x.x/2 = 0 M x = R A .x - w x.x/2 M x = 60kN x 3m – 20kN/m x 3m x 3m/2 Span, L = 6 m M x = 90kNm w = 20 kN/m A x = 3m say B Step 4: Draw Shear Force Diagram R A + V x – w .x = 0 V x = w .x – R A w V x = 20kN/m x 3m – 60kN M x A V x = 0kN x R = 60kN V x 32
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Beam and External Load Effects Bending Deflection = 5wL 4 / (384EI) M mid = wL 2 /8 Simply supported beam uniformly loaded w = PL 3 / (48EI) M mid = PL/4 Simply supported beam mid-span point load P = wL 4 / (8EI) M fixed-end = wL 2 /2 Cantilever uniformly loaded w = PL 3 / (3EI) Cantilever free-end point load P M free-end = PL = wL 4 / (384EI) M fixed-end = wL 2 /12 Fixed-ended beam (both sides) uniformly loaded w M mid = wL 2 /24 = PL 3 / (192EI) M fixed-end = PL/8 Fixed-ended beam (both sides) mid-span point load P M mid = PL/8 M fixed-end = wL 2 /8 sag = wL 4 / (185EI) Propped cantilever uniformly loaded w M sag = 9wL 2 /128 sag = 0.00932PL 3 /EI M fixed-end = 3PL/16 Propped cantilever mid-span point load P M sag = 5PL/32 i. 38
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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. 41
Nodes Element nt (Inc. Section n and Material) Loading ng Boundar ary Cond ndition Analysis 42
Select nodes and insert the element coordinates. 43
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After we define the nodes, now we need to connect the nodes with an element. 45
Connect nodes by using the Add Element function. Also add the section and material properties. 46
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