ACR 3413 BASIC STRUCTURAL ENGINEERING 3 Lecture 4 Univers rsit - PowerPoint PPT Presentation

ACR 3413 BASIC STRUCTURAL ENGINEERING 3 Lecture 4 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   Loading X  Scheme Design X  Analysis X  Design TODAY’S LECTURE X - Quality Control (QA) (V & H) - Do It All Again and Again 2

Shorter Span, l x = 6m; Longer Span, l y = 8m; Floor to Floor Height = 3m; 10 Floors Slab Thk = 250mm Screeding and Tiling = 50mm; Client Specified Future SDL Allowance = 2.5kPa Services = Centralised Air-Conditioning Ducting Etc Architectural = Ceiling and Lighting Architectural = 150mm Internal Brickwall 25m Long Per Panel 3 Architectural = External Cladding 250mm Stone

Span 1 Panel 1 LL: African Elephant = 5tonnes = 50kN  50kN/(6mx8m) = 1.1kPa Span 1 Panel 2 LL: Car Park = 3 cars x 2tonnes = 60kN  60kN/(6mx8m) = 1.25kPa But UBBL Says 2.5kPa Span 2 Panel 1 LL: Café With Fixed Seating = UBBL Says 4.0kPa Span 2 Panel 2 LL: Café Without Fixed Seating = UBBL Says 5.0kPa Span 3 Panel 1 LL: Library = UBBL Says 2.4kPa x Height Say 3m = 7.2kPa Span 3 Panel 2 LL: Residential Space = UBBL Says 1.5kPa 4

All Spans All Panels DL: Slab Thk Given as 250mm = 0.25m x 24kN/m 3 = 6.0kPa All Spans All Panels DL: Beam Say 400mm x 900mm = 0.4m x 0.9m x 24kN/m 3 / 6m = 1.5kPa All Spans All Panels DL: Total = 6.0+1.5 = 7.5kPa 5

All Spans All Panels SDL: Client Specified Future Allowance = 2.5kPa All Spans All Panels SDL: Screeding and Tiling Given as 50mm = 0.05m x 24kN/m 3 = 1.2kPa All Spans All Panels SDL: Centralised Air-Conditioning Ducting Given = 0.5kPa All Spans All Panels SDL: Ceiling and Lighting = 0.15kPa All Spans All Panels SDL: Brickwork Given As 25m Long Per Panel = 25m Long x 3m High x 19kN/m 3 x 0.150m = 215kN  215kN/(6mx8m) = 4.5kPa 6 All Spans All Panels SDL: Total = 2.5+1.2+0.5+0.15+4.5 = 8.85kPa

SLS: 1.0DL + 1.0SDL + 1.0LL ULS: 1.4DL + 1.4SDL + 1.6LL 7

Span 1 Panel 1 ULS : 1.4DL + 1.4SDL + 1.6LL = 1.4 x 7.5kPa + 1.4 x 8.85kPa + 1.6 x 1.1kPa = 25kPa Span 1 Panel 2 ULS : 1.4DL + 1.4SDL + 1.6LL = 1.4 x 7.5kPa + 1.4 x 8.85kPa + 1.6 x 2.5kPa = 27kPa Span 2 Panel 1 ULS : 1.4DL + 1.4SDL + 1.6LL = 1.4 x 7.5kPa + 1.4 x 8.85kPa + 1.6 x 4.0kPa = 30kPa Span 2 Panel 2 ULS : 1.4DL + 1.4SDL + 1.6LL = 1.4 x 7.5kPa + 1.4 x 8.85kPa + 1.6 x 5.0kPa = 31kPa Span 3 Panel 1 ULS : 1.4DL + 1.4SDL + 1.6LL = 1.4 x 7.5kPa + 1.4 x 8.85kPa + 1.6 x 7.2kPa = 35kPa Span 3 Panel 2 ULS : 1.4DL + 1.4SDL + 1.6LL = 1.4 x 7.5kPa + 1.4 x 8.85kPa + 1.6 x 1.5kPa = 26kPa 8

Span 1 Panel 1 ULS Tributary Line Loading : w ULS = 25kPa x 6m/2 = 75kN/m Span 1 Panel 2 ULS Tributary Line Loading : w ULS = 27kPa x 6m/2 = 81kN/m Span 2 Panel 1 ULS Tributary Line Loading : w ULS = 30kPa x 6m/2 = 90kN/m Span 2 Panel 2 ULS Tributary Line Loading : w ULS = 31kPa x 6m/2 = 93kN/m Span 3 Panel 1 ULS Tributary Line Loading : w ULS = 35kPa x 6m/2 = 105kN/m Span 3 Panel 2 ULS Tributary Line Loading : w ULS = 26kPa x 6m/2 = 78kN/m 9

Sum of Span 1 ULS Tributary Line Loading : Sw ULS,1 = 75kN/m + 81kN/m = 156kN/m Sum of Span 2 ULS Tributary Line Loading : Sw ULS,2 = 90kN/m + 93kN/m = 183kN/m Sum of Span 3 ULS Tributary Line Loading : Sw ULS,3 = 105kN/m + 78kN/m = 183kN/m 10

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 11

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ULS Bending Moment (Hogging), M ULS : Sw ULS,3 .L 2 /12 = 183kN/m x 8m 2 /12 = 1000kNm ULS Bending Moment (Sagging), M ULS : Sw ULS,3 .L 2 /24 = 183kN/m x 8m 2 /24 = 500kNm ULS Shear Force, V ULS : Sw ULS,3 .L/2 = 183kN/m x 8m/2 = 750kN ULS Axial Force, N ULS : No. of Floors x Sw ULS,3 .L/2 x 2 = 10 Floors x 750kN x 2 = 15000kN 16

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Definitions Scheme Design 1. Slab – Horizontal flat member supporting loads 1. RC Two-Way Slab With RC Beams 2. Beam - Horizontal member supporting slabs 2. RC One-Way Slab With RC Beams 3. Column / Wall – Vertical member supporting 3. RC Flat Slab beams and/or slabs 4. PT Flat Slab 4. Foundations – Vertical member supporting 5. ST Composite Slab With ST Beams columns Analys ysis Conc nceptua ual Design 1. ULS and SLS loading combinations 1. Discretization of Physical Model - Mechanism / 2. Structural analysis - mathematics Determinate / Indeterminate Structures 3. Force – internal distribution of effects bending moment (kNm) • Loading ng axial (kN) • shear (kN) • 1. Load – externally applied load torsion (kNm) • mass - kg / tonnes • 4. Deflections – externally displacements load – kN • pressure - kPa • Design 2. Dead load - externally applied v. DL (self- weight) 1. ULS Capacity 3. Superimposed dead load - externally applied v. - Stress SDL normal (direct) stress • 4. Live load - externally applied v. LL shear stress • 5. NHL load - externally applied h. NHL 2. SLS Capacity 6. Wind load - externally applied h. WL 7. EQ load - externally applied h. EQ 31