University Putra Malaysia

ACR 3413

BASIC STRUCTURAL ENGINEERING 3

Lecture 4

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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 Vertical Load {V} Horizontal Load {H}

Conceptual Design

Loading X

Scheme Design X

Analysis X

Design TODAY’S LECTURE X

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Shorter Span, lx = 6m; Longer Span, ly = 8m; Floor to Floor Height = 3m; 10 FloorsSlab Thk = 250mm

Screeding and Tiling = 50mm; Client Specified Future SDL Allowance = 2.5kPaServices = Centralised Air-Conditioning Ducting Etc

Architectural = Ceiling and LightingArchitectural = 150mm Internal Brickwall 25m Long Per Panel

Architectural = External Cladding 250mm Stone

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Span 1 Panel 1 LL: African Elephant = 5tonnes = 50kN 50kN/(6mx8m) = 1.1kPaSpan 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 4.0kPaSpan 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 = 7.2kPaSpan 3 Panel 2 LL: Residential Space = UBBL Says 1.5kPa

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All Spans All Panels DL: Slab Thk Given as 250mm = 0.25m x 24kN/m3 = 6.0kPaAll Spans All Panels DL: Beam Say 400mm x 900mm = 0.4m x 0.9m x 24kN/m3 / 6m = 1.5kPa

All Spans All Panels DL: Total = 6.0+1.5 = 7.5kPa

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All Spans All Panels SDL: Client Specified Future Allowance = 2.5kPaAll Spans All Panels SDL: Screeding and Tiling Given as 50mm = 0.05m x 24kN/m3 = 1.2kPa

All Spans All Panels SDL: Centralised Air-Conditioning Ducting Given = 0.5kPaAll 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/m3 x 0.150m = 215kN 215kN/(6mx8m) = 4.5kPa

All Spans All Panels SDL: Total = 2.5+1.2+0.5+0.15+4.5 = 8.85kPa

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SLS: 1.0DL + 1.0SDL + 1.0LLULS: 1.4DL + 1.4SDL + 1.6LL

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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

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Span 1 Panel 1 ULS Tributary Line Loading : wULS = 25kPa x 6m/2 = 75kN/mSpan 1 Panel 2 ULS Tributary Line Loading : wULS = 27kPa x 6m/2 = 81kN/mSpan 2 Panel 1 ULS Tributary Line Loading : wULS = 30kPa x 6m/2 = 90kN/mSpan 2 Panel 2 ULS Tributary Line Loading : wULS = 31kPa x 6m/2 = 93kN/m

Span 3 Panel 1 ULS Tributary Line Loading : wULS = 35kPa x 6m/2 = 105kN/mSpan 3 Panel 2 ULS Tributary Line Loading : wULS = 26kPa x 6m/2 = 78kN/m

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Sum of Span 1 ULS Tributary Line Loading : SwULS,1 = 75kN/m + 81kN/m = 156kN/mSum of Span 2 ULS Tributary Line Loading : SwULS,2 = 90kN/m + 93kN/m = 183kN/m

Sum of Span 3 ULS Tributary Line Loading : SwULS,3 = 105kN/m + 78kN/m = 183kN/m

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Analysis Methods Available

Statically Determinate Structures1. Use Statics – Practical to do by Hand2. Use Tabulated Coefficients – Practical to do by hand3. Use Stiffness Method – Not practical to do by hand, must

use computers

Statically Indeterminate Structures1. 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 hand3. Use Stiffness Method – Not practical to do by hand, must

use computers

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ULS Bending Moment (Hogging), MULS : SwULS,3.L2/12 = 183kN/m x 8m2/12 = 1000kNm

ULS Bending Moment (Sagging), MULS : SwULS,3.L2/24 = 183kN/m x 8m2/24 = 500kNm

ULS Shear Force, VULS : SwULS,3.L/2 = 183kN/m x 8m/2 = 750kNULS Axial Force, NULS : No. of Floors x SwULS,3.L/2 x 2 = 10 Floors x 750kN x 2 = 15000kN

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Definitions

1. Slab – Horizontal flat member supporting loads2. Beam - Horizontal member supporting slabs3. Column / Wall – Vertical member supporting beams and/or slabs 4. Foundations – Vertical member supporting columns

Conceptual Design

1. Discretization of Physical Model - Mechanism / Determinate / Indeterminate Structures

Loading

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. SDL4. Live load - externally applied v. LL5. NHL load - externally applied h. NHL6. Wind load - externally applied h. WL7. EQ load - externally applied h. EQ

Scheme Design

1. RC Two-Way Slab With RC Beams2. RC One-Way Slab With RC Beams3. RC Flat Slab4. PT Flat Slab5. ST Composite Slab With ST Beams

Analysis

1. ULS and SLS loading combinations2. Structural analysis - mathematics3. 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