acr 3413 basic structural engineering 3 lecture 3 · pdf file2 - communication - talk to...

University Putra Malaysia

ACR 3413

BASIC STRUCTURAL ENGINEERING 3

Lecture 3

2

- 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 X

Loading X

Scheme Design TODAY’S LECTURE X

Analysis TODAY’S LECTURE X

Design X X

Mechanism: Unstable structure, as < 0.

Statically Determinate: as = 0. The equilibrium

equations provide both the necessary and sufficientconditions for equilibrium. When all the forces in a structurecan be determined strictly from these equations.

Statically Indeterminate : as > 0. Structures having

more unknown forces than available equilibrium equationsaka Redundant.

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Statically Determinate Structures

- Simply Supported

- Cantilever

- 3 Pinned Arch

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Lessons From History

• 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) Mass (kg) (3-D)

The mass of an object is a measure

how heavy the object is. It is

measured in units of grams (g) or

kilograms (kg).

2) Pressure (kN/m2) (2-D)

Pressure is the force applied

perpendicular to the surface of an

object per unit area (kN/m2) over

which that force is distributed.

3) Uniformly Distributed Load (UDL) (kN/m) (1-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 Load (kN) (0-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) Mass (kg) (3-D)

Made up of the following types of

loading DL, SDL, LL, NHL, WL, EQ

2) Pressure (kN/m2) (2-D)



3) Uniformly Distributed Load (UDL) (kN/m) (1-D)



4) Point Load (kN) (0-D)



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MercedesOne car = 2500 kg

= 2.5 tonnes= 25kN

Proton SagaOne car = 1000 kg

= 1.0 tonnes= 10kN

Normal PersonAverage one person mass = 80 kg

= 0.8kN

Heavy PersonAverage one person mass = 100 kg

= 1.0kN

African Bush Elephant

[LL]

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Building Function 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 of 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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Burj Khalifa 171 Storeys (659m High) Concrete BuildingLoad = 9,460,000 kNArea = 280,000 m2

Pressure = 33.8 kN/m2

Sears Tower 113 Storeys (454.8m High) Steel BuildingLoad = 3,800,000 kNArea = 408,922 m2


Taipei 101 94 Storeys (405.8m High)Load = 3,650,000 kNArea = 187,110 m2


Petronas Twin Tower 93 Storeys (403.8m High)Load = 3,300,000 kNArea = 213,750 m2


Skyview Penang 43 Storeys (147.3m High)Load = 1,140,000 kNArea = 66,365 m2


Ampang Condo 18 Storeys (63.4m High)Load = 690,000 kNArea = 51,621 m2


Ipoh Hospital New Block 10 Storeys (46.1m High)Load = 1,830,000 kNArea = 103,717 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 Vertical Load {V} Horizontal 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 / Element

Vertical Load {V} Horizontal Load {H}

Effect Columns Beams Columns Beams

Acti

on

Axial Force

Primary Effect

NonePrimary Effect

Primary Effect

Shear Force

SecondaryEffect

Primary Effect

Primary Effect

Primary Effect

Bending Moment

SecondaryEffect

Primary Effect

Primary Effect

Primary Effect

Torsion Moment

NoneSpecial Cases

None None

Kin

em

ati

c

DeflectionPrimary Effect

Primary Effect

Primary Effect

Primary 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 Force

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Because of loading we apply on a member, the member will experiencebending. The bending will lead to compression and tension in member.Since our design will be in concrete and concrete is strong incompression, we interested to know how much tension the member isexperiencing.

This is because concrete is weak in tension, steel will be required at that zone.

Concrete is cracking at bottompart of beam due to tensionforce in member.

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Concrete is strong in compression.

Concrete is Weak in Tension.

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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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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 = 0RB = w.L – RA

RB = 20kN/m x 6m – 60kNRB = 60kN

A B

Span, L = 6 m

Step 1: Moment Equilibrium

RA.L - wL.L/2 = 0RA = wL/2 RA = 20kN/m x 6m / 2 = 60kN

For Statically Determinate StructuresUse 2 Moment Equilibriums

SMABOUT A = 0SMABOUT B = 0

Use 1 Moment Equilibrium and 1 Force Equilibrium

SMABOUT A = 0SF = 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 = 0Vx = w.x – RA

Vx = 20kN/m x 3m – 60kNVx = 0kN

A B

Span, L = 6 m

Step 3: Draw Bending Moment Diagram

Mx - RA.x + wx.x/2 = 0Mx = RA.x - wx.x/2Mx = 60kN x 3m – 20kN/m x 3m x 3m/2Mx = 90kNm

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Beam and External Load Effects Bending Deflection

Simply supported beam uniformly loaded w Mmid = wL2/8 = 5wL

4 / (384EI)

Simply supported beam mid-span point load P Mmid = PL/4 = PL3 / (48EI)

Cantilever uniformly loaded w Mfixed-end = wL2/2 = wL

4 / (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.

Nodes

Element (Inc. Section

and Material)

Loading

Boundary Condition

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 titles2) Load combination3) 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 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

acr 3413 basic structural engineering 3 lecture 3 · pdf file2 - communication - talk to...

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