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## Structural Engineering

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### Structural Engineering

### Forces in Structures

### Properties of Civil Engineering Materials

### Important Structural Properties

### Concepts in Equilibrium

Sergio F. Breña

STEM Education Institute

Saturday Workshop

September 30, 2006

University of Massachusetts Amherst

Outline

- Introduction to Structural Engineering
- Forces in Structures
- Structural Systems
- Civil Engineering Materials
- Some Definitions of Important Structural Properties

University of Massachusetts Amherst

Structural Engineering

- What does a Structural Engineer do?
- A Structural Engineer designs the structural systems and structural elements in buildings, bridges, stadiums, tunnels, and other civil engineering works (bones)
- Design: process of determining location, material, and size of structural elements to resist forces acting in a structure

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Engineering Design Process

- Identify the problem (challenge)
- Explore alternative solutions
- Research past experience
- Brainstorm
- Preliminary design of most promising solutions
- Analyze and design one or more viable solutions
- Testing and evaluation of solution
- Experimental testing (prototype) or field tests
- Peer evaluation
- Build solution using available resources (materials, equipment, labor)

University of Massachusetts Amherst

Design Process in Structural Engineering

- Select material for construction
- Determine appropriate structural system for a particular case
- Determine forces acting on a structure
- Calculate size of members and connections to avoid failure (collapse) or excessive deformation

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Examples of Typical Structures

University of Massachusetts Amherst

University of Massachusetts Amherst

Forces Acting in Structures

- Forces induced by gravity
- Dead Loads (permanent): self-weight of structure and attachments
- Live Loads (transient): moving loads (e.g. occupants, vehicles)
- Forces induced by wind
- Forces induced by earthquakes
- Forces induced by rain/snow
- Fluid pressures
- Others

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Forces Acting in Structures

Vertical: Gravity

Lateral: Wind, Earthquake

University of Massachusetts Amherst

University of Massachusetts Amherst

Example (English Units):

T = 1,000 lb (1 kip)

A = 10 in2.

Stress = 1,000/10 = 100 lb/in2

Stress = Force/Area

Section X

Example (SI Units):

1 lb = 4.448 N (Newton)

1 in = 25.4 mm

T = 1,000 lb x 4.448 N/lb = 4448 N

A = 10 in2 x (25.4 mm)2 = 6450 mm2

(1 in)2

Stress = 4448/6450 = 0.69 N/mm2 (MPa)

Section X

T

T

Definition of StressUniversity of Massachusetts Amherst

DL

Lo

T

Definition of StrainStrain = DL / Lo

Example:

Lo = 10 in.

DL = 0.12 in.

Strain = 0.12 / 10 = 0.012 in./in.

Strain is dimensionless!!

(same in English or SI units)

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Stress – Strain Behavior of Elastic Mats.

Stress

E

E = Modulus of Elasticity = Stress / Strain

Strain

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Stress

E

Strain

Strain

Stress

(a) Linear Elastic

Stress

(b) Non-linear Elastic

Strain

Strain

Plastic strain

Plastic strain

(c) Elastic-plastic

(d) Non-linear Plastic

Types of Stress-Strain BehaviorUniversity of Massachusetts Amherst

Materials Used in Civil Engineering

- Stone and Masonry
- Metals
- Cast Iron
- Steel
- Aluminum
- Concrete
- Wood
- Fiber-Reinforced Plastics

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Engineering Properties of Materials

- Steel
- Maximum stress: 40,000 – 120,000 lb/in2
- Maximum strain: 0.2 – 0.4
- Modulus of elasticity: 29,000,000 lb/in2
- Concrete
- Maximum stress: 4,000 – 12,000 lb/in2
- Maximum strain: 0.004
- Modulus of elasticity: 3,600,000 – 6,200,000 lb/in2
- Wood

Values depend on wood grade. Below are some samples

- Tension stress: 1300 lb/in2
- Compression stress: 1500 lb/in2
- Modulus of elasticity: 1,600,000 lb/in2

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Concrete Components

- Sand (Fine Aggregate)
- Gravel (Coarse Aggregate)
- Cement (Binder)
- Water
- Air

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Fiber-Reinforced Composites

Composite Laminate

Polyester

Polymer

Matrix

Epoxy

Vinylester

Glass

- Functions of matrix:
- Force transfer to fibers
- Compressive strength
- Chemical protection

Fiber Materials

Aramid (Kevlar)

Carbon

- Function of fibers:
- Provide stiffness
- Tensile strength

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University of Massachusetts Amherst

Tensile Failure

Engineering Properties of Structural Elements- Strength
- Ability to withstand a given stress without failure
- Depends on type of material and type of force (tension or compression)

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Engineering Properties of Structural Elements

- Stiffness (Rigidity)
- Property related to deformation
- Stiffer structural elements deform less under the same applied load
- Stiffness depends on type of material (E), structural shape, and structural configuration
- Two main types
- Axial stiffness
- Bending stiffness

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DL

Lo

T

Axial StiffnessStiffness = T / DL

Example:

T = 100 lb

DL = 0.12 in.

Stiffness = 100 lb / 0.12 in. = 833 lb/in.

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Bending Stiffness

Displacement

Force

Stiffness = Force / Displacement

Example:

Force = 1,000 lb

Displacement = 0.5 in.

Stiffness = 1,000 lb / 0.5 in. = 2,000 lb/in.

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Types of Structural Elements – Bars and Cables

Bars can carry either tension

or compression

Cables can only carry tension

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Racking Failure of Pinned Frame

Infilled Frame

Rigid Joints

Braced Frame

Providing Stability for Lateral LoadsUniversity of Massachusetts Amherst

University of Massachusetts Amherst

Equilibrium of Forces (Statics)

- Forces are a type of quantity called vectors
- Defined by magnitude and direction
- Statement of equilibrium
- Net force at a point in a structure = zero (summation of forces = zero)
- Net force at a point is determined using a force polygon to account for magnitude and direction

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Moment of Force = Force x Distance

To neutralize rotation about point A, moments from the two forces has to be equal and opposite:

100 lb x 3 ft = 50 lb x 6 ft

A

3 ft

6 ft

Moment (Rotational) EquilibriumUniversity of Massachusetts Amherst

10 ft

Side AC

Side BC

=

=

=

=

1.667

1.333

100

lb

6 ft

6 ft

Side AB

Side AB

A

10 ft

6 ft

Force BC

Force AC

1.333

1.667

=

=

Force AB

Force AB

C

B

Force AC = 1.667 x 100 lb = 166.7 lb

Force BC = 1.333 x 100 lb = 133.3 lb

8 ft

Force Calculation in Simple Structure36.9

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Force Transfer from Beams to Supports

Force, P

1/3 L

2/3 L

2/3 P

1/3 P

Span, L

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Force Transfer Example - Bridge

8,000 lb

32,000 lb

15 ft

45 ft

30 ft

30 ft

L = 60 ft

22,000 lb*

18,000 lb**

*Front axle: 8,000 lb x 45/60 = 6,000 lb

Rear axle: 32,000 lb x 30/60 = 16,000 lb

**Front axle: 8,000 lb x 15/60 = 2,000 lb

Rear axle: 32,000 lb x 30/60 = 16,000 lb

University of Massachusetts Amherst

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