Engineering 1 - Civil Engineering - Lecture 8 - Types of Beam Support and Critical Collapse Mechanisms PDF

Preview

Summary

Presented by Dr Luke Bisby: Types of beam supports and critical collapse mechanisms...

Description

Civil Engineers Example 6 “Envirofluid” Nerd Dr Leon Boegman

Numerical modelling of wave breaking in the coastal ocean

Numerical simulation

Modelling transport of contaminant discharges

New Orleans Levee Design

Computer model of levee failure

Shoreline protection and tsunami modelling

Back to Engineering 1… … example as promised!

Engineering 1 - Civil

Rules about the number of mechanisms which must be considered: 1.

In a beam, each span may collapse independently of the others

2.

A plastic hinge must form within the beam span (except for cantilevers), and may occur under any one of the loads

3.

For each new point load, another possible mechanism can occur

4.

Where there is a distributed load, the hinge may form anywhere under the load: only in special cases can we guess where it should be (not considered in Engineering 1)

5.

The critical mechanism, producing the lowest collapse load, can only be found by considering them all!!!

© School of Engineering, 2009

Engineering 1 - Civil

Rules about the number of plastic hinges needed in a mechanism: 1. A beam can collapse if there are three sources of rotation 2. The end of a beam, resting on a simple support allows a source of rotation 3. A plastic hinge at any point allows a rotation

Thus: 1. A beam simply supported at both ends has two sources at its ends, and needs only one extra source from a single plastic hinge within it 2. A beam built in at both ends needs three plastic hinges within it 3. A beam simply supported at one end and continuous at the other has one source of rotation at the simply supported end, and needs two plastic hinges to collapse © School of Engineering, 2009

Types of Beam Supports • Simple support at the end (1 hinge):

• Internal simple support on a continuous beam (2 hinges):

• Fixed supports at both ends (3 hinges):

Engineering 1 - Civil

Rules about where plastic hinges form: 1. A plastic hinge will form under a load 2. Plastic hinges to permit rotations at the ends of a beam form as far from the load as they can 3. A plastic hinge may form at a change in the crosssection of the beam: it always forms in the weaker part at a change of section (not considered in Engineering 1)

© School of Engineering, 2009

Engineering 1 - Civil

Rules about alternative mechanisms that must be considered: 1. It is vital that the weakest mechanism is discovered when a structure is analysed for collapse 2. In a single span, the only mechanisms which need to be considered are those which involve a load and the ends of the span, unless a change of section size occurs, when this change must also be included 3. In multiple spans, each span may generally be considered independently of the others 4. It is possible for parts of the beam to move upwards during collapse, if multiple spans are weaker in this condition © School of Engineering, 2009

Example: From the 2007 Exam… try it!

(i) Possible collapse mechanisms • We must assume a collapse mechanism… and in this case we have two choices (two point loads!): Mechanism A – 3 hinges 2W

L/3

Hinge

2W

W

L/3

L/3

θ

L/3

Φ δ1

Hinge

Mechanism B – mirror image

δ2 θ

Hinge

Hinge

W

L/3

L/3

Φ

θ δ1

Hinge

δ2 Hinge

Φ θ

Φ

(ii) Collapse load for each mechanism • Recall: We can determine the collapse load by equating the internal work done by the plastic hinges to the external work done by the loads as the beam deforms – The INTERNAL work during plastic deformation is given by:

UPi = ΣMPθ The EXTERNAL work is given by:

UPe = ΣW Pδ And CONSERVATION OF ENERGY dictates that:

UPi = UPe

L/3

Mechanism A

L/3

L/3 2W

• The deflection underneath the load 2W is:

W

θ

Φ δ1

• The rotation Φ of the hinge at the right hand end of the beam can then be deduced from:

δ2 θ

Φ

• Equating these two values of δ1 :

• The rotation of the plastic hinge under the load is found from the angle between parallel lines as (θ + Φ):

• It should be noted that the deflection under the second load of W is:   

  

• Work done by two loads and energy absorbed in the three plastic hinges 2W

δ1

W

δ2

– Work done by loads 2W and W is equal to the load times the distance they travel during the deformation, δ1 and δ2:

  

  

– All hinges carry a full plastic moment Mp – The energy absorbed in the plastic hinges is: • Left hand hinge rotates central hinge rotates ( right hand hinge rotates Φ

+ Φ),

– The total (internal) energy absorbed is then: Φ

θ

θ

Φ

 

 

L/3

Mechanism B

L/3

L/3

Φ

θ

• The deflection underneath the load W is:

δ1

• The rotation θ of the hinge at the right hand end of the beam can then be deduced from:

δ2

θ

Φ

• Equating these two values of δ1 :

• The rotation of the plastic hinge under the load is found from the angle between parallel lines as (θ + Φ):

• It should be noted that the deflection under the second load of W is:   

  

• Work done by two loads and energy absorbed in the three plastic hinges 2W

δ1

W

δ2

– Work done by loads 2W and W is equal to the load times the distance they travel during the deformation, δ1 and δ2:

  

  

– All hinges carry a full plastic moment Mp – The energy absorbed in the plastic hinges is: • Left hand hinge rotates central hinge rotates ( right hand hinge rotates Φ

+ Φ),

– The total (internal) energy absorbed is then: Φ

θ

  θ

Φ

 

(iii) Which is the critical collapse mechanism? – The minimum collapse load is the lower of the 2 loads: Summary Mechanism A: Mechanism B:

Wp 3.6Mp/L 4.5Mp/L

– The critical mechanism is Mechanism A; the internal hinge does occur under the bigger load! – The critical collapse mechanism at the end is: 2W

W

θ

Hinge

Φ δ2

δ1 Hinge

L/3

θ L/3

Hinge

Φ L/3

Let’s put some values into it… • What if I asked for the actual value of the collapse load, W, if: – The beam has a span of 3 m and is made of steel with a yield strength of σY = 400 MPa – The beam has an L-shaped cross-section as shown below 20 mm

100 mm

20 mm 100 mm

We have already found the collapse load in terms of Mp and L

We need to determine Mp for the beam’s cross-section… 20 mm

100 mm

PNA 20 mm 100 mm

20 mm

100 mm

σY

AC1

C1

AC2

a

PNA

AT

20 mm

y

σY

C2 b

T

100 mm

Stresses

Cross-Section

Forces

To find Mp we will take the moments of the two compressive forces, C1 and C2, about the tensile force, T 20 mm

100 mm

σY

AC1 A C2

100 mm

a

σY

PNA

AT

C1

20 mm

18 mm

T

C2 b

• From before:

• So now that we know L and Mp :

Things you need to know!!! 1. Equilibrium in 2-dimensions 2. Free body diagrams and solving for reactions 3. Internal Shear force (V), Axial force (N), and Bonding Moment (M) calculations for beams 4. Finding Mp for any given beam cross-section 5. Finding the collapse load (or collapse load factor) for a given beam subjected to a given loading condition

Collapse of Columns

Review – Factors causing collapse • Collapse can occur as a result of plastic collapse – Beams (plastic hinges; as studied in detail) – Tension members (tension failure) – stocky columns (compression failure; “crushing”)

• … or collapse can occur as a result of buckling – – – –

slender columns Shells Some kinds of beams etc.

Local bucking causing failure of a silo in Fife

Ground movement caused compressive loads in the rail and eventually resulted in global (Euler) buckling

Euler Buckling

Local buckling of steel beams due to fire exposure

Local Buckling

Column Collapse: Crushing vs. Buckling 



 

 



  

  

Next time we will look at: Collapse of Columns…

… See you next week...