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Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are horizontal PowerPoint PPT Presentation


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Beams. Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are horizontal. 1/39. devices for transferring vertical loads horizontally. Beams. action of beams involves combination of bending and shear. 2/39.

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Extremely common structural element In buildings majority of loads are vertical and majority of useable surfaces are horizontal

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Beams

  • Extremely common structural element

  • In buildings majority of loads are vertical and majority of useable surfaces are horizontal

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devices for transferring

vertical loads horizontally

Beams

action of beams involves combination of

bending and shear

2/39


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What Beams have to Do

  • Be strong enough for the loads

  • Not deflect too much

  • Suit the building for size, material, finish,

    fixing etc

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Checking a Beam

  • what we are trying to check (test)

  • stability - will not fall over

  • adequatestrength - will not break

  • adequatefunctionality - will not deflect too much

  • what do we need to know

  • span - how supported

  • loads on the beam

  • material, shape & dimensions of beam

  • allowable strength & allowable deflection

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?

?

Designing a Beam

  • what we are trying to do

  • determine shape & dimensions

  • what do we need to know

  • span - how supported

  • loads on the beam

  • material

  • allowable strength & allowable deflection

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Tributary Areas

  • A beam picks up the load halfway to its neighbours

  • Each member also carries its own weight

this beam supports the load that comes from this area

span

spacing

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Tributary Areas (Cont. 1)

  • A column generally picks up load from halfway to its neighbours

  • It also carries the load that comes from the floors above

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Length

Thickness

Load =

Surface area x

Wt per sq m,

or

volume x

wt per cu m

Height

Dead Loads on Elements

  • Code values per cubic metre or square metre

  • Multiply by the volume or area supported

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Live Loads on Elements

  • Code values per square metre

  • Multiply by the area supported

Area carried by one beam

Total Load =area x (Live load + Dead load) per sq m

+self weight

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Distributed Load

Point Load

Reactions

Loads on Beams

  • Point loads, from concentrated loads or other beams

  • Distributed loads, from anything continuous

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Bending

Shear

What the Loads Do

  • The loads (& reactions) bend the beam, and try to shearthrough it

11/39


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e

e

e

e

C

T

Bending

Shear

What the Loads Do (cont.)

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beams in building

designed for bending

checked for shear

Designing Beams

  • in architectural structures, bending moment more important

    • importance increases as span increases

  • short span structures with heavy loads, shear dominant

    • e.g. pin connecting engine parts

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How we Quantify the Effects

  • First, find ALL the forces (loads and reactions)

  • Make the beam into a freebody (cut it out and artificially support it)

  • Find the reactions, using the conditions of equilibrium

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W

vertical reaction, R = W

and moment reaction MR = - WL

MR= -WL

L

R = W

Example 1 - Cantilever Beam

Point Load at End

  • Consider cantilever beam with point load on end

  • Use the freebody idea to isolate part of the beam

  • Add in forces required for equilibrium

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W

M = -Wx

x

V = W

V = W

Shear Force Diagram

BM = Wx

BM = WL

Bending Moment Diagram

Example 1 - Cantilever Beam

Point Load at End (cont1.)

Take section anywhere at distance, x from end

Add in forces, V = W and moment M = - Wx

Shear V = W constant along length

(X = 0 -> L)

Bending Moment BM = W.x

when x = L BM = WL

when x = 0 BM = 0

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w /unit length

L

Total Load W = w.L

MR = -WL/2

= -wL2/2

L/2

L/2

R = W = wL

Example 2 - Cantilever Beam

Uniformly Distributed Load

Formaximum shear V and bending moment BM

vertical reaction,R = W = wL

and moment reactionMR = - WL/2 = - wL2/2

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wx

M = -wx2/2

X/2

X/2

V = wx

V = wL

= W

Shear Force Diagram

BM = wx /2

2

BM = wL2/2

= WL/2

Bending Moment Diagram

Example 2 - Cantilever Beam

Uniformly Distributed Load (cont.)

For distributed V and BM

Take section anywhere at distance, x from end

Add in forces,V = w.xand momentM = - wx.x/2

ShearV = wx

when x = LV = W = wL

when x = 0V = 0

Bending MomentBM = w.x2/2

when x = LBM = wL2/2 = WL/2

when x = 0BM = 0

(parabolic)

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“Positive” shear

“Negative” shear

R.H down

L.H down

L.H up

R.H up

Sign Conventions

Shear Force Diagrams

  • To plot a diagram, we need a sign convention

  • The opposite convention is equally valid,

    but this one is common

  • There is no difference in effect between

    positive and negative shear forces

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W3

W1

W2

R1

Diagram of loading

R2

W1

R1

W2

R2

W3

Shear Force Diagram

Plotting the

Shear Force Diagram

  • Starting at the left hand end, imitate each force you meet (up or down)

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Shape of the

Shear Force Diagram

  • Point loads produce

    a block diagram

  • Uniformly distributed loads

    produce triangular diagrams

Diagrams of loading

Shear force diagrams

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Split in timber beam

Reo in concrete beam

What Shear Force does

to the Beam

  • Although the shear forces are vertical, shear stresses are both horizontal and vertical

  • Timber may split

    horizontally along

    the grain

  • Shear is seldom critical for steel

  • Concrete needs

  • special shear reinforcement

  • (45o or stirrups)

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Hogging

NEGATIVE

Sagging

POSITIVE

-

+

Sign Conventions

Bending Moment Diagrams

  • To plot a diagram, we need a sign convention

  • This convention is almost universally agreed

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Saggingbending momentisPOSITIVE (happy)

+

Hoggingbending momentisNEGATIVE

(sad)

-

Sign Conventions

Bending Moment Diagrams (cont.)

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  • Cantilevers produce negative moments

-

-

Cantilevers

  • Simple beams produce positive moments

-

-

+

+

Simple beam

Built-in beam

  • Built-in & continuous beams have both, with negative over the supports

Positive and Negative Moments

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+

+

This way mimics the beam’s deflection

This way is normal coordinate geometry

Where to Draw the

Bending Moment Diagram

  • Positive moments are drawn downwards

    (textbooks are divided about this)

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Diagrams of loading

Bending moment diagrams

Shape of the Bending Moment Diagram

  • Point loads produce triangular diagrams

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UDL

UDL

Diagrams of loading

Bending moment diagrams

Shape of the Bending Moment Diagram (cont1.)

  • Distributed loads produceparabolic diagrams

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Maximum value

Shape of the

Bending Moment Diagram (cont.2)

  • We are mainly concerned

    with the maximum values

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-

-

-

-

+

+

+

Shape of the

Bending Moment Diagram (cont.3)

  • Draw theDeflected Shape(exaggerate)

  • Use the Deflected shape as a guide to where the sagging (+) and hogging (-) moments are

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Simply supported

Continuous

Can we Reduce the Maximum BM Values?

  • Cantilevered ends reduce the positive bending moment

  • Built-in and continuous beams also have lower maximum BMs and less deflection

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Total load = W

W

(w per metre length)

L

L

Central point load

Max bending moment

= WL/4

Uniformly distributed load

Max bending moment

= WL/8 or wL2/8

where W = wL

Standard BM Coefficients

Simply Supported Beams

  • Use the standard formulas where you can

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Total load = W

W

(w per metre length)

L

L

End point load

Max bending moment

= -WL

Uniformly Distributed Load

Max bending moment

= -WL/2 or -wL2/2

where W = wL

Standard BM Coefficients Cantilevers

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W

W

W /m

W

W

W

W

W

W

W

W

W /m

W

W

Standard BM Coefficients Simple Beams

Beam

Cable

BMD

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W

W

W = wL

W = wL

L

L

L

L

V = +W/2

V = +W/2

V = +W

V = +W

V = -W/2

V = -W/2

Mmax = WL/8

= wL2/8

Mmax = -WL

Mmax = -WL/2

= -wL2/2

Mmax = WL/4

SFD & BMD Simply Supported Beams

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How much compressive stress?

How much

tensile stress?

How much deflection?

What the Bending Moment

does to the Beam

  • Causes compression on one face and tension on the other

  • Causes the beam to deflect

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how big & what shape?

How to Calculate the

Bending Stress

  • It depends on the beam cross-section

  • We need some particular properties of the

    section

is the section we are using as a beam

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What to do with the

Bending Stress

  • Codes give maximum allowable stresses

  • Timber, depending on grade, can take 5 to 20 MPa

  • Steel can take around 165 MPa

  • Use of Codes comes later in the course

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we need to find the

Section Properties

Finding Section Properties

next lecture

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