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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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slide1

Beams

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

1/39

slide2

devices for transferring

vertical loads horizontally

Beams

action of beams involves combination of

bending and shear

2/39

slide3

What Beams have to Do

  • Be strong enough for the loads
  • Not deflect too much
  • Suit the building for size, material, finish,

fixing etc

3/39

slide4

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

4/39

slide5

?

?

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

5/39

slide6

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

6/39

slide7

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

7/39

slide8

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

8/39

slide9

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

9/39

slide10

Distributed Load

Point Load

Reactions

Loads on Beams

  • Point loads, from concentrated loads or other beams
  • Distributed loads, from anything continuous

10/39

slide11

Bending

Shear

What the Loads Do

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

11/39

slide12

e

e

e

e

C

T

Bending

Shear

What the Loads Do (cont.)

12/39

slide13

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

13/39

slide14

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

14/39

slide15

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

15/39

slide16

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

16/39

slide17

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

17/39

slide18

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)

18/39

slide19

“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

19/39

slide20

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)

2039

slide21

Shape of the

Shear Force Diagram

  • Point loads produce

a block diagram

  • Uniformly distributed loads

produce triangular diagrams

Diagrams of loading

Shear force diagrams

21/39

slide22

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)

22/39

slide23

Hogging

NEGATIVE

Sagging

POSITIVE

-

+

Sign Conventions

Bending Moment Diagrams

  • To plot a diagram, we need a sign convention
  • This convention is almost universally agreed

23/39

slide24

Saggingbending momentisPOSITIVE (happy)

+

Hoggingbending momentisNEGATIVE

(sad)

-

Sign Conventions

Bending Moment Diagrams (cont.)

24/39

slide25

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

25/39

slide26

+

+

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)

26/39

slide27

Diagrams of loading

Bending moment diagrams

Shape of the Bending Moment Diagram

  • Point loads produce triangular diagrams

27/39

slide28

UDL

UDL

Diagrams of loading

Bending moment diagrams

Shape of the Bending Moment Diagram (cont1.)

  • Distributed loads produceparabolic diagrams

28/39

slide29

Maximum value

Shape of the

Bending Moment Diagram (cont.2)

  • We are mainly concerned

with the maximum values

29/39

slide30

-

-

-

-

+

+

+

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

30/39

slide31

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

31/39

slide32

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

32/39

slide33

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

33/39

slide34

W

W

W /m

W

W

W

W

W

W

W

W

W /m

W

W

Standard BM Coefficients Simple Beams

Beam

Cable

BMD

34/39

slide35

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

35/39

slide36

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

34/37

slide37

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

37/39

slide38

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

38/39

slide39

we need to find the

Section Properties

Finding Section Properties

next lecture

39/39

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