Wk 7 Equilibrium

7/18/2019 Wk 7 Equilibrium

http://slidepdf.com/reader/full/wk-7-equilibrium 1/25

Universit of S dne – BDes Desi n Studies 1A - Structures

Peter Smith & Mike Rosenman

EQUILIBRIUMAs a whole A system of elements





Structural Systems

The building must be stable as a whole There must be enough structural elements

They must be in suitable places

They must be strong and stiff enough, and

suitably connected together

As a whole A system of elements

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Stability

Structure may be in equilibrium but not stable Arrangement of parts critical

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Qualitative and Quantitative

Understanding

First we will try to understand the structuralsystem qualitatively

If we don’t have the right kind of supports in theright kind of places, it won’t work

When we have a credible system, we can useprecedent and simple rules to get credible sizes

We need to understand the quantitative basis of afinal structural design





Equilibrium of the Whole Building

The building supports its loads

All loads are finally resisted by

the ground

Parts of the building intervenebetween loads and foundation

But first, let’s make sure the

building as a whole is stable

Loads

Reactions from

Foundation

?

Building





verall Equilibrium

What do !e "eed to #no!$

all the possible loads

Loads

?

Building

Reactions from

Foundation

where the building

can be supported

how big the reactions have to be for equilibrium





verall Equilibrium

What are the %roblems$

downward loadsjust need bigenough footings

Reactions from Foundation

Building

Loads

Biggerbuilding

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verall Equilibrium

What are the %roblems$ &cont'(

horizontal loads mightoverturn a tall building


Tallerbuilding

Building

Loads

wider base and heavy building are more stable

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Equilibrium of )orces

Newton’s Third law


‘to every force there is an equal

and opposite reaction’

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http://slidepdf.com/reader/full/wk-7-equilibrium 10/25Universit of S dne – BDes Desi n Studies 1A - Structures


Equilibrium &forces in line(

Every force is resisted by an equal andopposite one, exactly in line

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Equilibrium &forces out of line(

Tendency to overturn

The turning effect is amoment

Can be resisted by other

out-of-line forces

Maintains

equilibrium

Tries to

overturn

Tries to

restrain10/26




Equilibrium

forces on !hole structure and each

*art+ ,ust balance &sum equal -ero(

doesn’t move

not u*.and.do!n+ side!ays or s*in

V = 0 H = 0 M = 0

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P S ih Mik R




EQUILIBRIUM &cont' /(

w w w w w w

H

V = 0

H = 0 M = 0

w = 6w

F = 5kN

F = 5kN

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P S ih&Mik R




EQUILIBRIUM &cont' 0(

w w w w w w

H

w

V = 0

H = 0 M = 0

H

60kN 60kN

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Pt S ith&Mik R




EQUILIBRIUM &cont' 1(

w w w w w w

H

M

w

H

V = 0

H = 0 M = 0

M

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Pt S ith&Mik R




Equilibrium of Elements

total downward load = total upward reaction

if the load is symmetrical, so are the reactions

beam

5 units

2.5 units 2.5 units

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Pt S ith&Mik R




Equilibrium of )rame!or2

Total downward load is carried down by columns

beam

5 units

2.5 units 2.5 units

We can follow the ‘Load Path’

2.5 units 2.5 units

PeterSmith&MikeRosenman




Equilibrium of )rame!or2 &cont'(

If the load is off-centre, so are the reactions

5 units

beam

4L/5 L/5

1 unit 4 units

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Wor2 smarter+ not harder

Most beams are symmetrical

If the reactions are equal, don’t make hard work of it

W

R1 R2

Everything is symmetrical

R1 = R2 = W/2

R2R1

W is off centre

R1, R2 must be calculated

W (central)

19/26





Wor2 smarter+ not harder &cont'(

The M condition says the sum of moments aboutany point is zero

Pick a point that eliminates one of the unknowns,to make it easy

W (known)

R1 R2

Only W and R2 have a

moment about the dot20/26

R2R1

W (known)

R1 and R2 have a

moment about the dot





)inding the Reactions of

a 3antilever

A cantilever has oneV, oneH,and oneM reaction M

RH=0

W1 W2

d2

d1

Vertically:R = W1 + W2

Horizontally:H = 0 unless

there is a horizontal load

Moments:M = W1.d1 + W2.d2

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d1d2





)ree.bodies

&Loo2ing Inside the Elements(

can isolate any member or *art of it to study it

must *ut bac2 artificial forces to re*lace

!hatever su**orts !ere cut a!ay

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)ree.bodies

100kg 1kN 1kN

1kN

can ‘cut’ the wire at any point1kN

1kN

1kN

1kN

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)ree.bodies &cont/'(

20kg20kg20kg20kg20kg

100kg

1kN 1kN

0.2kN0.2kN

0.2kN0.2kN0.2kN

1kN

0.2kN0.2kN

0.2kN0.2kN0.2kN

2kN1kN

0.2kN0.2kN

0.2kN0.2kN0.2kN

2kN

1.4kN

2kN

0.2kN

0.2kN

0.2kN

1kN

0.2kN0.2kN

1.8kN

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)ree.bodies &cont0'(

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Wk 7 Equilibrium