axial members
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Axial Members. WORKSHEET 11. to answer just click on the button or image related to the answer. let's go !!. compression structures. tension structures. a. b. Question 1a. are tension structures or compression structures more efficient when loaded axially?. - PowerPoint PPT PresentationTRANSCRIPT
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Axial Members
WORKSHEET 11
to answer just click on the button or image related to the answer
![Page 2: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/2.jpg)
Question 1a
are tension structures or compression structuresmore efficient when loaded axially?
compression structuresa
tension structures b
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Question 1b
why are tension structures more efficient?
they are not subject to bucklinga
they are usually made of steelb
they pull straightc
they are strongerd
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Question 2a what is meant by a short column?
a column which looks shorta
a fat columnb
a column which will fail in true compressionc
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Question 2b what is meant by a long column?
a column which looks longa
a thin columnb
a column which buckles before its fullcompressive strength is reached
c
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Question 3 what affects the buckling load?
the maximum allowable compressive strengtha
the slenderness ratiob
the moment of inertia, I c
the modulus of elasticity, E d
the end fixing conditionse
b, c and df
b, d and eg
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Question 4a what are good sections for columns?
I-sectionsa
sections with similar radii of gyration
in all directions b
rectangular sectionsc
sections in which the major part of the material isas far from the Centre of Gravity as possible
d
b and de
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Question 4b
why are these good sections for columns?
so that they do not buckle in the weak directiona
they are cheaper b
they use the material more efficientlyc
a and cd
a, b and ce
similar radii of gyration in all directionsand material far away from the Centre of Gravity
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Question 5 what are two effects which can cause a pier
to overturn?
bad constructiona
a horizontal loadb
an eccentric vertical loadc
b and cd
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Question 6a does the middle third rule deal with?
the force acting on the pier being inthe middle third of the pier
a
the resultant reaction being inthe middle third of the base
b
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Question 6b if the middle third rule holds,
what does that tell you?
the pier will not overturn a
no tension will develop in the base of the pierb
the pier will not lift off its basec
b and cd
a and be
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Question 6c if the middle third rule holds, what does
that tell you about the safety factor?
the safety factor is > 3 a
the safety factor is 3b
the safety factor is 2c
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Question 7a
what is the overturning moment?
1.6 kNma
1.0 kNmb
1.3 kNmc
Pier600 x600mm
8kN 1kN
300 1000
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
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Question 7b
what is the stabilizing moment?
2.4 kNma
4.8 kNmb
8 kNmc
Pier600 x600mm
8kN 1kN
300 1000
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
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Question 7c
will the pier overturn andwhy or why not?
yes, the weight of the gate is eccentric a
no, the weight of the pier is greaterthan the weight of the gate
b
yes, the overturning moment is greater thanthe stabilizing moment
c
Pier600 x600mm
8kN 1kN
300 1000
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
no, the stabilizing moment is greater thanthe overturning moment
d
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Question 7d
what is the margin of safety?
2 : 1a
2.4 : 1b
3 : 1c
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
Pier600 x600mm
8kN 1kN
300 1000
> 3 : 1d
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Question 7e
what is the value of thevertical reaction, R?
8 kNa
9 kNb
2.4 kNc
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
Pier600 x600mm
8kN 1kN
300 1000
X
R
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Question 7f
what distance is the verticalreaction from X? h =?
144.4 mma
155.6 mmb
377.8 mmc
Pier600 x600mm
8kN 1kN
300 1000
X
h
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
R = 9 kN
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Question 7g
is the reaction in the middlethird of the base?
yesa
nob
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
a heavy steel gate is hung from a hollow brick pier as shown in the diagram
R = 9 kN144.4 mm
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Question 7h
is this what we would expectfrom our previous observations?
you can’t tella
nob
yesc
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The reaction isoutside the middle third.
R = 9 kN144.4 mm
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Question 7i
what is the stress distributionunder the pier?
(0.020 +/- 0.027) MPaa
(0.020 +/- 0.036) MPab
(0.025 +/- 0.036) MPac
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The reaction isoutside the middle third.
stress = P/A ±Pe/Z
Z = bd2/6
R = 9 kN144.4 mm
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Question 7j
is this what you would expect?
yesa
nob
not possible to tellc
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The pier isdeveloping tension on the left-hand side.
R = 9 kN144.4 mm
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Question 7k
will it be safer?
not possible to knowa
nob
yesc
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
a heavy steel gate is hung from a hollow brick pier as shown in the diagram. The pier isdeveloping tension on the left-hand side.
If you make the pier solid, i.e. it will be abouttwice as heavy
R = 9 kN144.4 mm
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Question 8a
what‘s the weight of the wall(for 1 m length)?
5.24 kNa
52.4 kNb
52.4 kPac
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
1200
mm
selfweight
wind0.5kPa
Rx
A
115115
h
600
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Question 8b
what‘s the total wind force(for 1 m length)?
0.5 kNa
0.6 kNb
0.6 kPac
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
1200
mm
selfweight
wind0.5kPa
Rx
A
115115
h
600
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Question 8c
what‘s the value of the reaction(for 1 m length)?
5.30 kNa
5.24 kNb
5.74 kNc
1200
mm
wind0.5kPa
Rx
A
115115
h
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
W = 5.24 kN
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Question 8d
what‘s the distance, h, of the reactionfrom the centre of the pier?
68.7 mma
46.3 mmb
183.8 mmc
1200
mm
wind0.5kPa
R = 5.24 kNx
A
115115
h
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
W = 5.24 kN
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h
Question 8e
is the reaction within the base?
maybea
yesb
noc
1200
mm
W = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
1200
mm
x
A
115115
h = 46.4 mm
600
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h
Question 8f
what does that mean?
tension develops on the RHSa
the pier will not overturnb
no tension develops on the RHSc
1200
mm
W = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
The reaction is within the base
1200
mm
x
A
115115
h = 46.4 mm
600
the pier will overturnd
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h
Question 8g
is the reaction within the middle third of the base?
noa
maybeb
yesc
1200
mm
W = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
1200
mm
x
A
115115
h = 46.4 mm
600
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h
Question 8h
what does that mean?
tension develops on the RHSa
the pier lifts off its baseb
a and bc
1200
mm
W = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
The reaction is not in the middle third
1200
mm
x
A
115115
h = 46.4 mm
600
the pier overturnsd
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h
Question 8i
how wide would the footing have to be for the reaction to fall within
the middle third of the base?
400 mma
450 mmb
410 mmc
1200
mm
W = 5.24 kN
wind0.6 kN
x
A
115115
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
1200
mm
x
A
115115
h = 46.4 mm
600
R = 5.24 kN
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h
Question 8j
how else could we increasethe stability of the wall?
no ideaa
make the wall thickerb
show mec
1200
mm
W = 5.24 kN
wind0.6 kN
x
A
115115
600
A freestanding garden brick wall is 230 mm thick and
1200 mm high. The density of brick is 19 kN/m3. Thewind load is 0.5 kPa
1200
mm
x
A
115115
h = 46.4 mm
600
R = 5.24 kN
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next question
enough !
tension structures are more efficient
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let me try again
let me out of here
what’s the problem with compression structures?
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next question
enough !
Yep! No buckling! Can use all its strength !!
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let me try again
they may be made of steel but then againthey may be made of other materials
let me out of here
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let me try again
they do pull straight but so what?
let me out of here
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let me try again
some materials are stronger in compressionsome materials are equally strong in compression and tension
so what is it about compression that may lead to early failure?
let me out of here
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next question
enough !
it does probably looks short and fat butthe technical answer is thatit fails in true compression
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but it’s not really the answer
let me try again
let me out of here
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next question
enough !
it does probably looks long and thin butthe technical answer is that
it fails in buckling before true compression
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but it’s not really the answer
let me try again
let me out of here
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the ease of buckling is a function of the slenderness ratio, the modulus of elasticity and the end restraints
the slenderness ratio takes into account the effective length and the radius of gyration which takesinto account the Moment of Inertia. The greater the slenderness ratio, the more likely buckling will occur.
Materials with a higher Modulus of Elasticity, E, will less likely buckle.The more restrained the ends, the less likely to buckle.
next question
enough !
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buckling is not a function of the maximum allowable compressive stress. Buckling doesn’t allow the element to reach its maximum allowable stress
let me try again
let me out of here
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but it is affected by other things too
let me try again
let me out of here
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the Moment of Inertia figures in it but as part of something else
let me try again
let me out of here
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next question
enough !
you’ve got it !!
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think again – why are I-sections or rectangular sections not good for columns
let me try again
let me out of here
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let me try again
let me out of here
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next question
enough !
with equal radii of gyration, columns will not buckle in a weak direction. Remember the studs and noggings.
With more material away from the Centre of Gravity, the same Radius of Gyration can be achieved with less material
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let me try again
let me out of here
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Where did cheaper come from?
let me try again
let me out of here
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next question
enough !
both an eccentric load or a horizontal load cause an overturning moment on a pier
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while bad construction may contribute it’s not the cause
let me try again
let me out of here
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there is something else which has the same effect
let me try again
let me out of here
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next question
enough !
the middle third deals with the reaction
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the middle third rule does not deal with the forces acting THINK !!
If there were several forces acting which one would you look at?
let me try again
let me out of here
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next question
enough !
since no tension tends to develop, the pier will not lift off its base
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the middle third rule doesn’t deal withthe overturning of the pier
let me try again
let me out of here
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let me try again
let me out of here
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next question
enough !
yes, if the middle third rule holds, the safety factor will be > 3
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let me try again
let me out of here
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next question
enough !
Taking moments about X1 (kN) x 1(m) = 1 kNm
Pier600 x600mm
8kN 1kN
300 1000
x
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taking moments about XWhat force is causing overturning?
What is the distance of this force to X?
Pier600 x600mm
8kN 1kN
300 1000
x
let me try again
let me out of here
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next question
enough !
taking moments about X8 (kN) x 0.3 (m) = 2.4 kNm
Pier600 x600mm
8kN 1kN
300 1000
x
the force contributing to the stabilizingmoment is the weight of the pier
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taking moments about XWhat force is resisting overturning?
What is the distance of this force to X?
Pier600 x600mm
8kN 1kN
300 1000
x
let me try again
let me out of here
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next question
enough !
the overturning moment, OTM, is 1 kNmthe stabilizing moment, SM, is 2.4 kNm
so SM > OTM so the pier will not overturn
Pier600 x600mm
8kN 1kN
300 1000
x
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a weight is a force. Forces DO NOT overturn
Moments overturn
let me try again
let me out of here
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WHAT?????the overturning moment is 1kNmthe stabilizing moment is 2.4 kNm
so which is bigger?
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next question
enough !
if the stabilizing moment is 2.4 kNm andthe restraining moment is 1 kNm
then the OTM can be 2.4 times greater before the pier will overturn
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how did you arrive at that?
what are the overturning moment and the stabilizing moment?By how much could the OTM increase before the pier would overturn?
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enough !
the total downward forces = 8 + 1 = 9 kNso the reaction must be 9kN
Pier600 x600mm
8kN 1kN
300 1000
x
R = 9 kN
![Page 74: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/74.jpg)
what is the sum ofthe downward forces?
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Pier600 x600mm
8kN 1kN
300 1000
x
![Page 75: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/75.jpg)
Taking moments about X clockwise moments = anti-clockwise moments
1 x 1000 + 9 x h = 8 x 300 9h = 1400h = 155.6
next question
enough !
Pier600 x600mm
8kN 1kN
300 1000
x
R = 9 kNh = 155.6 mm
![Page 76: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/76.jpg)
try again
take moments about X.what are the clockwise moments?
what is the anti-clockwise moment ?
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![Page 77: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/77.jpg)
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enough !
the reaction is 300 – 155.6 = 144.4 mm from the centre
the middle third is 100 mm from the centreso the reaction is outside the middle third
Pier600 x600mm
8kN 1kN
300 1000
x
R = 9 kNh = 155.6 mm
![Page 78: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/78.jpg)
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How far is the reaction from the centre?How wide is the middle third?
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enough !
If the middle third rule holds, the factor of safety would be > 3since the factor of safety is only 2.4 we would expect the reaction
to be just outside the middle third
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THINK !!
of course you can !what would the factor of safety be if the middle third rule holds?
what is the actual factor of safety?
![Page 81: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/81.jpg)
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THINK !!
what would the factor of safety be if the middle third rule holds?what is the actual factor of safety?
![Page 82: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/82.jpg)
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enough !
the compressive stress due to the weightP/A = 9000 / (600 x 600) = 0.025 MPa
The stress due to the eccentricity
Pe/Z = 1 x 1300 / (600 x 6002 /6) = 0.036 MPa
Total stress = 0.025 +/- 0.036LHS = 0.025 + 0.036 = 0.061 MPa (comp)
RHS = 0.025 – 0.036 = - 0.011 MPa (tension)
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
stress = P/A ±Pe/Z
Z = bd2/6
R = 9 kN144.4 mm
![Page 83: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/83.jpg)
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stress = P/A +/- Pe/Zcompression +/- compression / tension
what is P (total downward load)? what is A (area of the pier)?
what is P (the eccentric force)?what is e (the eccentricity)?
what is Z (the Section Modulus of the pier)?
Pier600 x600mm
8kN 1kN
300 1000
X
h = 155.6 mm
stress = P/A ±Pe/Z
Z = bd2/6
R = 9 kN144.4 mm
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enough !
since the reaction is outside the middle third, we would expect tension to develop
on the side away from the eccentric force.
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is the reaction in the middle third?so what does that mean?
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![Page 86: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/86.jpg)
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THINK !!
of course it is possible to tellis the reaction in or outside the middle third?
SO?
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enough !
since it is the weight of the pierwhich produces the stabilizing moment,
then it stands to reason thatanything that increases the weightwill increase the stabilizing moment
and hence make the pier safer
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THINK !!
of course it is possible to tellwill increasing the weight of the pier increase the stabilizing moment?
SO?
![Page 89: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/89.jpg)
THINK!!
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The greater the stabilizing moment the less likelythe pier is to overturn or lift off its base.
What contributes to the stabilizing moment?
![Page 90: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/90.jpg)
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enough !
for 1 m length of wall the volume, V of the wall is:
V = 1 x 1.2 x 0.23 m3
V = 0.276 m3
the weight of the wall, W, is volume x densityW = 0.276 x 19 kN
W = 5.24 KN
1200
mm
selfWeight = 5.24 kN
wind0.5kPa
Rx
A
115115
h
600
![Page 91: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/91.jpg)
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there must be an error in your calculations
![Page 92: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/92.jpg)
a weight is a force
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what are the units of force?
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enough !
for 1 metre length of wall, the area, A, of the wall is
A = 1 x 1.2 m2
A = 1.2 m2
for a wind load of 0.5 kPa, the total wind load, WL, isWL = 1.2 x 0.5 kN
WL = 0.6 kN
1200
mm
selfWeight = 5.24 kN
wind0.6 kN
Rx
A
115115
h
600
![Page 94: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/94.jpg)
let me try again
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there must be an error in your calculations
![Page 95: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/95.jpg)
a weight is a force
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what are the units of force?
![Page 96: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/96.jpg)
ΣV = 0
next question
enough !
if the total downward load is 5.24 kNthe total upward load must equal 5.24 kN
1200
mm
selfWeight = 5.24 kN
wind0.5kPa
R = 5.24 kNx
A
115115
h
600
![Page 97: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/97.jpg)
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THINK !!ΣV = 0
what is the total downward load?So what must the total upward load be
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enough !
taking moments about Aclockwise moments = anti-clockwise moments
W * 115 = R * x + 0.6 x 6005.24 * 115 = 5.24 * x + 0.6 x 600
602.6 = 5.24x + 360x = 242.6 / 5.24
x = 46.30
h = 115 – xh = 68.7 mm
1200
mm
selfWeight = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
h = 68.7 mm
600
for a graphic methodof solving for h
CLICK HERE
![Page 99: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/99.jpg)
let me try again
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taking moments about A,what are the clockwise moments? What are the anti-clockwise moments?
remember ΣM = 0(you can also solve this using a graphical method)
![Page 100: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/100.jpg)
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enough !
the base is 230 mm wide. So half the width is 115 mm.
The reaction is 68.7 mm from the centreSo the reaction is within the base.
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there’s no maybe about itit is or it isn’t
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try again
What is the width of the base?What is the half-width of the base?
How far is the reaction from the centre?
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1200
mm
selfWeight = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
h = 68.7 mm
600
![Page 103: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/103.jpg)
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enough !
as the reaction is within the basethe pier will not overturn
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we really don’t know yetwe have to know something first
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isn’t the reaction within the base ?
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![Page 106: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/106.jpg)
The base is 230 mm wide1/3 of that is 76.7 mm
so from the centre of the pier, the middle third is 38.3 mmthe reaction is 68.3 mm from the centre
So the reaction is outside the middle third
next question
enough !
38.3 mm
68.3 mm
230 mm
![Page 107: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/107.jpg)
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there’s no maybe about itit is or it isn’t
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![Page 108: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/108.jpg)
what is the distance of the middle third from the centre of the pier?what is the distance of the reaction from the centre of the pier
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enough !
Yes, if the reaction is outside the middle third thentension tends to develop on the RHS (in this case)
and if the pier does not stick to its base, it will lift off.
![Page 110: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/110.jpg)
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Firstly, the middle third rule doesn’t deal with overturningSecondly, didn’t we say before that the pier won’t overturn?
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![Page 112: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/112.jpg)
for mathematical answer
enough !
This is not the mathematical answer but it is the practical answerSince it is the lowest width of a back-hoe that would do
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enough !
This is the mathematical answer but in practicewe would make it 450 mm wide
Since it is the lowest width of a back-hoe that would do
distance of reaction from centre = 68.3 mmso middle third would have to be 68.3 mm from centreso width of base would have to be 6 x 68.3 = 410 mm
68.3 mm
136.6 mm
410 mm
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Think of how wide the middle third would have to beso that the reaction would be within it
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![Page 115: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/115.jpg)
that’s the end!
zig-zag plan
stepped plan
attached piers
heavy coping on
top
![Page 116: Axial Members](https://reader036.vdocuments.us/reader036/viewer/2022062518/5681430f550346895daf6104/html5/thumbnails/116.jpg)
you’re not trying
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It would be expensive
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back
we draw congruent triangles.one represents the forces andthe other represents distances
1200
mm
selfWeight = 5.24 kN
wind0.6 kN
R = 5.24 kNx
A
115115
h = 68.7 mm
600
5.24 kN
0.6 kN
600 mm
h mm
thenh / 600 = 0.6 / 5.24
h = (0.6 / 5.24) x 600 h = 68. 7 mm
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