Axial Members

WORKSHEET 11

to answer just click on the button or image related to the answer

Question 1a

are tension structures or compression structuresmore efficient when loaded axially?

compression structuresa

tension structures b

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

next question

enough !

tension structures are more efficient

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what’s the problem with compression structures?

next question

enough !

Yep! No buckling! Can use all its strength !!

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they may be made of steel but then againthey may be made of other materials

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they do pull straight but so what?

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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?

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next question

enough !

it does probably looks short and fat butthe technical answer is thatit fails in true compression

but it’s not really the answer

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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

but it’s not really the answer

let me try again

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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 !

buckling is not a function of the maximum allowable compressive stress. Buckling doesn’t allow the element to reach its maximum allowable stress

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but it is affected by other things too

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the Moment of Inertia figures in it but as part of something else

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next question

enough !

you’ve got it !!

think again – why are I-sections or rectangular sections not good for columns

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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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Where did cheaper come from?

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next question

enough !

both an eccentric load or a horizontal load cause an overturning moment on a pier

while bad construction may contribute it’s not the cause

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there is something else which has the same effect

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next question

enough !

the middle third deals with the reaction

the middle third rule does not deal with the forces acting THINK !!

If there were several forces acting which one would you look at?

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next question

enough !

since no tension tends to develop, the pier will not lift off its base

the middle third rule doesn’t deal withthe overturning of the pier

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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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next question

enough !

Taking moments about X1 (kN) x 1(m) = 1 kNm

Pier600 x600mm

8kN 1kN

300 1000

x

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

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

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

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

a weight is a force. Forces DO NOT overturn

Moments overturn

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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

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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next question

enough !

the total downward forces = 8 + 1 = 9 kNso the reaction must be 9kN

Pier600 x600mm

8kN 1kN

300 1000

x

R = 9 kN

what is the sum ofthe downward forces?

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Pier600 x600mm

8kN 1kN

300 1000

x

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

try again

take moments about X.what are the clockwise moments?

what is the anti-clockwise moment ?

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next question

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

try again

How far is the reaction from the centre?How wide is the middle third?

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next question

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?

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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?

next question

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

let me try again

let me out of here

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

next question

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?

let me try again

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let me try again

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THINK !!

of course it is possible to tellis the reaction in or outside the middle third?

SO?

next question

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?

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?

next question

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

let me try again

let me out of here

there must be an error in your calculations

a weight is a force

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what are the units of force?

next question

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

let me try again

let me out of here

there must be an error in your calculations

a weight is a force

let me try again

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what are the units of force?

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

let me try again

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THINK !!ΣV = 0

what is the total downward load?So what must the total upward load be

next question

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

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)

next question

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.

let me try again

there’s no maybe about itit is or it isn’t

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What is the width of the base?What is the half-width of the base?

How far is the reaction from the centre?

let me try again

let me out of here

1200

mm

selfWeight = 5.24 kN

wind0.6 kN

R = 5.24 kNx

A

115115

h = 68.7 mm

600

next question

enough !

as the reaction is within the basethe pier will not overturn

we really don’t know yetwe have to know something first

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let me try again

isn’t the reaction within the base ?

let me out of here

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

let me try again

there’s no maybe about itit is or it isn’t

let me out of here

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

let me try again

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next question

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.

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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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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

next question

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

Think of how wide the middle third would have to beso that the reaction would be within it

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that’s the end!

zig-zag plan

stepped plan

attached piers

heavy coping on

top

you’re not trying

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let me try again

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It would be expensive

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