WORKSHEET 7 AXIALLY LOADED MEMBERS

Q1

Are tension structures or compression structures more efficient when loaded axially ?

tension structures

Q2

why ?

tension structures are not subject to buckling they pull straight - can use thin cables can use: stress = Force / Area

Q3

What is meant by: (a) a short column a column which will fail in true compression (b) a long column a column which buckles before full compressive strength is reached

Q4

What affects the buckling load (or buckling stress)?

(a) the slenderness ratio l/r (or l/B rect. sections) the slenderness ratio takes into account the effective length, l and the stiffness ( radius of gyration ) of the X-section r =

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(I/A) buckling load inversely proportional to slenderness ratio i.e. greater the slenderness ratio - more will tend to buckle (b) the modulus of elasticity , E materials with higher E will buckle less (c) end fixing conditions restraints but included in slenderness ratio - effective length free end Eff Length = 2 x l, fixed ends Eff Length = 0.5 x l

Q5

What are good sections for columns?

(i) sections which have similar radii of gyration in all directions (ii) sections in which the major part of the material is a far from the Centre of Gravity as possible

Q6

Why?

(i) so that they do not buckle in a weak directions (ii) to use the material efficiently

Q7

What are two effects which can cause a pier to overturn?

(a) a horizontal load (b) an eccentric vertical load

Q8

What is the middle-third rule?

the middle-third rule tells you that as long as the resultant reaction falls in the middle-third of the base of the pier then no tension will develop in the pier.

if the middle-third rule holds the pier will not lift off its base there will be a factor of safety of >3 if the middle-third rule holds

Q9 (a)

The diagram shows a heavy steel gate hung from a hollow brick pier which weighs 8kN. Investigate the stability of the pier.

300 1000

(a) assume the pier is sitting on (but not stuck to) a strong concrete footing.

take moments about the point X (i) will the pier overturn?

No

Overturning Moment (clockwise): Pier 600 x 600mm M =1 x 1

= 1 kNm

Potential Stabilizing Moment (anticlockwise): M =8 x 0.3

= 2.4 kNm

8kN

R=9 X

h

(ii) what is the margin of safety?

2.4:1

OTM can increase up to 2.4kNm before overturning occurs Z = bd 1kN stress = P/A ±Pe/Z 2 /6

Q9 (b)

(b) where is the resultant of the two loads?

1 x 1000 + R x h = 8 x 300 9 x h = 2400 -1000 h = 1400 / 9 = 155.6

155.6mm from X 144.4mm from centre (i) is it within the middle third of the base?

No (ii) is this what you would expect from (a)?

In (a) we assumed that there would be no crushing of the leading edge of the pier. Under this condition the middle-third rule gives a factor of safety of >3.

For a factor of safety of 2.4 we would expect the reaction to be just outside the middle-third

Q9 (c)

(c) calculate the stress distribution under the pier

stress = P/A (compressive part) ± M/Z (bending part) M = P x e = 1 x 1300 (or 9 x 144.4) Z = bd 2 /6 = 600 x 600 2 / 6 stress = 9000 / (600 x 600) ± 1300000 / 36 x 10 6

= 1300kNmm = 36 x 10 6 mm 3 = 0.025 ± 0.036 MPa (i) is it trying to develop tension on one side?

yes

on the gate side the stress is 0.061 MPa on the other side it would be -0.011 MPa (61kPa) (11kPa) compression (tension) since the interface cannot develop tension, a redistribution would occur a different analysis is required to find that the maximum compressive stress would increase to 0.096 MPa

(ii) is this what you would expect from (b)?

yes

since the reaction is outside the middle third you would expect tension to tend to develop

Q9 (d)

(d) if you make the pier solid, it will be about twice as heavy.

Will this make it safer against overturning?

yes

The potential stabilising moment would double while the overturning moment remains the same.

No tension would develop (the reaction would be within the middle third) and it would be safer against overturning

Q10(a)

A freestanding garden wall is 230mm thick and 1200mm high.

The density of brick is 19kN/m3. The wind load in this location is 0.5 kPa (a) Find the location of the reaction on the base

self weight 115 115 weight of wall (of length 1m) = 1.2 x 0.23 x 1 x 19

= 5.244 kN

wind load on 1m length of wall = 0.5 x 1 x 1.2

taking moments about A: overturning moment = 0.6 x 600

= 0.6 kN = 360 kNmm

restraining moment = 5.244 x 115 5.244 x h + 360 = 603.1

h = 243.1/5.244

(i) is it within the base?

= 603.1 kNmm = 46.4 mm yes A

wind 0.5kPa

600 h

R = weight

x = 115-h

(i) is it within the middle third?

no

the reaction is 115 - 46.4 = 68.6mm from the centre of the wall.

the middle third is 38.3mm from the centre of the wall.

so the reaction is well outside the middle third.

Q10(a)

A freestanding garden wall is 230mm thick and 1200mm high.

The density of brick is 19kN/m3. The wind load in this location is 0.5 kPa (a) Find the location of the reaction on the base

self weight 115 115 weight of wall (of length 1m) = 1.2 x 0.23 x 1 x 19 wind load on 1m length of wall = 0.5 x 1 x 1.2

= 5.244kN

= 0.6kN

wind 0.5kPa

GRAPHICAL METHOD X/600 = 0.6 / 5.244

A

5.244kN

600 X = 0.6 x 600 / 5.244

h

R = weight

x X = 68.6

0.6kN

x the reaction is 68.6mm from the centre of the wall.

Q10(b, c)

(b) How wide would the footing have to be to bring the reaction within the middle third?

Forgetting the weight and depth of the footing, width of footing would have to be 6 x 68.6 = We would probably make it 450mm wide.

This is the width of a common backhoe bucket.

= 412mm (c) What other options are there for increasing the stability of the wall?

(i) put a heavy coping on top (has to be wide rather than high otherwise subject to wind) (ii) make the wall thicker (expensive) (iii) attached piers (iv) zigzag plan (similar to (iii) but more interesting

5.44kN

2 x 68.6

68.6

6 x 68.6