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 = (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)
8kN
1kN
a strong concrete footing.
take moments about the point X
Pier
(i) will the pier overturn?
No
600 x
Overturning Moment (clockwise):
600mm
M =1 x 1
= 1 kNm
X
R=9
Potential Stabilizing Moment (anticlockwise):
h
M =8 x 0.3
= 2.4 kNm
(ii) what is the margin of safety? 2.4:1
stress = P/A ±Pe/Z
OTM can increase up to 2.4kNm before
overturning occurs
Z = bd2/6
Q9 (b)
(b) where is the resultant of the two loads?
1 x 1000 + R x h = 8 x 300
155.6mm from X
144.4mm from centre
9 x h = 2400 -1000
h = 1400 / 9
= 155.6
(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)
= 1300kNmm
Z = bd2/6 = 600 x 6002 / 6
= 36 x 106 mm3
stress = 9000 / (600 x 600) ± 1300000 / 36 x 106
= 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 (61kPa) - compression
on the other side it would be -0.011 MPa (11kPa) (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
self
weight
(a) Find the location of the reaction on the base
wind load on 1m length of wall = 0.5 x 1 x 1.2
= 0.6 kN
taking moments about A:
= 360 kNmm
overturning moment = 0.6 x 600
restraining moment = 5.244 x 115 = 603.1 kNmm
5.244 x h + 360 = 603.1
h = 243.1/5.244
= 46.4 mm
(i) is it within the base?
(i) is it within the middle third?
yes
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.
115
wind
0.5kPa
1200mm
weight of wall (of length 1m) = 1.2 x 0.23 x 1 x 19 = 5.244 kN
115
600
A
R = weight
h x = 115-h
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
self
weight
(a) Find the location of the reaction on the base
wind load on 1m length of wall = 0.5 x 1 x 1.2
= 0.6kN
GRAPHICAL METHOD
X/600 = 0.6 / 5.244
5.244kN
600
X
X
0.6kN
= 0.6 x 600 / 5.244
= 68.6
x
the reaction is 68.6mm from the centre of the wall.
115
wind
0.5kPa
1200mm
weight of wall (of length 1m) = 1.2 x 0.23 x 1 x 19 = 5.244kN
115
A
R = weight
h x
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 =
= 412mm
We would probably make it 450mm wide.
This is the width of a common backhoe bucket.
(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