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Teaching schedule Week Lecture 1 Staff Lecture 2 Staff Tutorial Staff 1 Beam Shear Stresses 1 A P Beam Shear Stresses 2 A P --- --- 2 Shear centres A P Basic Concepts J E-R Shear Centre A P 3 Principle of Virtual
forces J E-R Indeterminate Structures J E-R Virtual Forces J E-R
4 The Compatibility Method
J E-R Examples J E-R Virtual Forces J E-R
5 Examples J E-R Moment Distribution -Basics
J E-R Comp. Method J E-R
6 The Hardy Cross Method
J E-R Fixed End Moments J E-R Comp. Method J E-R
7 Examples J E-R Non Sway Frames J E-R Mom. Dist J E-R 8 Column Stability 1 A P Sway Frames J E-R Mom. Dist J E-R 9 Column Stability 2 A P Unsymmetric Bending 1 A P Colum Stability A P 10 Unsymmetric Bending 2 A P Complex Stress/Strain A P Unsymmetric
Bending A P
11 Complex Stress/Strain A P Complex Stress/Strain A P Complex Stress/Strain
A P
Christmas Holiday
12 Revision 13 14 Exams 15 2
Mo@va@ons (1/5)
• Load-‐carrying structures may fail in a variety of ways, depending upon: – Type of structure (truss, frame, …) – Condi@ons of support (pinned, fixed, …) – Loads applied (sta@c, dynamic, …) – Materials used (briQle, duc@le, …)
• Failures are prevented by designing structures so that maximum stresses (strength criterion) and maximum displacements (s,ffness criterion) remain within admissible limits
3
Mo@va@ons (2/5)
4
• For the fans of The Big Bang Theory: – Sheldon and Howard have got this seriously wrong!
• You can’t use the Young’s modulus to quan@fy the strength of material, but its s,ffness!
Mo@va@ons (3/5)
• S@ffness and strength of materials – In the stress-‐strain curve for a duc@le material (e.g. steel), the Young’s modulus E defines the s@ffness, while the yield stress σy represents the strength
5
Mo@va@ons (4/5)
• S@ffness criterion: “Slender Column”
6
• Strength criterion: “Short Column”
Mo@va@ons (5/5)
7
Coventry Cathedral
ç Slender column
Detail of the support è
Learning Outcomes
• When we have completed this unit (2 lectures + 1 tutorial), you should be able to:
– Derive the Euler’s cri@cal load for slender pinned-‐pinned columns in compression
– Predict the mode of failure for both short and slender columns in compression
8
Further reading
• R C Hibbeler, “Mechanics of Materials”, 8th Ed, Pren@ce Hall – Chapter 13 on “Buckling of Column”
• T H G Megson, “Structural and Stress Analysis”, 2nd Ed, Elsevier – Chapter 21 on “Structural Instability” (eBook)
9
Short and Slender Struts
10
• Increasing the length of a strut reduces its buckling load – For instance, a matchs,ck is reasonably strong in compression (lek), but a longer s,ck, with the same cross sec@on and the same material, would be weaker and buckles in compression (right)
– The slenderness of a strut plays an important role in its mode of failure in compression
Buckling, i.e. Lateral Instability (1/2)
11
• That is, if a column is rela@vely slender, it may deflect laterally when subjected to a compressive force P (Fig (a)) and fail by bending (Fig (b)), rather than failing by direct compression of the material
Buckling, i.e. Lateral Instability (2/2)
12
• Pcrit is the so-‐called cri,cal buckling load – If the axial load P is less than Pcrit, bending is caused by lateral loads only – If P is greater than Pcrit, the ruler bends even without lateral loads
Euler’s Cri@cal Load for Pinned-‐Pinned Slender Columns
• One of the Learning Outcomes of this Unit is for you to become able to mathema@cally derive (and remember as well) the expression of the cri@cal load Pcrit for pinned-‐pinned slender column
• Pcrit=PE is oken called Euler’s buckling load – Aker the Swiss mathema@cian Leonhard Euler (1707-‐1783)
!!Pcrit =
π 2EImin
L2
13
Mathema@cal Deriva@on: Bending Equa@on
• What’s the equa@on ruling the beam’s downward deflec@on, uz(x), for a given bending moment diagram, My(x)?
• We used this second-‐order differen@al equa@on in part A to calculate the beam’s deflec@on under transverse loads…
• where, as usual: – E= Young’s modulus – Iyy= Second moment of area about the horizontal neutral axis
14
EIyyd2uz (x)dx2
= −My (x)
Mathema@cal Deriva@on: Sign Conven@on
• Do you remember from last year?
15
Mathema@cal Deriva@on: P-‐Delta (1/2)
• What’s the bending moment My in this circumstance?
• We don’t have transverse loads this @me – the column is subjected to the axial load P only
• How can we have a bending moment?
• In order to derive the expression of the Euler’s buckling load, we need to assume that – a disturbance/imperfec@on exists in the column, – therefore the buckling occurs – and My can be consistently evaluated by using the equilibrium
equa@ons in the deformed shape 16
Mathema@cal Deriva@on: P-‐Delta (1/2)
17
Deformed shape
Equilibrium condi@on P
P My= P uz
EIyy
z
uz
Mathema@cal Deriva@on: Buckling Equa@on (1/2)
• Knowing the bending moment My in the deformed shape:
• we can subs@tute it within the deflec@on equa@on:
• This equa@on can be rewriQen as:
• Where α is a posi@ve quan@ty, given by:
18
My = Puz
EIyyd2uzdx2
= −Puz
d2uzdx2
+α 2uz =0
α = PEIyy
Mathema@cal Deriva@on: Buckling Equa@on (2/2)
• What do we do in order to solve an ordinary differen@al equa@on?
• First, we find the general solu,on, which contains as many integra@on constants as the order of the differen@al equa@on (two, in this case)
19
uz= C
1cos(αx) + C
2sin(αx)
Mathema@cal Deriva@on: Boundary Condi@ons (1/2)
• Second, we apply the boundary condi,ons (BCs) to get the values of the integra@ons constants for the par@cular case
– For two unknown constants, C1 and C2, two BCs are needed!
• For a pinned-‐pinned column, the BCs read:
• uz=0 @ x=0 (i.e. the transverse transla@on is prevented at the lek-‐hand side end)
• uz=0 @ x=L (i.e. the transverse transla@on is prevented at the right-‐hand side end as well)
20
EIyy
z
Mathema@cal Deriva@on: Boundary Condi@ons (2/2)
21
• The applica@on of the first BC is quite straighrorward
uz(x) = C
1cos(αx) + C
2sin(αx)
uz= 0@ x = 0 1 20 1 0C C⇒ = × + ×
1 0C⇒ =
General solu5on
Boundary condi5on
Mathema@cal Deriva@on: Non-‐Trivial Solu@on (1/3)
22
• The second BC does require more effort
• Trivial solu@on: – It would follow y=0 for any value of the abscissa x – No transverse displacements would occur (straight column) – This solu@on is therefore unacceptable
• Non-‐trivial solu@on:
uz(x) = C
1cos(αx) + C
2sin(αx)
uz= 0@ x = L 20 sin( )C Lα⇒ =
sin( ) 0Lα⇒ = L nα π⇒ =
nnLπα α⇒ = =
2 0C⇒ =
Mathema@cal Deriva@on: Non-‐Trivial Solu@on (2/3)
23
• Recalling now the expression of the parameter a, one obtains:
• The associated modes of instability, for n= 1, 2, 3, …, are sinusoidal func@ons, having a total number n of peaks and valleys
PEIyy
= nπL
⇒ PEIyy
= n2 π 2
L2⇒ P
n= n2
π 2 EIyy
L2
n= 3
n= 2
n= 1
Mathema@cal Deriva@on: Non-‐Trivial Solu@on (3/3)
24
• Larger values of the buckling load are associated to more complicated modes of instability
• Theore@cally, these modes could be achieved if roller supports are applied at the points of contraflexure
• However, in prac@ce, the lower value P1 is never exceeded
Mathema@cal Deriva@on: Euler’s Buckling Load
• The actual cri,cal load, i.e. the so-‐called Euler’s buckling load, is the “engineering solu@on”, which is the minimum among the mathema@cal solu@ons P1, P2, P3, …, and is obtained for n=1
• Moreover, in order to be truly the minimum, you must use the minimum value of the second moment of area, which might not be Iyy
• The laQer expression is very important in Structural Engineering – You are requested to remember it – You must be able to derive this expression as well
25
Pcrit
= PE= P
1=π 2 EI
yy
L2
2min
E 2EIPL
π=
FEM-‐Computed Modes of Instability
26
• Euler’s buckling load (PE= P1= 181 kN)
• Higher buckling load in the orthogonal direc@on
• (P4= 3,518 kN)
XY X
Z
Y
ZXY X
Z
Y
Z
4min 330 cmI = = 4
max 6572 cmI
Horizontal sway
Ver0cal deflec0on
Effects of the Boundary Condi@ons (1/2)
27
The more the column’s ends are restrained, the higher is the buckling load
Similar sinusoidal shapes are
observed for different BCs
Effects of the Boundary Condi@ons (2/2)
28
(a) Pinned-‐pinned (b) Can@levered (c) Fixed-‐fixed (d) Propped
L0 is the distance between two consecu@ve crosses of the horizontal axis
Effec@ve Length • It is useful to introduce the concept of equivalent length, Le=k L as
the length of a pinned-‐pinned column having the same Euler’s cri@cal load
• We therefore must know the value of the coefficient k for different BCs
29
k=2
k=1 k=0.7 k=0.5
Can5levered Pinned-‐ pinned Propped
Fixed-‐ fixed
2min
E 2e
EIPL
π=
Stocky Columns (1/2) • If we divide the Euler’s cri@cal load PE by the cross sec@onal area A, we get
the so-‐called Euler’s cri@cal stress σcrit:
• This is the maximum normal stress which is allowable to prevent buckling instability, and is inversely propor@onal to the square of the equivalent length Le
• If we introduce the parameter ρmin as the minimum radius of gyra@on of the cross sec@on, and then the slenderness ra@o λ=Le/ρmin, the above equa@on can be rewriQen as:
30 σcrit
=π 2 E I
minA( )
Le2
=π 2 E ρ
min2
Le2
= π 2 E
Le
ρmin( )2
= π 2
λ2Eρ
min=Imin
A
2min
crit 2e
EP EIA AL
πσ = =
Stocky Columns (2/2)
31
• The s@ffer the material, i.e. the larger the Young’s modulus E, the higher is σcrit:
• The larger the slenderness ra@o λ, the lower is σcrit, i.e. very slender columns will have very low values of σcrit
• Conversely, stocky columns, with a small slenderness ra@o λ, will not experience the buckling failure, as the yielding of the material is likely to happen first:
2
crit 2 Eπσλ
=
crit y Material’s yield stressfσ > =
Strength and S@ffness Criteria (1/2)
• “Strength” criterion
• Stocky columns tend to fail because the elas@c limit of the material is reached
• The safety checks is:
• “S,ffness” criterion
• Slender columns tend to fail because the elas@c configura@on is unstable
• The safety check is:
32
2min
E 2e
EIP PL
π< =y yP P f A< =
Both must be sa0sfied
Strength and S@ffness Criteria (2/2)
• For briQle materials such as concrete, the yielding stress fy is replaced with the crushing stress fc
• The safety check then reads:
33
2min
E 2e
EIP PL
π< =P < Pc= f
cA
Both must be sa0sfied
fc
Strength and S@ffness Criteria
34
• The Rankine’s failure load PR combines these two different criteria, therefore taking into account both material and geometrical nonlineari@es
• PR=Py for λ=0
• PR approaches PE as λ goes to infinity
0 100 200 300 4000.0
0.5
1.0
1.5
2.0
l
PêPy Py
PE PR
PR=PyPE
Py+ P
Eλ
P/P y
Rankine (1820-‐1872) was a Scoush civil engineer, physicist and mathema@cian
Ul@mate Normal Stress
35
• … experimentally derived (dots) for wide-‐flange steel columns
• … as a func@on of the slenderness λ= k L/ρmin
λ
Key Learning Points 1. Columns in compression may fail because
– Insufficient bending s@ffness: è Lateral buckling – Insufficient axial capacity: è Yielding/Crushing
2. Euler’s buckling load PE depends on: – Minimum second moment of area, Imin
– Length of the column, L – Boundary condi@ons
3. Interac@on between lateral buckling and axial capacity can be taken into account through the (approximate) Rankine’s formula
36
è Effec@ve length, Le
PR=PyPE
Py+ P
E
PE =π 2EImin
Le2