Chapter 7
Bending of thin plates
1
Contents
0. Fundamentals of the theory of thin plates
1. Pure bending of thin plates
2. Thin plates under bending and twisting
3. Thin plates subjected to distributed transverse load:
Derivation of the governing differential equation for plates
4. Navier solution to simply supported rectangular plates
5. Combined bending and in-plane loading
6. Bending of thin plates with initial imperfection
7. Energy method
2
7. Energy method for the bending of thin plates
Two types of solution are obtainable for thin plate bending problems by the
application of the principle of the stationary value of the total potential energy(TPE) of
the plate and its external loading.
1.the form of the deflected shape of the plate is known, produces an exact solution;
2.the Rayleigh-Ritz method, assumes an approximate deflected shape in the form of
a series having a finite number of terms chosen to satisfy the boundary conditions of
the problem and also to give the kind of deflection pattern expected.
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The principle of the stationary value of the total potential energy (5.8)
y
TPE  U  V   Pdy  Py
Y
0
V-potential energy of external forces
V = - Py
U-strain energy
U = ky2/2 = Py/2
Equilibrium condition:
P
Fig. 5.23 (a) Potential energy of a spring–mass system; (b)
loss in potential energy due to change in position.
 U  V   0
Three types of equilibrium:
A: unstable  2(U+V) < 0
B: stable  2(U+V) > 0
C: neutral  2(U+V) = 0
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(1) Strain energy produced by bending and twisting - 1
Bending
1
 M x y  x x 
2
Twisting
1
M y x  y y 

2
See Px in Hw for another
approach of derivation
Total strain energy:
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1
M xy y   xy x 

2
1
M xy x   xy y 

2
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(1) Strain energy produced by bending and twisting - 2
Total strain energy:
For a plate of a×b:
For pure bending problem
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(2) Potential energy of a transverse load
Consider the potential energy of a “dead weight” subjected to gravity
 V   wq x y
V  
a
0
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
b
0
wqdxdy
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(3) Potential energy of in-plane loads —— Nx, Ny
Nx, Ny are assumed to be compressive
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(4) Potential energy of in-plane loads —— Nxy
Shear strain:
w w
x y
Work by Nxyδx:
1
1
w w
N xy xC1 D  N xy x y
2
2
x y
Work by Nxyδy:
1
w w
N xy x y
2
x y
Work over the complete plate:
Strain energy of Nxy:
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From Timoshenko’s Theory of
Plates and Shells
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(4) Potential energy of in-plane loading system
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(5) Solving w by using total potential energy - 1
Total potential energy (q0 only)
Double Fourier’s series expansion
TPE in terms of Amn:
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(5) Solving w by using total potential energy - 2
This is a huge leap. The number of degrees of freedom is “reduced” from the
cardinality (number of elements) of ℝ to that of ℕ.
Solving Amn and w
The result is the same with Navier solution!
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(6) Rayleigh-Ritz Method —— ‘approximate’ solution
Rayleigh-Ritz method: the appropriate infinite series is not known so that only an
approximate solution may be obtained.
If the ‘guessed’ series for w in a particular problem contains three different functions
of x and y, for example
w  A1 f1  x, y   A2 f 2  x, y   A3 f 3  x, y 
Substituting for w in the appropriate expression for the total potential energy. By
the principle of stationary value of of TPE, A1, A2 and A3 are solved by:
 U  V 
A1
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 0,
 U  V 
A2
 0,
 U  V 
A3
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0
14
Example 7.4
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Example 7.4
Solution of first four series:
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Summary
• Effect of small initial imperfections
• Principle of stationary value of TPE
• Potential energy for a plate associated with Mx, My, Mxy, Nx, Ny,
Nxy and q
• Exact solution by energy method
• Rayleigh-Ritz method
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Chapter 8
Structural instability: buckling of columns
Outline
•
•
•
•
•
Euler buckling of columns (柱的欧拉屈曲)
Inelastic buckling
Effect of initial imperfections
Stability of beams under transverse and axial loads
Energy method to calculate buckling loads in columns
21
Euler buckling of columns
1.
2.
3.
4.
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Introduction
Theoretical modeling
Calculation of critical load
How to avoid buckling
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1. Introduction
Buckling refers to the phenomenon that an axially compressed bar(column, beam) bends suddenly
when the compressive load reaches a certain critical load.
案例：国产大飞机用复合材料蜂窝夹芯板受压缩发生屈曲破坏
Example: a composite honeycomb sandwich panel subjected to axial compression failed due to buckling.
Buckling often occurs at stress levels far below the
material strength and leads to catastrophic failure.
Major mode of instability of slender bars.
Named after Euler, who first studied this topic
systematically.
This lecture tries to answer:
（1）What is the condition of buckling?
（2）How to avoid it?
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2.Theoretical modeling
Two questions: (1) What is the critical load of buckling?
(2) What shape would the column buckle into?
Consider a bar subjected to axial compressive force P. Its unbuckled configuration is
y
P
P
z
L
Assume that both ends are simply supported, which implies that the deflections of
the ends are zero. This is the most typical boundary condition.
It is observed in experiments that the configuration of the bar becomes the one
shown below when it buckles:
It reminds us to analyze buckling by beam theory!
y
P
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P
z
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2.Theoretical modeling
Solve the beam bending problem
y
P
P
v(z)
z
z
(1) Loads acting on the bar: only compressive force P at the simply supported ends, no other forces/moments.
(2) Focus on the cross section at an arbitrary coordinate z, where the deflection is denoted as v.
Internal loads: denote the internal axial force on the cross section as P and bending moment as M.
y
P
M
P
v
z
z
EI：bending rigidity
Equilibrium of the bar segment to the left of z, leads to M  Pv
E：Young’s modulus
2
d
v
(3) In beam theory
M   EI 2
I：moment of inertial of the cross section
dz
Combining the two equations, we get the differential equation about deflection v:
d 2v
EI 2  Pv  0
dz
or
d 2v
  2v  0
2
dz
where  
P
EI
The condition of buckling can be expressed as: the above equation has non-zero solutions
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2.Theoretical modeling
y
P
P
v(z)
z
General solution
d 2v
2


v0
2
dz
z

v( z )  A cos  z  B sin  z
v：deflection
E: Young’s moudlus
I：moment of inertia
L：length
where the constants A, B are determined by BCs at z = 0 and z = L
Recall the boundary conditions:
Left end
v(0)  0 →
A0
Right end
v( L)  0 →
B sin  L  0
thus
P
EI
v ( z )  B si n  z
B = 0 results in the trivial solution v ≡ 0，which implies that the bar does not buckle!
Buckling means B ≠ 0 and resultantly  must satisfy the characteristic equation:
Solution:
The n-th critical load
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 L  n ,
n 2 2 EI
Pn 
L2
n  1, 2,3,...
sin  L  0
The coefficient B is indeterminant!
The n-th buckling mode
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sin
n z
L
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2.Theoretical modeling
Critical loads
n=1
P1 
Buckling modes
 2 EI
L
n=2
n=3
9 2 EI
P3 
L2
n z
L
v：deflection
E: Young’s moudlus
I：moment of inertia
L：length
2
4 2 EI
P2 
L2
sin
Q: what buckling mode do we see in reality?
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3. Calculation of critical loads
Q: what buckling mode do we see in reality?
In practice, structures buckle at the first mode.
Obviously, the first critical load dominates the structural stability and thus we take
it as the critical load for design and denote it as Pcr
P
2
Pcr 
 EI
L2
Bifurcation point(分岔点)
cr = critical，临界
Pcr
E: Young’s modulus
I：cross-sectional moment of inertia
L：length
In reality, the bar must
buckle when the load
reaches Pcr.
|vmax|
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3. Calculation of critical loads
For the test on the sandwich panel，E = 1830 MPa, I = 6750 mm4 , L= 120 mm，the theoretical critical load is
Pcr 
 2 EI
L2
3.142 1,830  6750

 8466.3 N
1202
However, it is significantly different from the measured value 34078.2 N.
Why？
Note that both ends are clamped by steel blocks and cannot rotate.
 2 EI is based on simple supports
Pcr  2
y
L
P
P
z
y
P
P
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at both ends!
z
The BCs are different and the
buckling problem should be
solved again!
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3. Calculation of critical loads
y
P
P
z
Analysis steps:
(1) Analyze the loads on the bar as a whole:
M0
v(z)
P
R
M1
z
P
R
Both ends are clamped/fixed, which implies that the ends resist moments M0, M1 and reaction forces R, in
addition to axial foces P.
M
R
M0
P
(2) Consider a cross section at z
v
P
The internal forces and moment are as shown:
z
R
Equilibrium gives
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M  Pv  M 0  Rz
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3. Calculation of critical loads
y
P
P
(3) Combine it with
d 2v
EI 2  Pv  Rz  M 0
dz
d 2v
M   EI 2
dz
or
z
M0
R
and we will get
Rz  M 0
d 2v
  2v 
2
dz
EI
R
v

M
P
z
M  Pv  M 0  Rz
P
EI
1
General solution: v( z )  A cos  z  B sin  z   Rz  M 0 
P
Before proceeding, please think: how many constants are to be determined?
BC v(0) = 0 gives
The deflection
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A
v( z ) 
M0
P
BC v’(0) = 0 gives
B
R
P
M0
R
 cos  z  1    z  sin  z 
P
P
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3. Calculation of critical loads
y
Pz
P
v( z ) 
M0
R
 cos  z  1    z  sin  z 
P
P
BC v(L) = 0 gives
BC v’(L) = 0 gives
M0
R
 cos  L  1    L  sin  L   0
P
P
M
R
 0 sin  L  1  cos  L   0
P
P
matrix form:
1


cos

L

1
L

sin  L   M 0  0 




  R  0 
  sin  L 1  cos  L 
By linear algebra, existence of nontrivial solution requires the determinant to be zero, i.e.,
It can be organized into
Note that we may have
Their 1st eigenvalues are:
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sin
sin
L 
1
2


 1  cos  L    sin  L  L  sin  L   0



L L 

 tan
0
2 
2
2 
L
2
1 L
2
0
or
   3.1416
tan
L
1 L
2
2

L
2
 4.4934
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We should choose
1 L
2

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3. Calculation of critical loads
1 L
Combining
2
  
P
EI
we can get
4 2 EI
Pcr 
L2
Substituting E = 1830 MPa, I = 6750 mm4 , L= 120 mm，the estimated critical
load is
4 2 EI 4  3.142 1,830  6750
Pcr 
L2

1202
 33865.2 N
which is quite close to the measured value 34078.2 N with a relative error of
0.63%.
Buckling mode
1
2 z 
1  cos

2
L 
has a similar shape with the recorded deformation.
y
P
P
z
Boundary conditions have a dominant effect on
critical loads and buckling modes!
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3. Calculation of critical loads
Critical load for pressure bars can be summarized as
Pcr 
 2 EI
L2e
where Le is the effective length, and Le/L dependents on the boundary conditions:
BCs
Le/L
One fixed, one free
2.0
Both simply
supported (pinned)
1.0
One fixed, one
simply supported
0.6998
Both fixed
0.5
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Buckling mode
Trend:
The stronger the boundary
constraints are, the shorter the
effective length is.
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3. Calculation of critical loads
Critical stress is calculated to evaluate the risk of instability.
Critical load Pcr 
 2 EI
2
e
L
corresponds to the critical stress
Pcr  2 EI
 2E
 cr 


2
2
A
ALe
L
/
r
 e 
where
A：cross sectional area
E: Young’s moudlus
I：moment of inertia
L：length
r I/A
is called radius of gyration, which dependes on the geometry of the cross section.
The nondimensionalized number Le / r Is called slenderness ratio(长细比)
Buckling criterion for pressure bars
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P
  cr
A
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4. How to avoid buckling
P
Buckling criterion for pressure bars
  cr
A
For given working conditions, structural stability may be enhanced
by increasing the critical stress.
 2E
 cr 
2
 Le / r 
A：cross sectional area
E: Young’s moudlus
I：moment of inertia
L：length
r：radius of gyration
r I/A
Q：how to increase the critical stress?
1.Increase E：use material with larger E
BCs
Le/L
2.Increase r：optimize the cross sectional geometry
One fixed, one free
2.0
3.Reduce Le：
(1) use a shorter bar
(2) enhance the boundary constraints
(3) add lateral constraints (see Example 8.1)
Both simply supported (pinned)
1.0
One fixed, one simply supported
0.6998
Both fixed
0.5
4.Combination of the above.
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Example 8.1: lateral constraints increases critical load effectively
A uniform column of length L and flexural stiffness EI is simply supported at its ends, and has an additional elastic
spring with stiffness coefficient k. Derive the equation for solving the buckling load of the column.
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k/(EI/L3)
Pcr /(2EI/L2)
0
1
16
1.3267
100
2.9683
1000
7.8259
∞
8.183
37
Example 8.1
𝐵=
𝑣𝑐
𝑘𝐿
1−
𝑠𝑖𝑛(𝜆𝐿/2)
4𝑃
If
=(both fixed) or (both pinned and length= L/2)
=(one pinned, one fixed and L/2)
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Example
Show the boundary conditions and derive the equation for critical loads
2
d v
EI 2  Pv  Rz  0
dz
Boundary condition
K
PCR
A
v
z
R
B
L
z=0, v=0
z=L:
dv
d 2v
K  K ( )  M  RL   2 EI
dz
dz
(A transcendental equation)
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