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I. Introduction 1
truss, beam, frame (skeletal structures) ~Matrix Methods
panel, plate, shell (continua solids) ~
classic matrix mehtods
finite element methods
This course deals with the problems related with structures as well as other computational
mechanics or mathematics.
Variables: Displacement field, Temperature gradient, Head or Potential, etc.
Simultaneous Equation Problems, Eigenvalue Problems, and Propagation Problems.
(A ) General Concepts
Structure must be in equilibrium, with their displacements in a compatible state and
material laws satisfied.
External force ~ ~ Internal force ( Stress) ~ ~ Strain ~ ~ Displacement
Equilibrium Constitutive Law Compatibility
(i) DE Problems
Determine the displacement and stress of a rod
Q(x) : load/length = Qox / L under a surface force P and a body force Q(x).
N(x) Q(x)dx N +N
x
dx
Equilibrium:
- N(x) + Q(x)dx + ( N + N
x
dx ) = 0
N
x
+ Q = 0 (1)
Divided by cross section area A, it leads to x
x
+ f = 0 f : load/volume
P
Compatibility: strain and displacement 2
u(x) u(x) +u
dxx
u u
dx xx
(2)
dx
Constitutive law: Hooke’s law Ex x (3)
Solution Technique:
Using (2) and (3), N = x A = E x A = EAu
x
→ (1), we obtain
2
2
uE A Q 0
x
(4)
This is the displacement equation of equilibrium, a governing system equation based on the
so-called displacement method.
Boundary Conditions: u(0) = 0 , geometric boundary condition
N(L) = P , force (natural) boundary condition
u(0) = 0 EAu
x
︱L = P (5)
The solution of (4) and (5) is given as: 3
0Q L x x P xu ( x ) = ( )
A E 2 6 L A E (6)
(6) → (2) → (3) 2
0Q L x P= ( )
A 2 2 L Ax (6a)
Special case: Q(x) = 0 P x P L
u ( x ) = u ( L ) =A E A E
P
=A
x
Conclusion:
1. Although displacement method is used, the alternative force method is possible.
2. The problem associated with a differential equation with appropriate boundary values or
initial conditions is called the boundary value problem (BVP).
( B ) Classification of Structures 6
By geometry
*1-D beam flexural bending rod tension
column compression shaft torsion
truss tension and compression
frame tension, compression, shear, and moment
Matrix Method: designates investigations of structures composed of articulated or discrete
components (skeletal structures).
Note: This course covers the stress analysis rather than the design detail.
*2-D Surface structures are those which can be idealized to plane or curved surfaces.
Panel (in-plane load) plate (out-of-plane) shell
Finite Element Method: denotes analysis of continua (field problems).
*3-D Solid: neither of the above.
Note: Need complex mathematics and elasticity theory to deal with the latter two.
Linearity: material (Hooke’s law) and geometric (small deformation)
( C ) Classification of Loads
Static – mechanically applied loads (dead and live load) & thermal
1. Impulse, blast
a. determinate 2. periodic - harmonic and non-harmonic
Dynamic – 3. non-periodic
b. indeterminate (random) : wind, earthquake
( b ) principal stresses 12
For a 2-D problem, the stress state at a point is given as 11 12
21 22
.
11 1 1cos( , ) cosx x 1 2 1 2c o s ( , ) c o s ( ) s i n2
x x
21 2 1cos( , ) cos( ) sin2
x x
2 2 2 2c o s ( , ) c o sx x
Therefore,
11 1 1i j ij 11 11 11 11 12 12 12 11 21 12 12 22
2 2
11 12 22cos 2 cos sin sin
11 22 11 22 12
1 1( ) ( ) cos 2 sin 2
2 2 (a)
22 11 22 11 22 12
1 1( ) ( ) cos 2 sin 2
2 2 (b)
21 22 11 12
1( ) sin 2 cos 2
2 (c)
You have these equations in strength of material. Now, we can determine an orientation so
that 21 = 0 ( 11 is maximum). From equation (c), it yields
12 11 22tan 2 2 / ( ) (d)
(d) (a) get 2 211 22 11 2211 12( )
2 2
(a*)
(d) (b) get 2 211 22 11 2222 12( )
2 2
(b*)
(d) (c) get 21 = 0 (c*)
Construct a Mohr’s circle.
(iii) Equilibrium Equations 15
1111 11 1 2 3
1
( )dx dx dxx
+
2121 21 2 1 3
2
( )dx dx dxx
+
3131 31 3 1 2
3
( )dx dx dxx
+
1 1 2 3f dx dx dx = 0
After simplification,
3111 211
1 2 3
0fx x x
In general, , 0j i j if (9)
1111 1
1
dxx
2121 2
2
dxx
3131 3
3
dxx
1
2
3
Optional Derivation 18
Q ( i iX dX ) Q ( i ix dx )
P( iX ) P( ix )
2 2 2 2
1 2 3dS dX dX dX 2 2 2 2
1 2 3d s d x d x d x
The difference of 2 2ds dS is a measure of deformation, 2 2
i i i ids dS dx dx dX dX
Since 1 2 3( , , )i ix x X X X ,
1 2 3
1 2 3
i i ii
x x xdx dX dX dX
X X X
= ,i j jx dX ( or ,i k kx dX )
2 2ds dS = , ,i j j i k kx dX x dX - i idX dX = , ,i j j i k kx dX x dX - j k j kdX dX
= , ,( )i j i k j k j kx x dX dX
A measure of deformation in terms of the original coordinates j kdX dX is , ,i j i k j kx x . (a)
It is noted that such a measure is dimensionless. In addition, i i ix X u . Therefore,
, , , , , ,,i j i j i j i j i j i k i k i kx X u u x u
, ,i j i k j kx x = , ,( ) ( )i j i j i k i k j ku u
= , , , ,i j i k i j i k i j i k i j i k j ku u u u
= , , , ,j k j k k j i j i k j ku u u u = , , , ,j k k j i j i ku u u u 2 j kE
j kE = , , , ,
1( )
2j k k j i j i ku u u u (5)
j kE is the Lagrangian (material) strain tensor. Similarly, a measure of deformation in terms of
the deformed coordinates j kdx dx is , , , ,j k k j i j i ku u u u 2 j ke (b)
j ke = , , , ,
1( )
2j k k j i j i ku u u u (6)
j ke is the Eulerian (spatial) strain tensor. If the displacement gradients are small, both tensors are
identical and become the infinitesimal strain tensor, which is geometrically linear.
( ) ( )I u u I u > 0 (1*) ( ) ( )I I u u I u (2*) 25
I ( , )b
aF u u u u dx ( , )
b
aF u u dx (a)
By application of Eq.(5), we have
( , )F u u u u = ( , )F F
F u u u uu u
+
2 22 2
2 2
1[ ( ) 2 ( ) ]
2
F F F Fu u u u
u u u u
+ (b)
(b) → (a)
I 2 2
2 2
2 2
1( ) [ ( ) 2 ( ) ]
2
b
a
F F F F F Fu u u u u u dx
u u u u u u
+ (c)
Denote
the first variation of F, F F
F u uu u
the second variation of F, 2 2
2 2 2
2 2( ) 2 ( )
F F F FF u u u u
u u u u
As a result, (c) becomes 21
2
b b
a aI F dx F dx = I + 21
2I + (d)
Since u , (c) becomes
I 2 2
2 2 2
2 2
1( ) [ 2 ]
2
b
a
F F F F F Fdx
u u u u u u
+ (3*)
The higher order variations are negligible compared with the first term, i.e. I , when is
sufficiently small. Accordingly, a necessary condition for I to have a minimum, is that I vanish
identically.
0I b b
a a
F FI F dx u u dx
u u
= 0 (4*)
By integration by part, it leads to [ ] ( )b b
b
aa a
F F d Fu dx u u dx
u u dx u
(4*) yields
[ ( )] [ ]b
b
aa
F d F FI u dx u
u dx u u
= 0 (8)
( )F d F
u dx u
= 0 (9) At a and b, u = 0 or
F
u
= 0 (10)
Euler-Lagrange Equation, geometric boundary condition, natural boundary condition, respectively.
29
Now, apply (or )i iu q i j (satisfy the equation of compatibility). As a result,
E i i i iW T u dS f u dV or i iQ q (3)
i j i jV
U dV (4)
U is the work done by the actual stresses during the virtual distortion.
Theorem: Principle of virtual displacement
If a structure is in an equilibrium state and remains so while it is subject to virtual
displacements and strains (both satisfy compatibility), then
EW U (5)
Proof:
E ij j i i iW n u dS f u dV ,( )ij i j i iu dV f u dV
, ,i j j i i j i j i iu u dV f u dV
, ,( )i j j i i i j i jf u dV u dV (a)
, ,
1( )
2i j i j j iU u u dV , ,
1
2i j i j j i i ju u dV ,i j i ju dV (b)
Since ,i j j if = 0 (equilibrium), the first term in (a) is zero. Therefore, (a) = (b).
Note:
a. The inverse of this theorem is valid, i.e. if EW U , then equilibrium.
b. No constitutive law is used, so it can be applied to both elastic and inelastic solids.
c. Define EI U W . If EW U , then 0EI U W , i.e. I 0 implies equilibrium.
d. For rigid body system, 0U , therefore, 0EW .
Application of principle of virtual displacement: 30
Example: Given 1Q and 2Q , determine 1q , 2q and member stresses.
Consider member AB3
1 3 2 311
3 3
cos sinq q
L L
(geometry) (a)
1 1 1 1E (constitutive)(b)
Apply 1 2andq q , we have 1 3 2 311
3 3
cos sinq q
L L
11 11U dV = 1 3 2 3 1 3 2 3
3 3 3 3
cos sin cos sin( )( )q q q q
E dVL L L L
= 2 231 3 2 3 3 1 1 3 3 2 3 2
3
[ ( cos sin cos ) ( cos sin sin ) ]A E
q q q q q qL
(c)
Now, consider the total system.
32 2
1 2 1 1 2 2
1
[ ( cos sin cos ) ( cos sin sin ) ]i
i i i i i i
i i
A EU q q q q q q
L
(d)
But 1 1 2 2EW Q q Q q (e)
Let (d) = (e), we have 3
2
1 2 1 1
1
[ [ ( cos sin cos ) ]i
i i i
i i
A Eq q Q q
L
+
32
1 2 2 2
1
[ ( cos sin sin ) ]i
i i i
i i
A Eq q Q q
L
= 0 (f)
Through arrangement, the following simultaneous equations are obtained and 1q and 2q can be
solved. Then, use (a) to get strain and use (b) to get stress of each member.
3 32
1 2
1 1
( cos ) ( sin cos )i i
k i i
i ii i
EA EAq q
L L
= 1Q (f-1)
3 32
1 2
1 1
( cos sin ) ( sin )i i
i i i
i ii i
EA EAq q
L L
= 2Q (f-2)
B3 B2 B1
q1 , Q1
q2 , Q2
A
3 3
A RECIPE FOR STRUCTURES (II) 31
Displacement Force
Subject: Solid Mechanics (Elasticity Theory)
A. Differential Equation
Navier Equation of Elasticity Beltrami-Michell Equaiton
(Displacement Eq. of Equilibrium)
B. Energy Method Principle of Virtual Work (Displacement) Principle of Complementary Virtual Work (Force)
(elastic or inelastic) (elastic or inelastic)
Unit-Displacement Theorem Unit-Force Theorem*
(Dummy-Disp.) (Dummy-Force, Unit-Load, Dummy-Load)
Principle of Minimum PE Principle of Minimum Complementary PE
(elastic) (elastic)
First Castigliano Theorem Engesser Theorem
(elastic) (elastic)
Second Castigliano, Least Work Theorem
(linearly elastic)
*also nonlinear elastic for a determinate system
Subject: Classic Analysis of structures (Based on classic beam theory)
1. Deflections of Structures A. Solving the Governing Equation
Integration of Load-Deformation Eq., Moment-Area,
Conjugate-Beam
B. Energy Method
right side of above table
2. Stress Analysis
Slope-Deflection Consistent Deformation
Moment-Distribution Least Work
Subject: Modern Structural Analysis (Aided by digital computer)
1. Matrix Method (not wide sense)
2. Finite Element Method
(Direct) Stiffness Method Flexibility (Force) Method
Application of unit displacement: 33
Example: Given 1Q and 2Q , determine 1q , 2q and member stresses.
Consider member AB3
1 3 2 311
3 3
cos sinq q
L L
(geometry) (a)
1 1 1 1E (constitutive)(b)
Apply 1q =1 , 2q = 0, we have 311
3
cos
L
11 11 dV = 1 3 2 3 3
3 3 3
cos sin cos( )q q
E dVL L L
= 23
1 3 2 3 3
3
( cos sin cos )A E
q qL
(c)
Now, consider the total system.
32
11 11 1 2
1
( cos sin cos )i
i i i
i i
A EdV q q
L
(d)
By use of unit displacement theorem,
1Q =3
2
11 11 1 2
1
( cos sin cos )i
i i i
i i
A EdV q q
L
(f)
Similarly, by applying 2q = 1 and 1q = 0, we have
2Q =3
2
11 11 1 2
1
( cos sin sin )i
i i i
i i
A EdV q q
L
(f*)
Through arrangement, the following simultaneous equations are obtained and 1q and 2q can be
solved. Then, use (a) to get strain and use (b) to get stress of each member.
3 32
1 2
1 1
( cos ) ( sin cos )i i
i i i
i ii i
EA EAq q
L L
= 1Q (f-1)
3 32
1 2
1 1
( cos sin ) ( sin )i i
i i i
i ii i
EA EAq q
L L
= 2Q (f-2)
B3 B2 B1
q1 , Q1
q2 , Q2
A
3 3
Principle of minimum potential energy: 36
Of all compatible displacements satisfying given boundary conditions, those which satisfy the
equilibrium conditions make the total potential energy assume a stationary value.
** Elastic system only, is more limited than the virtual displacement method.
Example: Given 1Q and 2Q ,determine 1q and 2q .
Consider member AB3
.
i j i jU de dV
v
11 11de dV
v
11 11E e de dV
v
2
11 3 32
Ee A L (constitution) (a)
1 3 2 311
3 3
cos sinq qe
L L
(Strain-Displacement) (b)
Substituting (b) into (a), we have 3 2
1 3 2 3
3
cos sin )2
(EA
U q qL
(c)
Therefore, the total 3
2
1 2
1
cos sin )2
(k
k k
k k
EAU q q
L
(d)
1 1 2 2EV Q q Q q (e)
EI U V = 3
2
1 2
1
cos sin )2
(k
k k
k k
EAq q
L
1 1 2 2Q q Q q
0I 1
0I
q
2
0I
q
(Equilibrium)
3
1 2
1
cos sin )cos(k
k k k
k k
EAq q
L
= 1Q 3
1 2
1
cos sin )( sin )(k
k k k
k k
EAq q
L
= 2Q (f)
Through arrangement, 3 3
2
1 2
1 1
cos sin cosk k
k k k
k kk k
EA EAq q
L L
= 1Q (f-1)
3 32
1 2
1 1
cos sin sink k
k k k
k kk k
EA EAq q
L L
= 2Q (f-2)
Then, 1q and 2q are obtained by solving simultaneous equations of (f-1) and (f-2).
1x 1dx
(D) Classic Beam Theory 38
Make the assumptions about displacement field to reduce number of variables. Use a single
dependent variable, 3 1( )u x , deflection of centerline.
Checking the results for simple cases with those, which stem from the full theory of elasticity or
experiments. Also called as technical or elementary beam theory.
b
o
a
c
b
a
o
C
水平線
切線
Assumptions:
1. small deformation
2. plane sections remain plane and normal(neglect deflection due to shear
Displacement at b, o, a is 3 1( )u x , displacement at c is 3 1 1( )u x dx . When 1dx is small, c is
regarded on the tangential line from o and tan( ) . Thus, 3 1/du dx . Displacement at
b 1( , / 2)x h is / 2h , o 1( ,0)x is 0, a 1( , / 2)x h is - / 2h .
As a result of bending, 39
3
1 3
1
duu x
d x 2 0u 3 3 1( )u u x (2)
Based on (2), the infinitesimal strain is
2
3
11 3 2
1
d ux
d x 0i j (3)
Accordingly,
11 11 112 2 2 1 1 3 3 1 1 12 0 23 0 13 0 (4)
Modification 1:
22 and 33 are small and are set to be zero. In addition,
0 0 / 2E 1 1 1 1E (a)
Modification 2:
13 is not so small and is set to be nonzero. It is noted that we set 13 0 (3*)
13 132 (b)
This shear stress and the accompanying shear deformation will be small for a long beam ~ ( /h L ).
In summary, (4) becomes
11 11E 22 0 33 0 12 0 23 0 1 3 1 32 (5)
Those are stress formulas in the classic (elementary or technical) beam theory.
Some useful formulas:
By (3) & (a) →2
3
11 3 2
1
d uE x
d x (6)
/ 2
1 3 11 3/ 2
( )h
hM x x b d x
=
2/ 2 3
3 3 32/ 21
h
h
d ux E x b d x
d x =
233
2
112
d ubhE
d x =
2
3
2
1
d uEI
d x (7)
(6) → 2
3
11 3 2
1
d uE Ix
I d x =
3M x
I (by (7)) (8)
1x
3x
h
L
Strain energy based on the classic beam theory is obtained as follows. 40
11 11 13 13
1 1( ) 2
2 2VU dV 2 2
11 13
1 1
2 2VdV
E
2 2
3 2
2
1 1( )
2 2V
M x V QdV
E I I b
2 2 2
1 12 22 2 A
M V Qd x d x d A
EI I b b sU U (9)
Since M =
2
3
2
1
d uEI
d x , bU =
2
3 2
1201
( )2
L d uE Id x
d x (10)
Note: Equations (7) ~ (10) correspond to the so-called classic beam theory, and do not represent a
solution to the equations of elasticity. However, they are adequate for most engineering
application.
Example: A simple supported beam under a uniform load w.
bU =2 2 5
/ 2
10
22 240
L M w Ld x
E I E I
13 13sV
U dV 2
13
1
2VdV
2 2
12 22 A
V Qd x d A
I b
22 2 2 2
1 3 3 1
1 1( ) ( ) ( )
2 2 4
hw x x b d x d x
I
(Note: 2
2
1 3( )2 4
b hV w x Q x )
Let 2 (1 )
E
. →
sU 5
/ 22 2
1 12 0
12
4 30
L hw x b d x
E I
22 3
2
1
4 30
hw L I
E I
2 52[2(1 ) ( ) ]
240
w L h
E I L
2/ 2(1 ) ( )s b
hU U
L 0 ~ 0 . 5 / 0 . 1h L / 2 . 5 %s bU U
(E) Principle of Complementary Virtual Work (Force) 41
This is valid for the elastic & inelastic systems.
*
00
i j
i j i jU d
*
0
i j
i j i jV
U d dV
(1)
Recall 0i j
i j
U
, we have
*
0i j
i j
U
(2)
If it is linear, *U U (3)
Now, apply (or )i i iT f Q i j . As a result,
*
E i i i iW T u dS f u dV or i iQ q (4)
*
i j i jV
U dV (5)
Theorem: If a structure is in a compatible state of deformation and remains so while it is
subject to virtual forces and stresses (both satisfy the equation of equilibrium), then
* *
EW U (6)
Proof:
*
E ij j i i iW n u dS f u dV ,( )ij i j i iu dV f u dV
, ,i j j i i j i j i iu u dV f u dV
, ,( )i j j i i i j i jf u dV u dV = ,i j i ju dV (a)
If , ,
1( )
2i j i j j iu u
*
, ,
1( )
2i j j i i jU u u dV , ,
1
2i j i j j i i ju u dV ,i j i ju dV (b)
It is seen that (a) = (b).
*
0U
0U
i j
i j
3x
1x
w
q
Note: 42
a. The inverse of this theorem is valid, i.e. if * *
EW U , then the compatibility holds.
b. No constitutive law is used, so it can be applied to both elastic and inelastic solids.
c. Define * * *
EI U W . If * *
EW U , then * * * 0EI U W , i.e. *I 0 implies
compatibility.
(i) Unit-Force (Load) theorem, Dummy-Force (Load) method
In a discrete system, i iQ q i j i j
VdV (7)
Let 1kQ and all other 0iQ , then Equation (7) becomes
kqi j i j
VdV (8)
It is noted that i j in (8) is due to 1kQ , while i j in (7) is due to all iQ .
Example: A cantilever beam is under a uniform load w. Determine its rotation at mid-point.
q i j i jV
dV (a)
11 3M x
I 22 0 33 0 12 0 23 0 1 3 1 32
3
11
M x
EI & neglect the effect of shear, i.e. 13 13 / 2 0
11 3m x
I ( m is the bending moment produced by a unit virtual force at mid-point)
(a) becomes q3 3
V
M x mxdV
EI I 1
0
L M md x
EI (b)
1
43
m = -1 2
1 1
10 ( )
2 2
Lx M w L x . Therefore, (b) yields
q
231
10
1( ) ( 1)
72
48
Lw L x
wLd x
EI E I
(ii) Principle of Minimum Complementary Energy
Of all states of stress which satisfy the equation of equilibrium, the one which satisfies the
compatibility equation makes the total complementary energy a stationary value.
Since * *
EW U and * *
E EV W , we have
* *( ) 0EU V (9)
(iii) Castigliano’s second theorem (linearly elastic)
In a discrete system, *
E i iV Q q (10)
*( ) 0i iU Q q *
0i i i
i
UQ q Q
Q
From which, we have Engesser’s theorem: *
i
i
Uq
Q
(11)
For an linearly elastic system, (11) becomes the well known second theorem.
i
i
Uq
Q
(12)
44
Example: A cantilever beam is under a uniform load w. Determine its rotation at mid-point.
x
C
2
10 2
L MU d x
E I
2 2
/ 2
10
1( )
2
2
Lw x
d xE I
2 2
1/ 2
1( )
2
2
L
L
w x C
d xE I
2
1/ 2
1( ) ( 1)
2L
L
w x CU
d xC E I
2
1/ 2
1
2L
L
w x C
d xE I
3
/ 26
L
L
w x
E I
37
48
w L
E I ( C = 0)
(E) Settlement and Temperature 59
(i) Settlement
f ff f f s
s f s s s s
q QK K
K K q Q
Since 10 ( )s f f f f f s sq q K Q K q (1)
Example 1 2 515 20 27q mm q mm q mm
Determine fq ,
sQ and member forces.
3
4
6
f
q
q q
q
=
11.36 .48 .48
1.48 .64 .64
20000.48 .64 1.64
0 1 0 .36 .015
0 20000 0 0 .48 .020
0 0 1 .48 .027
=
.015
.011
.020
m
1
2
5
s
Q
Q Q
Q
=
1 0 0 .015
20000 0 0 1 .011
.36 .48 .48 .020
+
1 0 0 .015
20000 0 1 0 .020
0 0 .36 .027
=
0
0
0
12 1 0 1 0 .015 .020 .015 .011 0TAEP
L 1 3 2 3 0P P
It should be anticipated from the outset that there is neither support force nor member force
induced by the support movements. This is due to the fact that the system is a statically
determinate one.
(ii) Temperature effects and fabrication errors
1. Apply initial fixed-end forces to the members to prevent any nodal displacements introduced
by thermal changes and fabrication errors.
2. Remove the initial forces applied in Step 1 and introduce forces at all nodal points that are
equal in magnitude but opposite in direction to the initial fixed-end forces in Step 1.
Analyze the structure, i.e. calculate the nodal displacement and the internal forces
3. Suppose two steps.
5
6
5
6
1
2
1
21 2
3
3
4
3
4
Again, (7) must be partitioned as 62
0
0
f f ff f f s
s s f s s s s
Q q QK K
Q K K q Q
(7a)
1
0( )f f f f f f s sq K Q Q K q (8)
0s s f f s s s sQ K q K q Q (9)
Summary:
1. Get a for each element 2. Determine 0P for each element
3. Calculate 0 0
TQ a P for each element
4. Suppose the system equation 0Q K q Q (7)
5. Consider boundary condition to get fq , i.e. (8)
6. For each element, v a q 0P k v P i.e. 1 3( )i j
AEP v v AE T
L or
LAE
L
Example: Assume member 23 is 15 mm too long.
Determine fq ,
sQ and member forces.
For element 23, we have
34 6
0
1 300
0 015 105 10 200 10
1 3005
0 0
P KN
0 0
TQ a P
0.6 0.8 300 180
0.8 0.6 0 240
0.6 0.8 300 180
0.8 0.6 0 240
KN
5
6
5
6
1
2
1
21 2
3
3
4
3
4
For the total system, 63
0Q 0 0 180 240 180 240T
KN 0 0
180 0
240 0
240 180
f sQ Q
1
0( )f f f f f f s sq K Q Q K q 1
0f f fK Q
3
4
6
f
q
q q
q
=
11.36 .48 .48
1.48 .64 .64
20000.48 .64 1.64
180
240
240
=
0
.01875
0
m
1
2
5
s
Q
Q Q
Q
=
1 0 0 0
20000 0 0 1 .01875
.36 .48 .48 0
+
0
0
180
=
0
0
0
Member forces
12 1 0 1 0 0 0 0 .01875 0TAEP
L
For member 23,
0.6 0.8 0 .015
0.8 0.6 .01875 .01125
0.6 0.8 0 0
0.8 0.6 0 0
v
23 1 0 1 0AE
PL
.015 .01125 0 0T
+ (-300) = 0
13 0P
Again there is neither support force nor member forces induced by the fabrication errors. This is
due to the fact that the system is a statically determinate one.
Stiffness matrix of a truss element 64a
1
2
1 1
1 1
P AEP
P L
(1) 1P→ → 2P
(a) by definition
When 1 21 0v v , 1
1P L
AE so 1P
AE
L and 2P 1P
AE
L
1
2
1 .
1 .
P AEP
P L
(a)
Similarly, when 1 20 1v v 1
2
. 1
. 1
P AEP
P L
(b) (a) & (b) → (1)
(b) Previous derivation
2
11 11 11 11 11
1 1
2 2U d dV dV E AL Assume that 11 and 11 are constant.
11 2 1( ) /v v L (2) → 2
2 1( )2
AEU v v
L
1 2 1
1
( ) ( 1)U AE
P v vv L
(c) 2 2 1
2
( )U AE
P v vv L
(d) (c) & (d) → (1)
(c) Shape function
11
( )v x
x
1 1 2 2( )v x N v N v (3)
How to determine 1N and 2N ? Assume ( )v x a bx (4)
From(4), when 1 10x v a a v , when 2 2 1( ) /x L v a b L b v v L
Therefore, 2 11 1 2( ) ( ) (1 )
v v x xv x v x v v
L L L
. Compare this with (3), we have
1 ( 1 )x
NL
and 2
xN
L (5)
1N and 2N are called shape functions. As a result, 11
( )v x
x
1 21 2
1 1( )
v vv v
L L L
It is consistent with (2)
V. Analysis of Beam 65
(A) Stiffness Matrix by Shape-Function Concept
A beam element has 4 degrees of freedom.
The transverse displacement is denoted as v(x). Therefore, the strain energy due to bending is
2
0 2
L MU d x
E I =
2
0
( )
2
L E I vd x
E I
=
2
0
( )
2
L E I vd x
(a)
Let us assume that 1 1 2 2 3 3 4 4( ) k kv x N v N v N v N v N v (1)
Thus, ( ) k kv x N v (b)
(b) → (a)
2
0
( )
2
L k kE I N vU d x
(c)
According to Castigliano’s theorem, 0
( )L
i k k i
i
UP E I N v N dx
v
(d)
*Determination of iN
Note that if 1 2 3 41 0v v v v , then 1( )v x N .
A beam with this kind of boundary condition is shown
in the right.
Suppose that the reactions at the left end is A (force, upward) and B (moment, counterclockwise),
then
( ) i.e. ( )M x A x B E I v x A x B
By integration, 2
( )2
AxE I v x B x C →
32( )
6 2
Ax BE I v x x Cx D (e)
1 3
2 4
1
66
The boundary conditions are: (0) 1 (0) 0 ( ) 0 ( ) 0v v v L v L . On the basis of them, we
obtain the parameters A, B, C, and D. Therefore,
2 3
1( ) 1 3 ( ) 2 ( )x x
v x NL L
(2)
By the same manner, note that if 2 1 3 41 0v v v v , then 2( )v x N .
A beam with this kind of boundary condition is shown
in the right.
The appropriate boundary conditions are (0) 0 (0) 1 ( ) 0 ( ) 0v v v L v L . We have
2
2( ) (1 )x
v x x NL
(3)
Similarly,
2 3
3 3 ( ) 2 ( )x x
NL L
(4) 2
4 ( 1)x x
NL L
(5)
From (b), it leads to
12 3
6 12( )k k
xN v v
L L 22
4 6( )
xv
L L 32 3
6 12( )
xv
L L 42
6 2( )
xv
L L (f)
(f) → (d), with i = 1
1 1 2 3 42 3 2 2 3 20
6 12 4 6 6 12 6 2{ ( ) ( ) ( ) ( ) }
L x x x xP E I v v v v
L L L L L L L L
2 3
6 12 ( )
xdx
L L = 1 2 3 43
( 12 6 12 6 )E I
v Lv v L vL
Again, (f) → (d), with i = 2
2 1 2 3 42 3 2 2 3 20
6 12 4 6 6 12 6 2{ ( ) ( ) ( ) ( ) }
L x x x xP E I v v v v
L L L L L L L L
2
4 6 ( )
xdx
L L =
2 2
1 2 3 43( 6 4 6 2 )
E IL v L v L v L v
L
1
Finally, 3 4andP P are yielded. In a matrix form, 67
1 1
2 22 2
3
3 3
2 2
4 4
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
P vL L
P vL L L LE I
P L L vL
L L L LP v
(6)
It should be pointed out that sum of each column in no longer zero. When all elements within a
structure have a same length, (6) can be expressed as
1 1
2 2
3
3 3
4 4
12 6 12 6
/ 6 4 6 2
12 6 12 6
6 2 6 4/
P v
P L v LE I
P vL
P L v L
(7)
As a summary, in any case,
P k v (8)
(B) Formulation of System Equation 68
*Selection of a beam element:
(i) member is uniform, and (ii) shear is constant and moment varies linearly between two nodal
points.
*No transformation of element stiffness matrix is needed.
Example I = 1034 in4, L12 = L23 = L34 =15 ft, E = 29000 k/in2
Determine fq ,
sQ and member forces.
For member 12
k
12 6 12 6
6 4 6 25.142
12 6 12 6
6 2 6 4
/k in *k
The element stiffness matrices *k for member 23 and member 34 are identical. Thus, the stiffness
matrix of the unconstrained structure is
12 6 12 6
6 4 6 2
12 6 24 0 12 6
6 2 0 8 6 25.142
12 6 24 0 12 6
6 2 0 8 6 2
12 6 12 6
6 2 6 4
K
/k in
f ff f f s
s f s s s s
q QK K
K K q Q
1
2
7
8
s
q
q Lq
q
q L
and
3 3
4 4
5 5
6 6
/
/
f f
q Q
q L Q Lq Q
q Q
q L Q L
f f f fK q Q
30k 45k
1 2 3 4
69
24 0 12 6
0 8 6 25.142
12 6 24 0
6 2 0 8
3
4
5
6
30
0
45
0
q
q L
q
q L
3
4
5
6
1.170
1.242
1.260
1.188
q
q Lin
q
q L
Member forces:
Member 12
1
2
3
4
/
/
P
P L
P
P L
12 6 12 6
6 4 6 25.142
12 6 12 6
6 2 6 4
1
2
3 3
4 4
0
0
v
v L
v q
v L q L
=
33.89
23.33
33.89
10.56
k
1
2
3
4
P
P
P
P
33.89
350
33.89
158
k
k ft
k
k ft
Member 23
1
2
3
4
/
/
P
P L
P
P L
12 6 12 6
6 4 6 25.142
12 6 12 6
6 2 6 4
1 3
2 4
3 5
4 6
v q
v L q L
v q
v L q L
=
3.89
10.56
3.89
14.44
k
1
2
3
4
P
P
P
P
3.89
158
3.89
217
k
k ft
k
k ft
Member 34
1
2
3
4
/
/
P
P L
P
P L
12 6 12 6
6 4 6 25.142
12 6 12 6
6 2 6 4
1 5
2 6
3
4
0
0
v q
v L q L
v
v L
=
41.11
14.44
41.11
26.67
k
1
2
3
4
P
P
P
P
41.11
217
41.11
400
k
k ft
k
k ft
Moment and shear diagrams:
To convert to “beam” convection, the sign of P2 and P3 should be changed from “frame” one.
* Constant shear is observed in this example.
350
158217
400
33.89
3.89
41.11
(C) Equivalent Nodal Forces 70
→
Perform analysis
1 1
2
2 2
3 3
24 4
/ 2
/12
/ 2
/12
wLP v
P v wLk
P v wL
P v wL
(1)
1 1
2 2
3 3
4 4
/ 2
/ /12
/ 2
/12/
P v wL
P L v L wLk
P v wL
wLP L v L
(2)
0P k v P *
0Q k q Q (for each element) 0Q K q Q (system) (3)
(3) is rewritten as
0Q Q K q (4)
In the above equation, equivalent load 0Q is the fixed-end force with sign changed. Again,
0
0
f f ff f f s
s s f s s s s
Q q QK K
Q K K q Q
1
0( )f f f f f f s sq K Q Q K q (5)
Example: a simply supported beam is under a uniform load of w.
1 3 5 7 9
2 4 6 8 10
L
1 2 3 4 5
Member 12 71
k 3
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
E I
L
= *k 0 0
1/ 2
1/12
1/ 2
1/12
P wL Q
The above formulas are also applied to other members. Thus, the system matrices are:
3
12 6 12 6
6 4 6 2
12 6 24 0 12 6
6 2 0 8 6 2
12 6 24 0 12 6
6 2 0 8 6 2
12 6 24 0 12 6
6 2 0 8 6 2
12 6 12 6
6 2 6 4
E IK
L
0
1/ 2
1/12
1/ 2 1/ 2
1/12 1/12
1
0
1
0
1/ 2
1/12
Q wL
B.C. 1 9 2 10 4 8 3 7 60 8 DOF Symmetry : , , , 0 4 DOFq q q q q q q q q
2 1 2 3 43/ ( 6 4 6 2 )
E IQ L q q L q q L
L + wL /12
3 1 2 3 5 63( 12 6 24 0 12 6 )
E IQ q q L q q q L
L + wL
4 1 2 4 5 63/ ( 6 2 0 8 6 2 )
E IQ L q q L q L q q L
L
5 3 4 5 7 83( 12 6 24 12 6 )
E IQ q q L q q q L
L + wL 3 4 53
( 24 12 24 )E I
q q L qL
+ wL
2 2
3 3
3
4 4
5 5
/ 4 6 2 0 1/12
6 24 0 12 1
/ 2 0 8 6 0
0 24 12 24 1
Q L q
Q q LE IwL
Q L qL
Q q L
2
43
4
5
8 / 3
1 9 / 8
1 1 / 6
1 0 / 3
q L
q wL
q L E I
q
72
Member forces: 0P k v P
Member 12
1
2 2
3
3 3
4 4
012 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
v
v L q LE IP
v qL
v L q L
+
/ 2 2
/12 0
/ 2 1
/12 1.5
wL
wLwL
wL
wL
Member 23
1 3
2 4
3
3 5
4
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4 0
v q
v L q LE IP
v qL
v L
+
/ 2 1
/12 1.5
/ 2 0
/12 2
wL
wLwL
wL
wL
* The values of shears and moment are exact at the nodes.
2wLwL
wL2wL
2wL
2
1.5wL2
1.5wL2
shear diagram moment diagram
Element stiffness matrix: 82a
3 2 3 2
2 2
3 2 3 2
2 2
0 0 0 0
12 6 12 60 0
6 4 6 20 0
0 0 0 0
12 6 12 60 0
6 2 6 40 0
E A E A
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
E A E A
L L
EI EI EI EI
L L L L
EI EI EI EI
L L L L
Element stiffness matrix in the global coordinate:
2 2
3
2 2
3 3
2 2
2 2 2 2
3 3 2 3
2 2 2 2
3 3 2 3 3
2 2 2
12
12 12( )
6 6 4
12 12 6 12( )
12 12 6 12 12( ) ( )
6 6 2 6 6
E A EIC S
L L
E A EI EI E ACS C S
L L L L
EI EI EIS C
L L L
E A EI EI E A EI E A EIC S CS S C S
L L L L L L L
EI E A EI E A EI E A EI EI E ACS C S C CS C S
L L L L L L L L L
EI EI EI EI EIS C S C
L L L L
2
4EI
L L
The effective system stiffness matrix is superposed as: 84
f fK =
2
2 2
2
2 2
12 60 0 0
12 6 12 60 0
6 6 64 4 0 2
12 60 0 0
12 6 12 60 0
6 6 60 2 4 4
I IA A
L L
I I I IA
L L L L
I I II I I
EL L L
I I LA A
L L
I I I IA
L L L L
I I II I I
L L L
Substituting the numerical values, we have
f fK = 6
16574.75
0 16574.75
231 231 246410
16517 0 0 16574.75
0 57.75 231 0 16574.75
0 231 616 231 231 2464
E
L
4
5
6
7
8
9
q
q
q
q
q
q
= 1
f fK
fQ = 9
12406
26 61
935 6 5048
12376 26 931 12406200 10
26 0 6 26 61
931 6 38 935 6 504
0
0
0
140000
0
0
=
0.069306
0.000146
0.005214
0.069474
0.000146
0.005236
m
Member force Member 12
6
16517
0 57.75
0 231 123210
16517 0 0 16517
0 57.75 231 0 57.75
0 231 616 0 231 1232
EP k v
L
1
2
3
4 5
5 4
6 6
0
0
0
v
v
v
v q
v q
v q
=
60.12
69.95
319.95
60.12
69.95
239.66
KN
KN
KN m
Member 23
69.37
59.93
239.53
69.37
59.93
239.88
KN
KN
KN m
(B) Related Problems 91
A function ( , , )x y z is governed by a differential equation
( ) ( ) ( ) 0xx yy zzK K K Qx x y y z z
(1)
The boundary condition is
B on S1 and/or on S2 (S1 and S2 constitute a complete boundary)
( , , ) ( ) 0xx x yy y zz zK K K q x y z hx y z
(2)
**Heat transfer is temperature
Kxx : thermal conductivities Q : internal heat source or sink
q : heat flux over a portion of surface h : convection coefficient
**Solid mechanics is a stress function
Kxx = Kyy = 1 Q = 2G and B = 0 on the entire boundary.
2 2
2 22 0G
x y
(3)
G : material property : twisting angle
**Fluid mechanics
Irrotational flow of fluids Kxx = Kyy = 1 Q = 0 2 2
2 20
x y
(4)
= B and 0x yx y
If is specified on the fixed boundaries of region, (i.e. no flow can occur perpendicular to those
boundaries), Eq.(4) yields the “streamlines” for the irrotational flow field.
If is specified on those portions of the boundaries across which flow occurs, Eq.(4) yields the
“line of constant potential” that are perpendicular to the streamlines.
**Hydraulics is piezometric head 2 2
2 20xx yyK K Q
x y
(5)
= B and/or ( ) ( ) 0xx x yy yK K qx y
Kxx : permeability of the soil Q : water source (or sink)
q: seepage of water into or out of the aquifer along a portion of its boundary.
**Note: electrostatic fields, magnetostatics, and fluid-film lubrication.