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.