Fundamental-Finite-Element-Analysis-and-Applications

PROBLEMS 533

Figure 7.29. Finite element mesh around crack tip using triangular quarter-point elements

This expression clearly shows that the strain is proportional to 1/{T. Since the stress is

proportional to strain, it follows that the stress distribution near the cracletip is proportional
to the square root of the distance from the crack tip. Hence, defining the side node at
the quarter point of the side captures the desired stress singularity. Thus, for analyzing
fracture mechanics applications, we use the standard elements with the finite element mesh
as shown in Figure 7.28. There is no need to create specialized elements or to refine the
mesh excessively.

Elements with side nodes at quarter points are sometimes referred to as singularityele-
ments.However, it should be clear-from the discussion here that they are standard quadratic
elements except that the side nodes are placed at quarter points instead of near the middle.
Numerical experiments suggest even better performance with quarter-point triangular ele-
ments constructed by collapsing one side of an eight-node quadrilateral element. A typical
mesh around a crack tip using these elements is as shown in Figure 7.29. It is recom-
mended to have elements every 3D' to 40' in the circumferential direction. The radius of
the first row of elements around the cracle tip should be approximately one-eighth of the
craclelength.

PROBLEMS

Fundamental Concepts in Elasticity

7.1 Stresses at a point on planes perpendicular to the coordinate directions are given as
follows:

0;, = 100; a;, =-30; at =-10;

= = =T:<y 0;
TyZ 50; Tzx -50MPa

534 ANALYSIS OF ELASTIC SOLIDS

(a) Compute the stress vector, normal stress, and shear stress on a plane whose
equation is

x+2y-2z =-4

(b) Compute principal stresses and unit normals for principal planes.
(c) If the yield stress of the material is 250 MPa, compute the factors of safety

based on the maximum shear stress and the von Mises failure criteria.

7.2 Cartesian stress components at a point are given as follows:

0:, =-100; OJ, = 30; =iTz 10;
=TYe 50;
T.>:)' = 10; Tz.< = -50MPa

(a) Compute the stress vector, normal stress, and shear stress on a plane whose
unit normal is

n T = (0.15l99, -0.721676, -0.675339)

(b) Compute principal stresses and unit normals for principal planes.

(c) If the yield stress of the material is 200 MPa, compute the factors of safety
based on the maximum shear stress and the von Mises failure criteria.

7.3 A rectangular elastic solid with dimensions a x b x c is simply supported along the
three sides attached to the coordinate lines and is subjected to uniform pressure p
along the remaining three sides, as shown in Figure 7.30. (Note that the loading
is not concentrated but is applied uniformly over each face.) Thus the boundary
conditions on all six sides of the solid are as follows:

On xy plane at z = 0: =w(x, y, 0) 0; qxCx, y, 0) =q/x, y, 0) =0
q.,(x, y, c) =q/x, y, c) =0; qz(x, y, c) =-p
On xy plane at Z =c: vex, 0, z) =0; qxCx, 0, z) =qz(x, 0, z) =0
On xz plane at y =0:

xp y

Figure 7.30.

PROBLEMS 535

On xz plane at y = b: . qxCx, b, z) = qz(x, b, z) = 0; q/x, b, z) =-p

=On yz plane at x 0: =u(O, y, z) 0; = =q/a, y, z) qz<a, y, z) 0

On yz plane at x = a: q/a, y, z) = qz(a, y, z) = 0; qxCa, y, z) = -p

Show that the following displacements satisfy the governing differential equations
and all boundary conditions and hence represent the exact solution of the problem:

u(x, y, z) = -(l-2V)~X; w(x,y,z) = -(1-2V)~Z

where E = Young's modulus and v= Poisson's ratio.
7.4 A cylindrical shaft of diameter 30 mm is subjected to an axial force o~F = 10 kN, a

bending moment of M = 170 N . m, and a torque T = 220 N . m. The shaft is made
of steel that has a yield stress (]"/ = 300MPa. Determine the factor of safety against

yielding using von Mises theory. Note the following relationships for stresses in
shafts from the elementary mechanics of deformable bodies:

Axial stress = AF + TMr; Shear stress = JTr

where r is the radius of the shaft, A is its area of cross section, 1 is the moment of

inertia, and J is the torsional constant which is equal to the polar moment of inertia
for a circular section.

Plane Stress and Plane Str~in

7.5 A triangular cantilever plate 30 mm long, 20 mm high, and 10 mm thick is loaded
as shown in Figure 7.31. Use only one triangular element to determine the factor of
safety against stress failure (von Mises criterion). Use the following numerical data:

=E = 70 GPa; v = 0.3; p 100MPa; Yield stress = 350 MPa

Figure 7.31.

536 ANALYSIS OF ELASTIC SOLIDS

7.6 You are asked to determine stresses in a thin triangular plate that is subjected to a
temperature rise of 40°C. The plate is modeled with only one triangular element, as
shown in Figure 7.32. The height of the plate is 50 mm and the base width is 30 mm.
Other numerical data are as follows.

Thickness = I mm; E = 30,000N/mm2 ; V -- 14·'

3

2
Figure 7.32.

7.7 Determine stresses in the cantilever plate shown in Figure 7.33 due to a temperature
rise of 100°C. The plate is 5 mm thick, 50 mm long, 20 mm wide at the base, and
10 mm wide at the tip. Use the following material properties:

E =70 GPa; v =0.3; (l' = 23 x 1O-6/ oC

Consider a very coarse mesh/~ith only two triangular elements as shown. Use

N . mm units. Report displacements and effective stresses at element centroids.

3

10

o
o-----------~x

50
Figure 7.33.

7.8 The long diamond-shaped steel tube shown in Figure 7.34 is subjected to an internal
pressure p. Compute displacements and principal stresses. Since the tube is long, .
we can take a unit slice and model it as a plane strain problem. Thus we need to

PROBLEMS 537

create a finite element model of the planar cross section with thickness = 1. Use the

following numerical values.

E = 200 GPa; v = 0.3; Wall thickness = 15 mm;
d = 100mm; p= IMPa

T

pd

1

Figure 7.34.

Take advantage of symmetry and model one-quarter of the region using only a single
four-node element, as shown in Figure 7.35.

y

Figure 7.35.

7.9 A long steel pipe, shown in Figure 7.36, with outside diameter d and wall thickness t
is subjected to an internal pressure p. Compute displacements and principal stresses.
Use the following numerical values:

E = 30 X 106 psi; v = 0.3; t=12in; d = 120iri; p = 5000 psi

Figure 7.36.

538 ANALYSIS OF ELASTIC SOLIDS

Figure 7.37. X

345
Figure 7.38.

Since the cylinder is long, we can take a unit slice and model it as a plane strain
problem. Thus we need to create a finite element model of the planar region shown
in Figure 7.37. We can further take advantage of symmetry and model one-quarter
of the region. To show all calculations, only a single eight-node element is used, as
shown in Figure 7.38.

Computational Projects

7.10 Determine stresses in the bolted bracket shown in Figure 7.39. Material properties
are E = 29000 ksi and v = 0.3.

12 50k
10
8 20
6
4
2

o
o 5 10 15

Figure 7.39. Bolted steel bracket

Modeling question: To include bolt holes in the model or not? The answer de-
pends on the goal of the analysis. A reasonably accurate stress picture in the plate
can be obtained by ignoring the bolt holes and simply constraining the bolt centers
in the model. To get a more accurate stress distribution in the vicinity of the bolt
holes, one must consider the bolt holes in the model.

7.11 The cross section of a concrete darn is shown in Figure 7.40. Using a planar finite
element model, determine stresses in the dam due to self-weight of the dam and the
water pressure. The depth of water behind the dam is 18 m. The density of concrete

PROBLEMS 539

is 2400 kg/rrr' and that of water is 1000 kg/m'' . The modulus of elasticity of concrete
is 30GPa and Poisson's ratio is 0.15.

Figure 7.40. Concrete dam

7.12 The side view of a pry bar is shown in Figure 7.41. The cross section of the bar

!is rectangular with thickness = in and width (perpendicular to the plane of

=paper) 1.2in. The other dimensions (in inches) are shown in the figure. The

material properties are E = 29 X 106 psi and Y = 0.3. A load of P = 200 lb is applied

at the center of the handle. Using a plane stress model, determine the maximum
deflection and von Mises stress in the bar. To avoid stress concentration, assume the
load to be uniformly distributed over a 3-in length.

p

Figure 7.41. Pry bar

7.13 The shear wall shown in Figure 7.42 is subjected to a horizontal pressure p =
50 leN/m. The dimensions in meters are shown in the figure. The arches have a
radius of 0.5 m. The thickness of the wall is 0.25 m and the material properties are

E = 21 X 106 kN/m2 and y = 0.15. Using a plane stress model, determine the

maximum deflection and von Mises stress in the wall.

540 ANALYSIS OF ELASTIC SOLIDS

----7.8

6.4

4.4
P

--2.8
----0.8
----0

-1.5 0 1.5
Figure 7.42. Shear wall
7.14 A concrete frame with in-filled masonry wall is shown in Figure 7.43. The dimen-
sions in meters are shown in the figure. The beams and columns are 0.3 m thick and
the wall is 0.15 m thick. The material properties of concrete are E = 21 x 106l~/m2

and v = 0.15 and that of the masonry are E = 10 x 106l~/m2 and v = 0.15. The

horizontal load varies linearly from 0 at the top to 10l~/m at the bottom. The dis-
tributed load on the beams is 2kN/m. Using a plane stress model, determine the
maximum deflection and von Mises stress.

q

-4 -10 1 4
Figure 7.43. Frame with in-fill wall

7.15 Consider an aluminum machine part, as shown in Figure 7.44. A load of 1000lb is
applied near the tip. The other end can be considered fixed. The material properties

are E = 10.6 X 106psi and v =0.35.

(a) Frame model-hand calculations: From the geometry, loading, and support
conditions, it is clear that the structure is basically a cantilever beam. Thus a
very simple frame model of the part can be constructed by using the center
line dimensions, as shown by the solid line in Figure 7.45. Determine the
maximum stress using this model.

(b) Analyze the problem as a plane stress model. To avoid stress concentration
due to point load, assume the load to be applied over a length of 1 in.

PROBLEMS 541

10001b (" i\(~': j
5
4

3
2

o 2468 10
o Figure 7.44. Machine part

Ib
5
4
3
2
1

o

o 2 4 6 8 10

Figure 7.45.

7.16 A cast iron flywheel with spokes and rim is shown in Figure 7.46. The flywheel

operates at speeds up to n =4500 rpm. The cast iron weighs 0.2561b/in3 and has an

ultimate strength of 22,000 Ib/irr'. Young's modulus is 30x 106lb/in2 and Poisson's

ratio is 0.3. The goal is to determine the factor of safety against stress failure. The

dimensions of the flywheel can easily be seen from Figure 7.47, which shows the

front view without fillets and a sectional view:

=Shaft radius 2 in =Inside radius 3 in

Rim inner radius = lOin Rim outer radius = 12 in
Fillets near shaft = 1 in radius Fillets at the rim = 0.5 in radius
a = 6.404°

The solution can be approached from a variety of. different assumptions. Several
models are suggested, from a simple analytical model to a fully three-dimensional
finite element model. Your assignment is to use these models and compare solutions.
Your report should contain a discussion on the appropriateness of different models.
Comment on the costlbenefit of these models. Here the cost includes the time for
model preparation, interpretation of results, and computer resources needed for the
analysis. The benefit is the accuracy of the computed safety factor.

542 ANALYSIS OF ELASTIC SOLIDS

Figure 7.46.

8
4

-8
-12

-4-2024

/

Figure 7.47.

(a) Analytical solution based on spokes providing axial restraint to the rim: A
simple analytical solution is possible by considering the flywheel as essentially
a ring with mean radius r. The rim expands under the influence of centrifugal

= =force Fe mrw2lb/in, where m (P/g)Ar is the mass per unit length of the

rim, p Ib/in3 is the density, gin/s2 is the acceleration due to gravity, Ar in2

is the area of the cross section of the rim, and wrad/s is rotational speed of
the flywheel. The spokes provide axial restraint to the rim. The inertia forces
in the spokes are neglected. By considering the rim as a circular beam, it is
possible to derive analytical expressions for axial force (Fr ) and moment (M r )
in the rim and axial force in the spokes (Fs)' For a flywheel with six spokes,
the following expressions are obtained:

F;; = FerHlb

H= 2/3

(0.0203r2/h2) + 0.957 + (AlAs)

PROBLEMS 543

h = rim width in the front view
As = cross-sectional area of the spoke
F,. =Fcr(l - O.866H)lb; u, = O.0889F:,rin ·lb

The stresses can then be computed as follows:

Rim: a-=-FL+ _6M_r , Spoke:
Ar bh2 '

where it is assumed that the rim is of rectangular cross section with width li
(2 in) and thickness b (8 in).

(b) Finite element model using beam elements: A simple finite element model
of the flywheel can be created by using beam elements for both the spoke
and the rim. Create and analyze such a model and compare results with the
previous solutions. Take advantage of symmetry and model only one-fourth of
the flywheel using center line dimensions as shown by the dark lines in Figure
7.48. Use a reasonable number of beam elements to account for the circular
geometry of the rim.

Figure 7.48.

(c) Plane stress finite element model: With the beam element model it is not
possible to determine local stress concentrations at the junctions between the
spokes and the rim. A two-dimensional plane stress model is necessary if one
needs to determine a more complete stress distribution. Use a plane stress el-
ement to model the rim and the spokes. Make the model realistic by creating
fillets between lines meeting at common points, Compare your results with
those obtained from the previous simplified models. Take advantage of sym-
metry and model only one-fourth of the flywhee~; as shown in Figure 7.49.

(d) Three-dimensional finite element model: The plane stress model assumes that
the stresses in the thickness direction are negligible. A three-dimensional fi-
nite element model obviously does not need this assumption. Use an element
equivalent to A.NSYS Solid45 element to create a three-dimensional model

544 ANALYSIS OF ELASTIC SOLIDS

Figure 7.49.
of the flywheel. Compare your results with those obtained from the previous
simplified models. Take advantage of symmetry and model only one-eighth of
the flywheel (cut the wheel in half in the thickness direction also).

/

CHAPTER EIGHT ..... ? '*

W5IH6

TRANSiENT PROBLEMS

8.1 TRANSIENT FIELD PROBLEMS

A variety of transient field problems, such as heat and fluid flow, are governed by a differ-
ential equation of the following form:

where kx' ky' kz, p, q, and I1l are 'known functions of (x, y, z). The solution variable u is
a function both of space (x, y, z) and time i. Except for time dependence, this equation is
a direct extension to three dimensions of the two-dimensional boundary value problem
considered in Chapters 5 and' 6. The possible boundary conditions are as follows:

(i) Essential boundary condition: u specified.
(ii) Natural boundary condition-specified normal derivative along a boundary:

where a and (J are specified parameters along the boundary.

With respect to time, the differential equation is first order arid therefore one initial condi-
tion is also needed in the following form:

= =u(x, y, Z, t 0) uo(x, y, z) specified

545

546 TRANSIENTPROBLEMS

8.1.1 Finite Element Equations
The general form of finite element equations for the problem can easily be written using
exactly the same steps as those used for the steady-state problems in Chapters 5 and 6.
In order to use Galerkin's method, we need to develop an appropriate weak: form first.
Assuming V refers to volume of an arbitrary element, moving all terms in the differential
equation to the left-hand side, multiplying by the weighting functions Nj , and integrating
over the volume, the Galerkin weighted residual for the problem is

i = 1, 2, ...

=where dV dx dy dz is the differential volume. Using the Green-Gauss theorem on the

first three terms in the weighted residual, we have

where S is the surface of the element. On the surface there is a possibility of one of the
two types of boundary conditions to be specified. Designating the surface over which u
is specified as Se and that over which its normal derivative is specified as SII' the surface
integral can be written as follows:

,I

Requiring the assumed solution to satisfy the essential boundary conditions results in
weighting functions' that are zero over Se' Thus with admissible assumed solutions the
weak-form equivalent to the given boundary value problem is as follows:

I f f (-k -aaux a-aNxi - kY-aBuy a-aNyi -k'" aa_uz a-aN-zi + p u N . + q N - maau-t N ) dV
x
I I I

v

ffe+ au + j3)N; dS =0; i =1,2, ...

s,

TRANSIENT FIELDPROBLEMS 547

Rearranging by keeping all terms involving unknown solution on the left-hand side, the
weak form is

III II= qNtdV + f3NtdS; i = 1,2, ...
v 81/

The assumed solution over an element is written as follows:

where ul ' uz, ... are the unknown solutions at the element nodes that are functions of time.
Note the interpolation functions N, are functions only of space variables and are exactly
the same as those used in earlier chapters for steady-state problems. Differentiating the
assumed solution, we get

au =(aNI aNz ("']a;;) I~Z =B;d
ax ax ax u"

aauy =..(.aNayI aNz ("']aaNy,,)";: =BTd
ay Y

U"

au =(aNI aNz a~n) [U, ] =B~d
az az ai- ll{

u"
naaltl
=(N Nz ... N,,) i~l"Z =NT·~·
I

where an overdot indicates differentiation with respect to time. Substituting these into the
weak form, we have

548 TRANSIENT PROBLEMS

=III qNjdV + II f3Nl dS; i =1,2, ...

v s,

This system represents 11equations, one for each Ni' i = 1,2, ... , 11.Writing these equations

explicitly and arranging them in matrix form, we have

IIICI1lNNT)dV d+ IIICk.)1~; +kyByBJ +kzBzBDdV d
vv

- IIICPNNT)dV d - II aCNNl elSd

v s,

=III qNdV + II f3NdS

v8
/1

The three terms inside the second volume integral can be arranged in a more compact form
by using matrices as follows:

where

[~ax ax?!!.J.. ~] C' ~Jax

BT= Wi ay?!!.J.. ay'!!!JJ.
aN2 and c= ~ k0y
az'!!!JJ. 0
~az sz

+Thus the finite element equations are as follows:

IIIcmNN'ldV iJ+[JIIBCB'dV -IIfepNN'ldV -If aCNN)'

=III qNdV + II f3NdS

v s"

TRANSIENT FIELD PROBLEMS 549

Define n x n matrices lep = - III(PNNT)dV

III dV;kk = BeBT v.
v
III =III(JnNNT)dV
lea =- II a(NNl dS;
v
s,

Define n X 1 vectors I IrfJ = f3NdS

r, =III qNdV; A
v

Thus the element equations are a system of first-order ordinary differential equations

Note that, except for the first term, these equations are direct generalizations to three di-
mensions of the corresponding two-dimensional steady-state problem considered in Chap-
ters 5 and 6. The first term involves time derivatives of the nodal variables. Thus the
element equations now represent a system of ordinary differential equations. The assembly
and overall solution process remains the same. 'The global system of ordinary differential
equations must be solved using appropriate numerical schemes such as Runge-Kutta or
Gears methods.

In order to actually evaluate the integrals defining the element equations, we need to
assume an element shape and dev-elop appropriate interpolation functions. Any element
shape can be created as long as it is useful in modeling practical shapes and it is possible
to carry out required integrations arid differentiations. Some typical elements are presented
in the following sections.

8.1.2 TriangularElement

A triangular element is simple yet versatile for two-dimensional problems. Almost any
two-dimensional shape can be discretized into triangular elements. A typical three-node
triangular element is shown in Figure 8.1. The nodal coordinates are (XI' YI)' (xz' Yz), and
(x3' Y3)' Assuming unknown solutions. at the three comers as nodal degrees of freedom,
we have a total of three degrees of freedom, ul , uz, and u3• The following interpolation
functions for the element were developed in Chapter 5:

ue tN, ». N3)[::~J=NTd
u3
1
N I = 2A (xb t + yet + II);

550 TRANSIENTPROBLEMS

n

n
y

'-----------x

Figure 8.1. Three-node triangular element

I j = X2Y3 - X3Y2; 12 =X3Yj -XIY3; =I, XjY2 - X2YI;
b3 =YI - Y2;
b j = Y2 - Y3; b2 =Y3 -Yj;

Assuming a unit thickness, the volume integrals reduce to area integrals A and the surface
integrals to line integrals C. Except for the m matrix, all other matrices and vectors are
identical to those for the steady-state problem discussed in Chapter 5. To evaluate the m
matrix, the integral over the triangle is evaluated using the procedure explained in Chap-
ter 5. The final element equations are as follows:

md + (lck + lc,)d = rq + rp

In = fff(lnNNT)dV = ff(1nNNT)dA = I~ [ 21 211 1)
2

vA 1

and

fflck = BCBT dA = ABCBT

A

kxb jb2 + kyc j c2
??
kxbi + kyci

kxb2b3 + kyc2 c3

TRANSIENT FIELD PROBLEMS 551

[i ;.k =- pA 2i);
p 12 1 . 1

For a nonzero natural boundary condition on side 1,

Length of the side:

II i [2 ~l O~)k(1 = a(NNl dS = a(NNl de = - a~12 ~ -
s,

For a nonzero natural boundary condition on side 2,

Length of the side:

[~° ~k =- aLz3 °12 ) ;
6(1
1

For a nonzero natural boundary condition on side 3,

Length of the side:

[~~° °k =- aL31 1
(1 6 I );

2

The equations are assembled: in the usual manner to get a system of global first-order
ordinary differential equations. After adjusting for essential boundary conditions, the first-
order equations are then solved by one of the many numerical methods, such as Runge-
Kutta and Gears methods, available for solving such equations. The Mathematica function
NDSolve and MATLAB function ode use a variety of these methods to give solutions to
these equations.

8.1.3 Transient Heat Flow

Consider the problem of finding temperature distribution T in an arbitrary three-
dimensional body. Using conservation of energy on a differential volume, the follow-
ing governing differential equation can easily be established:

552 TRANSIENTPROBLEMS

where kx' ky, and k; are thermal conductivities in the x, y, and z directions and Q(x,y, z) is
specified heat generation per unit volume, p is density of the material, and cp is specific
heat of the material. The temperature T is a function both of space (x, y, z) and of time t.
The possible boundary conditions are as follows:

(i) Known temperature along a boundary: T specified.
(ii) Specified heat flux along a boundary:

-kea-nt == - (ka.-.aT-xnx + kys-yat ny + k-7 -aazTn-,) .= q specified

On an insulated boundary or across a line of symmetry there is no heat flow and

thus q = O. The sign convention for heat flow is that heat flowing into a body is

positive and that out of the body is negative.
(iii) Heat loss due to convection along a boundary surface:

where h is the convection coefficient, T is the unknown temperature on the bound-
ary, and Teo is the known temperature of the surrounding fluid.

Both types of natural boundary conditions (ii) and (iii) can be handled by specifying the

coefficients a and f3 in the following form:

If a heat flux is specified, then this boundary condition implies a = 0 and f3 = -q. If
= =convection is specified, then this boundary condition implies a -li and f3 hTeo•

The initial condition is of the following form:

= =T(x, y, Z, t 0) To(x, y, z) specified

Example 8.1 Transient Heat Flow The cross section of a long V-grooved ceramic
strip is shown in Figure 8.2. The sides are insulated while convection heat loss takes place

= =at the bottom with h 200 WIm2··C and Teo 50·C. The grooved surface is maintained at
a constant temperature of 300·C. For ceramic the thermal conductivity is k = 3 W/m' ·C,
= =density p 1600 kg/rn", and specific heat cp 0.8 kl/kg . ·C. Assuming the entire strip

to be at 50·C initially, determine the time history of temperature distribution. Estimate the
time it will take to reach steady state conditions.

Taking advantage of symmetry, we need to model only half of the cross section. In order
to show all calculations, a coarse mesh consisting of only two elements is used. Due to
symmetry, there is no heat flow across side 1-2. Nodes 2 and 4 have specified temperature

=of 300~C. At time t 0 the temperature at all other nodes is equal to 50·C.

The complete calculations are as follows:

TRANSIENT FIELD PROBLEMS 553

y (em) 4
4

3

o
n.r; o 0.5 1 1.5 2 x (em)

Figure 8.2. V-grooved strip and two-element model for half of the section

Global equations at the start of the assembly of element equations:

Equations for element l: q = 0; In = pCp = 1280000

C=(~ ~)

Nodal Coordinates:

Element Node GlobalNQdeNlllTIber .. x

T 1o o

2 3 1 o
3
50 1

4 1 25

50

-X2 _ I. x' - 510
-
50' 3-

Yl =0; Y2 =0; Y3 = -is

Using these values, we get

b- -2I5;' b- 2I5.' b3 = 0
C3 -- 510
1- 2- /3 =0

c1 = 0; C2 -- -5I0.'

/ _ 1. /2 =0;
1 - 1250'

554 TRANSIENTPROBLEMS

Element area A = 1
2500

:)BT=C~ I

25
-55I

['" 128 iza ] k.=[ ~ -315 _o:! ] . ,,=m
"3
=m li8 256 1;8 . 4" 4 '
3'
"3 -43 3

128 128 256 4

"3 "3 "3

NBC on side 1 (nodes {I, 3})-end nodal coordinates: ( (O, 0) (fa, 0)), giving side

length L = fa.

Specified parameter values:

a =-200; f3 =10000
liJ =( 110O0~)
42

!"3 "3

lea =[ ~

Complete equations for element 1:

2~6 128
128
"3
"3
256
[ 128
"3
"3
128

"3

The element contributes to (I, 3, 4) global degrees of freedom.
The global equations after assembly of this element are as follows:

T~ T
[128 0
256
"3
"3
128 0
128
"3
"3

Equations for element 2: q = 0; m =pCp = 1280000

C=(~ ~)

TRANSIENT FIELD PROBLEMS 555

, "Nodal Coordinates: Global Node Number x y
Element Node 1
4 o Q
2 2
3 1 1
23"
50 1

o 50

X2 _ I =X3 0
-
50 _1
Y3 - 50
Y2 =-is

Using these values, we get

b- 510.' b- 5I0.' b3---231"
C3 -- 5I0
1- 2-
13 =0
c- -5I0.' C2 = 0;

1-

1- I . 12 =0;
1 - 2500'

1
Element area A = 5000

-1)-k -kBT _ (
o- I 50 :.

-50 -3 "125

m=[~ l: ].64
'3

128

"3 3 '

64 64 128

'3 '3 "3

Complete equations for element 2:

1~8 64
64
'3
'3
128
( 64
"3
'3
-6Zl

'3

The element contributes to [I, 4, 2) global degrees of freedom.
The global equations after assembly of this element are as follows:

64 128 64 -z22 9 -37" 3

'3 "3 '3 Z
[m+ [~]{~O]128
64 128 0 64 -z9 15 0 -3

'3 "3 '3 "2

128 0 256 128 -37" 0 61 -43 T3 100

"3 "3 "3 T2 T4 0

64 M 128 128 3 -3 -43 9

3 "3 Z 4

Essential boundary conditions:

Node dof Value

2 T2 300

4 T4 300

556 TRANSIENT PROBLEMS

Delete equations (2,4}.

128 64 128 ~3 ) [3~0] =.(. 100)
128
( 3" "3 -- T3 100
"3
o 256 4 300

"3

Extract columns (2, 4), multiply by the corresponding known values, and move to the
right-hand side.

The final global system of equations after adjusting for essential boundary conditions
is as follows:

(¥ -1)(TN .1 281~8)(~Tl)+_L =1) (1000)

( 128 256 3 T3 325
"3
33

Solving the two ordinary differential equations using NDSolve, we get the following tem-

=peratures at the nodes. The results are computed for a time t 300 s. The table lists values

after every 30 s. The actual computations, however, are performed using a sophisticated

automatic adaptive time integration scheme.

Transient Response Analysis

=Initial values: T ~ for all nodes 50

Time history of nodal solution:

Time T1 T2 T3 T4

0 ~O. 300 50. 300
300 88.9721 300
~O 132.624 300 120.496 300
160.102 300 135.335 300
60 172.66 300 142.218 300
90 178.479 300 145.409 300
120 181.176 300 146.889 300
150 182.427 300 147.575 300
180 183.007 300 147.893 300
210 183.276 300 148.04 300
240 183.4 300 148.108 300
270 183.458
300

The temperatures at nodes 1 and 3 are shown in Figure 8.3. After about 200 s the tempera-
tures do not change much and thus a steady-state condition has been achieved.

For realistic results a finer mesh should be used. The problem is solved using ANSYS
with element size of approximately 0.002 m (AnsysFiles\Chap8\VGrooveFine.inp on the
book web site). The finite element mesh, showing the temperature distribution at t = 300 a,
is as seen in Figure 8.4. The time history of temperatures at the bottom is shown in the time
history plot.

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 557

TOe

180 - - Node
-------. Node 3
160 -_.._.._--_.._-_..__..__ _-_..~

"'

140 .-

120
100 / / /

80 /

60 !-¥--~-~-~-----,..-~ t, s

50 100 150 200 250 300

Figure 8.3. Temperature time history at nodes 1 and 3

AN

:l'l"I:.t·l ~:!&~OJI
'I1JioU 17.0J,~!

:::.0 IAWInlle-jOll

:;lUI .U'l.~Ol
1OI1l_JOO

~~ c, SOl ~G~.Jl~ :!B:!. Ul 100

Figure 8.4. Solution using a finer mesh

• MathematicafMATLAB .Implementatlon 8.1 on the Book Web Site:
Triangular element for transient 2D BVP

8.2 ELASTIC SOUDS SUBJECTED TO DYNAMIC LOADS

The stress equilibrium equations presented in Chapter 7 considered a body in static equi-

librium. If a body is in motion, then at any instant of time t Newton's law of motion implies

that the sum of all forces must be equal to the inertia force. Denoting the mass density of
the material by p and the accelerations in the coordinate directions by ii == a2u/at2, ii, and

w, the equations of motion can be written as follows:

-aa-o-x:" + irt + in: + b =pH

-a-.y2 -aEz. x

8i&aTyx + Ba; 8aTZyZ _ .•
+
+by -pv

8;Batr:; + Bt: az"80:,

+ -v b, =p"fil

558 TRANSIENT PROBLEMS

The weak form corresponding to the these differential equations can be obtained following
the same steps as those used in Chapter 7 for a static problem. Denoting the weighting
functions by Ii; ii, and W, multiplying each equation by its weighting function, integrating
over the volume, and adding all three terms, the total weighted residual is as follows:

Iff ( az_axo-+. ...a.T....:2B't..:+~+b ) TI+ (a-Tl::+_aoy-.+-1Bt:.;:+b ) iiy
v ax ay az x ax ay

Iff+ ( aaT;'+ aaT;+ aazoZ+-b.z)WdV- (piiTI+pvii+pivW)dV=O

v

Using the Green-Gauss theorem on each of the stress derivative terms, we get

I I (o;;n., + Txyny + Txznz)U + (TyxnX + O:,ny + TYZnz)ii + (Tzxnx + Tzyny + aznz)W dS
s

On the surface the applied forces are,qx =: o;;nx + T;C),ny + T.'Znz, etc. Substituting these and

rearranging terms, we get the following weak form:

III+ Tyz(:: + ~~)+Txz(: + ~~)dV+ (piiTI+pw+pwW)dV

v

== II(qxTI + qi + qzW)dS + III(bxU + bi + bzW)dV
sv

If we interpret the weighting functions as virtual displacements, then their derivatives are
virtual strains,

ali _ aw _

ay = Ey; az =Ez

av aw _ au aw
az + ax = Yzx
az + ay = 'YyZ;

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 559

Substituting these into the weak form, we have

J J J(O:l, + OJ,~ + O::EZ + TxyYxy + TyZ,YyZ+ TxZYZX) dV + J J J (pHu + pvv + pww)dV

vv

=JJ(qxU+ qyv + qzw)dS + JJJ(bxu + byv+ bzw)dV

Sv

Comparing this weak form to the one derived in Chapter 7 for static problems, we see
that the inertia forces simply add onemore body force term in the equations. By defining
several vectors, the weak form can be written in a compact matrix notation as follows:

where

=U (0;; OJ, 0:: T.'Y TyZ Txzf

€ = (Ex Ey Ez YX)' YYZ Yxzl

q = (qx qy qzl; b =(b, by bf
<

=It (u v w) T ; it = (ii v w{·, U= (u v w{

8.2.1 Finite Element Equations
The assumed solutions for displacements are as follows:

where HI' vI' WI' HZ"" are the nodal degrees of freedom. Here, Ni(x, y, z) are suitable in-
terpolation functions. The accelerations are written by differentiating the displacements.
The interpolation functions do not change with time. The nodal displacements are the only
quantities that are functions of time. Thus the accelerations. are expressed as follows:

)I 0 . ii]

ii=(:iVi)= (N~ NI0 Nz 0 ...... VI =NT(j
0 0 Nz
:wI

0 N] 0 0 Hz

560 TRANSIENTPROBLEMS

From the assumed solution the element strain vector can be computed by appropriate dif-
ferentiation as follows:

aaLxl ~ax 0 0 ~ax 0 0

Ex av 0 ~ay 0 0 ~ay 0 £11
Ey BY
alV 0 0 ~az 0 0 ~ VI
E= Ez 7fi.
'Yxy afy!!!.+a~x az WI =BTd
'YYZ
~az +~ay ~ay ~ax 0 ~ay aaN;, 0 U2
'Y;:.t
f!!!. + alV 0 ~az ~ay 0 ~az ~ay

az ax az~ez 0 ~ax aN, 0 ~ax

Using the constitutive matrix appropriate for the material, the element stress vector can be
written as follows:

a = C(E - EO) = CB Td - CEO

where EO is a vector of initial strains. The weighting functions in the weak form are deriva-
tives of the assumed solution with respect to the nodal degrees of freedom. Thus

adTi= -all =NT and

Substituting these into the weak form, we have

or

fff fff fff ff fffT
pNN
dV(i + T d = BCEOdV + Nq dS + Nb dV
BCB dV

v v v sv

Thus we get the following element equations:

where m is known as an element mass matrix

and all other terms are exactly the same as those for the static case. That is, k is the element
stiffness matrix, r; is the equivalent nodal load vector due to initial strains, rq is the equiv-

alent nodal load vector due to surface forces, and rb·is the equivalent nodal load vector due

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 561

to'body forces:

III IIIIe =
=T r, BCEodV

BCB dV;

vv

II IIIrq = NqdS; rb = NbdV

sv

The mass matrix term is the only new term in the element equations for dynamic problems.
Explicit mass matrices for commonly used structural elements are derived in the follow-
ing section. During assembly, the element mass matrices are assembled to form a global
mass matrix in exactly the same manner as the stiffness matrix. The final equations after
assembly are expressed as follows:

Md+Kd =R

This is a system of second-order ordinary differential equations. Any method for solving
ordinary differential equations can be used to solve these equations. Newmark's method,
generally discussed in courses on vibrations and structural dynamics, is one of the most
popular methods for solving for a large system of second-order ordinary differential equa-
tions. For details of this method refer to any textbook on structural dynamics.

8.2.2 Mass Matrices for Common Structural Elements

The element mass matrix for a general three-dimensional elastic solid is given by the fol-
lowing integral:

Explicit expressions for commonly used structural elements are derived in this section.

Axial Deformations A simple two-node axial deformation element, shown in Figure

8.5, was developed in Chapter 2. The element extends from XI to x2 and has a length

L =x2 - X I . For simplicity the element is assumed to have' a uniform area of cross section A.

The element is based on the following interpolation functions:

_ X-xI ---xx-22 _ -x--x-2 •
N1- - L'

Using these interpolation functions, the mass matrix can be written as follows:

III 1"2 ( ~=In pNNT dV =pA z -X--XI) dx
v~ L
x-x, ) ( - X X

L

562 TRANSIENTPROBLEMS

u

x

Figure 8.5. Axial deformation element

Carrying out integrations, we get the element mass matrix as follows:

(2 1)m=pATL 1 2

Plane Truss Element The two-node plane truss element shown in Figure 8.6 was
developed in Chapter 4 by transforming an axial deformation element. The transformation
works for the stiffness matrix since the strain energy expression for both the truss and the
axial deformation elements is based on deformations along the axis of the element. For
the mass matrix we must consider displacements both in the x and the y directions, since
motion along both axes generates inertia forces. Therefore the mass matrix for a truss
element is derived by writing the linear interpolation functions for x and y displacements.

Assuming a coordinate s along the axis of the element with s = 0 at node land s L,
at node 2 the interpolation functions are as follows:

)2Local element coordinates Global coordinates

/ . .. .....• ./ lv/~V2j
./
/ ~X
..... .i"
, ..;'./ (xl,yd

t /'

:/

Figure 8.6. Plane truss element

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 563

Using these interpolation functions, the mass matrix can be written as follows:

III LL -~" ](-J'm = pNNT dV =pA- Ts-L ~)dSO. ts:
V 0
-sT-t. 0
s

t.

0

Carrying out matrix multiplication and integrating each term, we get

2 0 1 0]

pAL 0 2 0 1

[m=-6- 1 0 2 0

o1 0 2

Space Truss Element The mass matrix for a three-dimensional space truss element,

shown in Figure 8.7, can be written using exactly the same arguments as those for the plane

truss.

The interpolation functions for x, y, and z displacements in terms of a coordinate s along

the axis of the element with s = 0 at node 1 and s =L at node 2 are as follows:

s-L Nz s
=-
N1-- - -L'' L

HI

!~, ~, ~, ~J ~ oN'd
Vz
Wz

Using these interpolation functions, the mass matrix can be written as follows:

200 100

IIJ LL 0 2 0 0 1 0
P~L=In 002001
pNNT dV =pA NNT ds = 100200

v 010020

00 1002

Local element coordinates Global coordinates

Figure 8.7. Space truss element

564 TRANSIENTPROBLEMS
Beam Element The two-node beam element shown in Figure 8.8 was developed in
Chapter 4. The interpolation functions are as follows:

Assuming a uniform cross-sectional area A and using these interpolation functions, the
mass matrix can be written as follows:

Carrying out matrix multiplication and integrating each term, we get

156 22L 54 -13L]
pAL 22L 4L2 13L -3L2
m = 420 [ 54 13L 156 -22L
-3L2 -22L 4L2
-13L

Plane Frame Element In terms oflocal coordinates, the plane frame element shown in
Figure 8.9 is a combination of beam and axial deformation elements. Thus the interpolation

91 q X

XI i'",

s=o w

r X2

s=L

L ..I

Figure 8.8. Beam element

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 565

Global coordinates
Vz

8z /.r--uz

/;:.~
/,~-,

./1,;./

Y t';;:;/;/

"r'/ X
1,.,:","/

81 ~'~

Figure 8.9. Plane frame element

functions are as follows:

dl

(~) (-t L~i)= 0 s 0 dz
0 d3
I d4 =NTd

uz.,3-[}3s2+ 1 L_Z?+s 0 3? z.,3 L2 L ds

L2 L L2 - U

d6

Assuming uniform cross-sectional area A and using these interpolation functions, the mass

matrix can be written as follows:

III LLm =
pNNT dV =pA NNT ds

v

Carrying out matrix multiplication and integrating each term, we get the mass matrix in
local coordinates as follows:

140 0 0 70 0 0

0 156 22L 0 54 -13L
pAr 0 22L 4Lz 0 13L -3Lz
m/ = 420 70 0
0 140 0 0

0 54 13L 0 156 -22L
0 -13L -3Lz 0 -22L 4Lz

The local-to-global transformation is given as

dl t, Ins 0 0 0 0 £II
dz -Ins i, 0 0 0 0 vi
°1d3 01 0 0 0
0 00 0 =? d/ = Td
d4 0 's Ins
Uz

ds 0 0 0 -Ins 's 0 Vz
°id6 0 0 0 0 0 1

l = cos a = Xz - XI . In = sin a = Yz -LYI
s
s L'

566 TRANSIENT PROBLEMS

Using the transformation matrix T, the mass matrix in the global coordinate system is
obtained as follows:

Triangular Element for Plane Stress/Strain The three-node triangular element

shown in Figure 8.10 was developed for plane stress and plane strain problems in Chap-

ter 7. The interpolation functions are as follows:

vUjj

~J(~) =(~j0N20N3 u2 =NTd
0 N2 0 v2
»,

u3

v3

= 11 =N3 1 + YC3 + 13 )
2A(xb3
N2 2A(xb2 + YC2 + 2 ) ;

12 =X3Yj -X1Y3; 13 = x j Y2 - x2Y l

,I

y

L----------x

Figure 8.10. Three-node triangular element

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 567

With constant thickness h and using these interpolation functions, the mass matrix can be

written as follows:

III IIm = T
pNNT dV =ph
NN dA

vA

Carrying out matrix multiplication and integrating each term over the triangle as explained
in Chapter 5, we get .

2 0 10 10

020 101

pM 1 0 2 0 1 0
m=- 0 0 2 0 1
i
12

10 1020

0 10 102

8.2.3 Free-Vibration Analysis

Most structural dynamics and vibration problems typically start with the analysis of the
free-vibration motion of the system. There is no externally applied load in this situation,
and thus the global system of equations for the system is of the following form:

Md+Kd= 0

where the global mass matrix M and the global stiffness matrix K are assembled from
the corresponding element matrices using the usual assembly procedure. This system of
equations is satisfied by a harmonic solution of the following form:

d =¢coswt

where ¢ and ware parameters that will be later identified as the mode shape and the natural
frequency. Substituting this.into the differential equation, we have

-M¢w2 cos on + K¢ cos wt = 0

Rearranging terms, we have what is known as the generalized eigenvalue problem

K¢ =ill¢

where i\. = w2 is introduced for convenience. From the solution of this system we get

.yr;n natural frequencies Wi = and corresponding free-vibration mode shapes ¢i' Several

methods are available for computing these frequencies and modes shapes and are discussed
in books on vibrations and structural dynamics. Both MATLAB and Mathematica have
built-in functions to solve these problems. In MATLAB the eigenvalues and corresponding
eigenvectors are obtained by using the eig command as follows:

=[V, Lam] eig(K, M)

568 TRANSIENTPROBLEMS

In the result Lam is a diagonal matrix of eigenvalues and V is a matrix whose columns are
the eigenvectors (mode shapes). Through Mathematica version 4.2 the Eigensystem func-
tion can solve only the standard eigenvalue problem in which the M matrix is an identity
matrix. When M is nonsingular, the generalized eigenvalue problem can easily be con-
verted to a standard eigenvalue problem by multiplying both sides by M-] as follows:

The function Eigensystem is then used to get the eigenvalues and the mode shapes:

{lam, V} = Ef.gensyat emjInversejrlj.K]

In the result lam is a vector of eigenvalues and V is a matrix whose rows are the eigenvec-
tors (mode shapes). The function returns the eigenvalues ordered in the descending order.
In most structural dynamics applications one is interested in lower frequencies and thus
preferred order is the ascending order. As demonstrated in the following example, using
the Reverse function, the order of the results can easily be changed.

Example 8.2 Modal Analysis of a Plane Truss (MATLAB) The following Transient-
PlaneTrussElement function returns the mass and the stiffness matrix of a plane truss ele-
ment. To generate the mass matrix the function needs the mass density (P).

MatlabFiles\Chap8\TransientPlaneTrussElement.m on the book web site

function [m, k] = TransientPlaneTrussElement(e, A,
rho, coord)

% [m, k] = TransientPlaneTrussElement(e, A, rho, coord)
%Generates mass & stiffness matrices for a plane truss
%element
%rho = mass density
%e = modulus of elasticity
%A = area of cross-section
%coord = coordinates at the element ends

x1=coord(1,1); y1=coord(1,2);
x2=coord(2,1)j y2=coord(2,2);

L=sqrt((x2-x1)~2+(y2-y1)~2)j

ls=(x2-x1)/Lj ms=(y2-y1)/Lj
k = e*A/L*[ls~2, ls*ms,-ls~2,-ls*msj

ls*ms, ms~2,-ls*ms,-ms-2j

-ls-2,-ls*ms,ls~2,ls*ms;
-ls*ms,-ms~2,ls*ms,ms-2]j

m ((rho*A*L)/6)*[2, 0, 1, OJ 0, 2, 0, 1j
1, 0, 2 , 0; 0, 1, 0, 2];

Using this function, we consider the free-vibration analysis of the plane truss structure
shown in Figure 8.11. All members have the same cross-sectional area and are made of the

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 569

15 0:1------+-----:::04

12.5

10

7.5

5

2.5

o
o- - - - - - - - - - ~(ft)

5 10 15 20

Figure 8.11. A plane truss

same material, p = 490 lb/ft", E = 30 x 1Q6 1blin2 , and A = 1.25 in2. The truss 'supports a
rotating machine weighing 2000 lb at its tip.

The weight of the machine is an added mass at node 4. This is defined as rna in the
following code and is added directly to the global mass matrix corresponding to degrees
of freedom 7 and 8 associated with node 4. The procedure for generating the global mass
and stiffness matrix is exactly the same as that for the static analysis used in the previous
chapters. Using in-lb units, the calculations are as follows:

MatlabFiles\Chap8\ModaITrussEx.m on the book web site

%Modal analysis of a plane truss

g = 386.4; e = 30*10-6; A = 1.25;

rho = (490/(12-3))/g; rna = 2000/g; .

nodes = 12 * [0,0; 10,0; -0, 15; 20, 15; 10, 10];

conn = [3, 4; 2, 5; 5, 3; 1, 5; 5, 4];

elems = size(conn,1);

Imm=[] ;

for i=1:elems

lmm = [Lmm ; [2*conn(i,1)-1, 2*conn(i,1),

2*conn(i,2)-1; 2*conn(i,2)]];

end

debe = [1:6]; ebcVals=zeros(length(debc),1);

dof=2*size(nodes,1); .

M=zeros(dof); K=zeros(dof);

%Add nodal masses to the global M matrix.

M(7,7)=ma; M(8,8) = rna;

%Generate equations for each element and asse~ble them.

for i=1:elems
con = conn(i,:);

1m = Imm(i,:);

Em, k] = TransientPlaneTrussElement(e, A, rho,

570 TRANSIENTPROBLEMS

nodes (con, .) j
M(1m, 1m) = M(lm, 1m) + m;
K(lm, 1m) = K(lm, 1m) + kj
end

%Adjust for essential boundary conditions

dof = 1ength(R)j
df = setdiff(1:dof, debe);
Mf = M(df, df)j
Kf = K(df, df)j

%Compute frequencies and mode shapes

[V, lam] = eig(Kf, Mf);
freq=sqrt (lam)
modeShapes = V

» Moda1TrussEx

freq =

55.835 °235.46 °°1598.4 °°°1971.6
°°
°°° °

modeShapes = .I

0.099847 -0.42131 -0.01827 -0.036118
-0.42305 -0.098538 0.0011071 -0.020645
-0.041882
-0.01476 -0.18022 1.6041 1.8044
-0.062462 -1. 8123 1.6046

Example 8.3 Modal Analysis of a Plane Frame (Mathematica) The following Tran-
sientPlaneFrarneElement function returns the mass and the stiffness matrix of a plane frame
element. To generate the mass matrix, the function needs the mass density (P).

TransientP1aneFrameElement[e_, L, a., p.; [qs., qt .], (n1..., n2_ll :=

Modu1e[{EI = e * i, EA = e * a, L, m1,k1, rl, ls, msJ,

L = Sqrt[(n2 - n1).(n2 - n1)];

{ls, ms] = (n2 - n1)/L;

T = {{ls, ms, 0, 0, 0, oj {-ms, Ls, 0, 0, 0, OJ, to, 0,1,0,0, OJ, to, 0, 0, ls, ms, OJ,
{O, 0, 0, -ms, 1s, OJ, to, 0, 0, 0, 0, ill; .

k1 = {{EAlL, 0, 0, -(EAlL), 0, OJ, to, (12 * EI)/L -3, (6 * EI)/L-2,0, -((12 *EI)/L -3),
(6 * EI)/L-2J, {O, (6 * EI)/L -2, (4 * EI)/L, 0, -((6 * EI)/L-2), (2 * EI)/L},
{-(EAlL), 0, 0, EAlL, 0, OJ, to, -((12 * EI)/L -3), -((6 * EI)/L-2),0,

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 571

2

T

L

13 4

I - - L -I' L --1

Figure 8.12. Plane frame

(12 * EI)/L -3, -((6 * EI)/L -2)}, [0, (6 * EI)/L-2, (2 * EI)/L, 0,
-((6 *EI)/L-2), (4 * EI)/L)};

=ml ((p * a * L)/420) * {{140, 0, 0, 70, 0, OJ, {O, 156,22 *L, 0, 54, -13 * L},

[0,22 * L, 4 * L-2,0, 13 * L, -3 * L-2}, {70, 0, 0, 140,0, O},
[0,54,13 * L, 0,156, -22 * L}, {O, -13 * L, -3 * L-2,0, -22 * L, 4 *L-2));
rl = [qs * (Ll2), qt * (Ll2), qt * (L-2112),
qs * (Ll2), qt * (Ll2), -qt * (L-2I12)};

[Transpose[T].ml.T, Transpose[T].kl.T, Transpose[T].rl}

Using these functions, we consider the modal analysis of the plane frame shown in Figure
8.12. The horizontal members carry a weight of 200 N/m in addition to their own weight.
The other numerical data are as follows (use N . mm units):

E =200GPa =200,000Nlmm2 ; = =p 7840kg/m3 7.84 x 10-6 kg/mar':
L = 600 mm; A = 240 mnr';
1= 2000mm4

Additional mass on horizontal members:

200 =20.3874kg/m =0.0203874kg/mm
9.81

L = 1000; e =200000;p =7.84 * 10-(-6);
a = 240; i = 2000;ma = 0.0203874;

nodes = [[0, OJ, {L, O}, {L, -L}, [2L, -L));

=Im[1] [1,2,3,4,5, 6};

Im[2] = {4, 5, 6,7,8, 9};

=Im[3] {7, 8, 9, 10, 11, 12};

[mjl], k[1], r[1]} = TransientPIaneFrameEIement[e, i, a, (p + mala), {O, O},

nodes[[{1,2}]]];

{m[2], k[2], r[2]} = TransientPIaneFrameEIement[e, i, a, (P), {O, OJ, nodes[[{2, 3}]]];

={m[3], k[3], r[3]} TransientPIaneFrameEIement[e, i, a, (p + mala), {O, OJ,

nodes[[{3,4}]]];

M= K = TabIe[O, {i2}, {12}];

Do[M[[lm[i], Im[i]]]+ =m[i];
= =K[[lm[i], Im[i]]]+ k[i]; R[[lm[i]]]+ r[i], [L, 1, 3}];

572 TRANSIENT PROBLEMS

The essential boundary conditions are as follows:

debc = (1,2,3,10,11,12);

ebcVals =Table[O, (6)];

df = Complement [Range[12], debc];

=Mf M[[df, df]]

8.12188 O. 98.56 0.24192 O. -58.24
O. 8.89854 -1166.47 O. 0.3136 O.
98.56 -1166.47 230006. 58.24 O.
0.24192 O. 8.12188 O. -13440.
O. 0.3136 58.24 O. 8.89854 -98.56
-58.24 O. O. -98.56 1166.47 1166.47

-13440. 230006.

Kf = K[(df, df))

-240s02-4 0 2400 - 524 0 2400
0 -240s02-4 -2400 0 -48000 0
-2400 3200000
2400 -2400 -2400 0 800000
- 524 0 -240s02-4 0 -2400
-48000 0 -24s002-4 2400
O 800000 0 2400 3200000
2400 0 -2400

Computation of frequencies arid mode shapes is as follows:

(lams, VsJ = Eigensystem[Inverse[Mf].Kf];

lam = Reverse[l'ams]

,I

{O.331293, 18.4509,19.7104,5921.42,6095.6, 41869.9J

The natural frequencies are the square roots of the eigenvalues:

Sqrt[lam]
{O.575581, 4.29546, 4.43964, 76.9508, 78.0743, 204.621J

The mode shapes are

v = Reverse[Vs]

6.64393 X 10- 7 0.707107 0.000613879 6.64393 X 10-7 0.707107 -0.000613879
-0.0585609 -0.137874 0.691058 0.0585609 0.137874 0.691058
0.00607015 0.000392188 0.707081
0.000392188 0.707081 0.00113752 -0.692288 0.143999 -0.00607015
-0.692288 0.143999 -0.000068896 -0.707018 -0.0112127 -0.00113752
0.0112127 0.00381233 0.0227638 -0.70673 -0.000068896
0.707018 0.70673
-0.0227638 0.00381233

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 573

8.2.4 Transient Response Examples

For structures subjected to dynamic loads the global system of equations for the system is
of the following form:

Md+Kd =R

where the global mass matrix 111, the global stiffness matrix K, and the global load vector

R are assembled from ·the corresponding element matrices using the usual assembly pro-

acedure. For a unique solution initial displacements and velocities at time at all degrees of

freedom must also be specified. Typically, these specified values are zero (the system is at
rest before application of the dynamic load):

d(O) =va

The second-order ordinary differential can be solved in Mathematica by using the NDSolve
function. In MATLAB the solution can be obtained using one of the ode solvers, such as
ode23. The following examples illustrate the use of these functions.

Example 8.4 Transient Analysis of a Plane Truss (Mathematica) Consider solution
of the plane truss structure shown in Figure 8.13. All members have the same cross-

= =sectional area and are of the same material, p 490 Ib/fr', E 30 x 1061b/in2, and

A= 1.25 in2. The truss supports \l rotating machine weighing 2000 lb at its tip. Due to
unbalance in the machine, a harmonic force P = 100 cos(71ft)lb is exerted on the truss.

The weight of the machine is an added mass at node 4. This is defined as rna in the
following code and is added directly to the global mass matrix corresponding to degrees of
freedom 7 and 8 associated with node 4. In defining the global load vector, a load of 1 unit is

=applied at degree of freedom 8. This global load vector is multiplied by 1(t) 100 cos(71ft)

before solving the differential equations of motion. The other procedure for generating the

f t)

15 D:::3-------l--------:::O 4

10

5

o 2 (ft)
o
5 10 15 20
Figure 8.13. A plane truss

574 TRANSIENT PROBLEMS

global mass, stiffness, and load vector is exactly the same as that for the static analysis.
Using in-lb units the calculations are as follows:

f[t] = 100 Cos[77lt];
= = = =g 386.4; e ,;, 30 * 10-6; A 1.25;p 490/123/g;rna 2000/g;
=nodes 12{{0, OJ, {10, OJ, to, 15J, {20, 15J, {10, 10}};
en[1] ={3,4};en[2] ={2, 5};en[3] ={5,3}; en[4] = (1, 5); en[5] = {5,4};
= = =lrn[1] (5, 6, 7, 8); lm[2] {3,4, 9, 10}; lrn[3] {9, 10,5, 6};lrn[4] = (1, 2, 9,10);

lrn[5] = {9,10,7, 8};

=Do[{rn[i], k[i]) TransientPlaneTrussElernent[e, A, p, nodes[[en[i]]]], [L, 1,5)];
M=K =Table[O, {10}, (10)]; M[[8, 8]] =rna;M[[7, 7]] =rna;
R =Table[O, {10lJ;R[[8]] = 1;
Do[M[[lrn[i], lrn[i]]]+ =rn[i]; K[[lm[i], lm[i]]]+ = k[i], [L 1, 5)];

The essential boundary conditions are as follows:

debe =Range[6];

ebeVals = Table[O, {6lJ;

=df Cornplernent[Range[10], debe];
Mf = M[[df, df]]

5.29039 0 0.0205121 o~.0205121].
5.29039
o 0 o 0.170634
0.0205121
[ 0.0205121 0.170634

o o

Kf =K[[df,df]]

379857. 111803. -223607.. -111803. ]
111803. 55901.7 -111803: -55901.7
[ -223607. -111803. 557699. 110485.
-111803. -55901.7 110485. 534789.

=Rf R[[df]] - K[[df, debe]].ebeVals

(O., 1.,0.,0.)

The equations can now be solved using the NDSolve function. To use this function, all
equations and initial conditions must be expressed in the form of a list. This list of equations
is generated as follows:

disp = (u4[t], v4[t], u5[t], v5[t]);
vel = D[df sp, t];

ace =D[vel, t];
eqns =Thread[Mf.aee + Kf.disp == Rf * f[t]]

(379857. u4(t) - 223607.u5(t) + 111803.v4(t) - 111803. v5(t) + 5.29039 u4"(t)

+ 0.0205121 u5"(t) == O. cos(7rct), 111803. u4(t) - 111803. u5(t) + 55901.7 v4(t)
- 55901.7v5(t) + 5.29039v4"(t) + 0.0205121 v5"(t) == 100.cos(7rct),

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 575

,,-223607. u4(t) + 557699. u5(t) -e- 111803. v4(t) + 110485. v5(t) + 0.0205121 u4"(t)
+ 0.170634 u5"(t) == O. cos(77ft), -111803. u4(t) + 110485. u5(t) - 55901.7 v4(t)
+ 534789. v5(t) + 0.0205121 v4"(t) + 0.170634 v5"(t) == O. cos(77Tt)}

initialVel = Thread[(vel/.t ~ 0) == 0]

{u4'(0) == 0, v4'(0) == 0, £15'(0) == 0,v5'(0) == O}

initialDisp = Thread[(disp/.t ~ 0) == 0]

{u4(0) == 0, v4(0) == 0, u5(0) == 0, v5(0) == 0)

sol = NDSolve[Flatten[{eqns, initialVel, initialDisp}], disp, It, 0,1),
MaxSteps ~ 10000]

{{u4(t) ~ InterpolatingFunction[( O. 1.), <>][t],
v4(t) ~ InterpolatingFunction[( O. 1.), <>][t],
u5(t) ~ InterpolatingFunction[( O. 1.), <>][t],
,y5(t) ~ InterpolatingFunction[( O. 1.), <>][t]}}

The NDSolve function returns time histories of nodal displacements in the form of inter-
polation functions. Any of the values can be plotted in the usual manner using the Plot
function, as shown in Figure 8.14. For example, the time history of vertical displacement
at node 4 is as follows:

Plot[v4[t]/.sol, (t, 0, 1), AxesLabel ~ {"t(s)", "disp(in)"},
PlotLabel ~ "Verticaldisplacementatnode4"];

Once the nodal displacements are known, any other quantity of interest, say axial stresses
in elements, can easily be computed using the same functions as those used in the static
analysis case. Since the displacements are functions of time, obviously the element quan-
tities are also functions of time. First we need the complete vector of nodal displacements,
in the original order established by the node numbering. This requires combining the com-

disp (in) ... Vertical displacement at node 4

0.01 t (s)
0.005

-0.005
-0.01

Figure 8.14. Displacement time history

576 TRANSIENTPROBLEMS

puted solution with those specified as essential boundary conditions and is exactly the same
process as used in the static analysis in earlier chapters.

=d Tab1e[0, (iO)];

d[[debc]] =ebcVa1s;
d[[df]] =displ.so1[[i]];

d

(0, 0, 0, 0, 0, 0, InterpolatingFunction[( 0. 1.), <>] [t],
InterpolatingFunction[( 0. 1.), <>][t], InterpolatingFunction[( 0. 1.), <> ][t],

InterpolatingFunction[( 0. 1.), <>llrll

Now, using the PlaneTrussResults function, the solution over any element can be com-
puted. For example, the time history of strain, axial stress, and axial force for the first
element is obtained as follows:

P1aneTrussResu1ts[e_, A_, ({xL., yL), (x2_, y2-ll, [ut., vi..., u2..., v2_)] :=
Modu1e[(ls, ms, L, T, ai, d2, sps},

=L Sqrt[(x2 - xW2 + (y2 - yi)-2];

ls = (x2 - xi)/L;

=ms (y2 - yi)/L;
=T ({ls, ms, 0, OJ, (0,0, Ls, ms]};

[dl, d2) = T.(ui, vi, u2, v2);
eps = (d2 - di)/L;
(eps, e * eps, e *A*eps)]

=e1emSo1 P1aneTrussResu1ts[e, A, nodes[[en[i]]], d[[lm[i]]]]

{2~0 InterpolatingFunction[( 0. 1) <> ][t],

125000 InterpolatingFunction[( 0. 1.), <> ][t],

][t]}156250. InterpolatingFunction[( 0. 1.), <>

Any of the response time histories can be plotted in a usual manner, as shown in Figure
8.15. For example, a plot of the axial stress time history for element 1 is as follows:

P1ot[e1emSo1[[2]], (t, 0, t], AxesLabe1 -7 (Ut(s)", ''IT(lb/in2) ''),
P1otLabe1 -7 "Axia1stressesine1ementi"];

Note that the dynamic displacements, and hence the resulting stresses, are due only to the
unbalance in the machine. For actualdesign of the truss, we must consider displacements
and stresses due to static load of the machine (2000 lb) as well. These quantities can be
calculated as usual (see Chapter 4).

Example 8.5 Transient Analysis of a Plane Frame (MATLAB) Consider solution
of the plane frame in Figure 8.16. The load P varies with time, as shown in the load-time
history. The horizontal members carry a weight of 200 N/m in addition to their own weight.

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 577

o:(lb/in 2) Axial stresses in element 1

300 t (s)
200
100

-100
-200
-300
-400

Figure 8.15. Axial stress time history

T P F(N) Load time history
2
L 50
3
1 20

f--- L 4 -50
L --I

Figure8.16. Plane frame

The other numerical data are as follows (use N . mm units):

= =E 200GPa 200,000N/mm2 ; p = 7840 kg/rrr' = 7.84 x 10-6 kg/mnr'
L =600mm; A ,,;; '240 mnr':
1= 2000mm4

Additional mass on horizontal members;

200 =.20.3874kglm =0.0203874kglmm
9.81

MatlabFiles\Chap8\TransientFrameEx.m on the book web site

% Transient analysis of a plane frame
global Mf Kf Rf
L = 1000; e = 200000; rho = 7.84*10-(-6); a = ~40;

inertia = 2000; rna = 0.0203874;

nodes = [0,0; L, 0; L, -L; 2*L, -L];

conn = [1,2; 2,3; 3,4];

elerns = size(conn,1);

578 TRANSIENTPROBLEMS

Imm=[] ;
for i=1:elems

Imm = [lmm; [3*conn(i,1)-2, 3*conn(i,1)-1,
3*conn(i,1), ... ,3*conn(i,2)-2,
3*conn(i,2)-1,3*conn(i,2)]];

end
debc [1,2,3,10,11,12]; ebcVals=zeros(length(debc),1);
dof=3*size(nodes,1);
M=zeros(dof); K=zeros(dof);
R = zeros(dof,1); R(5)=1;

%Generate equations for each element and assemble them.

for i=1:2:elems
con = conn(i,:);
1m = Imm(i,:);
em, k, r] = TransientPlaneFrameElement(e, inertia,
a, ... , rho+ma/a, 0, 0, nodes(con,:));
M(lm, 1m) = M(lm, 1m) + m;
K(lm, 1m) = K(lm, 1m) + k;
R(lm) = R(lm) + r;

end
for i=2

con = conn(i,:);
Lm = Imm(i,:);
em, k, r] = TransientPlaneFrameElement(e, inertia,

a, ... , rho, 0, 0, nodes(con,:));
M(lm, 1m) M(lm, 1m) + ill;
K(lm, 1m) = K(lm, 1m) +!k;
R(lm) = R(lm) + r;
end

%Adjust for essential boundary conditions
dof = length(R);
df = setdiff(1:dof, debc);
Mf = M(df, df);
Kf = K(df, df);
Rf = R(df) - K(df, debc)*ebcVals;

%Setup and solve the resulting first order differential
%equations

uo = zeros(length(Mf),1);
vO = zeros(length(Mf),1);
[t,d] = ode23('FrameODE',[0,10],[uO; vO]);
plot(t,d(:,2)); xlabel('time'); ylabel('Disp');
title('Vertical displacement at node 2');

ELASTIC SOLIDS SUBJECTED TO DYNAMIC LOADS 579

MatlabFiles\Chap8\FrameODE.m on the book web site

function ddot = FrameODE(t, d)

%ddot = FrameODE(t, d)
%function to set up equations for a transient frame,problem

global Mf Kf Rf
if t <= 0.6

ft = 50*t + 20;

elseif t <= .85

ft = 20;

elseif t <= 1.2
ft = 190 - 200*t;

elseif t <= 1.4
ft 250*t - 350;

else

ft O,'

end

end

end
end

n=length(d);
u = d(1 :n!2) ;
v = d(n!2+1:n);
vdot = inv(Mf)*(Rf*ft ~ Kf*u);
udot = v;
%format short g

%soln=[t, ft, u(2), udot(2), vdot(2)]

ddot ='[udot; vdot];

Executing the script file TransientFrameEx, the plot of vertical displacement at node 2
, shown in Figure 8.17 is obtained. The script file can be modified to plot any other quantity
of interest.

disp (mm) Vertical displacement at node 2

2

t (5)

2 4 6 8 10
-1

-2

Figure 8.17. Displacement-time history

580 TRANSIENTPROBLEMS

PROBLEMS

Transient Heat Flow

8.1 A square duct shown in Figure 8.18 is insulated by a layer of square fiberglass that

has a thermal conductivity of 0.035 W/m' °C, density p = 144 kg/m", and specific

=heat Cp 753 J/kg . °C. The convection coefficient of the outside surface of the

fiberglass is h = 45 W1m2 • °C. Initially the duct and the insulation are at the room
temperature of 2YC. A flow of hot gases suddently raises the inside duct temper"
ature to 1000°e. Determine what is the temperature distribution in the insulation
10 min after the flow of hot gases begins. Roughly how long will it take for the tem-
perature to reach asteady-state condition. Take advantage of symmetry and model
only one-eighth of the section, shown in dark shade. Assume that the temperature
on the inside surface is the same as that of the hot gases.

25°C 3

, Figure 8.18.

/"

8.2 Heating wires are embedded in a concrete slab, as shown in cross section in Figure
8.19, tokeep snow from accumulating on the slab. In a proposed design wires are
placed at 2 em from the top and at 4-cm intervals. The heat generated by each wire
is 10W1m length. The bottom of the slab is insulated. The top is exposed to a
convection heat loss with a convection coefficient of h = 30 W1m2 •°C. For concrete

=the thermal conductivity is l.046W/m· °C, density p 2300kg/m3, and specific
=heat Cp 656.9 J/kg . °C. Determine the slab surface temperature when the air

temperature is -YC. Assuming the entire slab is initially at the same temperature as
the air, how long will it take to raise the temperature of the surface to above freezing

T

6cm

1

Figure 8.19. Heating wires embedded in concrete

PROBLEMS 581

after starting the heating process? Note that, because of symmetry, we need to model
only the 2 em x 6 em dark shaded area. There will be no heat flow across the sides
of this area. The heating wire represents a concentrated point heat source.

8.3 The cross section of a brick masonry chimney is shown in Figure 8.20. The thermal

conductivity of brick is 0.711 W1m . ·e, density p = 2000 kg/m", and specific heat
=Cp 836.8 Izkg-"C.The convection coefficient of the outside surface of the chimney
is h = 15 W1m2 •"C, Initially the chimney is at the room temperature of 10·e. A flow

of hot gases suddently raises the inside chimney temperature to 250·e. Determine
what is the temperature distribution in the insulation 5 min after the flow of hot gases
begins. Roughly how long will it take for the temperature to reach a steady-state
condition. Take advantage of symmetry and model only one-eighth of the section,
shown in dark shade. Assume that the temperature on the inside surface is the same
as that of the'hot gases.

3

Figure 8.20.

Free Vibrations
8.4 Determine natural frequencies and mode shapes for an axially loaded bar fixed at
the left end and restrained by a spring at the right end, as shown in Figure 8.21.
Divide the bar into two equal-length elements and use the following numerical data:
E = 200 X 109 N/m2; p = 7500 kg/m''; A = 1.6 X 10-4 m2

L =2.5 m; =k 1 X 107 N/m

Assume the spring is massless.

Figure 8.21.

582 TRANSIENT PROBLEMS

8.5 Determine free-vibration frequencies and mode shapes for the three-bar pin-jointed

structure shown in Figure 8.22. All members have the same cross-sectional area and
are of the same material, p == 490lb/ft3, E == 30 x 106lb/in2, and A == 1 in2. The

dimensions in inches are shown in the figure.

72

oI()I----------7-,
o 108 (in)

Figure 8.22.

8.6 Determine free-vibration frequencies and mode shapes for the three-bar pin-jointed

structure shown in Figure 8.23. All members have the same cross-sectional area and

are of the same material, p == 490 lb/fr', E == 30 x 106lb/in2, and A == 2 in2 . The

attached mass is m == 77.6lb· s2/in. The dimensions in ~ are shown in the
-\QQ. t .
figure.

Mass,m Mass, m

6

6

,I

oI ~L}-------,Gll o
o o- - - - - - - - - (m)

Figure 8.23. 8
Figure 8.24.

8.7 Determine free-vibration frequencies and mode shapes for the truss shown in Figure
8.24. Note the diagonals are not connected to each other at their crossing. The cross-
sectional area of vertical and horizontal members is 30 x 10-4 m2 and that for the
diagonals is 10 x 10- 4 m2. All members are made of steel withp == 7850kglm3 and
E == 210 GPa. The attached masses are m == 500 kg. The dimensions in meters are
shown in the figure.

8.8 Determine free-vibration frequencies and mode shapes for the truss shown in Figure
8.25. The cross-sectional area of members is 15x 10- 4 m2. All members are made of