Solution to Final Exam

tobascothwackΠολεοδομικά Έργα

15 Νοε 2013 (πριν από 3 χρόνια και 11 μήνες)

124 εμφανίσεις

CEE 371: Modeling of Structural Systems


May 19, 2004



SOLUTION TO FINAL EXAMINATION


1.

[60 points total]

The 500
-
foot communications tower
1

shown in Figure 1 on the next page is
three
-
dimensional with a total of 4 guy cables at 90° intervals, but we will consider only the two
-
dimensional case illustrated.
In this view, the guy cables are uncharacteristically at unequal
angles from the horizontal because appropriate foundation conditions (near
-
surface bedrock) for
anchoring the cables are available for only the arrangement shown. The mast structure consists

of four (4) full
-
height legs that are circular tubes (pipes), as shown in Section A
-
A of Figure 1,
and these legs act together as truss chords because trusswork webs connect them. The outside
diameter of each cylindrical leg is 5 in., and the wall thickn
ess is ½ inch. All four legs are
securely bolted to a concrete foundation, so the base of the mast can be considered fully fixed
against translation and rotation. All members and cables are steel with
E

= 3 x 10
4

ksi. The total
dead load of the mast is
0.12 k/ft over the full 500
-
foot height. A wind loading is modeled by the
triangular distribution shown with an intensity of 0.10 k/ft at the top of the mast.
2



(a)

[10 points] We wish to convert the mast truss
a
-
b

to a single equivalent beam
-
column
elemen
t. Find the effective cross
-
sectional area,
A

=
A
eff
, and moment of inertia,
I
z

=
I
eff
.
Consider only the four tubular chords of the mast truss in these calculations. In Section A
-
A
of Figure 1, “O.C.” means “on centers.” [See also the circular
-
section

properties appended.]


(b)

[15 points] We decide to simplify our model by removing the top cantilever
b
-
c

and
replacing it with an equivalent set of forces applied at point
b
. Draw a free body diagram of
this cantilever, calculate its reactions, and draw th
e shear and bending moment diagrams.
Remember to include the uniformly distributed axial (dead) load as well as the wind load.


(c)

[5 points] In preparation for analysis by the Direct Stiffness method, sketch a structural
model with nodes at
a
,
b
,
d
, and
e
,

showing appropriate local (element) and global coordinate
systems. Use the global
x
-
y

coordinate system shown in Figure 1. Determine the degree of
kinematic indeterminacy of the structural model,
n
. [Remember, the cantilever
b
-
c

has been
removed.] Do
NOT neglect axial deformation of the guy cables and mast. Write the vector
of unknown displacements of the free degrees of freedom
, {Δ
f
}, with an appropriate symbol
to abbreviate each entry.


(d)

[10 points] Evaluate numerically the vector of applied loads, {
P
f
}. Remember to include the
equivalent loads from part (b). Use the global coordinate system shown in Figure 1.


(e)

[20 points]

Assemble the stiffness matrix [
K
ff
] in symbolic (not numerical) form. Use the
following symbols:
A
c

= the area of both cables,
L
bd

and
L
be

= the lengths of the cables,
L

=
the length of the mast
a
-
b
,
A

(=
A
eff
) = the effective area of the mast,
I

(=
I
z
=
I
eff
) = the
effective moment of inertia of the mast, and
E

= Young’s modulus of both the mast and the
cables. Be sure to show on a sketch of the model the local coordinate system for each
element. Use the global coordinate system shown in Figure 1.




1


This problem is in memory of “Climbing Hand” Nick Berg ’00, whose business, Prometheus Towers
, installed and
maintained communication structures.

2


One design objective is to determine what pretensioning is required in each cable to ensure that the guy cables never
slacken during a windstorm from any direction.

CEE 371: Modeling of Structural Systems

-
2

of
13
-

Solution to Final Exam, May 19, 2004



Figu
re 1 for Problem 1
:




L = 300'

200'

45°

60°

y

x

a

b

a


c

d

e

Cable
b
-
e

A
c
= 3 in
2

Length =
L
bc

Cable
b
-
d

A
c
= 3 in
2

Length =
L
bd

WL = 0.1 k/ft

DL = 0.12 k/ft

(over full height)

y

z

4' O.C.

4' O.C.

A

A

Sectio
n A
-
A



CEE 371: Modeling of Structural Systems

-
3

of
13
-

Solution to Final Exam, May 19, 2004





CEE 371: Modeling of Structural Systems

-
4

of
13
-

Solution to Final Exam, May 19, 2004




CEE 371: Modeling of Structural Systems

-
5

of
13
-

Solution to Final Exam, May 19, 2004





CEE 371: Modeling of Structural Systems

-
6

of
13
-

Solution to Final Exam, May 19, 2004



2.

[60 points total]

The reinforced
-
concrete “folded plate” roof shown in Figure 2 on the next page
is efficient for spanning large clear spaces. One way to model the behavior of this roof is
as a
series of parallel simply supported beams with the inverted V cross
-
section shown, and we shall
consider only one such interior unit spanning 60 feet, shown in bold lines and in cross section in
Figure 2.


You know how to calculate the moment of iner
tia of the inverted V cross section
shown in Figure 2 using the sectional properties appended here and the parallel axis theorem, and
the result of this calculation is
I
x

= 6.84 x 10
4

in
4
. For concrete,
E

= 3 x 10
3

ksi. Note that the
neutral axis (N.A.)
and the
x
-
axis coincide for the inverted V section in Figure 2.



(a)

[15 points] The uniformly distributed loads


on the beam are DL = 0.6 k/ft and LL = 0.4 k/ft
for a total uniformly distributed load of 1.0 k/ft. Find the reactions of the simply supported
beam and sketch the shear and bending moment diagrams, indicating their maximum
magnitudes.


(b)

[20 points] Find the mid
-
span deflection of this beam using the unit dummy load version of
the Principle of Virtual Forces. Include only bending (flexural) effec
ts,
i.e
., neglect shearing
deformations. Show your work clearly.


(c)

[10 points] Find the principal stress and the maximum shear stress at the crown of the cross
-
section at midspan of the simply supported beam. Use a sketch of Mohr’s Circle to aid your
cal
culations. [See Figure 2 for the location of the crown. Consistent with Euler
-
Bernoulli
beam theory, assume that normal stresses in the transverse direction,
σ
y
, are zero everywhere
in a cross section.]


(d)

[15 points] Find the shear flow, shear stress, an
d principal stresses at the neutral axis (N.A.)
very near to one end of the beam. Choose sections at the N.A. that are perpendicular to each
plate as shown in Figure 2. Use a sketch of Mohr’s Circle to aid your calculations.
[Consistent with Euler
-
Berno
ulli beam theory, assume that normal stresses in the transverse
direction,
σ
y
, are zero everywhere in a cross section.]








At each end of the 60
-
foot spa
n, the cross
-
sections are stiffened by a vertical membrane that prevents distortion of the
cross sections at the ends. Calculations of the stresses at various locations in this beam model can be used both to
check the adequacy of the dimensions and to hel
p design the reinforcement. Following this preliminary analysis with
the beam model, a more complete finite element analysis that includes 3D loadings and effects would normally be
carried out to verify or modify the design.



Obviously, other load cases
and combinations would need to be considered in the design of this structure.

CEE 371: Modeling of Structural Systems

-
7

of
13
-

Solution to Final Exam, May 19, 2004



Figure 2 for Problem 2
:



Crown

Sections

at N.A.

21.2"

21.2"

2 @ 42.4" = 84.8"

50"

5"

5"

5"

x

y

N.A.

45°

45°

5"

60'
-
0"



CEE 371: Modeling of Structural Systems

-
8

of
13
-

Solution to Final Exam, May 19, 2004



CEE 371: Modeling of Structural Systems

-
9

of
13
-

Solution to Final Exam, May 19, 2004



CEE 371: Modeling of Structural Systems

-
10

of
13
-

Solution to Final Exam, May 19, 2004



3.

[30 points total]

Note: the following solutions to the written questions


parts (a) and (c)


are somewhat more complete and detailed that would

be expected from a student during the heat
of an exam.


(a)

[6 points] Briefly describe and explain some of the
professional responsibilities and
obligations

of structural engineers in the structural design enterprise (including structural
modeling and stru
ctural analysis).

Structural engineers have a professional responsibility for life
safety

of those who use their
designs, making this the primary of the four main objectives of structural design. They have
an obligation to strive for the fulfillment of th
e other three objectives as well


serviceability
(can perform desired function), feasibility (economical and can be built), and aesthetics. In
the detailed aspects of structural design


including structural modeling and structural
analysis


they have t
he obligation to strive for accurate and realistic results. Models should
have a strong relation to reality and the bounds of various assumptions should be examined
and explored. Analyses should be carefully checked, computer programs should be used
prop
erly, and all numerical results from hand and computer calculations should be verified
because the engineer must take responsibility for them. In addition, the structural engineer
must continue to pursue life
-
long learning as the state of the art evolves
and new information
and techniques become available; and this obligation extends to monitoring the long
-
term
performance of designs one has made in order to judge their success and to learn from their
weaknesses. Finally, the structural engineer should be

aware of, and adhere to the overall
code of ethics applicable to his or her profession.

(b)

[6 points] What is the degree of static indeterminacy of the 4
-
node structural model used in
Problem 1, parts (c) through (e)? Choose an appropriate number of redund
ants and sketch a
primary (released) structure that you would use to analyze this structure by the Force
Method.

The degree of static indeterminacy,
d
s

= 2
, may be evaluated by any one of at least three
ways:



By inspection


seeing that by cutting the two
cables, one obtains a statically determinate
(vertical) cantilever beam


a logical choice also for a primary structure.



By observing that the structure is externally indeterminate and that there is only one
independent reaction at the ground support of ea
ch cable, yielding 5 unknown reactions
to be sought with the three equations of equilibrium.



By counting the total number of reaction components (
7
) plus the number of unknown
member internal forces (
2

cables @
1

+ one beam @ 3
= 5
) and comparing this sum
of
12

with the number of equations of equilibrium available at each joint, including
supports (
2

cable/truss supports with
2

equations of equilibrium +
2

beam ends @
3

equations
= 10
), yielding d
s

= 12


10 = 2
.

A common error on this portion is to use the

equation for determinacy of a pure truss
structure d
s

= (m + r)


2
j, which, depending on how one counts, may yield the correct
answer but by the wrong reasoning.


The primary (released) structures for this problem are shown on the next page.

CEE 371: Modeling of Structural Systems

-
11

of
13
-

Solution to Final Exam, May 19, 2004



CEE 371: Modeling of Structural Systems

-
12

of
13
-

Solution to Final Exam, May 19, 2004




(c)

[3 parts

at 6 points each] In a few sentences for each, explain and give the significance of
only three

of the following seven items (your choice).
Extra credit will be awarded if you
answer more than three of these!

(Be sure to re
-
state the chosen letter and i
tem name at the
beginning of each of your answers.)

(i)

Internal virtual work vs. External virtual work

The Principle of Virtual Work (PVW) may be stated concisely as “External Virtual
Work equals Internal Virtual Work.” There are two manifestations of the PV
W, the
Principle of Virtual Forces (PVF) and the Principle of Virtual Displacements (PVD).
The PVF was studied in this course and is used to compute displacements, especially
in the form of the Unit Dummy Load Method. In the PVF, the external virtual wor
k is
that done by an equilibrium external force/load system undergoing real
displacements, e.g., the product of a virtual (dummy) unit load and the real
displacement at the location and direction of the dummy load; and the internal virtual
work is the inte
gration over the total volume of the structure of the intensity of virtual
stresses (in equilibrium with the external virtual forces) times the intensity of real
strains (corresponding to the real displacements and usually expressed in terms of the
real st
resses or internal actions), i.e., the integration over the entire volume of the
virtual strain energy density.

[In the PVD, used to find forces or stiffnesses, the external virtual work is the real
external loads time the corresponding (compatible) virt
ual displacements, and the
internal virtual work is the integration over the volume of the virtual strain energy
density derived from the product of the intensity of virtual strains (compatible with
the external virtual displacements) and the real stresses
.]

(ii)

Allowable Stress Design (ASD) vs. Limit State Design (LSD or LRFD)

These are the two most common general approaches to structural design. ASD
computes stresses at working loads, usually by linear analysis, and compares them to
allowable stresses appr
opriate to the material and component type.


working allowable
 


The ASD factor of safety is approximated by the ratio of failure stress to allowable
stress. LSD applies factors generally greater than unity (γ
i
) to the loads (Q
i
) under
considera
tion in a load combination and compares them to the nominal resistance
(R
n
) factored by a capacity reduction factor (φ):


i i n
i
Q R
 



The LSD factors are based on statistical studies of uncertainty in loads and
resistance. The calculatio
n of nominal resistance often requires nonlinear analysis,
and the factor of safety is the ratio of the failure load to the applied load.

LSD is more modern and rational than the traditions ASD method.

(iii)

Plane stress vs. Plane strain

These are two modals of
planar structural behavior in the plane of the loads (x
-
y).
For both, loads are limited to occur only in the x
-
y directions. Plane stress considers
structures that have relatively small thicknesses in the z
-
direction and have loads
symmetric about the x
-
y plane; it is assumed that


0 0
z xz yz xz yz z
and
     
     

Plane strain considers a unit
-
thick slice of a structure long in the z
-
direction and with
x
-
y loadings not a function of z; it is assumed that


0 0
z xz yz xz yz z
and
     
     

CEE 371: Modeling of Structural Systems

-
13

of
13
-

Solution to Final Exam, May 19, 2004



For both,
0,0,0
x y xy
  
  
. Plane stress and plane strain have the same
equilibrium and strain
-
displacement (kinematic) equations but different constitutive
equations.

(iv)

Kinematics (strain
-
displacement relationships) vs. Equilibrium

These are two of the three basic types
of equations of structural mechanics (the third
are constitutive equations). An important distinction between the two is that
equilibrium is based on a
physical principle

(Newton’s Laws) while strain
-
displacement relations are
defined
, with different defi
nitions appropriate to different
kinematic models. Equilibrium equations, in their pure forms, deal with forces and
stresses; and they are necessary and sufficient for the analysis of statically
determinate structures. Kinematics equations, in their pure

forms, deal with
displacements and strains.

(v)

Strength limit states vs. Serviceability limit states

These are generic forms of limit states that must be considered in design. Strength
limit states deal with structural failure by such means as collapse, buc
kling, and
fatigue fracture; and their computation usually requires nonlinear analysis.
Serviceability limit states deal with insufficient ability of the structure to fulfill its
desired function such as by excessive deflection or vibration; and their com
putation
can often be accomplished by linear analysis.

(vi)

Load factors vs. Resistance (strength, capacity) factors

These are numerical factors that are used in LSD [see (ii) above]. They are based on
consideration of the uncertainties in loads and resistance

and are chosen statistically
to give an acceptably low probability of failure. The load factors

i
) are generally
greater than unity, the resistance factors (φ) are generally less than unity, and they
appear in the LSD relationship:


i i n
i
Q R
 



(vii)

Material nonlinear (inelastic) analysis vs. Geometric nonlinear (2
nd

order) analysis

Thes
e are the two fundamental forms of nonlinear structural analysis. Material
nonlinear analysis takes into account the nonlinear behavior of material such as
inelasticity, crushing, yielding, and fracture of materials. Geometric nonlinear
analysis takes in
to account equilibrium in the current displaced shape due to finite
displacements and rotations. Analysis that includes both is considered “fully
nonlinear.”