mech3300-1

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15 Νοε 2013 (πριν από 3 χρόνια και 9 μήνες)

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MECH3300

ASSIGNMENT ON ENERGY

METHODS OF STRUCTURA
L ANALYSIS

Submit questions 2 to 4. Questions 1 and 5 are easier practice questions you should attempt first.

1.

The terms describing bending in the stiffness matrix of a beam can be determined by applying

a unit transverse displacement or a unit rotation to either end. The forces and moments each
end of the beam reacting a unit imposed displacement give terms in the column of the matrix
that multiplies that displacement.

Similarly, forces and moments each e
nd reacting a unit rotation give terms in the column that
multiplies that rotation. In order to give a unit displacement, but no rotation, to give the right
-
hand end of the beam below, a force and a moment must be present. Use an energy method
to first fin
d
M
A

and hence show that the diagonal stiffness term relating
F
A

to
Δ
A

is
3
12
L
I
E



and the off
-
diagonal term relating
M
A

to
Δ
A

is
2
6
L
I
E



(see fig 1.1).

M
A
F
A

A
L
F
B
M
B
B
A
x
y

Fig. 1.1

Now consider the situation below (fig. 1.2). Show by an energy method that the stiffness
term relating
M
A

to
θ
A

is
L
I
E


4
. Show, using statics, that the term relating
M
B

to
θ
A

is
L
I
E


2
.

Hint:

find
F
A

first.

M
A
M
B
F
A
L
F
B
A
B
x
y

Fig. 1.2

Solution.

The case shown in the fig. 1.1.

The bending moment in each point is
A
A
b
M
x
F
M




and
1




A
b
M
M
.

F
rom which we have
















L
L
A
A
A
b
b
A
dx
x
F
M
I
E
dx
M
M
M
I
E
0
0
0
1
1

, therefore
0
2
2




L
F
L
M
A
A

or
2
L
F
M
A
A


.

The following expressions are valid in this case:






























L
A
A
A
A
A
L
A
b
b
A
A
I
E
L
F
I
E
L
M
I
E
L
F
dx
x
M
x
F
dx
F
M
M
I
E
F
U
0
3
2
3
2
0
12
2
3
1


and
3
12
L
I
E
F
A
A




,

A
A
A
L
I
E
L
F
M







2
6
2
.

Now let’s see the case

shown in the fig. 1.2.

There are
x
F
M
M
A
A
b




and
x
F
M
A
b




in this case.

From which we have



















L
L
A
A
A
b
b
A
dx
x
F
x
M
I
E
dx
F
M
M
I
E
F
U
0
0
2
0
1
1
,

therefore
3
2
3
2
L
F
L
M
A
A




or
L
M
F
A
A


2
3
.

The following expressions are valid in this case:




















L
L
A
A
A
b
b
A
A
dx
x
F
M
I
E
dx
M
M
M
I
E
M
U
0
0
1
1

, therefore

I
E
L
M
L
M
L
M
I
E
L
F
L
M
I
E
A
A
A
A
A
A































4
1
4
3
1
2
1
2


or
L
I
E
M
A
A



4

.

The sum of forces projections onto the
Y

axis is
0


B
A
F
F

in the considered static case.

The sum of moments about the point
B

is
0




L
F
M
M
A
B
A

in the considered static case.

But,
A
A
M
L
F



2
3
, so
A
A
B
L
I
E
M
M







2
2
1
.

The answer:

A
A
L
I
E
F





3
12
,
A
A
L
I
E
M





2
6
,
A
A
L
I
E
M





4
,
A
B
L
I
E
M





2

2.

A proving ring is commonly used to monitor large for
ces, by measuring the reduction in diameter

of a ring compressed in the direction of loading. Treating the ring a beam loaded mainly

in bending, develop an expression for the stiffness of the ring (i.e.
diameter
in
change
P


for the square cross
-
sectio
n shown in the fig. 2
-
1.

P
P
d
D

Fig. 2.1

A
B
O
M
b
P
2
P
2
M
M
b
R



Fig. 2.2


1



Hint:

exploiting symmetry, we can model
4
1

of the ring, exposed to half the applied load P.
There are no shear forces on the horizontal and vertical planes of symmetry, as these, if
pr
esent, would lead to unsymmetrical deflections. There is however an unknown bending
moment M, which must be found before the deflection due to
P

can be found. That is, first
use
0



M
U
= slope at
A

or
B

to relate
M

to
P
. Note that for the
w
hole

ring
P
U


= the sum of
deflections due to both loads
P

= diameter change.
P
1

= P /
2
.

Solution.

The basic solution is easier to perform, using a free body above the cut at angle

.

The bending moment in each point is
M
R
P
M
b





sin
2


and
1



M
M
b
,
1
1
P
P
P
M
P
M
b
b








.

From which we have
0
sin
2
2
0















d
R
M
R
P

therefore
0
2
2
2






R
M
R
P

and

R
P
M


.

Change in radius is the followin
g.


























d
R
R
M
R
P
I
E
P
U
R
2
0
1
sin
sin
2
1


































2
3
2
0
2
2
3
8
1
sin
sin
2
1
R
M
R
P
I
E
d
R
M
R
P
I
E



















1
8
3
I
E
R
P
.

Therefore, change in diameter is

















2
4
2
3
I
E
R
P
R
D
.

A check solution is performed for the other side of the cut.

The bending moment in each point is


1
1 cos
2
b
P
M R M

    

a
nd
1
1
b
M
M

 

,
1
1
P
P
P
M
P
M
b
b








.

Therefore



cos
1
2





R
P
M
b
,

0
2
0













d
R
M
M
M
M
U
I
E
b
b

then


2
1
0
1 cos 0
2
P R
M R d

 

 
    
 
 


or
1
0
2 4 2
M P R
P R
 
 
    
. Thus
1
1
4 2
2
P R
M


 
  
 
 


or
1
1 1
2
M P R

 
   
 
 
,





2 2
1
0 0
4 4 1 cos 1 cos
2 2
b
b
U M P R R
E I M R d M R d
P P
 
   
  
 
               
 
 
 
 
































2
0
2
2
2
cos
1
4
cos
1
1
2
1
2
4





d
R
R
P
R
P
































4
2
2
4
2
4
4
8
4
2
2
2
2
2




R
P
R
P
R
P
R
P
R
P
R

I
E
D
R
P
R
P
R
P






















3
3
3
4
2
4
2




,















2
4
3
I
E
R
P
D
.

Then the stiffness of the ring can be found from the expression:

3
3
3
721
.
6
1488
.
0
2
4
R
I
E
R
I
E
R
I
E
D
P



















.

But
12
4
d
I


and
2
D
R

, so
3
4
3
4
481
.
4
12
8
721
.
6
D
d
E
D
d
E
D
P









The answer:

3
4
481
.
4
D
d
E
D
P




.

3.

A portal frame is a common type of statically
-
indeterminate structure. Consider the frame
below, exposed first to a distributed vertical force
w

per length and then equal horizontal
forces
Q
, as shown below

in the fig. 3.1. Develop expressions for the worst bending moment
present in both cases, ignoring strain energy due to axial loading of the beams. All members
have the same
E

and
I
.

L
L
Q
Q
w

Fig. 3.1

Hint:

half the structure can be modelled, to find an internal
unknown loading on the plane of
symmetry. With the loading
w
, there is no shear force on the plane of symmetry. With the
loads
Q
, antisymmetric deformation occurs, so there is no bending moment on the plane of
symmetry (i.e. zero curvature there).

With sym
metric loading, there is no deflection at the plane of symmetry normal to it and no
slope there either. With anti
-
symmetric loading, there is no deflection in the plane of
symmetry. Sketch the deformed shapes to clarify this.

In the symmetric case, these c
onditions (




0
M
U
slope and




0
R
U
horizontal deflection;
see fig. 3.2) lead to compatibility equations of the form

I
E
L
w
F
E
L
R
M
D
C
B
A
I
E
L


























3
,

where

M

and
R

are the bending moment and axial load on the plane of symmetry,

A

to
F

are constants.

M
Q
w
L
L
2
R
x
y
A
B
C
V

Fig. 3.2

In the anti
-
symmetric case, there is only one compatibility equation to write (
0



V
U
, where
V

is the vertical shear force on the plane of symmetry).

Solution.

The case of symmetric loading due to
w

(see fig. 3.
2).

For the beam
BC
,
2
2
x
w
M
M
b



.

For the beam
AB
,
y
R
L
w
M
M
b





8
2
.

0
8
2
0
2
2
0
2































L
L
dy
y
R
L
w
M
dx
x
w
M
M
U
I
E

as
1



b
M
U

in both terms.

0
2
8
48
2
2
3
3










L
R
L
w
L
M
L
w
L
M

or
2
48
7
2
1
2
3
L
w
L
R
M








(1)

The integral over
AB

(only) gives the following

expression:

0
3
16
2
8
3
4
2
0
2


























L
R
L
w
L
M
dy
y
R
L
w
M
y
R
U
I
E
L
.

Therefore
2
16
1
3
1
2
1
L
w
L
R
M







,

(2)

or































16
1
48
7
3
1
2
1
2
1
2
3
2
L
w
L
R
M

from (1) and (2).

The solution of the latter system of linear equations gives

2
2
08333
.
0
06944
.
0
16
1
48
7
6
2
2
3
4
L
w
L
w
L
R
M









































At
B
,
2
2
05556
.
0
8
L
w
L
w
M
M
b







.

At
A
,
2
2
02777
.
0
8
L
w
L
R
L
w
M
M
b








.

Worst bending moment is that at
C
,
M
b

= 0.06944∙
w

L
2
.


Antisymetric loading due to Q
:

In
BC
,
M
b

=
V x
,
x
V
M
b



.

In
AB
,
2
L
V
y
Q
M
b




,
2
L
V
M
b




.

Therefore
0
2
2
0
2
0
2





















dy
L
V
y
Q
L
dx
x
V
V
U
I
E
L
L
,

and
0
4
4
24
3
3
3






L
V
L
Q
L
V
.

From the latter equation we have
Q
V



4
1
24
7

or
Q
V


7
6
.

At
B
,
L
Q
L
V
M
b





7
3
2
.

At
A
,
L
Q
L
Q
L
Q
L
V
L
Q
M
b












7
4
7
3
2
.

Maximal bending moment is that at
A
,
L
Q
M
b



7
4
.

The answer:

at
C

M
b

= 0.06944∙
w

L
2

and at

A

L
Q
M
b



7
4
.

4.

A solid steel shaft (fig. 4.1) of diameter 60 mm is clamped at one end. The other end is
attached to the middle of a cross
-
bar pinned at each end, of depth 60 mm, with 20 mm, and
length 800 mm. A torque of 3 kN∙m is applied m
idway along the shaft. Determine the
maximum bending stress present. Use
E

= 200 GPa and
G

= 77.5 GPa as properties of the
steel.

Hint:

you need to find either the reaction force F at the ends of the crossbar or the reaction
torque at the end of the shaft,

to make the problem statically determinate.

F
F
8
0
0

m
m
6
0

m
m
6
0
0

m
m
T
20
mm
60
mm

Fig. 4.1

M + T
L
2

1
L
2
T
L
1
M
x
y
z
A
B
C
D
E
F
F
F
=

Fig. 4.2

Solution.

The basic solution.

In
AB

or
BC

x
F
M
b


.

In
BD

1
2
L
F
M
tBD




(torque).

In
DE

T
L
F
M
tDE




1
2
.

















2
2
1
0
2
0
2
0
2
2
1
2
1
2
1
2
L
tDE
p
L
tBD
p
L
b
dy
M
I
G
dy
M
I
G
dx
M
I
E
U
.





























2
2
1
0
2
1
2
1
2
0
2
1
2
0
2
2
4
4
2
1
4
2
1
2
1
2
L
p
L
p
L
dy
T
T
L
F
L
F
I
G
dy
L
F
I
G
dx
x
F
I
E
U
.

dy
T
T
L
F
L
F
I
G
dy
L
F
I
G
dx
x
F
I
E
U
L
p
L
p
L



























2
2
1
0
2
1
2
1
2
0
2
1
2
0
2
2
2
2
2
1
.

0
2
4
4
2
2
2
1
2
0
1
0
2
1
0
0
2
1
2























L
p
L
p
L
L
p
dy
L
T
I
G
dy
L
F
I
G
dy
L
F
I
G
dx
x
F
I
E
F
U
.

0
2
8
3
2
2
8
2
2
1
2
2
1
3
1
2
1
2
2
1
0
2
1
































p
p
p
p
L
I
G
L
L
T
I
G
L
L
F
I
E
L
F
I
G
L
L
T
I
G
L
L
F
dx
x
F
I
E
F
U
.

0
2
8
3
2
2
1
2
2
1
3
1














p
p
I
G
L
L
T
I
G
L
L
F
I
E
L
F
.

F
L
L
I
E
I
E
L
I
G
L
T
p











2
1
2
1
3
12
.

Now

7
3
10
6
.
3
12
06
.
0
02
.
0





I

m
4

6
4
10
2724
.
1
32
06
.
0






p
I

m
4
;

L
1

= 0.4 m

L
2

= 0.3 m.

Substituting





















F
T
3
.
0
4
.
0
10
6
.
3
10
2
3
10
6
.
3
10
2
3
.
0
12
10
2724
.
1
10
75
.
7
4
.
0
7
11
7
11
6
10

= 1.8435 ∙
F

(N ∙ m).

3
3
1
max
10
6510
.
0
10
3
0.2170
0.2170
8435
.
1
4
.
0











T
T
L
F
M
b

(N ∙ m);

6510
.
0
max

b
M

(kN ∙ m).

7
7
3
max
max
max
10
5.425
10
6
.
3
03
.
0
10
6510
.
0









I
y
M
b

(Pa);

25
.
54
max


(MPa).

The check solution starts from another end of the shaft.

In rectangular section
x
L
T
M
M
b




1
2
.

In bar
M
M
tED


or
T
M
M
tDB


.

















2
2
1
0
2
0
2
0
2
2
1
2
1
2
1
2
L
tED
p
L
tDB
p
L
b
dy
M
I
G
dy
M
I
G
dx
M
I
E
U
.



























2
2
1
0
2
0
2
2
2
0
2
1
2
2
2
1
2
2
1
4
2
1
L
p
L
p
L
dy
M
I
G
dy
T
T
M
M
I
G
dx
x
L
T
T
M
M
I
E
U
.



0
1
1
2
1
2
2
1
0
0
0
2
2
1



















L
p
L
p
L
dy
M
I
G
dy
T
M
I
G
dx
x
L
T
M
I
E
M
U
.

0
6
2
2
2
1
1

















p
p
I
G
L
M
I
G
L
T
L
M
I
E
L
T
L
M
M
U
.

0
2
6
6
2
2
1
1


















p
p
I
G
L
T
I
G
L
M
I
E
L
T
I
E
L
M
M
U
.

T
I
G
L
I
E
L
M
I
G
L
I
E
L
p
p





























2
1
2
1
6
2
6
.





T
I
E
L
I
G
L
M
I
E
L
I
G
L
p
p
















2
1
2
1
6
12
.

T
I
E
L
I
G
L
I
E
L
I
G
L
M
p
p















2
1
2
1
12
6
.

T
T
M






























0.5660
10
6
.
3
10
2
3
.
0
12
10
2723
.
1
10
75
.
7
4
.
0
10
6
.
3
10
2
3
.
0
6
10
2723
.
1
10
75
.
7
4
.
0
7
11
6
10
7
11
6
10
.

.6981
1
3
5660
.
0





M

(kN ∙ m).

Reaction


T
0.5425
4
.
0
2
1
5660
.
0
2
1










T
L
T
M
F
.

F

= 0.5425 ∙ 3 = 1.6275 (kN).

651
.
0
4
.
0
6275
.
1
1
max





L
F
M
b

(kN ∙ m),

7
7
3
max
max
max
10
5.425
10
6
.
3
03
.
0
10
6510
.
0









I
y
M
b


(Pa)

54.25
max


(MPa)


5.

Use an energy method to develop an expression for

the worst bending moment in the beam
(fig. 5.1), supported by three identical springs of stiffness
k
, and loaded centrally as shown.

P
k
k
k
L
L

Fig. 5.1

P
k
k
k
L
L
R
1
R
2
R
1
A
B
C
x
y

Fig. 5.2

Hint:

due to symmetry, there is only one unknown redundant reaction.

Solution.

The sum of projection
s of forces onto the Y axis:

P



2 ∙
R
1

+
R
2
.

R
2

=
P



2 ∙
R
1
.

The bending moment is
M
b

=
R
1


x

for
A

to
B
.

The energy of deformation is



















L
k
R
P
k
R
dx
x
R
I
E
U
0
2
1
2
1
2
1
2
2
1
2
1
2
2
1
2

k
R
P
R
P
k
R
dx
x
R
I
E
L













2
4
4
1
2
1
1
2
2
1
0
2
2
1
.

The energy deformation has its minimum in static position, that’s why

0
2
4
2
2
1
1
0
2
1
1














k
P
k
R
k
R
dx
x
R
I
E
R
U
L
.

0
2
6
3
2
1
3
1








k
P
k
R
I
E
L
R
.

k
P
k
I
E
L
R













3
3
1
3
1
.

k
P
R
k
I
E
I
E
L
k









1
3
3
9
.

P
I
E
L
k
I
E
R








9
3
3
1
.

The maximal bending moment:
P
I
E
L
k
L
I
E
L
R
M
b











9
3
3
1
max
.