Plane Truss Stiffness Matrix · Lecture 13: Trusses & Grids –Stiffness Method Washkewicz College of Engineering 2 Consider an arbitrary member, i. isolated from a generalized plane
Post on 30-Apr-2020
27 Views
Preview:
Transcript
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Plane Truss – Stiffness Matrix
The distinguishing feature of a plane truss is that loads are applied in the plane of the
structure whereas in a space truss they are not.
We now wish to outline the procedure of formulating the joint stiffness matrix [SJ] for a
plane truss structure. The extension to a three dimensional space truss will be intuitively
obvious. 1
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
2
Consider an arbitrary member, i. isolated from a
generalized plane truss depicted below:
The joints at the end of truss member i are denoted j and k. The plane truss lies in the x-y
plane. The joint translations are the unknown displacements and these displacements are
expressed in terms of their x and y components.
We will relax the requirement that truss members are two force members. This allows for
loads that are applied between joints to a truss member, and it allows consideration of the
weight of the truss member.
i
k
j
x
y
Primary load carrying capabilities
are in the plane of the truss –
hence the need for shear walls.
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
The positive directions of the four displacement components (two translations at either
end) of member i are depicted in the figure below
It will be convenient to utilize the direction cosines associated with this arbitrary
member. In terms of the joint coordinates the direction cosines are
L
xx
C
jk
X
1cos
L
yy
C
jk
Y
2cos
22
jkjk yyxxL
with
3
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
The beam member stiffness matrix developed in the previous section of notes can be
easily adapted for use in the case of a plane truss. The member stiffness matrix [SM] for
an arbitrary truss member with member axes Xm and Ym oriented along the member and
perpendicular to the member can be obtained by considering Case #1 and Case #7 from
the previous section of notes.
Using the numbering joint numbering system and the member axes depicted in the
following figure
then the member stiffness matrix for
a truss member is as follows
4
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Note that [SJ] is based on axes oriented to the
structure. Truss member stiffnesses may be
obtained in one of two ways. Either the
stiffnesses are directly computed using the figure
to the left, or the second method consists of first
obtaining the stiffness matrix relative to the
member oriented axes and then imposing a
suitable matrix transformation that transforms
these elements to axes relative to the structure.
We will focus on the direct method first to help
develop an intuition of how the structure
behaves. Unit displacement in both the x and y
directions are applied at each end of the member.
If a unit displacement in the x direction is applied
to the j end of the member, the member shortens
and an axial compression force is induced. The
magnitude of the force is
xx C
L
EA
5
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Restraint actions at the ends of the truss member in the x and y directions are required.
They are equal to the components of the axial force induced in the member, and are
identified here as elements of the [SMD] matrix in order to distinguish them from elements
of the [SM] matrix. The numbering of these elements are shown in the previous figure.
Thus
yx
xMDMD
xx
MDMD
yxx
MD
xx
MD
CCL
EASS
CL
EASS
CCL
EAS
CL
EAS
2141
2
1131
21
2
11
6
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
In a similar fashion, a unit displacement in the y direction at the j end of the member yields
22242
1232
2
22
12
yx
MDMD
yxx
MDMD
yx
MD
yxx
MD
CL
EASS
CCL
EASS
CL
EAS
CCL
EAS
7
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
In a similar fashion, a unit displacement in the x direction at the k end of the member yields
yx
xMDMD
xx
MDMD
yxx
MD
xx
MD
CCL
EASS
CL
EASS
CCL
EAS
CL
EAS
2343
2
1333
23
2
13
8
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
In a similar fashion, a unit displacement in the y direction at the k end of the member yields
22444
1434
2
24
14
yx
MDMD
yxx
MDMD
yx
MD
yxx
MD
CL
EASS
CCL
EASS
CL
EAS
CCL
EAS
9
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
We have just developed the four rows of the [SMD] matrix, i.e.,
10
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Example 13.1
Consider the plane truss with four bars meeting at a common joint E. This truss
only has two degrees of freedom from a kinematic standpoint. It is a convenience
to identify the bars of the truss numerically. The bars have lengths L1, L2, L3 and
L4 and axial rigidities EA1, EA2, EA3 and EA4
The loads consist
of two
concentrated
forces P1 and P2
action at joint E.
We will consider
the bar weights
identified here as
w1, w2, w3 and w4
(force/length).
The unknown displacements at joint E are identified as D1 and D2. We seek to
calculate member end actions AM1, AM2, AM3 and AM4. 11
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Because the weight of each truss member is included, the axial forces at either end
of a truss member will be different at joints A, B, C and D then the axial force at
joint E. The axial forces at joint E could be computed as well as the shear stresses
at the end of each truss member, however they are omitted in this example for
simplicity.
The loads P1 and P2 correspond to unknown displacements D1 and D2, thus
We next consider the restrained
structure shown at the right. Here joint
E is fixed with a pin support that
produce loads ADL1 and ADL2 associated
with D1 and D2.
2
1
P
PAD
12
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Each truss member can be considered loaded as shown below. The points of
support are indicated as A and E for the purpose of discussion and do not
correspond to actual joints in labeled in the original truss. One could use the Greek
alphabet, but the nomenclature should be transparent given the context where it
used.
13
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Since the weights of the truss members produce no horizontal reactions, the actions
ADL1 must be zero and ADL2 must be equal to half the weight of all the truss elements,
i.e.,
2
0
2222
0
44332211 WLwLwLwLwADL
The quantity W is the total weight of the truss. For the purpose of calculating end
actions for the vector AML, consider that from the previous figure
iii
MLi
LwA sin
2
444
333
222
111
sin
sin
sin
sin
2
1
Lw
Lw
Lw
Lw
AMLor
14
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
The next step is formulating the stiffness matrix by imposing unit displacement
associated with D1 and D2 on the restrained structure as indicated below
To obtain the stiffness values it is necessary to compute the forces in the truss
elements when the unit displacements are applied to joint E.15
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
When the upper joint of the element
moves to the right, the lower joint stays
fixed.
When the upper joint of the element
moves up, again the lower joint stays
fixed. Both actions elongate the truss
elements. The geometry of the elongation
is determined by the translation of joint E.
cos
AE
PL
L
EAP cos
16
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
cosL
EA
sinL
EA
When joint E is subjected to a unit translation to the right the truss element elongates
an amount
When joint E is subjected to a unit translation vertically the truss element elongates
an amount
The formulas given above are suitable for use in analyzing this plane truss. In a later
lecture a more systematic approach to the development of member stiffnesses is
developed that works for trusses and all types of structures.
The stiffness S11 is composed of contributions from various elements of the truss.
Consider the contribution to S11 from member 3, i.e.,
3
2
3
311
3 cos L
EAS
17
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Thus
3
2
3
32
2
2
2
1
1
4
43
2
3
32
2
2
2
1
1
4
2
4
43
2
3
32
2
2
21
2
1
1
11
4
11
3
11
2
11
1
11
coscos
0coscos1
coscoscoscos
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
SSSSS
The final expression results from the fact that truss element 1 is horizontal and truss
element 4 is vertical.
18
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
21
4
21
3
21
2
21
1
21 SSSSS
Similarly the stiffness S21 is composed of contributions from various elements of the
truss. Consider the contribution to S21 from member 3, i.e.,
33
3
321
3 sincos L
EAS
Thus
33
3
322
2
2
4
433
3
322
2
2
1
1
44
4
433
3
322
2
211
1
1
sincossincos
10sincossincos01
sincossincossincossincos
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
19
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
By an analogous procedure S12 and S22 are
33
3
322
2
212 sincossincos
L
EA
L
EAS
4
43
2
3
32
2
2
222 sinsin
L
EA
L
EA
L
EAS
The two expressions on this page as well as the two from the previous page
constitute the stiffness matrix [S]. The next step would be inverting this matrix and
performing the following matrix computation to find the displacement D1 and D2.
DLD AASD 1
The vector {AD} and the matrix {ADL} were established earlier.
20
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Since the vector {AML} was determined earlier as well, we need only identify the elements
of the matrix {AMD}. This matrix contains the member end-actions due to unit
displacements associated with the displacements D1 and D2, but the end actions are
computed using the restrained structure. Thus for ith member using a previous figure
i
i
iMDi
L
EAA cos1 i
i
iMDi
L
EAA sin2
4
4
44
4
4
3
3
33
3
3
2
2
22
2
2
1
1
11
1
1
sincos
sincos
sincos
sincos
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
L
EA
AMD
thus
And we can now solve
DAAA MDMLM 21
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
22
Example 13.2
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
23
Structural Grids (Need a theory section on - Stiffness Method as it applies to
grids
Examples of two dimensional grids are depected below:
Grids behave more like frames. However, loads are typically applied perpendicular
to the plane of the structure. Moments are allowed in the plane of the structure and
perpendicular to the structure.
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
24
A space grid – the structure is not in a plane, but the loads are in essence perpendicular to
the structure.
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Example 13.3
The grid shown below consists of two members (AB and BC) that are rigidly joined at
B. Each member is assumed to have flexural rigidity EI and torsional rigidity GJ.
Kinematically, the only unknowns are the displacements at B. Since axial rigidities of
the members is assumed to be quite large relative to EI and GJ, the displacements at B
consist of one translation (D1) and two rotations (D2 and D3). Determine these
unknown displacements.
25
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
When analyzing a grid by the stiffness method, an artificial restraint is
provided at joint B, i.e.,
It is easier to see what the reactions are if we break the structure above into two
substructures such that
26
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
From the last figure it is easy to see that
or
L
PADL 0
4
8
80
2321
PLAA
PA DLDLDL
and in a matrix format
80
2321
PLAA
PA DLDLDL
000 321 DLDLDL AAA
27
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
0
0
0
DA
The vector {AD} represents actions in the unrestrained structure associated
with the unknown displacement D1, D2 and D3. Since there are no loads
associated with these displacements {AD} is a null vector and in a matrix
format
DLD AASD 1
We have {ADL} and {AD} the next step is the solution of the superposition
expression
for the unknown displacements. To do that we need to formulate the stiffness
matrix and find its inverse.
28
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
From the figures above
23121311
60
12
L
EISS
L
EIS
0612
31221311 SL
EIS
L
EIS
231221311
6624
L
EIS
L
EIS
L
EIS
The stiffness matrix is found by analyzing the restrained structure for the effects of unit
translations and rotations associated with the unknown displacements. In the following
figure the grid structure is once again split into two substructures.
29
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
To obtain the second column of the stiffness matrix utilize the following figure
00 322212 SL
GJSS
046
3222212 SL
EIS
L
EIS
046
3222212 SL
GJ
L
EIS
L
EIS
From the figures above
30
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
To obtain the third column of the stiffness matrix utilize the following figure
From the figures above
L
EISS
L
EIS
40
63323213
J
GJSSS
332313 00
L
GJ
L
EISS
L
EIS
40
63323213
31
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
Define
EI
GJ
2
2
3
406
046
6624
LL
LL
LL
L
EIS
2
2
2
1
2
21
1
406
046
66
24
1
LL
LL
LLCL
CEICS
then
and inverting this stiffness matrix leads to
25
1
4
3
2
1
C
C
C
where
32
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
DLD AASD 1
18
256
254
4196
2
L
EI
PLD
Solving
leads to
33
Lecture 13: Trusses & Grids – Stiffness Method
Washkewicz College of Engineering
34
Example 13.4
top related