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Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
ExampleConsider 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 f P d Pforces P1 and P2action at joint E. We will consider the bar weights gidentified 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.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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 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, thus1 2 p p 1 2,
{ }
=2
1
PP
AD
We next consider the restrained structure shown at the right. Here joint E i fi d ith i t th tE is fixed with a pin support that produce loads ADL1 and ADL2 associated with D1 and D2.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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
d l j i i l b l d i h i i l O ld h G kcorrespond 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.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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 ei.e.,
{ }
=
=
00
A{ }
=
+++
=
2222244332211 WLwLwLwLw
ADL
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
MLiLwA γsin
2−
= { }
−=333
222
111
sinsinsin
21
γγγ
LwLwLw
AMLor2
444
333
sin γγ
Lw
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
When the upper joint of the element moves to the right, the lower joint stays fixedfixed.
When the upper joint of the element moves up, again the lower joint stays fixed. Both actions elongate the truss gelements. The geometry of the elongation is determined by the translation of joint E.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
EA
When joint E is subjected to a unit translation to the right the truss element elongates an amount
γcosL
EA
When joint E is subjected to a unit translation vertically the truss element elongates
γsinL
EAan 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.,
A3
2
3
311
3 cos γL
EAS =
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
Thus
42
4
43
2
3
32
2
2
21
2
1
1
114
113
112
111
11
coscoscoscos γγγγL
EAL
EAL
EAL
EASSSSS
+++=
+++=
( ) ( )4
43
2
3
32
2
2
2
1
1
4321
0coscos1 γγ
EAEAEAL
EAL
EAL
EAL
EA+++=
32
3
32
2
2
2
1
1 coscos γγL
EAL
EAL
EA++=
The final expression results from the fact that truss element 1 is horizontal and truss element 4 is vertical.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
Similarly the stiffness S21 is composed of contributions from various elements of the truss. Consider the contribution to S21 from member 3, i.e.,
333
321
3 sincos γγL
EAS =
Thus
214
213
212
211
21 SSSSS +++=
Thus
EAEAEAEA
( )( ) ( )( )433
322
21
444
433
3
322
2
211
1
1
10sincossincos01
sincossincossincossincos
γγγγ
γγγγγγγγ
EAEAEAEAL
EAL
EAL
EAL
EA
+++=
+++=
( )( ) ( )( )
333
322
2
2
433
322
21
sincossincos γγγγ
γγγγ
LEA
LEA
LLLL
+=
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
By an analogous procedure S12 and S22 are
333
322
2
212 sincossincos γγγγ
LEA
LEAS +=
EAEAEA
4
43
2
3
32
2
2
222 sinsin
LEA
LEA
LEAS ++= γγ
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
Th {A } d h i {A } bli h d liThe vector {AD} and the matrix {ADL} were established earlier.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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
ii
iMDi L
EAA γcos1 = ii
iMDi L
EAA γsin2 =i i
11
11
1
1 sincos γγL
EAL
EAthus
{ }
=3
33
3
22
22
2
2
11
sincos
sincos
γγ
γγ
EAEAL
EAL
EA
AMD
4
4
44
4
4
33
33
sincos
sincos
γγ
γγ
LEA
LEA
LL
And we can now solve
{ } { } { }{ }DAAA MDMLM +=
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
ExampleThe 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 theseconsist of one translation (D1) and two rotations (D2 and D3). Determine these unknown displacements.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
From the last figure it is easy to see that
PLP8
02 321
PLAAPA DLDLDL −=′=′=′
or
000 321 =′′=′′=′′ DLDLDL AAA
80
2 321PLAAPA DLDLDL −===
d i t i f t
{ }
=
PADL 04
8
and in a matrix format
{ }
− L
DL 8
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
The vector {AD} represents actions in the unrestrained structure associated with the unknown displacement D1, D2 and D3. Since there are no loads
0
p 1, 2 3associated with these displacements {AD} is a null vector and in a matrix format
{ }
=
00DA
{ } [ ] { } { }{ }1
We have {ADL} and {AD} the next step is the solution of the superposition expression
{ } [ ] { } { }{ }DLD AASD −= −1
for the unknown displacements. To do that we need to formulate the stiffness matrix and find its inversematrix and find its inverse.
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
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 substructuresfigure the grid structure is once again split into two substructures.
From the figures above
231213116012LEISS
LEIS −=′=′=′
23 LL
061231221311 =′′=′′=′′ S
LEIS
LEIS
2312213116624LEIS
LEIS
LEIS −===
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
To obtain the second column of the stiffness matrix utilize the following figure
From the figures above
00 322212 =′=′=′ SL
GJSS
From the figures above
0463222212 =′′=′′=′′ S
LEIS
LEIS
0463222212 =+== S
LGJ
LEIS
LEIS
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
To obtain the third column of the stiffness matrix utilize the following figure
From the figures aboveFrom the figures above
LEISS
LEIS 406
3323213 =′=′−=′
JGJSSS =′′=′′=′′ 332313 00
LGJ
LEISS
LEIS +==−=
4063323213
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
Define
EIGJ
=η
[ ] ( )
+−
= 2 0466624
LLLL
EIS η
then
[ ] ( )( )
+−
+=2
3
406046
LLLL
LS
ηη
and inverting this stiffness matrix leads to
[ ]( )
( )
+−
=− 2
21
2
1 04666
241 LL
LLCL
CEICS η
( )
+− 221 406
24LL
CEICη
η4 +=Cwhere
ηηη
2514
3
2
1
+=+=+=
CCC
Lecture 13: TRUSSES & GRIDS – STIFFNESS METHOD
{ } [ ] { } { }{ }DLD AASD −= −1
Solving
{ } [ ] { } { }{ }DLD
( ) ( )
++− 254
2ηηL
leads to
{ } ( ) ( )
( ) ( )( )
−+
++=
18256
4196
2
ηηη
ηηEIPLD