question 1
DESCRIPTION
Frame stiffnessTRANSCRIPT
![Page 1: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/1.jpg)
LOCAL MATRICIES, ELEMENT A:
na⊗na=[0 0 00 1 00 0 0]; sa⊗sa=[1 0 0
0 0 00 0 0 ]; t a⊗ t a=[0 0 0
0 0 00 0 1 ]; t a⊗ sa=[ 0 0 0
0 0 0−1 0 0]; sa⊗ t a=[0 0 −1
0 0 00 0 0 ]
k fua= AE
Lana⊗ na+
12 E I ttLa3 sa⊗sa+
12E I ssLa3 t a⊗ t a
k fua= AE
La [0 0 00 1 00 0 0]+12 E IttLa
3 [1 0 00 0 00 0 0]+ 12E I ssLa
3 [0 0 00 0 00 0 1]=[
12E I ttLa3
0 0
0AELa
0
0 012E I ssLa3
]kmua =
6 E I ttLa2 t a⊗sa−
6 E I ssLa2 sa⊗ ta
kmua =
6 E I ttLa2 [ 0 0 0
0 0 0−1 0 0]−6 E I ssLa
2 [0 0 −10 0 00 0 0 ]=[ 0 0
6 EI ssLa2
0 0 0−6E I ttLa2 0 0 ]
kmua T=
6 E IttLa2 [0 0 −10 0 00 0 0 ]−6 E I ssLa
2 [ 0 0 00 0 0
−1 0 0]=[ 0 0−6 E IttLa2
0 0 06 E I ssLa2 0 0 ]
kmθa = JG
Lana⊗ na+
4 E I ttLa
t a⊗ t a+4 E I ssLa
sa⊗sa
kmθa = JG
La [0 0 00 1 00 0 0]+ 4 E I ttLa [0 0 0
0 0 00 0 1]+ 4 E I ssLa [1 0 0
0 0 00 0 0]=[
4 E I ssLa
0 0
0JGLa
0
0 04 E I ttLa
]
![Page 2: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/2.jpg)
k̂ amθ=−JGLa
na⊗na+2E I ttLa
t a⊗ ta+2E I ssLa
sa⊗ sa
k̂ amθ=−JGLa [0 0 0
0 1 00 0 0 ]+ 2E I ttLa [0 0 0
0 0 00 0 1]+ 2 E I ssLa [1 0 0
0 0 00 0 0]=[
2 E I ssLa
0 0
0−JGLa
0
0 02 E I ttLa
]LOCAL MATRICIES, ELEMENT B:
nb⊗nb=[1 0 00 0 00 0 0]; sb⊗sb=[0 0 0
0 1 00 0 0 ]; t b⊗ t b=[0 0 0
0 0 00 0 1 ]; t b⊗ sb=[0 0 0
0 0 00 −1 0]; sb⊗ t b=[0 0 0
0 0 −10 0 0 ]
k fub= AE
Lbnb⊗ nb+
12 E I ttLb3 sa⊗sa+
12E I ssLb3 t b⊗ t b
k fub= AE
Lb [1 0 00 0 00 0 0]+12 E IttLb
3 [0 0 00 1 00 0 0]+ 12E I ssLb
3 [0 0 00 0 00 0 1]=[
AELb
0 0
012E I ttLb3 0
0 012E I ssLb3
]kmub =
6 E I ttLb2 t b⊗sb−
6 E I ssLa2 sb⊗ tb
kmub =
6 E I ttLb2 [0 0 0
0 0 00 −1 0]−6 E I ssLa
2 [0 0 00 0 −10 0 0 ]=[
0 0 0
0 06 EI ssLa2
0−6 E I ttLb2 0 ]
kmub T=
6 E IttLb2 [0 0 00 0 −10 0 0 ]−6 E I ssLa
2 [0 0 00 0 00 −1 0]=[
0 0 0
0 0−6 E IttLb2
06 E I ssLa2 0 ]
![Page 3: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/3.jpg)
kmθb = JG
Lbnb⊗ nb+
4 E I ttLb
t b⊗ t b+4 E I ssLb
sb⊗sb
kmθb = JG
Lb [1 0 00 0 00 0 0]+ 4 E I ttLb [0 0 0
0 0 00 0 1]+ 4 E I ssLb [0 0 0
0 1 00 0 0]=[
JGLb
0 0
04 EI ssLb
0
0 04 E I ttLb
]k̂ bmθ=
−JGLb
nb⊗nb+2E I ttLb
t b⊗ tb+2E I ssLb
sb⊗ sb
k̂ bmθ=−JGLb [1 0 0
0 0 00 0 0 ]+ 2E I ttLb [0 0 0
0 0 00 0 1]+ 2 E I ssLb [0 0 0
0 1 00 0 0]=[
−JGLb
0 0
02 E I ssLb
0
0 02 E I ttLb
]
![Page 4: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/4.jpg)
GLOBAL MATRIX:
k=[ k11a k 12
a 0
k 21a k22
a +k22b k 21
b
0 k 12b k11
b ]k=[ [ k fu kmu
T
kmu kmθ]a
[−k fu kmuT
−kmu k̂mθ]a
0
[−k fu −kmuT
kmu k̂mθ ]a
[ k fu −kmuT
−kmu kmθ ]a
+[ k fu −kmuT
−kmu kmθ ]b
[−k fu −kmuT
kmu k̂mθ ]b
0 [−k fu kmuT
−kmu k̂mθ]b
[ k fu kmuT
kmu kmθ]b ]
![Page 5: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/5.jpg)
k=[ [[12E I ttLa3 0 0
0AELa
0
0 012E I ssLa3
] [ 0 0−6 E IttLa2
0 0 06 E I ssLa2
0 0 ][ 0 0
6 EI ssLa2
0 0 0−6 EI ttLa2 0 0 ] [
4 E I ssLa
0 0
0JGLa
0
0 04 E I ttLa
]]a
[−[12E I ttLa3 0 0
0AELa
0
0 012E I ssLa3
] [ 0 0−6 EI ttLa2
0 0 06 EI ssLa2
0 0 ]−[ 0 0
6E I ssLa2
0 0 0−6 E I ttLa2 0 0 ] [
2E I ssLa
0 0
0−JGLa
0
0 02 EI ttLa
]]a
[0 ]
[−[12E I ttLa3 0 0
0AELa
0
0 012E I ssLa3
] −[ 0 0−6 E IttLa2
0 0 06 E I ssLa2 0 0 ]
[ 0 06E I ssLa2
0 0 0−6E I ttLa2 0 0 ] [
2 E I ssLa
0 0
0−JGLa
0
0 02 E I ttLa
]]a
([[12E I ttLa3 0 0
0AELa
0
0 012E I ssLa3
] −[ 0 0−6 E I ttLa2
0 0 06 E I ssLa2 0 0 ]
−[ 0 06 EI ssLa2
0 0 0−6 E I ttLa2 0 0 ] [
4 E I ssLa
0 0
0JGLa
0
0 04 E IttLa
] ]a
+[ [AELb
0 0
012 E I ttLb3 0
0 012 E I ssLb3
] −[0 0 0
0 0−6 E I ttLb2
06 E I ssLa2 0 ]
−[0 0 0
0 06 E I ssLa2
0−6 EI ttLb2 0 ] [
JGLb
0 0
04 E I ssLb
0
0 04 E I ttLb
] ]b
) [−[AELb
0 0
012E I ttLb3 0
0 012 E I ssLb3
] −[0 0 0
0 0−6 E I ttLb2
06 E I ssLa2 0 ]
[0 0 0
0 06 E I ssLa2
0−6 E I ttLb2 0 ] [
−JGLb
0 0
02 EI ssLb
0
0 02E I ttLb
]]b
[0 ] [−[AELb
0 0
012E I ttLb3 0
0 012E I ssLb3
] [0 0 0
0 0−6 EI ttLb2
06 E I ssLa2 0 ]
−[0 0 0
0 06E I ssLa2
0−6 E I ttLb2 0 ] [
−JGLb
0 0
02 E I ssLb
0
0 02 EI ttLb
]]b
[[AELb
0 0
012E I ttLb3 0
0 012 E I ssLb3
] [0 0 0
0 0−6E I ttLb2
06 E I ssLa2 0 ]
[0 0 0
0 06 E I ssLa2
0−6 E I ttLb2 0 ] [
JGLb
0 0
04 E I ssLb
0
0 04 E I ttLb
] ]b ]
![Page 6: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/6.jpg)
Simplify the global matrix:
![Page 7: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/7.jpg)
k=[[12 E I ttLa3 0 0
0AELa
0
0 012 E I ssLa3
] [ 0 0−6 E I ttLa2
0 0 06 E I ssLa2 0 0 ]
[ 0 06 E I ssLa2
0 0 0−6 E I ttLa2 0 0 ] [
4 E I ssLa
0 0
0JGLa
0
0 04 E I ttLa
][−12E I ttLa3 0 0
0−AELa
0
0 0−12E I ssLa3
] [ 0 0−6 E IttLa2
0 0 06 E I ssLa2 0 0 ]
[ 0 0−6E I ssLa2
0 0 06 E I ttLa2 0 0 ] [
2 EI ssLa
0 0
0−JGLa
0
0 02 E IttLa
][ [0 ] [0 ]
[0 ] [0 ]]
[−12 E IttLa3 0 0
0−AELa
0
0 0−12 E I ssLa3
] [ 0 06 EI ttLa2
0 0 0−6 EI ssLa2 0 0 ]
[ 0 06 E I ssLa2
0 0 0−6 E I ttLa2 0 0 ] [
2E I ssLa
0 0
0−JGLa
0
0 02E I ttLa
][12 EI ttLa3 + AE
Lb0 0
0AELa
+12E I ttLb3
0
0 012 E I ssLa3 +
12E I ssLb3
] [ 0 06 E IttLa2
0 06 E IttLb2
−6 E I ssLa2
−6 EI ssLa2 0 ]
[ 0 0−6 E I ssLa2
0 0−6 E I ssLa2
6 E I ttLa2
6 E I ttLb2 0 ] [
4 E I ssLa
+ JGLb
0 0
0JGLa
+4 E I ssLb
0
0 04 EI ttLa
+4 E I ttLb
][−AELb
0 0
0−12 E IttLb3 0
0 0−12 EI ssLb3
] [0 0 0
0 06 E I ttLb2
0−6 E I ssLa2 0 ]
[0 0 0
0 06 E I ssLa2
0−6 EI ttLb2 0 ] [
−JGLb
0 0
02E I ssLb
0
0 02 E I ttLb
][ [0 ] [0 ]
[0 ] [0 ] ][−AELb
0 0
0−12 EI ttLb3 0
0 0−12E I ssLb3
] [0 0 0
0 0−6 E IttLb2
06 EI ssLa2 0 ]
[0 0 0
0 0−6E I ssLa2
06 E I ttLb2 0 ] [
−JGLb
0 0
02 E I ssLb
0
0 02 E IttLb
][AELb
0 0
012 E IttLb3 0
0 012 EI ssLb3
] [0 0 0
0 0−6 E I ttLb2
06 E I ssLa2 0 ]
[0 0 0
0 06 E I ssLa2
0−6 E I ttLb2 0 ] [
JGLb
0 0
04 E I ssLb
0
0 04 E IttLb
]]
![Page 8: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/8.jpg)
Simplify the global matrix futher:
![Page 9: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/9.jpg)
Substitute the global matrix into our system of equations:
{P1x¿P1 y¿P1 z
¿M 1x
¿M 1 y
¿M1 z
¿P2 x¿P2 y¿P2 z¿M 2 x
¿M 2 y
¿M2 z
¿P3 x¿P3 y¿P3 z
¿M 3x
¿M 3 y
¿M3 z
}=¿ {u1x¿u1 y¿u1 z¿θ1x¿θ1 y¿θ1 z¿u2x¿u2 y¿u2 z¿θ2x¿θ2 y¿θ2 z¿u3x¿u3 y¿u3 z¿θ3x¿θ3 y¿θ3 z
}12E I ttLa3 0 0 0 0
−6 E IttLa2
−12E I ttLa3 0 0 0 0
−6 E IttLa2 0 0 0 0 0 0
0AELa
0 0 0 0 0−AELa
0 0 0 0 0 0 0 0 0 0
0 012E I ssLa3
6 EI ssLa2 0 0 0 0
−12E I ssLa3
6 EI ssLa2 0 0 0 0 0 0 0 0
0 06 EI ssLa2
4 E I ssLa
0 0 0 0−6 E I ssLa2
2E I ssLa
0 0 0 0 0 0 0 0
0 0 0 0JGLa
0 0 0 0 0−JGLa
0 0 0 0 0 0 0
−6 E IttLa2 0 0 0 0
4 E I ttLa
6 EI ttLa2 0 0 0 0
2E I ttLa
0 0 0 0 0 0
−12E I ttLa3 0 0 0 0
6 EI ttLa2
12E I ttLa3 + AE
Lb0 0 0 0
6 EI ttLa2
−AELb
0 0 0 0 0
0−AELa
0 0 0 0 0AELa
+12 E I ttLb3 0 0 0
6 EI ttLb2 0
−12E I ttLb3 0 0 0
6 EI ttLb2
0 0−12E I ssLa3
−6 E I ssLa2 0 0 0 0
12E I ssLa3 +
12E I ssLb3
−6 E I ssLa2
−6 E I ssLa2 0 0 0
−12E I ssLb3 0
−6 E I ssLa2 0
0 06 EI ssLa2
2E I ssLa
0 0 0 0−6 E I ssLa2
4 E I ssLa
+ JGLb
0 0 0 0 0−JGLb
0 0
0 0 0 0−JGLa
0 0 0−6 E I ssLa2 0
JGLa
+4 E I ssLb
0 0 06 EI ssLa2 0
2E I ssLb
0
−6 E IttLa2 0 0 0 0
2E I ttLa
6 EI ttLa2
6 EI ttLb2 0 0 0
4 E I ttLa
+4 EI ttLb
0−6 E IttLb2 0 0 0
2E I ttLb
0 0 0 0 0 0−AELb
0 0 0 0 0AELb
0 0 0 0 0
0 0 0 0 0 0 0−12E I ttLb3 0 0 0
−6 E IttLb2 0
12E I ttLb3 0 0 0
−6 E IttLb2
0 0 0 0 0 0 0 0−12E I ssLb3 0
6 EI ssLa2 0 0 0
12E I ssLb3 0
6 EI ssLa2 0
0 0 0 0 0 0 0 0 0−JGLb
0 0 0 0 0JGLb
0 0
0 0 0 0 0 0 0 0−6 E I ssLa2 0
2E I ssLb
0 0 06 EI ssLa2 0
4 E I ssLb
0
0 0 0 0 0 0 06 EI ttLb2 0 0 0
2E I ttLb
0−6 E IttLb2 0 0 0
4 E I ttLb
![Page 10: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/10.jpg)
P1x12E I ttLa3 0 0 0 0
−6 E IttLa2
−12E I ttLa3 0 0 0 0
−6 E IttLa2 0 0 0 0 0 0 u1 x
P2 y 0AELa
0 0 0 0 0−AELa
0 0 0 0 0 0 0 0 0 0 u2 y
P3 z 0 012E I ssLa3
6 EI ssLa2 0 0 0 0
−12E I ssLa3
6 EI ssLa2 0 0 0 0 0 0 0 0 u3 z
M 1x 0 06 EI ssLa2
4 E I ssLa
0 0 0 0−6 E I ssLa2
2E I ssLa
0 0 0 0 0 0 0 0 θ1 x
M 1 y 0 0 0 0JGLa
0 0 0 0 0−JGLa
0 0 0 0 0 0 0 θ1 y
M 1 z
−6 E IttLa2 0 0 0 0
4 E I ttLa
6 EI ttLa2 0 0 0 0
2E I ttLa
0 0 0 0 0 0 θ1 z
P2x−12E I ttLa3 0 0 0 0
6 EI ttLa2
12E I ttLa3 + AE
Lb0 0 0 0
6 EI ttLa2
−AELb
0 0 0 0 0 u2 x
P2 y 0−AELa
0 0 0 0 0AELa
+12 E I ttLb3 0 0 0
6 EI ttLb2 0
−12E I ttLb3 0 0 0
6 EI ttLb2 u2 y
P2 z 0 0−12E I ssLa3
−6 E I ssLa2 0 0 0 0
12E I ssLa3 +
12E I ssLb3
−6 E I ssLa2
−6 E I ssLa2 0 0 0
−12E I ssLb3 0
−6 E I ssLa2 0 u2 z
M 2x
=0 0
6 EI ssLa2
2E I ssLa
0 0 0 0−6 E I ssLa2
4 E I ssLa
+ JGLb
0 0 0 0 0−JGLb
0 0 θ2 x
M 2 y 0 0 0 0−JGLa
0 0 0−6 E I ssLa2 0
JGLa
+4 E I ssLb
0 0 06 EI ssLa2 0
2E I ssLb
0 θ2 y
M 2 z
−6 E IttLa2 0 0 0 0
2E I ttLa
6 EI ttLa2
6 EI ttLb2 0 0 0
4 E I ttLa
+4 EI ttLb
0−6 E IttLb2 0 0 0
2E I ttLb
θ2 z
P3x 0 0 0 0 0 0−AELb
0 0 0 0 0AELb
0 0 0 0 0 u3 x
P3 y 0 0 0 0 0 0 0−12E I ttLb3 0 0 0
−6 E IttLb2 0
12E I ttLb3 0 0 0
−6 E IttLb2 u3 y
P3 z 0 0 0 0 0 0 0 0−12E I ssLb3 0
6 EI ssLa2 0 0 0
12E I ssLb3 0
6 EI ssLa2 0 u3 z
M 3x 0 0 0 0 0 0 0 0 0−JGLb
0 0 0 0 0JGLb
0 0 θ3 x
M 3 y 0 0 0 0 0 0 0 0−6 E I ssLa2 0
2E I ssLb
0 0 06 EI ssLa2 0
4 E I ssLb
0 θ3 y
M 3 z 0 0 0 0 0 0 06 EI ttLb2 0 0 0
2E I ttLb
0−6 E IttLb2 0 0 0
4 E I ttLb
θ3 z
From our boundary conditions we can reduce our system of equations as follows:
![Page 11: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/11.jpg)
After applying the boundary conditions we arrive at:
M 2 z
4 E I ttLa
+4 EI ttLb
0−6 E IttLb2
2E I ttLb
θ2 z
P3x=
0AELb
0 0 u3 x
P3 y−6 E IttLb2 0
12E I ttLb3
−6 E IttLb2 u3 y
M 3 z
2E I ttLb
0−6 E IttLb2
4 E I ttLb
θ3 z
Make the following substitutions to write the governing matrix in terms of the given variables:
La=L2
Lb=L1
M 2 z
4 E I ttL2
+4 EI ttL1
0−6 E IttL12
2E I ttL1
θ2 z
P3x=
0AEL1
0 0 u3 x
P3 y−6 E IttL12 0
12E I ttL13
−6 E IttL12 u3 y
M 3 z
2E I ttL1
0−6 E IttL12
4 E I ttL1
θ3 z
![Page 12: Question 1](https://reader036.vdocuments.us/reader036/viewer/2022083011/5695cf5e1a28ab9b028dc853/html5/thumbnails/12.jpg)
Applying the boundary conditions for the support at node 3:
This question differs from Homework 4, Question 1 at node 2 (referred to as node 1 in Homework 4). This problem has a pin at node 2 resulting in an additional unknown rotation. Homework 4 is fixed at this node and cannot rotate; this reduces the degrees of freedom. The end result is a stiffness matrix in homework 4 that is 4x4 and a stiffness matrix in this problem that is 5x5.
04 E I ttL2
+4 EI ttL1
0−6 E IttL12
2E I ttL1
0 θ2 z
−12
=
0AEL1
0 0−1√2
u3 x
−12
−6 E IttL12 0
12E I ttL13
−6 E IttL12
1
√2u3 y
02E I ttL1
0−6 E IttL12
4 E I ttL1
0 θ3 z
0 0−1√2
1
√20 0 −λ