Determine the stiffness matrix K for the assembly... Determine the horizontal and vertical displacementsat joint 3 of the assembly in Prob... Use the structure stiffness matrix of Prob..
Trang 160 D
0.64 0.48 -0.64 -0.480.48 0.36 -0.48 -0.36
14–1 Determine the stiffness matrix K for the assembly.
Take A = 0.5 in2and E = 29(103)ksi for each member
7
21
58
Trang 214–2. Determine the horizontal and vertical displacements
at joint 3 of the assembly in Prob 14–1
7
21
58
Trang 358
Trang 453
1
Trang 5Use the structure stiffness matrix of Prob 14–4 and applying Q = KD We have
Partition matrix
(1)(2)Solving Eq (1) and (2) yields:
XH
14–5. Determine the horizontal displacement of joint
ksi
E = 29(103)
A = 0.75 in22
4
53
1
Trang 614–6 Determine the force in member if its temperature
4
53
-77.8371392.42700
H
D1
D2
000000
XH
T = AED
0
-650 0 650
T(10- 4)D
Trang 7The origin of the global coordinate system will be set at joint
T
1234
lx = 2 - 4
2 22
= - 222
14–7 Determine the stiffness matrix K for the truss.
Take A = 0.0015 m2and E = 200GPa for each member
30 kN
10
9
33
5
35
71
2
28
Trang 80.5 0.5 -0.5 -0.5 0.5 0.5 -0.5 -0.5
-0.5 -0.5 0.5 0.5
-0.5 -0.5 0.5 0.5
T
3478
-lx = 0 - 2
2 22
= 222
-L = 2 22 m5
Trang 9T
78910
*14–8. Determine the vertical displacement at joint
5
35
71
2
28
Trang 10From the matrix partition, Q k=K11D u+K12D kis given by
Expanding this matrix equality,
(1)(2)(3)(4)(5)(6)
3 4 7 8
L = 2 22 m
ly =
-222
lx =
-2225
14–8 Continued
Trang 11The origin of the global coordinate system will be set at joint
D
0.64 0.48 -0.64 -0.48 0.48 0.36 -0.48 -0.36
-0.64 -0.48 0.64 0.48
-0.48 -0.36 0.48 0.36
T
1256
1
14–9 Determine the stiffness matrix K for the truss.
Take A = 0.0015 m2and E = 200GPa for each member
10
65
44
3 m
20 kN4
5
35
78
Trang 12For member , , and
T
5 6 9 10
14–9 Continued
Trang 13For member , , and
Trang 14Then applying Q = KD
Expanding this matrix equality,
(1)(2)(3)(4)(5)(6)(7)
1 2 3 4 5 6 7
Dk= C
000S
8 9 10
14–10. Determine the force in member Take
and E = 200GPa for each member
44
3 m
20 kN4
5
35
78
Trang 1514–11. Determine the vertical displacement of node
if member was 10 mm too long before it was fitted
into the truss For the solution, remove the 20-k load Take
and E = 200GPa for each member
0000
-0.75000.7500 (106)
Q = KD + Q0
Dk= C
000S
8910
T = D
-0.75
0 0.75
0 T
5687 (106)
44
3 m
20 kN4
5
35
78
Trang 16From the matrix partition,
Expanding this matrix equality,
(1)(2)(3)(4)(5)(6)(7)
W + G
0000
-0.7500W(106)
6 ft1
2
21
78
3 k4
Trang 1714–12 Continued
Trang 185678
D
0.64 0.48 -0.64 -0.48 0.48 0.36 -0.48 -0.36
-0.64 -0.48 0.64 0.48
-0.48 -0.36 0.48 0.36
T
1234
Trang 19Applying ,
From the matrix partition;
Expanding this matrix equality,
(1)(2)(3)(4)(5)Solving Eqs (1) to (5),
Q = KD
678
14–13. Determine the horizontal displacement of joint
and the force in member Take
ksi Neglect the short link at 2
6 ft1
2
21
78
3 k4
Trang 20For member , L = 8(12) = 96 in , and
Thus,
Also,
and
Applying Q = KD + Q0
From the matrix partition, Qk=K11Du+K12Dk+(Qk)0,
Expanding this matrix equality,
(1)(2)(3)(4)(5)Solving Eqs (1) to (5),
U + E
15.100000U
X + H
15.1000000
-15.100X
Dk = C
000
S
678
-15.100T
1278D
-1010
T =2[29(103)]( - 0.025)
96D
14–14. Determine the force in member if this member
was 0.025 in too short before it was fitted onto the truss Take
ksi Neglect the short link at 2
6 ft1
2
21
78
3 k4
Trang 21Force in member Here, L = 8(12) = 96 in., , and
Ans.
= 3.55 k (T)
D
-0.01912 0.003305
The origin of the global coordinate system is set at joint
For member , Referring to Fig a,
2
1
34
2
13
3 kN
45⬚
Trang 22The structure stiffness matrix is a matrix since the highest code number is 6 Thus,
From the matrix partition; Qk=K11Du+K12Dk,
Expanding this matrix equality,
(1)(2)(3)
Q = KD
Dk= C
000
S
456
14–15 Continued
*14–16. Determine the vertical displacement of joint
and the support reactions AE is constant.
2
1
34
2
13
3 kN
45⬚
Trang 23S + C
000S
D1 =
6.750(103)
AE
14–16 Continued