660
SOLUTION
k3=AE
D0.33333 0 –0.33333 0
0 0 0 0
–0.33333 0 0.33333 0
0 0 0 0 T5
6
1
2
14–7. Determine the stiffness matrix K for the truss. AE is
constant.
32
1
3 m
4 m
6
5
2
1
3
4
2
1
3
3 kN
458
661
Ans.
K
=AE
F0.40533 0.096 0.01697 –0.11879 –0.33333 0
0.096 0.128 0.02263 –0.15839 0 0
0.01697 0.02263 0.129 –0.153 0 0.17678
–0.11879 –0.15839 –0.153 0.321 0 –0.17678
–0.33333 0 0 0 0.33333 0
0 0 0.17678
–
0.17678 0 0.25
V
–0.11879 –0.15839 –0.153 0.321 0 –0.17678
–0.33333 0 0 0 0.33333 0
0 0 0.17678 –0.17678 0 0.25 V
663
SOLUTION
Member 1:
l
x=0
ly=
0–3
3
=–
1
k
1=AE
D1 2 3 4
0 0 0 0
0 0.3333 0 –0.3333
0 0 0 0
0–0.3333 0 0.3333 T1
2
3
4
Member 2:
l
x=
0–4
4
=–1
ly=
0
k
2=AE
D1 2 7 8
0.25 0 –0.25 0
0 0 0 0
–0.25 0 0.25 0
0 0 0 0 T
1
2
7
8
Member 3:
l
x=
0–4
4
=–1
ly=
0
k
3=AE
D3 4 6 5
0.25 0 –0.25 0
0 0 0 0
–0.25 0 0.25 0
0 0 0 0 T
3
4
6
5
Member 4:
l
x=0
ly=
3–0
3
=
1
k
4=AE
D6 5 7 8
0 0 0 0
0 0.3333 0 –0.3333
0 0 0 0
0–0.3333 0 0.3333 T6
5
7
8
Member 5:
l
x=
0–4
5
=–0.8
ly=
3–0
5
=0.
6
k
5=AE
D3478
0.128 –0.096 –0.128 0.096
–0.096 0.072 0.096 –0.072
–0.128 0.096 0.128 –0.096
0.096 –0.072 –0.096 0.072 T3
4
7
8
14–9. Determine the stiffness matrix K for the truss. AE is
constant.
1
2
6
4
8
473
2
1
5
5
10 kN
4
3 m
4 m
Member 6:
l
x=
0–4
5
=–0.8
ly=
0–3
5
=–0.
6
k
6=AE
D1265
0.128 0.096 –0.128 –0.096
0.096 0.072 –0.096 –0.072
–0.128 –0.096 0.128 0.096
–0.096 –0.072 0.096 0.072 T1
2
6
5
Structure stiffness matrix:
K=k1+k2+k3+k4+k5+k6
0.096 0.4053 0 –0.3333 –0.072 –0.096 0 0
0 0 0.378 –0.096 0 –0.25 –0.128 0.096
0–0.3333 –0.096 0.4053 0 0 0.096 –0.072
14–9. (Continued)
SOLUTION
D1Q
1
2
0
1
Q2
2
0
1
Q3
2
0
1
Q
420T
=
AE
1
0.01
2
3
D0
–1
0
1T1
2
3
4
=AE
D0
–0.003333
0
0.003333 T1
2
3
4
Use the structure stiffness matrix of Prob. 14-9.
H0
0
0
0
0
Q6
Q7
Q8X
=AE
H0.378 0.096 0 0 –0.096 –0.128 –0.25 0
0.096 0.4053 0 –0.3333 –0.072 –0.096 0 0
0 0 0.378 –0.096 0 –0.25 –0.128 0.096
0–0.3333 –0.096 0.4053 0 0 0.096 –0.072
–0.096 –0.072 0 0 0.4053 0.096 0 –0.3333
–0.128 –0.096 –0.25 0 0.096 0.378 0 0
–0.25 0 –0.128 0.096 0 0 0.378 –0.096
0 0 0.096 –0.072 –0.3333 0 –0.096 0.4053 XHD
1
D2
D3
D4
D5
0
0
0X
+AE
H0
–0.003333
0
0.003333
0
0
0
0X
0=0.378D1+0.096D2+0D3+0D4–0.096D5+0
(1)
0=0.096D1+0.4053D2+0D3–0.3333D4–0.072D5–0.003333
(2)
0=0D1+0D2+0.378D3–0.096D4+0D5+0
(3)
0=0D1=–0.3333D2–096D3+0.4053D4+0D5+0.003333
(4)
0=–0.096D1–0.072D2+0D3+0D4+0.4053D5+0
(5)
Solving the above equations yields:
D1=–0.0011111,
D2=0.005
D3=–0.0011111,
D4=–0.004375
D5=0.000625
The force in member 1:
lx=0,
ly=–1,
L=3
m
q
1
=
0.0015
1
200
21
109
2
3 [0
1
0
–1]
D–0.001111
0.0050000
–0.001111
–0.004375 T1
2
3
4
–
1
0.0015
21
200
21
109
21
0.003333
2
=–62.5
kN =62.5
kN
1C2
Ans. Ans.
q1=62.5
kN
1C2
14–11. Determine the force in member 1 if this member was
10 mm too long before it was fitted into the truss. For the
solution remove the 10-kN load. Take
A=0.0015
m2
and
E=200
GPa
for each member.
1
2
6
3
4
8
473
2
1
5
5
10 kN
16
4
3
3
m
4 m
2
668
Ans.
K
=AE
I
0.03536 –0.03536 0 0 –0.03536 0.03536 0 0 0 0
–0.03536 0.13536 0 –0.10 0.03536 –0.03536 0 0 0 0
0 0 0.10 0 –0.10 0 0 0 0 0
0–0.10 0 0.20 0 0 0 0 0 –0.10
–0.03536 0.03536 –0.10 0 0.17071 0 0 0 –0.03536 –0.03536
0.03536 –0.03536 0 0 0 0.17071 0 –10 –0.03536 –0.03536
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 –0.10 0 –0.10 0 0
0 0 0 –0.10 –0.03536 –0.03536 0 0 0.03536 0.03536
Y
Member 6:
l
x=0
ly=
0–10
10
=–
1
k
6=AE
D0 0 0 0
0 0.10 0 –0.10
0 0 0 0
0
–
0.10 0 0.10
T
Structure stiffness matrix:
K=k1+k2+k3+k4+k5+k6
0.03536 –0.03536 0 0 –0.03536 0.03536 0 0 0 0
–0.03536 0.13536 0 –0.10 0.03536 –0.03536 0 0 0 0
0 0 0.10 0 –0.10 0 0 0 0 0
0–0.10 0 0.20 0 0 0 0 0 –0.10
–0.03536 0.03536 –0.10 0 0.17071 0 0 0 –0.03536 –0.03536
0.03536 –0.03536 0 0 0 0.17071 0 –10 –0.03536 –0.03536
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 –0.10 0 –0.10 0 0
0 0 0 0 –0.03536 –0.03536 0 0 0.03536 0.03536
0 0 0 –0.10 –0.03536 –0.03536 0 0 0.03536 0.03536
14–12. (Continued)
670
Ans.
D
1=
933
AE
q1=11.3
k
1C2
D
932.548
D
160.00