649
SOLUTION
Member 1:
l
x=
1 – 0
12
=0.7071
ly=
1 – 2
12
=–0.707
1
k
1=AE
≥
1 2 3 4
0.3536 –0.3536 –0.3536 0.3536
–0.3536 0.3536 0.3536 –0.3536
–0.3536 0.3536 0.3536 –0.3536
0.3536 –0.3536 –0.3536 0.3536 ¥
1
2
3
4
Member 2:
l
x=
2 – 1
12
=0.7071
ly=
0 – 1
12
=–0.707
1
k
2=AE
≥
3 4 5 6
0.3536 –0.3536 –0.3536 0.3536
–0.3536 0.3536 0.3536 –0.3536
–0.3536 0.3536 0.3536 –0.3536
0.3536 –0.3536 –0.3536 0.3536 ¥
3
4
5
6
Member 4:
l
x=
0 – 1
12
=–0.7071
ly=
0 – 1
12
=–0.707
1
k
4=AE
≥
3 4 7 8
0.3536 0.3536 –0.3536 –0.3536
0.3536 0.3536 –0.3536 –0.3536
–0.3536 –0.3536 0.3536 0.3536
–0.3536 –0.3536 0.3536 0.3536 ¥
3
4
7
8
Member 3:
l
x=
0 – 2
2
=–1
ly=
0
k
3=AE
≥
5 6 7 8
0.5 0 –0.5 0
0 0 0 0
–0.5 0 0.5 0
0 0 0 0 ¥
5
6
7
8
14–1. Determine the stiffness matrix K for the truss. AE is
constant.
1
5
3
2 m
2 m
8
7
2
1
3
4
5
6
42
4
1
2
3
4 kN
908
Ans.
–0.3536 0.8536 0.3536 –0.3536 0 0 0 –0.5
–0.3536 0.3536 1.0607 –0.3536 –0.3536 0.3536 –0.3536 –0.3536
0.3536 –0.3536 –0.3536 1.0607 0.3536 –0.3536 –0.3536 –0.3536
Member 5:
0–2
0 0 –0.3536 0.3536 0.8536 –0.3536 –0.5 0
0 0 0.3536 –0.3536 –0.3536 0.3536 0 0
0 0 –0.3536 –0.3536 –0.5 0 0.8536 0.3536
0–0.5 –0.3536 –0.3536 0 0 0.3536 0.8536 X
14–1. (Continued)
0–0.5 –0.3536 –0.3536 0 0 0.3536 0.8536 X
652
Ans.
q1=5.66
kN
1C2
q5=4.00
kN
1T2
For member 1,
14–2. (Continued)
SOLUTION
The origin of the global coordinate system will be set at joint 1 .
For member
1
,
L=2
m
.
l
x=
0–2
2
=–1
ly=
0–0
2
=
0
2
L=2
m
l
2–4
0–0
0
2
2
14–3. Determine the stiffness matrix K for the truss. Take
A=0.0015
m2
and
E=200
GPa
for each member.
30 kN
10
9
3
3
6
54
4
2
m
2 m2 m
4
6
11
2
5
3
5
7
1
2
2
8
655
Ans.
0 0 0 0 0 0 0 0 0 0
10
14–3. (Continued)
000–150 0 150 0 0 0 0
0 0 –53.033 –53.033 –150 0 203.033 53.033 0 0
0 0 –53.033 –53.033 0 0 53.033 53.033 0 0
0 0 –150 0 0 0 0 0 150 0
0 0 0 0 0 0 0 0 0 0
6
7
8
9
10
656
SOLUTION
*14–4. Determine the vertical displacement at joint 2 and
the force in member 5. Take
A=0.0015
m2
and
E=200
GPa
.
30 kN
10
9
3
3
6
54
4
2
m
2 m2 m
4
6
11
2
5
3
5
7
1
2
2
8
657
D1=–0.0004
m
D2=–0.0023314
m
D3=0.0004
m
D4=–0.00096569
m
D5=–0.0002
m
D6=0.000966
mT
1q52F=–42.4
kN
SOLUTION
D
k=
F
0
0
0
0
0
0V
3
4
5
6
7
8
Qk=c0
–2d
1
2
Using the structure stiffness matrix of Prob. 14–5 and
applying
Q=KD,
H0
–2
Q3
Q4
Q5
Q6
Q7
Q8X
=
H845.3289 13.79611 –160.2039 160.2039 –453.125 0 –232 –174
13.79611 290.7039 160.2039 –160.2039 0 0 –174 –130.5
–160.2039 160.2039 160.2039 –160.2039 0 0 0 0
160.2039 –160.2039 –160.2039 160.2039 0 0 0 0
–453.125 0 0 0 453.125 0 0 0
0 0 0 0 0 0 0 0
–232 –174 0 0 0 0 232 174
–174 –130.5 0 0 0 0 174 130.5 XHD
1
D2
0
0
0
0
0
0X
Partition matrix:
0=845.3289D1+13.79611D2
–2=13.79611D1+290.7039D2
Solving the above linear equations:
D1=0.11237110 –32
in.;
D2=–6.8852110 –32
in. =6.8852110 –32
in.T
Ans.
To nd force in member 2:
l
x=
0–4
4
=–1;
ly=
3–3
4
=
0
q
2=
0.75
1
29
1
103
22
48 [1
0
–1
0]
1
10 –3
2D0.11237
–6.8852
0
0T1
2
5
6
q2=50.9
lb
1T2
Ans.
14–6. Determine the vertical deflection of joint 1 and the
force in member 2 of the truss in Prob. 14–5.
1
2
3
2
3
4
1
2 k
2
1
4 ft
3 ft
4 ft
3
5
6
47
8