SOLUTION
16–9. Determine the structure stiffness matrix K for the
frame. Assume ① and ③ are pins. Take
E=29(103)
ksi,
I=600
in4,
A=10
in2
for each member.
9
12
3
1
2
2
200 lb>ft
1
3
6
4
8 ft
12 ft
5
8
7
703
SOLUTION
See Prob. 16-9.
0
0
–1.2
16–10. Determine the internal loadings at the ends of each
member. Assume ① and ③ are pins. Take
E=29(103)
ksi,
I=600
in4,
A=10
in2
for each member.
9
12
3
1
2
2
200 lb>ft
1
3
6
4
8 ft
12 ft
5
8
7
SOLUTION
K=k1+k2
K
=
I
6123.248 0 5873.842 0 5873.842 –6041.667 0 –81.581 0
0 6123.248 5873.842 5873.842 0 0 –81.581 0 –6041.667
5873.842 5873.842 112 777.78 281 944.44 281 944.44 0 –5873.842 –5873.842 0
0 5873.842 281 944.44 563 888.88 0 0 –5873.842 0 0
5873.842 0 281 944.44 0 503 888.88 0 0 –5873.842 0
–6041.667 0 0 0 0 6041.667 0 0 0
0–81.581 –5873.842 –5873.842 0 0 81.581 0 0
–81.581 0 –5873.842 0 –5873.842 0 0 81.581 0
0–6041.667 0 0 0 0 0 0 6041.667
Y
D
k=
C
0
0
0
S
Qk=
F
5
–12
–288
288
0
0V
Substituting into
Q=KD,
partition, and solving for the displacements and loads yields
D1=0.63626 in.
D2=–1.159(10–3) in.
D3=–2.716(10–3) rad
D4=1.8809(10–3) rad
D5=–5.2697(10 –3) rad
D6=0.6363 in.
Q7=17.0 k
Ans.
Q8=–5.00 k
Ans.
Q9=7.00 k
Ans.
Ans.
Q7=17.0 k
Q8=–5.00 k
Q9=7.00 k
*16–12. Determine the support reactions at ① and ③. Take
E=29(103)
ksi,
I=700
in4,
A=30
in2
for each member.
Joint ① is pin connected.
12 ft
12 ft
5
k
2
1
3
2
7
1
4
6
9
8
5
2
1
3
2 k>ft
708
SOLUTION
12 –12
12 –0
0–0
–236 0 –11 328.125 –11 328.125 0 236 0 0 0
0–3020.833 0 0 0 0 3020.833 0 0
–2013.89 0 0 0 0 0 0 2013.89 0
0–69.927 5034.722 0 5034.722 0 0 0 69.927
16–13. Determine the structure stiffness matrix K for the
frame. Take
E=29(103)
ksi,
I=600
in4,
A=10
in2
for each
member. Assume joints ① and ③ are pinned; joint ② is fixed
connected. 1
2
9
8
3
2
1
3
8 ft
5 k
6 ft6 ft
1
2
4
7
6
5
SOLUTION
Member Stiffness Matrices. The origin of the global coordinate system
will be set at joint
○
1
. For member
1
,
L
=10 ft, l x=
0–0
10
=
0
and
l
y=
10 –0
10
=
1
AE
L
=
10[29(103)]
10(12) =2416.67 k
>
in.
12EI
L
3=
12[29(103)](300)
[10(12)]
3=60.4167 k
>
in
.
6EI
L
2=
6[29(103)](300)
[10(12)]
2=3625 k
4EI
L
=
4[29(103)](300)
10(12) =290 000 k #in
.
2EI
L
=
2[29(103)](300)
10(12)
=145 000 k #in
.
k
1=
F8 9 5 1 2 3
60.4167 0 –3625 –60.4167 0 –3625
0 2416.67 0 0 –2416.67 0
–3625 0 290 000 3625 0 145 000
–60.4167 0 3625 60.4167 0 3625
0–2416.67 0 0 2416.67 0
–3625 0 145 000 3625 0 290 000 V8
9
5
1
2
3
20 –0
10 –10
16–14. Determine the structure stiffness matrix K for the
frame. Take
E=29(103)
ksi,
I=300
in4,
A=10
in2
for each
member.
10 ft
20 ft
2
1
3
27
1
4
6
9
8
5
2
1
3
2 k>ft
L
10[29(103)]
>
L
12[29(103)](300)
[20(12)]
L
[20(12)]
L
20(12)
710
–1208.33 0 0 0 0 1208.33 0 0 0
0–7.5521 –906.25 –906.25 0 0 7.5521 0 0
–60.4167 0 –3625 0 –3625 0 0 60.4167 0
0–2416.67 0 0 0 0 0 0 2416.67
SOLUTION
16–17. Determine the structure stiffness matrix k
for each member of the two-member frame. Take
E=200
GPa,
I=350(106)
mm4,
A=20(103)
mm2
for each
member. Joints ① and ③ are pinned and joint ② is fixed
connected. 2
1
3
6
7
5
9
8
2
20 kN
12 kN>m
4
1
2
1
3
3 m 2 m 2 m
4 m
715
Ans.
K
=(106)
I
1292.3 380.77 13.44 0 13.44 –292.3 –380.77 –1000 0
380.77 527.544 16.17 26.25 –10.08 –380.77 –514.42 0 –13.125
13.44 16.17 126 35 28 –13.44 10.08 0 –26.25
0 26.25 35 70 0 0 0 0 –26.25
13.44 –10.08 28 0 56 –13.44 10.08 0 0
–292.3 –380.77 –13.44 0 –13.44 292.3 380.77 0 0
–380.77 –514.42 10.08 0 10.08 380.77 514.42 0 0
–1000 0 0 0 0 0 0 1000 0
0–13.125 –26.25 –26.25 0 0 0 0 13.125
Y
–292.3 –380.77 –13.44 0 –13.44 292.3 380.77 0 0
–380.77 –514.42 10.08 0 10.08 380.77 514.42 0 0
–1000 0 0 0 0 0 0 1000 0
0–13.125 –26.25 –26.25 0 0 0 0 13.125
SOLUTION
I
24 000
–28 000
15 000
10 000
–25 000
Q6 –24 000
Q7–18 000
Q8
Q9–10 000
Y
=(106)
I
1292.3 380.77 13.44 0 13.44 –292.3 –380.77 –1000 0
380.77 527.54 16.17 26.25 –10.08 –380.77 –514.42 0 –13.125
13.44 16.17 126 35 28 –13.44 10.08 0 –26.25
0 26.25 35 70 0 0 0 0 –26.25
13.44 –10.08 28 0 56 –13.44 10.08 0 0
–292.3 –380.77 –13.44 0 –13.44 292.3 380.77 0 0
–380.77 –514.42 10.08 0 10.08 380.77 514.42 0 0
–1000 0 0 0 0 0 0 1000 0
0–13.125 –26.25 –26.25 0 0 0 0 13.125
YI
D1
D2
D3
D4
D5
0
0
0
0
Y
24(10–3)=1292.3D1+380.77D2+13.44D3+13.44D5
–28(10–3)=380.77D1+527.54D2+16.17D3+26.25D4–10.08D5
15(10–3)=13.44D1+16.17D2+126D3+35D4+28D5
10(10–3)=26.25D2+35D3+70D4
–25(10–3)=13.44D1–10.08D2+28D3+56D5
D1=56.50(10–6) m
D2=–116.06(10–6) m
D3=244.01(10–6) rad
D4=64.38(10–6) rad
D5=–602.9(10–6) rad
Q6+24 000 =[–292.3(56.50)–380.77(–116.06)–13.44(244.01)
+0–13.44(–602.9)]
Ans.
Q6=8.50 kN
Q7–18 000 =[–380.77(56.50)–514.42(–116.06) +10.08(244.01)
+0+10.08(–602.9)]
Q7=52.6 kN
Ans.
Q8=[–1000(56.50) +0+0+0+0]
Q8=56.5 kN
Ans.
Q9–10 000 =[0–13.125(–116.06)–26.25(244.01)–26.25(64.38) +0]
16–18. Determine the support reactions at ① and ③. Take
E=200
GPa,
I=350(106)
mm4,
A=20(103)
mm2
for each
member. Joints ① and ③ are pinned and joint ② is fixed.
2
1
3
6
7
5
9
8
2
20 kN
12 kN>m
4
1
2
1
3
3 m 2 m 2 m
4 m