687
SOLUTION
Member 1.
l
x=
10 –0
10
=1
ly=
0
AE
L
=
20(29)(103)
10(12) =4833.33 k
>
in.
12EI
L
3=
12(29)(103)(650)
(10)
3
(12)
3=130.90 k
>
in
.
6EI
L
2=
6(29)(103)(650)
(10)
2
(12)
2=7854.17 k
4EI
L
=
4(29)(103)(650)
(10)(12) =628333.33 k #in
.
2EI
L
=
2(29)(103)(650)
(10)(12)
=314166.67 k #in
.
k
1=
F1 2 3 4 5 6
4833.33 0 0 –4833.33 0 0
0 130.90 7854.17 0 –130.90 7854.17
0 7854.17 628333.33 0 –7854.17 314166.67
–4833.33 0 0 4833.33 0 0
0–130.90 –7854.17 0 130.90 –7854.17
0 7854.17 314166.67 0 –7854.17 628333.33 V1
2
3
4
5
6
Member 2.
l
x=0
ly=
–12 –0
12
=–1
AE
L
=
(20)(29)(103)
(12)(12) =4027.78 k
>
in.
12EI
L
3=
12(29)(103)(650)
(12)
3
(12)
3=75.75 k
>
in
.
6EI
L
2=
6(29)(103)(650)
(12)
2
(12)
2=5454.28 k
4EI
L
=
4(29)(103)(650)
(12)(12) =523611.11 k #in
.
2EI
L
=
2(29)(103)(650)
(12)(12)
=261805.55 k #in
.
16–1. Determine the structure stiffness matrix K for the
frame. Take
E=29(103)
ksi,
I=650
in4,
A=20
in2
for each
member.
2
1
6
5
4
2
1
3
1
6 k
k
39
8
7
12 ft
688
Ans.
K
=
I
4833.33 0 0 –4833.33 0 0 0 0 0
0 130.90 7854.17 0 –130.90 7854.17 0 0 0
0 7854.17 628 333.33 0 –7854.17 314 166.67 0 0 0
–4833.33 0 0 4909.08 0 5454.28 –75.75 0 5454.28
0–130.90 –7854.17 0 4158.68 –7854.17 0 –4027.78 0
0 7854.17 314 166.67 5454.28 –7854.17 1151944.44 –5454.28 0 261805.55
0 0 0 –75.75 0 –5454.28 75.75 0 –5454.28
0 0 0 0 –4027.78 0 0 4027.78 0
0 0 0 5454.28 0 261805.55 –5454.28 0 523611.11
Y
k>in
.
0 7854.17 314 166.67 5454.28 –7854.17 1151944.44 –5454.28 0 261805.55
0 0 0 –75.75 0 –5454.28 75.75 0 –5454.28
0 0 0 0 –4027.78 0 0 4027.78 0
0 0 0 5454.28 0 261805.55 –5454.28 0 523611.11
SOLUTION
Member 1:
12 –0
16–3. Determine the structure stiffness matrix K for the
frame. Take
E=29(103)
ksi,
I=450
in4,
A=8
in2
for each
member. All joints are fixed connected.
50 k ?
ft
3
12 ft
8 ft
1
1
12
811
4
10
7
9
5
6
4
23
3
2
1
4
k
2
SOLUTION
0
0
0
4
0
0
*16–4. Determine the horizontal displacement of
joint ②. Also compute the support reactions. Take
E=29(103)
ksi,
I=450
in4,
A=8
in2
for each member. All
joints are fixed connected.
50 k ?
ft
3
12 ft
8 ft
1
1
12
811
4
10
7
9
5
6
4
23
3
2
1
4
k
2
693
Ans.
D1=0.0803 in.
Q7=–3.79 k
Q8=–8.15 k
Q9=23.6 k #ft
Q10 =–0.214 k
Q11 =8.15 k
Q12 =9.27 k #ft
D1=0.08029 in. =0.0803 in.
D2=0.005056 in.
D3=–0.0001121 rad
D4=0.08020 in.
D5=–0.005056 in.
D6=–0.001054 rad
Q7=–3.79 k
Q8=–8.15 k
Q10 =–0.214 k
Q11 =8.15 k
Q12 =9.27 k #ft
SOLUTION
Member 1:
12 –0
30(29)(103)
16–5. Determine the structure stiffness matrix K for each
member of the frame. Take
E=29(103) ksi, I=700 in4,
A=30 in2
for each member.
1
5
8
1
3
2
2
23
7
4
6
8 k
12 ft
1
6 ft 6 ft
9
2(29)(103)(700)
695
SOLUTION
16–6. Determine the internal loadings at the ends of each
member. Take
E=29(103) ksi, I=700 in4, A=30 in2
for
each member.
1
5
8
1
3
2
2
23
7
4
6
8 k
12 ft
1
6 ft 6 ft
9
697
Ans.
Q7=4.00 k;
Q8=0;
Q9=4.00 k
Support reactions:
Q7–4=0–81.58(–0.0006621) –5873.84(–0.0005062) –5873.84(0.0005153) +0+0
Q7=4.00 k
Ans.
Q8=–81.58(–0.07289) +0–5873.84(–0.0005062) –5873.84(–0.0005062)
Q8=0
Ans.
Q9=0–6041.67(–0.0006621) +0+0+0+0
Q9=4.00 k
Ans.
Check equilibrium:
+
c
ΣFy=0;
–8+4+4=0
(Check)
a+ΣM3=0;
4(12) –8(6) =0
(Check)
16–6. (Continued)
SOLUTION
For member 1,
5–0
16–7. Determine the structure stiffness matrix K for
the frame. Assume ③ is pinned and ① is fixed. Take
E=200
MPa,
I=300(106)
mm4,
A=21(103)
mm2
for each
member.
2
1
2
3
2
300 kN ?m
1
9
5 m
4 m
3
6
4
5
1
8
7
699
701
Ans.
Q5=36.3 kN d
Q6=46.4 kNT
Q7=36.3 kN S
Q8=46.4 kN
c
Q9=77.1 kN m B
Q5=–36.3 kN =36.3 kN d
Q7=36.3 kN S
Q9=77.1 kN m B