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Problem 11.1
Estimate the bending moment distribution for the cases listed below. Use qualitative reasoning based on relative stiffness. Assume I constant.
(a)
(b)
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2
Problem 11.2Estimate the bending moment distribution. Use qualitative reasoning based on relative stiffness.Assume I constant.
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Problem 11.3
Solve problem 11.1 cases (a) and (b) using moment distribution. Compare the approximate and
exact results.
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4
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Problem 11.4
Consider the multi-story steel frame shown below. Determine the maximum positive and
negative moments in the beams using the following approaches:
(i) Assume inflection points at .1L from each end of the beams.
(ii) Use a computer software system. Assume g cI 2I
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6
Moment diagram-
max
+
max
M 49 kN-m
M 27.6 kN-m
Problem 11.5
Estimate the axial force, shear force, and bending moment distributions. Assume g cI 2I
(a)
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7
(b)
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Assume shear is divided equally between columns.
Problem 11.6
Members AC and FD are continuous. Estimate the bending moment distribution in AC and FD,
and the axial forces in the pin ended members. Compare your results with results generated witha computer software system.
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9
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10
Axial forces & reaction
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Moment distribution
Problem 11.7
Repeat problem 11.6 assuming fixed supports at A and F.
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12
Axial forces & reaction
Moment distribution
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Problem 11.8
Consider the steel frame shown below. Assume g cI 3 I for all the members. Determine the
moment at each end of each member using
(a) The portal method.
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(b) The shear stiffness method.
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XX
C 1
FX 2.300E+00
XX
C 2FX 9.398E-01
XX
C 4
FX -9.398E-01
XX
C 3
FX -1.782E+00
XXG3
FX -6.667E+00
XX
C 5
FX 1.367E+00
XXG4
FX -3.040E+00
XX
C 6 FX -1.886E+00
XXG1
FX -7.289E+00
XXG2
FX -2.251E+00
Axial forces( kip)
XX
C 1FY -2.459E+00
XX
C 2FY -1.749E+00
XXG2
FY 9.398E-01
XX
C 4
FY -2.251E+00
XXG1
FY 1.360E+00
XXG3
FY 5.185E-01
XXG4
FY 1.886E+00
XX
C 6 FY -3.040E+00
XX
C 5
FY -3.627E+00
XX
C 3
FY -2.873E+00
Shear forces (kip)
Moment diagram (kip-ft)
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(a)
XX
C 1
FX 1.694E+00
XX
C 2FX 5.063E-01
XXG2
FX -4.362E+00
XXG1
FX -7.690E+00
XX
C 3
FX -1.010E+00
XXG3
FX -5.691E+00XX
C 4
FX -2.955E-01
XXG4
FX -3.073E+00
XXG6
FX -1.796E+00
XXG8
FX -4.256E-01
XX
C 1 0
FX -2.582E-01
XX
C 9
FX -8.080E-01
XXG7
FX -2.048E+00
XX
C 7
FX -7.199E-01
XX
C 8
FX -2.166E-01
XXG5
FX -3.822E+00
XX
C 5
FX 8.441E-01
XX
C 6FX 2.640E-01
Axial force (kip)
XX
C 1
FX 1.694E+00FY -2.948E+00
XX
C 2
FX 5.063E-01
FY -6.378E-01
XXG2
FX -4.362E+00FY 5.063E-01
XXG4
FX -3.073E+00FY 2.108E-01
XXG6
FX -1.796E+00FY 4.748E-01
XXG8
FX -4.256E-01FY 2.582E-01
XX
C 1 0
FX -2.582E-01FY -4.256E-01
XXG7
FX -2.048E+00FY 5.498E-01
XX
C 9
FX -8.080E-01FY -2.473E+00
XX
C 7
FX -7.199E-01FY -3.145E+00
XX
C 5
FX 8.441E-01FY -3.146E+00
XXG3
FX -5.691E+00
FY 4.730E-01 XX
C 6
FX 2.640E-01FY -1.277E+00 XX
C 8
FX -2.166E-01FY -1.370E+00XX
C 4
FX -2.955E-01FY -1.289E+00
XXG1
FX -7.690E+00FY 1.188E+00
XX
C 3
FX -1.010E+00FY -3.288E+00
Shear forces (kip)
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18
Bending moment diagram(kip-ft)
(c)
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XX
C 2FX 2.003E+00
XX
C 1
FX 4.911E+00
XXG1
FX -7.620E+00
XX
C 3
FX -1.818E+00
XXG3
FX -4.832E+00
XX
C 5
FX 1.808E+00
XXG5
FX -2.149E+00
XX
C 7
FX -4.900E+00
XX
C 8
FX -2.016E+00
XXG6
FX -3.178E+00XX
C 1 0
FX -9.926E-01
XXG7
FX -3.002E+00
XX
C 9FX 9.926E-01
XXG2
FX -4.884E+00
XX
C 4
FX -3.577E-02
XXG4
FX -4.042E+00
XX
C 6
FX 4.871E-02
Axial forces (kip)
XX
C 1
FX 4.911E+00FY -5.496E+00
XXG1
FX -7.620E+00FY 2.908E+00
XX
C 3
FX -1.818E+00FY -6.629E+00
XXG3
FX -4.832E+00
FY 1.125E+00
XX
C 5FX 1.808E+00
FY -6.548E+00
XX
C 7
FX -4.900E+00FY -5.327E+00
XX
C 8
FX -2.016E+00FY -3.178E+00
XXG5
FX -2.149E+00FY 2.884E+00
XXG6
FX -3.178E+00FY 2.016E+00
XXG4
FX -4.042E+00FY 9.750E-01
XX
C 6
FX 4.871E-02FY -3.866E+00
XXG7
FX -3.002E+00FY 9.926E-01
XX
C 9
FX 9.926E-01FY -2.998E+00
XXG2
FX -4.884E+00FY 2.003E+00
XX
C 4
FX -3.577E-02FY -3.841E+00
XX
C 2FX 2.003E+00
FY -3.116E+00
Shear forces (kip)
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20
Moment diagram (kip-ft)
Problem 11.10
For the steel frames shown, estimate the axial force, shear force, and moments for all of the
members. Take 4cI 480 in for all the columns and4
bI 600 in for all the beams. Use the
Stiffness method.(a)
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Problem 11.11
Estimate the column axial forces in the bottom story for the distribution of column areas shown.
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Problem 11.12
Estimate the column shears for cases (a) and (b). Compare your results with computer basedsolutions.
(a)
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25
(b)
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Problem 11.13
Consider the rigid steel frame with bracing shown below. Estimate the column shears and brace
forces. Take6 4 6 4 2
c g bI 40(10) mm I 120(10) mm A 650 mm E 200 GPa . Compare your
results with a computer based solution.
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Reactions and brace forces
Column shear
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1
Problem 12.1For the rigid frame shown below, use the direct stiffness method to find the joint displacements,
and reactions for the following conditions.
6 4
2
6 o
I 240(10) mm
A 3800 mm
E 200 Gpa
12(10) / F
Member m n n α
(1) 1 2 090
(2) 2 3 0
Member m α cos sin sin cos 2cos 2sin
(1) 090 0 1 0 0 1
(2) 0 1 0 0 1 0
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2
2 2
3 3 2
2 2
3 3 2m AA
2 2
2 2
3 3
m AB
AE 12EI AE 12EI 6EI( cos sin ) ( )sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cosL L L
AE 12EI AE 12EI( cos sin ) ( ) sin cos
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIsin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cos
L L L
2 2
3 3 2
2 2
m BA 3 3 2
2 2
2 2
3 3
m BB
AE 12EI AE 12EI 6EI( cos sin ) ( ) sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cos
L L L
AE 12EI AE 12EI( cos sin ) ( )sin co
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIs sin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cos
L L L
0 0
0
(1) AA (1) AB (1) AA (1) AB
(1) BA (1) BB (2) AA (2) AB (1) BA (1) BB (2) AA (2) AB
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00
K = k k 0 k k = k (k +k ) k
0 0 0 0 k k 0 k k
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3
1
11 E 12 I
E 21 22 I
( ) ( - - )
'' ''
E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''
21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K
K K
' '11 12
' '21 22
( )
0
(1) BB (2) AA (1) BA (2) AB
(1) AA(1) AB
(2) BA (2) BB
k + k k k
k k
k 0 k
K K
K K
1
1
2
1' ' '
2
3
2
3
3
u
uθ
uθ
θ
U U
1
x 1
y
1' '
x 3
y 3
3
R
R
M
R
R
M
EP
(a) The loading shown
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4
' ' ' '' '
E I I
0
020 kN 0
0
40 kN 45 kN 000 22.5 kN-m
45 kN
22.5 kN-m
P P P U
Using Mathcad:
2
1
2 11 E I
2
u 0.07056
( ) ( - ) -0.148
0.0000193θ
' ' ' 'U P PK
1
x 1
y
1
E 21 I
x 3
y 3
3
R -2.12
R 37.46
M 2.87
-17.87R
2.53R -4.11
M
'' ' ' ''P U PK
(b)
Member 2 experiences to a uniform temperature change throughout its span. The
temperature varies linearly through the depth, from 010 C at the top to 050 C at the
bottom.
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5
um u
62 6
F 6
3
F
62 6
F m 6
3
F m
T T (10 50)T 30 T T T 10 (50) 40
2 2ΔT (200)(240)(10) 40
M EIα (12)(10) ( ) 57.6 kN-mh (10) 0.4
ΔTM EIα 57.6 kN-m
h
(200)(240)(10)F EI α T (12)(10) (30) 17.28 kN
(10)
F EI α T 17.28 kN
l l
' ' ' '' '
E I I
0
017.28 kN
00 0 0
17.28 kN57.6 kN-m
0
57.6 kN-m
P P P U
Using Mathcad:
2
1
2 11 I
2
u -0.0556
( ) ( - ) -0.0556
0.00048θ
' ' 'U PK
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6
1
x 1
y
1
E 21 I
x 3
y 3
3
R -14.1
R 14.1
M 13.5
14.1R
-14.1R 71.1
M
'' ' ' ''P U PK
(c) Support 1 settles 12 mm.
' ' ' '' '
E I I
0
12 mm
00 0 0
0
0
0
P P P U
Using Mathcad:
2
1
2 11 12
2
u -0.342
( ) ( - ) -11.41
0.00294θ
' ' ' ''U UK K
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7
1
x 1
y
1
E 21 22
x 3
y 3
3
R -86.72
R -149.40
M 83.07
86.72R
149.40R -271.11
M
'' ' ' ' ''P U UK K
Problem 12.2
For the rigid frame shown below, use the direct stiffness method to find the joint displacements,
and reactions.4
2
2
I 400 in
A 10 inE 29,000 k/in
Member m n n α
(1) 1 2 - 0116.56
(2) 2 3 0
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8
2 2
3 3 2
2 2
3 3 2m AA
2 2
2 2
3 3
m AB
AE 12EI AE 12EI 6EI( cos sin ) ( )sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cosL L L
AE 12EI AE 12EI( cos sin ) ( ) sin cos
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIsin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cos
L L L
2 2
3 3 2
2 2
m BA 3 3 2
2 2
2 2
3 3
m BB
AE 12EI AE 12EI 6EI( cos sin ) ( ) sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cos
L L L
AE 12EI AE 12EI( cos sin ) ( )sin co
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIs sin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cos
L L L
B A
B A B A
0 0
0
(1) AB (1) AA (1) A (1) A
(1) B (1) B (2) AA (2) AB (1) B (1) B (2) AA (2) AB
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00
K = k k 0 k k = k (k + k ) k
0 0 0 0 k k 0 k k
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9
A B
BA
' '11 12
' '21 22
0
(1) B (2) AA (1) B (2) AB
(1) A(1) A
(2) BA (2) BB
(k + k ) k k
k k
k 0 k
K K
K K
1
11 E 12 I
E 21 22 I
( ) ( - - )
'' ''
E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K K K
1
1
2
1' ' '
2
3
2
3
3
u
uθ
uθ
θ
U U
1
x 1
y
1' '
x 3
y 3
3
R
R
M
R
R
M
EP
' ' ' '' 'E I I
6 kip
8 kip 0 0 0
0
P P P U
Using Mathcad:
2
1
2 11 E
2
u 0.00386
( ) ( ) -0.0062
0.000008θ
' ' 'U PK
1
x 1
y
1
E 21
x 3
y 3
3
R 3.33
R 7.54
M 2.12
-9.33R
0.46R -2.36
M
'' ' 'P UK
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Problem 12.3
For the rigid frames shown below, use the direct stiffness method to find the joint displacements,and reactions.
Member m n n α
(1) 1 2 0
(2) 2 3 - 053.13
2 2
3 3 2
2 2
3 3 2m AA
2 2
2 2
3 3
m AB
AE 12EI AE 12EI 6EI( cos sin ) ( )sin cos sinL L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cos
L L L
AE 12EI AE 12EI( cos sin ) ( ) sin cos
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIsin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cosL L L L L
6EI 6EI 2EIsin cos
L L L
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11
2 2
3 3 2
2 2
m BA 3 3 2
2 2
2 2
3 3
m BB
AE 12EI AE 12EI 6EI( cos sin ) ( ) sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cosL L L
AE 12EI AE 12EI( cos sin ) ( )sin co
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIs sin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cos
L L L
0 0
0
(1) AA (1) AB (1) AA (1) AB
(1) BA (1) BB (2) AA (2) AB (1) BA (1) BB (2) AA (2) AB
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00
K = k k 0 k k = k (k +k ) k
0 0 0 0 k k 0 k k
1
11 E 12 I
E 21 22 I
( ) ( - - )
' ' ''E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K K K
(a)
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12
1
x 1
2
y
2' ' ' ' ' ' '' '
1 E I I
2
x 33
y 3
R
u 20 kN
R 00 M 0 0
θ 0R
θ 0R
EP P P PU U
Using Mathcad:
Expand the matrix K
(1) AA (1) AB
(1) BA (1) BB (2) AA (2) AB
(2) BA (2) BB
k k 0
K = k (k + k ) k
0 k k
2
2 1
11 E
2
3
u 0.076
0.056 ( ) ( )
θ 0.0000015
-0.000029θ
' ' 'U PK
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13
1
x 1
y
1 E 21
x 3
y 3
R -19.78
R -0.16
M -0.41
-0.22R
0.16R
'' ' 'P UK
(b)
1
x 1
y
2
1' ' ' ' ' ' '' '
2 E I I
x 32
y 3
3
R 0
R 75kNu 0
M 62.5 kN0 0 75 kN
0R θ -62.5 kN
0R
0M
EP P P PU U
' '11 12
' '21 22
0
(1) BB (2) AA (1) BA (2) AB
(1) AA(1) AB
(2) BA (2) BB
(k + k ) k k
k k
k 0 k
K K
K K
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14
Using Mathcad:
2
1
2 11 I
2
u -0.22
( ) ( - ) -0.58
0.0012θ
' ' 'U PK
1
x 1
y
1
E 21 I
x 3
y 3
3
R 57.56
R 86.09
M 82.46 57.56
-57.56R
63.91R 11.49
M
'' ' ' ''P U PK
Problem 12.4
For the rigid frame shown, use partitioning to determine '11K ,'21
K 'EP , and '
IP for the loadings
shown. 6 4I 40(10) mm , A=3000 2mm , and E=200 GPa.
Member m n n α
(1) 1 4 0
(2) 2 4 0180
(3) 3 4 090
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15
2 2
3 3 2
2 23 3 2m AA
2 2
2 2
3 3
m AB
AE 12EI AE 12EI 6EI( cos sin ) ( )sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cosL L L L L
6EI 6EI 4EIsin cos
L L L
AE 12EI AE 12EI( cos sin ) ( ) sin cos
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIsin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cosL L L
2 2
3 3 2
2 2
m BA 3 3 2
2 2
2 2
3 3
m BB
AE 12EI AE 12EI 6EI( cos sin ) ( ) sin cos sin
L L L L L
AE 12EI AE 12EI 6EI= ( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 2EIsin cos
L L L
AE 12EI AE 12EI( cos sin ) ( )sin co
L L L L
=
k
k
2
2 2
3 3 2
2 2
6EIs sin
L
AE 12EI AE 12EI 6EI( )sin cos ( sin cos ) cos
L L L L L
6EI 6EI 4EIsin cos
L L L
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16
3 3
3 3
3 3
0 0 0 0 0 0 0
0 0
0 0 0 0
0
(1) AA (1) AB
(2) AA (2) AB
( ) AA ( ) AB
(1) BA (1) BB (2) BA (2) BB ( ) BA ( ) BB
(1) AA (1) AB
(2) AA (2) AB
( ) AA (
k 0 k 0 0
0 k 0 k 0 00 0 0 0K =
0 0 k k 0 0 0 0 0 0 0 0
k 0 k k 0 k k k
k 0 k
0 k 0 k
0 0 k k
2 3 3
( )
) AB
(1) BA ( ) BA ( ) BA (1) BB (2) BB ( ) BB
k k k k k k
3
2 3
3
3
'11
'12
'21
'22
( )
0
(1) BB (2) BB ( ) BB
(1) BA ( ) BA ( ) BA
(1) AB
(2) AB
( ) A B
(1) AA
(2) AA
( ) A A
k k k
k k k
k
k
k
k 0
0 k 0
0 0 k
K
K
K
K
' '
E I
0 0
0 84 kN
16 kN-m 84kN-m
P P
Problem 12.5
For the truss shown, use the direct stiffness method to find the joint displacements, reactions, and
member forces.2A 1200 mm
E 200 GPa
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Member m n
n
α L (m)(kN/mm)
AE
L
(1) 1 2 051.34 6.4 37.5
(2) 1 3 0 8 30
(3) 3 2 0128.66 6.4 37.5
3 3
3 3
3 3
3
0 0 0
0 0 0 0
0
( )
( )
(
(1) AA (1) AB (2) AA (2) AB
(1) BA (1) BB ( ) AA ( ) AB
(2) BA (2) BB( ) BA ( ) BB
(1) AA (2) AA (1) AB (2) AB
(1) BA (1) BB ( ) AA ( ) AB
(2) BA ( ) BA (2) B
k k 0 k k 0
K = k k 0 k k
k k 0 0 0 0 k k
k k k k
= k k k k
k k k 3
)
B ( ) BBk
2
( ) ( ) (m)
(m)AA (m)BB ( ) ( ) 2
(m) (m) ( )
2
( ) ( ) (m)
(m)AB (m)BA ( ) ( ) 2
(m) (m) ( )
cos α sinα cosαAE( )
sinα cosα sin αL
cos α sinα cosαAE( )
sinα cosα sin αL
m m
m m
m
m m
m m
m
k k k
k k k
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(m) (m)
F
B ( ) (m) (m) B-+ nn
AEF ( ) cos sin ( ) F
L m
U U
1
11 E 12 I
E 21 22 I
( ) ( - - )
' ' ''E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K K K
2' ' ' '
E I I
2
u 30 kN0 0 0
50 kN
' ''P P PU U
Using Mathcad:
1
2 1
11 E
2
x 1
y
E 21
x 3
y 3
u 0.04 ( )
-0.043
R 5
R 6.25
-35R
43.75R
' ' '
'' ' '
U P
P U
K
K
(1)
(2)
(3)
B (1) (1) (1)
B (2) (2) (2)
B (3) (3) (3)
-+
-+
-+
nn
nn
nn
AEF ( ) cos sin ( ) 8
L
AEF ( ) cos sin ( ) 0
L
AEF ( ) cos sin ( ) 56
L
U U
U U
U U
Problem 12.6
For the truss shown, use the direct stiffness method to find the joint displacements, reactions, and
member forces for
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2A 2 in
E 29,000 ksi
Member m n n α L (ft) (kip/in.) AE
L
(1) 1 2 053.13 20 241.67
(2) 4 2 0 16 302.1
(3) 3 2 0126.87 22.63 213.58
3 3
3 3
0 0 0 0 0 0 0
0 0
0
0 0 00 0 0
0
( )
(1) AA (1) AB
(2) AB (2) AA(1) BA (1) BB
( ) AB ( ) AA
(2) BB (2) BA ( ) BB ( ) BA
(1) AA (1) AB
(1) BA (1) BB (2) AB (2
k k 0 0 0
0 k 0 k 0 0k k 0 0K =
0 k k 0 0 0 00 0 0 0
k 0 k k k 0
k k 0
k k k 0 k
3 3
3 3
0
0 ( )
) AA
( ) AB ( ) AA
(2) BB ( ) BB ( ) BA (2) BA
0 k k
k k k k
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2
( ) ( ) (m)
(m)AA (m)BB ( ) ( ) 2
(m) (m) ( )
2
( ) ( ) (m)
(m)AB (m)BA ( ) ( ) 2
(m) (m) ( )
cos α sinα cosαAE( )
sinα cosα sin αL
cos α sinα cosαAE
( ) sinα cosα sin αL
m m
m m
m
m m
m m
m
k k k
k k k
(m) (m)
F
B ( ) (m) (m) B-+ nn
AEF ( ) cos sin ( ) F
L m
U U
1
11 E 12 I
E 21 22 I
( ) ( - - )
'' ''
E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K K K
(a) The loading shown
2' ' ' '
E I I
2
u 6 kip0 0 0
10 kip
' ''P P PU U
1
2 1
11 E
2
x 1
y
x 3
E 21y 3
x 4
y 4
u -0.03 ( )
-0.017
R
4.62R 6.16
R 1.37
R -1.37
0R 5.2
R
' ' '
'' ' '
U P
P U
K
K
(1)
(2)
(3)
B (1) (1) (1)
B (2) (2) (2)
B (3) (3) (3)
-+
-+
-+
nn
nn
nn
AEF ( ) cos sin ( ) -7.7
L
AEF ( ) cos sin ( ) -5.2
L
AEF ( ) cos sin ( ) 1.94
L
U U
U U
U U
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(b) A support settlement of 0.5 inch at joint 4
2' ' ' '
E I I
2
0
0
u 00 0 0
0
0
.5 in
' ''P P PU U
1
2 1
11 12
2
x 1
y
x 3
E 21 22y 3
x 4
y 4
u 0.0127 ( ) ( - )
-0.268
R
30R 40
R -30
R 30
0R -70
R
' ' ' ''
'' ' ' ' ''
U U
P U U
K K
K K
(1)
(2)
(3)
B (1) (1) (1)
B (2) (2) (2)
B (3) (3) (3)
-+
-+
-+
nn
nn
nn
AEF ( ) cos sin ( ) -50
L
AEF ( ) cos sin ( ) 70
L
AEF ( ) cos sin ( ) -42.4
L
U U
U U
U U
Problem 12.7For the truss shown, use the direct stiffness method to find the joint displacements, reactions, and
member forces due to (a) the loading shown, (b) a temperature decrease of 0C10 for all members,
(c) a support settlement of δ = 12 mm downward at node 4.2
-6 0
A 1200 mm
E 200 GPa
12x10 /C
h 3 m
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Member m n n α L (m)
(kN/mm)
AE
L (1) 2 1 045 4.24 56.6
(2) 3 1 0 3 80
(3) 4 1 0135 4.24 56.6
A 3 3
3 3
A 3 3
0 0 0 0
0 0 0 0
0 0 0
0 00 0 0 0 0 0
( )
(1) A (1) AB (2) AA (2) AB ( ) AA ( ) AB
(1) BA (1) BB
(2) BA (2) BB
( ) BA ( ) BB
(1) A (2) AA ( ) AA (1) AB (2) AB ( ) A
k k 0 k k k 0 k
k k 0 0 0 0 0 0K =
k 0 k 0 00 0 0 0
k k 0 0
k k k k k k
3 3
0
0 0
0 0
B
(1) BA (1) BB
(2) BA (2) BB
( ) BA ( ) BB
k k 0
k k
k k
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1
11 E 12 I
E 21 22 I
( ) ( - - )
'' ''
E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K K K
A 3
3
3
3
( )
0
0 0
0 0
11 (1) A (2) AA ( ) AA
'
12 (1) AB (2) AB ( ) AB
(1) BA
'
21 (2) BA
( ) BA
(1) BB
'
22 (2) BB
( ) BB
' k k k
k k k
k
k
k
k 0
k
k
K
K
K
K
2
( ) ( ) (m)
(m)AA (m)BB ( ) ( ) 2
(m) (m) ( )
2
( ) ( ) (m)
(m)AB (m)BA ( ) ( ) 2
(m) (m) ( )
cos α sinα cosαAE( )
sinα cosα sin αL
cos α sinα cosαAE( )
sinα cosα sin αL
m m
m m
m
m m
m m
m
k k k
k k k
(m) (m)
F
B ( ) (m) (m) B-+ nn
AEF ( ) cos sin ( ) F
L m
U U
(a)
1' ' ' '
E I I
1
u 27 kN0 0 0
45 kN
' ''P P PU U
Using Mathcad:
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1 1
11 E
1
x 2
y 2
x 3
E 21y 3
x 4
y 4
u 0.477 ( ) ( )
0.33
R
-22.82R -22.82
R 0
R -26.36
-4.18R 4.18
R
' ' '
'' ' '
U P
P U
K
K
(1)
(2)
(3)
B (1) (1) (1)
B (2) (2) (2)
B (3) (3) (3)
-+
-+
-+
nn
nn
nn
AE
F ( ) cos sin ( ) 32.27L
AEF ( ) cos sin ( ) 26.36
L
AEF ( ) cos sin ( ) -5.91
L
U U
U U
U U
(b)
F 6F EA α ΔT (200)(1200)(12)(10) (10) 28.8 kNl
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1' ' ' '
E I I
1
20.36 kN20.36 kN
u 0 00 0
49.18 kN 28.8 kN
20.36 kN
20.36 kN
' ''P P PU U
Using Mathcad:
1 1
11 E I
1
x 2
y 2
x 3
E 21 Iy 3
x 4
y 4
u 0 ( ) ( - )
-0.51
R
-5.96R
-5.96R
0
R 11.93
5.96R -5.96
R
' ' ' '
'' ' ' ''
U P P
P U P
K
K
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(1)
(2)
(3)
B (1) (1) (1)
B (2) (2) (2)
B (3) (3) (3)
-+
-+
-+
nn
nn
nn
AEF ( ) cos sin ( ) 8.43
L
AEF ( ) cos sin ( ) -11.93
L
AEF ( ) cos sin ( ) 8.43L
U U
U U
U U
(c)
1' ' ' '
E I I
1
0
0
u 00 0 0
0
0
12 mm
' ''P P PU U
Using Mathcad:
1 1
11 12
1
x 2
y 2
x 3
E 21 22y 3
x 4
y 4
u 6.0 ( ) ( - )
-2.485
R
-99.4R -99.4
R 0
R 198.8
99.41R -99.41
R
' ' ' ''
'' ' ' ' ''
U U
P U U
K K
K K
(1)
(2)
(3)
B (1) (1) (1)
B (2) (2) (2)
B (3) (3) (3)
-+
-+
-+
nn
nn
nn
AEF ( ) cos sin ( ) 140.59L
AEF ( ) cos sin ( ) -198.82
L
AEF ( ) cos sin ( ) 140.59
L
U U
U U
U U
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Problem 12.8
For the truss shown, determine '11K and'21
K .2A 2000 mm
E 200 GPa
Member m n n α L (m) (kN/mm) AE
L
(1) 1 2 0 6 66.67
(2) 4 2 090 5 80
(3) 3 2 0135 7.07 56.58
2
( ) ( ) (m)
(m)AA (m)BB ( ) ( ) 2
(m) (m) ( )
2
( ) ( ) (m)
(m)AB (m)BA ( ) ( ) 2
(m) (m) ( )
cos α sinα cosαAE( )
sinα cosα sin αL
cos α sinα cosαAE( )
sinα cosα sin αL
m m
m m
m
m m
m m
m
k k k
k k k
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3 3
3 3
3
0 0 0 0 0 0 0
0
0
00 0 0 0 0 0 0
0
(
(1) AA (1) AB
(2) AA (2) AB ( ) AA ( ) AB(1) BA (1) BB
( ) BA ( ) BB
(2) BA (2) BB
(1) AA (1) AB
(1) BA (1) BB (2) AA ( )
k k 0 0 0
0 k 0 k 0 k k k k 0 0K =
0 0 0 0 0 k k 0 0 0 0
k 0 k 0
k k 0
k k k k 3
3 3
)
0
0
AA ( ) AB (2) AB
( ) BA ( ) BB
(2) BA (2) BB
k k
0 k k
k 0 k
3 3
3 3
0
( )
0
0
(1) AA (1) AB
(1) BA (1) BB (2) AA ( ) AA ( ) AB (2) AB
( ) BA ( ) BB
(2) BA (2) BB
k k 0
k k k k k k
0 k k
k 0 k
3
3
( )
11 (1) BB (2) AA ( ) AA
(1) AB'
21 ( ) BA
(2) BA
'k k k
k
k
k
K
K
Problem 12.9
For the truss shown, determine '11K and'22K .
2A 3 in
E 29,000 ksi
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Member m n n α L (ft) (kip/in.) AE L
(1) 1 2 037.57 32.8 221
(2) 2 3 - 0116.56 22.36 324.2
(3) 1 3 0 16 453
3 3
3 3
3 3
3
0 0 0
0 0
00
( )
( )
(
(1) AA (1) AB ( ) AA ( ) AB
(1) BA (1) BB (2) AA (2) AB
( ) BA ( ) BB(2) BA (2) BB
(1) AA ( ) AA (1) AB ( ) AB
(1) BA (1) BB (2) AA (2) AB
( ) BA (2) BA (2) B
k k 0 k 0 k
K = k k 0 0 k k 0
k k 0 0 0 k k
k k k k
k k k k
k k k 3
)
B ( ) BBk
Expand the K matrix.
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' ' '
' ' '
' ' '
11 12 13
21 ( 22 k) 23
31 32 33
k k k
k k k
k k k
11
'K
Problem 12.10
For the beam shown, use the direct stiffness method to find the joint displacements, reactions,
and member forces for the loading shown.
4I 300 in
E 29,000 ksi
3 2 3 2
AA AB
2 2
3 2 3 2
BA BB
2 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -
L L L L
6EI 2EI 6EI 4EI-
L L L L
k k
k k
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F
A AB B AA A A
F
B BB B BA A B
= +
= +
P k U k U P
P k U k U P
(1) AA (1) AB (1) AA (1) AB
(1) BB (2) AA (2) AB (1) BB (2) AA (2) AB(1) BA (1) BA
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00 0 0
K k k 0 0 k k k (k + k ) k
0 k k 0 0 0 0 k k
1
11 E 12 I
E 21 22 I
( ) ( - - )
'' ''
E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P|U P U P
= ___ | ___ __ + __ P U U P
|P U P
K K K K
K K K K
y 1
1
' ' ' ' ' '' '
2 E E y 2 I I
y 3
3
'
R 34.6 kip
M 260 kip-ft
θ 0 0 R -185 kip-ft 39.6 kip
5 kipR
75 kip-ftM
P P P PU U
1
11 I
E 21 I
( ) ( - )
' ' '
'' ' ' ''
U P
P U P
K
K
Problem 12.11
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For the beam shown, use the direct stiffness method to find the joint displacements, reactions,
and member forces for
a) The loading shown
b) A support settlement of 12 mm at joint 2
6 4I 120(10) mm
E 200 GPa
3 2 3 2
AA AB
2 2
3 2 3 2
BA BB
2 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -
L L L L6EI 2EI 6EI 4EI
-L L L L
k k
k k
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F
A AB B AA A A
F
B BB B BA A B
= +
= +
P k U k U P
P k U k U P
3 3
3 3
0 0 0 0
0 0
( )
(1) AA (1) AB
(2) AA (2) AB(1) BB(1) BA
( ) AA ( ) AB(2) BA (2) BB
( ) BA ( ) BB
(1) AA (1) AB
(1) BB (2) AA (2) AB(1) BA
(
k k 0 0 0 0 0 0
0 k k 0 0 0k k 0 0K
0 0 k k 0 k k 00 0 0 0
0 0 k k 0 0 0 00 0 0 0
k k 0 0
k k k k 0
0 k 3 3
3 3
( )
2) BA (2) BB ( ) AA ( ) AB
( ) BA ( ) BB
k k k
0 0 k k
1
1
2
2
3
3
θ
=θ
θ
U 2
3
' θ=θ
U 'E 0P
Problem 12.12
Given the system matrix listed below. Determine '11K ,'21K ,
''U , ''I'I ,P P ,
'
EP and ''
EP for the
loading and displacement constraint shown. L= 30 ft, 4I 300 in , and E=29,000 ksi.
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34
3 2 3 2
2 2
3 2 3 3 2
2 2
3 2 3 2
2 2
12EI 6EI 12EI 6EI0 0
L L L L6EI 6EI
- 0L L
12EI 6EI 24EI 12EI 6EI- - 0 -
L L L L L
6EI 6EI0 -L L
12EI 6EI 12EI 6EI0 0 - - -
L L L L
6EI 6E
4EI 2EI0
L L
2EI 8EI 2EI
L L L
2EI 4EI0 I0L L
-L L
K
1
1
2
2
3
3
θ
=θ
θ
U
'' ''
EI11 12
___
' ''' '' ''21 22E I
'P U P|
= ___ | ___ __ + __
|P U P
K K
K K
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4EI 2EI0
L L
2EI 8EI 2EI
L L L
2EI 4EI0L L
11
'K
2 2
2 2
2 2
6EI 6EI0
L L
6EI 6EI- 0
L L
6EI 6EI0 - -L L
'
21K
1
2
3
θ 0
θ 0
0.25 inθ
' ''U U
0
10 kip-ft
0
'
EP
I
51.84 kip-ft
0
-51.84 kip-ft
'P I
5.184 kip
13.632 kip
5.184 kip
''P
Problem 12.13
Investigate the effect of varying the spring stiffness on the behavior (moment and deflected
profile) of the structure shown below. Consider a range of values of k v.
6 4
18 kN/mmm
I 120(10) mm , E 200 GPa and k 36 kN/mm
90 kN/mm
v
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37
(1) AA (1) AB (1) AA (1) AB
(1) BB (2) AA (2) AB (1) BB (2) AA (2) AB(1) BA (1) BA
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00 0 0
K k k 0 0 k k k (k + k ) k
0 k k 0 0 0 0 k k
2
2
θ0
' ''U U
'E = 14 kN-mP
Problem 12.14 Determine the bending moment and deflection profiles for the following structures.
Take 4I 300 in , 2CableA 3 in , and2E 29,000 k/in .
2
(2) (2) (2)Cable
(2)AA (2)BB (2) 2 2(2) (2) (2)Cable
2
(2) (2) 2)Cable(2)AB (2)BA (2) 2 2
(2) (2) (2)Cable
cos α sinα cosαA E( )
sinα cosα sin αL
cos α sinα cosαA E( )
sinα cosα sin αL
k k k
k k k
3 2 3 2
AA AB
2 2
3 2 3 2
BA BB
2 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -
L L L L
6EI 2EI 6EI 4EI-
L L L L
k k
k k
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F
A AB B AA A A
F
B BB B BA A B
= +
= +
P k U k U P
P k U k U P
1
11 E 12 I
E 21 22 I
( ) ( - - )
'' ''E ' ' ' ' '' 'I11 12
___
'' ' ' ' '' ''' ''' '' ''21 22E I
'P U P| U P U P= ___ | ___ __ + __
P U U P|P U P
K K K K
K K K K
Case (a)
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3 2 3 2
2 2
3 2 3 2
2 3 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -L L L L
6EI 2EI 12EI 6EI-
L L L L
K
y 1
' ' ' ''
E 1 I I
2
R 9 kip
0 M -90 kip-ft 60 kip-ft
21 kipR
''U P P P
2
6EI-
L
11
'K
Using Mathcad:
12 11 Iθ ( ) ( - ) .011 ' ' 'U PK
y 1
1 E 21 I
3
R 13.5
M 105
16.5M
'' ' ' ''P U PK
Case (b)
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3 2 3 2
2 2
3 2 3 2
2 3 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -L L L L
6EI 2EI 12EI 6EI-
L L L L
K
y 1
' ' ' ''
2 E 1 I I
2
R 9 kip
M 21 kip 60 kip-ft
90 kip-ftM
'U P P P 0''U
Using Mathcad:
12 11 I( ) (- ) -9.38 ' ' 'U PK
y 1
1 E 21 I
2
R 30
M 375
225M
'' ' ' ''P U PK
Case (c)
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3 2 3 2
2 2
3 2 3 2
2 3 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -L L L L
6EI 2EI 12EI 6EI-
L L L L
K
y 12 ' ' ' ''
E I I
2 1
R θ 21 kip 9 kip
-90 kip-ft 60 kip-ftM
'U P P P 0''U
Using Mathcad:
2 1
11 I
2
θ -0.27 ( ) (- )
0.01
' ' 'U PK
y 1
E 21 I
1
R 13.65
109.57M
'' ' ' ''P U PK
Case (d)
(d) Cantilever beam + Cable
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(1) (1)
(1) (1)
3 2 3 2
AA AB
2 2
3 2 3 2
BA BB
2 2
12EI 6EI 12EI 6EI-
L L L L
6EI 4EI 6EI 2EI-
L L L L
12EI 6EI 12EI 6EI- - -
L L L L
6EI 2EI 6EI 4EI-
L L L L
k k
k k
2
(2) (2) (2)Cable(2)AA (2)BB (2) 2 2
(2) (2) (2)Cable
2
(2) (2) 2)Cable(2)AB (2)BA (2) 2 2
(2) (2) (2)Cable
cos α sinα cosαA E( )
sinα cosα sin αL
cos α sinα cosαA E( )
sinα cosα sin αL
k k k
k k k
(1) AA (1) AB (1) AA (1) AB
(1) BB (2) AA (2) AB (1) BB (2) AA (2) AB(1) BA (1) BA
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00 0 0
K k k 0 0 k k k (k + k ) k
0 k k 0 0 0 0 k k
y 1
2 ' ' ' ''
E 1 I I2
y 3
R θ 21 kip 9 kip
M-90 kip-ft 60 kip-ft
R
'U P P P 0''U
Using Mathcad:
1
11 I( ) (- )' ' 'U PK
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E 21 I '' ' ' ''P U PK
Problem 12.15Consider the guyed tower defined in the sketch. The cables have an initial tension of 220 kN.
Determine the horizontal displacements at B, C, the change in cable tension, and the bending
moment distribution in member ABC. Treat the cables as axial elements. Develop a computer based scheme to solve this problem. Take 6 4
tower I 200(10) mm 2
CableA 650 mm ,2
tower A 6000 mm , and the material to be steel.
strudl ' 2012 Problem 12.15 Guyed tower metric GTstrudl input file Connor & Faraji'type plane frameunits kN meter
Joint Coordinates
1 0.0 0.0
2 15.0 60.03 30.0 0.0
4 15.0 0.0
5 15.0 30.0type plane truss
member incidences
1 1 2
2 2 3
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3 3 5
4 1 5
type plane framemember incidences
5 4 5
6 5 2status support joints 1 3 4 joint releases
4 moment z
unit mmconstants
E 200.0
cte 12E-06
Member properties1 2 3 4 AX 650
5 6 AX 6000 IZ 200000000
units m
Loading 1
member tempreture loads
1 2 axial -1823 4 axial -187
Loading 2
joint load2 force x 45.
5 force x 90.
Loading 3
joint load2 force x 45.
5 force x 90.
member tempreture loads1 2 axial -182
3 4 axial -187
stiffness analysislist reactions
list forces
unit mm
list displacements
_____________________________________
Analysis results:
{ 54} > list reactions
--- LOADING - 1 ---
----------------------------------------------------------------------------------------------------------------------------- ------
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RESULTANT JOINT LOADS SUPPORTS
JOINT /---------------------FORCE---------------------//--------------------MOMENT--------------------/
X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z MOMENT
1 GLOBAL -152.0266571 -410.8855896
3 GLOBAL 152.0266571 -410.8855896
4 GLOBAL 0.0000000 821.7711792 0.0000000
-----------------------------------------------------------------------------------------------------------------------------------
--- LOADING - 2 ---
-----------------------------------------------------------------------------------------------------------------------------------
RESULTANT JOINT LOADS SUPPORTS
JOINT /---------------------FORCE---------------------//--------------------MOMENT--------------------/
X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z MOMENT
1 GLOBAL -67.5721207 -180.0000000
3 GLOBAL -67.5721207 180.0000000
4 GLOBAL 0.1442504 0.0000000 0.0000000
----------------------------------------------------------------------------------------------------------------------------- ------
--- LOADING - 3 ---
----------------------------------------------------------------------------------------------------------------------------- ------
RESULTANT JOINT LOADS SUPPORTS
JOINT /---------------------FORCE---------------------//--------------------MOMENT--------------------/
X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z MOMENT
1 GLOBAL -219.5987854 -590.8856201
3 GLOBAL 84.4545364 -230.8856049
4 GLOBAL 0.1442504 821.7711792 0.0000000
{ 55} > list forces
ACTIVE UNITS M KN RAD DEGF SEC
----------------------------------------------------------------------------------------------------------------------------- ------
--- LOADING - 1 ---
----------------------------------------------------------------------------------------------------------------------------- ------
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MEMBER FORCES
MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------/
AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING Z
1 2 220.2403717
2 3 220.2403717
3 5 220.4998169
4 5 220.4998169
5 4 821.7711792 0.0000000 0.0000000
5 5 -821.7711792 0.0000000 0.0000000
6 5 427.3291016 0.0000000 0.0000000
6 2 -427.3291016 0.0000000 0.0000000
----------------------------------------------------------------------------------------------------------------------------- ------
--- LOADING - 2 ---
----------------------------------------------------------------------------------------------------------------------------- ------
MEMBER FORCES
MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------/
AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING Z
1 2 92.47249602 3 -92.4724960
3 5 -100.9456024
4 5 100.9456024
5 4 0.0000000 -0.1442504 0.0000000
5 5 0.0000000 0.1442504 -4.3275118
6 5 0.0000000 0.1442504 4.3275118
6 2 0.0000000 -0.1442504 0.0000000
----------------------------------------------------------------------------------------------------------------------------- ------
--- LOADING - 3 ---
----------------------------------------------------------------------------------------------------------------------------- ------
MEMBER FORCES
MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------/
AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING Z
1 2 312.7128906
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2 3 127.7678833
3 5 119.5542145
4 5 321.4454041
5 4 821.7711792 -0.1442504 0.0000000
5 5 -821.7711792 0.1442504 -4.3275118
6 5 427.3291016 0.1442504 4.3275118
6 2 -427.3291016 -0.1442504 0.0000000
{ 56} > unit mm
{ 57} > list displacements
ACTIVE UNITS MM KN RAD DEGF SEC
-----------------------------------------------------------------------------------------------------------------------------------
--- LOADING - 1 ---
-----------------------------------------------------------------------------------------------------------------------------------
RESULTANT JOINT DISPLACEMENTS SUPPORTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
1 GLOBAL 0.0000000 0.0000000
3 GLOBAL 0.0000000 0.0000000
4 GLOBAL 0.0000000 0.0000000 0.0000000
RESULTANT JOINT DISPLACEMENTS FREE JOINTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
2 GLOBAL 0.0000000 -31.2275105 0.0000000
5 GLOBAL 0.0000000 -20.5442810 0.0000000
----------------------------------------------------------------------------------------------------------------------------- ------
--- LOADING - 2 ---
----------------------------------------------------------------------------------------------------------------------------- ------
RESULTANT JOINT DISPLACEMENTS SUPPORTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
1 GLOBAL 0.0000000 0.0000000
3 GLOBAL 0.0000000 0.0000000
4 GLOBAL 0.0000000 0.0000000 -0.0014003
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RESULTANT JOINT DISPLACEMENTS FREE JOINTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
2 GLOBAL 181.3883820 0.0000000 -0.0046460
5 GLOBAL 58.2378464 0.0000000 -0.0030231
-----------------------------------------------------------------------------------------------------------------------------------
--- LOADING - 3 ---
-----------------------------------------------------------------------------------------------------------------------------------
RESULTANT JOINT DISPLACEMENTS SUPPORTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
1 GLOBAL 0.0000000 0.0000000
3 GLOBAL 0.0000000 0.0000000
4 GLOBAL 0.0000000 0.0000000 -0.0014003
RESULTANT JOINT DISPLACEMENTS FREE JOINTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
2 GLOBAL 181.3883820 -31.2275105 -0.00464605 GLOBAL 58.2378464 -20.5442810 -0.0030231
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Problem 12.16
Consider the rigid frame shown above. Investigate how the response changes when 1A is varied.
Use computer software. Vary 1A from2
2 in to 2
10 in . Take 4I 600 in , 2A 5 in , L 200 ft ,4
1I 300 in , 1 k/ftw . Material is steel.
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Analysis Results for 1A 22 in :
{ 37} > list reactions
ACTIVE UNITS FEET KIP RAD DEGF SEC
RESULTANT JOINT LOADS SUPPORTS
JOINT /---------------------FORCE---------------------//--------------------MOMENT--------------------/
X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z MOMENT
1 GLOBAL 0.2750448 26.2271576 0.0000000
4 GLOBAL 0.0000000 26.3188362 0.0000000
5 GLOBAL -0.2750448 147.4540253 -0.0000033
{ 38} > list forces
ACTIVE UNITS FEET KIP RAD DEGF SEC
MEMBER FORCES
MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------/
AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING Z
1 1 0.2750448 26.2271576 0.0000000
1 2 -0.2750448 40.4394569 -473.7429199
2 2 -75.6282272 33.4004898 421.63412482 3 75.6282272 33.2659035 -417.1477966
3 3 0.0000000 40.3481750 467.6471252
3 4 0.0000000 26.3188343 0.0000000
4 5 105.8844528 -1.4583676 -16.6390419
4 2 -105.8844528 1.4583676 -52.1087914
5 5 105.5302505 -1.4242343 -16.6390438
5 3 -105.5302505 1.4242343 -50.4993286
{ 39} > unit in
{ 40} > list displacements
ACTIVE UNITS INCH KIP RAD DEGF SEC
RESULTANT JOINT DISPLACEMENTS SUPPORTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
1 GLOBAL 0.0000000 0.0000000 -0.0604358
4 GLOBAL 0.4157400 0.0000000 0.0615096
5 GLOBAL 0.0000000 0.0000000 -0.0006194
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RESULTANT JOINT DISPLACEMENTS FREE JOINTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
2 GLOBAL -0.0015175 -1.4619952 0.0132183
3 GLOBAL 0.4157400 -1.8713145 -0.0138291
____________________________________________________________________
Analysis Results for 1A 210 in :
{ 37} > list reactions
ACTIVE UNITS FEET KIP RAD DEGF SEC
RESULTANT JOINT LOADS SUPPORTS
JOINT /---------------------FORCE---------------------//--------------------MOMENT--------------------/
X FORCE Y FORCE Z FORCE X MOMENT Y MOMENT Z
MOMENT
1 GLOBAL 0.2801770 26.0890694 0.0000000
4 GLOBAL 0.0000000 26.1824589 0.0000000
5 GLOBAL -0.2801770 147.7284851 -0.0000043
{ 38} > list forces
ACTIVE UNITS FEET KIP RAD DEGF SEC
MEMBER FORCES
MEMBER JOINT /-------------------- FORCE --------------------//-------------------- MOMENT --------------------
/
AXIAL SHEAR Y SHEAR Z TORSIONAL BENDING Y BENDING
Z
1 1 0.2801770 26.0890694 0.0000000
1 2 -0.2801770 40.5775414 -482.9487305
2 2 -76.3793488 33.4019623 420.8519592
2 3 76.3793488 33.2644310 -416.2675171
3 3 0.0000000 40.4845543 476.7389832
3 4 0.0000000 26.1824589 0.0000000
4 5 106.5178833 -1.8944246 -27.2068596
4 2 -106.5178833 1.8944246 -62.0968056
5 5 106.1567535 -1.8599567 -27.2068634
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5 3 -106.1567535 1.8599567 -60.4714508
{ 39} > unit in
{ 40} > list displacements
ACTIVE UNITS INCH KIP RAD DEGF SEC
RESULTANT JOINT DISPLACEMENTS SUPPORTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
1 GLOBAL 0.0000000 0.0000000 -0.0581310
4 GLOBAL 0.4198557 0.0000000 0.0592253
5 GLOBAL 0.0000000 0.0000000 -0.0006280
RESULTANT JOINT DISPLACEMENTS FREE JOINTS
JOINT /-----------------DISPLACEMENT-----------------//-------------------ROTATION-------------------/
X DISP. Y DISP. Z DISP. X ROT. Y ROT. Z ROT.
2 GLOBAL -0.0015458 -0.2953888 0.0129835
3 GLOBAL 0.4198557 -0.7126988 -0.0136053
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Problem 12.17
(a) Develop a computer code to automate the generation of the member stiffness matrices
defined by Equation (12.6). Assume A, E, I, L given.(b) Develop a computer code to carry out the operations defined by Equation (12.13).
(c) Develop a computer code to carry out the operation defined by Equation (12.20) and
(12.21).
BB 3 2
2
AE0 0
L
12EI 6EI= 0 -
L L
6EI 4EI0 -
L L
k BA 3 2
2
AE- 0 0
L
12EI 6EI0 - -
L L
6EI 2EI0
L L
k
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AA 3 2
2
AE0 0
L
12EI 6EI= 0
L L
6EI 4EI0 L L
k AB 3 2
2
AE- 0 0
L
12EI 6EI0 -
L L
6EI 2EI0 - L L
k
g
cosα -sinα 0
= sinα cosα 0
0 0 1l
R
cosα sinα 0
= -sinα cosα 0
0 0 1
g l
R
BB BA
T T
BB g BA gg g
T T
AA AA g AB AB gg g
=
=
l l l l
l l l l
k R k R k R k R
k R k R k R k R
B A
B A B A
0 0
0
(1) AB (1) AA (1) A (1) A
(1) B (1) B (2) AA (2) AB (1) B (1) B (2) AA (2) AB
(2) BA (2) BB (2) BA (2) BB
k k 0 k k 00
K = k k 0 k k = k (k + k ) k
0 0 0 0 k k 0 k k
'' ''
EI11 12
___
' ''' '' ''21 22E I
'P U P|
= ___ | ___ __ + __
|P U P
K K
K K
1
11 E 12 I
E 21 22 I
( ) ( - - )
' ' ' ' '' '
'' ' ' ' '' ''
U P U P
P U U P
K K
K K
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Problem 12.18
For the space truss shown, use the direct stiffness method to find displacements at joint1. 2A 3 in
E 29,000 ksi
X-Y Plan
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X-Z Plan
Member l x l y l z l xβ yβ zβ
(1) 12 6 18 22.45 0.53 0.27 0.8
(2) 10 6 18 21.45 0.47 0.28 0.84
(3) 2 12 18 21.72 0.092 0.55 0.83
Using Mathcad:
Lm
X2m
X1m
2
Y2m
Y1m
2
Z2m
Z1m
2
2
( ) ( ) (m)
(m)AA (m)BB ( ) ( ) 2
(m) (m) ( )
2
( ) ( ) (m)
(m)AB (m)BA ( ) ( ) 2
(m) (m) ( )
cos α sinα cosαAE( )
sinα cosα sin αL
cos α sinα cosαAE( )
sinα cosα sin αL
m m
m m
m
m m
m m
m
k k k
k k k
E
0
10 kip
20 kip
'P
1
1
1 11 E
1
u .011 in
( ) .066 in
.03 inw
' ' 'U PK
Problem 12.19For the space frame shown, use the direct stiffness method to find displacements at joint B. The
load P is applied perpendicular to the plane ABC. The cross section is circular tube. Take L=4 m,
P=30 kN, E = 200 GPa , and G=77 GPa.
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Cross section
Member m n n 1θ (xg ,x )l 2θ (xg ,x )l 3θ (xg ,x )l
(1) 1 2 π/2 -π/2 0
(2) 2 3 0 π/2 -π/2
i i
11 12 13
21 22 23
31 32 33
i i i
i i i
i
α α α
α α α
α α α
'
iR
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i
lg
ii
0
0
R'R
R'
T
global gl g k R k R
1 AA 1 AB
1 BA 1 BB 2 AA 2 AB
2 BA 2 BB
k k 0
K = k (k + k ) k
0 k k
(1) (1)T
global lg (1)BB lg
1 BBk R k R
( 2) (2)T
global lg (2)AA lg2 AAk R k R
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11 1 BB 2 AA'K = (k +k )
1joint2
2 joint2
3joint2'
1joint2
2 joint2
3joint2
u
u
u
θ
θ
θ
U E
0
0
30
0
0
0
'P
1
11 E
0
023.55
( ) ( ).003
.003
0
' ' 'U K P
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1joint 2 2 joint 2 3 joint2
1joint 2 2 joint 2 3 joint 2
23.55 mm
.003 rad
u u 0 u
θ θ θ 0
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Problem 13.1
(a)
(b) Use a software package to determine:
(i) The maximum values of3
R , 2M , and 1-1M cause by a uniformly distributed dead
load of 2 k/ft.
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(ii) The maximum value of 2M caused by a uniformly distributed live load of 1k/ft.
-
2 maxM 146.7 kip-ft
Problem 13.2
8(28) 32(14)x 9.33 ft
72e
e 14 9.33 4.66 ft 2.33 ft2
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maxM 1703 kip-ft
Moment envelop
maxM 1700 kip-ft
Problem 13.3
Lane load: 10 kN/mw
uniformTruck load:
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(a) 1 2L L 30 m , EI constant
Global moment envelope-Lane
Global moment envelope-Truck
(b) 1L 15 m 2L 30 m , EI constant
Global moment envelope-Lane
Truck
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(c) 1 2L L 30 m , EI constant
Moment diagram-Lane
Global moment envelope-Truck
(d) Compare the global moment envelopes for the structure shown below with the
envelopes generated in part (a). Is there any effect of varying I?
Global moment envelope-Lane
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Global moment envelope-Truck
Problem 13.4
Consider the multi span bridge shown below. Suppose the bridge is expected to experience a
temperature change of ∆T over its entire length. Where would you place a hinge support: at A orat B? Determine the end movement corresponding to your choice of support location.
Case (a)
u α T L
Case (b)
u α T L
Case (b) is better because there is less movement of span.
Problem 13.5
Most design codes limit the deflection due to live loading to some fraction of length, say L/α
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where α is on the order of 500. Generate the global “deflection” envelope for the multi-span
beam and truck loading shown below. Take E = 29,000ksi and I = 60,000 in4.
maxmax req
L L 90(12)2.16 in I (I)
L500 500( )
Problem 13.6
Investigate convergence of the internal forces for the parabolic arch shown as the discretizationis refined. Take the interval as 2.4 m, 1.2 m, and .6 m.
Axial Force
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Bending Moment
Axial Force
Bending Moment
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Axial Force
Bending Moment
Problem 13.7
Suppose a uniform loading is applied to span ABC. Investigate how the response changes as x
varies from L/2 to L. Take h=L/2, 2 2CA 50 in , A 2 in ,2
21k/ft, f 1
L
x
w
.
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Deformation profile (
L
, f 22 x
)
Deformation profile ( L, f 5 x )
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Problem 13.8
Determine the structural response (forces and displacements) of the idealized tied arch shown
below under a uniformly distributed gravity load of 30 kN/m.
Assume 2 6 4 6 2arch Arch hanger A 26000 mm , I 160(10) mm ,A 2(10) mm
Note: roadway girder and arch are pined to together at pints A and B.
Deformation profile
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Axial force
Bending moment
Problem 13.9
Determine the distribution of internal forces and displacements for the cable stayed structureshown below. Member AB acts as counterweight for loading applied on member BC. The 2
members are connected by 9 parallel equally spaced cables. Self weight of members AB and BC
is 16 kN/m and 8 kN/m respectively. Assume 2CableA 50000 mm and E=200GPa. Consider the
following cases:
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(a) 9 4AB BCI I 40(10) mm
Deformation profile
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Axial force
Bending moment
(b) 9 4AB BC BCI 4I I 40(10) mm
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Deformation profile
Bending moment
(c) Uniform Live load of 2 kN/m applied to member BC in addition to self weight.9 4
AB BC BCI 4I I 40(10) mm
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Deformation profile
Bending moment
Problem 13.10
Consider the symmetrical cable structure shown below. Determine a set of cable areas ( 1 10C C )
such that the maximum vertical displacement is less than 375 mm under a uniformly distributed
live load of 10 kN/m. Assume 6 4 3 2girder girder I 400(10) mm ,A 120(10) mm . Take the allowablestress as 700MPa.
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maxδ 371 mm
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Cable A,2mm Tension, kN
C1 100,000 325.5
C2 500,000 62.7
C3 50,000 167
C4 50000 132C5 50000 129
C6 50000 120
C7 50000 113
C8 50000 108
C9 50000 103
C10 50000 104
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Problem 14.1 Consider the plan view of one story rigid frames shown below. Determine the center of twist
corresponding to the brace stiffness patterns shown.(a)
(b)
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(c)
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Problem 14.2
The one story frame shown has an unsymmetrical mass distribution.
Distribution of mass
Stiffness distribution
(a) Determine the center of mass.
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(b) Determine the stiffness parameters1k and
2k such that the center of stiffness coincides
with the center of mass.
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(c) Determine the earthquake floor loads corresponding toa
S =0.3g. Consider both direction,
i.e., N-S and E-W.
Problem 14.3
For the rigid frame shown below, determine (a) The center of twist (T
C ) (b) the seismic floor
loads applied at the center of mass (M
C ) for a N-S earthquake witha
S =0.3g. Assume
properties are equal for each floor.
Elevation
Typical floor plan
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Problem 14.4 Consider the two story rigid frame defined below. Assume the weight of the floor slabs are
equal to floor w . Concentrated masses are located on each floor as indicated.
Plan-Floor 1
Plan –
Floor 2
(a) Determine the position of the center of mass for each floor.
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(b) Assume the structure is subjected to an earthquake acting in the east direction. Determine
the earthquake forces for the individual floors. Assume floor w =1000 kips,
1 2w w 1000 kips , and aS = 0.3g.
(c)
Suppose the story stiffness distribution shown below is used. Describe qualitatively howthe structure will displace when subjected to an earthquake. Consider the stiffness
distribution to be the same for each floor.
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Problem 14.5 Consider the single story multi frame structure shown below. Determine the lateral force in
the frames due to a global load P. Consider both wind and earthquake loading. Assume theslab is rigid.
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Problem 14.6 Consider the stiffness distribution for the one story rigid frame shown below. Determine the
displaced configuration under the action of the loading shown. Assume the slab is rigid.
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Problem 14.7
(a) Determine the center of twist.
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(b) Using Equation (14.47), determine oK .
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Problem 14.8 Consider the plan view of a one story frame shown below. Using the matrix formulation
presented in section 14.3.5, generate the equations of motion for the story. Take 1m 1000 ,
2m 500 , and k=10 kips/in.
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Problem 14.9
Consider the one story plan view shown below.
(a) Locate the center of mass.
(b) Locate the center of twist. Take 1 2 3 4k k k k k .
(c) Take 1 3k k 10 . Suggest values for 2k and 4k such that the center of mass
coincides with the center of twist.
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Problem 14.10 Consider the floor plan shown below. Assume the mass is uniformly distributed over the
floor area. Establish the equations of motion referred to point o.
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Problem 14.11 Consider the roof plan for a one story structure shown below. Assume the shear walls haveequal stiffness and the roof dead load is uniform.
(a) Determine the center of mass and