lecture method of consistent deformation.pdf
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method of consistent deformation, structural analysis, civil engineering, construction engineering,theory of structuresTRANSCRIPT
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13. Method of Consistent Deformations 1
13. Method of Consistent Deformations
By James C. Maxwell in 1864
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13. Method of Consistent Deformations 2
13.1 Structures with a Single Degree of Indeterminacy
AB
C
32kE = 30,000 ksi
I = 512 in4
10 10
1. Free-Body Diagram
32 k
AxAy Cy
MA
)yAyx C,M,A,(A
)0,0,0( === MFF yxUnknown variables = 4
3Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 4 -3 =1
= Number of Redundancy
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13. Method of Consistent Deformations 3
32 k
Ax
Ay Cy
Ma
2. Selecting Cy as a redundant
32k
Axo
Ayo
MAO Determinate Structure
Primary Structure
CO
0= xF 0=xoA= kAyo 320 y =F
kfkM AO = 32001032 =AOM0= AM
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13. Method of Consistent Deformations 4
32k
32k
321 k-ft
CO
Bending Moment Diagram
A B C
-320
Determining CO using conjugate-beam method,
EI320
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13. Method of Consistent Deformations 5
EI320
)101032
2110320 += (
EICO
EI67.26666=
512300001267.26666 3
=
3ftk in0.3=
= in.CO 033. Applying the redundant
Axc
MACCCf
yC1 k
Ayc fCC : Flexibility Coefficient
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13. Method of Consistent Deformations 6
Axc
Ayc
MacCCf
1kyC
0=xcA0= xF= kAyc 10 y =F
kfkM AC = 200201 =+ ACM0= AM
Bending Moment Diagram
A B C
1020
yC
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13. Method of Consistent Deformations 7
Determining fCC using conjugate-beam method,
EI20
)2032(
212020 =
EIfCC
EIftk 367.2666 = 51230000
1267.2666 3
= = in3.0
yCCCC Cf =
4. Compatibility Condition
0=+= yCCCOC CfyCCCO Cf=
Unknowns : Forces Force Method
Flexibility MethodFlexibility
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13. Method of Consistent Deformations 8
=== kf
CCC
COy 103.0
0.3
5. Reaction Forces
Using the equilibrium equations,
0A ,0 == xxF==+= k22A0,1032 ,0 yyy AF
120 ,020101032 ,0 ftkMMM AAA ==+=
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13. Method of Consistent Deformations 9
Using the principle of superposition,
32 k
32k
321 k-ft
in0.3
1k x 10
20 k-ft x 10103.0 in
32 k
Ay = 22 k
MA = 120 k-ft
Ax = 0
Cy = 10 k
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13. Method of Consistent Deformations 10
6. Shear Force Diagram & Bending Moment Diagram
Using the reaction forces & the applied loads,
Using the principle of superposition,A B C
-320
A B C
20 x 1010 x 10
A
B
C
-120
100
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13. Method of Consistent Deformations 11
Moment as a redundantPrimary Structure Statically Determinate & Stable, ()Redundant
AxAy Cy
MA
Selecting Ax as a redundant,
Ay Cy
MA
Unstable Structure
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13. Method of Consistent Deformations 12
AxAy Cy
MA
32k
32k
Selecting MA as a redundant,
Ayo Cyo
32k
Axo
Statically Determinate & Stable Structures
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13. Method of Consistent Deformations 13
2. Selecting MA as a redundant
Axo
Ayo Cyo
32k
AO
0=xoA0= xF= kCyo 160= AM= kAyo 160 y =F
EIPL
16
2
= rad0075.0=AO
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13. Method of Consistent Deformations 14
3. Applying the redundant
AxA
AyACyA
ftk 1AAf
AM
0=xAA0= xF= kCyA 20
10= AM
= kAyA 201
0 y =F
EIL
3=AAf ftkrad = /0000625.0
AAAAA Mf =
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13. Method of Consistent Deformations 15
4. Compatibility Condition
0=+= AAAAOA MfAAAAO Mf=
0000625.00075.0=
AA
AOA f
M = ftk =120
5. Reaction ForcesUsing the equilibrium equations,
Using the principle of superposition,
6. Shear Force Diagram & Bending Moment Diagram
Using the reaction forces & the applied loads,
Using the principle of superposition,
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13. Method of Consistent Deformations 16
Example 13.1
AB
L
M
EI = constant
1. Free-Body Diagram
AxAy Cy
MAM
Unknown variables = 43Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 4 -3 =1
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13. Method of Consistent Deformations 17
AxAy By
MAM
2. Selecting By as a redundantMMAO
BO
Determinate Structure
Primary Structure
=BO
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13. Method of Consistent Deformations 18
3. Applying the redundant
MABBBf
yB1
AyB
=BBf4. Compatibility Condition
0=+= yBBBOB Bf== BBBOy fB /
5. Reaction Forces
6. Shear Force Diagram & Bending Moment Diagram
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13. Method of Consistent Deformations 19
Example 13.2
15 kN/m60 kN
AB C
D
10 m 5 m 5 m
E = 200 GPa
I = 700106 mm4
1. Free-Body Diagram
AxAy
By
15 kN/m60 kN
Dy
Unknown variables = 43Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 4 -3 =1
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13. Method of Consistent Deformations 20
AxAy
By
15 kN/m60 kN
Dy
2. Selecting By as a redundant
15 kN/m60 kN
AxoAyo Dyo
BO
=BO
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13. Method of Consistent Deformations 21
3. Applying the redundant
1yBAxB
AyB DyB
BBf
=BBf4. Compatibility Condition
0=+= yBBBOB Bf== BBBOy fB /
5. Reaction Forces
6. Shear Force Diagram & Bending Moment Diagram
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13. Method of Consistent Deformations 22
Example 13.3
25 k25 k
28 k
3 @ 20 ft
15 ft
A B C D
E F
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13. Method of Consistent Deformations 23
1. Free-Body Diagram
25 k25 k
28 k
AxAy Dy
Dx
m + r Unknown variables = = 13
Equations of Equilibrium = 2j = 12
Statically Indeterminate Structure
Degree of Indeterminacy = 13 12 = 1
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13. Method of Consistent Deformations 24
25 k25 k
28 k
AxAy Dy
Dx
2. Selecting Dx as a redundant
25 k25 k
28 k
Axo
Ayo Dyo
DO
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13. Method of Consistent Deformations 25
= FAELFvOF
Du
== ODDO FAELu
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13. Method of Consistent Deformations 26
3. Applying the redundant
1AxD
AyD DyD
DDf
xD
= FAELFv
Du
== AELuf DDD 2
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13. Method of Consistent Deformations 27
4. Compatibility Condition0=+= xDDDOD Df
== DDDOx fD /
5. Reaction Forces
6. Shear Force Diagram & Bending Moment Diagram
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13. Method of Consistent Deformations 28
Example 13.4
3 k/ft
30 ftA
B CEI = constant
20 ft
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13. Method of Consistent Deformations 29
1. Free-Body Diagram3 k/ft
AxAy
CyCx
Unknown variables = 43Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 4 -3 =1
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13. Method of Consistent Deformations 30
2. Selecting Ax as a redundant3 k/ft
Ay
Cy
Cx
AO
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13. Method of Consistent Deformations 31
= dxEIMM vOM
Am
== dxEIMm OAAO
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13. Method of Consistent Deformations 32
3. Applying the redundant
1
AyA
CyA
CxA
AAf
= dxEIMM v
Am xA
== dxEImf AAA2
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13. Method of Consistent Deformations 33
4. Compatibility Condition0=+= xAAAOA Af
== AAAOx fA /
5. Reaction Forces
6. Shear Force Diagram & Bending Moment Diagram
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13. Method of Consistent Deformations 34
13.2 Internal Forces and Moments as Redundants
10 k
A B C
10 k 12 k
6 ft 8 ft 6 ft 20 ft 10 ft
EI = constant
1. Free-Body Diagram
10 k 10 k 12 k
AyBy Cy
Ax
Unknown variables = 43Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 4 -3 =1
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13. Method of Consistent Deformations 35
BM
B
BRBL =BL
BR Brel BLBR = 0=
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13. Method of Consistent Deformations 36
2. Selecting Internal Moment (Bending Moment) MB as a redundant
10 k 10 k 12 k
BOL BOR
Determinate Structure
Primary Structure
BORBOL BOrel BOLBOR = 0
EIftk
BOR
233.533 =EI
ftkBOL
2420 =
EIftk
BOrel
233.953 =
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13. Method of Consistent Deformations 37
3. Applying the redundant1 1
BBLf BBRf
BM
EIftkftk
fBBR= /10
2
EIftkftk
fBBL= /67.6
2
EIftkftk
fBBrel= /67.16
2
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13. Method of Consistent Deformations 38
4. Compatibility Condition0=+= BBBrelBOrelBrel Mf
ftkfM BBrelBOrelB == 19.57/5. Reaction Forces
- Equilibrium Equations of Members & Joints
57.19 57.1910 10 57.19 1257.19
Ay = 7.14 ByAB = 12.86 Cy = 6.09By
BC = 5.9112.86 5.91
By = 18.77
6. Shear Force Diagram & Bending Moment Diagram
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13. Method of Consistent Deformations 39
Internally Indeterminate StructuresSelecting a reaction as a redundant
OR Selecting an internal force or moment as a redundant
Externally determinate, but Internally indeterminate structures
Selecting a reaction as a redundant ()
Selecting an internal force or moment as a redundant
A B
CD
P
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13. Method of Consistent Deformations 40
1. Static Determinacy
Externally Static Determinacy
r = 3
Statically Determinate Externally
A B
CD
P
Internally Static Determinacy
m = 6, r = 3, j = 4
m + r > 2j
Statically Indeterminate
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13. Method of Consistent Deformations 41
2. Selecting Internal Force FAD as a redundant
= FAELFv
== OADADO FAELu
A B
CD
P
ADOOF
A B
CD
1ADu
1
FO : Internal Forces due to External Forces
of the Primary System
uAD : Internal Forces due to unit axial Forces
of the Primary System
ADOF 0=ADADu , 0.1=
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13. Method of Consistent Deformations 42
3. Applying the redundant
A B
CD
1
1
ADADf ,
= FAELFv
ADu
ADF
== AELuf ADADAD 2,4. Compatibility Condition
0, =+= ADADADADOAD Ff== ADADADOAD fF ,/
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13. Method of Consistent Deformations 43
5. Member Forces
ADADO FuFF +=
6. Reaction Forces
ADADO FrRR +=
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13. Method of Consistent Deformations 44
Example 13.5
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13. Method of Consistent Deformations 45
Example 13.6
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13. Method of Consistent Deformations 46
13.3 Structures with Multiple Degrees of Indeterminacy
w
A B C D E
1. Free-Body Diagram
AxAy
By Cy
w
Dy Ey
Unknown variables = 63Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 6 - 3 =3
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13. Method of Consistent Deformations 47
AxAy
By Cy
w
Dy Ey
2. Selecting By, Cy and Dy as the redundants
w
AxOAyO EyO
BO CO DO
=BO=CO=DO
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13. Method of Consistent Deformations 48
3. Applying the redundant
1yBBBfAxB
AyBEyB
CBf DBf
=BBf =CBf =DBf
1yCBCfAxC
AyCEyC
CCf DCf
=BCf =CCf =DCf
1yDBDfAxD
AyDEyD
CDf DDf
=BDf =CDf =DDf
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13. Method of Consistent Deformations 49
4. Compatibility Condition
0=+++ yBDyBCyBBBO DfCfBf0=+++ yCDyCCyCBCO DfCfBf0=+++ yDDyDCyDBDO DfCfBf
No. of Unknown Redundants = 3
No. of Compatibility Equations = 3
By, Cy, Dy
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13. Method of Consistent Deformations 50
5. Reaction Forces
6. Shear Force Diagram & Bending Moment Diagram
Deflections to be calculated for determining unknown reducdants
=DO=CO=BO=DBf=CBf=BBf=DCf=BCf =CCf
=BDf =CDf =DDf
12
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13. Method of Consistent Deformations 51
According to the Maxwells Law of Reciprocal Deflections,
=BO =CO =DO=BBf =CBf =DBf=BCf =CCf =DCf=BDf =CDf =DDf
9Homework 4 (10 Points)Definition & Proof
- Bettis Law
- Maxwells Law of Reciprocal Deflections
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13. Method of Consistent Deformations 52
Example 13.7
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13. Method of Consistent Deformations 53
Example 13.8
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13. Method of Consistent Deformations 54
Example 13.9
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13. Method of Consistent Deformations 55
Example 13.10
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13. Method of Consistent Deformations 56
Example 13.11
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13. Method of Consistent Deformations 57
13.4 Support Settlements and Temperature Changes
Support Settlementsw
A B C DB C
2. Selecting By and Cy as the redundantsw
BO CO
=CO=BO
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13. Method of Consistent Deformations 58
3. Applying the redundant
1
BBf CBfyB
=CBf=BBf
1
BCf CCf
=BCf =CCfyC
4. Compatibility Condition
=++ yBCyBBBO CfBf B=++ yCCyCBCO CfBf C
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13. Method of Consistent Deformations 59
A B C D
AB
C D
=++ yBCyBBBO CfBf=++ yCCyCBCO CfBf
Rigid-Body Displacement
Chord of Primary BeamBR
CR
BRCR
Relative Displacement
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13. Method of Consistent Deformations 60
Rigid-Body Displacement
A B C D
No Bending Deformations No Member Forces (Bending Moments) No Stresses
Support Settlements in a Externally Determinate Structure No Effects
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13. Method of Consistent Deformations 61
Example 13.12
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13. Method of Consistent Deformations 62
Temperature Changes
Support Settlements or Temperture Changes in a Determinate Structure
No EffectsSupport Settlements or Temperture Changes in a Indeterminate Structure
Resulting Stresses
Externally Indeterminate, and Internally Indeterminate
Reactions 0 Internal Forces 0
Externally Determinate, and Internally Indeterminate Reactions = 0 Internal Forces 0
Externally Determinate, and Internally Determinate
Reactions = 0 Internal Forces = 0
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13. Method of Consistent Deformations 63
Example 13.13
A B
C D1T
2T
1. Static Determinacy
Externally Static Determinacy
r = 3
Statically Determinate Externally
Internally Static Determinacy
m = 6, r = 3, j = 4
m + r > 2j
Statically Indeterminate
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13. Method of Consistent Deformations 64
2. Selecting Internal Force FAD as a redundant
A B
C D1T
2TADO
= LTFv )(
A B
C D
1
1
ADu
== LTuADADO )(
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13. Method of Consistent Deformations 65
3. Applying the redundant
A B
C D
1
1ADF
ADADf ,
= FAELFv
ADu
== AELuf ADADAD 2,4. Compatibility Condition
0, =+= ADADADADOAD Ff== ADADADOAD fF ,/
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13. Method of Consistent Deformations 66
5. Member Forces
ADADFu=ADADO FuFF +=
6. Reaction Forces
ADADFr=ADADO FrRR += 0=
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13. Method of Consistent Deformations 67
13.5 Method of Least Workw
A B CAxAy By Cy
Selecting By as the redundantw
By
w
By
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13. Method of Consistent Deformations 68
=U ),( yBwf)(wf=UAccording to the Castiglianos Second Theorem,
=
yBU
0 Compatibility Condition
For the value of the redundant that satisfies the equilibrium equation and compatibility,the strain energy of the structure is a maximum or minimum.
MINIMUM The magnitude of the redundants of a statically indeterminate structure
must be such that the strain energy (internal work) is a minimum (the least).
Principle of Least Work
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13. Method of Consistent Deformations 69
For structures with n redundants,
),,,,,( 321 nRRRRwf L=U0
1
=RU 0=
nR
U02
=RU 0
3
=RU L
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13. Method of Consistent Deformations 70
Example 13.1430 kN/m 80 kN
10 m 5 m 5 mB C
EI = constant
A D
AxAy
By Dy
Free-Body Diagram30 kN/m 80 kN
Unknown variables = 43Equations of Equilibrium =
Statically Indeterminate Structure
Degree of Indeterminacy = 4 - 3 = 1
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13. Method of Consistent Deformations 71
Selecting By as the redundant
30 kN/m 80 kN
Ax
Ay Dy
By
yy BD 5.0135=0=xA yy BA 5.0245=
L dxEIM02
2=U
L
y
dxEIM
BM
00==
yB
U
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13. Method of Consistent Deformations 72
30 kN/m 80 kN
245 0.5By 135 0.5By
ByA
B
C D
(m)Limit
scoordinate XSegment
215)5.0245( xxBy xBy )5.0135(
)5(80)5.0135( xxBy
yBM /MOriginx5.0100AABx5.0D 50 DCx5.0105CB D
=L
y
dxEIM
BM
0=
yB
U 100 dxL + 50 dxL + 105 dxL 0=)(5.242 = kNBy
=yA =yD
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13. Method of Consistent Deformations 73
Example 13.15
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13. Method of Consistent Deformations 74
Example 13.16
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13. Method of Consistent Deformations 75
Homework 5 (10 Points)
Why cannot the method of least work be used for analyzing the effects of support settlements and temperature changes?
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13. Method of Consistent Deformations 76
Homework 6 (40 Points)Solve the following problems.
13.1 ~ 13.12 : 2 problems13.13 ~ 13.25 : 2 problems13.26 ~ 13.36 : 2 problems13.37 ~ 13.45 : 2 problems13.46 ~ 13.47 : 1 problem13.48 : 1 problem13.49 ~ 13.61 : 2 problems
12 problems