06_InfluenceLineMax

7/31/2019 06_InfluenceLineMax

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120

Maximum Influence at a Point Due to a Series of Concentrated Loads

F2

x1 x2

F1

F3

x

MC

ab/L

VC

x

(b/L)

(-a/L)

AB

Ca b


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121

AB

Ca b

x

MC

ab/L

F2

x1 x2

F1

F3

'3V

'2V

'3M'2M

VC

x

(b/L)

(-a/L)

)()()/()'( '33'221 VFVFLbFVCR ++=

)()()/()'( '33'221 VFVFLaFVCL ++=

)()()/(' '33'221 MFMFLabFM ++=


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122

AB

Ca b

VC

x

(b/L)

(-a/L)

x

MC

ab/L

F2

x1 x2

F1

F3

''1V

''1M

''3V

''3M

)()/(')'( ''332''11 VFLbFVFVCR ++=

)()/(')'( ''332''11 VFLaFVFVCL ++=

)()/('' ''332''11 MFLabFMFM ++=


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123

AB

Ca b

VC

x

(b/L)

(-a/L)

x

MC

ab/L

F2

x1 x2

F1

F3

'''2V

'''2M

)/('')'( 3''22 LbFVFVCR +=

)/('')'( 3'''22 LaFVFVCL +=

)/(''' 3'''22 LabFMFM +=


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124

VC

x

0.75

-0.25

4 kN

1 m 3 m

1 kN

6 kN Shear

AB

C2 m 6 m


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125

Case 1

VC

x

0.75

-0.25

4 kN

1 m 3 m

1 kN

6 kN

0.625

0.25

(VC)1= 1(0.75) + 4(0.625) + 6(0.25) = 4.75 kN

AB

C2 m 6 m


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126

Case 2

VC

x

0.75

-0.25

4 kN

1 m 3 m

1 kN

6 kN

-0.125

0.375

AB

C2 m 6 m

(VC)2= 1(-0.125) + 4(0.75) + 6(0.375) = 5.125 kN

V1-2= 5.125 - 4.75 = 0.375 kN

(VC)1= 1(0.75) + 4(0.625) + 6(0.25) = 4.75 kN


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127

Case 3

VC

x

0.75

-0.25

4 kN

1 m 3 m

1 kN

6 kN

(VC)3= 6(0.75) = 4.5 kN

AB

C2 m 6 m

V2-3

= 4.5 - 5.125 = - 0.625 kN

(VC)max= (VC)2 = 5.125 kN

(VC)2= 1(-0.125) + 4(0.75) + 6(0.375) = 5.125 kN


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128

The critical position of the loads can be determined in a more direct manner by finding

the change in shear, V, which occurs when the load are moved from Case 1 to Case 2,

then from Case 2 to Case 3, and so on.If the slope of the influence line iss, then

(y2 -y1) =s(x2 -x1), and therefore

V=Ps (x2

-x1

)

Sloping Line

----------(6-1)

If the load moves past a point where there is a discontinuity or jump in the influence

line, as point C, then the change in shear is simply

V=P(y2 -y1)

Jump

----------(6-2)


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129

AB

C2 m 6 m

4 kN

1 m 3 m

1 kN

6 kN

VC

x

0.75

-0.25

V1-2= 1(-0.25 - 0.75) + 1(1/8)(1) + 4(1/8)(1) + 6(1/8)(1) = 0.375 kN

(VC)2 = (VC)1 + V1-2 = 4.75 + 0.375 = 5.125 kN

s = 1/8

(VC)1= 1(0.75) + 4(0.625) + 6(0.25) = 4.75 kN


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4 kN

1 m 3 m

1 kN

6 kN

VC

x

0.75

-0.25

V2-3= 1(1/8)(1) + 4(-0.25-0.75) + 4(1/8)(2) + 6(1/8)(3) = -0.625 kN

(VC)3 = (VC)2 + V2-3 = 5.125 -0.625 = 4.5 kN

s = 1/8

AB

C2 m 6 m


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131

4 kN

1 m 3 m

1 kN

6 kN

Moment

MC

x

(2)(6)/8 = 1.5

AB

C2 m 6 m


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132

Case 1

MC

x

1.5 1.250.5

(MC)1= 1(1.5) + 4(1.25) + 6(0.5) = 9.5 kNm

4 kN

1 m 3 m

1 kN

6 kN

AB

C2 m 6 m


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133

Case 2

MC

x

1.5

(MC)2= 1(0.75) + 4(1.5) + 6(0.75) = 11.25 kNm

0.75 0.75

4 kN

1 m 3 m

1 kN

6 kN

AB

C2 m 6 m

M1-2= 11.25 - 9.5 = 1.75 kNm

(MC)1= 1(1.5) + 4(1.25) + 6(0.5) = 9.5 kNm


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134

Case 3

MC

x

1.5

(MC)max = (MC)2 = 11.25 kNm

4 kN

1 m 3 m

1 kN

6 kN

AB

C2 m 6 m

M2-3= 9 - 11.25 = -2.25 kNm

(MC)3= 6(1.5) = 9 kNm


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135

We can also use the foregoing methods to determine the critical position of a series of

concentrated forces so that they create the largest internal moment at a specific point in

a structure. Of course, it is first necessary to draw the influence line for the moment at

the point and determine the slopess of its line segments. For a horizontal movement

(x2 -x1) of a concentrated forceP, the change in moment, M, is equivalent to the

magnitude of the force times the change in the influence-line ordinate under the load,

that is

M=Ps (x2 -x1)

Sloping Line----------(6-3)


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136

MC

x

1.5

4 kN

1 m 3 m

1 kN

6 kN

AB

C

2 m 6 m

mkNM =++=

75.1)1)(6

5.1)(64()1)(

2

5.1(121

mkNMMM CC =+=+= 25.1175.15.9)()( 2112

(MC)1= 1(1.5) + 4(1.25) + 6(0.5) = 9.5 kNm


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137

MC

x

1.5

4 kN

1 m 3 m

1 kN

6 kN

AB

C

2 m 6 m

mkNM =++=

25.2)3)(

6

5.1(6)2)(

2

5.1(4)1)(

2

5.1(132

mkNMMM CC ==+= 925.225.11)()( 3223

(MC)2= 11.25 kNm


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138

Example 6-9

Determine the maximum shear created at pointB in the beam shown in the figure

below due to the wheel loads of the moving truck.

4 kN9 kN 15 kN 10 kN

1 m 2 m 2 m

AC

B

4 m 4 m


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AC

B

4 m 4 m

SOLUTION

-0.5

0.5VB

x

4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

VBR= 4(0.5) + 9(0.375) + 15(0.125) = 7.25 kN

0.1250.375

VBL= 4(-0.5) + 9(0.375) + 15(0.125) = 3.25 kN

Case 1


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140

-0.5

0.5VB

x

4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

-0.375

0.25

VBR= 4(-0.375) + 9(0.5) + 15(0.25) + 10(0) = 6.75 kN

VBL= 4(-0.375) + 9(-0.5) + 15(0.25) + 10(0) = -2.25 kN

Case 2

AC

B

4 m 4 m


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-0.5

0.5VB

x

4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

0.25

-0.25-0.125

VBR= 4(-0.125) + 9(-0.25) + 15(0.5) + 10(0.25) = 7.25 kN

VBL= 4(-0.125) + 9(-0.25) + 15(-0.5) + 10(0.25) = -7.75 kN

Case 3

AC

B

4 m 4 m


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-0.5

0.5VB

x

4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

-0.25

VBR= 9(0) + 15(-0.25) + 10(0.5) = 1.25 kN

VBL= 9(0) + 15(-0.25) + 10(-0.5) = -8.75 kN

The maximum shearcreated at pointB is -8.75 kN

Case 4

AC

B

4 m 4 m


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Example 6-10

Determine the maximum positive moment and negative moment created at point

B in the beam shown in the figure below due to the wheel loads of the crane.

A C

B

2 m 2 m3 m

3 kN8 kN4 kN

2 m3 m


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x

MB 2(3)/5 = 1.2

-0.8

A CB

2 m 2 m3 m

3 kN

8 kN4 kN

2 m3 m

SOLUTION

MB = 3(1.2) + 8(0) = 3.6 kNm

Positive moment

CASE I


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145

A CB

2 m 2 m3 m

3 kN

8 kN4 kN

2 m3 m

x

MB 1.2

-0.8

MB = 8(1.2) + 3(0.4) = 10.8 kNm

0.4

CASE II


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A CB

2 m 2 m3 m

3 kN

8 kN4 kN

2 m3 m

x

MB 1.2

-0.8

MB = 4(1.2) + 8(0) + 3(-0.8) = 2.4 kNm

The maximum positive momentcreated at pointB is 10.8 kNm

CASE III


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A CB

2 m 2 m3 m

3 kN

8 kN4 kN

2 m3 m

x

MB 1.2

-0.8

MB = 8(-0.8) + 4(0.4) = -4.8 kNm

Negative moment

0.4

CASE I


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3 kN

8 kN4 kN

2 m3 m

x

MB 1.2

-0.8

MB = 4(-0.8) = -3.2 kNm

A CB

2 m 2 m3 m

The maximum negative momentcreated at pointB is - 4.8 kNm

CASE II


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149

Absolute Maximum Shear and Moment

L

CL

L/2 L/2

FR

x

F1 F2 F3

d1 d2

2'x

xx '

A B

Ay By

)]'(2

)[(1:0 xxLFL

AM RyB ==


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150

A

(L/2 -x)

F1d1

V2

M2

)]'(2

)[(1

xxL

FL

A Ry =

For maximumM2 we require

2

L

CL

L/2 L/2

x

F1

F2

F3

d1 d2

2

'x

xx '

A B

Ay By

FR

1122 )2

(:0 dFxLAMM y ==

11)2

)]('(2

)[(1

dFxL

xxL

FL

R =

11

2

2

'

2

'

4dF

L

xxF

L

xFxFLFM RRRR +=

0'22=+=

L

xF

L

xF

dx

dM RR

2'xx =


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151

L

A B

CL

L/2 L/2

F1 F2 F3

x1 x2

xx 2

FR

x

xx +1


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152

FR

F1 F2 F3

x1 x2

b

b

2

x

FR

F1 F2 F3

x1 x2

b

b

2

x

a b

M x

''2/ MLab =

''3M''1M

''33''22''111 MFMFMFMS ++=

L

A B

CL


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153

FR

F1 F2 F3

x1 x2

a

a

2

2 xx

a b

'''33'''22'''112 MFMFMFMS ++=

M x

'''3/ MLab =

The absolute maximum momentis comparison by MS1and MS2.

'''1M'''2M

L

A B

CL


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154

1200 kg 400 kg

8 m

Example 6-11

Determine the absolute maximum moment on the simply supported beam cased

by the wheel loads.

30 m

AB


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155

CL

15 m 15 m

M@1200 kg = 0 :

)8(400)0(12001600 +=x

mx 2=

AB

b

b

3 m 3 m

1600 kg

1200 kg 400 kg

6 mmx 2=


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156

CL

MS = (1200)(7.47) + (400)(3.74)

= 10 460 kgm

14 m 16 m

M1200x

(14)(16)/30 = 7.47

3.74

+ MB= 0:Global:

Ay= 746.67 kg

1600 kg

1200 kg 400 kg

6 m2 m

a

a

1 m

1 m

1

A

14 m

1

Ay= 746.7 kg

V1

M1

+ M1= 0:

M1= 10 460 kgm

Or using equilibrium conditions:

AB

0)30()14(1600 = yA

CL


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CL

MS = (1200)(4) + (400)(7.2) = 7680 kgm

By comparison, Mmax = 10 460 kgm

M400

(18)(12)/30 = 7.2

x

4

18 m 12 m

b

b

3 m 3 m

1600 kg

1200 kg 400 kg

6 m2 m

AB


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158

4.6 T 8.2T 8.2T

4.2 m 1.2 m

Example 6-12

Determine the absolute maximum moment and maximum shear on the simply

supported beam cased by the wheel loads.

20 m

AB

F 21 T


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4.6 T 8.2T 8.2T

4.2 m 1.2 m

M@ 4.6 T = 0 :

)4.5(2.8)2.4(2.8)0(6.421 ++=x

mx 75.3=

FR=21 T

x

45.02.4 =x

20 m

AB

Absolute maximum moment

F =21 T

CL


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160

AB

4.6 T 8.2T 8.2T

FR=21 T

4.2 m 1.2 m

1.875 m10 m 10 m

M4.6T x

a = 8.125 m b = 11.875 m

4.82

3.12 2.63

MS = (4.6)(4.82) + (8.2)(3.12) = 69.32 Tm+ (8.2)(2.63)

FR=21 T


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161

4.6 T 8.2T 8.2T

4.2 m 1.2 m

0.225 m

CL

10 m 10 m

M8.2T x

a = 10.225 m b = 9.775 m

52.95 4.39

MS = (4.6)(2.95) + (8.2)(5) = 90.57 Tm+ (8.2)(4.39)

By comparison, Mmax = 90.57 Tm

+ MB= 0:0)20()225.10(21 = yA

Global:

Ay= 10.74 T

Or using equilibrium conditions:2

+ M2= 0:

M2= 90.50 Tm

A

10.225 m2

Ay= 10.74 T

V2

M2

4.6 T

4.2 m

AB

Ab l t i h


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162

20 m

A B

Absolute maximum shear

4.6 T 8.2T 8.2T

4.2 m 1.2 m

(Vmax)1 = 4.6(0.73) + 8.2(0.94) + 8.2(1.0) = 19.27 T

xVB

10.940.73


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163

20 m

A B

xVA

1

4.6 T 8.2T 8.2T

4.2 m 1.2 m

0.79 0.73

(Vmax)2 = 4.6(1) + 8.2(0.79) + 8.2(0.73) = 17.06 T


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20 m

A B

xVA

1

4.6 T 8.2T 8.2T

4.2 m 1.2 m

0.94

(Vmax)3 = 8.2(1.0) + 8.2(0.94) = 15.91 T

By comparison, Vmax

= 19.27 T


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165

Example 6-13

Determine the absolute maximum moment on the simply supported beam cased

by the wheel loads.

4 kN9 kN

15 kN 10 kN

1 m 2 m 2 m

AB

8 m

SOLUTION 9 kNFR=38 kN


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166

SOLUTION

M@ 4 kN = 0 :

4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

x

1.74 m 0.26 m

AB

8 m

)5(10)3(15)1(9)0(438 +++=x

mx 74.2=

9 kN k k

FR=38 kNCL


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167

a = 3.13 m b = 4.87 m

x

M9 kN (3.13)(4.87)/8 = 1.91

1.130.34

1.30

MS = 4(1.30) + 9(1.91) + 15(1.13) + 10(0.34) = 42.74 kNm

4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

0.87 0.87

AB

4 m 4 m

F9 kN 15 kN 10 kN

CL


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FR

0.13 m4 kN

9 kN 15 kN 10 kN

1 m 2 m 2 m

0.13 m

a = 4.13 m b = 3.87 m

x

M15 kN (4.13)(3.87)/8 = 2.01.03

0.55 0.97

MS = 4(0.55) + 9(1.03) + 15(2.0) + 10(0.97) = 51.17 kNm

By comparison, Mmax

= 51.17 kNm

3

+ MA= 0:Global:

By= 18.38 kN

Or using equilibrium conditions:

B2 m

15 kN 10 kN

3

18.38 kN

3.87 mV3

M3

+ M3= 0:

-M3 -10(2) + 18.38(3.87) = 0

M3= 51.13 kNm

AB

4 m 4 m 0)8()87.3(38 =+ yB

Date post:	05-Apr-2018
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