frames_mf asdfg asdfg

8/21/2019 frames_mf asdfg asdfg

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Gh. Asachi Technical University of Iai Structural Statics Hyperstatic Structures

Department of Structural Mechanics Cezar Aanici, Dr. Eng.

- 1 -

FRAMES

FORCES METHODEXTERNAL LOADS

Using the forces method, let solve the following frame subjected to the indicated external loads.

Numerical data:

3,8 ; 3 ; 30 / ; 48 ; 64L m H m p kN m F kN M kN m= = = = =

Solution:

4I0

I0

2I0

F

4I0

p

L L/3

3H/4

H

1. Establishing the static indeterminacy degree:The degree of static indeterminacy is given by the following relationship:

3 ( ) 3 1 (0 6) -3n c l r s = + = + =

The frame is three times statically indeterminate.

2. Choosing the primary system:The forces methodwas originally developed by J.G. Maxwell and refined by Otto Mohr and Muller-

Breslau. This method was one of the first available for the analysis of statically indeterminate structures.

The force method consists of writing equations that satisfy the compatibility and forces displacement

requirements for the structure and involve redundant forces as the unknowns. The coefficients of these

unknowns are called flexibility coefficients. Since compatibility forms the basis for this method, it has

sometimes been referred to as the compatibility methodor the method of consistent deformations.Once the redundant forces have been determined, the structure are determined by satisfying the

equilibrium requirements for the structure. The fundamental principles involved in applying this method

are easy to understand and develop.


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- 2 -

First, a primary systemmust be chosen and attached to the hyperstatic frame. The primary system must

be statically determinate and loaded with the external loads (forces, temperature variation, different

settlements) and with the internal efforts and/or reactive forces or bending moments which correspond

and must be compatible to all supplementary removed links, identified as 1, 2 ,...X X . the unknowns to

be solved.Theoretically, in the forces method an infinite number of primary systems could be attached to the

original hyperstatic system. Usually, the most suitable or the optimum primary system will be chosen

considering the numerical analysis the simplest way to draw the bending moment diagrams, the

minimum number of beams involved in the multiplications, the structure and/or the load symmetry etc.

For the above frame, some examples of primary systems are shown in Fig. 2.b, 2.c, 2.d, 2.e, and theoptimum one was considered the frame presented in Fig. 2.a

4I0

I0

2I0

F

4I0

p

3,8 1,27

2,25

3

X 1

X 2

X 3

a)

3,8 1,27

2,25

3

A

B C

D E

X3

X2

X1

3,8 1,27

2,25

3

A

B C

D E

X3

X1

X2

b) c) Mechanism!Fig. 2


3/24



- 3 -

3,8 1,27

2,25

3

A

B C

D E

X1 X1

X2

X2

X3

X3

3,8 1,27

2,25

3

A

B C

D E

X1

X2

X2

X2

d) e) Critical system!Fig. 2(continued)3. Elastic equilibrium equations

are obtained writing the compatibility relationships on each removed link, which means a

null displacement (translation or rotation) on each unknown directioni

X ,

and, if expanded, the following equations system results:

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

0

0

0

p

p

p

X X X

X X X

X X X

+ + + =

+ + + =

+ + + =

4. Find out the unknowns coefficients and the corresponding free terms4.1.The main and secondary coefficients

are obtained with the following simplified Mohr-Maxwell relationships (the axial and shear forceeffects are neglected):

0

l

i iM M dxEIii

0 0

l l

i jj iM M M M

dx dxEI EIij ji

= = where Mi si Mj are the bending moments diagrams on the primary system which is loaded only

withi

X =1, andj

X =1 respectively, representing the real load,i

M andj

M are the bending moments

diagrams on the primary system loaded only with a virtual unit force 1 , representing the virtual load.

Of course, by the geometrical point of view, the real bending moments diagrams are identical with the

virtual bending moments diagrams, consequently only one diagram will be drawn.

0i

a

X =


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- 4 -

- Primary system loaded with 1X = 1, and with 1 , respectively ( 1 1M M ):

3,8 1,27

2,25

3

A

B C

D E

5,25

3

M1X =11

1

M1

0; 1 3 3 ; 0; 3 ;

1 5,25 5,25 ; 5,25 ; 5,25 .

AB CB BD

A B B B

BD DE

D D E

M M kNm M M kNm

M kNm M kNm M kNm

= = = = =

= = = =

( )

2 21 111

0 00

0 0

1 1 2 2, 253 3 3 2 3 2 5, 25 3 5, 25 5, 25 3

2 2 3 6

1 78,67 5,25 5.07 5.25

4

l

b

M Mdx

EI E I E I

E I EI

= = + + + + +

+ =


3,8 1,27

2,25

3

A

B C

D

E

5,07

M2

X =121

M2


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- 5 -

0A B C DM M M M= = = =

5,07E

M kNm=

2 222

0 00

1 1 2 10,865,07 5,07 5,07

4 2 3

l

b

M Mdx

EI E I EI = = =


3,8 1,27

2,25

3

A

B C

D

E

3,8

3,8

1,27

M3

X =13

1

M3

0

AM = ;

0; 1 3,8 3,8 ; 0; 3,8 ; 3,8 ;CB AB BC BDC B B B DM M kNm M M kNm M kNm= = = = = =

3,8 ; 1,27DED E

M kNm M kNm= =

( )

3 333

0 00

2 2

0 0

1 1 2 13,8 3,8 3,8 3,8 2, 25 3,8

4 2 3

5,07 41,8

2 3,8 2 1, 27 3,8 1, 27 1,27 3,86 4

l

b

M Mdx

EI E I E I

E I EI

= = + +

+ + =

The secondary coefficients are:

1 2 1 212 21

0 00 0

1 5,07 5,07 16,875,25

4 2

l l

b b

M M M Mdx dx

EI EI E I EI

= = = = =

1 313 31

0 0 00

1 (3 5,25) 2, 25 1 1 1 43,693,8 3,8 5.07 5.25 1.27 5.07 5.25

2 4 2 2

l

b

M Mdx

EI EI E I EI

+ = = = + =

2 3 3 2

23 320 0

0 0

5,07 1,35(2 5, 07 1, 27 5, 07 3,8)

6 4

l l

b b

M M M Mdx dx

EI EI E I EI

= = = = =


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- 6 -

The sum of the unknowns coefficients is:

, 1

78, 67 10,86 41,8 2 (43, 69 16,87 1,35) 182, 27sn

I

SS ij

i j

=

= = + + + =

4.2. The free terms are obtained applying also the simplified Mohr-Maxwell equation:

0

l

p i

ip

b

M Mdx

EI

where

pM is the bending moment diagram on the primary system produced by the external loads (and,

of course, it represents the real load).

1

1

00

0 0

2,25(2 216, 6 3 2 108, 6 5, 25 216, 6 5, 25 108, 6 3)

6

5,07 263, 24( 2 108, 6 5, 25 2 469, 36 5, 25 108, 6 5, 25 469, 36 5, 25)

6 4

l

p

p

b

M Mdx

EI E I

E I EI

= = + + + +

+ + + =

( )220 0

0

5,07 889,092 469, 36 5, 07 108, 6 5, 07

6 4

l

p

p

b

M Mdx

EI E I EI = = + =

3

3

0 00

0 0

1 1 3 1 (216,6 108,6) 2.253,8 216,6 3,8 3,8

4 3 4 2

5,07 1606( 2 108, 6 3,8 2 469,36 1, 27 108, 6 1, 27 469, 36 3,8)

6 4

l

p

p

b

M Mdx

EI E I E I

E I EI

+ = = +

+ + + =

48 kN

30 kN/m

3,8 1,27

2,25

3

A

B C

D

E

3,8 1,27

2,25

3

216,6

108,6

469,36

Mp


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- 7 -

The sum of all free terms is:

1

263, 24 889, 09 1606 2758.33sn

I

Sp ip

i=

= = =

5. Checking the unknown coefficients and the free termsThe coefficients are correctly solved if the following condition is checked:

2

1 10

s s

ln n

I II sSS ij SS

i j b

Mdx

EI

= =

= = =

3,8 1,27

2,25

3

Ms

3

3,8

9,05

1,09

6,8

X =13

X =12

X =11

( )( )

2

0 00

2 2

0

2 2

0 0

1 1 2 1 1 23 3 3 3,8 3,8 3,8

2 2 3 4 2 3

2,252 6,8 2 9,05 6,8 9,05 6,8 9,05

6

5,07 182,272 9,05 2 1,09 9,05 1,09 9,05 1,09

6 4

l

II sSS

b

Mdx

EI E I E I

E I

E I EI

= = + +

+ + + + +

+ + =

I II

SS SS =

Condition checked, the unknown coefficients are correctly solved.

The free terms are correctly solved if the following condition is checked:

10

s

ln

p sI IISp ip Sp

i b

M M dxEI=

= = =


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- 8 -

( )

( )

00

0

0 0

1 1 3216,6 3,8 3,8

4 3 4

2,252 216,6 6,8 2 108, 6 9,05 108, 6 6,8 216, 6 9,056

5,07 2758.332 108, 6 9, 05 2 469, 36 1, 09 463, 36 9, 05 108, 6 1, 09

6 4

l

p sII

sp

b

M Mdx

EI E I

E I

E I EI

= =

+ + + +

+ + + =

I II

Sp Sp =

Condition checked, the free terms are correctly solved.

6.

Let solve the system of equations to find out the unknownsXi

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

0

0

0

p

p

p

X X X

X X X

X X X

+ + + =

+ + + =

+ + + =

1 2 3

0 0 0 0

1 2 3

0 0 0 0

1 2 3

0 0 0 0

78,67 16,87 43,69 263,240

16,87 10,86 1,35 889,09 0

43,39 1,35 41,8 16060

X X XEI EI EI EI

X X XEI EI EI EI

X X XEI EI EI EI

+ =

+ =

+ =

or

1 2 3

1 2 3

1 2 3

78,67 16,87 43,69 263, 24

16,87 10,86 1,35 889,09

43,39 1,35 41,8 1606

X X X

X X X

X X X

+ =

+ =

+ =

having the following solution:

1

2

3

6.293

77.99

47.472

X kN

X kN

X kN

=

=

=


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- 9 -

7. Drawing the final internal efforts diagrams:

3.8 1.27

2.25

3.0

48 kN

30 kN/m

6.29 kN

47.47 kN

77.99 kN

-

N

-

77.99 kN

11.46 kN

41.71 kN

-

+

-

6.29 kN

41.71 kN

Q

47.47 kN

66.53 kN+

-

11.46 kN

+


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- 10 -

M

39.37 kNm

19.34 kNm

54.48 kNm

18.87 kNm

35.61 kNm37.55 kNm

1.58 m

8. Checking the diagrams8.1 Static equilibrium checks:

- Joint equilibrium of the node B:

41.71 48 6.29 0

77.99 11, 46 66.53 0

54.48 18.87 35.61 0

X

Y

node

F

F

M

= + =

= =

= + + =


11/24



- 11 -

The final bending moment diagram Mf could be checked applying the virtual work method on the

following floor mechanism or on any other local or global mechanism:

M

39.37 kNm

54.48 kNm

18.87 kNm

48 kN

1= 2 25

2= 3

( )1 2( ) (39,37 54, 48) 48 18,87 39,37 54,48 48 18,87 0

2, 25 3VW

= + + = + + =

The bending moment diagram is checked.

8.2 Elastic checkings of the final bending moment diagram Mf:

The final bending moment diagram Mf is correctly drawn if the following condition is

checked:

0

0i

l

f i

X

b

M M

dxEI

= =

which means that the displacement on any unknown direction must be null.

( )

( )

1

1

0

0 0

5,072 39.37 5.25 2 19.34 5.25 39.37 5.25 19.34 5.25

6 4

2.25 1 1 22 39.37 5.25 2 3 54.48 39.37 3 54.48 5.25 18.87 3 3 0

4 2 3

l

f

X

b

M Mdx

EI E I

E I E I

= = + +

+ + =


12/24



- 12 -

SUPPORT SETTLEMENTS

Applying the forces method, let draw the internal efforts diagrams caused by indicated

support settlements.

Numerical data:

03,8 ; 3 ; 30 / ; 48 ; 1 ; 1, 4L m H m p kN m F kN v cm = = = = = =

3,8m 1,27m

2,25m

3m

A

B

C

D E

v

1. The same primary system as before will be used:4I0

I0

2I0

4I0

3,8 m 1,27 m

2,25 m

3 m

X 1

X 2

X 3


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- 13 -

2. The elastic equilibrium system of equations is:11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

c

c

c

X X X L

X X X L

X X X L

+ + =

+ + =

+ + =

The main and the secondary coefficients are the same, and they were calculated and checked in

the above application.

3. The free terms are calculated by the following virtual work relationships:( )

1i

kic i k L R=

where:

1 is the virtual unit load (concentrated force/bending moment or a couple of concentratedforces/bending moments, respectively) applied on the direction of the unknownX

i;

i - the corresponding support settlement on the direction of the unknownXi;

( )i

kR the reactive forces in the kexternal link caused by the virtual unit load1.

The primary system is consecutively loaded with a virtual unit force1 on eachXidirection, and

the reactive forces in the external links of the primary system will be calculated.

3,8 1,27

2,25

3

A

B C

D E

Rx1,1

Rx1,2

1

1,1 1 5,25 5,25xR = =

1,1 1 1xR = =

1

1,41 0 1 0 5,25 0,128

180c

L

= + =


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- 14 -

2,1 1 5,07 5,07xR = = 3,1 1 1, 27 1, 27xR = =

2,1 1 1xR = = 3,1 1xR =

2

1,41 0 1 0 5,07 0,124

180c

L

= + + = 31,4

1 0,01 1 0 1,27 0,021180

cL

= + + =

The sum of all free terms is:

1

0,128 0,124 0, 021 0, 017sn

I

sc ic

i

L L=

= = + + =

4. Checking the free termsThe free terms are correctly solved if the following condition is checked:

( )

( 1 ) ( )s

IIksc i k

L R= +

The primary system is loaded with all unit forces applied on the correspondingXidirections, and the

reactive forces are calculated.

3,8 1,27

2,25

3

A

B C

D E

Rx2,1

Rx2,2

1

3,8 1,27

2,25

3

A

B C

D E

Rx3,2

Rx3,1

1


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- 15 -

3,8 1,27

2,25

3

A

B C

D E

Rs3

Rs2

Rs1

1

1

1

1 1sR =

2 2sR =

3 1 1,27 1 5,25 1 5,07 1,09sR = + =

Consequently,

1,4 3,14151 0,01 1 0 2 0 1,09 0,017

180

II

scL

= + + + =

Condition checked, the free terms are correctly solved.

5. Solving the system of elastic equilibrium equations ( )7 4 3 20 2,1 10 72 10 151,2 10E I kNm = = 11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

c

c

c

X X X L

X X X LX X X L

+ + =

+ + = + + =

31 2 3

31 2 3

31 2 3

78,67 16,87 43,69 19,3536 10

16,87 10,86 1,35 18,7488 10

43,39 1,35 41,8 3,1752 10

X X X

X X X

X X X

+ =

+ =

+ =

having the solution

{ }

391.395

2302258.775

X

=


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- 16 -

6. Drawing the final internal efforts diagrams:The primary system is loaded with allthe unknown reactive and/or internal efforts, considering

the resulted sign, and the diagrams will be drawn as for a statically determinate system.

3,8 1,27

2,25

3

A

B C

D E

258,775

2302

391,395

3,8 1,27

2,25

3

A

BC

D E

2302

2043,225

391,395

N Diagrama finala eforturilor axiale


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- 17 -

Q Diagrama finala forelor tietoare

3,8 1,27

2,25

3

A

B C

D E

983,345

1071,478

1174,185190,84

9287,672

1071,478

M

M Diagrama finala momentelor ncovoietoare


18/24



- 18 -

Checkingthe final bending moment diagramMfapplying the virtual work method for the following

floor mechanism:

M

1071.48 kNm

190.84 kNm

1174.18 kNm

1= 2 25

2= 3

1 2( ) ( 1071.48 190.84) 1174.18 ( 1071.48 190.84) 1174.18 0

2.25 3LMV

= + + = + + =

The bending moment diagram is checked.


19/24



- 19 -

TEMPERATURE VARIATION

Let solve the following frame to the indicated non-uniform temperature variation

applying the forces method

Numerical data:3

5 5 2 40

40 603,8 ; 3 ; 10 ; 2,1 10 / ;

12tL m H m E daN cm I cm

= = = = =

3,8 1,27

2,25

3

A

BC

D

E

100

100

100 20

0

80

Considering for the beam with the reference moment of inertia a rectangular cross-section3

4

0

40 60

12I cm

=

and adopting the hypothesis that the width of all structural elements is the same (b = constant), it

results:

- for the beams with the moment of inertia of 02I

0 0

0

3 3

2

22 7512 12

I I

I

b h b hh cm

= =

- for the beams with the moment of inertia 04I

0 0

0

3 3

4

44 9512 12

I I

I

b h b hh cm

= =

where 30 0nI Ih h n= .

1. Primary system and solving the unknown coefficients

The primary system which will be used is the same, consequently the main and secondary

coefficients are the same as before, and are already solved.


20/24



- 20 -

3,8 1,27

2,25

3

A

BC

D

E

100

100

100 20

0

80

X1

X2

X3

2. The compatibility equation system is:

11 1 12 2 13 3 1

21 1 22 2 23 3 2

31 1 32 2 33 3 3

0

0

0

t

t

t

X X X

X X X

X X X

+ + + =

+ + + = + + + =

3. The free terms will be calculated with the following relationship

i i

oo

it t ax t N M

tt

h

=

where:

t is the linear variation coefficient -

5 1

10t oC

(quite the same for the concrete and steel);

o

axt is the average temperature in the longitudinal axis of the beam:2

o oo i eax

t tt

+= ;

ot is the temperature difference between the extreme fibers of the cross-sectional area (usually

the temperature difference between the internal side of the beam and external one,

respectively): o o oi et t t = ;

h is the height of the beam cross-section;

,i iN M

are the corresponding axial effort and bending moment diagrams area, respectively,

of the beams.

The bending moment diagrams were already drawn before, in the first solved application, and

now, only the axial effort diagrams will be drawn, consecutively loading the primary system with

virtual unit forces1on eachXidirection.


21/24



- 21 -

A

BC

D E

X =11

N1

1

A

BC

D E

X =12

N2

1

A

BC

D E

X =12N3

1

The average temperatures in the longitudinal axis of the beams are:

2

o oo o o oi e

ax i e

t tt t t t

+= =

BARA ti te Dt tax

AB 20 10 10 15

BC 20 10 10 15

BD 10 10 0 10

DE 10 8 2 9

The tensioned fibers due the temperature variationis drawn in the figure below with a dashed

line:


22/24



- 22 -

A

BC

D E

100100

100 20

0

80

X1

X2

X3

The primary system and tensioned fibers

due to the temperature variation

The free terms values are:

1 1 1 111 , , , , ,

5 10 1 2

10 9 5,07 3 3 0 0 5,07 5,25 156,410,75 2 0,95

oo o o

o BCAB BD DEt t ax DE t N M AB M BD M BC M DE

AB BD BC DE

tt t tt

h h h h

= + + + =

= + + + =

5

10

( )

( )

2 2 2 22,2 , 2, , , , , ,

5 2 1 10 15 3 10 2,25 0 0 0 5,07 50,95 2

oo o oo o BCAB BD DE

t tax BDt ax AB N AB N BD M AB M BD M BC M DE BD DEAB BC

tt t tt t

h h h h

= + =

= + + + + 57 40,84 10,0

=

( )

3 3 3 33 , 3, , , , ,

5 510 1 2 1 1 10 10 2,25 0 0 3,8 3,8 3,8 3,8 1,27 1,27 39,086 100,95 2 0,95 2 2

oo o o

o BCAB BD DE

t t ax BD t N BD M AB M BD M BC M DE

AB BD BC DE

tt t tt

h h h h

= + =

= + + + + + =

5 5( 156,41 40,84 39,086) 10 158,164 10Ist it = = + =

4. Checking the free terms:

The free terms are correctly solved if the following condition is checked:o

II o

st it t ax t Ns Ms

tt h

= =


23/24



- 23 -

A

BC

D E

Ns

1

2

1

Mss

3

3,8

9,05

1,09

6,8

( )5

5

10 1 10 110 15 3 10 4,5 9 5,07 3 3 3,8 3,8

0,75 2 0,95 2

2 1 14,25 9,05 1,09 0,545 158,14 10

0,95 2 2

II

st

I

st

= + + +

+ + = =

The free terms are correctly calculated.

5. Solving the equation system ( )7 4 3 20 2,1 10 72 10 151,2 10E I kNm = =

1 2 3

1 2 3

1 2 3

78,67 16,87 43,69 236,5

16,87 10,86 1,35 61,75

43,69 1,35 41,8 59,1

X X X

X X X

X X X

+ =

+ = + =

with the following solution:

{ }

35.177

55.809

36.379

X

=


24/24



6. Drawing the final internal efforts diagrams (on the original statically indeterminate system):

3,8 1,27

2,25

3

A

BC

D

E

100

100

100 20

0

80

36,379

55,809

35,177

105,53

138,24

52,07

46,44

M

32,71

55,809kN

19,43kN

35,177kN

N

Q

35,177kN

19,43kN

36,379kN

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