Structural Theory 2 (Part 5)1

THEORY2 Structural Theory 2 Chapter 1

Statically Indeterminate Structures;- when a structure has too many external reactions and/or internal forces to be

determined with the equations of statics (including any equation of condition), it is statically indeterminate. A load placed on one part of a statically indeterminate or continuous structure will cause shears, moments and deflections in the other parts of the structure. In other words, loads applied to a column affect the beams, slabs and other columns and vice versa.

Advantages of Statically Indeterminate Structures;- in comparing statically indeterminate structures with statically determinate

ones, the first consideration, to most, would pertain to cost. However, it is impossible to make a statement favoring one type, economically, without reservation. Each structure presents a different situation and all factors must be considered – economic or otherwise.

- Savings in Materials;- the smaller moments developed permit the use of smaller members. For bridges,

the material saving could possibly as high as 10 to 20%. The number of force reversals occurring in railroad bridges keeps their maximum saving nearer to 10%.

A structural member of a given size can support more loads if it is part of a continuous structure than it is simply supported. The continuity permits the use of smaller members for the same loads and spans or increased spacing of supports for the same size members. The possibility of fewer columns in buildings or fewer piers in bridges may permit a reduction in overall costs.

Continuous structures of concrete or steel are cheaper without the joints, pins and so on required to make them statically determinate, as was frequently the practice in recent years. Monolithic reinforced-concrete structures are erected so that they are naturally continuous and statically indeterminate. To install the hinges and other devices necessary to make them statically determinate would not only be a difficult problem but also very expensive. Furthermore, if a building frame consisted of columns and simple beams, it would necessary to have objectionable diagonal bracing between the joints to make the frame stable and rigid.

- Larger Safety Factors;- statically indeterminate structures often have higher safety factors than

statically determinate ones. Structural designers know very well that when portions of the structures are overstressed they will often have the ability to redistribute portions of those stresses to less-stressed areas. Statically determinate structures generally do not have this ability. Should the moment in a statically determinate beam or frame reach the ultimate moment capacity of the structure at a particular point, the structure will fail. This is not the case for statically indeterminate structures. It can be clearly shown that a statically indeterminate beam or frame normally will not collapse when its ultimate moment capacity is reached at just one section. Instead, there is a redistribution of the moments in the structure.

- More Attractive Structures;- it is difficult to imagine statically determinate structures having the

gracefulness and beauty of many statically indeterminate structures, such as arches and frames, being erected today.

- Greater Rigidity and Smaller Deflections;- statically indeterminate structures are more rigid than statically

determinate ones and have smaller deflections. Because of their continuity,

Engr. I.R. Bonzon 1


they are stiffer and have greater stability against all types of loads (horizontal, vertical, moving, etc).

- Adaption to Cantilever Erection;- the cantilever method of erecting bridges is of particular value where

conditions underneath (probably marine traffic or deep water) hinder the erection of formworks. Continuous statically indeterminate bridges and cantilever-type bridges are conveniently erected by the cantilever method.

Disadvantages of Statically Indeterminate Structures;- a comparison of statically determinate and statically indeterminate structures

shows the latter have several disadvantages that make their use undesirable on many occasions.

- Support Settlement;- statically indeterminate structures are not desirable where foundation

conditions are poor, because seemingly minor support settlements or rotations may cause major changes in the moments, shears, reactions and bar forces. Where statically indeterminate bridges are used despite the presence of poor foundation conditions, it is occasionally felt necessary to physically measure the dead-load reactions. The supports of the bridge are jacked up or down until the calculated reaction is obtained, after which the support is built to that elevation.

- Difficulty of Analysis and Design;- the forces in statically indeterminate structure depend not only on their

dimensions but also on their properties (moduli of elasticity, moments of inertia and cross-sectional areas). This situation presents a design difficulty. The forces cannot be determined until the members’ sizes are known and the member sizes cannot be determined until their forces are known. The problem is handled by assuming member sizes and computing the forces, designing the members for these forces and computing the forces for the new sizes and so on, until the final design is obtained. Design by this method – the method of successive approximations – takes more time than the design of a comparable statically determinate structure, but extra cost is only a part of the total cost of the structure.

- Development of Other Stresses;- support settlement is not the only condition that causes stress variations in

statically indeterminate structures. Variation in the relative positions of members caused by temperature changes, poor fabrication or internal deformation of members of the structure under load may cause serious changes throughout the structure.

- Stress Reversals;- generally, more force reversals occur in statically indeterminate structures

than in statically determinate ones. Additional material may be required at certain sections to resist the different force conditions and to prevent fatigue failures.

Method of Consistent Deformation;- statically indeterminate structures can be analyzed by the direct use of the

theory of elastic deformations. Any statically indeterminate structure can be made statically determinate and stable by removing the extra restraints called redundant forces, that is, the force elements which are more than the minimum necessary for the static equilibrium of the structure. The statically

Engr. I.R. Bonzon 2


determinate and stable structure that remains after the removal of the extra restraints is called the primary structure. The original structure is then equivalent to the primary structure subjected to the combined action of the original loads plus the unknown redundant. The conditional equations for geometric consistencies of the original structure at redundant points called the compatibility equations, are then obtained from the primary structure by superposition of the deformations caused by the original loads and redundant. This method known as consistent deformations is generally applicable to the analysis of any structure, whether it is being analyzed for the effects of loads, support settlements, temperature changes or any other cases. However, there is only one restriction on the use of this method: the principle of superposition must hold.

EX. Analyze the propped beam shown.Given: E = 15.0 GPa I = 2.0 x

assume as the redundant force,

considering the applied loads,

(+) = 0;

50(3) - = 0; = 150.0 kN-m

(+) = 0;

- 50 = 0; = 50.0 kN

Engr. I.R. Bonzon 3

"AR

CR"AM

"

50 kN

BCA

3 m 2 m

50 kN

'50

150

50 kN

CRAR

AM

28.4

21.642

43.2

150 150

'CV

'CM

CR5

"CV

"CM

CR5


considering the redundant force,

(+) = 0;

- (5) – 0; = 5

(+) = 0;

- = 0; =

using conjugate beam method,


= = - = - kN-


= = =

kN-

from compatibility requirements,

+ = 0;

- + = 0; = 21.6 kN

considering the whole structure,

(+) = 0;

50(3) – 21.6(5) - = 0; = 42.0 kN-m

(+) = 0;

+ 21.6 - 50 = 0; = 28.4 kN

EX. Draw the shear and moment diagram for the beam shown.

Engr. I.R. Bonzon 42 m 2 m 4 m

50 kN

BCA D

10 kN/m

"AR

"DR

CR

50 kN

10 kN/m

77.5 kN 52.5 kN




(+) = 0;

50(2) + 10(8)(4) - (8) = 0; = 52.5 kN

(+) = 0;

(8) – 50(6) – 10(8)(4) = 0; = 77.5 kN


(+) = 0;

(4) - (8) = 0; = 0.50

(+) = 0;

(8) - (4) = 0; = 0.50

using the applied loads,

for segment AB; 0 x 2,

M = 77.5x - = -5 + 77.5x

Engr. I.R. Bonzon 5

35.3115.31

34.6954.69

29.69

10.31

5.33

50.64

38.74

1

0.5 0.5 1

0.5 0.5

CR

50 kN

10 kN/m

AR DR


m = 0.5x

Mm = -2.5 + 38.75

for segment BC; 2 x 4,

M = 77.5x - - 50(x – 2) = -5 + 27.5x + 100

m = 0.5x

Mm = -2.5 + 13.75 + 50x

for segment CD; 4 x 8,

M = 77.5x - - 50(x – 2) = -5 + 27.5x + 100

m = 0.5x – 1(x – 4) = -0.5x + 4

Mm = 2.5 - 33.75 + 60x + 400

=

= +

+

= +

+

= kN-

using the redundant force,

Engr. I.R. Bonzon 6


for segment AC; 0 x 4,

M = -0.5 x

m = -0.5x

Mm = 0.25


M = -0.5 x + (x – 4) = 0.5 x - 4

m = -0.5x + 1(x – 4) = 0.5x – 4

Mm = 0.25 - 4 x + 16

=

= +

= +

= kN-

from the compatibility requirements,

- = 0

- = 0; = 84.375 kN


(+) = 0;

50(2) + 10(8)(4) – 84.375(4) - (8) = 0; = 10.3125 kN

(+) = 0;

(8) + 84.375(4) - 50(6) - 10(8)(4) = 0; = 35.3125 kN

Check;

(+) = 0;

Engr. I.R. Bonzon 7


35.3125 + 84.375 + 10.3125 – 50 - 10(8)(4) = 0; ok

EX. Analyze completely the beam shown.

assume and as the redundant forces,


(+) = 0;

120(4) - = 0; = 480.0 kN-m

(+) = 0;

- 120 = 0; = 120.0 kN

Engr. I.R. Bonzon 8

BA

120 kN

4 m 2 m

120 kN

BRAR

AM BM

120 kN

480 kN-m

120 kN480

'BV

'BM480

BR

''AR

''AM

BR6

''BV

''BM

BR6

BM'''AM

BM

'''BV

'''BMBM

88.89

31.1171.44

106.67

53.33


considering the redundant force, ,

(+) = 0;

- (6) = 0; = 0.167

(+) = 0;

- = 0; =


(+) = 0;

- = 0; =

= = - = - kN-

= = = kN-

= = = kN-

= = - = -

kN-

= = - = - kN-

= = = kN-


+ + = 0;

-960 + 18 - 6 = 0; eqn. 1

+ + = 0;

4480 - 72 + = 0; eqn. 2

solving simultaneously,

Engr. I.R. Bonzon 9


= 88.89 kN = 106.67 kN-m


(+) = 0;

120(4) + 106.67 - - 88.89(6) = 0; = 53.33 kN-m

(+) = 0;

(6) – 53.33 + 120(2) + 106.67 = 0; = 31.11 kN

Check;

31.11 + 88.89 – 120 = 0; ok

EX. Analyze completely the given beam shown.

assume and as the redundant forces,


Engr. I.R. Bonzon 10

50 kN 50 kN 100 kN

DCBA

2 m2 m2 m 2 m 2 m2 m

100 kN50 kN 50 kN

DRBR CRAR100 kN50 kN50 kN

116.67 kN83.33 kN

166.67300

233.33

'AV

'DV

166.67300

233.33

BR

''AR

''DR

BR67.2

''AV

''DV

BR67.2

CR67.2

'''AR '''

DR

CR

'''DV

'''AV

CR67.2

18.87

31.13

18.44

31.56

62.69

37.31

37.74

24.52

12.36

50.76

74.62


(+) = 0;

50(2) + 50(6) + 100(10) - (12) = 0; = 116.67 kN

(+) = 0;

(12) – 50(10) – 50(6) – 100(2) = 0; = 83.33 kN

= ½(166.67)(2) = 166.67 kN-

= ½(133.33)(4) = 266.67 kN-

= (166.67)(4) = 666.67 kN-

= ½(66.67)(4) = 133.33 kN-

= (233.33)(4) = 933.33 kN-

= ½(233.33)(2) = 233.33 kN-

= 2400.0 kN-

(+) = 0;

166.67(1.33) + 266.67(4.67) + 666.67(4) + 133.33(7.33) + 933.33(8)

+ 233.33(10.67) - (12) = 0; = 1255.56 kN-

(+) = 0;

(12) – 166.67(10.67) – 266.67(7.33) – 666.67(8) – 133.33(4.67)

- 933.33(4) – 233.33(1.33) = 0; = 1144.44 kN-

for the deflection at B,

= = 1144.44(4) – ½(166.67)(20(2.67) – ½(66.67)(2)(0.67)

- 166.67(2)(1) = kN-

for the deflection at C,

= = 1255.56(4) – ½(233.33)(20(2.67) – ½(33.33)(2)(0.67)



- 233.33(2)(1) = kN-


(+) = 0;

(12) - (4) = 0; = 0.33

(+) = 0;

(8) - (12) = 0; = 0.67

= ½(2.67 )(4) = 5.33 kN-

= ½(2.67 )(8) = 10.67 kN-

= 16.0 kN-

(+) = 0;

(12) – 5.33 (2.67) – 10.67 (6.67) = 0; = 7.11 kN-

(+) = 0;

5.33 (9.33) + 10.67 (5.33) - (12) = 0; = 8.89 kN-


= = ½(2.67 )(4)(1.33) – 8.89 (4) = - kN-


= = ½(1.33 )(4)(1.33) – 7.11 (4) = - kN-


(+) = 0;

(12) - (8) = 0; = 0.67

(+) = 0;



(4) - (12) = 0; = 0.33

= ½(2.67 )(8) = 10.67 kN-

= ½(2.67 )(4) = 5.33 kN-

= 16.0 kN-

(+) = 0;

(12) – 10.67 (5.33) – 5.33 (9.33) = 0; = 8.89 kN-

(+) = 0;

5.33 (2.67) + 10.67 (6.67) - (12) = 0; = 7.11 kN-


= = ½(1.33 )(4)(1.33) – 7.11 (4) = - kN-


= = ½(2.67 )(4)(1.33) – 8.89 (4) = - kN-


+ + = 0;

3755.56 – 28.44 - 24.89 = 0; eqn. 1

+ + = 0;

3911.11 – 24.89 - 28.44 = 0; eqn. 2

solving simultaneously,

= 49.57 kN, = 94.25 kN


(+) = 0;



50(2) + 50(6) + 100(10) – 49.57(4) – 94.25(8) - (12) = 0;

= 37.31 kN

(+) = 0;

(12) + 49.57(8) + 94.25(4) – 50(10) – 50(6) – 100(2) = 0;

= 18.987 kN

Check;

= 0;

18.87 + 49.57 + 94.25 + 37.31 – 50 – 50 – 100 =0; ok

EX. Analyze and the draw the shear and moment diagrams for the beam shown.



(+) = 0;

30(3) + 60(6) - (9) = 0; = 50.0 kN

(+) = 0;

(9) – 30(6) – 60(3) = 0; = 40.0 kN


30 kN 60 kN

A

BC D

E

6 m

3 m3 m 3 m

EVAH

EH

AV

30 kN 60 kN

50 kN

40 kN

30 kN 60 kN

1

1

0.67

0.67

70.23

40.23

19.7745.35

61.41

59.28

272.1

272.1


for the horizontal deflection at A,


M = 0

m = 0

Mm = 0

for segment BC; 0 x 3,

M = 40x

m = 0.67xMm = 26.67


M = 40x – 30(x – 3) = 10x + 90

m = 0.67x

Mm = 6.67 + 60x

for segment DE; 6 x 9,

M = 40x – 30(x – 3) – 60(x – 6) = -50x + 450

m = 0.67x

Mm = -33.33 + 300x

=

= + +

= + +

= kN-

considering a redundant force, ,

(+) = 0;

(9) – 1(6) = 0; = 0.67



(+) = 0;

(9) – 1(6) = 0; = 0.67

for the horizontal deflection at A,


M = - x

m = -x

Mm =for segment BC; 0 x 9,

M = 0.67 x

m = 0.67x

Mm = 0.44

=

= +

= + = kN-


- = 0

- = 0; = 45.35 kN


= 0;

45.35 - = 0; = 45.35 kN

(+) = 0;

30(3) + 60(6) – 45.35(6) - (9) = 0; = 19.77 kN



(+) = 0;

(9) – 45.35(6) – 30(6) – 60(3) = 0; = 70.23 kN

Check;

= 0;

70.23 + 19.77 – 30 – 60 = 0; ok


Structural Theory 2 (Part 5)1

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Transcript of Structural Theory 2 (Part 5)1