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Structural Analysis- Text Book by: R. C. HIBBELER

Chapter 10Force Method

Analysis of Statically Indeterminate Structures

Iqbal Marie2018-2019

Structures

Determinate Indeterminate

https://www.youtube.com/watch?v=Ff0GsQpw0.Xc

No effect due settlement

Serious effect due settlement

No effect due to Rise in temperature

Serious effect due to rise in temperature

Not economicaleconomical

Deflection

Deflection

Formation of plastic hinges results unstable structure

Formation of plastic hinges results stable structure

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There are two different Methods of analysis

Force method • known as consistent deformation, unit load

method, flexibility method• The primary unknowns are forces

Displacement method• Known as stiffness method• The primary unknowns are displacements

The deflection or slope at any point on a structure as a result of a number of forces, including the reactions, is equal to the algebraic sum of the deflections or slopes at this particular point as a result of these loads acting individually

Indeterminate to the first degree ( 4 unknowns)

• Three equilibrium equations and one compatibility equation is needed

• Choosing one of the support reaction as a redundant (say By)

• The structure become statically determinate and stable ( primary structure)

• Calculate Downward displacement B at B (load action)• Apply unit load at B and calculate fBB (upward deflection

)- correction structure• Apply the Compatibility equation: actual beam = B+ Byf BB

to find By

• Now the structure is statically determinate and the equilibrium equations can be used to find the other reactions

• Final moment or reaction can be calculated as:

10.2 Analysis Procedure

10.1 – 10-3 Force Method of Analysis: Beams

Correction structure

actual beam = 0 for no settlement or spring

By

By

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Indeterminate to the 2nd degree

• 2 Compatibility equations are needed • Select two redundant reaction B & C • Determine displacement B &C caused by loads • fBB & fBC Deflection per unit force at B are determined • fCC & fCB Deflection per unit force at C are determined • Compatibility equations are applied to find reactions at

B and C• 0 = B+ By fBB + Cy fBC

• 0 = C+ By fCB + Cy fCC

• Then find the other reaction by applying the equilibrium equation

By

By

Cy

Cy

10.3 Maxwell’s Theorem

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300

12

12

6

Draw bending moment diagrams for the beam

B=1/6(6+2x12)(6x-300)/EI= -999kN.m3/EI

fBB =1/3(12x12x12) /EI= 576 /EI

compatibility equation: 0 = B+ By f BB

-999+ By (576) = 0By = 15.6 kNApply equilibrium equations Ay = 34.4 kNMA= 112.8 kNm

Draw shear and bending moment diagrams for the beam. The support at B settles 1.5 in ( 40mm). E= 29(103) ksi, I = 750 in4

m

6M

120

B= 1/3(180x6x12) + 1/3(120x12x24)+1/6(2x180x6 + 180x12 + 120x6 +2x120x12)12 = 31680/EI

fBB = 1/3( 12x12x48) = 2304/EI

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Determine the support reactions on the frame shown EI is Constant

20

20

25kN.m

4

4

Mm

B =1/3(25x4x5) + 0 = 166.7/EI

fBB =1/3(4x4x5 )+1/3(4x4x4) = 48/EI

Draw bending moment diagram

a= -1/3(5x50x5)-1/6(2x50x5+50x8+55x5+2x8x55)x5 –(1/6)(5+2x8)x55x5 =- 3091.67/EI

a= 1/3(5x5x5)x2 +2 (1/6)(2x5x5+5x8+ 8x5+ 2x8x8)x5= 513.33/EI

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+

16

4 a =[0.25(16x4)(4)/2+(6-R)/6(16)(4) = [32+(6-R)*10.66]/EI

3

2.25

a =[(1/3)3 x3x 12 R +[(1/3)x1.5x3x6 +(1/3)x3x1.5x6 +(1/6)(2x3*1.5+3x2.25+1.5x1.5+2*2.25x1.5)(3)]x10 =[36R +247.5]/EI

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10.6 Force Method of Analysis: Trussa. Externally indeterminate

The degree of indeterminacy of a truss : b + r >2j.

The force method is quite suitable for analyzing trusses that arestatically indeterminate to the first or second degree

b. Internally indeterminate

Procedure for analysis:a. Chose the redundant ( support reaction or redundant

member)- [X]b. Convert the indeterminate structure to determinate and

stable primary structure by removing the redundantc. Calculate the reactions ( R0)and (N) forces in each truss

member of the primary structure due to external loadsd. Apply a unit load in the direction of redundant reaction or

a pair of unit loads in direction of redundant member ( always assumed as tensile)

e. Calculate the (n) forces in the truss members due to unit load

f. Tabulate the results and calculate:g. Calculate the redundant as: +x = 0h. Calculate the final reactions and the final member forces :

member L A N n NnL/EA nnL/EA Nf

n

Example: For the truss shown, find the force in each member, EA 200000 kN

N

Statically indeterminate to the first degree ( externally)

b + r >2j.Primary structure ( determinate and stable

Structure with unit load in the direction of the redundant

* X

+x = 0 244.5 +23.59 X = 0X= -10.36kNThe final force in each member ( N final = N+ Xn) = N+(-10.36 n)Reactions = RN + XRn

n

-2.26

-4.18

-7.64

10

.36

1.183

n

NnL/EA nnL/EAn

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Example : For the truss shown, find the force in each member, EA 200000 kN

Statically indeterminate to the first degree (internally)

Primary structure ( determinate and stable

Structure with unit load in the direction of the redundant

N n

+ x = 0 146.7+13.5 X = 0X= -10.87 kNThe final force in each member ( N final = N+ Xn) = N+(-10.87 n)Reactions = RN + XRn

NnL/EA nnL/EA

n