Analysis of statically indeterminate structures2- Statically indeterminate structures can...

THEORY OF STRUCTURES -------------------- BY THAAR AL-GASHAM

1 WASSIT UNIVERSITY – ENGINEERING COLLEGE- CIVIL ENGINEERING DEPARTMENT

Analysis of statically indeterminate structures

Structure of any type is classified as statically indeterminate when the number of

unknown reactions or internal forces exceeds the number of equations available for its

analysis.

Advantages of statically indeterminate structures

1- For a given loading the maximum stress and deflection of an indeterminate

structure are generally smaller than those of its statically determinate

counterpart.

2- Statically indeterminate structures can redistribute its load to its redundant

supports in cases where faulty design or overloading occurs.

Although from these advantages of selecting statically indeterminate structures, also;

there are some disadvantages such as deformations caused by relative support

displacement, or changes in member length caused by temperature or fabrication errors

will introduce additional stresses in the structure, which must be considered when

designing indeterminate structures.

Methods of analysis

When analyzing any indeterminate structures, it is necessary to satisfy equilibrium,

compatibility, and force-displacement requirements for the structure. Equilibrium is

satisfied when the reactive forces hold the structure at rest, and compatibility is satisfied

when the various segments of the structures fit together without intentional breaks or

overlaps. the force-displacements requirements depends upon the way of material

responds. Generally, there are two different methods to satisfy these requirements: the

force or flexibility method, and the displacement or stiffness method.

In this stage, the following methods will be studied:

1- Consistent method

2- Slope-deflection method

3- Moment distribution method

4- Stiffness matrix method



1-consistent method

1- Determine the degree of indeterminacy of structures.

The above structure is statically indeterminate to third degree.

2- Remove redundant to convert structure from statically indeterminate to stable

statically determinate structure , this structure is named as primary structure.

Primary structure

3- Determine internal moment at each part of primary structure (M)

4- Apply unit load in position of removed redundant, then internal moment at each

part of structure is calculated as shown in figures.

m1

1

m2

1

m3

1

X1 X2

X3



5- Determine redundant by solving the following simultaneous equations

∆10 + 𝑠11𝑥1 + 𝑠12𝑥2 + 𝑠13𝑥3 = 𝑜 𝑜𝑟 𝑠𝑒𝑡𝑡𝑙𝑚𝑒𝑛𝑡 𝑎𝑡 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 1-----1

∆20 + 𝑠21𝑥1 + 𝑠22𝑥2 + 𝑠23𝑥3 = 𝑜 𝑜𝑟 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑡 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 2---2

∆30 + 𝑠31𝑥1 + 𝑠32𝑥2 + 𝑠33𝑥3 = 𝑜 𝑜𝑟 𝑠𝑒𝑡𝑡𝑙𝑚𝑒𝑛𝑡 𝑎𝑡 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 3---3

Where

∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼

∆20= ∫𝑀𝑚2 𝑑𝑥

𝐸𝐼

∆30= ∫𝑀𝑚3 𝑑𝑥

𝐸𝐼

𝑠11 = ∫𝑚1. 𝑚1 𝑑𝑥

𝐸𝐼

𝑠12 = ∫𝑚1. 𝑚2 𝑑𝑥

𝐸𝐼

𝑠13 = ∫𝑚1. 𝑚3 𝑑𝑥

𝐸𝐼

𝑠22 = ∫𝑚2. 𝑚2 𝑑𝑥

𝐸𝐼

𝑠33 = ∫𝑚3. 𝑚3𝑑𝑥

𝐸𝐼

S12=S21 , S31=S13, S32=S23

Applications of consistent method

1-beams

Example(1):- determine the reaction at the supports

A B L

W



Solution

The beam is statically indeterminate to first degree, the roller is considered as redundant

and is removed

𝑙𝑜𝑎𝑑 𝑎𝑡 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑥 𝑓𝑟𝑜𝑚 𝑏 =𝑊𝑋

𝐿

𝑀 = −𝑋𝑊𝑋

𝐿.1

2.𝑋

3= −

𝑊𝑋3

6𝐿

Apply unit load at position of removed support

𝑚1 = −𝑋

∆10 + 𝑠11𝑥1 = 𝑜

∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼= ∫

−𝑤𝑥3

6𝑙

𝑙

𝑜

∗−𝑥𝑑𝑥

𝐸𝐼=

𝑤𝑙4

30𝐸𝐼

𝑠11 = ∫𝑚1. 𝑚1 𝑑𝑥

𝐸𝐼= ∫ −𝑥

𝑙

0

−𝑥𝑑𝑥

𝐸𝐼=

𝑙3

3𝐸𝐼

𝑤𝑙4

30𝐸𝐼+

𝑙3

3𝐸𝐼𝑥1 = 𝑜

A L

W

X

M

A L

X

m1 1



𝑥1 = −𝑤𝑙

10=

𝑤𝑙

10↑

After the reaction at roller is known , remaining reactions can be determined by applying

equilibrium equations.

𝑅𝑦𝐴 =𝑤𝑙

2−

𝑤𝑙

10=

2

5𝑤𝑙 ↑

𝑀𝐴 =𝑤𝑙

10𝑙 −

𝑤𝑙

2∗

1

3= −

𝑤𝑙2

15=

𝑤𝑙2

15 𝑐𝑜𝑢𝑛𝑡𝑒𝑟𝑐𝑙𝑜𝑐𝑘 𝑤𝑖𝑠𝑒

H.W: repeat previous example and choose end moment at A as redundant

Example (2): analysis the following beam

The beam is statically indeterminate to first degree due to symmetry

𝑀 =𝑊𝐿

2𝑋 −

𝑊𝑋2

2=

𝑊

2[𝐿𝑋 − 𝑋2]

A L B

w

A L B

w M

X

X

A L B

m 1 1



m=1

∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼= ∫

𝑤

2𝐸𝐼[𝑙𝑥 − 𝑥2]

𝑙

0

𝑑𝑥 =𝑤𝑙3

12𝐸𝐼

𝑠11 = ∫𝑚1. 𝑚1 𝑑𝑥

𝐸𝐼= ∫

𝑑𝑥

𝐸𝐼

𝑙

0

=𝑙

𝐸𝐼

∆10 + 𝑠11𝑥1 = 𝑜

𝑤𝑙3

12𝐸𝐼+

𝑙

𝐸𝐼𝑥1 = 𝑜

𝑥1 = −𝑤𝑙2

12

Example (3):- determine the internal moments acting in the beam at support B and c .

the wall at A moves upward 30mm. take E=200GPa,I=90(106)mm4.

The structure is statically indeterminate to first degree.

MAB=0

MBC=-20X

MCD=-40X

10 m 10 m 5 m

40 kN A B C D

M M M

60

10 m 10 m 5 m

A B C D 40 kN

X X X

20



mAB=-X

mBC=-(10+X)+2X=-10+X

mCD=0

∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼[∫ −20𝑥(−10 + 𝑥)𝑑𝑥

10

0

] =10000

3𝐸𝐼𝑘𝑁2. 𝑚3

𝑠11 = ∫𝑚12 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼[∫ 𝑥2𝑑𝑥 +

10

0

∫ (−10 + 𝑥)2𝑑𝑥10

0

] =2000

3𝐸𝐼𝑘𝑁2. 𝑚3

∆10 + 𝑠11𝑥1 = 𝑠𝑒𝑡𝑡𝑙𝑒𝑚𝑒𝑛𝑡

10000

3 ∗ 200 ∗ 90+

2000

3 ∗ 200 ∗ 90𝑥1 = −30 ∗ 10

−3

𝑥1 = −5.81 𝑘𝑁

RA=5.81 KN upward

RB=-20+2*-5.81=-31.62=31.62 downward

RC=60-1*5.81=65.81 upward

MB=5.81*10=58.1 kN.m

MC=40*5=200 kN.m

10 m 10 m 5 m

A B C D

1

X X X

2 1



2-frame

Example(4):- Analysis the frame shown. There is rotational slip of 0.003 rad counter

clockwise at support B. take EI=104 kN.m2

The structure above is statically indeterminate to 2nd degree, there is part of structure is

statically determinate

4kN/m

3m

4m

4m 2m 1m 1m

2kN

4kN/m

3m

4m

4m

1kN

2kN.m



Primary structure

∆10 + 𝑠11𝑥1 + 𝑠12𝑥2 = 𝑜

∆20 + 𝑠21𝑥1 + 𝑠22𝑥2 = 0.003

∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼[∫ (7.6𝑋 − 1.6𝑋2)(0.8𝑋)𝑑𝑥

5

0

] =53.33

𝐸𝐼

4kN/m 3m

4m

4m

1kN

2kN.m

11.5

9.5

0

X

M

M=0

X

M

M=9.5(0.8x)-

4x(0.4x)=7.6x-

1.6x2

3m

4m

4m 1.75

1.75

1

X

M

M=X

X

M

M=1.75(0.8X)-

1(0.6X)=0.8X

1

3m

4m

4m 0.25

0.25

1 X

M

M=-1

X

M

M=-0.25(0.8X)=-

0.2X

0



∆20= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼[∫ (7.6𝑋 − 1.6𝑋2)(−0.2𝑋)𝑑𝑥

5

0

] =−13.33

𝐸𝐼

𝑠11 = ∫𝑚12 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼[∫ (0.8𝑋)2𝑑𝑥 +

5

0

∫ (𝑥)2𝑑𝑥4

0

] =48

𝐸𝐼

𝑠12 = ∫𝑚1. 𝑚2 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼(∫ (0.8𝑋)(−0.2𝑋)𝑑𝑥 + ∫ −𝑋𝑑𝑥) =

4

0

5

0

−14.67

𝐸𝐼

𝑠22 = ∫𝑚22 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼[∫ (−0.2𝑋)2𝑑𝑥 +

5

0

∫ (−1)2𝑑𝑥4

0

] =5.67

𝐸𝐼

53.33

𝐸𝐼+

48

𝐸𝐼𝑋1 −

14.67

𝐸𝐼𝑋2 = 0

48𝑋1 − 14.67𝑋2 = −53.33 − − − − − 1

−13.33

𝐸𝐼−

14.67

𝐸𝐼𝑋1 +

5.67

𝐸𝐼𝑋2 = 0.003

−14.67𝑋1 + 5.67𝑋2 = 43.33 − − − − − 2

Solving Eqs 1&2,gives

X1=5.85kN , X2=22.78 kN.m

Final result

4kN/m

3m

4m

4m 2m 1m 1m

2kN

1

6.96

14.04

5.85

5.85

22.78



Example(5):- using method of consistent deformation, analysis the frame shown

EI=104 kN.m2, settlement at support A=0.002m ↓and 0.003m← , rotational slip at

A=0.003 rad counter clock wise , settlement at B=0.004m ↓

The structure is statically indeterminate to 1st degree, the deformations of support

A must be included by using virtual work

3m

4m

4m

A

B

3m

4m

4m

A

B

1

1

4

X

m=0.8X

X m=4

∆equ



1𝑥∆𝑒𝑞= 4𝑥0.003 − 1𝑥0.002 + 0𝑥0.003 = 0.01

∆𝑡𝑜𝑡𝑎𝑙= 0.01 + 0.004 = 0.014𝑚

∆10 + 𝑠11𝑥1 = 0.014

∆10= 𝑜 𝑛𝑜 𝑙𝑜𝑎𝑑 𝑎𝑝𝑝𝑙𝑖𝑒𝑑

𝑠11 = ∫𝑚12 𝑑𝑥

𝐸𝐼=

1

𝐸𝐼(∫ 16 𝑑𝑥 + ∫ (0.8𝑥)2𝑑𝑥) =

90.67

𝐸𝐼

5

0

4

0

90.67

𝐸𝐼𝑥1 = 0.014

𝑥1 = 1.54 ↓ 𝑘𝑁

3-Arch

Example(6):- Analysis the semi-circular arch shown in figure by the method of consistent

deformation. Radius=R, EI constant

3m

4m

4m

A

B

1.54

1.54

6.16

P



The arch is statically indeterminate to 1st degree

∆10 + 𝑠11𝑥1 = 0

∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼= 2 ∫

𝑃

2𝑅(1 − 𝑐𝑜𝑠𝜃)(−𝑅𝑠𝑖𝑛𝜃)𝑅𝑑𝜃

𝐸𝐼

0.5𝜋

0

=−𝑃𝑅3

𝐸𝐼[−𝑐𝑜𝑠𝜃 −

𝑠𝑖𝑛2𝜃

2]0

0.5𝜋

=−𝑃𝑅3

2𝐸𝐼

𝑠11 = ∫𝑚12 𝑑𝑥

𝐸𝐼

= 2 ∫(−𝑅𝑠𝑖𝑛𝜃)2𝑅𝑑𝜃

𝐸𝐼=

2𝑅3

𝐸𝐼

0.5𝜋

0

[∫(1 − 𝑐𝑜𝑠2𝜃)

2𝑑𝜃]

0.5𝜋

0

=2𝑅3

𝐸𝐼[𝜃

2− 0.25𝑠𝑖𝑛2𝜃]0.5𝜋 =

𝜋𝑅3

2𝐸𝐼

−𝑃𝑅3

2𝐸𝐼+

𝜋𝑅3

2𝐸𝐼𝑥1 = 0

𝑥1 =𝑃

𝜋

0.5P

P

0.5P

M

θ

𝑀 =𝑃

2𝑅(1 − 𝑐𝑜𝑠𝜃)

m

θ

𝑚 = −𝑅𝑠𝑖𝑛𝜃

1 1

P

0.5P 0.5P

P/π P/π



4-Trusses

Same procedure for beam and frame can be followed for truss, the following equations

are used in the analysis of trusses. The truss may externally indeterminate or internally

indeterminate or both cases

∆10 + 𝑠11𝑥1 + 𝑠12𝑥2 + 𝑠13𝑥3 = 𝑜

∆20 + 𝑠21𝑥1 + 𝑠22𝑥2 + 𝑠23𝑥3 = 𝑜

∆30 + 𝑠31𝑥1 + 𝑠32𝑥2 + 𝑠33𝑥3 = 𝑜

Where

∆10= ∑𝑛1𝑁𝐿

𝐸𝐴+ ∑ 𝑛1 ∝ ∆𝑇𝐿 + ∑ 𝑛1 𝑒𝑟𝑟𝑜𝑟

∆20= ∑𝑛2𝑁𝐿


∆30= ∑𝑛3𝑁𝐿


𝑠11 = ∑𝑛1

2𝐿

𝐸𝐴

𝑠12 = ∑𝑛1𝑛2𝐿

𝐸𝐴

𝑠13 = ∑𝑛1𝑛3𝐿

𝐸𝐴

𝑠22 = ∑𝑛2

2𝐿

𝐸𝐴

𝑠33 = ∑𝑛3

2𝐿

𝐸𝐴

𝑠33 = ∑𝑛2𝑛3𝐿

𝐸𝐴

S12=S21 , S31=S13, S32=S23



Example(7):- Analysis the truss shown. EA constant

The truss is statically indeterminate to 2nd degree; it is internally and externally

indeterminate

b+r 2j 6+4 2(4) 10 8

4m

3m

3m

A

B

D C

E

10kN

4m 10

3m

A

B

D C 13.33

10

13.33

10

13.33

10

0

-16.67

N

4m

3m

A

B

D C 1.6

0.6

0.8

0

-1.6

-0.6

0

1

1 0.6

0.8

n1

n2

4m

3m

A

B

D C 0

0

0

-0.6

-0.8

-0.6

-0.8

1 1

1

0.6

0.8



∆10= ∑𝑛1𝑁𝐿

𝐸𝐴=

−186.66

𝐸𝐴

∆20= ∑𝑛2𝑁𝐿

𝐸𝐴=

−162.01

𝐸𝐴

𝑠11 = ∑𝑛1

2𝐿

𝐸𝐴=

21.32

𝐸𝐴

𝑠12 = ∑𝑛1𝑛2𝐿

𝐸𝐴=

11.2

𝐸𝐴

𝑠22 = ∑𝑛2

2𝐿

𝐸𝐴=

17.28

𝐸𝐴

21.32𝑥1 + 11.2𝑥2 = 186.66 − − − −1

11.2𝑥1 + 17.28𝑥2 = 162.01 − − − −2

𝑥1 = 5.8 , 𝑥2 = 5.6

Force in each member=N+n1X1+n2X2

member L(m) N n1 n2 N n1 L N n2 L n12 L n22 L n1 n2 L DC 4 13.33 -1.6 -0.8 -85.312 -42.66 10.24 2.56 5.12

AB 4 0 0 -0.8 0 0 0 2.56 0 AD 3 10 -0.6 -0.6 -18 -18 1.08 1.08 1.08

BC 3 10 0 -0.6 0 -18 0 1.08 0

AC 5 -16.67 1 1 -83.35 -83.35 5 5 5 EC 5 0 1 0 0 0 5 0 0

DB 5 0 0 1 0 0 0 5 0 ∑ -

186.66 -162.01 21.32 17.28 11.2



member F

DC -0.43 AB -4.48

AD 3.16

BC 6.64 AC -5.27

EC 5.8 DB 5.6

Example(8):- repeat example 7, if member BC subjected to rise of temperature 40o take

α=12(10-6) and member DC has error of 1cm too short.

∆10= ∑𝑛1𝑁𝐿


∆10=−186.66

𝐸𝐴+ 0𝑥40𝑥3𝑥12𝐸 − 6 − 1.6𝑥(−0.01) =

−186.66

𝐸𝐴+ 0.016

∆20= ∑𝑛2𝑁𝐿


∆20=−162.01

𝐸𝐴− 0.6𝑥40𝑥3𝑥12𝐸 − 6 − 0.8𝑥(−0.01) =

−162.01

𝐸𝐴+ 0.00713

Take EA=105

21.32𝑥1 + 11.2𝑥2 = 186.66 − 0.016𝑥10𝐸5 − − − −1

11.2𝑥1 + 17.28𝑥2 = 162.01 − 0.00713𝑥10𝐸5 − − − −2

X1=-75.11 X2=16.801

Then find the force in each bar by same procedure of Example 7



5-composite structure

Example(8):- Analysis the composite structure shown in figure. The beam has moment

of inertia of 240x106mm4,the member AB&BC each have a cross-sectional area 0f 1250

mm2, and BD has across-sectional area 0f 2500 mm2. Take E=200 GPa.

The structure is statically indeterminate to 1st degree

A

B

D C

0.9m

0.9m

1.2m 150kN

A

B

D C

0.9m

0.9m

1.2m 150kN

200kN 350kN

M=-150x

x x

M=-200x 0 0

A

B

D C

0.9m 1.2m

0.00033

m=0.53x

x x

m=0.7067x

1

0.707

0.707

0.707

0.53

0.884

1.24

0.53

0.707

0.707 0.707

1.24

0.00033



∆10= ∫𝑀𝑚1 𝑑𝑥

𝐸𝐼+ ∑

𝑛1𝑁𝐿

𝐸𝐴

=1

200𝑥240[∫ (−150𝑋)(0.53𝑋)𝑑𝑋 + (−200𝑋)(0.7067𝑋)𝑑𝑋]

1.2

0

= −1.67𝑥10−3

𝑠11 = ∫𝑚12 𝑑𝑥

𝐸𝐼

+ ∑𝑛1

2𝐿

𝐸𝐴

=1

200𝑥240[∫ (0.53𝑋)2𝑑𝑋 + (0.7067𝑋)2𝑑𝑋] +

0.8842𝑥1.5

200𝑥1250+

1.242𝑥0.9

200𝑥2500

1.2

0

+12𝑥1.27

200𝑥1250= 1.84𝑥10−5

∆10 + 𝑠11𝑥1 = 0

𝑥1 = 90.76 𝑇

FAB=90.76T

FBD=1.24x90.76=112.5 C

FBC=0.884x90.76=80.2 T

A

B

D C

0.9m

-112.5

1.2m 150kN

90.76 80.2

200.03 350.03

Analysis of statically indeterminate structures2- Statically indeterminate structures can...

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