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(1)

건설안전역학

(Constructional Safety Mechanics)

토목안전환경공학과 안전트랙

옥승용

LN11: Deflection Computation

by using Energy Method (3)

(2)

Class Schedule

Week Topics Remarks

01 Introduction to class (1) & (2)

02 Analysis of Truss Structures (1) Homework #01

03 Analysis of Truss Structures (2)

04 Analysis of Horizontal Beams (1) Quiz #01

05 Analysis of Horizontal Beams (2)

06 Analysis of Frame Structures (1) Quiz #02

07 Analysis of Frame Structures (2)

08 Mid-Term Exam

09 Deflection Computation by using Energy Method (1) 10 Deflection Computation by using Energy Method (2)

11 Deflection Computation by using Energy Method (3) Quiz #03 12 Analysis of Indeterminate Structures (1)

13 Analysis of Indeterminate Structures (2) Quiz #04

14 Analysis of Indeterminate Structures (3)

(3)

Review on Principle of Virtual Work

U_e = 1 ×  = U_i = ∫ u dL

Real deformation to compute when the structure is subjected to real loads P_i

Internal forces caused by virtual unit load that is applied at the same point as  & in the same direction as 

Real internal deformation caused by real loads

U = 1 × q = U = ∫ u L

Unit load can be unit moment

(4)

• The principle of virtual work can be

applied to deflection problems for beam &

frame.

– In fact, the subsequent equation applies to the frame structure as well.

• Strains due to bending are the primary cause of beam & frame deflections.

– Here, we will see the deflection by the bending moment.

• Shear, axial and torsional loadings, and temperature may also provoke deflections in beam and frame, which will be

Principle of Virtual Work: Beams/Frames

dq

(5)

• To compute 

• A virtual unit load acting in the direction of  is placed on the beam at A.

• The internal virtual moment m is

determined by the method of sections at an arbitrary location x from the left

support.

• When real loads act on the beam, point A is displaced , the element dx deforms or rotates

dq = (M/EI)dx

M = the internal moment caused by real loads

Principle of Virtual Work: Beams/Frames

dq

(6)

• To compute ,

Principle of Virtual Work: Beams/Frames

dq e 1

U   

i

dU m d m M dx q EI

   

External Work

Internal Work for dx

1 0^L

e

U mM dx

   



EI

1 = external virtual unit load in the direction of 

 = external displacement caused by real loads

m = internal virtual moment caused by external virtual load M = internal moment in beam or frame caused by real loads

(7)

• To compute q,

Principle of Virtual Work: Beams/Frames

q

1 e 1

U  q

i

dU m d m M dx

q q q EI

   

External Work

Internal Work for dx

1 0^L

e

U m M dx

EI q q

  



1 = external virtual unit moment in the direction of q q = external rotation caused by real loads

m_q = internal virtual moment caused by external virtual load M = internal moment in beam or frame caused by real loads

(8)

• Center deflection of a simple beam subject to an uniform load

Example 4: Beam

q l

EI dx mM



L





 0

1

EI dx M m x l



l







0

0 )

( 2 1

1 q

(9)

Example 4: Beam

Moment of virtual load Moment of real load

Deflection at the center of the span:

0 0

( ) 1 2

l l

l mM

dx dx

EI EI

 







q 1

ql²/8 l/4

q 1

ql²/8 l/4

위 식의 의미: 보의 각 위치(x)별 두 모멘트의 곱을 전체 보 의 길이에 대하여 더한다(적분한다).

(10)

• Tabular method for integration of moment product

Principle of Virtual Work: Beams/Frames

Do not forget to divide by EI

(11)

Example 4: Beam

Moment of virtual load Moment of real load

Deflection at the center of the span:

1 3

2 0

2 4

( ) 1 (1 )

2 3

1 5

(1 )

3 4 4 8 384

l l mM l ab

dx M M

EI EI l

l l ql ql

EI EI

    

  



q 1

ql²/8 l/4

q 1

ql²/8 l/4

(12)

• Determine the displacement of point B of the steel beam (E = 200GPa; I = 50010⁶mm⁴)

Example 5: Cantilever Beam

EI dx mM



L





 0

1

10 2

0 0

( 1 ) ( 6 )

1 kN _B ^L mM x x

dx dx

EI EI

 

  







3 3 3 3 3

6 4

3 3 3

9 2 6 12 4

15 10 15 10 10

200 500 10

15 10 10

200 10 / 500 10 10

15 0.15 150

20 5

B

kNm Nm

EI GPa mm

Nm

N m m

m m mm



  

  

 

    

  



(13)

• Determine the tangential rotation at point A of the steel beam (E=200GPa;

I=6010⁶mm⁴)

Example 6: Cantilever Beam

EI dx M m



L



 0

1 q ^q

x dx dx EI

EI M m

L

A ^



^



^ ^_^{ } ^_^

 ³

0

3

0 3

) 1 kNm (

1 q ^q

rad 00563 .

0 qA

(14)

Principle of Virtual Work

1 ×  = u × L 1 × q = u_q × L

Virtual loadings

Real deformations

T R U S S

B E A M

&

F R A M

1 NL

n AE

 

     

 

1   n  T L

 

1   n L

0 L b

U m M dx

EI

 





 

0 L n

U n N dx

AE

 





 

0 L s

U v K V dx GA

 





 

0 L t

U t T dx GJ

 

  

 



 

 

(15)

Example 7

Bending:

Axial load:

Shear:

EI dx U_b ^



0^L mM

0 L n

U nN dx





AE

νV dx K

U ^



^L ^^ ^^

• Apply a unit horizontal load at C

• Free body diagrams for the real & virtual loadings

• Determine the horizontal displacement of point C on the frame

E = 200GPa G = 80GPa

I =23510⁶mm⁴ A = 5010³mm²

• Include the internal strain energy due to axial & shear forces

(16)

Example 7

E = 200GPa G = 80GPa I =23510⁶mm⁴ A = 5010³mm²

A

B C

60kN/m

A

B C

1kN

60kN/m

112.5kN 1.25kN 1kN

Real load Virtual unit load

• Include the internal strain energy due to axial load & shear

Moment?

Axial Force Shear Force?

(17)

Example 7

E = 200GPa G = 80GPa I =23510⁶mm⁴ A = 5010³mm²

270kNm M(x)=112.5 x2

x1

x2

Real load

60kN/m

180kN 112.5kN

112.5kN

1kN 1.25kN

1.25kN 1kN

EI dx U_b ^



0^L mM

m(x)=1.25 x2

x2

x₁

 

¹^.²⁵ ¹¹²^.⁵ 

30 180

1

4 .

2 2 2

1 3

0

2 1 1

1



 



x dx x

EI dx

x x

U_b x Virtual unit load

Bending:

(18)

Example 7

E = 200GPa G = 80GPa I =23510⁶mm⁴ A = 5010³mm²

Real load

60kN/m

180kN 112.5kN

112.5kN

1kN 1.25kN

1.25kN 1kN

Virtual unit load



 AE U_n nNL

N=112.5kN

n=1.25kN

n=1kN

^1.25  ^112.5   ³

0 0.0421875

kN kN

kNmm Un

 AE 



Axial Force:

(19)

Example 7

E = 200GPa G = 80GPa I =23510⁶mm⁴ A = 5010³mm²

Real load

60kN/m

180kN 112.5kN

112.5kN

1kN 1.25kN

1.25kN 1kN

Virtual unit load

GA dx K νV

U_s ^



0^L ^_^ ^_^

V=112.5kN

v=1kN

v=1.25kN

) 5 . 112 )(

25 . 1 ( 2 . 1

) 60 180 )(

1 ( 2 . 1

4 .

2 2

3 0

1 1



 

 



dx GA

dx U_s x

Shear Force:

(20)

Example 7

E = 200GPa G = 80GPa I =23510⁶mm⁴ A = 5010³mm²

1

35.3 kN

kNmm 0.0421875kNmm 0.182kNmm 35.53mm

h

C b n s

C

U U U

    

  

 

negligible

(21)

where  is the displacement in the direction of the applied unit concentrated load or moment in the virtual system.

Frame Problems

 

^



^







e

l l l

s

e e e

EA dx F dx F

GA V f V

EI dx M M

0 0 0

) (

l l

H_A H_B=P/4

R_A=P/2 R_B=P/2 P

(22)

Frame Problems

- Pl/4

- + + -

P/4 P/2

- -

- P/4 P/2

Moment Shear Axial

EI Pl l

Pl l

l Pl l

dx EI EI dx

l l

M ) 16

4 4 6 4 4 (3

) 2

2 ( ^/² ³

0 0















  

GA Pl f P

l l P

GA dx f

GA dx

f_s ^l ^l _s _s

S 8

) 3 2 1 2 2 4 1 ( 4

) 2

2 ( ^/²

0 0















  



Pl/4 l/4



Pl/4 l/4



P/4 1/4



P/2 1/2

 

2 / 2 2 1 1 9

( )

2 2 2 4 4 16

l l

A

P l P Pl

dx dx l

EA EA EA

 







      

(23)

Frame Problems

- Pl/4

- + + -

P/4 P/2

- -

- P/4 P/2

Moment Shear Axial

EI Pl

M 16

 3

 GA

Pl f_s

S 8

 3

 EA

Pl

A 16

 9



) ) ( 75 . 0 )

( 56 . 1 1 16 (

) 9

6 1 16 (

2 2

3

2 2

3

l h l

h EI

Pl

EAl EI GAl

EI f EI

Pl _s

A S

M























 ^^ 

 21 G E

3 .

0



2 .

1 fs

(24)

• In most cases, the deformation caused by the shear force and the axial force is negligibly small compared to that caused by the bending moment.

• If this is the case, the displacement of a frame can be approximated by considering only the bending moment.

Frame Problems

 

^



^







e

l l l

s

e e e

EA dx F dx F

GA V f V

EI dx M M

0 0 0

) (

 





e l_e

EI dx M M

0