Beam Deflection-Singularity Functions

Singularity Functions

Beam Deflection

Wawan Hermawan Department of Mechanical and Biosystem Engineering Faculty of Agricultural Engineering and Technology Bogor Agricultural University

Introduction

• By using double integration

method, in the case of two or more differential functions are needed to represent the bending moment

over the entire length of the beam, additional constants and a

corresponding number of additional equations would be required.

A B

P₁ w P₂

1 2 3 4

Needs 4 differential functions

Lengthy computations

These computations will be

simplified through the use of the

singularity functions

Singularity function

⟨

 

� − � ⟩

^�

=¿

( � − � )

^�

 

0

 

when x ≥ a

when x < a

∫

 

^{⟨ � − � ⟩}

^�

^{�� =} _� ¹ ₊₁ ^{⟨ � − � ⟩}

^{�+1 for n ≥ 0}

a

Loading Bending Moment

M_o

O x  � (�)=�_�

⟨

�−�

⟩

⁰

a

O x

P

� ( � ) =− � ⟨ � − � ⟩

¹

 

a

O x

w_o

� ( � ) =− 1

2 �

_�

⟨ � − � ⟩

²

 

a

O x

Slope=k

� ( � ) =− �

6 ⟨ � − � ⟩

³

 

�  (�)=�

⟨

�−�

⟩

¹

a

O x

� ( � ) =− �

(� + 1 )( �+ 2) ⟨ � − � ⟩

^�+2

 

�  (�)=�

⟨

�−�

⟩

^�

L A

P

D

3�  4

� 4  y

x B Consider the beam and loading as

shown in the figure.

3�

 4 �

4  Using the appropriate singularity

function, to represent the contribution to the shear of concentrated load P,

� ( �)=3 �

4 − �

⟨

^{�− 14 �}

⟩

⁰

 

Integrated in x,

� (�)=3 �

4 �− �

⟨

^{�− 1}4 �

⟩

 

�� �²�

�� =3 �

4 � −�

⟨

^{�− 14 �}

⟩

�� �²�  

�� =�(_�)  

Integrated in x,

�� ��

��=�� �=3

8 � �²− 1

2 �

⟨

^{�− 14 �}

⟩

^2+�1

 

���=1

8 � �³− 1

6 �

⟨

^{�− 14 �}

⟩

^{3+�1 �+�2}

 

and can be determined from the boundary conditions

 

� =0, � =0

 

0=0− 1

6 �

⟨

^{0− 14 �}

⟩

^3+0+�2

  = 0 

�

₂

= 0

 

� = � , � =0

 

0=1

8 � �³− 1

6 �

⟨

^{�− 14 �}

⟩

^{3+�1 �}

 

�₁=− 7 � �² 128

  1

2

0=1

8 � �³− 1

6 �

⟨

^{34 �}

⟩

^3+�1�

 

�₁=

− 1 6

128 � �³+ 9

128 � �³

�

 

Example 1

�� ��

��=�� �=3

8 � �²− 1

2 �

⟨

^{�− 14 �}

⟩

^2+�1

 

���=1

8 � �³−1

6 �

⟨

^{�− 14 �}

⟩

^{3+�1 �+�2}

 

�₁=− 7 � �² 128

  �� ��

�� =�� �=3

8 � �²− 1

2 �

⟨

^{� − 14 �}

⟩

^{2− 7128� �2}

 

���=1

8 � �³− 1

6 �

⟨

^{�− 14 �}

⟩

^{3− 7128� �2 �}

�

₂

= 0

   

For x  L

 

⟨

 ^{� − 14 �}

⟩

^=0❑

^���

 

^{= 1} ₈ ^{� �}

³

^{− 7} ₁₂₈ ^{� �}

²

^{� �}

 

⁼ _�� ¹ ( ¹ ₈ ^{� �}

³

^{− 7} ₁₂₈ ^{� �}

²

^� )

For x  L  x  L

 

⟨

 ^{� − 14 �}

⟩

^{0❑ ���=1}8 � �³− 1

6 �

(

^{�− 1}4 �

)

^{3− 7}128^{� �2} �

 

�= 1

��

(

^{18 � �3− 16 �}

⁽

^{�− 14 �}

⁾

^{3−7128� �2 �}

)

 

AD AD segment^segment

DB DB segmentsegment

Example 2

Consider the beam and loading as shown in the figure. Determine the deflection equation and the deflection at the mid point C. EI = 50 kN.m²

w=2 kN/m

A B

5 m

3 m 2 m

C

5

 

�

_�

− ( 4 �� × 1 )= 0 �

  _�

=0.8 �� =800 �

2.5 m

R_A

R_B

� ( � ) = �

_�

� − 1

2 �

_�

⟨ � − � ⟩

²

 

� ( � ) = 800 � − 1

2 ( 2000 ) ⟨ � − 3 ⟩

²

 

� ( � ) = 800 � − 1000 ⟨ � − 3 ⟩

²

 

�� �=400 �²−3 33.3

⟨

�−3

⟩

³+�₁

 

�� �=133.3 �³−8 3.3

⟨

�−3

⟩

⁴+�₁�+�₂

 

Boundary conditions

Integrating twice Reaction

Moment

� =0, � =0

 

0 = 0 − 0 + 0 + �

₂

   

�

₂

= 0

� =5 � , � = 0

 

0  =133.3(5³)−8 3.3

⟨

5 −3

⟩

⁴+�₁(5)+0

�₁=−133.3

(

5³

)

⁺83.3

⟨

5− 3

⟩

⁴

5 =− 3066

 

� = 1

�� ( ^{133.3 �}

³

^{− 83.3 ⟨ � − 3 ⟩}

⁴

^{− 3066 �} )

 

� = 1

�� ( ^{133.3 �}

³

^{− 83.3 ⟨ � − 3 ⟩}

⁴

^{− 3066 �} ) ^{133.3 �}

³

 

EI = 50 kN.m²

� = 1

50000 ( ^{133.3 �}

³

^{− 83.3 ⟨ � − 3 ⟩}

⁴

^{− 3066 �} )

  (m)

Deflection at point CDeflection at point C

x = 2.5 m

� = 1

50000 ( ^{133.3 × 2.5}

³

^{− 83.3} ⟨ 2.5 − 3 ⟩

⁴

− 3066 × 2.5 )

 

- 0.11 m = - 11 cm

 

Determine deflection at x=4 m

Determine deflection at x=4

m

Untuk batang dan pembebanan seperti pada Gambar (a), dengan menggunakan metode fungsi singularitas:

(a)Nyatakan persamaan slope dan defleksi sebagai fungsi dari jarak x dari titik A,

(b)Hitung defleksi pada titik tengah D.

Gunakan E= 200 Gpa, I= 6.87 x 10^-6 m⁴

Example 3

Beban terdistribusi (C-D) diganti dengan beban terdistribusi C-B dan D-B (penyeimbang)

Beban terdistribusi (C-D) diganti dengan beban terdistribusi C-B dan D-B (penyeimbang)

Shear and Bending Moment:

Diintegrasikan dua kali:

Menggunakan kondisi batas untuk mencari C₁ dan C₂

Untuk x=0, y=0, pada persamaan (9.49) maka diperoleh C₂ =0 Untuk x=3,6, y=0, dan C₂=0 maka:

Diperoleh C₁= -2,692 Untuk titik D (x = 1,8 m), maka dapat persamaan :

Bila nilai dalam kurung negative, maka menjadi NOL, sehingga diperoleh:

Dengan menggunakan data E dan I, maka diperoleh defelksi pada titik D:

Problem 1

A

5 kN

C 2  �

x B

Consider the beam and loading as shown in the figure. Determine

the deflection equation and the deflection at point C. EI = 50

kN.m

²

3 �

4  �

Hint:

 Determine reaction at A

 Compose the moment equation  a=2 m 

 Double integration  EI y

 Determine constants (using boundary condition)

 Compose deflection equation  y= ...

 Determine deflection at C (x= 4 m)

 

Problem 2

Determine magnitude and location of the maximum deflection.

E= 50 GPa and I = 210^-6 m⁴

P=5 kN

A B

5 m 2 m

C

Hint:

 Determine reaction at A

 Compose the moment equation  a=2 m 

 Integration  EI 

 Double integration  EI y

 Determine constants (using boundary condition)

 Compose slope equation   = ...

 Determine the location of the maximum deflection   =