Slide Beam deflection-Moment-Area Method

Moment-Area Method

Beam Deflection

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

Introduction

• The moment-area method, developed by Otto Mohr in 1868, is a powerful tool for finding the deflections of structures primarily subjected to bending.

• Mohr’s Theorems also provide a relatively easy way to derive many of the classical methods of

structural analysis.

Otto C. Mohr (1835-1918) A German civil engineer

• The moment-area method is a semi graphical method

utilizing the relations between successive derivatives of the deflection y and the moment diagram.

• For problems involving several changes in loading, the area-moment method is usually much faster than the double- integration method; consequently, it is widely used in

practice.

Moment-Area Theorems

A B

P

C

D

C B

A

�

��

 

x

C D B

A

D

_D

_D

_C

C D B

A _D/C

Except for a difference in the scale of ordinates, the diagram will be the same as the bending moment diagram, if the flexural rigidity EI is constant.

 

deflection diagram  

  � �= �

�� ��

 

Considering two arbitrary points C and D, and integrating both equations from C to D

∫

�_�

�_�

� � = ∫

�_�

� _�

�

�� ��

 

slope

= area under (M/EI) diagram between C and D

= area under (M/EI) diagram between C and D

  

The first moment-area theorem

�

_�

− �

_�

= ∫

�_�

�_�

�

�� ��

 

Area under (M/EI) diagram between C and D

�

  _�/�

P D

B A

C P’

dx x₁

dt d

D P

A C P’

x₁ E

dx

x

Consider two points P and P’ located between C and D, and a distance dx from each other.

�� =�

₁

� �

 

�� = �

₁

�

�� ��

� �= �  

�� ��

 

�

��

 

Since the slope  at P and the angle d

are small quantities,  dt equal to the arc of circle of radius x₁ subtending the angle d

B D

A C

x

�

 ��  

� ´

₁

B

D A

C

C’

t_C/D

Tangential deviation of C with respect to D Tangential deviation

�

_�/�

= ∫

�_�

�_�

�

₁

�

�� ��

 

The tangential deviation t

_C/D

of C with respect to D is equal to the first

moment with respect to a vertical axis through C of the area under the (M/EI) diagram between C and D

The tangential deviation t

_C/D

of C with respect to D is equal to the first

moment with respect to a vertical axis through C of the area under the (M/EI) diagram between C and D

The second moment- area theorem

(area between C and D)

 

is the distance from the centroid of the area to the vertical axis through C

 

Areas and centroid

of common shapes

Example 1

Determine the slope and deflection at end B of the prismatic cantilever beam AB when it is loaded as shown in the figure, knowing EI= 10

 

50 kN

90 kN.m

A B

3 m

50 kN

A B

R_A =50 kN M_A =60

kN.m 1 Free body diagram

2 Bending-moment diagram

+90 kN.m

A B

M

x

- 60 kN.m D x_D

3 m - x_D

Determine the point D where M = 0

�_�

60 =3−�_�

90 = 3 150

  x_D = 1.2 m

90 kN.m

3 M/EI diagram +910^-3 m^-1

A B

x - 610^-3 m^-1

D

1.8 m

�

 ��

1.2

EI =10 MN.m² m M = -60 kN.m = 

M = 90 kN.m = 

0.8 m

2.6 m

0.6 m

4 Slope at B (using first moment-area theorem)

negatif

positif

A₁

A₂

�

_�/�

= �

_�

− �

_�

= area ¿ � �� �= �

₁

+ �

₂

 

¿  −

(

¹2 (1.2m)(6×10⁻³m⁻¹)

)

⁺

(

¹2 (1. 8m)(9×10⁻³m⁻¹)

)

¿ − 3.6 × 10

⁻³

+ 8.1 × 10

⁻³

 

¿ + 4.5 × 10

⁻³

rad

 

_A = 0

�

_�

=+ 4.5 × 10

⁻³

rad

 

�  _�=+4.5×10⁻³rad

B A

5 Deflection at B (using second moment- area theorem)

t_B/A = A₁ () + A₂ ()

 

t_B/A = A₁ () + A₂ ()

  A B

x D

1.8 m

�

 ��

1.2 m

0.8 m

2.6 m negatif

positif

A₁

A₂ 0.6 m

t_B/A = ) () + () ()

 

t_B/A = + 4.86 )

 

t_B/A = =

 

_A = 0, the reference is horizontal

y

_B

= t

_B/A

=

 

�  _�=+4.5×10⁻³rad

B A

y_B = - 4.5 mm

� ´

₂

 

� ´

₁

  Reference tangent

_A = 0

Bending moment by parts

For the prismatic beam AB and the loading shown in the figure, determine the slope at a support (_A)and the maximum

deflection. E= 50 GPa and I = 0.810^-6 m⁴

w=2 kN/m

A B

4 m

1 m 2 m 1 m

C

1 Sketch the deflected beam A B

C

Reference tangent

�

  _�

max= t_A/C  

�

  _�/�

=�

_�

− �

_�

=− �

_�

_�

�

=0

�

  _�

= − �

_�/�  

or

A B

C

4 kN

R_A= 2 kN

R_B= 2 kN 2 Free body diagram

�

_�

= �

_�

=2 kN

 

As the maximum deflection at point C, the bending moment diagram may be drawn for the portion AC.

Since the loading is symetrical, the midpoint C can be used as the reference

Example 2

A B C

4 kN

R_A= 2 kN R_B= 2 kN

3 Bending moment diagram by parts

M

D

A C

D

4 kN.m

-1 kN.m

x

Moment by R_A

Moment by distributed load 2 kN/m 4 diagram by parts and their centroids 

�

��

 

A C

D

0.1 m^-1

-0.025 m^-1

x

E= 50 GPa and I = 0.8 10^-6 m⁴ EI= (50 × 10⁹ N/m²) (0.8 10^-6 m⁴)

EI= 40 × 10³ Nm²

a. Moment by R_A  Triangle

b. Moment by distributed load 2 kN/m

 Parabolic spandrel

A₁ A₂

2 m m  

m   A₁ triangle A₁= (2 m)(0.1 m^-1)=

0.1  

A₂ parabolic

spandrel A₂= (1 m)(-0.025 m^-1)= - 0.0083

 

0.1 m^-1 2 m

A₁

-0.025 m^-1 1 m

Slope at A )

rad = 5.25^o

  A₂

�

��

 

A C

D

x A₁

2 m m  

m   Maximum deflection (at

C)

= 0.1333 m – 0.0145= – 0.1188 m

 

� ´

₁

 

� ´

₂

 

0.1 m^-1

-0.025 m^-1A₂

max

= t

_A/C

= – 0.1188 m = 11.88 cm

 

A B

C

5.25o

 

11.88

Problem: cm

For the example 2, determine moment of the inertia I if the desired deflection does not exceed 5 cm, using the same material (E= 50 GPa)

�_�/�=

∫

�_�

�_�

�₁ �

�� ��

 

constant

I = 0.8 10^-6 m⁴ max  = t_A/C = 11.88 cm

max= t_A/C = 5 cm

  I =  0.8 10  ^-6 m⁴

I = 1.9 10^-6 m⁴

New:

Problem:

Determine E if the

desired deflection does not exceed 5 cm, using the same beam size (I = 0.8 10^-6 m⁴)

For the prismatic beam and loading shown in the figure, determine the deflection at

point A. E= 100 GPa and I = 210^-6 m⁴ A C

4.5 m 1 kN/m

1.5 m

Example 3

M

B

-2 kN.m

-0.01 m^-1

�

��

  _{3 m}

2 m

B

A₁ A₂

A₂ triangle A₁= (4.5 m)(0.01 m^-1)

= 0.0225  

A₁ parabolic

spandrel A₁= (2 m)(-0.01 m^-1)= - 0.0067

 

diagram by parts and their centroids

 

2 m

-0.01 m^-1

4.5 m

-0.01 m^-

1

at B  =- 0.01 m^-1

 

B

B

4.5 m 2 m

A”

C’

C A’

Reference tangent

t_A/B y_A

t_C/B

1.5 m

- 0.01 m^-1 3 m

A₁ ^B A₂

�

��

 

A₂= - 0.0225 A₁= - 0.0067

 

� ´

₂

´

 

�

₁

 

C

From the similar triangles A”A’B and CC’B,

A

�

 

� {� {� � ′= } rsub / } left ({2} over {4.5} right ) =−0.0675× {2} over {4.5} =−0.03

Using the second moment area theorem,

Using the second moment area theorem,  

 

�

_�

= 4 cm

 

w=2 kN/m

A B

5 m

3 m 2 m

C

Problem 1

For the prismatic beam and loading shown in the figure, determine the deflection at point C.

E= 100 GPa and I = 210^-6 m⁴

Problem 2

For the prismatic beam and loading

shown in the figure, determine magnitude and location of the maximum

deflection.

E= 100 GPa and I = 210^-6 m⁴

P=5 kN

A B

5 m 2 m

C

w=2 kN/m

A B

5 m

3 m 2 m

C

Problem 1Problem 1

�_�=

2 ��

� ×2�×1�

5� =0.8��

 

R_A=0.8 kN

A₂ A₁

Maximum moment by R_A at point B

Maximum moment by R_A at point

B

�

  _�−��

=0.8 �� × 5 � = 4 �� . �

E= 100 GPa and I = 210^-6 m⁴

� � =100 ��� × 2 × 10

⁻⁶

�

⁴

= 200 �� . �

²

 

�

�� = 4 ��.�

200�� .�²=0.02�⁻¹  

�

 ��

0.02

 

�

⁻¹

−

 

0.02 �

⁻¹

�₁=5×0.02

2 =0.05

 

�₂=2× −0.02

3 =−0.05

 

C

A B

C t_C/B

Reference tangent

A₂ A₁

�

 ��

0.02

 

�

⁻¹

−

 

0.02 �

⁻¹

C

D E y_C

Problem 2

For the prismatic beam and loading

shown in the figure, determine magnitude and location of the maximum

deflection.

E= 100 GPa and I = 210^-6 m⁴

P=5 kN

A B

5 m 2 m

C

�_�=

5 ��

� ×3�

5� =3��

  Reaction at AReaction at A

Maximum moment at CMaximum moment at C

E= 100 GPa and I = 210^-6 m⁴

�

_�−��

= 3 �� × 2 � =6 �� . �

 

� � =100 ��� × 2 × 10

⁻⁶

�

⁴

= 200 �� . �

²

 

�

�� = 6�� .�

200�� .�²=0.03�⁻¹  

0.03 m^-1

C B

A

�

��

 

A B

C