3 Moment Area

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2101-310 Structural Analysis I Method of Moment Area Equations Note provided by JRR Page-1

Method of Moment Area Equations

Introduction

Perform deformation analysis of flexure-dominated structures

Beams Frames

Provide equations to determine

Displacement Rotation

Basic Assumptions

Small displacement (u,v) and small rotation (θ)

|u/L|, |v/L|, |θ| << 1; L ~ characteristic dimension of the structure

Rotation is approximated by θ = dv/dx

Curvature is approximated by κ = d2v/dx2

Kinematics of the cross section

Plane section remains plane

No shear deformation Plane section always normal to Neutral Axis

NA NA

Deformed stateUndeformed state

Y

X Deformed state

Undeformed state

Deformed state

X

Y

Undeformed state


Basic Assumptions (Cont.)

No axial deformation (axially rigid members) no change in length of all members

L = Lo

uA = uB

Material behavior

Linearly elastic, i.e. linear stress-strain relation

Isotropic, i.e. material properties are directional independent

Equilibrium of the structure

Equilibrium equations are set up on undeformed state

ε

σ

E = Young Modulus

1

A

B

L

Lo

uA

vA

uB

vB Deformed state

Undeformed state




Moment Area Equations

Basic Equations

Kinematics

Plane sections remain plane

Material behavior

Equilibrium of the cross section

Static equilibrium

Moment-Curvature Relation

BMD in terms of applied loads

EI

M

dx

d

dx

v d 2

2

==θ

BMD

L

x

y

A

B

u A

u B

v A

v B

θ A

θ B

u

v

2

2

dx

v d

dx

d ==κ

θ

y κ−=ε

ε=σ E

κ= EI M

qdx

dV = ; V

dx

dM =

I is the moment of inertia of the cross section


First Moment Area Equation

EI

M

dx

d =

θ

L

A

B

dx

d θ

x

y

M/EI diagram

dx EI

M

dx

Change of angle over the portion dx

Moment-Curvature Relationship

dx EI

M d =θ

Graphical Interpretation




First Moment Area Equation (Cont.)

Change of angle over the portion dx dx EI

M d =θ

Total change of angle over the portion AB dx EI

M d

L

0

B

A

∫ ∫ =θθ

θ

AB

M/EI B/A AB Area =θ=θ−θ

Sign Convention and Remarks

θ A, θB are positive when they aremeasured CCW from undeformed

state

θB/A is positive when it is measuredCCW from tangent line from A

Sign convention of bending momentM follows the local coordinate x-y

y

x

M+

y

x

M-


A

B

M/EI diagram

L

θB/A

θB

θ A x

y

Assume no discontinuity of slopewithin portion AB (e.g. no hinge )

Area under M/EI diagram over portion AB

Total change of angle over the portion AB

1st

MOMENT AREA EQUATION


Second Moment Area Equation

EI

M

dx

d =

θ

L

A

B

dx

d θ

x

y

M/EI diagram

dx EI

M

dx dt

x

x

L-x

L-x

Change of angle over the portion dx

Moment-Curvature Relationship

dx EI

M d =θ


dt = deviation of element dx measured

on a normal line at point B

( ) ( ) ⎟

⎠

⎞⎜

⎝

⎛ −=θ−= dx

EI

M x Ld x Ldt

Assume small rotation






Length Constraint Equation

No Axial Deformation

Small Displacement and Rotation

No Change in Length of Member

Longitudinal Displacement are Constant

L L =′

B A u u u ==


L A

B

x

y

u A

u B

u

C

C´ A´

B´ L´

Sign Convention and Remarks

L´ is the length of the member AB measured in the longitudinal directionof the undeformed state of the member

u is the longitudinal component of thedisplacement at any point with themember AB

u A, u A, and u are positive when theydirect in positive x -direction

The real length of the deformed statecan be approximated by the projectedlength L´ provided that displacementand rotation of the member are small


Application of Moment Area and Length Constraint Equations

Summary of Equations

3 independent equations per member

Remark: The superscript or subscript “AB” is used to emphasize that quantities are associated with the member AB

Kinematical Unknowns

3 unknowns at point A : AB

Au , AB Av , and AB

Aθ

3 unknowns at point B : ABBu , AB

Bv , and ABBθ

Total 6 unknowns per member

The rotations { AB

Aθ , AB

Bθ } and the transverse components of the displacement { AB

Av , AB

Bv }

are related by the 1st and 2nd moment area equations

The longitudinal components of the displacement { AB

Au , AB

Bu } are related by length

constraint equation

AB

M/EI

AB

A

AB

BB/A Area =θ−θ=θ

( ) ( ) B

AB

M/EI AB

AB

A

AB

A

AB

BB/A x Area Lθ v v t ⋅=+−=

( ) ( ) A

AB

M/EI AB

AB

B

AB

B

AB

A A/B x Area Lθ v v t ⋅=−−=

AB

B

AB

A u u =

L AB

A

B

x

y

B x

t B/A

A x

t A/B

M/EI diagram

Centroid 1st Moment Area Equation

2 nd Moment Area Equation


AB

Aθ

AB

Bθ

AB

Au

AB

Av

AB

Bu

AB

Bv




Useful Remarks

Remark1: If one of { AB Au , AB

Bu } is known, the other can be obtained from length

constraint equation

Remark2: If one of { AB

Aθ , AB

Bθ } and one of { AB

Av , AB

Bv } are known, the other two of

{ AB

Aθ , AB

Bθ , AB

Av , AB

Bv } can be computed from the 1st moment area equation and then

follow by the 2 nd moment area equations

Remark3: If { AB

Av , AB

Bv } are known, one of rotations { AB

Aθ , AB

Bθ } is computed from 2 nd

moment area equation and the other rotation is obtained from 1st moment area equations

Remark4: If all three unknowns are known at one end, other three unknowns at theother end of the member can be computed from 1st and 2 nd moment area equations and the length constraint equation

One of { AB Au , AB

Bu } is known Length Constraint Equation

One of { AB

Aθ , AB

Bθ } is known 1st & 2 nd Moment Area

Equations One of { AB

Av , AB

Bv } is known

2 nd Moment Area Equations { AB Av , AB

Bv } are known

1st Moment Area Equations

{ AB

Au , AB

Av , AB

Aθ } are known 1st & 2 nd Moment Area Equations

or

{ AB

Bu , AB

Bv , AB

Bθ } are known Length Constraint Equation


Useful Remarks (Cont.)

Remark5: If there exists a point within the structure where the rotation and the twocomponents of the displacement are known (e.g. a point at a fixed supports or a pointwhere their rotation and displacement were already computed), the rotations anddisplacements at all other points can be determined from 1st and 2 nd moment area

equations and the length constraint equation

Remark6: A member or a segment used in the calculation must not contain hinge except at its ends

Segments AC and CE (contain hinge inside) are not allowed

Segments AB, BC,CD, and DE (contain hinge at the ends) can be used

A

A

u A=v A=θ A=0

A

u A=v A=θ A=0

B

C D

A

B

C D E

F




Example 1 Given a statically determinate frame subjected to a concentrated load P asshown in the figure. The Young’s modulus and moment of inertia of the cross section areconstant and denoted by E and I, respectively. Determine the displacements and rotations atA, B, C, D, and E.

Solution

Define local coordinate systems for the segments ABC, BD and DE

Compute support reactions from static equilibrium equations (i.e. ΣF X = ΣF Y = ΣM = 0 )

L

A X B

E

Y

L L

L

C

D

P

x

x

x,y

y

y

AB

E

C

D

AB

E

C

D

X

Y

P

R AX =0

R AY =2P/3 R CY =P/3


Obtain BMD and M/EI diagram based on the local coordinate systems

Sketch Qualitative Elastic Curve

Start at segment AC since 3 quantities are already known (u A = v A = 0 & v C = 0)

AB

E

C

D

X

Y

P

R AX =0

R AY =2P/3 R CY =P/3

-PL/EI

4PL/3EI

PL/3EI

-PL/EI

A B

E

C

D

Length Constraint Equation 0 u u AC AAC C ==

2 nd Moment Area Equation ( ) ( ) C

AC M/EI AC A

AC A

AC C C/A x Area Lv v t ⋅=+−= θ

( ) ( )

( )3EI

7PL

3

2LL

3EI

PL

2

1

3

2LL2L

3EI

4PL

2

1 3L0 0

3

AC A

=⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ +

⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ =+− )(θ

9EI 7PL

9EI 7PL

2 2 AC A =−=θ

1st Moment Area Equation AC M/EI

AC A

AC C C/A Area =θ−θ=θ ( ) ( )

2EI

3PLL

3EI

PL

2

12L

3EI

4PL

2

1

9EI

7PL 2 2 AC C =⎟

⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−θ

18EI

13PL2 AC C =θ

x

y

A B

E

C

D

θC/A

t C/A

AC Aθ AC

C θ




Move to segment AB displacement and rotation at point A are already known

Move to segment BD displacement and rotation at point B are already known


1st Moment Area Equation

( ) ( ) B AB M/EI AB

AB A

AB A

AB B B/A x Area Lv v t ⋅=+−= θ ( )

9EI

8PL

3

2L2L

3EI

4PL

2

1 2L

9EI

7PL0 v

3

2 AB B

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+− )(

3EI

2PL

3EI

2PL v

3 3 AB B =−=


AB M/EI

AB A

AB B B/A Area =θ−θ=θ

( )3EI

4PL2L

3EI

4PL

2

1

9EI

7PL 2 2 AB B =⎟

⎠

⎞⎜⎝

⎛ =⎟⎟

⎠

⎞⎜⎜⎝

⎛ −−θ

9EI

5PL2 AB B =θ

A B

E

C

D

θB/A t B/A

x

y 0 u u AB

AAB B ==

0 v AB

A =

0 u AB A =

9EI

7PL

2 AC A

AB A −=θ=θ

0 u v AB

B

DB

B ==

3EI

2PL v u

3 AB B

DB B =−=

9EI

5PL

2 AB

B

DB

B =θ=θ

AB

Aθ AB B θ


Move to segment BD displacement and rotation at point B are already known


1st

Moment Area Equation


DB M/EI

DB D

DB B B/D Area =θ−θ=θ

( )EI

PLL

EI

PL

9EI

5PL 2 DB D

2

−=⎟ ⎠

⎞⎜⎝

⎛ −=θ−

9EI

14PL2 DB D =θ

A B

E

C

D

θB/D t D/B

x

y

3EI

2PL u u

3 DB B

DB D ==

( ) ( ) D

DB

M/EI DB

DB

B

DB

B

DB

D D/B x Area Lv v t ⋅=−−= θ

( ) 2EI

PL

2

L

LEI

PL

L9EI

5PL

0 v

3 2 DB

D −=

⎟ ⎠

⎞

⎜⎝

⎛

⎟ ⎠

⎞

⎜⎝

⎛ −=

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛

⎟⎟ ⎠

⎞

⎜⎜⎝

⎛ −−

)(

18EI

19PL

18EI

19PLv

3 3 DB D =−=

3EI

2PL u v

3 DB D

ED D −=−=

18EI

19PL v u

3 DB D

ED D −==

9EI

14PL

2 DB D

ED D =θ=θ


1st Moment Area Equation ED M/EI

ED E

ED D D/E Area =θ−θ=θ

( )2EI

PLL

EI

PL

2

1

9EI

14PL 2 ED E

2

−=⎟ ⎠

⎞⎜⎝

⎛ −=θ−

18EI

37PL2 ED E =θ

A B

E

C

D

θD/E

t E/D

x

y

18EI

19PL

18EI

19PL u u

3 3 ED D

ED E =−==

2 nd Moment Area Equation ( ) ( ) E

ED M/EI ED

ED D

ED D

ED E E/D x Area Lv v t ⋅=−−= θ

( )

3EI

PL

3

2LL

2EI

PL L

9EI

14PL

3EI

2PLv

3

2 3 ED E

−=

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ −=⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−− )(

9EI

23PL

9EI

23PLv

3 3 DB D =−=

DB B θ DB

D θ

ED D θ

ED E θ




Example 2 Given a statically determinate beam subjected to external loads as shown in thefigure. The Young’s modulus E is constant and the moment of inertia of the cross section isdenoted by 2I for a segment AB and by I for segments BC and CD . Determine the relativerotation at the hinge and the displacements at the end point D .

Solution

Local coordinate systems for all segments are the same as the global coordinate system



2L

AX

B

Y

2L L

C D

2q

R AX =0

R AY =3qL/2 R CY =3qL/2

qL

AX

B

Y

C D

2q qL

M A=3qL2

-qL2 /2EI

R AX =0

R AY =3qL/2 R CY =3qL/2

AX

B

Y

C D

2q qL

M A=3qL2

X

-2qL2 /EI

-qL2 /EI



Start at segment AB displacement and rotation at point A are already known

A

B C D




AB A

AB A

AB B B/A x Area Lv v t ⋅=+−= θ ( )( ) ( )

( ) ⎟ ⎠

⎞⎜⎝

⎛ ⎟⎟ ⎞

⎜⎜⎝

⎛ +

⎟

⎠

⎞⎜

⎝

⎛ ⎟⎟

⎠

⎞⎜⎜

⎝

⎛ −=+−

3

4L2L

2EI

qL

2

1

2

3L2L

EI

2qL

3

1 2L0 0 v

2

2 AB B )(

3EI

4qL

3EI

4qL v

4 4 AB B =−=


AB M/EI AB AAB B B/A Area =θ−θ=θ

( ) ( )2L2EI

qL

2

12L

EI

2qL

3

1 0

2 2 AB B ⎟⎟

⎠

⎞⎜⎜⎝

⎛ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=−θ

6EI

5qL

6EI

5qL 3 3 AB B =−=θ

A B C D

θB/A

t B/A AB B θ

x

y 0 u u AB

AAB B ==

0 v AB A =

0 u AB A =

0 AB A =θ




Move to segment BC 3 quantities at both ends of segment BC are already known


2 nd Moment Area Equation ( ) ( ) C

BC M/EI BC

BC B

BC B

BC C C/B x Area Lv v t ⋅=+−= θ

( ) ⎟ ⎠

⎞⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=⎟⎟

⎠

⎞⎜⎜⎝

⎛ θ+−−

3

2L2L

EI

qL

2

1 2L

3EI

4qL0

2 BC B

4

)(

2 nd Moment Area Equation BC M/EI

BC B

BC C C/B Area =θ−θ=θ ( )2L

EI

qL

2

1

EI

qL 2 3 BC C ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=−θ

0 BC C =θ

A B

C

D

θC/B t C/B

BC B θ

x

y 0 u u BC

B BC C ==

3EI

4qL v v

4 AB B

BC B −==

0 u u AB B

BC B ==

0 v BC C =

EI

qL3 BC B =θ

Relative rotation at hinge

AB

B

BC

B B

θ−θ=θ∆

⎟

⎟

⎠

⎞

⎜

⎜

⎝

⎛ −−=θ∆

6EI

5qL

EI

qL 3 3

B

6EI

11qL3

B =θ∆

BC C θ

AB B θ

B θ∆


Move to segment CD displacement and rotation at point A are already known


2 nd Moment Area Equation ( ) ( ) D

CD M/EI CD

CD C

CD C

CD D D/C x Area Lv v t ⋅=+−= θ

( ) ( ) ⎟ ⎠

⎞⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −=+−

3

2LL

EI

qL

2

1 L0 0 v

2 CD D )(

2 nd Moment Area Equation CD M/EI

CD C

CD D D/C Area =θ−θ=θ ( )L

EI

qL

2

1 0

2 CD D ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −=−θ

A B

C

D θD/C = CD

D θ

t D/C x

y

0 u u CD

C

CD

D ==

0 v CD

C =

0 u u BC C

CD C ==

0 BC C

CD C =θ=θ

3EI

qL

3EI

qLv

4 4 CD D =−=

2EI

qL

2EI

qL

3 3 CD D =−=θ

2101 310 St t l A l i I M th d f M t A E ti

2101 310 St t l A l i I M th d f M t A E ti




Example 3 Given a statically determinate frame subjected to external loads as shown in thefigure. The Young’s modulus and moment of inertia of the cross section are constant anddenoted by E and I , respectively. Determine the displacements and rotations A, B, C , and D .

Solution

Define local coordinate systems for the segments ABC, BD and DE


R AX =-P

R AY =-2P

R DY =2P

L 2L

2L

P 2PL

A X

B

Y

C

D

A

B

C

D x

x y

y

P 2PL

AX

B

Y

C

D




Two components of displacement at point A and the transverse component of displacement at point D are prescribed equal to zero while the displacement and rotation at point B and C are still unknowns. Thus, the segments BC and BD contain too many unknowns to be solved by the moment area and length constraint equations.

By applying Remark1 to the segment AB , we obtain

4PL/EI

A

B C D

2PL/EI

2PL/EI

A

B C D


0 u u AB

A

AB

B ==

2101 310 Structural Analysis I Method of Moment Area Equations

2101 310 Structural Analysis I Method of Moment Area Equations




Move to segment BD The transverse components of displacement at both ends of this segment are already known. By using Remark3, we obtain

Return to segment AB Now, four kinematical quantities of the segment BD are already known.


( ) ( ) B BD M/EI BD

BD D

BD D

BD B B/D x Area Lv v t ⋅=−−= θ

( ) ( )3EI

8PL

3

2L2L

EI

4PL

2

1 2L0 0

3 BD D =⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =θ−− )(


BD M/EI

BD B

BD D D/B Area =θ−θ=θ ( )

EI

4PL2L

EI

4PL

2

1

3EI

4PL 2 BD B

2

=⎟ ⎠

⎞⎜⎝

⎛ =θ−

3EI

8PL

3EI

8PL 2 2 BD B =−=θ

A

B

C

D

θD/B

t B/D

x

y

0 v BD

D =

0 v BD B =

BD B θ BD

D θ

3EI

4PL

2 BD D =θ

0 v AB A =

0 u AB

A =

0 u AB

B =

3EI

8PL

2 BD B

AB B −=θ=θ


Return to segment BD The longitudinal component of the displacement at point B of the segment BD is already known; thus, the longitudinal component of the displacement at point D can be readily computed from the length constraint equation.



AB A

AB A

AB B B/A x Area Lv v t ⋅=+−= θ ( )

3EI

4PL

3

2L2L

EI

2PL

2

1 2L

3EI

14PL0 v

3

2 AB B

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −+− )(


AB M/EI

AB A

AB B B/A Area =θ−θ=θ

( )EI

2PL2L

EI

2PL

2

1

3EI

8PL 2 AB A

2

=⎟ ⎠

⎞⎜⎝

⎛ =θ−−

3EI

14PL

3EI

14PL 2 2 AB A =−=θ A

B

C

D

θB/A

t B/A

x

y

AB Aθ

AB B θ

EI

8PL

EI

8PL v

3 3 AB B =−=


EI

8PLv u u

3 AB B

BD B

BD D =−==

2101-310 Structural Analysis I Method of Moment Area Equations



2101 310 Structural Analysis I Method of Moment Area Equations Note provided by JRR Page-25

Finally move to segment CB The displacement and rotation at point C are already known


( ) ( ) C CB M/EI CB

CB B

CB B

CB C C/B x Area Lv v t ⋅=−−= θ

( )

EI

PL

2

LL

EI

2PL L

3EI

8PL0 v

3

2 CB C

=

⎟ ⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ =⎟

⎟ ⎠

⎞⎜⎜⎝

⎛ ⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −−− )(


CB M/EI

CB C

CB B B/C Area =θ−θ=θ ( )

EI

2PLL

EI

2PL

3EI

8PL 2 CB C

2

=⎟ ⎠

⎞⎜⎝

⎛ =θ−−

3EI

14PL

3EI

14PL 2 2 CB C =−=θ

A

B C D

θB/C

t C/B

x

y

0 v v BD B

CB B ==

EI

8PL

u u

3 BD

B

CB

B ==

CB B θ

CB C θ

3EI 11PL v

3

CB C =

3EI

8PL

2 BD

B

CB

B −=θ=θ


EI

8PL u u

3 CB B

CB C ==

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