Beam Deflection Review (4.3-4.5)

Beam Deflection Review (4.3-4.5)

MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering

Beam Deflection Review1


Deflection Due to Bending (4.3)

ydx

dy

ds

θ

tandx

dy

cosds

dx

sinds

dydsd

1dcurvature of beam

ds

Slope of the deflection curve



Assumption 1: θ is small. 1. 2.

Assumption 2: Beam is linearly elastic. Thus, the differential equation for the deflection curve is:

dxdsdx

d

ds

d

1

tandx

dy

2

2

dx

yd

dx

d2

21

dx

yd

EI

M

1

EI

M

dx

yd

2

2



Recall: So we can write: Deflection curve can be found by integrating

Bending moment equation (2 constants of integration) Shear-force equation (3 constants of integration) Load equation (4 constants of integration)

Chosen method depends on which is more convenient.

wdx

dV V

dx

dM

wdx

ydEI

4

4

Vdx

ydEI

3

3

Mdx

ydEI

2

2


Method of Superposition (4.5)

Deflection and slope of a beam produced by multiple loads acting simultaneously can be found by superposing the deflections produced by the same loads acting separately. Reference Appendix A-9 (Beam Deflections and Slopes) Method of superposition can be applied to statically determinate and statically indeterminate beams.

Consider the following example: Find reactions at A and C. Method 1: Choose MC and RC asredundant.

Method 2: Choose MC and MA as redundant.Beam Deflection Review6

Method of Superposition (4.5)

Beam Deflection: Method of Superposition

7

Example Problem

For the beam and loading shown, determine (a) the deflection at C, and (b) the slope at end A.

8

Example Problem

Loading I: Case 6 in Appendix D.

4 35

384 24C AwL wL

yEI EI

Loading II: Case 7 in Appendix D.

3 2

2 1

6 2 2 16 3A A A

C AM M L M LL L

y LEIL EI EI

with 2 4 31 1

12 192 36A C AwL wL wL

M yEI EI

Loading III: Case 7 in Appendix D.

31(using Loading II result)

16

6

BC

BA

M Ly

EIM L

EI

with 2 4 31 1

12 192 72B C AwL wL wL

M yEI EI

(a) Deflection at C.

4 4 4 45 1 1 1

384 192 192 384CwL wL wL wL

yEI EI EI EI

41

384CwL

yEI

(b) Slope at A.

3 3 31 1 1

024 36 72A

wL wL wL

EI EI EI

0A

Beam Deflection: Method of Superposition

9

Example Problem

For the beam shown, determine the reaction at B.

10

Example Problem

Beam is second degree indeterminate. Choose BR and BM as redundant reactions.

Loading I: Case 1 of Appendix D.

3 2

I I( ) , ( )3 2B B

B BR L R L

yEI EI

Loading II: Case 3 of Appendix D.

2

II II( ) , ( )2B B

B BM L M L

yEI EI

Loading III: Case 3 applied to Portion AC.

2 20 0

III

0 0III

( /2)( )

2 8( /2)

( )2

C

C

M L M Ly

EI EIM L M L

EI EI

Portion CB remains straight.

20

III III III

20

III III

3( ) ( ) ( )

2 8

1( ) ( )

2

B C C

B C

M LLy y

EI

M L

EI

Super position and constraint:

I II III

23 20

( ) ( ) ( ) 0

30

3 2 8

B B B B

B B

y y y y

M LL LR M

EI EI EI

(1)

I II III

20

( ) ( ) ( ) 0

10

2 2

B B B B

B B

M LL LR M

EI EI EI

(2)

Solving (1) and (2) simultaneously,

03

2B

MR

L

01

4BM M

Castigliano’s Theorem(4.7-4.10)

MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering

Castigliano’s Theorem11

Castigliano’s Theorem (4.8)


This method is a powerful new way to determine deflections in many types of structures – bars, beams, frames, trusses, curved beams, etc.

We can calculate both horizontal and vertical displacements and rotations (slopes).

There are actually two Castigliano’s Theorems. The first can be used for structures made of both

linear and non-linear elastic materials. The second is restricted to structures made of linear

elastic materials only. This is the one we will use.



“When forces act on elastic systems subject to small displacements, the displacement corresponding to any force, in the direction of the force, is equal to the partial derivative of the total strain energy w.r.t. that force.”

ii F

U

Where:Fi = Force at i-th application pointδi = Displacement at i-th point in the direction of Fi

U = Total strain energy



We can also use this method to find the angle of rotation (θ).

Where:Mi = Moment at i-th application pointθi = Slope at i-th point resulting from Mi

U = Total strain energy

ii M

U



General case

U stored in structure

F1

F2

F3

Fn

δ1

δ2

δ3

δn

11 F

U

22 F

U

33 F

U

nn F

U

Strain Energy in Common Members (4.7)


Spring

kFδ

2

2

1 kU

kFkF /

k

F

k

FkU

22

2

1

2

1

k

F

F

U

Note:

Check:



Bar subject to axial load

2 2

0

1 ( )

2 2 ( ) ( )

LF L F xU or U dx

AE A x E x

F F

L

A,E



Shaft subject to torque

2 2

0

1 ( )

2 2 ( ) ( )

LT L T xU or U dx

GJ G x J x

L

J,G

T T

Beam subject to bending



2 2

0

1 ( )

2 2 ( ) ( )

LM L M xU or U dx

EI E x I x

L

I,E

M M

Beam in direct shear



2 2

0

( )

2 2 ( ) ( )

LF L F xU or U dx

AG A x G x

Beam in transverse shear



2

0

( )

2 ( ) ( )

L CV xU dx

A x G x

Correction factor for transverse shear (see table 4-1 in textbook)

L

I,E

V V

For structures with combined loading (or multi-component structures) add up contributions to U.



2 2 2

2 2

0

( ) ( ) ( )

2 ( ) ( ) 2 ( ) ( ) 2 ( ) ( )

( ) ( )

2 ( ) ( ) 2 ( ) ( )

L

F x T x M xU dx dx dx

A x E x G x J x E x I x

F x CV xdx dx

A x G x A x G x

For the structure and loading shown below, determine the vertical deflection at point B. Neglect axial force in the column.

Castigliano’s Theorem - Frame (4.8)


P

B

A

L1

L2

E, I

For the structure and loading shown below, determine the vertical and horizontal deflection at point B. Neglect axial force in the column.

Castigliano’s Theorem - Frame (4.8)


w

B

A

L1

L2

E, I

For the structure and loading shown below, determine the vertical and horizontal deflection at point B. Consider the effects of bending only.

Castigliano’s Theorem – Curved Beam (4.9)


B

A

E, I

Fv

Fh

R

For the structure and loading shown below, determine the vertical deflection at D and horizontal deflection at C. Let L = 16 ft, h = 6 ft,E = 30 x 103 ksi, P = 18 kips, Atens = 2.5 in2, and Acomp = 5 in2.

Castigliano’s Theorem - Trusses


P

A C

B

D

L/2L/2

h

For the structure and loading shown below, find the fixed end reactions.

Statically Indeterminate Problems (4.10)


L

A B

w

A curved frame ABC is fixed at one end, hinged at another, and subjected to a concentrated load P, as shown in the figure below. What are the horizontal H and vertical F reactions? Consider bending only.

Statically Indeterminate Problems (4.10)


Find the tip deflection for the structure and loading shown below.

Special Cases: Hollow Tapered Beam


P

dA dB= 2dA

t

L

x

For the beam-spring system below, find the deflection at end C.

Special Cases: Beam With Spring


L

P

AC

B

k

a

x

E, I

Beam Deflection Review (4.3-4.5)

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