Lecture Mechanics of materials (Third edition) - Chapter 9: Deflection of beams

The following will be discussed in this chapter: Deformation of a beam under transverse loading, equation of the elastic curve, direct determination of the elastic curve from the load di, statically indeterminate beams, application of superposition to statically indeterminate, moment-area theorems,...

Deflection of Beams

Deformation of a Beam Under Transverse

Loading

Equation of the Elastic Curve

Direct Determination of the Elastic Curve

From the Load Di

Statically Indeterminate Beams

Bending Moment Diagrams by Parts Sample Problem 9.11

Application of Moment-Area Theorems to Beams With Unsymme

Maximum Deflection Use of Moment-Area Theorems With Statically Indeterminate

• Relationship between bending moment and curvature for pure bending remains valid for general transverse loadings.

• Curvature varies linearly with x

• At the free end A, = A = ∞

=

ρ 0,

1

• Overhanging beam

• Reactions at A and C

• Bending moment diagram

• Curvature is zero at points where the bending

moment is zero, i.e., at each end and at E.

• Maximum curvature occurs where the moment magnitude is a maximum

• An equation for the beam shape or elastic curve

is required to determine maximum deflection and slope

• From elementary calculus, simplified for beam parameters,

2

2 2

3 2 2 2

1

1

dx

y d dx

dy dx

y d

=

ρ

• Substituting and integrating,

( ) ( )

0 0

1 0

2

2

1

C x C dx x M dx y

EI

C dx x

M dx

dy EI EI

x

M dx

y d EI EI

x x

x

+ +

( ) 1 2

0 0

C x C dx x M dx y

EI

x

x

+ +

• For a beam subjected to a distributed load,

dx

dV dx

M d x

V dx

2 2

• Equation for beam displacement becomes

( )x

w dx

y d EI dx

M

4

4 2

2

4 3

2 2 2 1

3 1 6

1C x C x C x C

dx x w dx dx dx x

y EI

+ +

+ +

−

= ∫ ∫ ∫ ∫

• Integrating four times yields

• Constants are determined from boundary conditions

• Consider beam with fixed support at A and roller support at B.

• From free-body diagram, note that there are four unknown reaction components

• Conditions for static equilibrium yield

0 0

C x C dx x M dx y

EI

x

x

+ +

= ∫ ∫

• Also have the beam deflection equation,

which introduces two unknowns but provides three additional equations from the boundary conditions:

0 ,

At 0

0 ,

0

At x = θ = y = x = L y =

ft 4 ft

15 kips

50

psi 10 29 in

723 68

P

E I

W

For portion AB of the overhanging beam,

(a) derive the equation for the elastic curve,

(b) determine the maximum deflection,

(c) evaluate y max

SOLUTION:

• Develop an expression for M(x) and derive differential equation for elastic curve

• Integrate differential equation twice and apply boundary conditions to obtain elastic curve

• Locate point of zero slope or point

of maximum deflection

• Evaluate corresponding maximum deflection

R L

a P

M = − 0 < <

x L

a P dx

y d

2 2

- The differential equation for the elastic curve,

PaL C

L C

L L

a P y

L x

C y

x

6

1 6

1 0

: 0 ,

at

0 :

0 ,

0

at

1 1

3

2

= +

3

1 2

6 1 2 1

C x C x

L

a P y

EI

C

x L

a P dx

dy EI

+ +

a P dx

y d

2 2

x EI

PaL y

PaLx

x L

a P y

EI

L

x EI

PaL dx

dy PaL

x L

a P dx

dy EI

6

1 6

1

3

1 6

6

1 2

1

3

2 2

−

=Substituting,

• Locate point of zero slope or point

x EI

PaL y

L

L x

L

x EI

PaL dx

dy

m

3 3

1 6

=

EI

PaL y

EI

PaL y

6 0642

0

2 max =

2 max

in 723 psi

10 29 6

in 180 in

48 kips 50 0642

max =

y

For the uniform beam, determine the

reaction at A, derive the equation for

the elastic curve, and determine the

slope at A (Note that the beam is

statically indeterminate to the first

degree)

SOLUTION:

• Develop the differential equation for the elastic curve (will be functionally

dependent on the reaction at A).

• Integrate twice and apply boundary

conditions to solve for reaction at A

and to obtain the elastic curve

• Evaluate the slope at A.

• Consider moment acting at section D,

L

x w x R M

M

x L

x w x

R M

A

A D

6

0 3

2 1 0

3 0

2 0

M dx

y d

6

3 0 2

x w x R

M dx

y d

6

3 0 2

5 0 3

1

4 0 2

120 6

1

24 2

1

C x

C L

x w x

R y

EI

C L

x w x

R EI

dx

dy EI

A

A

+ +

6

1 : 0 ,

at

0 24

2

1 : 0 ,

at

0 :

0 ,

0

at

2 1

4 0 3

1

3 0 2

2

= +

w L

R y

L x

C L

w L

R L

x

C y

1 3

x L w L

x w x

L w y

3 0

120

1 120

10

1 6 1

=

θ

• Differentiate once to find the slope,

at x = 0,

Principle of Superposition:

• Deformations of beams subjected to

combinations of loadings may be

obtained as the linear combination of

the deformations from the individual

For the beam and loading shown, determine the slope and deflection at

point B.

SOLUTION:

Superpose the deformations due to Loading I and Loading II as shown.

( )

EI

wL

L EI

wL EI

wL

y B II

384

7 2

48 128

4 3

=

Combine the two solutions,

wL

II B I

B B

48 6

3 3

+

−

= +

wL y

y

y B B I B II

384

7 8

4 4

+

−

= +

=

• Method of superposition may be

applied to determine the reactions at

the supports of statically indeterminate

beams

• Designate one of the reactions as

redundant and eliminate or modify

deformations compatible with the original supports

For the uniform beam and loading shown, determine the reaction at each support and

the slope at end A.

SOLUTION:

• Release the “redundant” support at B, and find deformation

• Apply reaction at B as an unknown load to force zero displacement at B.

( )

EI wL

L L

L L

L EI

w

y B w

4

3 3

4

01132

0

3

2 3

2 2 3

2 24

L EIL

R

2 2

01646

0 3

wL y

Slope at end A,

( )

EI

wL EI

wL

w A

3 3

04167

L

L EIL

wL

R A

3

2 2

03398

0 3

3 6

0688

wL

R A w

A A

3 3

03398

0 04167

.

−

= +

θ

• Geometric properties of the elastic curve can

be used to determine deflection and slope

• Consider a beam subjected to arbitrary loading,

• First Moment-Area Theorem:

area under (M/EI) diagram between

C and D.

• Second Moment-Area Theorem:

The tangential deviation 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.

• Tangents to the elastic curve at P and P’ intercept

a segment of length dt on the vertical through C.

= tangential deviation of C with respect to D

• Cantilever beam - Select tangent at A as the

reference

• Simply supported, symmetrically loaded

beam - select tangent at C as the reference.

• Determination of the change of slope and the tangential deviation is simplified if the effect of each load is evaluated separately.

• Construct a separate (M/EI) diagram for each

• Bending moment diagram constructed from

individual loads is said to be drawn by parts.

For the prismatic beam shown, determine

the slope and deflection at E.

SOLUTION:

• Determine the reactions at supports

• Construct shear, bending moment and

(M/EI) diagrams.

• Taking the tangent at C as the

reference, evaluate the slope and

tangential deviations at E.

• Determine the reactions at supports

wa R

EI

wa A

EI

L wa

L EI

wa A

6 2

3 1

4 2

2

3 2

2

2 2

wa EI

L wa A

A

C E C

E C

E

6 4

3 2

2

1 + = − −

=

= +

wa EI

L

wa EI

L wa

L A

a A

L a A

t t

y E E C D C

16 8

16 4

4 4

3 4

2 2 4

2 2 3

1 2

With Unsymmetric Loadings

• Define reference tangent at support A Evaluate θA

by determining the tangential deviation at B with respect to A.

• The slope at other points is found with respect to reference tangent

A D A

• The deflection at D is found from the tangential deviation at D.

• Maximum deflection occurs at point K

where the tangent is horizontal

• Point K may be determined by measuring

an area under the (M/EI) diagram equal

Indeterminate Beams

• Reactions at supports of statically indeterminate beams are found by designating a redundant constraint and treating it as an unknown load which satisfies a displacement compatibility requirement

• The (M/EI) diagram is drawn by parts The

resulting tangential deviations are superposed and related by the compatibility requirement

• With reactions determined, the slope and deflection are found from the moment-area method