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

• An equation for the beam shape or elastic curve is required to determine maximum deflection.[r]

MECHANICS OF MATERIALS

Ferdinand P Beer

E Russell Johnston, Jr.

John T DeWolf

Lecture Notes:

J Walt Oler Texas Tech University

CHAPTER

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

Sample Problem 9.1

Sample Problem 9.3

Method of Superposition

Sample Problem 9.7

Application of Superposition to Statically

Indeterminate

Sample Problem 9.8 Moment-Area Theorems Application to Cantilever Beams and Beams With Symmetric

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

EI

x

1 =

ρ

• Cantilever beam subjected to concentrated load at the free end,

EI

Px

−

=

ρ

1

• Curvature varies linearly with x

• At the free end A, = A = ∞

A

ρ

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.

EI

x

1 =

ρ

• Beam is concave upwards where the bending moment is positive and concave downwards where it is negative

• Maximum curvature occurs where the moment magnitude is a maximum

• An equation for the beam shape or elastic curve

• 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,

( )

( ) 1 0

2

2

1

C dx x

M dx

dy EI EI

x

M dx

y d EI EI

x

+

=

≈

=

=

∫

θ ρ

( ) 1 2 0

0

C x C dx x M dx y

EI

x

x

+ +

= ∫ ∫

• Constants are determined from boundary conditions

• Three cases for statically determinant beams, – Simply supported beam

0 ,

A y y

– Overhanging beam

0 ,

A y y

– Cantilever beam

0 ,

A

• More complicated loadings require multiple integrals and application of requirement for

Load Distribution

• For a beam subjected to a distributed load,

( ) w( )x

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

• 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

=

The beam is statically indeterminate

( ) 1 2 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

=

=

=

×

=

=

×

a L

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

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

- Reactions:

↑

⎟

⎠

⎞

⎜

⎝

⎛ +

=

↓

=

L

a P

R L

Pa

- From the free-body diagram for section AD,

( x L)

x L

a P

M = − 0 < <

x

a P y

d

- The differential equation for the elastic curve,