# Beam Deflection 1 1

By Touhid Ahamed

Introduction

• In this chapter rigidity of the beam will be considered

• Design of beam (specially steel beam) base on strength

consideration and deflection evaluation

Introduction

Different Techniques for determining beam deflection

• Double integration method

• Area moment method

• Conjugate-beam method

• Superposition method

• Virtual work method

Double Integration Method

The edge view of the neutral surface of a deflected beam is

called the elastic curve

1 M ( x)

EI

ρ

Double Integration Method

• From elementary calculus, simplified

for beam parameters,

d2y

2

2

1

d

y

dx

2

2 3 2

dx

dy

1

dx

• Substituting and integrating,

1

d2y

EI EI 2 M x

dx

x

dy

EI EI

M x dx C1

dx

0

x

x

EI y dx M x dx C1x C2

0

0

Double Integration Method

Boundary conditions for statically determinate beam

x

x

0

0

EI y dx M x dx C1 x C2

Solved Problem

SOLUTION:

• Develop an expression for

M(x) and derive differential

equation for elastic curve.

W 14 68

I 723 in 4

P 50 kips L 15 ft

E 29 106 psi

a 4 ft

For portion AB of the overhanging

beam, (a) derive the equation for

the elastic curve, (b) determine

the maximum deflection,

(c) evaluate ymax.

• 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.

Solved Problem

SOLUTION:

• Develop an expression for M(x) and

derive differential equation for

elastic curve.

- Reactions

:

Pa

a

RA

L

RB P1

L

- From the free-body diagram for

section AD,

a

M P

L

x

0 x L

- The differential equation for the

elastic curve,

EI

d2y

a

P

x

2

L

dx

Solved Problem

• Integrate differential equation twice

and apply boundary conditions to

obtain

dy elastic

1 a 2 curve.

EI

dx

2

P

L

x C1

1 a 3

P x C1x C2

6 L

at x 0, y 0 : C2 0

EI y

EI

d2y

a

P

x

2

L

dx

Substituting,

dy

1 a

1

EI

P x 2 PaL

dx

2 L

6

EI y

1 a 3 1

P x PaLx

6 L

6

at x L, y 0 : 0

dy PaL

1

dx 6 EI

x

3

L

2

3

PaL2 x x

y

6 EI L L

1 a 3

1

P L C1L C1 PaL

6 L

6

Solved Problem

• Locate point of zero slope or

point of maximum

deflection.

2

dy

PaL

x

0

1 3 m

dx

6 EI

L

3

PaL2 x x

y

6 EI L L

xm

L

0.577 L

3

• Evaluate corresponding

maximum deflection.

PaL2

ymax

0.577 0.577 3

6 EI

PaL2

ymax 0.0642

6 EI

ymax

50 kips 48 in 180 in 2

0.0642

6 29 106 psi 723 in 4

ymax 0.238 in

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