EMA 3702 Mechanics & Materials Science (Mechanics of ... · w M x wx wLx 2 1 2 1 M x( ) wx R 2x D 2...

EMA 3702

Mechanics & Materials Science

(Mechanics of Materials)

Chapter 9 Deflection of Beams

Introduction • Beams will deform under load, and the extent of

deformation, especially the transverse deformation or

displacement is of great interest for structure design, and

the transverse displacement is named “deflection”

• Goal: understand deformation/deflection under transverse

loading/bending moment and design appropriately

Deformation under Transverse Loading (1)

For bending of beam with moment

M, bending curvature (Chapter 4)

E Elastic modulus

I Moment of inertia

M Bending moment

Generally, M = M (x), therefore,

On the other hand, when and y and

dy/dx are very small, mathematically:

Therefore,

dx

dx

dyd

dx

yd

x

)(

)(

12

2

EI

xM

dx

yd )(2

2

1

𝜌=𝑀

𝐸𝐼

1

𝜌(𝑥)=𝑀(𝑥)

𝐸𝐼

Governing differential equation for

transverse deflection of a beam, i.e., for the

y(x) curve, also known as the “elastic curve”

EI Flexural rigidity


EI

xM

dx

yd )(2

2

Integrate the differential

equation in x, we have

Define as the angle

between elastic curve local

tangent and x axis

tandx

dywhen y

very small

in radian

𝐸𝐼𝑑𝑦

𝑑𝑥= න

0

𝑥

𝑀 𝑥 𝑑𝑥 + 𝐶1

Therefore, for , we have


EI

xM

dx

yd )(2

2

tandx

dy

From previous

To get y(x), integrate again

in x, and we have

or

𝐸𝐼𝑑𝑦

𝑑𝑥= න

0

𝑥


𝐸𝐼𝜃(𝑥) = න0

𝑥


𝐸𝐼𝑦 = න0

𝑥

𝑑𝑥න0

𝑥

𝑀 𝑥 𝑑𝑥 + 𝐶1𝑥 + 𝐶2

𝑦 =1

𝐸𝐼න0

𝑥

𝑑𝑥න0

𝑥


C1 & C2 are integration

constants and could be

determined from boundary

conditions.

Statically determinate beams

• M (x) easily determined

• 2 unknowns of C1 & C2

• 2 boundary conditions

Deflection for Statically

Determinate Beams

Deflection y(x) is given as

𝐸𝐼𝑦 = න0

𝑥

𝑑𝑥න0

𝑥


Free body diagram for the

entire beam DE is drawn.

Equilibrium for DE gives:

∑Fy = 0 RD + RE = wL

∑M = 0 wL • 0.5L= REL

RD = RE = 0.5wL

Class Example: Simply Supported Beam (1)

For a simply supported beam DE

with vertical distributed load w

For arbitrary point G, FBD

for DG is drawn

The bending moment at G is:

w

L

D E

D E

RD RE

w

wLxwxxM2

1

2

1)( 2 xRwxxM D 2

2

1)( Therefore,

D G

RD V

w

M

wx0.5x

x

Therefore, w

L

D EwLxwxxMdx

ydEI

2

1

2

1)( 2

2

2

Integrating wrt. x

Integrating wrt. x again

1

23

4

1

6

1CwLxwx

dx

dyEI

21

34

12

1

24

1CxCwLxwxEIy


02 C

012

1

24

121

44 CLCwLwL

To solve C1 and C2, consider

two boundary conditions:

when x = 0, y = 0

when x = L, y = 0

21

34

12

1

24

1CxCwLxwxEIy

3

124

1wLC 02 C

xLLxxEI

wy 334 2

24Therefore


Therefore,

Class Exercise: Cantilever Beam (1)

For cantilever beam DE with

vertical load P at free end D,

determine the y deflection

P L

D E

For section DG with length x,

free body diagram is drawn

Equilibrium for DG gives

∑Fy = 0 V = P

∑MD = 0 M = Vx = -Px

(Minus sign for M as it is to

bending the beam concave down)

V

M(x)

D G

x

P

Therefore, PxxMdx

ydEI )(

2

2

x

y

PxxMdx

ydEI )(

2

2

Integrating in x

1

2

2

1CPx

dx

dyEI

Integrating in x again

21

3

6

1CxCPxEIy

P L

D E

V = P

M(x)= -Px

D G

x

P

To solve C1 and C2, consider

boundary conditions:

When x = L, y = 0 & = dy/dx = 0

Class Exercise: Cantilever Beam (2)

P

D E

x = L, y = 0

x = L, dy/dx = 0

x = L, y = 0

02

11

2 CPLdx

dyEI

P L

D E

V = P

M(x)= -Px

D G

x

P

0)2

1(

6

12

23 CLPLPLEIy

Solve for C2, we have

3

23

1PLC

Therefore,

323 236

LxLxEI

Py

Class Exercise: Cantilever Beam (3)x = L, = dy/dx = 0

21

3

6

1CxCPxEIy

1

2

2

1CPx

dx

dyEI 2

12

1PLC

Example: Statically Indeterminate Beam (1)For a prismatic beam AB with A

side fixed and B side supported on

a roller as illustrated, what are the

reactions at A and B?

Draw FBD

Reactions of Ay, By, Ax, MA.

Equilibrium gives 3 equations

∑Fx = 0 Ax = 0

∑Fy = 0 Ay + B - wL = 0

∑MA = 0 MA+BL–0.5wL2=0

3 remaining reactions Ay, B, MA

2 equations

Therefore, statically indeterminate (to the 1st degree, in this

case)

Solving such a problem needs

to consider geometry or

deformation.

Integration for the differential

equation for the elastic curve

will give 2 additional

parameters of C1 and C2

Example: Statically Indeterminate Beam (2)

Boundary conditions will provide:

x = 0, dy/dx = 0; x = 0, y = 0; x = L, y = 0

All considered:

(3 + 2) = 5 variables & (2+3) = 5 equations

Problem can be solved

𝐸𝐼𝑦 = න0

𝑥

𝑑𝑥න0

𝑥



Integrating wrt. x

FBD for AC is drawn

Equilibrium of M around C gives:

A CAy V

w

M

wx

0.5x

MA

x

xAwxMxM yA 2

2

1)(

Ay MxAwxxMdx

ydEI 2

2

2

2

1)(

1

23

2

1

6

1CxMxAwx

dx

dyEI Ay

21

234

2

1

6

1

24

1CxCxMxAwxEIy Ay

Therefore,

Remember

Consider 3 boundary conditions:

1

23

2

1

6

1CxMxAwx

dx

dyEI Ay

21

234

2

1

6

1

24

1CxCxMxAwxEIy Ay

x = 0, dy/dx = 0

x = 0, y = 0

x = L, y = 0

01 C

02 C

2

24

1

2

1

6

1wLMLA Ay

∑Fy = 0 wLBAy

∑MA = 0 2

2

1wLBLM A

wLAy8

5

2

8

1wLM A

wLB8

3


Problem on right is statically

indeterminate to the 2nd degree:

Equilibrium gives 3 equations

∑Fx=0 Ax = 0

∑Fy=0 Ay + B - wL = 0

∑MA=0 MA+MB+BL=0.5wL2

4 remaining reactions

2 equations

Integration for elastic curve

gives 2 variables of C1 and C2

Consider boundary conditions

4 more equations

(4+2) = 6 variables

(2+4) = 6 equations

Statically Indeterminate Beam to 2nd Degree

Method of Superposition for Total DeflectionFor beams subject to more than one loads, local slope

(dy/dx or ) and local deflection y could be calculated

separately for each simple load (often by referring to tables

in engineering handbooks), and they could then be

combined following superposition principle at the same

position as long as:

• The stresses involved (even after combination) are well

within the proportional limit of the material

• None of the loadings will produce excessive

deformation that will impact the determination of any

other loadings

• Section considered for analysis is not too close to

loading points

Example: Superposition for Deflection

w

L

D E

L

D E

Px

xLLxxEI

wy wG

334 224

)( EIL

xLPxy PG

3

)()(

22

w

L

D E

Px

G

PGwGG yyy )()( At the same position G

EMA 3702 Mechanics & Materials Science Zhe Cheng (2018) 8 Deflection of Beams

Homework

Read textbook section 9.1, 9.2, and 9.4 and give a statement

confirm reading


Homework 9.1

For a simply supported prismatic beam DE with local moment at D,

please step-by-step determine (a) the equation of the elastic curve,

(b) the maximum deflection, (c) the slope at the two ends.

L

D EM


Homework 9.2

For a simply supported prismatic beam DE with concentrated load at G,

please step-by-step determine (a) the elastic curve y(x) for 0 < x < a and

(b) the deflection at point G.

L=a+b

D E

Pa b

G


Homework 9.3

For a prismatic cantilever beam DE with D side on roller and subject

to moment M0 and E side fixed, please step-by-step determine (a) the

elastic curve y(x) and (b) reactions at D and E

L

DE

M0


Homework 9.4

For cantilever beam with both distributed and concentrated load as

illustrated, please determine the slope and deflection at D end. (Hint: to

treat concentrated load at G alone, section GE will behave like a

cantilever beam under vertical load at the end (case 1 of textbook

Appendix D), while section DG remains straight.)

L

DE

w=P/L

P

G

0.5L

EMA 3702 Mechanics & Materials Science (Mechanics of ... · w M x wx wLx 2 1 2 1 M x( ) wx R 2x D 2...

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