1 classes #11 & #12 civil engineering materials – cive 2110 bending fall 2010 dr. gupta dr....

11

Classes #11 & #12Classes #11 & #12Civil Engineering Materials – CIVE 2110Civil Engineering Materials – CIVE 2110

BendingBending

Fall 2010Fall 2010

Dr. GuptaDr. Gupta

Dr. PickettDr. Pickett

ASSUMPTIONS OF BEAM BENDING THEORY

Beam Length is Much Larger Than Beam Width or Depth.

so most of the deflection

is caused by bending,

very little deflection is caused by shear Beam Deflections are small. Beam is Perfectly Straight,

With a Constant Cross Section (beam is prismatic).

Beam has a Plane of Symmetry. Resultant of All Loads

acts in the Plane of Symmetry. Beam has a Linear Stress-Strain

Relationship.

1

1

E

ASSUMPTIONS OF BEAM BENDING THEORY

Beam Material is Homogeneous. Beam Material is Isotropic. Beam is Loaded ONLY by a

Moment about an axis Perpendicular to the long axis of Symmetry.

Thus Moment is CONSTANT

across the

Length of the Beam.

There is NO SHEAR.

d

ASSUMPTIONS OF BEAM BENDING THEORY

Plane Sections Remain Plane.

No Warping

(no buckling, no rotation about vertical axis).

Motion is only in Vertical Plane.

Beam Cross Sections originally

Perpendicular to Longitudinal Axis

Remain Perpendicular.

d

BEAM BENDING THEORY

When a POSITVE moment is applied, TOP of beam is in COMPRESSION BOTTOM of beam is in TENSION.

NEUTRAL SURFACE: - plane on which NO change in LENGTH occurs.

BEAM BENDING THEORY

When a POSITVE moment is applied, (POSITIVE Bending)

TOP of beam is in COMPRESSION

BOTTOM of beam is in TENSION.

NEUTRAL SURFACE:

- plane on which

NO change

in LENGTH occurs.

Cross Sections

perpendicular to

Longitudinal axis

Rotate about the

NEUTRAL (Z) axis.

BEAM BENDING THEORY

When a POSITVE moment is applied, (POSITIVE Bending)

TOP of beam is in COMPRESSION

BOTTOM of beam is in TENSION.

NEUTRAL SURFACE:

- plane on which

NO change

in LENGTH occurs.

Cross Sections

perpendicular to

Longitudinal axis

Rotate about the

NEUTRAL (Z) axis.

BEAM BENDING THEORY

For M = +

Any line segment, Δx :

- shortens,

if located

above Neutral Surface.

BEAM BENDING THEORY

For M = +

Any line segment, Δx :

- does not change length,

if located

at Neutral Surface.

BEAM BENDING THEORY

For M = +

Any line segment, Δx :

- lengthens,

if located

below Neutral Surface.

BEAM BENDING THEORY

y

yThus

ysand

xand

RadiansTan

smallarensdeformatioSince

TanxndeformatioAfter

xsndeformatioBefore

s

ss

LengthOriginal

LengthinChange

StrainNormalmembering

s

s

)(lim

)('

)(

)]([

'lim

Re

0

0

BEAM BENDING THEORY

max

max

var

c

yThus

cand

y

SurfaceNeutral

beloworabove

ywithlinearilyies

DirectionalLongitudinthein

StrainNormalThus

Flexural Bending Equation

We assumed: Cross Sections remain constant

However,

do to the Poisson’s Effect;

there will be strains

in the 2 directions

perpendicular to the

Longitudinal Axis.

xz

xy

and

Axial Compressive Strain

Axial Tensile Strain

BEAM BENDING THEORY

For material that is: Homogeneous Isotropic Linear-Elastic

We can conclude for

STRESS, σ

xz

xy

and

max

max

max

maxmax

maxmax

sin

c

y

cy

cy

EEthus

cy

Ethen

c

yEthen

c

yceand

EandEthusE

BEAM BENDING THEORY

For material that is: Homogeneous Isotropic Linear-Elastic

We can conclude for

STRESS, σ

max

maxsin

c

ythen

c

yceand

EthusE

Compressive Stress

Tensile Stress

Compressive Strain

Tensile Strain

BEAM BENDING THEORY

Internal Moment must resist

External Moment.

Internal Resisting Moment: Caused by an Internal Force resisting an External force

Can find Neutral Axis by balance

of Forces:

Σ Internal Forces must = ZERO

A

A

AAx

dAyc

dAc

y

dAdFF

)(0

)(0

)(0

max

max

Compressive Stress

Tensile Stress

Neutral Axis

BEAM BENDING THEORY

Can find Neutral Axis by balance

of Forces:

Σ Internal Forces must = ZERO

Neutral Axis = Centroidal Axis

0 =1st Moment of Area

about Neutral Axis

AdAy )(0

Compressive Stress

Tensile Stress

Neutral Axis

BEAM BENDING THEORY

Internal Moment must resist

External Moment.

M = (Lever Arm)x(Internal Force) M = (Lever Arm)x(Stress x Area)

AExternalZ

AExternalZ

AExternalZ

AExternalZ

InternalZExternalZ

dAyc

M

dAc

yyM

dAyM

dFyM

MM

)(

)(

)(

)(

2max

max

Compressive Stress

Tensile Stress

Neutral Axis

BEAM BENDING THEORY

Internal Moment must resist

External Moment.

M = (Lever Arm)x(Internal Force) M = (Lever Arm)x(Stress x Area)

I

cMorI

cM

axisZaboutAreaofMomentI

dAyI

dAyc

M

MM

externalZ

ExternalZ

nd

A

AExternalZ

InternalZExternalZ

maxmax

2

2max

2

)(

)(

Compressive Stress

Tensile Stress

Neutral Axis

BEAM BENDING THEORY

Flexural Bending Stress Equation:

For Stress in the Direction of the

Long Axis (X),

At any location, Y,

above or below the

Neutral Axis

I

yMexternalZ

x

Compressive Stress

Tensile Stress

Neutral Axis

Beam Bending

2nd Moment of AreaCalculation

A Rectangular Cross Section

PARALLEL AXIS THEOREM FOR 2nd MOMENTS OF AREA

2nd MOMENTS OF COMPOSITE AREASB & J 8th,9.6, 9.7

233

3332

22

3222

11

311

321

121212YA

hbYA

hbYA

hb

IIII ZZZZ

Y2

Z Z

PARALLEL AXIS THEOREM

FOR 2nd MOMENTS OF AREA

QUESTION: Why are I-beams shaped like I ???????? ANSWER: In order to achieve maximum strength (and least deflection) for the least weight, by maximizing the Second Moment of Area, I, of a beam. This is achieved by maximizing the distance between beam material in the flanges and the beam mid-height. For Example: Construct an I-beam from three pieces of balsa wood, With each piece of balsa wood 3”x3/8”x36”

4444 _26.7_21.3_844.0_21.3 ininininIIII CBAbeam

4

44

2

3

23

_21.3

2.3013.0

8

"3

2

1"5.1

8

"3"3

12

8

"3"3

12

inI

ininI

I

yHBAHB

I

A

A

A

AAAAAA

A

4

44

2

3

23

_844.0

0844.0

0"38

"3

12

"38

"312

inI

ininI

I

yHBAHB

I

B

B

B

BBBBBB

B

Sign Convention for Diagrams

V=-

Tension

Tension

Compression

Free End

Or

Pinned End

Free End

Or

Pinned End

V=+

MIntrnl=+

Compression

Tension

Tension

Compression

TensionMExtrnl=-

Compression

Tension

Fixed End Fixed End

MIntrnl=+

MIntrnl=-

MIntrnl=-

MExtrnl=+

MIntrnl=+

MIntrnl=-

MIntrnl=-

MIntrnl=-

Steps for V and BM diagrams

1.Draw FBD

2.Obtain reactions:

M (@left support) to obtain reaction at right;

M (@Right support) to obtain reaction at left;

Check Fy = 0

3. Cut a section ;

Obtain internal F (or P), V, M at cut section ;

M, Fy, Fx

4. Record, draw internal F (or P), V, M on both sides of cut sections ;

- magnitude

- units

- direction on both sides of cut

BEAM END CONDITIONS

Roller Pin - Pin

Fixed - Free

Fixed - ?

VL=RLY

BEAM END CONDITIONS

Pin VL=RLY

BEAM END CONDITIONS

Roller Pin

VL=RLY