lecture 5-6 beam mechanics of materials laboratory sec. 3-4 nathan sniadecki university of...

Lecture 5-6Beam

Mechanics of Materials Laboratory Sec. 3-4

Nathan SniadeckiUniversity of Washington

Mechanics of Materials Lab

$100

2x

A

I y dA

Answer: Answer: What is the moment of inertia (Second What is the moment of inertia (Second Moment of Area) with respect to the Moment of Area) with respect to the xx axis axis

$100

2y

A

I x dA

Answer: Answer: What is the moment of inertia (second moment What is the moment of inertia (second moment of area) with respect to the of area) with respect to the yy axis axis

$200

31

12I bh

Answer: Answer: What is the moment of inertia for a rectangleWhat is the moment of inertia for a rectangle

$200

41

4I r

Answer: Answer: What is the moment of inertia for a circleWhat is the moment of inertia for a circle

$300

• The straight line that defines a surface where x and x are zero

Answer: Answer: What is the neutral axisWhat is the neutral axis

$300

y

x

y

M z

I

Answer: Answer: What is the bending stress in the What is the bending stress in the xx-direction at a -direction at a distance distance yy from the origin of the coordinate from the origin of the coordinate system due to the loading of a couple vectorsystem due to the loading of a couple vector M Mxx

acting in acting in xx-direction-direction

$400

• The point on the stress-strain curve where the material no longer deforms elastically, but also plastically.

Answer: Answer: What is the proportional limitWhat is the proportional limit

$400

• The theorem that expresses that the moment of inertia Ix of an area with respect to an arbitrary x axis is equal to the moment of inertia Ixc with respect to the centroidal x axis, plus the product Ad2 of the area A and the square of the distance d between the two axis?

Answer: Answer: What is the Parallel Axis Theorem What is the Parallel Axis Theorem IIxx = I = Ixcxc + Ad + Ad22

$500

• The principle that states that the effect of a given combined loading on a structure can be obtained by determining the effects of each load separately and then combining the results obtained together as long as 1) each effect is linearly related to the load that produces it and 2) the deformation resulting from any given load is small and does not affect the conditions of application of the other loads

Answer: Answer: What is the Principle of SuperpositionWhat is the Principle of Superposition

$500

• The principle that states that except in the immediate vicinity of the point of loading, the stress distribution may be assumed to be independent of the actual mode of loading, i.e. for axial loading, at a distance equal to or greater than the width of a member, the distribution of stress across a given section is the same.

Answer: Answer: What is Saint-Venant’s PrincipleWhat is Saint-Venant’s Principle

FINAL JEOPARDYFINAL JEOPARDY

FINAL JEOPARDY

arctan y z

z y

M I

M I

Answer: Answer: What is the What is the angleangle of the neutral axis for an of the neutral axis for an asymmetrically loaded beamasymmetrically loaded beam

Inclined Load

Notice the sign convention: positive Mz compress upper part, negative stress; positive My extend front part, positive stress!

Inclined Load

z

z

y

yx I

yMI

zM

Stress

Neutral axis

0z

z

y

yx I

yMI

zM

yz

zy

IM

IM

zy tan

Example

Stress Distribution

The centroid of the area A is defined as the point C with coordinates (yc, zc) which satisfies

Asymmetrical Beam

If the origin of y and z axes is placed at centroid C

(orientation is arbitrary.)

c

A

ydA Ay c

A

zdA Az

0A

ydA 0A

zdA

( is inverse of )

x

x

x x

L y

L L L y y

L y y

L

y

E yE

• Consider of beam segment AB of length L• After deformation, length of neutral surface

DE remains L, but JK becomes

Pure Bending

Asymmetric Beam

zy

yxz

EI

dAyEydAM

2

yzy

yxy

EI

yzdAEzdAM

If z is a principal axis (symmetry), the product of inertia Iyz is zero My = 0, bending in x-y plane, analogous to a symmetric beam

z

yz

z

y

I

I

M

M

When z axis is the neutral axis;

Asymmetric Beam

yzz

zxz

EI

yzdAEydAM

yz

zxy

EI

dAzEzdAM

2

y

yz

y

z

I

I

MM

If y is a principal axis, the product of inertia Iyz is zero Mz = 0, bending in x-z plane, analogous to a symmetric beam

When y axis is the neutral axis;

Asymmetric Beam

• When an asymmetric beam is in pure bending, the plane in which the bending moments acts is perpendicular to the neutral surface if and only if (iff) the y and z axes are principle centroidal axes and the bending moments act in one of the two principle planes. In such a case, the principle plane in which bending moment acts becomes the plane of bending and the usual bending theory is valid

Analysis of Asymmetric Beam

• Locating the centroid, and constructing a set of principal axes

• Resolving bending moment into My and Mz

• Superposition

z

z

y

yx I

yMI

zM

tantany

z

yz

zy

II

IM

IM

zy

Principle Axes

dAyIx2 dAxI y

2

xydAIxy

1 1, cos2 sin 22 2

x y x y

x y xy

I I I II I I

2cos2sin211 xy

yxyx I

III

yx

xyp II

I

22tan

Analysis of Asymmetric Beam

A channel section C 10 15.3

M = 15 kips-inIy=2.28 in4, Iz=67.4 in4

Location of Point C c=0.634 in

Location of Point A yA=5.00 in zA=-2.6+0.634=-1.966 in

Calculate bending stress

Locate neutral axis

Analysis of Asymmetric Beam

ink605.2sin MM y

ink77.14cos MM z

psi3340z

z

y

yx I

yMI

zM

o

y

z

II

zy

1.79

212.5tantan

Normal Stress in Beam

yy

E y

Curved Beams

• What if the beam is already ‘bent’?

• Where will the beam likely fail?

Bending Stress for Curved Beam

• #1: Neutral surface remains constant: • #2: Deformation at JK:

R R

L r r

R y R y

y y

Bending Stress for Curved Beam

• #3: Strain:

• #4: Stress:

x

y

R y

x

E y

R y

E R r

r

r R y

r R y

Bending Stress for Curved Beam

• #5: Neutral Axis:

• #6 Centroid:

0 1x

AdA R

dAr

1

r rdAA

e r R • #7: N.A. Location:

xy dA M

E M

A r R

Since > 0 for M > 0

R r

Aside: R = rn

Location of N.A. in Curved Beam

• Cross-sectional dimensions define neutral axis location for a curve beam about C

Curved Beams

Positive M decreases curvature

i

ii Aer

Mc

o

oo Aer

Mc

1/

( )

n

n

r A dAr

My

Ae r y

Neutral axis is no longer the centroidal axis

Curved Beam

1/

3.64in

n

o

i

bhr A dA

br drr

hr

lnr

( )n

F My

A Ae r y

Curved Beams

rr

IMs

ArI

e

yrAeMy

c

c

n

)(

rrs c

Curvature is small, e is small, rn is close to rc

Recover to straight beam

Curved Beam

sbsrb

rc

srsb

sbsr

s

e

c

c

Pay attention to the sign of s

Curved Beam

srsb

sbsr

s

e

c

c

222 sRb

Pay attention to the sign of s

• Read Mechanics of Materials Lab Sec. 4

• 4.26(e), 4.72 posted online

Assignment