AERSP 301 Bending of open and closed section beams Dr. Jose Palacios

AERSP 301Bending of open and closed section beams

Dr. Jose Palacios

Today’s ItemsToday’s Items

• HW 2 due today!• HW 3: Extra Credit – 50% of 1 HW worth towards HWs grade• HW 4: will be assigned on Monday. HW helper session coming up. • Bending of Beams -- Megson Chapter 16,

Reference: Donaldson – Chapters Lambda and Mu– Direct stress calculation

– Bending deflections

TodayToday

• HW 3 (EXTRA CREDIT, 50% OF 1 HW): DUE WEDNESDAY• HW 4: ASSIGNED TODAY• EXAM: OCTOBER 20 – 26 HOSLER – 8:15 – 10:15 PM• REVIEW SESSION: OCTOBER 19 – 220 HAMMOND – 6 – 9 PM• WEDNESDAY AFTER CLASS: VOTING REGISTRATION

• Direct stress at a point in the c/s depends on:– Its location in the c/s– The loading– The geometry of the c/s

• Assumption – plane sections remain plane after deformation (No Warping), or cross-section does not deform in plane (i.e. σxx, σyy = 0)

• Sign Conventions!

Megson pp 461

Direct stress calculation due to bending

M – bending moment

S – shear force

P – axial load

T – torque

W – distributed load

Direct stress calculation due to bending (cont’d)

• Beam subject to bending moments Mx and My and bends about its neutral axis (N.A.)

• N.A. – stresses are zero at N.A.

• C – centroid of c/s (origin of axes assumed to be at C).

Neutral Surface Definition

In the process of bending there is an axial line that do not extend or contract. The surface described by the set of lines that do not extend or contract is called the neutral surface. Lines on one side of the neutral surface extend and on the other contract since the arc length is smaller on one side and larger on the other side of the neutral surface. The figure shows the neutral surface in both the initial and the bent configuration.

The axial strain in a line element a distance y above the neutral surface is given by:

• Consider element A at a distance ξ from the N.A.

• Direct Stress:

• Because ρ (bending radius of curvature) relates the strain to the distance to the neutral surface:


0

0

l

llz

EE zzz

First Moment of Inertia Definition

• Given an area of any shape, and division of that area into very small, equal-sized, elemental areas (dA)

• and given an Cx-Cy axis, from where each elemental area is located (yi and xi)

• The first moment of area in the "X" and "Y" directions are respectively:

dAxAxI

ydAAyI

y

x

• IF the beam is in pure bending, axial load resultant on the c/s is zero:

• 1st moment of inertia of the c/s about the N.A. is zero N.A. passes through the centroid, C

• Assume the inclination of the N.A. to Cx is α


• Then

The direct stress becomes:

AA

z dAdA 00

cossin yx

cossin yxEE

z


• Moment Resultants:

• Substituting for σz in the above expressions for Mx and My, and using definitions for Ixx, Iyy, Ixy

dAxM

ydAM

zy

zx

dAxyI

dAxI

dAyI

xy

yy

xx

2

2

cos

sin

cossin

cossin

xyyy

xxxy

y

x

xyyyy

xxxyx

II

IIEM

M

IE

IE

M

IE

IE

M


• Using the above equation in:

• Gives:

From Matrix Form

y

x

xyyy

xxxy

M

M

II

IIE1

cos

sin

y

x

xyyy

xxxy

xyyyxxM

M

II

II

III

E2

1

cos

sin

cossin yxE

z

yIII

IMIMx

III

IMIM

xyyyxx

xyyyyx

xyyyxx

xyxxxyz

22


• Or, rearranging terms:

• If My= 0, Mx produces a stress that varies with both x and y. Similarly for My, if Mx=0.

• If the beam c/s has either Cx or Cy (or both) as an axis of symmetry, then Ixy = 0.

• Then:

22xyyyxx

xyxxy

xyyyxx

xyyyxz III

yIxIM

III

xIyIM

yy

y

xx

xz I

xM

I

yM

• Further, if either My or Mx is zero, then:

• We saw that the N.A. passes through the centroid of the c/s. But what about its orientation α?

At any point on the N.A. σz = 0


xx

xz I

yM

yy

yz I

xMor

tan

022

xyyyyx

xyxxxy

xyyyxx

xyyyyx

xyyyxx

xyxxxyz

IMIM

IMIM

x

y

yIII

IMIMx

III

IMIM

Example Problem

The beam shown is subjected to a 1500 Nm bending moment in the vertical plane.

Calculate the magnitude and location of max σz.

0

8 mm

40 mm 80 mm

80 mm

8 mm

1st: Calculate location of Centroid

mm 528808120

)880(40)8120(60

xx

xxxx

A

Axxc

mm 4.668808120

)880(40)8120(84

xx

xxxx

A

Ayyc

x

y

Example Problem (cont’d)• Calculate Ixx, Iyy, Ixy, with respect to Cxy:

4623

23

23

mm 1009.1)404.66(80812

808

)4.6684(812012

8120

12

cxx dAbt

I

4623

23

23

mm 1031.1)4052(80812

808

)5260(812012

8120

12

cyy dAtb

I

Example Problem (cont’d)

46 mm 1034.0)4.6640()5240(808

)4.6684()5260(8120

A

xy xydAI

Mx = 1500 Nm, My = 0

22xyyyxx

xyxxy

xyyyxx

xyyyxz III

yIxIM

III

xIyIM

mm]in y x,N/mmin [

39.05.12

z

xyz By inspection, MAX at y = -66.4 mm and x = -8 mm(Max stress always further awayFrom centroid)

0x

y

Deflections due to bendingDeflections due to bending

• From strength of materials, recall that [Megson Ch. 16.2.5]:

• Beam bends about its N.A. under moments Mx, My.

– Deflection normal to N.A. is ζ Centroid C moves from CI (initial) to CF (final).

– With R as the center of curvature and ρ as the radius of curvature

Mx

My

wx

w

wy

Distributed Load

Deflections due to bending (cont’d)

• Further,

• Because


• Inverse relation:

• Clearly Mx produces curvatures (deflections) in xz and yz planes even when My = 0 (and vice-versa)

• So an unsymmetrical beam will deflect vertically and horizontally even when loading is entirely in vertical (or horizontal) plane.

What if I have something symmetric?? Like NACA 0012 airfoil?


• If Cx or Cy (or both) are axes of symmetry then Ixy = 0. Then the expressions simplify to:

• Starting with the general expression:

and integrating twice you can calculate the disp. u in the x-direction


• Consider the case where a downward vertical force, W, is applied to the tip of a beam. What is the tip deflection of the beam?

• Integrating,

• Integrating again


• Using b.c.’s: @ z=0 u = 0, u’ = 0– Gives: A = B = 0

• Thus,

– Tip deflection:

• If the c/s has an axis of symmetry, Ixy = 0

You should do this on your own

(z = L)

Simplifications for thin-walled sections

• Thin-walled t << c/s dimensions. – Stresses constant through thickness

– Terms in t2, t3, etc… neglected

• In that case Ixx reduces to:

• What about Ixy for this c/s

• What about Iyy for this c/s

horizontal members

vertical members

Example:

You should do this on your own

Doubly Symmetrical Cross-Section Beam Bending

Beam has a flexural rigidity: EI

)()('''' zpzEIv EI v’’’’ = distributed load

ApzzvEI )(

BpLzpz

zvEI 2

)(2

EI w’’’ = Shear Force

EI w’’ = Moment

CzpLpLzpz

zvEI 226

)(223

EI w’ = Slope

yp(z)

z

pLA

LvEI

0)(

Bc’s

2

0)(2pL

B

LvEI

0

0)0(

C

vEI

DzpLpLzpz

zEIv 4624

)(2234

EI w’ = Displacement

0

0)0(

D

EIv

Doubly Symmetrical Cross-Section Beam Bending

Beam has a flexural rigidity: EI

0)()('''' zpzEIvEI w’’’’ = distributed load

AzvEI )(

BFzzvEI )(

EI w’’’ = Shear Force

EI w’’ = Moment

CFLzFz

zvEI 2

)(2

EI w’ = Slope

y

z

FA

FLvEI

)(

Bc’s

FLB

LvEI

0)(

0

0)0(

C

vEI

DFLzFz

zEIv 26

)(23

EI w’ = Displacement

0

0)0(

D

EIv

F

Concentrated and Partial Span Loads

• Diract delta function:

Example vertical force of magnitude F0 locater at L/2

• Heaviside step function:

Example vertical distributed force of magnitude f0(z) over the second part of the beam only

)/()()( Lzzstpyfzf oy

)2/( LzF

z0

f

Concentrated and Partial Span Loads Example

z

y

L/3L/3 L/3

f0

Fo

z

x

L/3L/3 L/3

f0

Mo

3

2

33)(''''

Lzstp

Lzstpf

LzFzvEI ooxx

3

2

3

23

3)(''''

Lzstp

Lz

L

fLzMxuEI o

oyy

Bc’s

w(0) = w’(0) = v(0) = v’(0) = w(L)= w’(L) = v(L) = v’(L) = 0

Integrating Diract Delta and Heaviside Function

3

2

33)(''''

Lzstp

Lzstpf

LzFzvEI ooxx

3

2

3

2

3324

33662)(

3

2

3

2

336

3322)(

3

2

3

2

33233)(

3

2

3

2

333)(

44

331

22

34

33

221

23

22

12

1

Lzstp

Lz

Lzstp

Lz

f

Lzstp

Lz

FzCzCzCCzvEI

Lzstp

Lz

Lzstp

Lz

f

Lzstp

Lz

FzCzCCzvEI

Lzstp

Lz

Lzstp

Lz

fLzstp

LzFzCCzvEI

Lzstp

Lz

Lzstp

Lzf

LzstpFCzvEI

o

oxx

o

oxx

ooxx

ooxx

Integrating Diract Delta and Heaviside Function

3

2

3

23

3)(''''

Lzstp

Lz

L

fLzMzuEI o

oyy

3

2

3

2

40

33262)(

3

2

3

2

8332)(

3

2

3

2

23)(

3

2

3

2

2

3

3)(

5

231

22

34

421

23

3

12

2

1

Lzstp

Lz

L

f

Lzstp

Lz

MzCzCzCCzuEI

Lzstp

Lz

L

fLzstp

LzM

zCzCCzuEI

Lzstp

Lz

L

fLzstpMzCCzuEI

Lzstp

Lz

L

fLzMCzuEI

o

oyy

ooyy

ooyy

ooyy

OPTIONAL: Macauley’s Method Read Megson Chp 16.)

y

za a a a

W W 2W

AB C D F

Ra Rf

Determine the position and magnitude of the maximum upward and downwarddeflection of the beam:

(upward) 4

3WRA (downward)

4

3WRF

]3[2]2[][ azWazWazWzRM A

The bending moments around the left hand side at any section Z between D and F is:

Documents

AERSP 301 Bending of open and closed section beams Dr. Jose Palacios