Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10)

Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10)

MAE 316 – Strength of Mechanical ComponentsNC State University Department of Mechanical and Aerospace Engineering

Skew Loads & Non-Symmetric XSections1

Introduction


Will perform advanced stress and deflection analysis of beams with skew loads and non-symmetric cross sections.

Challenge: Need to calculatemoments of inertia – Iyy, Izz,and Iyz – and principal moments of inertia.

x

y

P

z

y

Pα

z

y

P

Skew load

Non-Symmetric

Moments of Inertia


For any cross-section shape

A

yz

A

zz

A

yy dAyzIdAyIdAzI ,, 22

z

y

CdA

Moments of Inertia


The moments of inertia can be transformed to y1-z1 coordinates by

Does this look familiar??

z

y

C

y1

z1

θ

dA

2cos2sin2

2sin2cos22

2sin2cos22

11

11

11

yzzzyy

zy

yzzzyyzzyy

zz

yzzzyyzzyy

yy

III

I

IIIII

I

IIIII

I

Moments of Inertia


Similar to transformation of stress, principal angle (angle to the principal axes of inertia) can be found from

Where θP is the angle at which Iyz is zero.

z

y

C

y1

z1

θ

dA

zzyy

yzP II

I

22tan

Example


Find Iyy and Izz for a rectangleb

h

y

z

dA = dydz

C

Example


Find Iyy and Izz for a Z-section (non-symmetric about y-z) Let b = 7 in, t = 1 in, and h = 16 in.

b

h/2

y

z

b

h/2

t (all)

C

Example


Find Iyy and Izz for an L-section (non-symmetric about y-z) Let b = 4 in, t = 0.5 in, and h = 6 in. t

h

y

z

b

t

C

Skew Loads (3.10)


Skew loads for doubly symmetric cross sections

Beam will bend in two directions Py = P cos α Pz = P sin α

C

y

z

x(origin of x-axis at fixed end)

Pz

P

Py α

Skew Loads (3.10)


Find bending moments Side view

From statics: Mz = Py(L-x) = P cos α (L-x)

Why is Mz positive? beam is curving in direction of positive y

C

y

z

x

Pz

P

Py α

Sidex

y

z (in)

Py

L-x

Mz Mz

Skew Loads (3.10)


Find bending moments Top view

From statics: My = Pz(L-x) = P sin α (L-x)

Why is My positive or negative?

C

y

z

x

Pz

P

Py α

Top

x

z

y (in)

Pz

L-x

My My

Skew Loads (3.10)


Bending stress From side view

From top view

Combine to get

C

y

z

x

Pz

P

Py α

TopSide

zz

zxx I

yM

yy

yxx I

zM

zz

z

yy

yxx I

yM

I

zM

Skew Loads


What about the neutral axis?

When there is only vertical bending, σxx=0 because y=0 at the neutral axis.

y

zC

N.A. (y=0)

no stress on this line

00 yatI

Myxx

Skew Loads


But with a skew load:

It turns out deflection will be perpendicular to this line.

tan

0

yy

zz

z

y

zz

z

yy

yxx

I

I

M

M

z

y

I

yM

I

zM

y

zC

no stress on this line

β

Skew Loads


Curvature due to moment From side view

From top view

Where vy and vz are deflections in the positive y and z directions, respectively.

C

y

z

x

Pz

P

Py α

TopSide

2

2

)(dx

vdEIxM y

zzz

2

2

)(dx

vdEIxM z

yyy

Skew Loads


Find deflection at free end.

Apply B.C.’s: vy(0)=0 & vy’(0)=0

The tip deflection in the y-direction is

C

y

z

x

Pz

P

Py α

)(cos)(2

2

xLPxMdx

vdEI z

yzz

1

2

2cos c

xLxP

dx

dvEI y

zz

21

32

62cos cxc

xLxPvEI yzz

00)0(6

)0(

2

)0(cos)0( 221

32

ccc

LPvy

002

)0()0(cos)0( 11

2

ccLP

dx

dvy

zzy EI

PLLv

3

cos)(

3

Skew Loads


Continued…

Apply B.C.’s: vz(0)=0 & vz’(0)=0

The tip deflection in the z-direction is

C

y

z

x

Pz

P

Py α

)(sin)(2

2

xLPxMdx

vdEI y

zyy

1

2

2sin c

xLxP

dx

dvEI z

yy

21

32

62sin cxc

xLxPvEI zyy

00)0(6

)0(

2

)0(sin)0( 221

32

ccc

LPvz

002

)0()0(sin)0( 11

2

ccLP

dx

dvz

yyz EI

PLLv

3

sin)(

3

Skew Loads


The resultant tip deflection is

C

y

z

x

Pz

P

Py α

22zy vv

2

2

2

23 sincos

3 yyzz IIE

PL

y

zC

N.A.

βvz

vyδ

Example


Consider a cantilever beam with the cross-section and load shown below. Find the stress at A and the tip deflection when α = 0o and α=1o. Let L = 12 ft, P = 10 kips, E = 30x106 psi and assume an S24x80 rolled steel beam is used.

y

zC

A (z=3.5 in, y=-12 in)

P α

Non-Symmetric Cross-Sections


Bending of non-symmetriccross-sections Iyz ≠ 0

Iyy & Izz are not principal axes

Use generalized flexure formula

z

y

C

My

Mz

2

)()(

yzzzyy

yzyyyzyzzzzyx

III

yIMIMzIMIM



Generalized moment-curvature formulas

z

y

C

My

Mz

)(

)(

22

2

22

2

yzzzyy

yzzzzyz

yzzzyy

yzyyyzy

IIIE

IMIM

dx

vd

IIIE

IMIM

dx

vd



A special case – which we discussed previously – is whenIyz = 0 and y & z are the principal axes.

yy

yz

zz

zy

zz

z

yy

yxx

EI

M

dx

vd

EI

M

dx

vd

I

yM

I

zM

2

2

2

2

Example


Analysis choices Work in principal coordinates – simple formulas Work in arbitrary coordinates – more complex formulas

Calculate the stress at A and the tip deflection for the beam shown below.

L = 10 ft

y

zC

A (z=-0.99 in, y=-4.01 in)

Mz = 10,000 in-lbs(pure bending)

Cross-section dimensions:6 x 4 x 0.5 in

Iyy = 6.27 in4

Izz = 17.4 in4

Iyz = 6.07 in4

E = 30 x 106 psi

Skew Loads and Non-Symmetric Cross Sections (Notes + 3.10)

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