Home Page Address

l http://www-personal.engin.umich.edu/~kikuchi/GMBodyStructure.html

Enhanced Handout #3

MEAM599-08Automotive Body Structure

1/27/97

This Week’s Terminology

This Week’s Subject

l Review of Basic Structural Mechanics for Automotive Bodies

– Frames (Open and Closed Cross Section) – Panels (Shear Panels, Plates, and Shells) – Joints (Welded Components)

l Reference– J.T. Oden, Mechanics of Elastic Structures,

McGraw-Hill, New York, 1967– Despite of out of print, this is the best textbook

on elementary structural mechanics

Forces and Stresses

l Based on the law of statics

F F F

M M M

x y z

x y z

∑ ∑ ∑

∑ ∑ ∑

= = = ( )

= = = ( )

0 0 0

0 0 0

, ,

, ,

force

moment

l External forces are independent of the deformation and the material properties of the body, though they may depend on the mass distribution

l body and surface forces (traction)

Stresses (1)

l Stress is a measure of the internal force per unit area within a body

l normal stressl shear stress

σ = =→

lim ,∆

∆∆

∆ ∆A

F

AF A

0normal to

τ = =→

lim ,∆

∆∆

∆ ∆A

F

AF

0parallel to A

normal

shear 1

shear 2

Stress (2)

l Three normal planes to the coordinate axes x, y, and z

σ x = normal stress acting in the x direction

τxy = shear stress on a plane normal to the x axis

acting in the y direction

Σ = stress tensor

σ τ ττ σ ττ τ σ

x xy xz

yx y yz

zx zy z

=

Equilibrium (1)*

∂σ∂

∂τ∂

∂τ∂

∂τ∂

∂σ∂

∂τ∂

∂τ∂

∂τ∂

∂σ∂

x yx zxb

xy y zyb

xz yz zb

x y zX

x y zY

x y zZ

+ + + =

+ + + =

+ + + =

0

0

0

Summation of all the forces in the x direction

∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆σ τ τx yz zx by z x z x y X x y z+ + + = 0

Dividing by the differential volume, and taking the limit

* See Additional Note #1 for details

Equilibrium (2)*Moment equilibrium yields

τ τ τ τ τ τyz zy zx xz xy yx− = − = − =0 0 0, ,

Equilibrium of the Surface Element

Xl l l

S X S S S Sbx y z

s x x

l S

yx y

l S

zx z

l Sx y z

4

30

3∆ ∆ ∆ ∆ ∆

∆ ∆ ∆

( ) + − − − == = =

σ τ τ

X l l l

Y l l l

Z l l l

s x x yx y zx z

s xy x y y zy z

s xz x yz y z z

= + +

= + +

= + +

σ τ τ

τ σ τ

τ τ σ

Remarks (1)

l In the same stressed configuration, values of the stresses components in a given coordinate system are different from the ones in a different coordinate system

x

y

z

x

y

z

Remarks (2)

l Thus, we need coordinate free stress measures to make comparison of the stress state

– principal stresses– Mises equivalent stress– .........– quantities which are scalor

Principal Stresses

l The eigenvalues of the stress tensor are called the principal stresses

l The associated eigenvectors are the three orthogonal directions of the principal axes

σ σ σ1 2 3 are the solution of

det detΣΣ −( ) =−

−−

= − + − + =λσ λ τ τ

τ σ λ ττ τ σ λ

λ λ λIx xy xz

yx y yz

zx zy z

I I I31

22 3 0

example

ΣΣ =

2 1 1

1 0 1

1 1 3

MPa

S=[2,1,1;1,0,1;1,1,3]

S = 2 1 1 1 0 1 1 1 3

[X,L]=eig(S)

X = 0.3005 -0.7928 0.5302 -0.9365 -0.1398 0.3217 0.1810 0.5932 0.7844

L = -0.5141 0 0 0 1.4280 0 0 0 4.0861

Given stress tensor

Using MATLAB

exercise

Check the following relation using the above example

I

I

I

1 1 2 3

2 1 2 2 3 3 1

3 1 2 3

= + += + +=

σ σ σσ σ σ σ σ σσ σ σ

These three quantities are called the three stress invariants:

first, second, and third invariants

which are independent of the coordinate systems

Equivalent Stress σ

σ σ σ σ σ σ σm x y z= + +( ) = + +( )13

13 1 2 3

mean or hydrostatic stress

stress deviator

ΣΣ ΣΣD m

x m xy xz

yx y m yz

zx zy z m

= − =−

−−

σσ σ τ τ

τ σ σ ττ τ σ σ

I

det

σ σ λ τ ττ σ σ λ ττ τ σ σ λ

λ λ λx m xy xz

yx y m yz

zx zy z m

J J J

− −− −

− −

= − + − +31

22 3

exercises

Show the following relation

J

J x y z y z z x x z yz zy zx xz xy yx

1

22 2 2

0

13

=

= + + − − −( ) + + +σ σ σ σ σ σ σ σ σ τ τ τ τ τ τ

Mises Equivalent Stress

σ = 3 2J

Show that

J2 1 2

2

2 3

2

3 1

216

= −( ) + −( ) + −( ) σ σ σ σ σ σ

Stress Resultants in Bars (1)

stress resultants are the internal forces and momentsdue to a given stress distribution over the area of across-sectional plane

geometric axis

cross-sectional planes

Stress Resultants in Bars (2)

N dA M y z dA

N dA M zdA

N dA M ydA

x xA x xz xyA

y xyA y xA

z xzA z xA

= = −( )= =

= =

∫ ∫∫ ∫∫ ∫

σ τ τ

τ σ

τ σ

,

,

,

two shearing forces and a normal forcetwo bending moments and a torque or twisting moment

6 Stress Resultants in Bars

Exercise 1

Find the equilibrium equation of a straight prismatic bar loadedonly in the xy plane shown in the following figure.

Exercise 2

Show the equilibrium equations inpolar coordinates for a two-dimensional element are

∂σ∂

∂τ∂α

σ σ

∂τ∂

∂σ∂α

τ

α α

α α αα

r r rr

r r

r r rF

r r rF

+ + − + =

+ + + =

10

12 0

F

Fr

α

= body forces

σστ

α

α

r

r

= stress components

where

Saint-Venant’s Principle

Two different but statically equivalent force systems actingon a small portion of the surface of an elastic body producethe same stress distribution at distance large in comparisonwith the linear dimensions of the portion of the surface onwhich these forces act.

Special Note

However, in the case of thin-walled bars under general loading, applications of the principle may lead tosignificant errors

Typical Explanation

Except the loading points, the stressdistribution becomes almost the same

See, also

Axial stress rapidly becomes uniformin cross sections

However, Hoff’s Example

For a cantilever applied atorque at the free end, Hoffcalculated the ratio of themaximum stress over the cross-section and that at the fixed end.

If Saint-Venant’s principleholds, this must decay rapidlyas the distance from the fixedend increases.

Hoff’s Example

Details can be found inY.C. Fung, Foundations of Solid MechanicsPrentice-Hall, Inc., Englewood Cliffs, 1965

Although it is correct for the rectangular cross-section, but it is quite in correct for the thin-walled cross-section.

Very similar phenomenon was also found for spaceframe structures such as

Exercise 3

Examine the Saint-Venant effect by using a bit morerealistic automotive body frame for

(a) torsion(b) axial forces(b) bending in two orthogonal directions

by fixing at the two rear wheel portions.

Strains (1)

Strains are the measure ofdeformation, and is defined to bethe ratio of the change and itsoriginal length of the differential line element.

lim∆

∆ ∆∆l

l l

l→

−0

2 2

2

axial strain of a bar ε = =∆L

L

elongation/contractionlength of the bar

Strains (2)

∆ ∆

∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆

∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆ ∆

l l

x u y v z w x y z

u x v y w z u v w

2 2

2 2 2 2 2 2

2 2 22 2 2

differential line element after deformation differential element before deformation −

= +( ) + +( ) + +( ) − + +( )= + + + + +

Noting that ∆ ∆ ∆ ∆

∆ ∆ ∆ ∆

∆ ∆ ∆ ∆

uu

xx

u

yy

u

zz

vv

xx

v

yy

v

zz

ww

xx

w

yy

w

zz

= + +

= + +

= + +

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

Strains (2)

∆ ∆ ∆l lu

x

u

x

v

x

w

xx

v

y

u

y

v

y

w

y

2 22 2 2

2

2 2 2

212

212

− = +

+

+

+ +

+

+

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+ +

+

+

+ + + + +

∆

∆

∆ ∆

y

w

z

u

z

v

z

w

zz

v

z

w

y

u

z

u

y

v

z

v

y

w

z

w

yy

2

2 2 222

12

2

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

zz

w

x

u

z

u

x

u

z

v

x

v

z

w

x

w

zz x

u

y

v

x

u

y

u

x

v

y

v

x

w

y

w

zx y

x yx y z

+ + + + +

+ + + + +

= + +

2

2

2 2 22 2

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

ε ε ε

∆ ∆

∆ ∆

∆ ∆ ∆zz y z z x x yyz zx xy2 2 2 2+ + +γ γ γ∆ ∆ ∆ ∆ ∆ ∆

Strain Components

ε ∂∂

∂∂

∂∂

∂∂

ε ∂∂

∂∂

∂∂

∂∂

ε ∂∂

∂

x

y

z

u

x

u

x

v

x

w

x

v

y

u

y

v

y

w

y

w

z

= +

+

+

= +

+

+

= +

12

12

12

2 2 2

2 2 2

uu

z

v

z

w

z

v

z

w

y

u

z

u

y

v

z

v

y

w

z

w

y

w

x

u

z

u

x

u

z

v

x

v

z

w

x

yz

zx

∂∂∂

∂∂

γ ∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

γ ∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

+

+

= + + + +

= + + + +

2 2 2

∂∂∂

γ ∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

w

zu

y

v

x

u

y

u

x

v

y

v

x

w

y

w

zxy = + + + +

ε ∂∂

ε ∂∂

ε ∂∂

γ ∂∂

∂∂

γ ∂∂

∂∂

γ ∂∂

∂∂

x

y

z

yz

zx

xy

u

xv

y

w

zv

z

w

y

w

x

u

zu

y

v

x

=

=

=

= +

= +

= +

Linearlized StrainLagrangian Strain

Infinitesimal StrainsIf the gradient of the displacement is sufficiently small, the nonlinear terms in the Lagrange strain can be neglected,and the linearized strain components can be obtained.

ε x x=

rate of change in length per unit length of fibers

originally parallet to the axis

γ γ ∂∂

γ γ ∂∂

γ γ γ ∂∂

∂∂

1 1 2 2

1 2

≈ = ≈ =

= + = +

tan tanu

y

v

x

u

y

v

xxy

and

Shearing Strains

Extensional/longitudinal Strains

Example (1)

P

L

rE

u x zdw

dxx

v x

w xP

EIx Lx

( ) = − ( )

( ) =

( ) = −

0

16

12

3

ε x x y zP

EIx L z

P

EIx L z

P

EIx Lx, ,( ) = − −( ) + − −( )

+ −

12

12

22

2

E GPa

d r cm

L m

== ==

200

2 1

1

ε ∂∂x

u

x

P

EIx L z P N≈ = − −( ) = ⇒ =0 2 39 3699. % .

δ = 13 3. cm

Example (2)

ε x xL

z rat and= = ±2

P

∂∂u

x

12

2 2 2∂∂

∂∂

∂∂

u

x

v

x

w

x

+

+

P Px x

linear nonlinear

Remarks

Example shown, represents that nonlinear strainscan be large, although no yielding occurs in linearanalysis, and this implies possibility of instabilityof a structure

Geometric Nonlinear Problems

Another note is the effect of stiffening effect of the non-linear strains, see

softer responsein linear analysis

Recommendation

l Not only– Mises Equivalent Stress– Principal Stresses and Their Principal Aexes– in order to check possibility of plasticity, i.e,

material nonlinearity

l But Also– nonlinear parts of the Lagrange Strain– in order to check possibility of geometric

nonlinearity in (FE) analysis

Three Different Problems

l Large Strain and Large Displacement– metal forming & crashworthiness study– large plastic deformation (material nonlinearity)– buckling (global and local)

l Small Strain and Large Displacement– buckling (global)– but mostly in material linearity

l Small Strain and Small Displacement– standard elastic FE analysis or elasticity

Deformation of Bars

Displacements normal to the bar’saxis is said to be deflections, and

we refer to angular displacementsas rotations

Rotations transverse to the axis of a bar are called slopes

that about the axis is called twist

θθθ

x

y

z

x y z

= rotation about , , and axes, respectively

Stress-Strain Relation

σσστττ

εεεγγγ

x

y

z

yz

zx

xy

x

y

z

yz

zx

xy

a a a a a a

a a a a a

a a a a

a a a

a a

a

=

11 12 13 14 15 16

22 23 24 25 25

33 34 35 36

44 45 46

55 56

66

Hook’s Law ( Constitutive Relation/Equation )

contracted notation σσ εε= Eε σ νσ νσ γ τ

ε σ νσ νσ γ τ

ε σ νσ νσ γ τ

x x y z yz yz

y y z x zx zx

z z x y xy xy

E G

E G

E G

= − −( ) =

= − −( ) =

= − −( ) =

1 1

1 1

1 1

,

,

,

Isotropic Material

GE=+( )2 1 ν

Elastic Stability

The elastic stability of a structure is dependent of thegeometry and the material properties of the structure as well as the nature of the applied loading.

The term of elastic stability pertains to the phenomenaof structural bodies possessing more than one possibleequilibrium configuration for certain loads.

The changes in geometry due to deformation enter theequilibrium considerations and there is no longer a uniquesolution to a given problem.

Mechanics of Frames

l Axial ...... can be neglectedl Torsion

– noncircular open/closed sections

l Bending– unsymmetric possibly curved beams

l Impact– effect of corners in the cross section

Torsion of Circular Bars

Coulomb’s Torsion / Circular Bars

γ φ γ φ τ φα α αx x xx r r

xGr

d

dx∆ ∆ ∆

∆= ⇒ = ⇒ =

rate of change is constant

M r dA Grd

dxdA G

d

dxr dA G

d

dxIx x

A A A= = = =∫ ∫ ∫τ φ φ φ

α α2 2

I

d

dx

α

θ φ=

=

polar moment of inertia

= angle of twist per unit length

τ

θ

αα

α

xx

x

M

Ir

M

GI

=

=

Torsion of Noncircular Bars

For noncircular prism in torsion

σ σ σ τx y z yz= = = = 0

∂τ∂

∂τ∂

∂τ∂

∂τ∂

xy xz

xy xz

y z

x x

+ =

= =

0

0

ε ε ε γ∂τ∂x y z yz

xy

x= = = = ⊕ =0 0

u f y z

v xz

w xy

= ( )= −=

,

θθ

Equilibrium

Derivation

ε ∂∂

ε ∂∂

ε ∂∂

x

y

z

u

xu f y z

v

yv g x z

w

zw h x y

= = ⇒ = ( )

= = ⇒ = ( )

= = ⇒ = ( )

0

0

0

,

,

,

∂τ∂

γ ∂∂

∂∂

∂∂

∂∂

∂∂

θxyxyx

u

y

v

x

f

yy z

e

xx z l y z

e

xx= ⇒ = + = ( ) + ( ) = ( ) ⇒ ( ) = − =0 , , constant

γ ∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

∂∂

yz

v

z

w

y

g

zx z

h

yx y

g

zx z

h

yx y

g

zx z

h

yx y e x

g x z e x z

h x y e x y

= + = ( ) + ( ) = ⇒ ( ) = − ( )

⇒ ( ) = − ( ) = ( ) ⇒( ) = ( )( ) = − ( )

, , , ,

, ,,

,

0

u f y z

v xz

w xy

= ( )= −=

,

θθ

τ γ ∂∂

∂∂

∂∂

θ

τ γ ∂∂

∂∂

∂∂

θ

xy xy

xz xz

G Gu

y

v

xG

f

yz

G Gu

z

w

xG

f

zy

= = +

= −

= = +

= +

Saint-Venant’s Stress Function

τ ∂∂

τ ∂∂xy xzz y

= = −Φ Φand

Compatibility Equation

d

ds

Φ Φ= =0 0that is on the boundary

Twisting Moment

M z y dA dA GJx xy xzA A

= − +( ) = =∫ ∫τ τ θ2 Φ

torsional rigidity

∂∂

∂∂

θ2

2

2

2 2Φ Φ

y zG+ = −

Derivation (1)

Stress functions satisfy the equilibrium equation

∂τ∂

∂τ∂

∂∂

∂∂

∂∂

∂∂

xy xz

y z y z z y+ =

+ −

=Φ Φ0

Also, on the boundary ∂∂

∂∂

∂∂

τ τΦ Φ Φ

Φs

ny

nz

n nz y z xz y xy= − + = + =

⇒ =

0

constant on the boundary

wheren =

0

unit vector normal to the boundaryyn

nz

=

Compatibility Condition

∂∂

τ γ ∂∂

∂∂

∂∂

θ

∂∂

τ γ ∂∂

∂∂

∂∂

θ

∂τ∂

∂τ∂

θz

G Gu

y

v

xG

f

yz

yG G

u

z

w

xG

f

zy

z yG

xy xy

xz xz

xy xz

= = +

= −

= = +

= +

⇒ − = −2

Derivation (2)

M z y dAz

zy

y dA

n n dsz

z

y

ydA

dA

x xy xzA A

z yA A

A

= − +( ) = − −

= − +( ) + +

=

∫ ∫

∫ ∫

∫

∂

τ τ ∂∂

∂∂

∂∂

∂∂

Φ Φ

Φ Φ Φ Φ

Φ2

Twisting moment, i.e. moment about the x-axis

Here we applied the divergence ( Green’s ) Theorem

fg

ydA fgn ds g

f

ydA

fg

zdA fgn ds g

f

zdA

A yA A

A zA A

∂∂

∂∂

∂∂

∂∂

∫ ∫ ∫

∫ ∫ ∫

= −

= −

∂

∂

andΦ = 0 on the boundary

Variational PrincipleBoundary Value Problem

∂∂

∂∂

θ2

2

2

2 2

0

Φ Φ

Φy z

G+ = −

= on the boundary

is equivalent to the minimization problem

min= onΦ

Φ Φ Φ0

2 212

2∂

∂∂

∂∂

θA A Ay z

dA G dA

+

+∫ ∫

which corresponds to the principle of the minimumcomplementary energy

Torsional constant J of an arbitrary shaped cross sectioncan be obtained by solving this approximately

Special Case (1)Elliptical Cross Section

Φ = + −

⇒ = −+

cy

a

z

bc

a b G

a b

2

2

2

2

2 2

2 21θ

Ja b

a b=

+π 3 3

2 2

stress function

torsional constant

τπmax = = ± <2

2M

abz b b ax at ifmaximum shear stress

Saint-Venant’s approximation

JA

Ip≈ 0 025 4.

Special Case (2)

Thin-Walled Open Cross Section

∂∂

∂∂

θ θ2

2

2

2

2

22 2Φ Φ Φ

y zG

d

dzG+ = − ⇒ = −

Φ = − −

G ztθ 22

4

τ τ ∂∂

θxz xy zG z= = = −0 2and

Φ

J bt= 13

3

z

y

b

t

Thin-walled Sections

J b tM

J

Jt

J

M t

Ji ii

n

i

xi

i

i

x i= ( ) =( )

=( )

=∑1

33

1

& max

maxmaxτ

Stress Concentration

At the corners, we makeround (fillets), and thenthe maximum shear stressmust be corrected :

τ τf

t

r= max .1 743

Trefftz (1922)

where r is the radius of the fillet. Concentration becomesless severe when the angle is greater than a right angle.

Exercise

Evaluate and the section shown.maxJ τ

Rectangular Cross Section

Assuming the solution form

Φ y z

a my

bn

z

t

yb

zt

a y z

mnm n

mnm n

m n

,

cos cos,

,

( ) =

−( ) −( )

−

−

=

∞

− −

=

∞

∑

∑

2 12

22 1

22

4 4

1

22

22

1 1

1

π π

we can obtain

Jbt

b

b

tc bt

M t

J

nb

tn

cM

btx

n

x

≈ −

=

≈ −+( )

+( )

==

∞

∑

3

5 13

2 20

2 2

31

1922

18 2 1

2 1

ππ

τπ

π

tanh

max

sech

Constants c1 and c2

0 20 40 60 80 1000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

b/t

c1 a

nd c

2

Open to Closed

J rt I trp≈ ⇒ ≈13

2 23 3π π

The shearing stress is uniform over the thickness of the wall and is directedtangent to the boundary curve s describing the center line of the wall.

Closed Cross Sections (1)

Force Equilibrium yields

τ ττ

1 1 2 2t x t x

q t

q

∆ ∆=⇒ = =

=

constant

shear flow

M qr s ds qx = ( ) =∫ 2Ω

Ω = ( ) =∫12

r s ds total area enclosed

by the center line of the tube

Closed Cross Section (2)

Now we shall look at the deformation. Assuming that the wallis really thin, we have two non-zero displacements :

(1) warping in the x direction (2) tangential component in the s direction η θ= r x

τ γ ∂∂

∂η∂

∂∂

θxsxsG

u

s x

u

sr= = + = +

τ ∂∂

θ θ θxs

Gds

u

sr ds du rds∫ ∫∫ ∫= +

= + = 2Ω

Integrating along the closed curve

Closed Cross Section (3)

Noting that τxs

q

t=

q

Gtds

q

G tds∫ ∫= ⇒ =2

21Ω

Ωθ θ

we have

Thus the torsional constant J can be defined by

JM

Gt

ds

x= =∫θ41

2Ω

J

tds

b ti ii

n

= +∫

∑=

41

13

23

1

Ω

Exercise 1

(a) valuate the torsional constant and themaximum stress for the closed thin-walledtube shown.

(b) A longitudinal cut is made along thedashed line indicated so that the clrosssection is open. Calculate J and Tmax.

(c) If the allowable stress is 10,000psi,compute the maximum torque that eachof the sections in (a) and (b) can carrysafely.

This can be extended tothe cross sections in a car body

Exercise 2

Evaluate J and Tmax for the section shown.

A Special Note

For closed cross section case, we may apply thevariational principle to find out the torsional constant J :

min= onΦ

Φ Φ Φ0

2 212

2∂

∂∂

∂∂

θA A Ay z

dA G dA

+

+∫ ∫

for . After solving this we computeGθ = 1

J dAA

= ∫2 ΦΦ = constant

Φ = constantΦ = constant

Φ = constant

Φ = constant

Φ = 0

Multiple Cells (1)

In this case, the problem becomesstatically indeterminant. That is,we must consider displacement.

M qx i ii

n

==∑2

1

Ω

θ jj

j s i s k sGq

ds

tq

ds

tq

ds

tj ji jk

= − −

∫ ∫ ∫1

2 Ω

θ jj

j s l sl

j

Gq

ds

tq

ds

tj jl

m

= −

∫ ∫∑

=

12 1Ω

Twist of angles of sections

Multiple Cells (2)

Noting that although cross sections warp, they do not distort in their own plane, that is, the entire cross sectionand each cell rotate at the same rate of twist :

θ θ θ θ1 2= = = =...... n

That is, solving the n+1 equations :

M q

Gq

ds

tq

ds

tj n

x i ii

n

jj s l s

l

j

j jl

m

=

= −

=

=

=

∑

∫ ∫∑

2

12

1 2

1

1

Ω

Ωθ , , ,....,

we can determine

q q qn1 2 ..... θ

Example

b b b b

Ω1 = 3b2 Ω2 = 2b 2 Ω3 = π2

b2

θ

θ

θ

= −

= + −

= − −

= − −

=

∫ ∫

∫ ∫ ∫

12

16

4 2 2 2

12

14

6 2 2

12

11 2 2 1 2

22 1 3 2 2 1 3

1 12

2 21 23

Gq

ds

tq

ds

t Gbq

b b

tq

b

t

Gq

ds

tq

ds

tq

ds

t Gbq

b

tq

b

tq

b

t

G

s s

s s s

Ω

Ω

ΩΩ33 2 2 3 2

3 32

1 2 2q

ds

tq

ds

t G bq

b r

tq

b

ts s∫ ∫−

= + −

π

π

M q q q b q b q b qx = + +( ) = + +

2 2 3 2

21 1 2 2 3 32

12

22

3Ω Ω Ω π

Solving these q q qM

bt

1 2 3 20 0746 0 0851 0 0676 = . . .

θ = 0 0566 3.M

Gtbx J

Gq tbj j

j

n

= ==

∑217 6984

1

3

θΩ .

MATLAB

»A=[4+2*sqrt(2),-2,0;-2,6,-2;0,-2,2+pi]A = 6.8284 -2.0000 0 -2.0000 6.0000 -2.0000 0 -2.0000 5.1416

»b=[6;4;pi]b = 6.0000 4.0000 3.1416

»inv(A)*bans = 1.3196 1.5054 1.1966

That is, we have

q Gtb

q Gtb

q Gtb

1

2

3

1 3196

1 5054

1 1966

===

.

.

.

θθθ

M b q q q Gtbx = + +( ) =21 2 3

36 4 17 6984π θ.

J tb= 17 6984 3.

Exercise

The hollow section shown is designed to withstand a maximum shearing stressof 4,000 psi. The wall thickness is a constant.

(a) Find the largest twisting moment that can besafely carried by the section and the angle of twistdeveloped in a 100-in length

(b) Repeat (a) assuming that both cells are closed.

(c) Compare the two solutions.

t

a a

a