Formulary ”Fundamentals of Lightweight Design”

2004-10-11

1. General formulas

• standard box for the determination of cutting forces:

• differential equations for the equilibrium of beams:

dQz

dx= −p(x)

dMy

dx= Qz

• degree of redundancy:

– beam structures:U = A − 3 n + V

A: number of reaction forcesn: number of beamsV : valence of joint

– trusses / stiffened shear webs

U = n + A − 2ke (two-dimensional)

U = n + A − 3kr (three-dimensional)

n: number of bars (trusses) or number of bars and panels (stiffened shear webs)A: number of reaction forceske: number of two-dimensional nodeskr: number of three-dimensional nodes

1

2. Bending of beams

• statical moments of an area:

Sy =

∫

A

z dA Sz =

∫

A

y dA

• moments of inertia of an area:

Iy =

∫

A

z2 dA Iz =

∫

A

y2 dA

• product of inertia of an area:

Iyz =

∫

A

yz dA

• stress distribution for an arbitrary coordinate system in the center of gravity:

σx(z, y) =N

A+

MyIz + MzIyz

IyIz − I2yz

z − MzIy + MyIyz

IyIz − I2yz

y

• stress distribution for a system of principal axes in the center of gravity:

σx(z, y) =N

A+

My

Iyz − Mz

Izy

• coordinate transformation for a rotation of the coordinate system about ϕ:

z = −y′ sin ϕ + z′ cos ϕ

y = y′ cos ϕ + z′ sin ϕ

• theorem of Steiner:

Iη = Iy′ + ζ2

s A Iζ = Iz′ + η2

sA Iηζ = Iy′z′ + ηsζsA

• the moments and the product of inertia for the rotation of the coordinate system:

Iy(ϕ) =1

2

(

Iy′ + Iz′)

+1

2

(

Iy′ − Iz′)

cos 2ϕ − Iy′z′ sin 2ϕ

Iz(ϕ) =1

2

(

Iy′ + Iz′)

− 1

2

(

Iy′ − Iz′)

cos 2ϕ + Iy′z′ sin 2ϕ

Iyz(ϕ) =1

2

(

Iy′ − Iz′)

sin 2ϕ + Iy′z′ cos 2ϕ

• location of the principal axes:

tan 2α = − 2Iy′z′

Iy′ − Iz′

• differential equation of the deflection line (valid only for a system of principal axes):

w′′′′IyE = p(x) ⇒ w′′′ = − Qz

IyE⇒ w′′ = − My

IyE

2

3. Plastic bending

• in the case of pure bending it applies for:

– ideal plastic material behaviour:

Mplastic = K · Melastic

– Cozzone’s approximation of real plastic material behaviour:

Mplastic =

[

1 + (K − 1)σo

σM

]

Melastic

• K depends on the shape of the cross section:

– rectangular solid cross section: K = 1.5

– circular solid cross section: K = 1.698

• definitions:

ν =N

A · σBµ =

M

W · σB

• interaction curves:

– elastic material behaviour:ν + µ = 1

– ideal plastic material behaviour (valid only for a rectangular cross section):

µ = 1.5(

1 − ν2)

4. Transverse shear - open thin-walled cross section

• shear flow formula:

t =QzSy

Iy

• shear center:

∫

s

t(s)ρ(s)ds = Qz · ey

3

5. Transverse shear - closed thin-walled cross section

• Bredt’s first formula:Mt = 2t0A

• calculation of shear flow :

– superposition of shear flow of the open cross section t′

and the constantshear flow t0:

– determination of the constant shear flow by using equality of twisting moments:

Qza =

∮

t′

ρds + 2At0

• determination of shear center :

– determination of constant shear flow t0,SMP :

ϑ′

= 0 ⇒∮

t′

+ t0,SMP

Ghds = 0 ⇒ t0,SMP = −

∮

t′

Ghds

∮

dsGh

– equality of twisting moments yields the shear center e:

Qze =

∮

t′

ρds + 2t0,SMP A

4

6. Saint Venant’s torsion

• definitions:

· rate of the twist angle / torsional moment of an area :

ϑ′

=T

GIT(IT 6= Ip)

· maximum shear stress / torsional section modulus:

τmax =T

WT

• closed thin-walled cross sections:

– Bredt’s first formula:Mt = 2t0A

– Bredt’s second formula :

GIT =4A2

∮

dsGh

– torsional section modulus (hmin: minmum thickness of wall):

WT = 2Ahmin

• prismatic bars with solid cross section:

– circular solid section:IT =

π

2R4 WT =

π

2R3

– elliptical solid section:

IT = πa3b3

a2 + b2

WT =πab2

2

– rectangular solid section:

IT = η1

B3H

3WT = η2

B2H

3

H/B η1 η2

1 0.4217 0.62451.25 0.5152 0.66361.5 0.5873 0.69292 0.6860 0.73763 0.7900 0.80164 0.8424 0.84505 0.8745 0.87406 0.8951 0.89508 0.9212 0.921210 0.9370 0.9370∞ 1.0000 1.0000

5

• approximations:

– narrow rectangular section:

IT =1

3B3H

WT =B2H

3

τ = τmax2 y

B

– open thin-walled cross section:

IT = η3

1

3

∑

i

h3

i li WT =It

hmax=

η3

3hmax

∑

i

h3

i li

Section shape L U Z T I X

η3 0.99 1.12 1.12 1.12 1.30 1.17

– thick-walled hollow section:

ϑ′ = ϑ′

B = ϑ′

i

T = TB +∑

i

Ti

⇒ IT = ITB+

∑

i

ITi

τi =3Ti

η2ih2

i liτBi =

TB

2HBhi

6

7. Warping torsion

• twisting moments of Saint Venant’s torsion MSV and of bending torsion MB:

MSV = ϑ′GIT MB = −ϑ′′′CT E

• differential equation of warping torsion:

d3ϑ

dξ3− α2

dϑ

dξ= −µ

with

ξ =x

lα2 =

GIT l2

ECTµ =

T l3

ECT

• solution for a clamped cantilever beam subjected to a constant twisting moment:

ϑ = A1 + A2 cosh αξ + A3 sinhαξ +µ

α2ξ

withA1 = −A2 = − µ

α3tanh α A3 = − µ

α3

8. Stiffened shear web with parallel flanges

• shear flow:

t =Q

a

• shear center:

e =2A

a

9. Two-dimensional stiffened shear webs

• rectangular webs:

t = t1 = t2 = t3 = t4 = const.

• parallelogram webs:

t = t1 = t2 = t3 = t4 = const.

nx = 2t tan ϕ

7

• trapezoidal webs:

– definition of mean shear flows:

t =

∫ l

0t(x)dx

l

– determination of the mean shear flows:

t3t1

=

(

a1

a3

)2 t1t2

=t4t3

=a3

a1

t2 = t4

– the shear flows along the edges 1 and 3 are constant:

t1 = t1 = const. t3 = t3 = const.

– shear flow distribution along the edges 2 and 4:

t(x) = t3

(

a3

a(x)

)2

= t3

(

d

d + x

)2

t(x) = tma1a3

a(x)2tm =

√

t2t4 =√

t1t3

– longitudinal force in the bars 2 and 4:

N2(x) = N20 +t3

cos α

x

1 + xl

(

a1

a3− 1

) N4(x) = N40 −t3

cos β

x

1 + xl

(

a1

a3− 1

)

8