Esercizi Theory of structures

C

ÒØ 2 ∙ 10 ∙ 3.18 ∙ 10 Ùœœ œœ

j -

 + 1 1 1  − 1

ÃÄ ÃÄ

Ï(‡›) = D−1 + ‡› E = 6.790 ; Õ(‡›) = D − E = 0.372

−1 + 1

 2 ‡› Â

ÃÄ ÃÄ

œ › Õ(‡›) 5ÚÙ ∙ 25œ 0.372

ÊÇ = ∙ = ∙ = 3.42ÚÙ ∙ œ

‘

2 Ï(‡›) 2 6.790

ÊÇ 3.42 ∙ 10 œ 5 ∙ 10

Ùœœ Ùœœ

H

= = 2.22 ∙ 10 =

‘

C (1.56)

¦ 8 ∙ 10 ∙ 1.93 ∙ 10 Ùœœ ‡ ¦ ∙ 10 ∙ 8 ∙ 10 ∙ 1.93 ∙ 10 Ùœœ

- C -

= 1.33 ∙ 10

COMPUTING THE INTERNAL ACTIONS

ÊÇ Â ÃÄ

LEFT SPAN

¾ É = ∙ = 9.17 ∙ 10

C

¦ Â − 1

Æ

¼ ÃÄ

ÊÇ

¼ Â ÃÄ

Ê = − ∙ = −9.17 ∙ 10 °

C

¦ Â −1

Æ ÃÄ

½ ÊÇ

1

¼ Å = − ∙ = −4.43 ∙ 10 C

› ¦

¼ Æ

» Á = 0

Æ (¨) (¨) (¨)

q = q + q

RIGHT SPAN Ö

(¨)

q œ q

Here we have the superposition of two solutions: ‘ ª

(¨)

q is the solution corresponding to the beam doubly supported with distributed distribuito ( )

is the solution corresponding to the beam doubly supported with concentrated bimoment at the

Æ

left end z=0. It can be inferred from the solution (q ) with the same restraints but bimoment located at

(¨) (›

q = q − ¨) → q

the end z=l, due to the symmetry of the function.

Æ

(¨) (›

q = −q − ¨) → q

z z z

is a symmetric function

Æ

(¨) (›

q = q − ¨) → q

zz zz zz is a anti-symmetric function

Æ

(¨) (›

q = −q − ¨) → q

zzz zzz zzz

is a symmetric function

Æ

œ 1 is a anti-symmetric function

= ∙ = 5.5155 ∙ 10

¾É ‘ C

Â

‡ ¦ + 1

ª ÃÄ

¼

¼ œ 1

¼ Ê = ∙ Ó1 − Ô = 0.0133

‘

‡ ¦ Â +1

ª °

ÃÄ

œ › 1

½ Å = = 8.0959 ∙ 10

‘ Cj

¼ 2¦ œœ

ª

¼ œ

¼ Á = − = −0.0133

‘

» ‡ ¦

ª

PLOTS

The torsion is null on the flexural-torsional restraints, and in the left span the beam rotates in opposite

£ = £ + £

direction with respect to the distributed torsional moment, then in the same direction in the right span

‘ ‘ ‘ = œ

©ª

The torsional moment is constant in the left span and linear in the right span. This

« ‘

©š .

result is in accordance with the external loads, since

PROBLEM 3

£ = 2000ÚÙœ = 2 ∙ 10 Ùnœ

,

The thin-walled multi-cell section in the above figure is loaded with the following torsional moment

‘

¦ = 18000£¤¥

The shear modulus of the material is Û = • ∙ ˜(p)

š™

τ , we assume a shear tension flux constant on

In order to compute the shear tensions zs

each cell. In the portions of material belonging to two adjacent cells flow the fluxes of the two adjacent

cells; therefore in the compatibility equations the flux in these portions will be equal to the algebraic

sum of the two fluxes coming from the two adjacent cells (thus if these two fluxes have opposite

direction, they will be subtracted).

-

The equilibrium equations are

¾ £ = Ü 2Û Ω → ÂÛßr›r˜àrßœ

¼ ‘ Ý Ý °

½ Û

á ·p = 2Ω Ï¦ → noœâ¥¯r˜r›r¯v

¼ ˜(p) Ý

»

For the section in figure, these become:

£ = 2Û Ω + 2Û Ω + 2Û Ω + 2Û Ω

¾ ‘ - -

¼ ·p ·p ·p

Û á − Û : −Û : = 2Ω Ï¦

¼ ˜(p) ˜(p) ˜(p)

¼

¼

¼ ·p ·p ·p ·p

¼ −Û : +Û á − Û : −Û : = 2Ω Ï¦

˜(p) ˜(p) ˜(p) ˜(p)

- °

½ -

·p ·p ·p ·p

¼

−Û : − Û : + Û á −Û : = 2Ω Ï¦

¼ ˜(p) ˜(p) ˜(p) ˜(p)

-

¼

¼ -

¼ ·p ·p ·p

−Û : −Û : +Û á = 2Ω Ï¦

¼ ˜(p) ˜(p) ˜(p)

- -

» - - -

We therefore have 5 equations with 5 variables, i.e. the fluxes q and the angle of twist per unit length

i

β.

Considering the symmetry of the section, we can simplify the computation of the coefficients as

follows: ·p ·p ·p ·p ·p ·p

Ω = Ω , á =á , : = : ,: = :

˜(p) ˜(p) ˜(p) ˜(p) ˜(p) ˜(p)

- - - -

1

Ω = Ω = ∙ 350 ∙ 180 nœ = 31500nœ

2

-

Ω = 500 ∙ 90nœ = 4500 nœ = Ω = 45000nœ + 180 = 393.57nœ

√350

Length of the inclined portion of the section:

·p ·p 393.57 180 350

á =á = + + = 50.94

˜(p+ ˜(p+ 15 25 20

-

·p ·p 90

: = : = = 3.6

˜(p+ ˜(p+ 25

-

·p ·p 90

: = : = = 3.6

˜(p+ ˜(p+ 25

-

·p 500 90 500 90

á = + + + = 65.53

˜(p+ 15 25 20 25

·p 500 90 500 90

á = + + + = 73.87

˜(p+ 15 25 15 25

·p 500

: = = 33.33

˜(p+ 15 Û

31500 45000 45000 31500 0 10

,

å è

å è

å è Û

50.94 −3.6 −3.6 0 −2 ∙ 31500 ∙ 18000 0

ä ç

ä ç ä ç

Û

−3.6 65.53 −33.33 −3.6 −2 ∙ 45000 ∙ 18000 ∙ = 0

ä ç ä ç ä ç

Û

−3.6 −33.33 73.87 −3.6 −2 ∙ 45000 ∙ 18000

ä ç ä ç ä 0 ç

-

ã 18000æ ã æ

Ï ã æ

0 −3.6 −3.6 50.94 −2 ∙ 31500 ∙ 0

Ù

Û = 459

¾ nœ

¼ Ù

¼ Û = 821.7

¼ nœ

¼ Ù °

Û = 757.8 nœ

½ Ù

¼ Û = 459

¼ nœ

-

¼ 1

¼

Ï = 1.56 ∙ 10 Cj

» nœ • = êℎÂà • ¥à ÂuâàÂpp· ¥p

• ž

š™ é(™) ¡Ÿ h

We now compute the tangential tensions

LOOP of cells 1 e 4

Û

• = = 30.6

15 nœ

Û

• = = 18.36

25 nœ

Û

• = = 22.95

20 nœ

LOOP of cell 2

Û

• = = 54.78

15 nœ Û

• = • = = 32.87

25 nœ

-

Û

• = = 41.09

20 nœ

LOOP of cell 3

Û

• =• = = 50.52

15 nœ

Û

• = • = = 30.31

25 nœ

-

1 − 2: • = 18.36 − 32.87 = −14.51 (p¥œÂ ¥p ¯ℎ âoà¯ros ˜Â¯êÂÂs n››p 2 − 4)

Portions common to two adjacent cells

1 − 3: • = 18.36 − 30.31 = −11.95 (p¥œÂ ¥p ¯ℎ âoà¯ros ˜Â¯êÂÂs n››p 3 − 4)

2 − 3: • = 54.78 − 50.52 = 4.26

PROBLEM 4

The thin walled multi-cell section in the figure above has a uniform thickness of 40 mm. The

dimensions in the figure are in mm. Find the position of the shear center.

Solution

É = >400 + 1200 + 400 + 500 ∙ 4 + 400 ∙ 3 + 400√2 ∙ 2B ∙ 40 = 253.2548 ∙ 10 œœ

Geometric quantities u = 0 œœ

³

c400 c400

ìc500 ∙ 4 ∙ 250d ∙ 4 + ∙ 40 ∙ 500d ∙ 2 + ∙ 40 ∙ 900d + `400√2 ∙ 40 ∙ 700b ∙ 2í

v = 253.2548 ∙ 10

³ v = 324.1 œœ

³

1 1 1

(40 + (40 + (500 +

î = ∙ ∙ 2000 + ∙ 400 + ð ∙ 40 + 500 ∙ 40 ∙ 1000 ñ ∙ 2

12 12

12

ï 1 1

(500 + (40 +

+ ð ∙ 40 + 500 ∙ 40 ∙ 600 ñ ∙ 2 + ð ∙ 400 + 40 ∙ 400 ∙ 800 ñ ∙ 2

12 12

1

+ð 40 ∙ >400√2B ! + 40 ∙ 400√2 ∙ 400 ñ ∙ 2;

24 î = 1.100415 ∙ 10 œœ -

ï

We compute now the shear tension fluxes q* for the section assumed without closed loops as in the

(p) (p)

Û = ∙ ô

-

following figure. V is the shear force.

∗ ∗

¬

ó Jourawski formula

(p)

ô →

∗

Where:

î → first moment of area

ï second moment of area with respect to x axis

p (p ) (p cœœ d

ô = 20 ∙ p ∙ − 2000)

∗

Line :

p (p ) d

cœœ

= −24000 ∙ p

ô ∗

Line :

p (p ) cœœ d

ô = −40000 ∙ p

∗

:

Line p

- (p ) (p ) cœœ d

ô = 20 ∙ − 2000 ∙ p − 10

∗

:

Line - - -

p

j (p ) cœœ d

ô = 10√2 ∙ p − 24000 ∙ p − 448 ∙ 10

Line : ∗ j

j j j

p (p + cœœ d

ô = 20 ∙ >p − 400 ∙ p − 320000 ∙ − 2240000B

√2

Line : ∗

p

" (p + cœœ d

ô = 10√2 ∙ p + 8000