Takehome Theory of structures

Z 1

− −71.34 ·

Ψ = ρ ds = s

1 1 1

0 √ s

Z 2

−71.34 · − −10088.86 − ·

Ψ = d 2 ρ ds = 30.89 s

2 2 2

0 s

Z 3 −14413.27 − ·

−10088.88 − · − ρ ds = 71.34 s

Ψ = 30.89 h 3 3

3 0

h o

− − · −170 ·

x = d + s sin(45 ) = + 0.71 s

s1 1 1

2

h −70

− + s = + s

x = 2 2

s2 2

h o

· ·

x = + s sin(45 ) = 70 + 0.71 s

s3 3 3

2 Z

− · ·

Ψ y dA

s

A

x = =0

C I x

Z · ·

Ψ x dA

s

A

y = =

C I

y

√ √

d 2 h d 2

Z Z Z

· · · · · · · · ·

Ψ x b ds + Ψ x b ds + Ψ x b ds

1 s1 1 1 2 s2 2 2 3 s3 1 3

0 0 0

= =

I

x

√

"Z d 2

1 · · · · ·

·

= (−71.34 s ) (−170 + 0.71 s ) 4 ds +

1 1 1

6

·

18.378 10 0 h

Z − · · · ·

+ (−10088.86 30.89 s ) (−70 + s ) 5 ds +

2 2 2

0 √ #

d 2

Z − · · · · ·

+ (−14413.27 71.34 s ) (70 + 0.71 s ) 4 ds =

3 3 3

0

√

"Z d 2

1 21

· · − · ·

= (48510.55 s 201.78 s ) ds +

1 1

6

·

18.38 10 0 2

h

Z 22

· − · ·

+ (−154.44 s 39633.29 s + 3531102.16) ds +

2 2

0 √ #

d 2

Z 23

· − · − ·

+ (−201.78 s 60741.82 s 4035716.15) ds =

3 3

0 √ √

1 h 2 3

· · − ·

24255.27 (d 2) 67.26 (d 2) +

= 6

·

18.38 10 3 2

−51.48 · − · ·

h 19816.65 h + 3531102.16 h +

√ √

√ i

3 2

− · − ·

−67.26 · 2) 30370.91 (d 2) 4035716.15 d 2 =

(d

1 6 6 6

· · −60.34

− · − ·

= 294.87 10 = mm

35.32 10 1368.39 10

6

·

18.378 10

−y − −

e = e = 60.34 30.89 = 29.45 mm

C C G

Now let’s calculate the warping function about the shear center to calculate the warping rigidity.

s

Z

∗ ∗

−

Ψ = Ψ ρ ds

0 0 h

∗ ∗ o o

· − ·

ρ > 0 ; ρ = cos(45 ) e sin(45 ) = 28.68 mm

C

1 1 2

∗ ∗ −e −29.45

ρ < 0 ; ρ = = mm

C

2 2

∗ ∗ ∗

ρ > 0 ; ρ = ρ = 28.68 mm

3 3 1

s

Z 1

∗ ∗

− − ·

Ψ = Ψ ρ ds = Ψ 28.68 s

0 0 1

1 1

0 √ s

Z 2 ∗

∗ − − ·

− · 2 ρ ds = Ψ 4055.31 + 29.45 s

Ψ = Ψ 28.68 d 0 2

0 2

2 0 s

Z 3

∗ ∗

− · − − ·

Ψ = Ψ 4055.31 + 29.45 h ρ ds = Ψ + 67.27 28.68 s

0 0 3

3 3

0

∗

To find Ψ , the average value of Ψ must be zero on the whole area.

0 √ √

d 2 h d 2

Z Z Z Z

∗ ∗ ∗ ∗

⇒ · · · · · · ⇒

Ψ dA = 0 Ψ b ds + Ψ b ds + Ψ b ds = 0

1 1 2 2 1 3

1 2 3

A 0 0 0

√ √

d 2 h d 2

Z Z

Z −28.68·s −4055.31+29.45·s ⇒

(Ψ )·4·ds + (Ψ )·5·ds + (Ψ +67.27−28.68·s )·4·ds = 0

0 1 1 0 2 2 0 3 3

0 0 0

3

√ √ √ √ √

2 2 2

−57.35·(d −20276.53·h+4·d −57.35·(d ⇒

4·d 2·Ψ 2) +5·h·Ψ +73.62·h 2·Ψ 2) +269.06·d 2 = 0

0 0 0

· − · − · − ⇒

565.69 Ψ 1147013.77 + 700 Ψ 1395814.25 + 565.69 Ψ 1108962.73 = 0

0 0 0

2

· − ⇒

1831.37 Ψ 3651790.75 = 0 Ψ = 1994.02 mm

0 0

∗

Now we can recover the expression of Ψ .

∗ − ·

Ψ = 1994.02 28.68 s

1

1

∗ − · · −

Ψ = 1994.02 4055.31 + 29.45 s = 29.45 s 2061.29

2 2

2

∗ − · − ·

Ψ = 1994.02 + 67.27 28.68 s = 2061.29 28.68 s

3 3

3

Warping rigidity. √

√ 2 h d 2

d Z Z

Z Z ∗2 ∗2 ∗2

∗2 · · · · · ·

Ψ b ds + Ψ b ds + Ψ b ds =

Γ= Ψ dA = 1 1 2 2 1 3

1 2 3

0 0 0

A √ √

d 2 2

h d

Z Z Z

2 2

2

·b ·ds ·b ·ds

−2061.29) ·b ·ds

= (1994.02−28.68·s ) + (2061.29−28.68·s ) =

(29.45·s +

1 1 1 3 1 3

2 2 2

0 0

0

√ √

√ 3 2 6 3 2 6

− · · · · − · · ·

· 2) 228716.88 (d 2) + 15.9 10 d 2 + 1445.20 h 303492.78 h + 21.24 10 h +

= 1096.37 (d √ √

√ 3 2 6

− · · ·

· 2) 236432.31 (d 2) + 17 10 d 2 =

+1096.37 (d 6

6 6

· ·

= 775.89 10 + 991.41 10 + 775.89 = 2543.19 mm

Torsional stiffness.

√

1

31 32 4

· · ·

J = 2 d 2 b + h b = 11867.31 mm

3

Young and Shear modulus. 2

E = 210000 MPa = 210000 N/mm

E 210000 2

G = = = 80769.23 N/mm

·

2 (1 + ν) 2 (1 + 0.3)

DIFFERENTIAL EQUATION.

The structure behavior is governed by:

00

IV −

EΓθ GJθ = m , and doing

t

r

r 4

GJ 11867.31 mm −1

α = = = 0.00133968 mm , remains:

6

·

EΓ 2.6 2543.19 mm 4

m

00 t

IV 2

−

θ α θ = EΓ

where: ·

1200 N m/m 4

· · ·

m = z (mm) = z N mm/mm , and the equation remains:

t 4500 mm 15

4

00

IV 2

−

θ α θ = z

15EΓ

the complete solution is in the form:

θ = θ + θ

h p

where θ is a particular solution of the inhomogeneous equation, and θ is the solution of the homogeneous

p h

equation.

PARTICULAR SOLUTION.

4

00

2

IV − α θ = z

θ p p 15EΓ

we will look for a solution of the form:

3 4 3

θ = z (az + b) = az + bz

p 000

00

0 IV

2

3 2 = 24a

= 24az + 6b , θ

= 12az + 6bz , θ

= 4az + 3bz , θ

θ p

p

p

p

replacing:  24a = 0 ( a =0



4 2

 −12α a = 0

2 2 −2

− ⇒

⇒

24a α (12az + 6bz) = z 4 b =

15EΓ 2

−6α b = 2

 45α EΓ

 15EΓ

and the particular solution will be:

−2 3

θ = z

p 2

45α EΓ

HOMOGENEOUS SOLUTION.

00

IV 2

−

θ α θ = 0

h h  D = α



4 2 2 2 −α

− → − → D =

(D α D )θ = 0 D (D + α)(D α) = 0

h D = 0 (double root)



then the homogeneous solution will be of the form:

−αz −αz

αz 0z αz

θ = Ae + Be + (Cz + D)e = Ae + Be + Cz + D

h

COMPLETE SOLUTION

The complete solution will be θ = θ + θ , that is:

h p 5

2

−αz

αz 3

−

θ = Ae + Be + Cz + D z

2

45α EΓ

the boundary conditions given by the problem, since the supports do not allow any rotation and there are no

external bimoments applied at the ends, are:

00 00

θ(0) = 0 , θ (0) = 0 , θ(L) = 0 , θ (L) = 0

deriving θ: 6

0 −αz

αz 2

− − z

θ = αAe αBe + C 2

45α EΓ

12

00 −αz

2 αz 2 −

θ = α Ae + α Be z

2

45α EΓ

and therefore:

⇒

θ(0) = 0 A + B + D =0

00 2 2

⇒ ⇒

θ (0) = 0 α A + α B = 0 A + B =0

2

−αL 3

αL ⇒

⇒ − L = 0

θ(L) = 0 Ae + Be + CL + D 2

45α EΓ

⇒ 415.1127A + 0.0024B + 4500 C + D = 4.2253

12

00 −αL

2 αL 2

⇒ − ⇒

θ (L) = 0 α Ae + α Be L =0

2

45α EΓ

⇒ 745.018A + 0.00432B = 1.2519

and solving: −4

−0.00168 ·

A = 0.00168 B = C = 7.8394 10 D =0

and the complete solution is: −0.00134z −4 −11

0.00134z 3

− · − ·

θ = 0.00168e 0.00168e + 7.8394 10 z 4.6368 10 z

6

VLASOV AND SAINT VENANT TORSIONAL MOMENTS, BIMOMENT.

0 −6 −6 −0.00134·z −10 −

0.00134z 2

· · · · − · · ·

θ = 2.251 10 e + 2.251 10 e 1.391 10 z + 7.839 10 4

00 −9 −9 −0.00134·z −10

0.00134z

· · − · · − · ·

θ = 3.0159 10 e 3.0159 10 e 2.782 10 z

000 −12 −12 −0.00134·z −10

0.00134z

· · · · − ·

θ = 4.04 10 e + 4.04 10 e 2.782 10

The Saint Venant torsional moment is:

0 −0.00134·z

0.00134·z 2

· · − ·

M = GJθ = 2157.83 e + 2157.83 e 0.133 z + 751417.38

t1

The Wagner-Vlasov torsional moment is:

000 −0.00134·z

0.00134·z

−EΓθ −2157.83 · − ·

M = = e 2157.83 e + 148582.62

t2

The total torsional moment, or torque, can be obtained as the sum of the two preceding:

2

−0.1333z

M = M + M = + 900000

t t1 t2

The bimoment is:

00 −0.00134·z

0.00134·z

· − · − ·

B = EΓθ = 1610708.716 e 1610