Takehome Theory of structures

Z

−

ψ = ψ ρ ds

0 0

since ψ does not influence the determination of the shear center, we will assume ψ = 0

0 0

We calculate the distances with sign from G to s , s and s .

1 2 3

ρ < 0 (s is clockwise from G)

1 1

ρ < 0 (s is clockwise from G)

2 2

ρ > 0 (s is counterclockwise from G)

3 3

−e −32.72

ρ = = mm

1 G

−75

ρ = mm

2

ρ = 75 mm

3 s

Z 1

− ·

ψ = ρ ds = 32.72 s

1 1 1

0 s

Z 2

· − ·

ψ = 32.72 75 ρ ds = 2453.71 + 75 s

2 2 2

0 s

Z 3

· − −2453.71 − ·

ψ = 32.72 (−75) ρ ds = 75 s

3 3 3

0

y = s

s1 1

y = 75

s2 −75

y =

s3 235 160 160

Z Z Z

Z · · · · · · · · ·

ψ y b ds + ψ y b ds + ψ y b ds

− · ·

ψ y dA 1 s1 2 1 2 s2 1 2 3 s3 1 3

s −235 0 0

A −

x = = =

C I I

x x

235 160

Z

−1 Z

· · · · · · · · ·

= (32.72 s ) s 8 ds + (2453.71 + 75 s ) 75 8 ds +

1 1 1 2 2

6

·

83.27 10 −235 0

160

Z − · · · ·ds

+ (−2453.71 75 s ) (−75) =

3 3

0

235 160

Z

−1 Z

· · · · · ·

= 261.73 s ds + 2 (1472225.06 + 45000 s ) ds =

1 1 2 2

6

·

83.27 10 −235 0

−1 −3887.57

6 6

· · · · −46.69

= 2264.45 10 + 2 811.56 10 = = mm

6

·

83.27 10 83.27

2

−x − −

e = e = 46.89 32.72 = 13.97 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 →

− →

− →

−

We calculate the distances with sign from C to s , s and s .

1 2 3

∗

ρ > 0 (s is counterclockwise from C)

1

1

∗

ρ < 0 (s is clockwise from C)

2

2

∗

ρ > 0 (s is counterclockwise from C)

3

3

∗

ρ = e = 13.97 mm

C

1

∗ −75

ρ = mm

2

∗

ρ = 75 mm

3 s

Z 1

∗ ∗ ∗ ∗

− − ·

ψ = ψ ρ ds = ψ 13.97 s

1

1 0 1 0

0 s

Z 2

∗ ∗ ∗ ∗

− · − − ·

ψ = ψ 13.97 75 ρ ds = ψ 1047.75 + 75 s

2

2 0 2 0

0 s

Z 3

∗ ∗ ∗ ∗

− · − − ·

ψ = ψ 13.97 (−75) ρ ds = ψ + 1047.75 75 s

3

3 0 3 0

0

∗ ∗

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

0 235 160 160

Z Z Z Z

∗ ∗ ∗ ∗

⇒ · · · · · · ⇒

ψ dA = 0 ψ b ds + ψ b ds + ψ b ds = 0

2 1 1 2 1 3

1 2 3

−235

A 0 0

235 160 160

Z Z Z

∗ ∗ ∗

− · · · − · · · − · · · ⇒

(ψ 13.97 s ) 8 ds + (ψ 1047.75 + 75 s ) 8 ds + (ψ + 1047.75 75 s ) 8 ds = 0

1 1 2 2 3 3

0 0 0

−235 0 0 ∗ 2

∗ ∗ ∗ ∗

· · · − ⇒ · ⇒ ψ = 0 mm

3760 ψ + (1280 ψ + 6338880) + (1280 ψ 6338880) = 0 6320 ψ = 0 0

0 0 0

0 ∗

Now we can recover the expression of ψ .

∗ 2

−13.97 ·

ψ = s mm

1

1

∗ 2

−1047.82 ·

ψ = + 75 s mm

2

2 3

∗ 2

− ·

ψ = 1047.82 75 s mm

3

3

Warping rigidity. 235 160 160

Z Z Z Z

∗2 ∗2 ∗2 ∗2

· · · · · ·

Γ= ψ dA = ψ b ds + ψ b ds + ψ b ds =

2 1 1 2 1 3

1 2 3

−235

A 0 0

235 160 160

Z Z Z

2 2 2

· · · · · · − · · ·

= (−13.97 s ) 8 ds + (−1047.82 + 75 s ) 8 ds + (1047.82 75 s ) 8 ds =

1 1 2 2 3 3

−235 0 0

−7 6

10 10 11 6 ·

· · · · 1.07 10 m

= 1.35 10 + 2 4.675 10 = 1.07 10 mm =

Torsional stiffness.

1 −7

3 3 5 4

· · · ·

J = 2 156 8 + 470 8 = 133461 mm = 1.33 10 m

3

Young and Shear modulus. 6

·

210 10 kPa

E = 210000 MPa = 6

·

E 210 10 6

·

G = 80.769 10 kPa

= =

·

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 −7 4

·

GJ 1.33 10 m −1

= = 0.69 m , remains:

α = −7 6

· ·

EΓ 2.6 1.07 10 m

m

00 t

IV 2

−

θ α θ = EΓ

where: ·

1.5 kN m/m · · ·

m = z (m) = 0.25 z kN m/m , and the equation remains:

t 6m 1

00

IV 2

−

θ α θ = z

4EΓ

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. 4

PARTICULAR SOLUTION.

1

00

IV 2

−

θ α θ = z

p p 4EΓ

we will look for a solution of the form:

3 4 3

θ = z (az + b) = az + bz

p 00 000

0 3 2 2 IV

θ = 4az + 3bz , θ = 12az + 6bz , θ = 24az + 6b , θ = 24a

p p p p

replacing:  24a = 0 ( a =0



1 2

 −12α a = 0

2 2 −1

−1

⇒ ⇒

− z

24a α (12az + 6bz) = 1 b = =

4EΓ 2

−6α b = 2

 24α EΓ 24GJ

 4EΓ

and the particular solution will be:

−1 3

θ = z

p 24GJ

HOMOGENEOUS SOLUTION.

00

2

IV − = 0

α θ

θ 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

1

−αz

αz 3

−

θ = Ae + Be + Cz + D z

24GJ

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: 3

0 −αz

αz 2

− −

θ = αAe αBe + C z

24GJ

6

00 −αz

2 αz 2 −

θ = α Ae + α Be z

24GJ

and therefore: 5

⇒

θ(0) = 0 A + B + D =0

00 2 2

⇒ ⇒

θ (0) = 0 α A + α B = 0 A + B =0

1

−αL −0.69·6

αL 3 0.69·6 3

⇒ − ⇒ · − · ⇒

θ(L) = 0 Ae + Be + CL + D L = 0 Ae + Be + C 6 + D 0.039 6 = 0

24GJ

⇒ −

62.8A + 0.016B + 6C + D 8.4 = 0

6

00 −αL −0.69·6

2 αL 2 2 0.69·6

⇒ − ⇒ − · ⇒

θ (L) = 0 α Ae + α Be L = 0 0.69 Ae + 0.69 0.023 6 = 0

24GJ

⇒ −

29.9A + 0.007B 0.14 = 0

and solving: −0.0044

A = 0.0044 B = C = 0.09 D =0

and the complete solution is:

−0.69z

0.69z 3

− −

θ = 0.0045 e 0.0045 e 0.0038z + 0.09z

0 −0.69z −1

0.69z 2

−

θ = 0.0031 e + 0.0031 e 0.011z + 0.09 m

00 −0.69z −2

0.69z − −

θ = 0.0022 e 0.0022 e 0.023z m

000 −0.69z −3

0.69z −

θ = 0.0015 e + 0.0015 e 0.023 m

VLASOV AND SAINT VENANT TORSIONAL MOMENTS, BIMOMENT.

6

The Saint Venant torsional moment is:

0

M = GJθ

t1 2

· ·

G J = 10.78 kN m −0.69z

0.69z 2

· − ·

M = 10.78 0.0031 e + 0.0031 e 0.011z + 0.09 kN m

t1

The Wagner-Vlasov torsional moment is:

000

−EΓθ

M =

t2 4

· ·

E Γ = 22.47 kN m −0.69z

0.69z

−22.47 · − ·

M = 0.0015 e + 0.0015 e 0.023 kN m

t2

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

2

− ·

M = M + M = 1.5 0.125 z kN m

t t1 t2 7