Takehome problems

G G

1 G1 2 G2

0 0

0 1

−y −25 −y

· −48.81 · −25

25

s : ρ = = cm s : ρ = = cm

G G

3 G3 4 G4

1 0

1 0

−25 − −

· −25 · −18.81

30 y 25 30 y

s : ρ = = cm s : ρ = = cm

G G

5 G5 6 G6

0 1

0

− ·

25 80 y

s : ρ = = 31.19 cm

G

7 G7 1 s

s

cos θ 8

8

− · −

R cos θ R sin θ + 80 y

s : ρ = = R + (80 y ) sin = 25 + 31.19 sin cm

G

8 G8 G

sin θ R 25

Theory of Structures 7

2. Problem T2

In order to compute the center of shear we must define the Jourawsky stress flux as:

T

T y

x ∗ ∗

∗ −

− S (s) S (s)

q (s) = y x

I I

y x

∗

Since we already know x we need only to define S (s).

C y

∗

S (s ) = 25 b s = 100 s

1 2 1 1

y

∗ ∗

S (s ) = S (s = 50) + 25 b s = 5000 + 100 s

2 1 2 2 2

y y s

3

∗ ∗ 23

−

− b s = 8000 + 100 s 2 s

S (s ) = S (s = 30) + 25 4 3 3

3 2

y y 2

∗ ∗ − −

S (s ) = S (s = 50) 25 b s = 8000 100 s

4 3 3 4 4

y y

∗ ∗ − −

S (s ) = S (s = 30) 25 b s = 5000 100 s

5 4 3 5 5

y y s

6

∗ ∗ 26

− −

S (s ) = S (s = 50) + 25 b s = 5000 + 150 s 3 s

6 1 5 6 6

y y 2

s

7

∗ 27

− −

S (s ) = 25 b s = 100 s 2 s

7 6 7 7

y 2

R+r θ R+r s

Z Z Z Z s ds

∗ 2 2

S (s ) = ρ cos θ dρ dθ = ρ cos dρ =

8

y R R

R−r 0 R−r 0

s b s s

1 h i 8 1 8 8

3 3 2 2

− − ≃

(R + r) (R r) sin = b + 12 R sin 2505.33 sin

= 1

3 R 12 R 25

In order to obtain the coordinate y of the center of shear we must satisfy the following system of

C

equations:  ∗

H

−

T y + 2 q Ω + 2 q Ω + 2 q Ω = q (s) ρ ds

x C 1 1 2 2 3 3 G

1+2+3









 ∗

q

ds ds

 H H

R

− −

q q = ds

 1 2

 b b b

1 1

l

 12

 ∗

q

ds ds ds H

H R R

− − −

q q q = ds

2 1 3

 b b b b

2

2 l l

12 23







 ∗

 q

ds

ds R H

H − −

q = ds

q



 3 2

b b b

3 l 3

 23



Let us compute the constant terms:

50 30 50

Z

I Z Z

T

x

∗ ∗ ∗ ∗

−

q (s) ρ ds = S (s ) ρ ds + S (s ) ρ ds + S (s ) ρ ds +

G 1 G1 1 2 G2 2 3 G3 3

y y y

I

y

1+2+3 0 0 0 #

30 50 50 50 πR

Z Z Z Z Z

∗ ∗ ∗ ∗ ∗

+ S (s ) ρ ds + S (s ) ρ ds + S (s ) ρ ds + S (s ) ρ ds + S (s ) ρ ds =

4 G4 4 5 G5 5 6 G6 6 7 G7 7 8 G8 8

y y y y y

0 0 0 0 0

50 30 50

Z Z Z

T

x 23

− −25 − − −

= (100 s ) ds 25 (5000 + 100 s ) ds 48.81 8000 + 100 s 2 s ds +

1 1 2 2 3 3

I

y 0 0 0

30 50 50

Z Z Z 26

−25 − − − − −

(8000 100 s ) ds 25 (5000 100 s ) ds 18.81 5000 + 150 s 3 s ds +

4 4 5 5 6 6

0 0 0 #

50 πR

Z Z s

s

8

8

27

− 25 + 31.19 sin ds =

+31.19 100 s 2 s ds + 2505.33 sin

7 7 8

25 25

0 0

T 35938126.62

x

− (−35938126.62) = T = 55.6771 T N cm

x x

I 645474.18

y

Theory of Structures 8

2. Problem T2 50 30 50 30

∗ Z

I Z Z Z

T ds ds ds ds

q x 6 4 3

∗ ∗ ∗ ∗

− − − −

S

ds = (s ) S (s ) S (s ) S (s ) =

6 4 3 2

y y y y

b I b b b b

y 5 3 4 2

0 0 0 0

1 50 30 50

Z Z Z

T ds ds ds

x 6 4 3

26 23

− − − − − −

= 5000 + 150 s 3 s (8000 100 s ) 8000 + 100 s 2 s +

6 4 3

I 6 4 4

y 0 0 0

30

Z ds T 155833.33

x

− − ≃

(5000 + 100 s ) = (−155833.33) = T 0.2414 T N/cm

2 x x

4 I 645474.18

y

0 50 50 50 50

∗ Z

I Z Z Z

ds

T

q ds ds ds

7

x 5 6 1

∗ ∗ ∗ ∗

− − − −

S (s )

ds = S S S =

(s ) (s ) (s )

7 5 6 1

y y y y

b I b b b b

y 6 3 5 2

0 0 0 0

2 50 50 50

Z Z Z

T ds ds ds

x 7 5 6

27 26

− − − − − −

= 100 s 2 s (5000 100 s ) 5000 + 150 s 3 s +

7 5 6

I 4 b 6

y 3

0 0 0

50

Z T 104166.67

ds x

1 − ≃

− = (−104166.67) = T 0.1614 T N/cm

(100 s ) x x

1 4 I 645474.18

y

0 25π 50

∗ Z

I Z

T

q ds ds

x 8 7

∗ ∗

− −

ds = S S

(s ) (s ) =

8 7

y y

b I b b

y 1 6

0 0

3 50

25π

Z

Z

ds ds

T s T

8 7

x 8 x

27

− −

− −

100 s 2 s

= 2505.33 sin = (20900.00) =

7

I 25 4 4 I

y y

0

0

20900.00 −0.0324

− = T N/cm

= x

645474.18

Computation of the coefficients: 2

·

π 25

2 2 2

· · · · ·

2 Ω = 2 30 50 = 3000 cm 2 Ω = 2 50 50 = 5000 cm 2 Ω = 2 = 1963.50 cm

1 2 3 2

I

I 50 30 50 30 ds 50 50 50 50

ds = + + + = 35.83 = + + + = 45.83

b 6 4 4 4 b 4 4 6 4

2

1

I Z Z

ds 25π 50 ds 50 ds 50

= + = 32.13 = = 8.33 = = 12.5

b 4 4 b 6 b 4

3 l l

12 23

Now we solve the linear system:

     

−55.6771

3000 5000 1963.50 T q

x 1

−8.33 −0.2414

35.83 0 0 q 2

     

= T

x

     

−8.33 −12.5 −0.1614

45.83 0 q 3

     

−12.5

0 32.13 0 y 0.0324

C

We obtain: ′

−3.52 −→

y = cm y = y + y = 45.29 cm

C G C

C

′

Where y is the position of the shear center with respect to the horizontal axis passing through lamina

C

4. Once we have the shear center coordinates we can compute ρ :

C

−1 −1

− −

· −25 · −25

25 80 (y + y ) 25 30 (y + y )

s : ρ = = cm s : ρ = = cm

G C G C

1 C1 2 C2

0 0

0 1

−(y −25 −(y

· −45.29 · −25

25 + y ) + y )

s : ρ = = cm s : ρ = = cm

G C G C

3 C3 4 C4

1 0

1 0

−25 − −

· −25 · −15.29

30 (y + y ) 25 30 (y + y )

s : ρ = = cm s : ρ = = cm

G C G C

5 C5 6 C6

0 1

Theory of Structures 9

2. Problem T2

0

− ·

25 80 (y + y )

s : ρ = = 34.71 cm

G C

7 C7 1 s s

cos θ 8 8

− · −

R cos θ R sin θ + 80 (y + y )

s : ρ = = R + (80 (y + y )) sin = 25 + 34.71 sin cm

G C

8 C8 G C

sin θ R 25

∗ 3

Figure 2.2: The distribution S (s) [cm ], the position of the centroid and the center of shear, the

y

convention adopted for computing the closed loop integrals

Now we compute the effective stress field by the system:



M = 2 Ω q + 2 Ω q + 2 Ω q

t 1 1 2 2 3 3









 ds ds

 H R

− −

q q 2 Ω G β = 0

 1 2 1

 b b

l

1

 12

 ds ds

ds R R

H − − −

q q 2 Ω G β = 0

q 1 2 2

2

 b b b

l l

2 12 23









 ds

ds

H R

− −

q q 2 Ω G β = 0



 3 2 3

b b

3 l

 23



The shear modulus is: 6

·

E 3.5 10 6 2

·

G = = = 1.4583 10 N/cm

2 (1 + ν) 2 (1 + 0.2)

6

·

And the torque is equal to 9 10 N cm. The new terms are:

11 11 11

· · ·

2 Ω G = 4.3750 10 N 2 Ω G = 7.2917 10 N 2 Ω G = 2.8634 10 N

1 2 3

In the end we have: 6

     

·

3000 5000 1963.50 0 q 9 10

1

11

−8.33 −4.3750 ·

35.83 0 10 q 0

2

     

=

     

11

−8.33 −12.5 −7.2917 ·

45.83 10 q 0

3

     

11

−12.5 −2.8634 ·

0 32.13 10 β 0

The results are: −6 −1

·

q = 763.47 N/cm q = 1034.07 N/cm q = 783.92 N/cm β = 4.2835 10 cm

1 2 3

Once the flux is known we can compute the stress field in the section:

q

2 2

−258.5184 −2.59

− = N/cm = M P a

s : τ =

1 zs b

2

q

1 2

− −190.8666 −1.91

s : τ = = N/cm = M P a

2 zs b

2

q

1 2

− −190.8666 −1.91

s : τ = = N/cm = M P a

3 zs b

4

Theory of Structures 10

2. Problem T2 q 1 2

− −190.8666 −1.91

s : τ = = N/cm = M P a

4 zs b 3

q 2 2

− −258.5184 −2.59

s : τ = = N/cm = M P a

5 zs b 3

−

q q

1 2 2

−45.1012 −0.45

= N/cm = M P a

s : τ =

6 zs b 5

−

q q

2 3 2

= 62.5372 N/cm = 0.63 M P a

s : τ =

7 zs b 6

q 3 2

s : τ = = 195.9812 N/cm = 1.96 M P a

8 zs b 1

Once we have the stress field in the section we can compute the warping function with the formula:

a

Z τ zs − ρ ds + Ψ

Ψ (x, y) = C 0

C Gβ

0

The value of the constant Ψ is determined by imposing the normalization condition:

0 Z Ψ dA = 0

C

A

We can avoid this computation by simply observing that we have a symmetric section and, since the

warping function in antisymmetric, we can claim that the function is null when it crosses the vertical