tutorial #1 - cive. 205 name: i.d: · pdf file- draw the shear force diagram (s.f.d.) and...

Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II

Tutorial #1 - CivE. 205 Name: __________________________ I.D:_____________ _______________________________________________________________________________ Exercise 1: For the Beam below:

- Calculate the reactions at the supports and check the equilibrium of point a - Define the points at which there is change in load or beam shape - Draw the Shear Force Diagram (S.F.D.) and Bending Moment Diagram (B.M.D.). Drawing the diagrams to-scale and

show the compression side.

10.5 10.5

0.5 0.5

11.5

+

0

21 22

17

0

22.04

10 KN

a

2 m

3 KN / m

2 m 4 m

S.F.D.

B.M.D.


Exercise 2: - Draw the (N.F.D.), (S.F.D.), and (B.M.D.) for the following frame.

13

13

3

3 1

1

13

9

1 1 10

10

2 2

The column has zero shear.

38

38

42

2 2

20 18

2

0

2

18

20

C sum of moments = 0

10 KN

4 m

2 m

4 m 1.5 m

a

b

1 m

2 m

1 KN / m 2 KN.m

10 KN

2 KN

2 m

c

N.F.D.

S.F..D.

B.M.D.


Tutorial 3:

7.33

3.33 3.33

6.67

S.F.D.

6.67 4

20

6.66

4

10

B.M.D

1

Beam cross section

0.5 m

0.3 m

2

3

4

5

6

1

2

3

4

5

6

M =10

M =10 V = 6.7

Properties of area: Ix = b.t3/12 = 0.3*0.53/12 = 0.003125 m4 Q6, 4 = A.y’ = 0.125*0.3*0.1875 = 0.00703125 m3 Q3 = A.y’ = 0.25*0.3*0.125 = 0.009375 m3 Q1, 2, 5 = 0 m3

Forces: Mx = 10 KN.m Vy = -6.67 KN

Stresses: σ1, 2 = Mx. y / Ix = -10 * 0.25 / 0.003125 = -800 KN/m2 σ5 = Mx. y / Ix = 10 * 0.25 / 0.003125 = 800 KN/m2 σ6 = Mx. y / Ix = -10 * 0.125 / 0.003125 = -400 KN/m2 σ4 = Mx. y / Ix = 10 * 0.125 / 0.003125 = 400 KN/m2 σ3 = Mx. y / Ix = 10 * 0 / 0.003125 = 0 KN/m2 τ1, 2, 5 = Vy. Q1 / t . Ix = -6.67 * 0 / 0.3*0.003125 = 0 KN/m2

τ6, 4 = Vy. Q6 / t . Ix = -6.67 * 0.00703125 / 0.3*0.003125 = -50 KN/m2

τ3 = Vy. Q3 / t . Ix = -6.67 * 0.009375 / 0.3*0.003125 = -66.67 KN/m2

800 Element1, 2

66.67 Element3

800 Element5

Element6 50 400

Element4

50 400


σx = - 800, σy = 0, τxy = 0 tan 2θp = 2τxy/ (σx – σy) θp = 0o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy

2 σ max, min = 0, - 800 σx = 0, σy = 0, τxy = + 66.67 tan 2θp = 2τxy/ (σx – σy) θp = 45o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy

2 σ max, min = +/- 66.67 σx = - 400, σy = 0, τxy = + 50 tan 2θp = 2τxy/ (σx – σy) θp = - 7.02o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy

2 σ max, min = 6.16, - 406.16 σx = 400, σy = 0, τxy = + 50 tan 2θp = 2τxy/ (σx – σy) θp = 7.02o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy

2 σ max, min = 406.16, - 6.16 σx = 800, σy = 0, τxy = 0 tan 2θp = 2τxy/ (σx – σy) θp = 0o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy

2 σ max, min = 800, 0

800 Element 1 & 2

66.67 Element 3

45o

66.67 66.67

Element 6 50 400

7.02o

406.16

6.16

Element 4 50 400

7.02o

406.16

6.16

800 800 Element 5


Element Mohr’s Circle principle plane max. shear

Tutorial 4 - Ex.1:

5

5 5

5

5 5

45ο

σmax σmax = 10 σmin

= 0

x

σ

τ x

y = (5, -5)

σmax σmin

2θp= 90ο

From point x to σmax

R = 5

τ

σmax

45ο

σmax = 5

σmin = -5

x

σ

y

x

σmax σmin

2θp = 90DFrom point x to σmax

� R = 5

5 5

x θp = 31.7ο θs = 76.7ο x

σmax

τ max

5

5

σ

τ

y

x 2θ from point x to σmax

(σav, τmax)

(σav, -τmax)

(σmax, 0)

2θs from point x to σav & τmax

(σmin, 0)

R = 5.6

τ max = 5.6 σ av = 2.5

x

5

5 σ

τ

x = (0, 5)

y = (5, -5)

2θp from point x to σmax

(σav, -τmax)

(σmax, 0)


(σmin, 0)

θp = 59.3ο θs = 13.3ο

x

σmax

τ max

R = 5.6

τ max = 5.6 σ av = 2.5

σ max = 8.1 σ min = -3.1

σ max = 8.1 σ min = -3.1

σ max = 5 σ min = -5


Element Mohr’s Circle principle plane max. shear

5

5

5

σ

τ

y

x

σmax σmin

2θp = 45ο From point x to σmax

R = 7.07 2θs from point x to σav & τmax

σmax

22.5ο

σmax = 7.07

σmin = -7.07

x

τ max

22.5ο x

τ max = 7.07

5

3

4 σ

τ

y

x

2θp from point x to σmax

(σav, τmax)

(σav, -τmax)

(σmax, 0)


(σmin, 0)

R = 5.025

σmax

θ p= 42.1ο

σmax = 8.5

σmin = -1.5

x

τ max

θs = 2.9ο x

τ max = 5.025 σ av = 3.5

5

5 R = 0

τ

σ x = y = (5, 0)

θp = θs = 0 (one point)

5

5

5

5

5

5

τmax

45ο x

σ y x

σmax σmin

R = 5


5

5

τ max = 5 σ av = 0

σ max = 7.07 σ min = -7.07

σ max = 8.5 σ min = -1.5

τ


Exercise 2:

S.F.D.

B.M.D

41 kips

12.2 kips

7.8 kips

43 kips

16 kips

239.4 kip.ft

40 kip.ft

I

I

1

Beam cross section

24 in

12 in

2

3

Properties of area: Ix = b.t3/12 = 1*23/12 = 0 .667 ft4 Q2 = A.y’ = 1*1*0.5 = 0.5 ft3 Q1 =Q3 = 0

Forces: Mx = 239.4 kip.ft Vy = 12.2 kips

Stresses: σx1 = Mx. y / Ix = -239.4 * 1 / 0.667 = -359.1 kip/ft2 σx2 = Mx. y / Ix = -239.4 * 0 / 0.667 = 0 kip/ft2 σx3 = Mx. y / Ix = 239.4 * 1 / 0.667 = 359.1 kip/ft2 τy1 = Vy. Q1 / t . Ix = 12.2 * 0 / 1*0.667 = 0 kip/ft2 τy2 = Vy. Q2 / t . Ix = 12.2 * 0.5/ 1*0.667 = 9.15 kip/ft2 τy3 = Vy. Q3 / t . Ix = 12.2 * 0 / 1* 0.667 = 0 kip/ft2

9.15

359.1

359.1

Element1

Element2

Element3

Element1, R = 179.55 Element3, R = 179.55 Element2, R = 9.15

2θp from point x to σmax x

y = (0, -9.15)

σ max = 9.15 σ min = -9.15

9.15

-9.15

45ο x

σmax

σ min = -359.1 σ max = 0 σ max = 359.1 σ min = 0

x y y x

τ τ

τ

σ σ σ


Tutorial #5

The state of strain at a point on a wrench has components εx = 150 (10-6), εy = 200 (10-6), and γxy = -700 (10-6).

- Use Mohr’s circle to determine the equivalent in-plane strains on an element oriented at an angle of θ = 30o clockwise from the original position.

- Sketch the deformed elements at the original and the new orientation. - Sketch the elements at the principal plane and the maximum shear plane. - Determine the absolute maximum shear strain.


Tutorial # 6

1. From the given strains, calculate the state of plain stress (draw the element).

Strain Values εx = + 350 (10-6), εy = + 600 (10-6), γxy = - 400 (10-6)

τxy = G.γxy = 11.28*106 * (-400)* 10-6 = - 4512 psi

E. εx = σx - υ. σy

29*350 = σx - .285 σy (1)

29*600 = σy - .285 σx (2)

By solving Eq. (1) and (2) σx = 16444.3 psi & σy = 22085.2 psi 2. Draw Mohr’s circles for stress and for strain. 3. Calculate the required yield stress σyield for the material so that to prevent failure with respect to both

Von Mises and TRESCA criteria, with a factor of safety of 2.5. Von Mises: For 2-D: σ1

2 - σ1. σ2 + σ2

2 < σyield

2 σ12

- σ1. σ2 + σ22

< (σyield / F.S)2

24585.752 -24585.75*13943.75 + 13943.752 = 456069715.5625 = (σyield / F.S)

2 σyield = 53389.5 psi TRESCA: τmax (3-D) = σmax / 2 = 12292.875 psi τmax = σyield / 2 τmax = σyield / 2 F.S σyield = 2*2.5*12292.875 σyield = 61464.375 psi Select a material with σyield = 61464.375 psi answer

22085.2

4512

16444.3

R= 5321 x

y y σmax = 24585.75

σmin = 13943.75

σmax = 24585.75

tutorial #1 - cive. 205 name: i.d: · pdf file- draw the shear force diagram (s.f.d.) and...

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