Tutorials 1 & 2 - CivE. 205 · 2008-08-01 · Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II Tutorial 6 1. Calculate the required yield stress σyield for the material so

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

Tutorials 1 & 2 - CivE. 205 _______________________________________________________________________________

Exercise 1: Exercise 2:

2

18

20

C sum of moments = 0

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

13

-

13

-

3

3

1

1

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.

13

9

1

1

10

10

2

2

-

- -

-

-

The column has

zero shear.

-

38

38

42

2

2

-

-

-

-

20

18

2

B.M.D.

S.F.D.


Tutorial 3 - CivE. 205 _______________________________________________________________________________ Exercise 1

Section I-I

Section II-II:

Forces:

N = -7.5 KN; Mx = -4 KN.m , Vy = 7.33 KN

1

2

3

4

5

6

M =10

M =10 V = 6.7

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 m

3

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

M= -4

M= -4 V = 7.33

N = - 7.5


Element 1, 2

370 Element 5

Element 3

73.3

50

270

Element 4

55

210

Element 6

55

110

Normal Stresses:

2

2,1 /270003125.0

25.0*4

15.0

5.7mKN

I

yM

A

N

x

x=

−−

−=−=σ

2

5 /370003125.0

)25.0*4(

15.0

5.7mKN−=

−−−

−=σ

2

6 /110003125.0

125.0*4

15.0

5.7mKN=

−−

−=σ

2

4 /210003125.0

)125.0*4(

15.0

5.7mKN−=

−−−

−=σ

2

3 /50003125.0

0*4

15.0

5.7mKN−=

−−

−=σ

Shear Stresses:

0003125.0*3.0

0*33.7

.

. 1

5,2,1 ===

x

y

It

QVτ

21

4,6 /55003125.0*3.0

00703125.0*33.7

.

.mKN

It

QV

x

y===τ

21

3 /3.73003125.0*3.0

009375.0*33.7

.

.mKN

It

QV

x

y===τ


Exercise 2:

Circle the correct answer from the given choices

a

P

Section a

1

2

3

4

P = 2

2 m

16

20

- Expected B.M.D.

(b)

Reactions directions

P

L

Normal stresse at point 1 in the beam.

1 (c) 2

Shear stresses at point 2 in the beam.

(a)

Hinge support

a a a a

If the shear on Element 1 is

Then, the expected element under normal stress is

1

(a)

- For section a, the maximum shear stress will be at point (4)

My

Mx

X

Y

1 2

3 4

- The quarter that is sure under tension (3)


Tutorial # 4 Exercise 1 For the following Elements, calculate and draw the elements at maximum in-plane normal & shear stresses.

Exercise 2


Tutorial 5

2. The state of strain at a point on a wrench is: εx = 150 (10-6), εy = 200 (10

-6), and γxy = -700 (10-6).



Tutorial 6 1. 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: σ12 - σ1. σ2 + σ2

2 < σyield

2 � σ1

2 - σ1. σ2 + σ2

2 < (σyield / F.S)

2

24585.752 -24585.75*13943.75 + 13943.75

2 = 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

2. Draw Mohr’s circles for stress and strain for the following element

σmax = 24585.75

-

+


Tutorial 7


Tutorial 8


Question 2

:


Tutorial 10 Exercise 1



Exercise 2:

Documents

Tutorials 1 & 2 - CivE. 205 · 2008-08-01 · Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II Tutorial 6 1. Calculate the required yield stress σyield for the material so