# Tutorial #1 - Mechanics of Materials II

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18 Ιουλ 2012 (πριν από 6 χρόνια και 7 μέρες)

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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.

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

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

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

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

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:

I
x
= b.t
3
/12 = 0.3*0.5
3
/12 = 0.003125 m
4

Q
6, 4
= A.y’ = 0.125*0.3*0.1875 = 0.00703125 m
3

Q
3
= A.y’ = 0.25*0.3*0.125 = 0.009375 m
3

Q
1, 2, 5
= 0 m
3

Forces:

M
x
= 10 KN.m
V
y
= -6.67 KN
Stresses:

σ
1, 2
= M
x
. y / I
x
= -10 * 0.25 / 0.003125 = -800 KN/m
2

σ
5
= M
x
. y / I
x
= 10 * 0.25 / 0.003125 = 800 KN/m
2

σ
6
= M
x
. y / I
x
= -10 * 0.125 / 0.003125 = -400 KN/m
2

σ
4
= M
x
. y / I
x
= 10 * 0.125 / 0.003125 = 400 KN/m
2

σ
3
= M
x
. y / I
x
= 10 * 0 / 0.003125 = 0 KN/m
2

τ
1, 2, 5
= V
y
. Q
1
/ t . I
x
= -6.67 * 0 / 0.3*0.003125 = 0 KN/m
2
τ
6, 4
= V
y
. Q
6
/ t . I
x
= -6.67 * 0.00703125 / 0.3*0.003125 = -50 KN/m
2
τ
3
= V
y
. Q
3
/ t . I
x
= -6.67 * 0.009375 / 0.3*0.003125 = -66.67 KN/m
2
800
Element1, 2
66.67
Element3
800
Element5
Element6
50
400
Element4
50
400

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

σ
x
= - 800, σ
y
= 0, τ
xy
= 0
tan 2θ
p
= 2τ
xy
/ (σ
x
– σ
y
)  θ
p
= 0
o

σ
max, min
= (σ
x
+ σ
y
) / 2 +/- ((σ
x
– σ
y
) / 2)
2
+ τ
xy
2
 σ
max, min
= 0, - 800

σ
x
= 0, σ
y
= 0, τ
xy
= + 66.67
tan 2θ
p
= 2τ
xy
/ (σ
x
– σ
y
)  θ
p
= 45
o

σ
max, min
= (σ
x
+ σ
y
) / 2 +/- ((σ
x
– σ
y
) / 2)
2
+ τ
xy
2
 σ
max, min
= +/- 66.67

σ
x
= - 400, σ
y
= 0, τ
xy
= + 50
tan 2θ
p
= 2τ
xy
/ (σ
x
– σ
y
)  θ
p
= - 7.02
o

σ
max, min
= (σ
x
+ σ
y
) / 2 +/- ((σ
x
– σ
y
) / 2)
2
+ τ
xy
2
 σ
max, min
= 6.16, - 406.16

σ
x
= 400, σ
y
= 0, τ
xy
= + 50
tan 2θ
p
= 2τ
xy
/ (σ
x
– σ
y
)  θ
p
= 7.02
o

σ
max, min
= (σ
x
+ σ
y
) / 2 +/- ((σ
x
– σ
y
) / 2)
2
+ τ
xy
2
 σ
max, min
= 406.16, - 6.16

σ
x
= 800, σ
y
= 0, τ
xy
= 0
tan 2θ
p
= 2τ
xy
/ (σ
x
– σ
y
)  θ
p
= 0
o

σ
max, min
= (σ
x
+ σ
y
) / 2 +/- ((σ
x
– σ
y
) / 2)
2
+ τ
xy
2
 σ
max, min
= 800, 0

800
Element 1 & 2
66.67
Element 3
45
o

66.67
66.67
Element 6
50
400
7.02
o

406.16
6.16
Element 4
50
400
7.02
o

406.16
6.16
800
800
Element 5

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

Element Mohr’s Circle
principle plane max. shear
Tutorial 4 - Ex.1
:
5
5
5
5
5
5
4
5
ο

σ
max

σ
max

= 10
σ
min

= 0
x
σ
τ
x
y
=
(
5
,
-5
)

σ
max

σ
min

2
θ
p
=
90
ο

From point x to
σ
max

R = 5
τ
σ
max

4
5
ο

σ
max

= 5
σ
min

= -5
x
σ
y
x
σ
max

σ
min

2
θ
p

= 90
D
From point x to σ
max

?
R
=
5
5
5
x
θ
p

= 31.7
ο

θ
s

= 76.7
ο

x
σ
max

τ
max

5
5
σ
τ
y
x

from point
x to σ
max

av
, τ
max
)

av
, -τ
max
)

max
, 0)
2
θ
s
from point
x to σ
av
& τ
max

min
, 0)
R = 5.6
τ

max
= 5.6
σ
av
= 2.5

x
5
5
σ
τ
x = (0, 5)
y
=
(
5
,
-5
)
2
θ
p
from point
x to σ
max

av
, -τ
max
)
(
σ
max
, 0)

s
from point
x to σ
av
& τ
max

min
, 0)
θ
p

= 59.3
ο

θ
s
= 13.3
ο

x
σ
max

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

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

Element Mohr’s Circle
principle plane max. shear
5
5
5
σ
τ
y
x
σ
max

σ
min

2
θ
p
?
?
?
?
R
?
From point x to
σ
max

R = 7.07

s
from point
x to σ
av
& τ
max

σ
max

2
2.
5
ο

σ
max

= 7.07
σ
min
=
-7.07
x
τ

max

22.5
ο

x
τ
max
= 7.07

5
3
4
σ
τ
y
x
2
θ
p
from point
x to σ
max

av
, τ
max
)

av
, -
τ
max
)
(
σ
max
, 0)

s
from point
x to σ
av
& τ
max

min
, 0)
R = 5.025
σ
max

θ
p
=
42.1
ο

σ
max

= 8.5
σ
min
=
-1.5
x
τ

max

θ
s
= 2.9
ο

x
τ

max
= 5.025
σ
av
= 3.5

5
5
R = 0
τ
σ
x = y = (5, 0)
θ
p
=
θ
s

= 0

(
one
p
oint
)
5
5
5
5
5
5
τ
max

45
ο

x
σ
y
x
σ
max

σ
min

R = 5
2
θ
s
from point
x to σ
av
& τ
max

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

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

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:

I
x
= b.t
3
/12 = 1*2
3
/12 = 0 .667 ft
4

Q
2
= A.y’ = 1*1*0.5 = 0.5 ft
3

Q
1
=Q
3
= 0
Forces:

M
x
= 239.4 kip.ft
V
y
= 12.2 kips
Stresses:

σ
x1
= M
x
. y / I
x
= -239.4 * 1 / 0.667 = -359.1 kip/ft
2

σ
x2
= M
x
. y / I
x
= -239.4 * 0 / 0.667 = 0 kip/ft
2

σ
x3
= M
x
. y / I
x
= 239.4 * 1 / 0.667 = 359.1 kip/ft
2

τ
y1
= V
y
. Q
1
/ t . I
x
= 12.2 * 0 / 1*0.667 = 0 kip/ft
2

τ
y2
= V
y
. Q
2
/ t . I
x
= 12.2 * 0.5/ 1*0.667 = 9.15 kip/ft
2

τ
y3
= V
y
. Q
3
/ t . I
x
= 12.2 * 0 / 1* 0.667 = 0 kip/ft
2

9.15
359.1
359.1
Element1
Element2
Element3
Element1, R = 179.55 Element3, R = 179.55
Element2, R = 9.15
2
θ
p
from point
x to σ
max

x
y
=
(
0
,
-9.15
)
σ
=

‽‹=ㄵ1
σ
min
= -9.15
9.15
-9.15
4
5
ο

x
σ
max

σ
min
= -359.1
σ
max
= 0
σ
=

‽″㔹=ㄠ
σ
=

‽‰=

†=†=†=†=
y

y
=
†††=†=†=†=†=†==
τ
τ

τ
σ
σ

σ

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

Tutorial #5

The state of strain at a point on a wrench has components
ε
x
= 150 (10
-6
), ε
y
= 200 (10
-6
), and γ
xy
= -700 (10
-6
).

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

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

Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial # 6

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

Strain Values  ε
x
= + 350 (10
-6
), ε
y
= + 600 (10
-6
), γ
xy
=

- 400 (10
-6
)
τ
xy
= G.γ
xy
= 11.28*10
6
* (-400)* 10
-6
= - 4512 psi
E. ε
x
= σ
x
- υ. σ
y

29*350 = σ
x
- .285 σ
y
(1)
29*600 = σ
y
- .285 σ
x
(2)
By solving Eq. (1) and (2)  σ
x
= 16444.3 psi & σ
y
= 22085.2 psi

2. Draw Mohr’s circles for stress and for strain.

3. Calculate the required yield stress σ
yield
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: σ
1
2

-

σ
1.
σ
2
+

σ
2
2

<

σ
yield
2
 σ
1
2

-

σ
1.
σ
2
+

σ
2
2

<

yield
/ F.S
)
2

24585.75
2
-24585.75*13943.75 + 13943.75
2
= 456069715.5625 =

yield
/ F.S
)
2

σ
yield
= 53389.5 psi

TRESCA:

τ
max (3-D)
= σ
max
/

2 = 12292.875 psi

τ
max
=

σ
yield
/ 2 
τ
max
=

σ
yield
/ 2 F.S

σ
yield
= 2*2.5*12292.875

σ
yield
= 61464.375 psi

Select a material with
σ
yield
22085.2
4512
16444.3
R= 5321
x
y y
σ

‽′=㔸㔮75=
σ

‽‱=㤴㌮75=
σ
max
= 24585.75

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