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.
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
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
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
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
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
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)
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial # 4 Exercise 1 For the following Elements, calculate and draw the elements at maximum in-plane normal & shear stresses.
Exercise 2
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
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).
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
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
-
+
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial 7
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial 8
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Question 2
:
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial 10 Exercise 1
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Exercise 2: