ii
O
F J]
i
2.119 The steel rod ABC is attached to rigid supports and is unstressed s: a"
PROBLEM 2-119 temperature of 38°F. The steel is assumed elastoplastic, with as = 36 ksi and E = 29
x 10 psi. The temperature of bot
2.128 The uniform rods AB and BC are made of steel and are loaded as shown.
PROBLEM 2-123 Knowing that E = 29 x 106 psi, determine the magnitude and direction of the
deection of point B when t9= 22°.
SOLUTION
Pa: = Pcusé = (2.531053cm 22"= 23.13,? JL.
_ P
2.109 Rod AB consists of two cylindrical portions AC and BC, each with a cross-
sectional area of 1750 mmz. Portion AC is made of a mild steel with E = 200 GPa and
as = 250 MPa, and portion is made of a high-strength steel with E = 200 GPa and
0, = 345 MP
*2.91 Show that for any given material, the ratio G/E of the modulus of rigidity over
the modulus of elasticity is always less than 3; but more than é [Hint Refer to Eq.
(2.43) and to Sec. 2.13.]
PROBLEM 2.91
SOLUTION
E .
' 5U:I+u) W a 2 20»
Assume 72,20
'2
AA. ECJ; -
[As = "Er-Jae
ES
0
PROBLEM 2.55
52}.
(0h.-
(GA No" MAT) =
E
a
8
2.55 The assembly shown consists of an aluminum shell (E3 = 10.6 X 106 psi, 5; =
12.9 X 10'6/°F) fully bonded to a steel core (E5 = 29 X 106 psi, ccs = 6.5 X 10/°F) and
C1 n i f7 lT m i 1 twat [IWJRC]
l l
2.65 A 2-m length of an aluminum pipe of 240-1an outer diameter and 10-min wall
thickness is used as a short column and carries a centric axial load of 640 kN.
Knowing that E = 73 GPa and v= 0.33, determine (a) the
i
7} T l 1 Al
PROBLEM 2.81
PROBLEM 2.82
2.81 An elastomeric bearing (G = 0.9 MP3) is used to support a bridge girder as
shown to provide exibility during earthquakes. The beam must not displace more
than 10 mm when a 22 EN lateral load is appli
Solution of Quiz #1
Mechanics of Materials ME330
Instructor: Dr. Peyman Honarmandi
Full Name:
Department:
Time: 80 minutes
Date: 09/ 27/2011
1- Each of the four vertical links has an 8 36-mm uniform rectangular cross section and each
of the four pins has
i
2.35 The 4.5ft concrete post is reinforced with six steel bars, each with a 13];- -in.
PROBLEM 2.35
diameter. Knowing that E8 = 29 X 106 psi and .15'c = 4.2 X 106 psi, determine the normal
stresses in the steel and in the concrete when a 350-kip axial c
1
«i
m 1
Kiri
mmi:
iplit!
2.45 The rigid bar AD is supported by two steel wires of 1'; -in. diameter (E = 29 X
10 psi) and a pin and bracket at D. Knewing that the wires were initially taught,
determine (a) the additional tension in each wire wh
i": I
1.53 Each of the two vertical links CF connecting the two horizontal members AD
and EG has 3 to K 40 mm uniform rectangular cross section and is made of a steel with
an ultimate strength in tension of400 MPa, .while each of the pins at C and F ha
Quiz #2
Mechanics of Materials ME330
Instructor: Dr. Peyman Honarmandi
Full Name:
Department:
Time: 40 minutes
Date: 04/ 28/2011
1- Shafts A and B are made of the same material and have the same cross-sectional area, but A
has a circular cross section and
l
R
r:
T
DI
A
H
C F_|
l 4ng
4_-_J
(61
(b)
1.1 Two solid cylindrical rods AB and BC are welded together at B and loaded as
shown. Knowing that d1 = 30 mm and d2 = 50 mm, nd the average normal stress in
the mid section of (a) rod AB, (.15) rod
2
DH
E
T:
CHAP : iA-grm-ruarua w1q;-:m a" u.- - W A - -~-.A.A._-_-. ma.- 7 :.- nun-"virvwm- r_ _ _r._.;=7:~7.gvknwminauwvr .-.-~ - - - -
.- A 2.1 A steel rod is 2.2 m long and must not stretch more than 1.2 mm when a 8.5 kN
[ PROBLEM 2-1 load is applied t
l
l
(7 "rf 71
1 l m
7]
\_.
K,
2.22 For the steel truss (E = 200 GPa) and loading shown, determine the deformations
of members AB and AD, knowing that their cross-sectional areas are 2400 mm2 and
1800 mm, respectively.
PROBLEM 2.22
SOLUTION
S
1.68 Link AC has a uniform % x %- in. uniform rectangular cross section and is
made of a steel with a 60-ksi ultimate normal stress. It is connected to a support at A
and to member BCD at C by '3" in.-diameter pins, while member BCD is connected to
PROBLE
pROBLEM 1.26 1.9 Two horizontal S-k'tp forces are applied to pin B of the assembly shown Knowing
that a pin ofO.8-in. diameter is used at each connection, determine the maximum value
of the average normal stress (a) in link AB, (b) in link BC
[.26 For the
LII For the Pratt bridge truss and loading shown, determine the average normal
PROBLEM 1.11 . . . A 2
stress m member BE, knowmg that the cross-secttonal area of that member 15 5.87 in .
SOLUTION
Use entire. +Nss as PN-t Lady
2MH = 0
(N20) «moon (
r mmwVVVtVVr; 7. 7 :7. ,vnyvrwrrzr myrmn VNwVAg: 7~;.-~;~ ue ~-ygAA~-:_F\L;wrvyw j- n, -t-~- 1- t -v-.- - n - Ad. 7 ,7
[.42 Members AB and AC of the truss shown consist of bars of square cross section
made of the same alloy. It is known that a 20-mmsquare
Solution of Assignment #2
ME 330
Dr. Peyman Honarmandi: The City College of New York
Chapter 1, Solution 31.
= 90 45 = 45
P = 11 kN = 11 103 N
A0 = (150)(75) = 11.25 103 mm 2 = 11.25 103 m 2
=
P cos 2
(11 103 ) cos 2 45
=
= 489 103 Pa
A0
11.25 103
= 4
Solution of Assignment #1
ME330
Dr. Peyman Honarmandi: The City College of New York
SOLUTION 1.3
AAB
AB
(2) 2 3.1416 in 2
4
P
P
AAB
3.1416
0.31831 P
ABC
BC
(3)2 7.0686 in 2
4
(2)(30) P
AAB
60 P
8.4883 0.14147 P
7.0686
Equating AB to 2 BC
0.31831 P 2
Solution of Assignment #14
ME 330
Dr. Peyman Honarmandi: The City College of New York
SOLUTION 7.85
P 200 kN 200 103 N
c
1
d 18 mm 18 103 m
2
A c 2 (18 103 ) 2 1.01788 103 m 2
200 103
P
196.488 106 Pa
A
1.01788 103
196.488 MPa
y
x 0
ave
1
1
( x y ) y
Solution of Assignment #15
ME 330
Dr. Peyman Honarmandi: The City College of New York
Chapter 8, Solution 53.
Equivalent force-couple system at section containing Points a and b.
Fx = 9 kN,
Fy = 13 kN,
Fz = 0
3
M x = (0.400)(13 10 ) = 5200 N m
3
M y = (0.
Solution of Assignment #
ME 330
Dr. Peyman Honarmandi: The City College of New York
SOLUTION 2.93
(a)
At hole A:
r
1 1 1
in.
2 2 4
d 3
1
2.50 in.
2
1
Anet dt (2.50) 1.25 in 2
2
non
P
6.5
5.2 ksi
Anet 1.25
2 1
2r
4
0.1667
D
3
From Fig. 2.60a,
K 2.5
Solution of Assignment #13
ME 330
Dr. Peyman Honarmandi: The City College of New York
Chapter 7, Solution 1.
Stresses
+
Areas
Forces
F = 0: A 15 A sin 30 cos 30 15A cos 30 sin 30 + 10A cos 30 cos 30 = 0
= 30 sin 30 cos 30 10 cos 2 30
= 5.49 ksi
+
F = 0:
Solution of Assignment #10
ME 330
Dr. Peyman Honarmandi: The City College of New York
Chapter 5, Solution 6.
Calculate reactions after replacing distributed load by an equivalent concentrated
load.
Reactions are
A= D=
From A to B:
1
w ( L 2a )
2
0< x<a
Fy
Solution of Assignment #7
ME 330
Dr. Peyman Honarmandi: The City College of New York
Chapter 4, Solution 5.
Allowable stress
=
U 450
=
= 150 MPa
F.S .
3
= 150 106 Pa
Moment of inertia about z-axis.
1
(16)(80)3 = 682.67 103 mm 4
12
1
I 2 = (16)(32)3 = 43.