Problem Set 1 - Solution
Problem 1
M C =
0
1500 R1 + 300(5) + 1200(9) =
0
R1 = 8.2 kN Ans.
Fy =
0
8.2 9 5 + R2 =
0
R2 = 5.8 kN
Ans.
= 8.2(300) 2460 N m Ans.
M1
=
M 2 = 0.8(900) = N m Ans.
2460
1740
M 3 = 5.8(300) =checks!
1740
0
_
Problem 2
0
Fy =
RO = +

Problem Set #5
ME-322 Mechanical Design I
Mechanical and Mechatronics Engineering Department
University of Waterloo
Problem 1: The cantilevered bar in the gure is made from a ductile material and is statically loaded
with Fx = 75 lbf, Fy = 200 lbf, and Fz

Problem Set 1
ME-322 Mechanical Design I
Mechanical and Mechatronics Engineering Department
University of Waterloo
Solutions will be posted after tutorials
Problems 1 to 4: For the beams shown, nd the reactions at the supports and plot the shear-force and

Problem Set #4
ME-322 Mechanical Design I
Mechanical and Mechatronics Engineering Department
University of Waterloo
Solutions will be posted after tutorials
Problem 1 (3-34): For each section illustrated, find the second moment of area, the location of th

Problem Set #3
ME-322 Mechanical Design I
Mechanical and Mechatronics Engineering Department
University of Waterloo
Solutions will be posted after tutorials
3
4
Problem 1 (3-23): A -in-diameter steel tension rod is 5 ft long and carries a load of 15 kip.

Problem Set 2
ME-322 Mechanical Design I
Mechanical and Mechatronics Engineering Department
University of Waterloo
Solutions will be posted after tutorials
Problem 1: For each of the plane stress states listed below, draw a Mohr's circle diagram properly

Problem Set #8 Solution
P-1
(a) Thread depth= 2.5 mm Ans.
Width = 2.5 mm Ans.
dm = 25 - 1.25 - 1.25 = 22.5 mm
dr = 25 - 5 = 20 mm
l = p = 5 mm Ans.
(b) Thread depth = 2.5 mm Ans.
Width at pitch line = 2.5 mm Ans.
dm = 22.5 mm
dr = 20 mm
l = p = 5 mm Ans.

Problem Set #9 Solution
Problem 1
2 (100 ) (103 )
2F
= =
= 141 MPa
Ans.
hl
5 2 ( 50 + 50 )
_
Problem 2
b = d =50 mm, c = 150 mm, h = 5 mm, and allow = 140 MPa.
(a) Primary shear, Table 9-1, Case 2 (Note: b and d are interchanged between
problem figure an

Assignment #6-2 Solution
P-1
From a free-body diagram analysis, the bearing reaction forces are found to be 2.1 kN at the left
bearing and 3.9 kN at the right bearing. The critical location will be at the shoulder fillet
between the 35 mm and the 50 mm di

Problem Set #7 Solution
P-1
F cos 20(d / 2) = TA, F = 2 TA / ( d cos 20) = 2(340) / (0.150 cos 20) = 4824 N.
The maximum bending moment will be at point C, with MC = 4824(0.100) = 482.4 Nm.
Due to the rotation, the bending is completely reversed, while th

Problem Set #4 Solution
Problem 1:
(a)
Let a = total area of entire envelope
Let b = area of side notch
A = 2b =
a
40(2)(37.5) 25 ( 34 ) =
2150 mm 2
1
1
3
3
I = a 2 I b = ( 40 )( 75 ) ( 34 )( 25 )
I
12
12
6
4
I = 1.36 (10 ) mm
Ans.
Dimensions in mm.
(b)
A

Problem Set #5 Solution
Problem 1:
(a)
Rod AB experiences constant torsion and constant axial tension throughout its length, and
maximum bending moments at the wall from both planes of bending. Both torsional shear stress and
bending stress will be maximu

Problem Set #3 Solution
Problem 1:
F
=
A
=
15000
= 33 950 psi 34.0 kpsi
=
( 4 ) ( 0.752 )
FL
L
60
= = 33 950
= 0.0679 in
AE
E
30 (106 )
=
1
=
=
L
0.0679
= 1130 (106 = 1130
)
60
Ans.
Ans.
Ans.
From Table A-5, v = 0.292
2 = 1 =
v
0.292(1130) =
330
Ans.
d

A
2010 2011 1
_ _
Production and Operations Analysis
_
_
1. A simple forecasting method for weekly sales of DVD derives used by a local computer dealer
is to form the average of the two most recent sales figures. Suppose sales for the drives for the