Solution shaft problem
Question 1
The tangential force on the gear is 15000*cos(200) = 14095 N
The radial force is 15000*sin(200) = 5130.3 N
The other gear has half the diameter, so the loads are twice (28190 N and 10260.6 N)
Summing moments and forces th
Solution MECH2100 tutorial 8 2011
1. The standard reliability is 90%. An application factor of 1.5 is on the boundary between light and
moderate. 30000 hours is the maximum expected of a machine on eight hour shifts. 90 million
revolutions is the standard
Solution for Tutorial Four MECH2100, 2011
The University of Queensland
School of Engineering
MECH2100 - Machine Element Design
Solution for Tutorial Four 2011
Q1:
This figure shows the Constant Life Fatigue Diagram. The Y-axis is Alternating stress
and th
Solution for Tutorial Two MECH2100, 2011
The University of Queensland
School of Mechanical & Mining Engineering
MECH2100 - Machine Element Design
Solution for Tutorial Two 2011
Q1:
50 mm
25 mm
10 mm
400 MPa
Nominal area A = (50-10)25 = 1000 mm2; nom = 250
MECH2100 Practice Final Semester Examination, Semester Two 2007
VENUE:
SEAT NUMBER:
STUDENT NUMBER:
Practice EXAMINATION
Second Semester, 2007
St Lucia Campus
Internal
MECH2100- Machine Element Design
PERUSAL TIME
10 Mins. During perusal, write only on th
Solution MECH2100 Tutorial 11 2011
Question 1
Aluminium plates are fastened together with brass rivets, the
Lead sheet is pop-riveted with special bronze fasteners, the
Galvanized steel is riveted with copper rivets, the
Aluminium will corrode.
Lead will
Solution tutorial 9 gears
1.
2.
3.
4.
Pitch diameter
mm (US units uses pitch not module)
The angle between the normal force and the tangential force this is always 900
The average number of teeth in contact
5. Centre distance if mounted correctly is (20 +
Solution: tutorial 7, plain bearings and fracture
1. p = N/(DL) hence L = N/(Dp) = 0.2MN/(0.25m x 4MPa) = 0.2 m = 200 mm
2. Power dissipated = frictional torque x angular velocity = f N R
= 0.015 x 0.2E6 N x 0.125 m x (150/60 x 2) rad/s
= 5890.48 Nm/s =
MECH2100 2011 Solution Tutorial 6 bolts and screw threads
1. L/(d) =2/(30) = 0.02122 This is more than the coefficient of friction of 0.03.
Hence the thread will lock.
2. The proof stress from Table 10.5 is 225 MPa. The 2.5 mm pitch indicates a
coarse thr
Tutorial Five Solutions
Q1 (0.5 point):
Raman spectroscopy
.
Q2 (0.5 point):
Fatigue stress is reduced near welded joints. X
Q3 (2 points):
The load will be shared equally, as top and bottom weld lengths are proportional to
distance from the line of actio
Solution for Tutorial Three MECH2100, 2010
The University of Queensland
School of Mechanical & Mining Engineering
MECH2100 - Machine Element Design
Solution for Tutorial Three 2011
Q1: Answer: C & D
Q2 & Q3:
Cross sectional area of the rod A= 302 mm 2 , I
Solution for Tutorial One MECH2100, 2011
The University of Queensland
School of Mechanical & Mining Engineering
MECH2100 - Machine Element Design
Solution for Tutorial One 2011
Q1:
There are four assessment items in this course. The computer-based assessm