Figure 3.2.6
The coil-spring clutch has a series of coil springs set in a circle.
At high rotational speeds, problems can arise with multi coil spring clutches
owing to the effects of centrifugal forces both on the spring themselves and the
lever of the r
of ongoing open source
software projects listed in http:/freshmeat.net. This conveys the relative
importance of various languages
(a mixed bag of newcomers, going-strongs, have-beens and never-was) in the
present environment.
Lang Projects Perc Lang Proje
analysis for highly
redundant structures that is particularly suited to the use of high-speed digital
computing machines. . . .
The stiffness matrix for the entire structure is computed by simple summation of of
the stiffness matrices
of the elements of t
improved asymptotically by continued
refinement of the finite element mesh. The conclusions drawn from that summers
work were presented in a paper
given by Jon Turner at the annual meeting of the Institute of Aeronautical Sciences
in January 1954. However
following are required.
1. The torque transmitted
2. The actuating force.
3. The energy loss
4. The temperature rise
FRICTION CLUTCHES
As in brakes a wide range of clutches are in use wherein they vary in their are in
use their working principle as well t
(C) Assuming r = b/d = 0.5, b = 0.5d = 0.5(15.5 in) = 7.75 in, use b = 8 in.
(D) Compute Mu and Mn. With our estimated beam height and width, we can
calculate the
self-weight, SW , and determine the total dead load, wD.
SW = bhc
= (8 in)(18 in)(150 lb/ft3
time on the stress analysis question until I went on my sabbatical leave to
Trondheim, Norway in September 1956.
Then, when I arrived in Norway all I could do was to outline the procedures for
carrying out the analysis, and to do
calculations for very sma
(G) Choose bars, make sure they fit. Based on the required As = 1.88 in2, the
following options
are available:
Bar Size Ab (in2) # Bars As (in2)
6 0.44 5 2.20
7 0.60 4 2.40
8 0.79 3 2.37
9 1.00 2 2.00
10 1.27 2 2.54
11 1.56 2 3.12
Based on these options,
One-way slab
Beam
L/20
L/16
L/24
L/18.5
L = span length
For steel with f
y not equal 4,000 kg/cm2 multiply with 0.4 + fy/7,000
Example 3.2: Working Stress Design of Beam
w = 4 t/m
5.0 m
Concrete: f
c = 65 kg/cm2
Steel: f
s = 1,700 kg/cm2
From table: n = 1
USE section 40 x 80 cm d = 75 cm
Revised Design due to Self Weight
From selected section 40 x 80 cm
Beam weight wbm = 0.4 0.8 2.4(t/m3) = 0.768 t/m
Required moment M = (4 + 0.768) (5)2 / 8 = 14.90 < 18.36 t-m OK
Revised Design due to Support width
5.0 m s
BC = 67.5 k (T) FHG = 75 k (C) FHC = 12.5 k (T)
Method of Sections
Example: Determine the forces BC, BG, HG, and CG in the following
truss.
CIVL 3121 Trusses - Method of Sections 1/2
500 lb. 800 lb. 500 lb.
10 ft.
10 ft. 10 ft.
A
BCD
E
GF
Method of Sectio
Thomas Telford, 2002
GULVANESSIAN, H., CALGARO, J.A., FORMICHI P. and HARDING, G.
Designers' guide to EN 1991-1-1, 1991-1-3 and 1991-1-5 to 1-7 Eurocode 1:
Actions on structures: General rules and actions on buildings
Thomas Telford (to be published in 20
Intermediate level: Hughes [142]. It requires substantial mathematical expertise on
the part of the reader.
Recently (2000) reprinted as Dover edition.
Mathematically oriented: Strang and Fix [228]. Still the most readable
mathematical treatment for engin
disc called the clutch plate and a pressure plate. When the engine is running
and the flywheel is rotating, the pressure plate also rotates as the pressure plate
is attached to the flywheel. The friction disc is located between the two. When
Machine Desig
calculations on a triangular
net for determining the torsional stiffness of a hollow shaft. He used piecewise
linear interpolation over
each triangle as Rayleigh-Ritz trial functions, and called his idea generalized
finite differences.
15 If you go too fa
=am
T is the torque (Nm).
N is the number of frictional discs in contact.
f is the coefficient of friction
Fa is the actuating force (N).
Rm is the mean or equivalent radius (m).
Note that N = n1 + n2 -1
Where n1= number of driving discs
n2 = number of dr
Homework Exercises for Chapter 1
Overview
EXERCISE 1.1 [A:15] Work out Archimedes problem using a circumscribed
regular polygon, with n =
1, 2, 4, . . . 256. Does the sequence converge any faster?
EXERCISE 1.2 [D:20] Select one of the following vehicles:
1.128 in.
Adding it all up:
bmin = 2(1.5 in) + 2(0.375 in) + 3(1.128 in) = 7.13 in < b = 8 in (6)
The minimum width, bmin, is less than our actual width so the bars will fit.
(H) Analyze the section, check Mn Mu.
(1) Factor loads and compute required Mu.
War paranoia. Can you imagine
defense funds pouring into hypercircles? Once all pieces were in place, synergy
transformed the method into
a product, and FEM took off.
1.7.7. Recommended Books for Linear FEM
The literature is vast: over 200 textbooks and m
F
rd
i
dr
lining
A single-Surface Axial Disk Clutch
Figure 3.2.5
Now the torque that can be transmitted by this elemental are is equal to the
frictional force times the moment arm about the axis that is the radius r
i.e. T = dF. r = f.dN. r = f.p.A.r
= f.
difference methods, in which the
exact equations of the actual physical system are solved by approximate
mathematical procedures.
Discuss critically the contents of this paragraph while placing it in the context of
time of writing (early 1960s).
Is the la
17 Curiously this book does not mention, even in passing, the use of digital
computers that had already been commercially
available for several years. The few numerical examples, all in 2D, are done by
hand via relaxation methods.
116
117 1. Notes and Bib
unit was Mr. M. J. Turner a very capable man in dealing with problems of
structural vibrations and flutter.
When I arrived for the summer of 1952, Jon Turner asked me to work on the
vibration analysis of a delta wing
structure. Because of its triangular p
most important development during the winter was that Jon suggested we try to
formulate the stiffness property
of the wing by assembling plane stress plates of either triangular or rectangular
shapes. So I developed stiffness
matrices for plates of both s
(4) Check minimum strain. Because we only have one row, d = dt.
c dt
=
cd
=
4.75 in
15.56 in
= 0.305 < 0.375 (16)
The ratio c
dt is less than 0.375, thus the section is tension-controlled. A tensioncontrolled
section has t > 0.005, so it follows that t >
The beam is strong enough (Mn > Mu) and all other code checks were
satisfactory.
Thus, the beam is adequate for the loads.
(I) Sketch.
d=
b = 8"
15.6"
18"
h=
2 - #9
3Machine Design II Prof. K.Gopinath & Prof. M.M.Mayuram
Indian Institute of Technology Mad
3 (r r )
=
Uniform Wear Condition
According to some established theories the wear in a mechanical system is
proportional to the PV factor where P refers the contact pressure and V the
sliding velocity. Based on this for the case of a plate clutch we can s
JOHNSON, R .P. and ANDERSON D.
Designers guide to EN 1994-1-1 Eurocode 4: Design of composite steel and
concrete structures Part 1.1: General rules and rules for buildings
Thomas Telford, 2004
7.3 Sources of electronic information
Sources of electronic in
perpendicular to the surface of the product. Technical delivery conditions
BS EN 10210-1:2006 Hot finished structural hollow sections of non-alloy and
fine grain structural steels Part 1: Technical delivery requirements
BS EN 10219-1:2006 Cold formed holl
Floor slab and material properties
BS EN
Construction
Total depth of slab h = 130 mm
1992-1-1
stage:
Corus profiled steel sheeting CF60
Table 3.1
gk = 0.11
Thickness of profile t = 1.0 mm
BS EN
kN/m 2
Depth of profile hp = 60 mm
1991-1-1
Composite
Span L
Maximum moment at mid span :
M
y,Ed = 11.4 8 / 8 = 91.2 kNm
Design moment diagram
Bending moment
91.2 kNm
Calculation of maximum bending stress:
1610.9
91.21000
=
el,y
y,Ed
Ed
W
M
= 56.6 N/mm
2.3.2
BS EN
10025-2
Table 7
Stress level as a proportion of no
Subject Composite slab
Date Nov
Made by ALS
Checked by MEB
Client
2007
Date Jan 2008
63
Unless
Composite slab
stated
Introduction
otherwise
This example demonstrates the design of the composite
all
floor slab
references
on the second storey that is suppor
Unless
stated
otherwise all
references
are to BS EN
1993-1-10
Choosing a steel sub-grade
Introduction
Determine the steel sub-grade that may be used for the simply
supported restrained beam (UKB 457 191 82 steel grade
S275).
The example follows the proced
BS EN
Partial factors for resistance
1993-
Structural steel M0 = 1.0
1-1
Concrete C = 1.5
NA 2.15
BS EN
1992-1-1
NA 2
Table NA.1
BS EN
1992-1-1
NA 2
Table NA.1
BS EN
1993-1-3
Reinforcement S = 1.15
Longitudinal shear VS = 1.25
Design values of material st