Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Dr. Konstantinos A. Sierros
Problem 1.8-6
A suspender on a suspension bridge consists of
a cable that passes over the main cable (see
figure) and supports the bridge deck, which is far
below. The suspender is held in position by a
metal tie that is prevented from sliding downward
by clamps around
the suspender cable. Let P represent the load in
each part of the suspender cable, and let θ
represent the angle of the suspender cable just
above the tie. Finally, let σallow represent the
allowable tensile stress in the metal tie.
(a) Obtain a formula for the minimum required
cross-sectional area of the tie.
(b) Calculate the minimum area if P 130 kN,
75°, and allow 80 MPa.
Problem 1.8-9
A pressurized circular cylinder has a sealed cover plate fastened with
steel bolts (see figure). The pressure p of the gas in the cylinder is 290
psi, the inside diameter D of the cylinder is 10.0 in., and the diameter db of
the bolts is 0.50 in. If the allowable tensile stress in the bolts is 10,000 psi,
find the number n of bolts needed to fasten the cover.
Homework Set 1
4) Answer true or false and justify each answer in one
sentence.
(a) You can measure stress directly with an instrument the way
you measure temperature with a thermometer
FALSE. Stress is an internal quantity that can only be inferred but cannot be
measured directly
(b) If a normal stress component puts the left surface of an
imaginary cut in tension, then the right surface will be in compression
FALSE. Tension implies that the normal stress pulls the imaginary surface
outward, which will result in opposite directions for the stresses on the two
surfaces produced by the imaginary cut
(c) The correct way of reporting positive axial stress is σ = +15 MPa
FALSE. The normal stress should be reported as tensile
Homework Set 1
4) Answer true or false and justify each answer in one
sentence.
(d) 1 GPa equals 106 Pa
FALSE. 1 GPa equals 109 Pa
(e) 1 psi is approximately equal to 7 Pa
FALSE. 1 psi nearly equals 7 kPa and not 7 Pa
(f) A common failure stress value for metals is 10000 Pa
FALSE. Failure stress values are in millions of pascals for metals
(g) Stress on a surface is the same as pressure on a surface
as both quantities have the same units
FALSE. Pressure on a surface is always normal to the surface.
Stress on a surface can be normal or tangential to it
2.1 Introduction
Chapter 2: Axially Loaded Members
• Axially loaded members are structural components subjected only to
tension or compression
• Sections 2.2 and 2.3 deal with the determination of changes in lengths
caused by loads
• Section 2.4 is dealing with statically indeterminate structures
• Section 2.5 introduces the effects of temperature on the length of a bar
• Section 2.6 deals with stresses on inclined sections
• Section 2.7: Strain energy
• Section 2.8: Impact loading
• Section 2.9: Fatigue, 2.10: Stress concentration
• Sections 2.11 & 2.12: Non-linear behaviour
2.2 Changes in lengths of axially loaded members - Springs
• Spring stretching or shortening is analogous to
the behaviour of a bar under axial loading
• If the material of the spring is linearly elastic,
the load (P) and elongation (δ) are proportional
P = k δ and δ = f P
stiffness
FIG. 2-1
Spring subjected to an axial load P
flexibility
Copyright 2005 by Nelson, a division of Thomson Canada Limited
Therefore;
k = 1/f and f = 1/k
FIG. 2-2
Elongation of
an axially
loaded spring
Copyright 2005 by Nelson, a division of Thomson Canada Limited
http://www.youtube.com/watch?v=xqANn3sYv18
2.2 Changes in lengths of axially loaded members –Prismatic bars
• A prismatic bar is a structural member having a straight longitudinal axis
and a constant cross-sectional area throughout its length
Elongation of a
prismatic bar in tension
FIG. 2-5
FIG. 2-4
Typical cross
sections of
structural
members
Prismatic bar of
circular cross section
FIG. 2-3
Copyright 2005 by Nelson, a division of Thomson Canada Limited
Copyright 2005 by Nelson, a division of Thomson Canada Limited
• However, we can have a variety of cross-sections
• If we assume that material is homogeneous and linearly elastic then
elongation δ is:
axial rigidity
δ = (PL)/(EA)
Stiffness = (EA)/L and flexibility = f = (L)/(EA)
Copyright 2005 by Nelson, a division of Thomson Canada Limited
2.2 Changes in lengths of axially loaded members – Cables
• Cables are used transmit large tensile forces
(eg. lifting heavy objects, supporting
suspension bridges etc)
• They cannot resist compression and have
little resistance to bending
• Cross sectional area of a cable = total cross
sectional
area of individual wires
• Total cross sectional area of individual wires
is called effective area or metallic area
2.2 Changes in lengths of axially loaded members – Cables
• When under the same tension, the elongation of a cable is greater than that of
a solid bar since the wires in the cable ‘tighten up’
• The effective modulus of the wire rope is less than the modulus of the material
it is made
• In practice the properties of the cables are obtained from the manufacturers
• For solving problems in this book, use Table 2.1
Typical arrangement of strands and
wires in a steel cable
FIG. 2-6
Copyright 2005 by Nelson, a division of Thomson Canada Limited
Please study example 1.8 (pages 46-48)
• Next time we will discuss about it.
Therefore, you need to make sure that you have comments/questions
ready for discussion
Example 1-8.
Two-bar truss ABC
supporting a sign of
weight W
FIG. 1-33
Copyright 2005 by Nelson, a division of Thomson Canada Limited
Monday 4 February 2008 during class…
Quiz covering Chapter 1
Duration: 20 mins
Solve 1 out of 2 Questions