Mechanics of Materials – MAE 243 (Section 002)
Spring 2008
Dr. Konstantinos A. Sierros
Problem 1.4-4
A circular bar of magnesium alloy is 800 mm long. The stress-strain diagram for
the material is shown in the figure. The bar is loaded in tension to an elongation
of 5.6 mm, and then the load is removed.
(a) What is the permanent set of the bar?
(b) If the bar is reloaded, what is the proportional limit?
(Hint: Use the concepts illustrated in Figs. 1-18b and 1-19.)
Problem 1.5-6
A tensile test is performed on a brass specimen 10 mm in diameter using a
gage length of 50 mm (see figure). When the tensile load P reaches a value of
20 kN, the distance between the gage marks has increased by 0.122 mm.
(a) What is the modulus of elasticity E of the brass?
(b) If the diameter decreases by 0.00830 mm, what is Poisson’s ratio?
1.6: Shear stress and strain
• Shear stress acts tangential to the surface
of the material and not perpendicular
perpendicular
tangential
• Consider the bolted connection of fig.
1-24 where A is a flat bar, C a clevis and
B a bolt
• When load P is applied, the bar and
clevis will press against the bolt and
bearing stresses will be developed
• The bar and clevis tend to shear the bolt
Bolted connection in which the bolt
is loaded in double shear
FIG. 1-24
Copyright 2005 by Nelson, a division of Thomson Canada Limited
This tendency is resisted
by shear stresses in the
bolt
1.6: Shear stress and strain
• If we have a closer look from the side
view (fig 1-24b) and draw a FBD (fig 124c)
• Bearing stresses exerted by the clevis
against the bolt appear on the left-hand
side (1 and 3)
• Stresses from the bar are on the righthand side (2)
Bolted connection in which the bolt
is loaded in double shear
FIG. 1-24
• Based on the assumption of uniform
stress distribution we can calculate an
average bearing stress σb
Copyright 2005 by Nelson, a division of Thomson Canada Limited
total bearing force
bearing area
1.6: Shear stress and strain
• The bearing area Ab is defined as the projected area of the curved
bearing surface. For example (for stresses labeled 1) the projected area
on which the stresses act is a rectangle with height equal to the thickness
of the clevis and width equal to the diameter of the bolt
• The bearing force Fb (for stresses labeled 1) is equal to P/2
• The same area and force apply for stresses labeled 3
• For bearing stresses labeled 2 the bearing area is a rectangle with height
equal to the thickness of the flat bar and width equal to the diameter of
the bolt. The force is equal to P
Bolted connection in which the bolt
is loaded in double shear
FIG. 1-24
Copyright 2005 by Nelson, a division of Thomson Canada Limited
1.6: Shear stress and strain
• The FBD (c) shows that there is a tendency to shear the bolt along the
cross sections mn and pq
• From the FBD (d) of the portion mnqp of the bolt we see that the shear
forces V act over the cut surfaces of the bolt. There are two planes of
shear (plane mn and plane pq). Therefore, the bolt is in double shear
• The shear stresses acting on the cross section mn are shown (e)
• Shear stresses are denoted by τ
1.6: Shear stress and strain: Single shear
• The axial force P in the metal bar is
transmitted to the flange of the steel
column through a bolt
• A cross section of the column (fig 1-25b)
shows more details
• Fig 1-25c shows the assumed
distribution of the bearing stresses acting
on the bolt
• Cutting through the bolt at section mn
(fig 1-25d) we see the shear force V (equal
to load P). V is the resultant of the shear
stresses that act over the cross-sectional
area of the bolt
FIG. 1-25
Bolted connection
in which the bolt is
loaded in single
shear
Copyright 2005 by Nelson, a division of Thomson Canada Limited
1.6: Shear stress and strain: Single shear
• Fig 1-26 shows the deformation of a bolt
loaded almost to fracture in single shear
FIG. 1-26
Failure of a bolt in single shear
Copyright 2005 by Nelson, a division of Thomson Canada Limited
1.6: Shear stress and strain
• Discussing about bolted connections we disregard friction which is
produced by tightening the bolts
• Average shear stress on the cross section of a bolt is obtained by
dividing the total shear force V by the area A of the cross section on
which it acts:
• Shear stresses have the same units as normal stresses
• The two previous examples (double and single shear) are examples of
direct shear
• Direct shear arises in the design of bolts, pins, rivets, keys, welds and
glued joints
1.6: Equality of shear stresses on perpendicular planes
• Consider a small rectangular
parallelepiped element
• Assume that a shear stress τ1 is uniformly
distributed over the right-hand side area bc
(τ1bc)
• For equilibrium in the y direction the τ1bc
must be balanced by an equal and of
opposite direction shear force on the lefthand side
• The forces τ1bc acting on the right-hand
and left-hand side faces form a couple
having a moment about z-axis equal to τ1 abc
(counterclockwise direction)
Small element of material
subjected to shear stresses
FIG. 1-27
Copyright 2005 by Nelson, a division of Thomson Canada Limited
1.6: Equality of shear stresses on perpendicular planes
• Similarly, in order to have equilibrium of
the element, we have a shear force τ2ac and
consequently a clockwise couple of
moment τ2abc
• It is therefore evident that for moment
equilibrium we have:
τ1 = τ2
Small element of material
subjected to shear stresses
FIG. 1-27
Copyright 2005 by Nelson, a division of Thomson Canada Limited
1. Shear stresses on opposite and parallel faces of an element are equal
in magnitude and opposite in direction
2. Shear stresses on adjacent and perpendicular faces of an element are
equal in magnitude and have directions such that both stresses point
toward, or both point away from, the line of intersection of the faces
1.6: Shear strain
• Shear stresses acting on an element of
material (fig 1-28a) are accompanied by
shear strains
• The lengths of the sides of the element
do not change but, the shear stresses
produce a change in the shape of the
element
• Rectangular parallelepiped becomes
oblique parallelepiped. Front and rear
faces become rhomboids
• The angle γ (fig 1-28b) is a measure of
distortion of the element and is called
shear strain
Element of material subjected to
shear stresses and strains
FIG. 1-28
Copyright 2005 by Nelson, a division of Thomson Canada Limited
1.6: Sign conventions for shear stresses and strains
• We refer to the faces oriented toward the
positive directions of the axes as the positive
faces of the element
• Right-hand, top and front faces are the positive
x, y and z faces and the opposite faces are the
negative ones
A shear stress acting on a positive face of an
element is positive if it acts in the positive
direction of one of the coordinate axes and
negative if it acts in the negative direction of an
axis. A shear stress acting on a negative face of
an element is positive if it acts in the negative
direction of an axis and negative if it acts in a
positive direction
Shear strain in an element is positive when the
angle between two positive faces, or two
negative, is increased
1.6: Hooke’s law in shear
• We can plot shear stress-strain diagrams
• Hooke’s Law in shear:
shear stress
shear modulus of elasticity
τ = Gγ
shear strain
• G has the same units as E (Young’s modulus)
• G and E are also related by:
Poisson’s ratio
G = E / (2(1+ν))
• For mild steel G = 75 GPa and for aluminum alloys G = 28 GPa
• Additional values can be found in Table H-2, Appendix H
Please do not forget to correct your Statics quiz and return it on
Wednesday 30th of January during class