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(a) The distribution of shearforce on a sectioning plane
(b) The resultant shear force on thesectioning plane
Chapter 2 SHEAR STRESS AND STRAIN p1
Fx
:V =
A
dA
2.
1
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FIGURE 2.1 Shear force on a sectioning plane
avg =
V
A
s
Average Shear Stress
2.2
avg =
V
A
s
=P
(d)t
Chapter 2 SHEAR STRESS AND STRAIN p2
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(c) A pair of pliers (d) Direct shear of pin
FIGURE 2.2 Examples of direct shearSingle Shear
Chapter 2 SHEAR STRESS AND STRAIN p4
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(a) A lap splice (b) The free-bodydiagram
(c) The average-shear-stress distribution
FIGURE 2.3 An illustration of direct shear a lap splice.
avg =V
As=
P
Lsw
Chapter 2 SHEAR STRESS AND STRAIN p5
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FIGURE 2.4 Direct single shear (a) and (b) Bolted lap joint, (c) and (d)
Glued lap joint
Chapter 2 SHEAR STRESS AND STRAIN p6
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Double Shear
FIGURE 2.4 Direct double shear (a) and (b) Bolted double lap joint, (c) and
(d) Glued double lap joint
So, the shear load, V for double shear is given by
V = F/2
Chapter 2 SHEAR STRESS AND STRAIN p7
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EXAMPLE 2.1
The wooden strut shown in figure below is
suspended from a 10-mm-diameter steel rod,
which is fastened to the wall. If the strut
supports a vertical load of 5 kN, compute the
average shear stress in the rod at the wall and
along the two shaded planes of the strut, oneof which is indicated as abcd.
FIGURE 2.5
Chapter 2 SHEAR STRESS AND STRAIN p8
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2.2 Shear Deformation
FIGURE 2.7 Three-dimensional state of stress
Chapter 2 SHEAR STRESS AND STRAIN p9
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FIGURE 2.8 Three-dimensional state of stress
Chapter 2 SHEAR STRESS AND STRAIN p10
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Force ForceStres
s
x Area Stress x Area
Fx = 0: yx (xz) - `xy (xz) = 0yx = `yx
And in a similar manner, force of equilibrium in the y-direction yield
xy = `xy
Finally, by taking moment about z-axisMoment Moment
Force x Distance Force x DistanceStres
sx Area x Arm Stress x Area x Arm
Mz = 0: yx (xz) (x) - xy (xy) (z) = 0
yx = xy
So, yx = `yx = xy = `xy = . Thus Figure 2.8 can be replaced by Figure 2.9.
Chapter 2 SHEAR STRESS AND STRAIN p11
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FIGURE 2.9 Pure shear deformation
Allowable Stress
Chapter 2 SHEAR STRESS AND STRAIN p12
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Recall from previous chapter,
F.S
= Ffail
Fallow
So, for a body that is subjected to shear stress,
F.S =fail
allow
We can design the dimension of the body to sustained the allowable shear
stress, allow, to be within the range of the decided factor of safety which is
generally bigger than 1.
From the calculated allowable shear stress,allow, we can determine the area
and hence the dimension as well.
Chapter 2 SHEAR STRESS AND STRAIN p13
A = Vallow
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See Figure 2.10 below
FIGURE 2.10 A bolt subjected to shear stress
EXAMPLE 2.2
Chapter 2 SHEAR STRESS AND STRAIN p14
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EXAMPLE 2.3
Chapter 2 SHEAR STRESS AND STRAIN p15
The two members
are pinned
together at B as
shown. Top views
of the pin
connections at A
and B are alsogiven in the figure.
If the pins have anallowable shear stress ofallow = 37 MPa and the allowable tensile stress of rod
CB is (t)allow = 100 MPa, determine the smallest diameter of pinsA and B and
the diameter of rod CB necessary to support the load.
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Simple Shear Strain
Chapter 2 SHEAR STRESS AND STRAIN p16
The suspender rod is supported at
its end by a fixed-connected circular
disk as shown. If the rod passes
through a 40-mm-diameter hole,
determine the minimum required
diameter of the rod and the
minimum thickness of the diskneeded to support the 20-kN load.
The allowable normal stress for therod is allow = 60 MPa, and the allowable shear stress for the disk is allow = 35
MPa.
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(a) Original (undeformed) element (b) Pure Shear DeformationFIGURE 2.5 Illustrations for a definition of shear strain
Chapter 2 SHEAR STRESS AND STRAIN p17
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=
-
*
Shear Strain 2.42
=
-
*
ta
n(
-
*) =
s
2 2L
s
Where sand Ls are defined in Figure 2.5 (b).
= G Hookes Law for Shear 2.5
Chapter 2 SHEAR STRESS AND STRAIN p18
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EXAMPLE 2.4
The plate is deformed into the dashed
shape as shown. If in this deformed
shape horizontal lines on the plate
remain horizontal and do not change
their length, determine
(a) the average normal strain
along the sideAB, and
(b) the average shear strain in the
plate relative to thexand yaxes.
Chapter 2 SHEAR STRESS AND STRAIN p19