2 Shear Stress and Strain Transparency

8/14/2019 2 Shear Stress and Strain Transparency

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DMV 4343JAN ~ JUN 07

(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



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(c) A pair of pliers (d) Direct shear of pin

FIGURE 2.2 Examples of direct shearSingle Shear



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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



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FIGURE 2.4 Direct single shear (a) and (b) Bolted lap joint, (c) and (d)

Glued lap joint



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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



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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



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2.2 Shear Deformation

FIGURE 2.7 Three-dimensional state of stress



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FIGURE 2.8 Three-dimensional state of stress



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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.



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FIGURE 2.9 Pure shear deformation

Allowable Stress



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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.


A = Vallow


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See Figure 2.10 below

FIGURE 2.10 A bolt subjected to shear stress

EXAMPLE 2.2



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EXAMPLE 2.3


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


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



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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



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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.


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2 Shear Stress and Strain Transparency