2 Shear Stress and Strain

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DMV 4343JAN ~ JUN `07

DEPARTMENT MANUFACTURING / PRODUCT DESIGN /MOULD / TOOL AND DIE

SEMESTER 4 / 6

COURSE MECHANICS OF MATERIALS DURATION 8 hrsCOURSE CODE DMV 4343 / DMV 5343 REF. NO.

VTO’S NAME MISS AFZAN BINTI ROZALIMR RIDHWAN BIN RAMELI

PAGE 13

TOPICSHEAR STRESS AND STRAIN

SUB TOPIC2.1 Simple Shear Stress2.2 Shear Deformation2.3 Simple Shear Strain

Chapter 2 SHEAR STRESS AND STRAIN p1

INFORMATION SHEET





REF NO. :

PAGE :




2.1 Simple Shear Stress

In the preceding sections of Chapter 1, you were introduced to stress and strain through a

discussion of normal stress and extensional strain. We turn now to a discussion of shear

stress and shear strain, which are used, respectively, to quantify the distribution of force

acting tangent to a surface and the angle change produced by tangential forces.

Definition of Shear Stress.

Shear stress is the intensity of force, or force per unit area, acting tangent to ΔA. It is

denoted by lower Greek later, τ (tau).

Referring to Figure 2.1a, we define the shear stress at a point by the equation1''

τ =

lim

Δ A

→0

( ΔV

Δ A

) Shear

Stress

where ΔV is the tangential (shear) force acting on an infinitesimal area Δ A at the point

where the shear stress is to be determined. As in the case of normal stress σ, the units of

shear stress are force/area; hence, usually psi or ksi in the USCS units, and kPa or MPa in

the SI system.

The resultant shear force, shown in Figure 2.1b, is obtained by summing the ΔV ‘s over the

cross section, giving

(a) The distribution

of shear force on a

sectioning plane

(b) The resultant

shear force on the

sectioning planeFIGURE 2.1 Shear force on a sectioning plane


∑ F x : V = ∫ A τ dA 2.1








Average Shear Stress.

Even when the exact shear stress distribution on a surface cannot be readily determined, it

is sometimes useful to calculate the average shear stress on the surface. This is given by

τ avg =

V

As

Average Shear

Stress 2.2

where V is the total shear force acting on area As. In order to determine τ avg we must first

determine what area has shear stress acting on it, and then, using a free-body diagram,

determine the value of the shear force, V, acting on this area.

Direct Shear.

The average shear stress can be readily calculated in the case of direct shear, examples of

which are shear in bolts, pins, and rivets, and shear in welds and lap splices.

Direct shear (or simple shear) is caused by forces that act parallel to a particular surface of some part,

with the direct result of shearing, or lending to shear (i.e., sever), the material at that surface.

For example, in the case of the sheet-metal punch in Figure 2.2a, the force P of the punch

rod shears the sheet metal on a cylindrical surface, producing a coin-shaped metal slug

(Figure 2.2b). The shear stress for this example of direct shear is

τ avg =

V

As

=P

(Πd )t

where V = P is determined from the free-body diagram in Figure 2.2b, As is the area of the

cylindrical surface, t is the thickness of the sheet metal, and d is the diameter of the punch.

When the pliers in Figure 2.2c are gripped, the pin that holds the two arms of the pliers

together is subjected to direct shear on the pin cross section indicated in Figure 2.2d .

(a) A sheet-metal punch (b) Free-body diagram of

sheet-metal slug





(c) A pair of pliers (d) Direct shear of pinFIGURE 2.2 Examples of direct shear

Single Shear

Many of the circumstances that can be characterized as direct shear may be further

classified as single shear or as double shear. This applies particularly to connections such

as pinned, bolted, or welded joints. A single-shear connection is one where there is a single

plane on which shear stress acts to transfer load from one member to the adjacent member.

The pin of the pliers in Figure 2.2d is one example of a single-shear connection.

As another example of single shear, consider the lap joint, or lap splice, in Fig. 2.3a, where

two rectangular bars are glued together to form a tension member. (Assume that, because

the bars are very thin, the moment Pt caused by misalignment of the tensile forces, P , may

be neglected.)

(a) A lap splice (b) The free-body diagram (c) The average-shear-stress distributionFIGURE 2.3 An illustration of direct shear – a lap splice.

Applying ∑F = 0 to the free-body diagram in Fig. 2.3b, we get V = P . The area on which the

shear V acts is As = Lsw. Therefore, the average shear stress on the splice area is

τ avg =

V

As

=P

Lsw





Although the transfer of load from one member to another through a riveted or bolted joint is

not easily analyzed when there are several rows of bolts or rivets at the joint, the case of

load transfer through a single bolt or a single pin can be treated as direct shear.

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

Double Shear

While in case of members subjected to double shear, the shear force, V is half of the direct

load applied to it. This is illustrated in the following Figure 2.5.

FIGURE 2.5 Direct double shear (a) and (b) Bolted lap joint, (c) and (d) Glued lap joint

So, the shear load, V for double shear is given by

V = F/2

The following example illustrates the calculation of direct shear stress in single-shear and

double-shear joints.





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, one of which is indicated as abcd.

FIGURE 2.5





2.2 Shear Deformation

In order to be able to determine stresses on an infinite number of plane passing through a

point O (Figure 2.6), thus defining the state of stress at the point, we need only specify the

stress components on three mutually perpendicular planes passing through the point. These

planes, perpendicular to the coordinate axes, contain three sides of an infinitesimal cubic

element.

FIGURE 2.6

This three-dimensional state of stress acting on an isolated element within a body is shown

in Figure 2.7.

FIGURE 2.7 Three-dimensional state of stress

Stresses are considered to be identical at points O and O' and are uniformly distributed on

each face. They are indicated by a single vector acting at the centre of each face. We

observe total of nine components of stress that compose three groups of stresses acting on

the mutually perpendicular planes passing through O.

FIGURE 2.8 Three-dimensional state of stress





Now lets consider a volume element of material taken at a point located on the surface of

Figure 2.8. If we consider force equilibrium in the x-direction, then

Force Force

Stress

x Area Stress x Area

∑Fx = 0: τyx (ΔxΔz) - τ`xy (ΔxΔz) = 0τyx = τ`yx

And in a similar manner, force of equilibrium in the y-direction yield

τxy = τ`xy

Finally, by taking moment about z-axis

Moment MomentForce x Distance Force x DistanceStres

s

x

Areax Arm Stress

x

Areax Arm

∑Mz = 0: τyx (ΔxΔz) (Δx) - τxy (ΔxΔy) (Δz) = 0τyx = τxy

So, τyx = τ`yx = τxy = τ`xy = τ. Thus Figure 2.8 can be replaced by Figure 2.9.

FIGURE 2.9 Pure shear deformation

We shall employ here a sign convention that applies to both normal and shear stresses and

that is based upon the relationship between the direction of an outward normal drawn to

surface and the direction of the stress components on the same surface. When both theouter normal and the stress component point in a positive (or negative) direction relative to

the coordinate axes, the stress is positive. When the normal points in a positive direction

while the stress points in a negative direction (or vice versa), the stress is negative.

Accordingly, tensile stresses are always positive and compressive stresses always negative.

It is clear that the same sign and the same notation apply no matter which face of a stress

element we choose to work with. Figure 2.9 depicts positive normal and shear stresses. This

sign convention for stress, which agrees with that adopted for internal forces and moments.





Allowable Stress

Recall from previous chapter,

F.S =F fail

F allow

So, for a body that is subjected to shear stress,

F.S =τ fail

τ allow

We can design the dimension of the body to sustained the allowable shear stress, τ allow , to be

within the range of the decided factor of safety which is generally bigger than 1.

From the calculated allowable shear stress, τ allow , we can determine the area and hence the

dimension as well.

A =V

τ allow

See Figure 2.10 below

FIGURE 2.10 A bolt subjected to shear stress





EXAMPLE 2.2


The two members are pinned

together at B as shown. Top views

of the pin connections at A and B

are also given in the figure. If the

pins have an allowable shear

stress of τallow = 37 MPa and the

allowable tensile stress of rod CB

is (σt)allow = 100 MPa,determine the smallest diameter of pins A and B and the diameter of rod CB necessary to

support the load.




EXAMPLE 2.3


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 disk needed to support the

20-kN load. The allowable normal stress for the rod is

σallow = 60 MPa, and the allowable shear stress for the

disk is τallow = 35 MPa.




2.3 Simple Shear Strain

Shear strain is the change in angle that occurs between two line segments that were

originally perpendicular to one another. It is denoted by lower Greek letter, γ (gamma) and is

measured in radians (rad). Shear strain on a body subjected to shear stress is illustrated inFigure 2.5.

(a) Original (undeformed)element

(b) Pure Shear Deformation

FIGURE 2.5 Illustrations for a definition of shear strain

Referring to Figure 2.5 let us now consider the shear strain that is associated with shear

stress. As a result of the shear stress τ, the original right angle at A becomes an acute angle

θ * . The shear strain γ (lowercase Greek letter gamma) at A is defined as the change in angle

between two originally perpendicular line segments that intersect at A. Thus,

γ =

π

-

θ

* Shear Strain 2.42

where π / 2 is the angle at A before deformation, and θ * is the angle at A after deformation.

The corresponding shear stresses point toward the two corners where the original right

angle is decreased by γ, and they point away from the two corners where the angle is

increased by γ. Although γ is dimensionless, it is frequently dated in the same

"dimensionless units" as extensional strain, that is, in./in., and on, or, since shear strain is an

angle, it may be stated in radians. Since shear strains, like extensional strains, are usually

very small in magnitude, we can use the small-angle approximations tan(-γ) = γ and sin(γ) =

γ. Then, γ can be computed using the formula

γ =

π

-θ

* ≈

ta

n(

π

-θ

* ) =

δs

2 2L

s

Where δ s and Ls are defined in Figure 2.5 (b).





Material Properties in Shear.

Material properties relating to shear, like those for normal-stress-extensional-strain

behaviour, must be determined experimentally. The material properties in shear, such as

yield stress in shear, shear modulus of elasticity, and so on, may be obtained from a torsion

test, which will be discussed later in Section 4.4. For example, linearly elastic behavior in

shear is described by Hooke's Law for shear, which

τ = G γ Hooke’s Law for Shear 2.5

The constant of proportionality, G, is called the shear modulus of elasticity, or, simply, the

shear modulus. Like E, the shear modulus G is usually expressed in units of ksi or GPa. The

shear properties are closely related to the extensional properties through equations of

equilibrium and geometry of deformation.





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 side AB,

and

(b) the average shear strain in the plate relative

to the x and y axes.


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2 Shear Stress and Strain