1_Load and Stress Analysis

Load and Stress AnalysisDESIGN OF MACHINE ELEMENT

Lec Amir HamzaDepartment of Mechatronics Engineering

Spring - 2011

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Load and Stress Analysis

Equilibrium Free Body Diagram Shear Force and Bending Moment in beams Stress Cartesian Stress Components Plane Stress Mohr’s Circle for Plane Stress General Three Dimensional Stresses Elastic Strain

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Equilibrium

If we assume that the system to be studied is motionless or, at most, has constant velocity, then the system has zero acceleration. Under this condition the system is said to be in equilibrium

The phrase static equilibrium is also used to imply that the system is at rest. For equilibrium, the forces and moments acting on the system

balance such that

which states that the sum of all force and the sum of all moment vectors acting upon a system in equilibrium is zero.

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Free Body Diagram

We can greatly simplify the analysis of a very complex structure or machine by successively isolating each element and studying and analyzing it by the use of free-body diagrams

Thus, free-body diagramming is essentially a means of breaking a complicated problem into manageable segments, analyzing these simple problems, and then, usually, putting the information together again

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Shear Force and Bending Moment in beams

If the beam is cut at some section located at x = x1 and the left-hand portion is removed as a free body, an internal shear force V and bending moment M must act on the cut surface to ensure equilibrium

Shear force and bending moment are related by the equation

dMV

dx=

Fig 1

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Shear Force and Bending Moment in beams

Sometimes the bending is caused by a distributed load q(x)

q(x) is called the load intensity with units of force per unit length and is positive in the positive y direction

Fig 2

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Stress

When an internal surface is isolated as in the last F.B.D, the net force and moment acting on the surface manifest themselves as force distributions across the entire area

The force distribution acting at a point on the surface is unique and will have components in the normal and tangential directions called normal stress and tangential shear stress, respectively

Normal and shear stresses are labeled by the symbols σ and τ

If the direction of σ is outward from the surface it is considered to be a tensile stress and is a positive normal stress. If σ is into the surface it is a compressive stress and commonly considered to be a negative quantity

Units: U.S Customary units are pounds per square inch (psi)SI units, stress is in newtons per square meter (N/m2): 1N/m2 = 1 Pa.

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Cartesian Stress Components

The Cartesian stress components are established by defining three mutually orthogonal surfaces at a point within the body

The normals to each surface will establish the x, y, z Cartesian axes

In general, each surface will have a normal and shear stress The shear stress may have components along two axes

Fig 3

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The state of stress at a point described by three mutually perpendicular surfaces is shown in Fig a

As the dimensions of the cube in Fig. 4a approach zero, the stresses on the hidden faces become equal and opposite to those on the opposing visible faces

Fig 4

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Thus, in general, a complete state of stress is defined by nine stress components

σx , σy , σz, τxy ,τxz, τyx , τyz, τzx , and τzy

For equilibrium, in most cases, “cross-shears” are equal, hence

τyx = τxy τzy = τyz τxz = τzx

This reduces the number of stress components for most 3-D states of stress from nine to six quantities

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A very common state of stress occurs when the stresses on one surface are zero. When this occurs the state of stress is called plane stress

Figure b shows a state of plane stress, arbitrarily assuming that the normal for the stress-free surface is the z direction such that

σz = τzx = τzy = 0

Here it is assumed that the cross-shears are equal such that τyx = τxy , and τyz = τzy = τxz = τzx = 0


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Plane Stress

Suppose the dx dy dz element of Fig 4b is cut by an oblique plane with a normal n at an arbitrary angle φ counterclockwise from the x axis as shown in Fig. 5

We are concerned with the stresses σ and τ that act upon this oblique plane

By summing the forces caused by all the stress components to zero, the stresses σ and τ are found to be

plane-stress transformation equations

cos 2 sin 22 2

x y x yxy

s s s ss f t f

+ -= + +

sin 2 cos 22

x yxy

s st f t f

-= +

Fig 5

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Differentiating the first Eq. with respect to φ and setting the result equal to zero gives

This Equation defines two particular values for the angle 2φp, one of which defines the maximum normal stress σ1 and the other, the minimum normal stress σ2. These two stresses are called the principal stresses, and their corresponding directions, the principal directions

It is important to note that the above Eq. can be written in the form

Comparing, we see that τ = 0, meaning that the surfaces containing principal stresses have zero shear stresses.

Plane Stress

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In a similar manner, we differentiate Eq. of shear stress, set the result equal to zero, and obtain

It defines the two values of 2φs at which the shear stress τ reaches an extreme value

Also,

Plane Stress

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Formulas for two principal stresses and two extreme valued shear stresses are found to be

Plane Stress

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Mohr’s Shear Convention The relationship between σ and τ is that of a circle plotted in the σ, τ plane,

where the center of the circle is located at C = (σ, τ ) = [(σx + σy)/2, 0] and has a radius of R = ([(σx − σy)/2]2 + τ2

xy )1/2. Shear stresses tending to rotate the element clockwise (cw) are plotted

above the σ axis Shear stresses tending to rotate the element counterclockwise (ccw) are

plotted below the σ axis.

Mohr’s Circle for Plane Stress

17Fig 6

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A stress element has σx = 80 MPa and τxy = 50 MPa cw, as shown in Fig. 7

(a) Using Mohr’s circle, find the principal stresses and directions, and show these on a stress element correctly aligned with respect to the xy coordinates. Draw another stress element to show τ1 and τ2, find the corresponding normal stresses, and label the drawing completely.

(b) Repeat part a using the transformation equations only.

Mohr’s Circle for Plane Stress

Fig 7

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General Three-Dimensional Stresses

As in the case of plane stress, a particular orientation of a stress element occurs in space for which all shear-stress components are zero

When an element has this particular orientation, the normals to the faces are mutually orthogonal and correspond to the principal directions, and the normal stresses associated with these faces are the principal stresses

Since there are three faces, there are three principal directions and three principal stresses σ1, σ2, and σ3. For plane stress, the stress-free surface contains the third principal stress which is zero

In our studies of plane stress we were able to specify any stress state σx , σy and τxy and find the principal stresses and principal directions

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Six components of stress are required to specify a general state of stress in three dimensions, and the problem of determining the principal stresses and directions is more difficult

In plotting Mohr’s circles for three-dimensional stress, the principal normal stresses are ordered so that σ1 ≥ σ2 ≥ σ3. Then the result appears as in Fig. 8

The stress coordinates σ , τ for any arbitrarily located plane will always lie on the boundaries or

within the shaded area


Fig 8

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Fig 8 also shows the three principal shear stresses τ1/2, τ2/3, and τ1/3

Each of these occurs on the two planes, one of which is shown in Fig. 9

The figure shows that the principal shear stresses are given by the

equations

Of course, τmax = τ1/3 when the normal principal stresses are ordered (σ1 > σ2 > σ3), so always order your principal stresses


Fig 9

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For the stress states listed below, find all three principal normal and shear stresses. Draw a complete Mohr’s three-circle diagram and label all points of interest

(a) σx = 10, σy = −4

(b) σx = 10, τxy = 4 ccw


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Solution (a)

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Solution (b)

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Elastic Strain

Normal strain ε is defined for the tensile specimen and is given by ε = δ/l

Hooke’s law for the tensile specimen is given byσ = Eε

where the constant E is called Young’s modulus or the modulus of elasticity

When a material is placed in tension, there exists not only an axial strain, but also negative strain (contraction) perpendicular to the axial strain. Assuming a linear, homogeneous, isotropic material, this lateral strain is proportional to the axial strain

If the axial direction is x, then the lateral strains are εy = εz = −νεx

The constant of proportionality v is called Poisson’s ratio

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If the axial stress is in the x direction, then

For a stress element undergoing σx , σy , and σz simultaneously, the normal strains are given by

Shear strain γ is the change in a right angle of a stress element when subjected to pure shear stress, and Hooke’s law for shear is given by

(G is the shear modulus of elasticity or modulus of rigidity)

The three elastic constants are related to each other by

Elastic Strain

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1_Load and Stress Analysis