1 Review Stresses Pages2

8/10/2019 1 Review Stresses Pages2

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Equilibrium of Deformable Body


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Review Static Equilibrium

external forces, the Newtons Second Law provides us

six scalar equations of equilibrium

0; 0; 0x y

F F Fz 0; 0; 0

x y zM M M

The sum of all the forces in x, y, and z coordinate

directions are zero

The sum of moments of all the forces about axes x,y,

and z are zero


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Considering Static Equilibrium of the element, we have

* Normal and shear stress components acting on oppositesides of an element must be equal in magnitude and

opposite in direction

* Shear stress components satisfy moment equilibrium


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PRINCIPAL STRESSESState of stress at a point in a material is

completely defined by the stresses acting on

Often it is im ortant to determine the state of stress on a lane

edges are parallel to coordinate directions.

at some angle to the coordinate axes. Mostly, the state of stress

on a plane on which no shear stresses act is important for design

.

A plane, where no shear stresses act is called a principal

plane and the normal stress that acts on such a plane is

called principal stress.


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10/16

Mohrs Circle - Its Use and Limitations

coordinate plane, there are no shear stresses and only normal

stress (may be zero or not) is present. That normal stress is one

o e pr nc pa s ress as s own n gure.

You can see that on an plane perpendicular to z-axis only sz ( may

be tensile, compressive or zero) is present. Then remaining two

rinci al stresses can be determined usin biaxial stress field and

Mohr Circle.

Biaxial stress field in xy-plane can be shown as above.

Using stress transformation equation, the normal stress alongx

x

.

be written as.

1 1

cos s n2 2

1sin 2 cos2

2

x x y x y xy

x y x y xy

(A)

will become principal stress if = 0. Using this condition,

the remaining principal stresses can be determined as.x

x y

2

21 1

2 2p x y x y xy

and the orientation of the principal planes with respect to x-

axis is given by

tan2 xy

x y


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The same results can be expressed graphically using Mohrs

Circle. Rearranging the equations (A), we obtain

1 1

cos2 sin 22 2

1sin 2 cos2

x x y x y xy

Squaring the above equations, then adding, gives

2

2 2

2 2

2 2x x y x y x y xy

This is the equation of a circle whose center is at 1

,02

x y

an w ose ra us s g ven y

2

21

2 x y xy

Every point on the circle defines the stress state acting on

planes at any angle q from the original x or y axis.

For the correct construction of Mohrs circle, certain rules are followed and a consistent

handling of positive and negative stress is essential, only if proper orientation of planes isdesired. No such concern is re uired if onl the ma nitudes of the rinci al stresses are

sought.

Although various conventions are in use, we follow the convention given in Hibbler Book.

1.Normal stresses are plotted to scale along the abscissa (horizontal axis) with tensile

stresses considered positive and compressive stresses negative.

2.Shear stresses are plotted along the ordinate (vertical axis) with positive direction

downward to the same scale as used for normal stresses.

A shear stress that would tend to cause counter-clock wise rotation of the stress element in

rotation.


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3.Angle between lines of direction on the Mohr plot are twice the

indicated angle on the physical plane.

The angle 2' on the Mohr circle is measured in the same directionas the angle for the orientation of the plane in physical plane.

According definition, the values of corresponding to the

points D and B (where = 0) are the principal stresses. Theradius of the Mohrs circle gives the maximum in-plane shear

stress

Three -Dimensional Mohrs Plot

As mentioned previously, Mohrs circle can be drawn to determine principal stresses

only if one of the three principal stresses is known.

Since the known principal stress is also a normal stress, it can be plotted on s axis and

circles can be drawn between all the principal stresses as shown.

then the maximum absolute shear stress is equal to radius of largest Mohrs circle.

Absolute maximum shear stress is given as

max min1

2 pmax p


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Static Failure Theories


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2

4 3 3

*( / 2) *( / 2) 32 32*600018108 /

/ 64 *(1.5)x

M d M d Mlb in

I d d

2

4 3 3

( / 2) 16 16*800048288 /

/ 32 1.5zx

Tr T d T lb in

J d d

2

2 2

4 4 4*1000754.5 /3 3 / 4 3* * 1.5 / 4

v

V V

lb inA d

At Point A

2

2 2

1 2, 58184, 40076 /2 2

x z x zxz

lb in

2

max 49130 /lb in


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248288 754 49042 /lb in

At point B

21 22

max

,

49042 /lb in

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