lecture-4_moher_circle.ppt

7/23/2019 lecture-4_moher_circle.ppt

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MOHR'S CIRCLE

The formulas developed in the preceding article may be used for any case of plane

stress. A visual interpretation of them, devised by the German engineer Otto Mohr in

1882, eliminates the necessity for remembering them. !n this interpretation a circle is

used" accordingly, the construction is called Mohr#s, circle. !f this construction is

plotted to scale, the results can be obtained graphically" usually, ho$ever, only a rough

s%etch is dra$n, analytical results being obtained from it by follo$ing the rules given

later.

&e can easily sho$ that '(s. )1* and )2* define a circle by first re$riting them as

follo$s+

2sin2cos22

xy

yxyx

n

+

+

= )1*


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2sin2cos

22xy

yxyx

n

=+

2cos2sin

2

xy

yx +

=

)*

-e$riting the e(uation )1*

)2*

Ta%ing s(uares of e(uations )2* )*

22

/2cos2sin20/0

xyyx

+

=

22 /2sin2cos2

0/2

0

xyyxyx

n

=+

)*

)*


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3y adding e(u.)* )*, and simplifying, $e obtain

( )22

2

2

22xy

yxyx

n

+

=+

+ )4*

-ecall that 56, 5y, and 76y are %no$n constants defining the specified state of stress,

$hereas 5nand 7 are variables. onse(uently, )569 5y*:2 is a constant, say, h, and the

right;hand member of '(. )4* is another constant, say, r.


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The e(uation )=* is similar to e(uation of ircle i.e.,

222

*)*) rkyhx =+


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enter of circle is

2

yx

hC

+==

>rom the origin.

>igure ?;1 represents Mohr#s circle for the state of plane stress that $as analy@ed in

the preceding article. The center is the average of the normal stresses, and the radius

( )22

2xy

yxrR

+

==

>rom figure

2

yxa

=


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is the hypotenuse of the right triangle A. Bo$ do the coordinates of points ', >, and

G compare $ith the e6pressions derived for 51,52 ,7ma6C&e shall see that Mohr#s circle

is a graphic visuali@ation of the stress variation given by '(s. )1* and )2*. The

follo$ing rules summari@e the construction of Mohr#s circle.

Figure 9-14 Mohr#s circle for general state of plane stress.


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Rules for Applying Mohr's Circle to Combined Stresses

1. On rectangular 5;7 a6es, plot points having the coordinates )56, 76y* and )5y, 7y6*.

These points represent the normal and shearing stresses acting on the 6 and y faces of

an element for $hich the stresses are %no$n. !n plotting these points, assume tension as

plus, compression as minus, and shearing stress as plus $hen its moment about the

center of the element is cloc%$ise.

2. Doin the points Eust plotted by a straight line. This line is the diameter of a circle

$hose center is on the a a6is.

. As different planes are passed through the selected point in a stressed body, the

normal and shearing stress components on these planes are represented by the

coordinates of points $hose position shifts around the circumference of Mohr#s circle.


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. The radius of the circle to any point on its circumference represents the a6is directed

normal to the plane $hose stress components are given by the coordinates of that point.

. The angle bet$een the radii to selected points on Mohr#s circle is t$ice the angle

bet$een the normal to the actual planes represented by these points, or to t$ice the

space angularity bet$een the planes so represented. The rotational sense of this angle

corresponds to the rotational sense of the actual angle bet$een the normal to the

planes" that is, if the n a6is is actually at a countercloc%$ise angle F from the 6 a6is,

then on Mohr#s circle the n radius is laid off at a countercloc%$ise angle 2F from the 6

radius.


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2!2 2!1

56, 6y

5y, ;6y

"2

"1

6;a6is

v, v1plane

6

y;a

6is

B,

B1

plane y

2

yx

"#

+2

yx

"y

2s1

m$#

min

2s2


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%#$mple !roblem 1

!t has been determined that a point in a load;carrying member is subEected to the

follo$ing stress condition+

56HHMIa 5y;HHMIa 76y2HHMIa)&*

Ierform the follo$ing

)a* ra$ the initial stress element.

)b* ra$ the complete MohrJs circle, labeling critical points.

)c* ra$ the complete principal stress element.

)d* ra$ the ma6imum shear stress element.


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Solution

The 1;step Irocedure for dra$ing Mohr#s circle is used here to complete the problem.

The numerical results from steps 1;12 are summari@ed here and sho$n in >igure 11;12.

Ktep 1. The initial stress element is sho$n at the upper left of >igure 11;12.

Ktep 2. Ioint 1 is plotted at a6 HH MIa and 76y 2HH MIa in (uadrant 1.

Ktep . Ioint 2 is plotted at ay ;HH MIa and 7y6 ;2HH MIa in (uadrant .

Ktep . The line from point 1 to point 2 has been dra$n.

Ktep . The line from step crosses the 5;a6is at the average applied normal stress,

called O in >ig 11;12, is computed from any,


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( ) [ ] MPayxavg H*HH)HH2121 =+=+=

Ktep 4. Ioint H is the center of the circle. The line from point O through point 1 is

labeled as the 6;a6is to correspond $ith the 6;a6is on the initial stress element.

Ktep =. The values of G, b, and - are found using the triangle formed by the lines

from point H to point 1 to 56 HH MIa and bac% to point O.

The lo$er side of the triangle,

( ) [ ] MPaa yx H*HH)HH2121 ===


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FIG 11-12 Complete Mohrs circle


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The vertical side of the triangle, b, is completed from

MPab xy 2HH==

The radius of the circle,R, is completed from+

MPabaR H*2HH)*H) 2222 =+=+=Ktep 8. This is the dra$ing of the circle $ith point H as the center at 5 avg H MIa

and a radius of - H MIa.

Ktep ?. The vertical diameter of the circle has been dra$n through point O. The

intersection of this line $ith the circle at the top indicates the value of 7ma6 H MIa,

the same as the value of -.

Ktep 1H. The ma6imum principal stress, 51, is at the right end of the hori@ontal

diameter of the circle and the minimum principal stress, 52, is at the left.

Ktep 11. The values for al and a2 are


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MPaRO

MPaRO

HH

HH

2

1

====+=+=

Ktep 12. The angle 2L is sho$n on the circle as the angle from the 6;a6is to the 51;a6is,

a cloc%$ise rotation. The value is computed from

o=.2?H

2HHtan2 1 ==

ote that 2L is & from the 6;a6is to 51on the circle.

oo

8=.1

2

=.2?==

Ktep 1. igure 11;1)b*. The element is rotated 1.8=H& from the original 6;a6is

to


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FIG 11-13 Results for Example Problem 11-2

the face on $hich the tensile stress 51 MIa acts. The compressive stress 2 ;

MIa acts on the faces perpendicular to the al faces.

Ktep 1. The angle 2LJ is sho$n in >igure 11;12 dra$n from the 6 ;a6is & to the

vertical diameter that locates 7ma6at the top of the circle. !ts value can be found in either

of t$o $ays. >irst using '(uation 11;8 and observing that the numerator is the same as

the value of a and the denominator is the same as the value of b from the construction

of the circle. Then


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CCWoba 24.4H*)tan*)tan#2 2HHH11

===

Or, using the geometry of the circle. $e can compute

CCWoooo 24.4H=.2??H2?H#2 ===

Then the angle LJ is one;half of 2LJ.

o

o

1.H224.4H# ==

Ktep 1. The ma6imum shear stress element is dra$n in >igure 11;1)c*, rotated H.1N

& from the original 6;a6is to the face on $hich the positive 7ma6acts. The ma6imum

shear stress of H MIa is sho$n on all four faces $ith vectors that create the t$o pairs

of opposing couples characteristic of shear stresses on a stress element. Also sho$n is

the tensile stress 5ma6 H MIa acting on all four faces of the element.


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Summ$ry of Results for %#$mple !roblem 1 Mohr's Circle

Given 56HMIa 5y ;HHMIa 76y2HHMIa &

-esults >igures 11;12 and 11;1.

51MIa 52 ;MIa L1.8=o& from 6;a6is

7ma6HMIa 5avgHMIa LJH.1o& fron 6;a6is


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%#$mple !roblem 2

Given 56;12HMIa 5y 18HMIa 76y8HMIa &

-esults >igures 11;1.

512HHMIa 52 ;1HMIa L=.?4o&

7ma61=HMIa 5avgHMIa LJ?.Ho

&





Solution&


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Figure 11-1 Result for %#$mple !roblem 11-4( )-$#is in the third *u$dr$nt+


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%#$mple !roblem ,

Given 56;H%si 5y2H %si 76yH %si &

-esults >igures 11;.

512.1= %si 5

2 ;2.1= %si L41.Ho&

7ma6=.1= %si 5avg;.H %si LJ14.Ho&

omments The 6;a6is is in the fourth (uadrant.





Solution&


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Figure 11-1 Result for %#$mple !roblem 11-( )-$#is in the fourth *u$dr$nt+

%#$mple !roblem4

Given 5622HMIa 5y;12HMIa 76yHMIa

-esults >igures 11;1=.

5122HMIa 52 ;12HMIa LHo

7ma61=HMIa 5avgHMIa LJ.Ho&

Solution&


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Fig 11-1. Result for %#$mple !roblem 11-(Speci$l c$se of bi$#i$l stress /ith no

she$r


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%#$mple !roblem &

Given 56H %si 5yH %si 76yH%si

-esults >igures 11;18.

51H %si 52H %si LHo

7ma62H %si 5avg2H %si LJ.Ho&

Solution&






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Fig 11-10 Results of %#$mple !roblem 11-.+ Speci$l c$se of uni$#i$l tension

%#$mple !roblem

Given 56H %si 5yH %si 76yH%si &

-esults >igures 11;1?.

51H %si 52;H %si Lo &

Solution&


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7ma6H %si 5avgH %si LJHo


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Fig 11-19 Results of %#$mple !roblem 11-0( Speci$l c$se of !ure she$r+


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%#$mple !roblem .&

At a certain point in a stressed body, the principal stresses are 56 8H MIa and 5y ;H

MIa. etermine 5 and 7 on the planes $hose normal are at 9HN and 9 1 2HN $ith the 6

a6is. Kho$ your results on a s%etch of a differential element.


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Solution&The given state of stress is sho$n in >ig. ?; 1 a. >ollo$ing the rules given

previously, dra$ a set of rectangular a6es and label them a and r as sho$n in >ig. ?;

1b. )ote that, for convenience, the stresses are plotted in units of MIa.* Kince the

normal stress component on the x face is 8H MIa and the shear stress on that face is

@ero, these components are represented by point A $hich has the coordinates )8H, H*.

Kimilarly, the stress components on they face are represented by pointB );H, H*.

According to rule 2, the diameter of Mohr#s circle is AB. !ts center C, lying mid$ay

bet$eenA andB, is 2H MIa from the origin O. The radius of the circle is the distance

CA 8H ; 2H 4H MIa. >rom rule , the radius CA represents thex a6is. !n accordance

$ith rules and , point D represents the stress components on the face $hose normal

is inclined at 9HN to the x a6is, and point E represents the stress components on the

perpendicular face. Observe that positive angles on the circle are plotted in a

countercloc%$ise direction from the x a6is and are double the angles bet$een actual

planes.


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This special rule of sign for shearing stress ma%es x= yx in Mohr#s circle. >rom

here on, $e use this rule to designate positive shearing stress. Bo$ever, the

mathematical theory of elasticity uses the convention that shearing stress is positive$hen directed in the positive coordinate direction on a positive face of an element, that

is, $hen acting up$ard on the right face or right$ard on the upper face. This other rule

ma%es xy= yx, $hich is convenient for mathematical $or% but confusing $hen applied

to Mohr#s circle.


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Figure 9-1


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>rom rule , the coordinates of pointD represent the re(uired stress components on the

HN face. >rom the geometry of Mohr#s circle, these values are

MPaCFOCOF

o

H4Hcos4H2H =+=+==

MPaDF oo H.24Hsin4H ===

On the perpendicular 12HN face $e have

MPaCGOCOG o 1H4Hcos4H2H# ====

MPaGE o H.24Hsin4H# ===3oth sets of these stress components are sho$n on the differential element in >ig. ?;14.

Observe the cloc%$ise and countercloc%$ise moments of and , respectively, relative

to the center of the element )see rule 1*. >inally, note that a complete s%etch of a

differential element sho$s the stress components acting on all four faces of the element

and the angle at $hich the element is inclined.


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Figure 9-1

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lecture-4_moher_circle.ppt