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Unformatted text preview: Corresponding Variables in the Transformation Equations for Plane Stress and Plane Strain Comparison of Equations
Plane Strain Plane Stress
Transformation Equations 8 +8 8 —s 7 . 0' +0" 0 ‘0 
gx = x y + x y00326+ism26 ox = x y + x y00326+rxys1n26
l 2 2 2 I 2 2
y 6‘ ”8 _ 7 0' “0' .
x'y' =— x y sm2t9+—xy 00326 rm =— x y 3111 26+er 00326 Plane Strain 1 Note: >The shear strains are zero on the principal planes. >At a given point in a stressed body, the principal strains and principal stresses occur in the same directions. Mohr’s Circle for Plane Strain 4 ~61 '9 B(0= 90°) Strain Rosette yxlyl
2 ea = 3x cos2 (49a ) + 8), sin2 (6“) + yxy sin(6a)cos(6la) 8b = 8x cos2 (6,, ) + By sin2 (49b ) + 7n sin(6,, )cos(6b) so = 8x cos2 ((90)+ 8y sin2 (66) + yxy sin (66)cos(66) c< Plane Strain x. SECTION 7.7 Plane Strain 595 An element of material in plane strain undergoes the following strains:
6, : 340 x 10’6 67,. : 110x 10’“ y“, 2 180 x 10’6 These strains are shown highly exaggerated in Fig. 737a, which shows the
deformations of an element of unit dimensions. Since the edges of the element
have unit lengths, the Changes in linear dimensions have the same magnitudes as
the normal strains EX and 6y. The shear strain 7“, is the decrease in angle at the
lowerleft corner of the element. Determine the following quantities: (a) the strains for an element oriented at
an angle 6 : 30°, (b) the principal strains, and (C) the maximum shear strains.
(Consider only the in—plane strains, and show all results on sketches of properly
oriented elements.) (/ 6: 300 (C) ((1) FIG. 737 Example 7—7. Element of
material in plane strain: (a) element
Oriented to the x and y axes, (b) element
Oriented at an angle 0 = 30°, (C) pﬁncipal strains, and ((1) maximum
shear strains. (Note: The edges of the
Clements have unit lengths.) continued 595 CHAPTER 7 Analysis of Stress and Strain Solution (3) Element oriented at an angle 6 : 30°. The strains for an element oriented
at an angle 0 to the x axis can be found from the transformation equations (Eqs.
7713 and 7—71b)1 As a preliminary matter, we make the following calculations: er :3: : (340 +110)10'6 :22 x1""
2 2 5 0
_ _ is
3 e), : (340 110m) :115 ><10‘6
2 2
h:90x10”6
2 Now substituting into Eqsi (77713) and (7~7 lb), we get 6, + 6, ex —/ 6‘, , .
6" ~— 3 y 1 ’~ ’ cos 26 l '7“ sm 20 : (225 x 10”") + (115 X 10’6)(cos60°) + (90 x 10’6)(sin 60°)
: 360 x 10 6 YA r Ex 7 61 . xv
1L : ,,,, ’ 511126+ 1; C05 26
2 2 (— : ~e(l [5 X 1(l76)(sin 60°) + (90 X l076)(COS 60") : ~55 x 10’“ Therefore. the shear strain is 7/1612): i110>< 10’6 The strain 5),] can be obtained from Eq. (772), as follows: 6, : ex + are“ = (340 +110 — 360)10e = 90 x1043 )1 The strains 6X1, Eyv and Wm are shown in Fig. 7—37b for an element oriented at
6 : 300. Note that the angle at the lower—left corner of the element increases
because 75m is negative.
(b) Principal strains. The principal strains are readily determined from
'1 Eq. (7~74), as follows: 6X+E. 5X75.2 yx 2
ELZA 2 ) ~ /( 2 'y)+<2y) : 225 x 10’6 t \/(115 X 10*)2 + (90 x 1056)2 + : 225 x 10’6 i 146 >< 10s6 Thus, the principal strains are 61 z 370 X 10‘6 62 : 80 x 10% HG. 737c and d (Repeated) (C) SECTION 7.7 Plane Strain 5Q? 290 X 1043 «l (d) larger principal strain and 62 denotes in which 61 denotes the algebraically
Recall that we are considering only the algebraically smaller principal strain (
inaplane strains in this example.) The angles to the principal directions can be obtained from Eq. (7—73): tan 2a,, : y” : ”lg—Or : 0.7826
6‘ , 6). 340 — llO The values of 20,, between 0 and 360C are 38.00 and 2180", and therefore the angles to the principal directions are
Hr.» : 19,00 and 109.00 ociated with each principal strain, we sub— To determine the value of (9‘, ass
miation equation (Eq. 7—7 121) and solve for stitute 0,, : 19.0O into the first transfo the strain: + r 6t + F ' '
’12.; + ”7%," cos 26 + h SUI 20
2 7 2 = (225 X in 6) + (115 >< l()”6)(cos 38.00) + (90 X 10’6)(sin 380°) : 370 ><10’6 This result shows that the larger principal strain 61 is at the angle 0p] : 190°.
The smaller strain 62 acts at 90° from that direction (sz : 109.00). Thus, el : 370 x 10*6 and 6,,x = 19.00 an x 104‘ and 9,,2 : 109.0o H 62 Note that 61 + 62 : ex + 5y,
The principal strains are portrayed in Fig. shear strains on the principal planes. 7—37c. There are, of course, no continued LP! 598 CHAPTER 7 Analysis of Stress and Strain (c) Maximum shear strain. The maximum shear strain is calculated from
Eq. (7—75): ‘ X 1' 7V l i;V 2
MW : K? 6’) + (Vi : ‘46 X W‘ mm : 290 >< inn6
2 \ t 2 , v / The element having the maximum shear strains is oriented at 45° to the principal
directions; therefore, 0‘. : 19.00 + 45° ': (34.00 and 26K : 128.00. By substitut—
ing this value of 20‘ into the second transformation equation (Eq. 7—7lb), We can
determine the sign of the shear strain associated with this direction. The calcula— tions are as follows: . 6‘. " 6, , ,x‘
11 :: 7 7'7iL SI“ 26 + 1“ COS 20
2 2 2 : 7(115 >< 10"*)(sin 128.00) + (90 x 10’6)(c05128.0°) : 7 l4() >< 10"" This result shows that an element oriented at an angle 652 : 64.00 has the maxi— mum negative shear strain. ‘
We can arrive at the same result by observing that the angle 951 to the direce tion of maximum positive shear strain is always 450 less than 8“. Hence, 8M : 6pl * 45O : 19.0O , 45D 2 26.0° 9,2 : 65, + 9 0 : 64.00 The shear strains corresponding to 6r; and (95,2 are 'ymax : 290 X 1043 and
7min : 7290 X 10“], respectively. The normal strains on the element having the maximum and minimum
shear strains are + t
: 6‘ 6’ : 225 x 10*6 Eaver /) A sketch of the element having the maximum in—plane shear strains is shown in
Fig. 737d. ln this example, we solved for the strains by using the transformation equa—
tions. However, all of the results can be obtained just as easily from Mohr’S circle. , as ___’ SECTION 7.7 Plane Strain 599 Example 73 ' (b) HG 9’38 Example 78. (a) 45° strain
rosette, and (b) element oriented at an
angle (9 to the xy axes A 45° strain rosette (also called a rectangular rosette) consists of three electrical
resistance strain gages arranged to measure strains in two perpendicular direc—
tions and also at a 45° angle between them, as shown in Fig. 7—38a. The rosette
is bonded to the surface of the structure before it is loaded. Gages A, B, and C
measure the normal strains ea, 6b, and EC in the directions of lines 0a, Oh, and
()c, respectively. Explain how to obtain the strains EX], 6),], and 7mm associated with an ele—
ment oriented at an angle 6 to the xy axes (Fig. 738b). Solution
At the surface of the stressed object, the material is in plane stress. Since the strain—transformation equations (Eqs. 747la and 7771b) apply to plane stress as
well as to plane strain. we can use those equations to determine the strains in any desired direction.
Strains associated with the xy axes. We begin by determining the strains associated with the xy axes. Because gages A and C are aligned with the x and y
axes, respectively, they give the strains er and 6), directly: 6" :» g, e. 2 6‘. ' (7—77a,b) To obtain the shear strain y”. we use the transformation equation for normal
strains (Eq. 7—7121): ﬂ ’6‘ + 6' 7‘ Ex A, E)‘ ‘ y 359' sin 26 For an angle 0 : 450, we know that ex} : 6,, (Fig. 77383); therefore, the pre—
ceding equation gives ea + 6C ea  6(
eb : T— + 2 7x y (cos 90°) + (sin 90") Solving for y”. we get
ny : 26/7 V 6a 7 ft (778) Thus, the strains EX, 6)., and y“, are easily determined from the given straingage
readings. Strains associated with the xly, axes. Knowing the strains 6X, 6y, and y”,
we can calculate the strains for an element oriented at any angle 0 (Fig. 7—38b)
from the straiirtransformation equations (Eqs. 7—7la and 7—71b) or from Mohr’s
circle. We can also calculate the principal strains and the maximum shear strains ' from Eqs. (7774) and (7775), respectively. 52" ...
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 Spring '06
 Elghandour

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