A fied using the will define these STRAIN various types of Normal Strain. If we define the normal strain as the change in length of a line per unit length, then we will not have to specify the actual length of any particular line segment. Consider, for example, the line AB, which is contained within undeformed 66 C H Athe PTER 2 S T R A I N body shown in Fig. 2–1a. This line lies along the n axis and has an original length of ¢s. After formed body deformation, points A and B are displaced to A¿ and B¿, and the line Strain (a) becomes a curve having a length of ¢s¿, Fig. 2–1b. The change in length BİRİM DEFORMASYON Strain of to thedescribe line is therefore ¢s¿ - ¢s. we define the average normal In order the deformation of If a body by changes in length of strain using the symbol P (epsilon), then avg Deformasyon: ısı etkisiinvs.the altında şekilwe vedescribe boyutlarında meydana of gelen line segments and Yük, the changes angleselemanların betweenInthem, will order to the deformation a body by changes in len ange the body’s develop the concept of strain. measured by değişikliklere deformasyon denir. Strain is actually line segments and the changes in the angles between them, w n and they experiments, and once the strain is obtained, it will be shown in the next ation, develop the concept of strain. Strain is actually measur For example, a chapter how 2it can be related to the stress acting within the body. n experiments, and once the strain is obtained, it will be shown in th œ ¢s - ¢s B¿ when stretched, Pavg = chapter how it can be related to the stress acting within the body. (2–1) s occur B ¢s !s¿ when a ation of a body Normal Strain. If we define the normal strain as the change in ¿ 2.2 STRAIN 67 67 nged. A typical length of a line per unit length, then we!swill not have to specify the actual 2.2 STRAIN Normal Strain. If we define the normal strain as the cha A of caused by the length of any particular line segment. Consider, for example, the line AB, lengthe.g., of a line per unit length, then we will not have to specify the for experimental work, strain is expressed astoshown a percent, As point Bwithin is chosen and closer point A, the length of the line is contained the closer undeformed body inlength Fig. 2–1a. This for experimental work, 660.001which C H A P T E Rstrain 2 S T RisA I Nexpressed as a percent, e.g., of any particular line segment. Consider, for example, the li -6 Strain m>m = 0.1%. 480110 2 As an example, a normal strain of can be -6 not0.001 be uniform will become shorter and shorter, such that ¢s : 0. Also, this causes to line lies along the na axis andstrain has of an480110 original2 length of ¢s. After B¿within m>m = 0.1%. As an example, normal can be body which is contained the undeformed body shown in Fig. 2–1 -6 rmed body 2 480110 in.>in. 480 mm>m, as , or 0.0480%. Also, one can state -6 tryreported of any reported line approach A¿, such that ¢s¿ : 0. Consequently, in the limit the normal deformation, points A andorB0.0480%. are displaced to A¿ B¿, and the line the n axis and has an original length of ¢s. 2 in.>in. 480110 mm>m, as , 480 Also, of one canand state line lies along to describe the deformation of a body by changes in length (b) Undeformed 480 mA(480 this answer as simply “micros”). s length. Hence, strain at point and in theof direction n is The change in length becomes am curve having a length ¢s¿,body Fig.of2–1b. # 480 this and answer simply (480 “micros”). deformation, points A and B are displaced to A¿ and B¿, and t ments theaschanges in the angles between them, will (a) we Strain manner, g. 2–1 we will of the line is therefore ¢s¿ - ¢s. If we define the average normalastrain becomes curve having a length of ¢s¿, Fig. 2–1b. The change in the concept of strain. Strain is actually measured by Note the before and after positionsthen of three Birim ocated in the using 2 theDeformasyon: symbol Pavg (epsilon), 2 of of the line is therefore ¢s¿ ¢s. If we define the average norma ents, and once the strain is obtained, it will be shown in the next different line segments on this rubber In only order cause to describe the deformation of a body by -changes in length Shear Strain. Deformations not line segments to hanges will also Shear Strain. not only line segments to 2.2 STRAIN 67 membrane which isacting subjected to tension. The the Pavg (epsilon), then how it can beelongate related toDeformations the stress within thecause body. line segments the using changes in the angles between them, we will or contract, but they also cause them todirection. change If wesymbol nt.elongate For example, ¢s¿and - direction. ¢s or contract, also cause to change If we verticalbut linethey is lengthened, the them horizontal line P = lim (2–2) develop the concept of strain. Strain is actually measured by select two line segments that are originally perpendicular to one another, tion, whereas it B : A along 2 two line segments is shortened, inclined perpendicular line changes select thatand arethe originally ton one¢s another, n experiments, and once the strain is obtained, it will be shown in the next œ xperimental work, strain is expressed as a percent, e.g., then the change in angle that occurs between these two line segments is its length and rotates. ¢s two --6 ¢s then the change in anglethe thatnormal occurs strain between these lineinsegments is lB¿ Strain. IfAs wean define as the change Pangle (2–1) chapter how itbe can be related to the stress acting within the body. m>m = 0.1%. 480110 2 can example, a normal strain of = g referred to as shear strain. This is denoted by (gamma) and is avg B angle is denoted by¢s gB¿actual referred to-6as shearthen strain. This (gamma) and is ¢sœ - ¢s a line per unit will not have towhich specify thedimensionless. 2length, in.>in. mm>m, ed as 480110 , 480 we or 0.0480%. Also,are one can state always measured in(rad), radians (rad), For example, Pavg = always measured in radians which are dimensionless. For example, any particular line Consider, for line AB,from the same point ¢s !s¿ 480segment. mthe nswer as simply (480 “micros”). consider line segments ABexample, and) is AC the originating !s andP AC Hence, when (or Poriginating the initial will elongate, whereas if considerwithin the line segments AB from the sameline point avg in positive contained the undeformed body shown Fig. 2–1a. This Strain. If we define the normal strain as the change in A inand a body, directed along perpendicular n and A¿Normal Pand isAnegative linethe contracts. A in athe body, directed thethe perpendicular n¢s. andAfter t axes, Fig.t axes, 2–2a.Fig. 2–2a. along n axisdeformation, and hasalong anthe original length of length of a line per unit length, then we will not After ends of both lines are displaced, and the lines 2 the actual normal strain isdisplaced, a dimensionless since it isline a ratio of have to specify point B isthat chosen closer and closer to line point the length of the After deformation, theNote ends of both lines are and A, thequantity, lines ion, points A Deformations and As B are displaced to A¿ and B¿, and the length of any particular line segment. Consider, for example, the line AB, r themselves Strain. not only cause line segments to u¿, themselves become curves, such that the angle between them at A is two lengths. Although thissuch is thethat case, isat sometimes stated inB¿ of a closer and closer become shorter andthe shorter, ¢sit: 0. Also, thisAs causes tochosen become curves, such that angle between them A is u¿, point Bterms is to point a curve having awill length of ¢s¿, Fig. 2–1b. The change inis length which contained within the undeformed body shown in Fig. 2–1a. ThisA, the length of t ate or contract, but they also cause them to change direction. If we Fig. 2–2b. Hence the shear strain at point A associated with the n and t dy ratio ofstrain length units. IfA the systemwith is used, for length approach A¿, such that ¢s¿ : 0.SI Consequently, limit the unit normal Fig. 2–2b. Hence the shear at point associated theinthen nthe andthe t basic " " will become shorter and shorter, such that ¢s : 0. Also, this cause etwo is therefore ¢s¿ ¢s. If we define the average normal strain line lies along the n axis and has Pan of ¢s. After line segments that are originally perpendicular tofor one another, axes becomes is point the meter (m). Ordinarily, most applications willoriginal be ¢s¿length Deformed body strainUndeformed at Abody and in the direction of n is engineeringapproach axes becomes A¿, such that : 0. Consequently, symbol P (epsilon), then points Amicrometers and B gelen are displaced to A¿ andBirimsizdir B¿, and the line in the limit the he changeavg in angle thatvery occurs between these twodeformation, lineBirim segments is in (a) Normal birim deformasyon: boyda meydana boy değişimidir. small, so measurements of strain are per meter (b) 65 strain at point A and in the direction of n aand curve having a length of ¢s¿, Fig. 2–1b. The change inislength ed to as shear strain. This angle is denoted by =g becomes (gamma) -6 1mm>m2, where 1 mm 10 m. In theis Foot-Pound-Second system, Fig. 2–1 ancak # gibi birimlerle gösterilmesi uygundur. of the is therefore - ¢s. we define the average normal strain s measured in radians (rad), are dimensionless. For example, strainwhich is often stated in units ofline inches per inch¢s¿ ( in.>in. ). If Sometimes the symbol der the line segments AB andœ AC originating fromusing the same point Pavg (epsilon), then p - ¢s p gnt = lim u¿¢s¿ -and ¢s -= ¢s body, and directedPalong the tA axes, Fig. 2–2a. = nu¿ lim - Plim gnt perpendicular (2–3)(2–2) 2 BA : along n(2–1) avg = B : along n (2–3) ¢s 2 B : A alongCn : A along t ¢s¿ - ¢s ¢s lines deformation, the ends of both are displaced, and the lines P = lim C : A along t # # B : A along n ¢s elves become curves, such that the angle between them at A is u¿, œ ¢s ¢s B¿ –2b. Hence the shear strain at point A associated with the n and tbirim deformasyon ise kısalmayı gösterir. Pozitif birim deformasyon uzamayı,negatif Pavg = (2–1) ecomes ¢s dik doğru parçaları !s¿ Kayma birim deformasyonu: Şekil değiştirmeden önce birbirine Hence, when P (or P ) is positive the initial line will elongate, whereas if avg A, the length of the line nt B is chosen closer and closer to point Notice that if u¿ smaller than p>2 the shear strain olarak is positive, whereas ifBirimi radyandır. Deformasyon A¿ is açıda is negative the ¢s line contracts. meydana değişim tanımlanır. that u¿Parasındaki smaller p>2 strain is positive, if meNotice shorter andif shorter, such than that :the 0. shear Also,gelen this causes B¿ to whereas Hence, when P (or Pavg) is positive the initial line will elongate, wh u¿ is larger than p>2 the shear strain is negative. Note that normal strain is a dimensionless quantity, since is a ratio oftaktirde u¿ issuch larger than p>2 theConsequently, shear strainbüyük isinnegative. sonrası dik açıdan açıthe meydana aksithe pozitiftir. h A¿, that ¢s¿ : 0. thebir limit normal geliyorsaPnegatif, isitnegative line contracts. two lengths. Although this is the case, it is sometimes stated in terms of a point A and in the direction of n is Note that normal strain is the a dimensionless since it is a r As point B is chosen closer and closer to point A, length of thequantity, line p lim u¿ ratio If the SI system is used, then the basictwo unit for length = length - units. gnt of lengths. Although this is the case, it is sometimes stated in ter will become(2–3) shorter and shorter, such that ¢s : 0. Also, this causes B¿ to 2 (m). B : A along n is theDeformed meter Ordinarily, for most engineering applications Plength will beunits. If the SI system is used, then the basic unit for Cbody : A along t ratio of approach A¿, such that ¢s¿ : 0. Consequently, in the limit the normal # very small, meter (b)so measurements of strain are in micrometers is theper meter (m). Ordinarily, for most engineering applications P -6 strain at point A and in the direction of n is 1mm>m2, where 1 mm = 10 m. In the Foot-Pound-Second system, ¢s¿ ¢s very small, so measurements of strain are in micrometers per Fig. 2–1 P strain = lim (2–2) is often stated in units of inches per inch ( in.>in. ). Sometimes 1mm>m2, where 1 mm = 10-6 m. In the Foot-Pound-Second s B : A along n ¢s strain is often stated in units of inches per inch ( in.>in.). Som e that if u¿ is smaller than p>2 the shear strain is positive, whereas if n u¿ n t u¿ ¢s¿ - ¢s t p>2 the shear strain is negative. arger than B¿ P = lim (2–2) C¿ B¿ C¿ B : A along n ¢s p hen P (or Pavg) is positive the initial line p B will elongate, whereas if 2 A¿ tive the line contracts. 2 C B C A¿ hat normal strain is a dimensionless quantity, since it is a ratio of A A hs. Although this is the case, it is sometimes statedHence, in terms of aP (or Pavg) is positive the initial line will elongate, whereas if when ength units. If the SI system is used, then the basic unit lengththe line contracts. P is for negative # for most engineering applications # ter (m). Ordinarily, P will be Note that normal strain is a dimensionless quantity, since it is a ratio of Deformed body Undeformed body Deformed body Undeformed body all, so measurements ofn strain are in micrometers per meter two lengths. Although this is the case, it is sometimes stated in terms of a (b) (a) u¿ (b) t (a)m. In the Foot-Pound-Second where 1 mm = 10-6 system, ratio of length units. If the SI system is used, then the basic unit for length B¿ often stated in units of inches per inchC¿( in.>in. ). Sometimes Fig. 2–2 is the meter (m). Ordinarily, for most engineering applications P will be Fig. 2–2 p very small, so measurements of strain are in micrometers per meter B 2 C A¿ 1mm>m2, where 1 mm = 10-6 m. In the Foot-Pound-Second system, #23 A strain is often stated in units of inches per inch ( in.>in.). Sometimes 2.2 2.2 2.2 deformation of the body in Fig. 2–3a. ToCartesian do so, imagine the Components. body is Strain subdivided into small elements such as the one shown in Fig. now show how Cartesian Cartesian Strain Components. Strain Components. Using the definitions Using theofdefinitions normal of normal and shear strain, we will2–3b. ¢x, ¢y, and This element is rectangular, has undeformed dimensions Cartesian Strain Components. Using the definitions of normal deformation of the body in Fig. 2–3a and shear strain, and we shear will strain, now show we will how now they show can how be used they to can describe be used the to describe the z RAIN 68 C H A P68 TER 2 S T R ACI H N A P T E R 2¢z,S T and is To located in imagine the neighborhood point in the body, Fig.elements 2–3a. 2deformation and shear strain, we will now how they canbody be used to the into small such deformation ofshow the body of in the Fig. 2–3a. in ToFig. dodescribe 2–3a. so, imagine do the so, body is the body of is asubdivided If the element’s dimensions are very 2–3b. small,This thenelement its deformed shape will has undef deformation of the body in Fig. 2–3a. To do so, imagine the body is is rectangular, subdivided into subdivided small elements into small such elements as the such one shown as the in one Fig. shown 2–3b. in Fig. 2 STRAIN bein aundeformed parallelepiped, Fig. since very small will remain Cartesian Strain Components. Using the definitions of Cartesian Components. Using the segments definitions ofthe normal subdivided into elements such asisthe one shown Fig. 2–3b. ¢z, line and is located in neighborhood ¢x, ¢y,2–3c, ¢y, Thissmall element This is rectangular, element rectangular, hasUsing undeformed has dimensions dimensions and¢x, and 2 Cartesian Strain Components. the definitions of Strain normal approximately straight after the body is deformed. In order to achieve Birim Deformasyon Kartezyen Eksenlerdeki Bileşenleri and shear strain, we will now show how they can be used descr y n Components. Using definitions shear strain, we will now show they be used todimensions describe the ¢x, This element isstrain, rectangular, has undeformed dimensions and If can the element’s areto very sm ¢z,the and we is located and in is thelocated neighborhood innormal the neighborhood of point in of the a body, point in the 2–3a. body, Fig.how 2–3a. and shear¢z, will now show howof they can be aand used to¢y, describe theFig. this deformed shape, we will first consider how the normal strain sian Strain Components. Using the definitions of normal deformation of the body in Fig. 2–3a. To do so, imagine the z Cartesian Strain Components. Using the definitions of normal e will now show how they can be used to describe the deformation of the body in Fig. 2–3a. To do so, imagine the body is ¢z, and is located in the neighborhood of a point in the body, Fig. 2–3a. be a parallelepiped, Fig. 2–3c, since ver If the element’s If the dimensions element’s are dimensions very small, are then very its small, deformed then its shape deformed will shape will z deformation of the body in Fig. 2–3a. To do so, imagine the body is changes the lengths ofremain the sides of theaselements rectangular element, and then ar strain, weFig. will now show how they can be used to2–3c, describe the subdivided into small asstraight theFig. oneafter shown in Fig and shear we will now show how they can be used to describe the e the body in 2–3a. To do so, imagine the is small subdivided into small elements such the one such shown in 2–3b. If element’s are very small, then its deformed shape will approximately the body bedimensions ainto parallelepiped, be astrain, parallelepiped, Fig. 2–3c, since Fig. very since line very segments small line will segments will remain subdivided small elements such as body the one shown in Fig. 2–3b. y ¢x how the shear strain changes the angles of each side. For example, ation of the body in Fig. 2–3a. To do so, imagine the body is ¢x, c¢ This element is rectangular, has undeformed dimensions of the body in Fig. To do so, imagine theorder body mall elements such as the one shown instraight Fig. 2–3b. ¢x, ¢y,weand Thisbody element is and rectangular, has undeformed be a parallelepiped, Fig. 2–3c, since very small line segments will remain deformed shape, will first approximately approximately straight after the body after is2–3a. deformed. the isIn deformed. order to achieve In tois achieve this dimensions ¢x, ¢y, This rectangular, has undeformed dimensions x is y element ydeformation P ¢x ¢x + P ¢x elongates , so its new length is . Therefore, the ded into small elements such as the one shown in Fig. 2–3b. ¢z, and is located in the neighborhood of a point in the body, Fi 2deformed x shown xthe subdivided into small elements such as theachieve inthe Fig. 2–3b. strain ¢x, ctangular, has undeformed dimensions and ¢z, and isone located in the neighborhood of achanges point in thelengths body, Fig. approximately straight after the body is deformed. In order to 2is of 2–3a. the sides of th this deformed this shape, we will shape, first we consider will first how consider the normal how strain normal ¢z, and located in the neighborhood of¢y, a point in the body, Fig. 2–3a. (a) approximate lengths of the three sides of the parallelepiped are ¢x, ¢y, ment is rectangular, has undeformed dimensions and If the element’s dimensions are very small, then its deformed sha ¢x, ¢y, This isthe rectangular, has dimensions in ofthe aelement point body, Fig.how 2–3a. Ifnormal the dimensions areand very thenthe its deformed shape will the angle histhe deformed shape, we will in first consider the shear strain changes changes changes lengths of the the lengths sides of theundeformed rectangular ofelement’s thestrain element, rectangular element, then and small, then how If neighborhood the element’s dimensions are very small, then itssides deformed shape willand x a point in the is located inlengths the neighborhood ofdeformed achanges point inshape thechanges body, Fig. 2–3a. be a parallelepiped, Fig. 2–3c, since very small line segments will ¢z, and is located in the neighborhood of body, Fig. 2–3a. mensions are very small, then its will be a parallelepiped, Fig. 2–3c, since very small line segments will remain changes the of the sides of the rectangular element, and then P ¢x elongates , so its new length ¢x ¢x how the shear how strain the shear strain the angles of the each angles side. of For each example, side. For example, be a parallelepiped, Fig. 2–3c, since very small line segments will remain x ement’s dimensions are very small, then its deformed shape will approximately straight after the body is deformed. In order to Ifstraight the dimensions are very small, deformed d, Fig.approximately 2–3c, since very small line segments remain yto approximately straight after the body is to of achieve how the shear strain changes angles ofwill example, approximate lengths the three sides Pelement’s ¢x ¢x + then Px ¢x ¢x elongates ,the so itsPbody new , iseach so length its new isFor is its .¢x Therefore, .shape the Therefore, yside. after the deformed. Inlength order achieve 11 ++ PPx(a) 2¢x ¢x 11will + Py2 the ¢ydeformed. 11 + In Pz2order ¢z xelongates x ¢x x allelepiped, Fig. 2–3c, since very small line segments will remain this deformed shape, we will first consider how the normal adeformed. parallelepiped, Fig. 2–3c, small line segments will will remain aight afterdeformed body isits In to achieve this deformed shape, first consider how the normal strain (a) Pthe ¢x +sides Pthe elongates , sobeshape, new is .very Therefore, the approximate approximate lengths oforder the lengths three ofsince ofthree the parallelepiped sides of the parallelepiped are we are this we length will first consider how the normal strain x ¢x x ¢x mately straight after the body is sides deformed. In order to changes achieve changes the lengths of rectangular the sides of element, the rectangular element, an approximately straight after the body is deformed. In order to achieve ape, we will first consider how the normal strain y the lengths of the sides of the and then approximate lengths of the three of the parallelepiped are changes the lengths of the sides of rectangular element, and then 11 + P 2 ¢x 11 + Py2 ormed shape, we will first consider how the normal strain x how the shear strain changes the angles of each side. this deformed shape, weand will first consider how strain the strain hs ofhow the the sidesshear of the rectangular element, then how theexample, shear changes the angles of sides each are side. For example, ¢x For exam ¢x+normal strain changes angles of each side. For And approximate angles between these xthe x sides of 11 + P 2 ¢x 11 + 11 P + 2 ¢x P 2 ¢y 11 + 11 P + 2 ¢y P 2 ¢z 11 P 2 ¢z the lengths of the the rectangular element, and then x x y y z z P ¢x ¢x + P ¢x elongates , so its new length is . changes lengths the sides the ain changes the Pangles of each side. For of example, Px ¢xelement, + Px ¢x. Therefore, xthe Therefo elongates , the so its and new length is ¢x uzama miktarı: # xthen Böylece deformasyon ¢x+# ¢x +Pof2Pboyundaki ¢x. rectangular elongates , ¢x so itsthe new length is 11 Therefore, x" ¢x x¢z 11 + P 2 11 + P 2 ¢y ¢x esoshear strain changes the angles of each side. For example, x y z (a) approximate lengths of the three sides of the parallelepiped are how strain changes of each For example, ¢x + Pthree its new length is the . sides Therefore, theangles (a) approximate lengths the three¢x are x ¢x p of psides of thepparallelepiped approximate lengths of shear the of thethe parallelepiped are side. And the approximate angles between t g g g P ¢x ¢x + P ¢x es , so its new length is . Therefore, the xy yz xz sırasıyla x new Px ¢x# , so its Px ¢x. 2Therefore, elongates length Herbir is ¢x +eksendeki the hs ofxthe three sides of the parallelepiped are sonrası uzunluk olacaktır. deformasyon sonrası 2 2boylar And approximate And the approximate angles between angles these between sides are these sides are mate lengths of thethe three sides of the parallelepiped are approximate lengths of the three sides of the parallelepiped are 11 + P 2 ¢x 11 ++ PPzy22 ¢z ¢y p 11 + Pz2 ¢z p 11 + Px2 ¢x 11 + Pyx2 ¢y 11 And the approximate between 11 +angles Px2 ¢x 11 these + Py2 sides ¢y are11 + Pz2 ¢z olacaktır. # Kenarlar arasındaki yaklaşık açılar da - gxy - gy Px2 ¢x 11 + Py2 ¢y 11 +pPz2 ¢z p p p p normal strains cause a change in volume of the2 element, 2 Notice that the g g g g g 11 + Px2 ¢x p 11 + Py2 ¢y 11 + P 2 ¢z xy yz yz xz11 + P 2 ¢zxz + Pywhereas 2 ¢y p11 2 -+gPxxy2 ¢xp22z - 11 2 2 z strains cause a change in its shape. Of course, both of the shear - gxy gxz yz And the approximate angles between the approximate angles between these are these sides are 2 2 between these 2 sides areAnd And the approximate angles these effects occur simultaneously during thesides deformation. # these sides are olacaktır. Görülmektedir ki normal birim deformasyonlar Notice that the normal strains cause a ate angles between Ininsummary, then, the state ofelement, strain atwhereas a pointthe in shear a body requires strains cause a chan Notice that the Notice normal that strains the normal cause strains a change cause volume a change of in the volume element, of the e approximate angles between these sides are between p p And these sidesthree are normal p deformasyonları pPx , p p pthe approximate p angles p olurken P , P , gxzxy , specifying strains, and three shear strains, g g g hacimsel değişikliğine sebep kayma birim biçim y of z xy yz g g g Notice that the normal strains cause a change in volume of the element, these effects occur simultaneously durin whereas the shear whereas strains the shear cause strains a change cause in its a change shape. Of in its course, shape. both Of of course, both - gyz - gxz xy p p p gxy 2 xz deformation 2 2 2 2 yz 2 the gcourse, , during . These gxzboth strains completely describe ofstate a of strai 2pcause 2p occur 2 Ofthe yzdeformation. - gxythepshear - strains geffects - a geffects yz xz whereas change in its shape. of In summary, then, the these these occur simultaneously simultaneously during the deformation. p -olmaktadır. p Bir p herhangi 2 2- gdeğişikliğine 2- gyz sebep cismin bir noktasındaki birimlocated deformasyon element of requires material thenormal point strains, and P , P , ggxzxy -the -atrectangular gyz -atavolume gaxzbody hese effects2occur during deformation. specifying at three Inxysimultaneously summary, then, summary, the then, of strain the2state aofpoint strain in pointrequires in a body x y 2 In 22 state 2 so oriented that its sides are originally parallel to the x, y, z axes. Notice that the normal strains cause a change in volume of the ed Notice that the normal strains cause a change volume of the element, In summary, then, state of strain atiçin# achange point body . kayma These strains completely Px , Pystrains, ,inPvolume Px ,three Prequires ,birim Pzshear , and gxy , strains, gxy , # gyz , ingxz three specifying normal three strains, and strains, three shear Notice specifying that the the normal strains cause anormal in of element, za, normal ythe durumunu belirlemek deformasyonları ile # Provided these are defined at points the body, whereas strains cause ainchange in itsthen shape. course, ormal strains volume of thestrains element, whereas the strains cause a change its shape. Ofvolume course, boththe ofOf of Pcompletely gshear , strains specifying three strains, shear strains, element mate gcause ,shear . strains gyzin , cause gaxzchange These strains gPxz These describe completely thedescribe deformation the deformation ofthe a shear of allainrectangular whereas the inthree its shape. Of course, both of x ,. Pa y ,change z , and xy yznormal deformed shape of the body can be determined. e that the normal strains cause a change in volume of the element, these effects occur simultaneously during the deformation. Notice that the normal strains cause a change in volume of the element, cause a change in its shape. Of course, both of these effects occur simultaneously during the deformation. gstrains , . g These strains completely describe the deformation of a oriented so that its sides are origina rectangular rectangular volume element volume of element material of located material at the located point at and the point and effects birim occur simultaneously during bilinmesi the deformation. deformasyonlarının gereklidir. yz these xz srectangular the shear strains cause a change its that shape. Of course, both ofitstoshape. In then, theatProvided state of strain at a point a bodyatr whereas thestate shear cause a at change Ofy, course, both simultaneously during the deformation. Inain summary, then, the state strain a point in a body requires volume element of material the point these strains areindefined oriented so oriented that itsin sides so arelocated its originally areparallel the parallel x, to zsummary, axes. the x, of y,of z axes. In summary, then, the ofstrains strain at sides a point inoriginally body and requires fects occur simultaneously during the deformation. P , P , P , specifying three normal strains, and three shear effects occur simultaneously during en, the state of strain atthese aProvided point in these body x shape y z of Deformasyon öncesi eleman: Pz ,deformed gxy ,can bestrai specifying three strains, and three shear strains, oriented so that itsthese sides are originally the x, the y,at axes. the body det Provided strains strains are attothree all defined points inzdeformation. all the points body, then the the body, Pthen Paare , Prequires gnormal specifying three normal strains, shear strains, x , Py ,the x , Pdefined yparallel z , and xy ,in ( p $ gxy) mmary, then, the state of strain at a point in a body requires g , . g These strains completely describe the deformatio In summary, then, the state of strain at a point in a body requires P , P , P , g , ormalg strains, and three shear strains, yz xz g , . g These strains completely describe the deformation of a x y z xy 2 Provided these strains are defined at all points in the body, then the deformed shape deformed of the shape body can of the be determined. body can be determined. deformation of a xz p yz , gxz . These strains completely describe the yz Px , can Pythree , Pbe gxyand , three ng three normal strains, and three shear rectangular element of material locatedand at the poi z ,deformation Pastrains, gofxy ,material specifying normal strains, shear strains,volume strains completely describe the of volume element located at the point x , Prectangular y ,atPz ,the deformed shape of the body determined. 2 point rectangular volume element of material located and . These strains completely describe the deformation of a oriented so that its sides are originally parallel to z gyz ,itsgxz . These strains completely describe deformation a me element located the point and to oriented that its sides areoforiginally parallel to the x, y, z axes.the x, y, orientedofso material that sides areatoriginally parallel the x, so y,the z axes. p p p (1 " # )!z ular volume element of material located at the point and z Provided these strains are defined at all points in the body, th element of material at point and at all points its sides are originally parallel to the x, axes. ( $!z ( strains defined in the body, ) these ) are gxylocated $ gpthe Provided theserectangular strains are volume defined at y, allzpoints inProvided the body, then p then the 2axes. 2 the2xy 2 p ( p p drains so that its sides are originally parallel to the x, y, z $ gxz) deformed shape of the body can be determined. so in that itsbody, sides are parallel to of the x, y, z axes. are defined atoriented all points thebe ( then )originally $ gxythe deformed shape 2 deformed shape determined. ! x the body can be determined. p 2the body, 2of the body 2can p dthe these strains are defined at all points in then the Provided these strains are defined at all 2points body can be determined. (1 " #p (1 " #z x)!x !yin the body, then the ( p $ g ) 2 of the body can be determined. # yz ed shape (1 "!z#y)!y 2 deformed shape of the body(1can " #zbe (1 " #z)!z )!zdetermined. 2 z p !zp ! x( p $ gxy) ( pDeformed Undeformed $pgxy) ( p $eleman: ( p $ gxz) ( p $Deformasyon (1 " #z)!z gxz)p gxy) 2 2 2 sonrası 2 p !z 2 2 !y 2 p element 2 element x !x p ( $! gp xy) ( p $ gxz) 2 p 2 p p p (1 " # (1 " # )!x )!x 2 2 ( ( $$ (!y$ gxy) x x 2 !y 2 (c) ( $ gyz) p g (b) g) ) 2 2 2#y)!y 2 yz xy (1 " #y)!y Undeformed (1 (1 "" #x)!x p !y 2 ( p $ gyz) (1element " #z)!z (1 "p#z)!z (1 " #y)!y 2 (1 " #z)!z !z 2 p Deformed Deformed Undeformed Undeformed p !z Fig. 2–3 !z p p (1 " # )!z 2 element (b) ( $ gxz) element ( $ gxz) element 2 p 2 Deformed Undeformed z element !x p 2 p 2 ! x (1 " # )!z p ! x z (1 " #z)!z (c) p p 2 element (c) 2 (b) (b) xz) p ( 2 $ gelement !y !z p p #x)!x ( p $ g(1yz" x p p (1 " #x!y )!x ) 2 ( ( 2 2 !y ) $ g ) $ g ( ( $ gxz) xz (1 " #y)!y gyz) (c) 2 yz 2(1 "$ (1 " #y)!y2 2 p # )!x (1 " # )!y ! x(b) 2 2 x y ! x ( $ gyz)p Fig. 2–3 Fig. 2–3 (1 " # )!y Deformed p Undeformed 2 (1 " # )!x y !y Deformed Deformed Undeformed 2 ( !y $ gyz) (x p $ gyz) Undeformed (1 " #x)!x element element (1 " #y)!y Fig. 2–3 2 (1element " #y)!y element 2 Deformed element element # 2.2 S TRAIN 69 (c) (b) element Deformed Undeformed (c) Undeformed (c) 2.2Deformed S(b) TRAIN 69 (b) element element element (c) Deformasyon Analizi: element Küçük Birim Fig. 2–3 (c) (b) (c) (b) Fig. 2–3 !z p p 2 ! x2 Fig. 2–3 engineering design involves Fig. 2–3 tasarımları alysis. Most engineering design involves Small Straindeformasyonlara Analysis. Most Most engineering design involves Small Analysis. design için Strain küçük izin engineering verilir. Bu derste bir involves cisimde tions are allowed.Çoğu In thismühendislik text, Fig. text, 2–3applications Fig. 2–3 small ly small deformations are allowed. In this applicationsfor forwhich which only smalldeformations deformationsare areallowed. allowed.InInthis thistext, text, only ations that take place withingelen a meydana deformasyonların hemen sonsuz oldukları varsayılmaktadır. that the deformations that take place withintherefore, atherefore, wewill willhemen assumethat thatthe theküçük deformations that takeplace placewithin withina a we assume deformations that take r, the normal strains occurring mal. In particular, the normal strains occurring bodyare arealmost almostinfinitesimal. infinitesimal.InInparticular, particular,the thenormal normalstrains strainsoccurring occurring body red to 1, so that # P V 1. This 1 denare çok çoksmall küçük) ery small compared to 1, (normal so that P birim This withinthe thematerial material are very small comparedtoto1,1,sosothat thatP PV This V 1. deformasyonlar V1.1.This within very compared in engineering, and it is often 2 ctical application in engineering, and it is often assumptionhas haswide widepractical practicalapplication applicationinin engineering,and andititisisoften often 2engineering, assumption varsayım can be used, forBu example, to mühendislikte çok yaygındır ve genellikle küçük birim deformasyon analizi rain analysis. It can be used, for example, to referredtotoasasa asmall smallstrain strainanalysis. analysis.ItItcan canbe beused, used,for forexample, example,toto referred u = u, provided u is very small. os u = 1, and tan uolarak = u, provided u is very small. approximate sin u = u, cos u = 1 , and tan u = u , provided u is verysmall. small. approximate sin u = u, cos u = 1 , and tan u = u , provided u is very bilinir. # çok küçük olduğunda # ve # olarak yaklaşımı kullanılabilir. e rubber bearing support under this Theisrubber bearing ncrete bridge girder subjected to support under this concrete bridge girder is subjected to h normal and shear strain. The bothbynormal and shear strain. The rmal strain is caused the weight normal strain is caused by the weight Therubber rubberbearing bearingsupport supportunder underthis this The concretebridge bridgegirder girderisissubjected subjectedtoto concrete bothnormal normaland andshear shearstrain. strain.The The both normalstrain strainisiscaused causedbybythe theweight weight normal #24 (p $ 2 (p $ g 2 (1 " #x)! 2.2 S 2.2 od shown in Fig. 2–4 is subjected to an increase of 70 CC HA TPFig. ETnormal RE R22–4 RTstrain A IANI N 70rod creates HP Aa 2 S TSis Rsubjected slender shown in to an of long The its axis, which in the rodincrease of slender rodrod shown in in Fig. 2–42–4 is is subjected to anan increase of the he shown Fig. subjected to increase of rod of 1>2slender temperature along its axis, which creates a normal strain in zitsis-3 measured inin meters. the zmperature , where The slender rod which shown Fig. 2–4 Determine isstrain subjected to(a) an increase of 1>2which perature along axis, creates a normal in in the rod of of along its creates a normal the rod P-3 z iswhich measured in strain meters. Determine (a)rod theof = 1>2 40110 2zaxis, , where ztemperature along its axis, creates a normal strain in the -3 1>2 of40110 end the rod due to temperature increase, zthe is inthe meters. Determine (a)(a) thetheincrease, 2z 2zB , where where z measured isend measured in meters. Determine = the 40110 ,of EXAMPLE 2.1 EXAMPLE 2.1 displacement of B of the rod due to the temperature -3 1>2 Pzof=of z due is to measured in meters. Determine 40110 , where # end acement the end B2zof the rodrod due the temperature increase, erage normal strain in the rod. splacement the B of the to the temperature increase, (a) the and (b) the average normal strain in the rod. displacement of the end B of the rod due to the temperature increase, (b) thethe average normal strain in in thethe rod. d (b) average normal strain rod. The ininFig. 2–4 ananincrease Theslender slender rodshown shown Fig.artışına 2–4isissubjected subjectedtotoBu increaseofof Şekildeki narin çubuk ekseni rod boyunca sıcaklık maruzdur. and (b) the average normal strain in the rod. A 22 z dz temperature temperaturealong alongitsitsaxis, axis,which whichcreates createsa anormal normalstrain strainininthe therod rodofof -3-3 1>2 1>2 P where z is measured in meters. Determine (a) the = 40110 2z , P where z is measured in meters. Determine (a) the = 40110 2z , z z değişim çubukta # normal birim deformasyonu meydana A displacement displacementofofthe theend endBBofofthe therod roddue duetotothe thetemperature temperatureincrease, increase, z average ininthe and(b) (b)the the averagenormal normal strain therod. rod. getirmektedir. (a) and Sıcaklık artışından dolayıstrain B ucunun deplasmanını A A A z z dz dz z dz hesaplayınız. (b) çubuktaki ortalama normal birim deformasyonu AA 200200 mmmm dz 200 mm 200 mm 200 mm hesaplayınız. B B Fig.Fig. 2–42–4 zz dzdz B 200 200mm mm B 2–4 Fig. B Fig. 2–4 SOLUTION UTION OLUTION Part (a). Since Fig. the normal strain is reported at each point along the SOLUTION 2–4is reported BB (a).(a).rod, Since the normal strain at at each point along thethe art Since the normal strain isdz, reported each point along Çözüm a differential segment located at position z, Fig. 2–4, has the a Part (a). Since the normal strain is reported at each point along a differential segment dz, located at position z, Fig. 2–4, has a d, a differential segment at position z, Eq. Fig.2–1; 2–4,that has a deformed length thatdz, canlocated be determined from Fig. Fig.2–4 2–4 rod, athat differential segment dz, located at position z, Fig.is,2–4, has a Deformasyon sonrası dz uzunluğu dz' rmed length cancan bebe determined from Eq.Eq. 2–1; that is, formed length that determined from 2–1; that is, olacaktır. length that can=be Eq. 2–1; that dzdetermined + at Pz dz SOLUTION SOLUTION ce the deformed normal dz¿ strain is +dz¿ reported eachfrom point along theis, = dz P+z Pdz dz¿ = dz dz z -3 1>2 Part Since Part(a). (a). a Sincethe thenormal normalstrain strainisisreported reportedatateach eachpoint pointalong alongthe the dz¿=at=C 1position dz Pz dz ntial segment dz, located z, has dz¿ ++ 40110 2z Fig. D dz2–4, -3 -31>2 1>2 dz¿dz¿ = = 1 + 40110 2z dz C 1 + 40110 2zD D dz-3 1>2 rod, C a differential segment dz, located at position z, Fig. 2–4, has aa rod, a differential segment dz, located at position z, Fig. 2–4, has th that determined 2–1;yields is, deformed length dz¿ from = C 1 +Eq. 40110 2zthat dz D the Thecan sumbe of# these segments along the axis deformed length that can be determined from Eq. 2–1; that is, deformed length that can be determined from Eq. 2–1; that is, sum ofofof these segments along thethe axis yields thethe deformed length he sum these segments along axis yields deformed length the rod, i.e.,these The along the axis yields the deformed length Deformasyon dz¿sum = of dz + Psegments ethe rod, i.e., rod, i.e., z dz sonrası çubuk boyu: dz¿ dz¿==dzdz++PzPdz z dz 0.2 m of the rod, i.e., m 0.2 m -3-3 1>2 1>2 z¿ = -32z 1 + 1>2 40110-32z1>2 D dz dz¿ 2z2z1>2D dz dz¿==C 1C 1++40110 40110 D dz dz¿ = C 1 +0.240110 0.2 Cm -3 D-3dz z¿ z¿ = = 2z 2z1>2 C 1 C+1L040110 D dzD dz-3 1>2 + 40110 dz sum C 1 + 40110 2z The D The L0 L0 z¿ = sumofofthese thesesegments segmentsalong alongthe theaxis axisyields yieldsthe thedeformed deformedlength length 2 3>2 ese segments along the= axis deformed length -3the 0.2 m zL +0 2yields 40110 2 z C -3 D ƒ of the rod, i.e., 0 of the rod, i.e., 2 3>2 0.2 m 0.23m = = C z C+z 40110 + 401102-332z3>2 3 z D ƒ 0 D ƒ 0-3 2 3>2 0.2 m 0.20.2 mm = z + 40110 2 3 z D ƒ0 C = 0.20239 m -3-3 1>2 0.2 m = 0.20239 m z¿ = 2z2z1>2D dz z¿ = 40110 = 0.20239 m C 1C 1++40110 D dz # 0.20239 m 0 L The displacement of the=end of the rod is therefore 0 L -3 1>2 z¿ = of of + of 40110 2z C 1end D dz displacement thethe end of the rodrod is therefore he displacement the is therefore ucunun deplasmanı: -3-3 2 2 3>2 mm 0.2 The displacement of the end of the rod is therefore L0¢B Ans. ==C zC z++40110 2 23 z3 z3>2D ƒ 0.2 B = 0.20239 m - 0.2 m = 0.00239 m = 2.39 mm T 40110 0D ƒ 0 ¢ B¢=B 0.20239 mm - 0.2 mm = 0.00239 mm = 2.39 mm T T Ans. Ans. = 0.20239 - 0.2 = 0.00239 = 2.39 mm Ans. ¢ =average 0.20239 - 0.2 2.39 mm T from Part (b). #The strain the rodmis=determined 2m3>2 -3 normal 0.2m m =in0.00239 ==0.20239 0.20239mm =average zwhich + B40110 2that C average ƒrod (b). The normal strain in Din the rod is determined from 0 the art (b). The normal strain rod is determined from 3 z the Eq. 2–1, assumes or “line segment” has an original Part (b). The average normal strain in the rod is determined from 2–1, which assumes that thethe rod or or “line segment” hashas an original q. 2–1, which assumes that rod “line segment” an original The displacement The displacementofofthe theend endofofthe therod rodisistherefore therefore Çubuktaki normal deformasyon: length of 200 mm and a ortalama change in length of birim 2.39 mm. Hence, Eq. 2–1, which assumes that the rod or “line segment” has an original = 0.20239 m h of 200 mm and a change in length of 2.39 mm. Hence, ngth of 200 mm and a change in length of 2.39 mm. Hence, ¢¢ Ans. 0.20239mm--0.2 0.2mm==0.00239 0.00239mm==2.39 2.39mm mmT T Ans. length of 200 mm¢s¿ and-a2.2 change in length B B==0.20239 ¢s STRAIN 2.39 mm of 2.39 71 mm. Hence, P Ans. = = = 0.0119 mm>mm 2.2 S TRAIN 71 ¢s¿ ¢s 2.39 mm avg ¢s¿ ¢s 2.39 mm ent ofPthe end of the=rod is therefore mm Part (b). Part (b). The Theaverage averagenormal normalstrain strainininthe therod rodisisdetermined determinedfrom from Ans. = = # 0.0119 mm>mm Pavg Ans. =¢s - ¢s =200 = 0.0119 mm>mm ¢s¿ 2.39 mm avg ¢s 200200 mm mm= P¢s Ans. that = 0.0119Eq. mm>mm avg = 2–1, 2–1,which whichassumes assumes thatthe therod rodoror“line “linesegment” segment”has hasananoriginal original ¢s 200= mm Ans. = 0.20239 m - 0.2 m = 0.00239 m 2.39 mmEq. T length of 200 mm and a change in length of 2.39 mm. Hence, length of 200 mm and a change in length of 2.39 mm. Hence, EXAMPLE 2.2 # average normal strain in the rod is determined from ¢s¿ ¢s¿--¢s ¢s 2.39 2.39mm mm ne Fig. 2–5a, the Pavg =in = ==0.0119 P = = 0.0119mm>mm mm>mm When force P is applied to the rigid lever arm ABC Fig. 2–5a, the avg assumes that the rod or “line segment” has an original angle of 0.05°. ¢s 200 ¢s 200mm mm arm rotates Dcounterclockwise about pin A through an angle of 0.05°. mm and a changeDetermine in lengththeofnormal 2.39 mm. D strainHence, developed 2 in wire BD. P ¢s¿ - ¢s SOLUTION 2.39 mmI 300 mm P Ans. = = = 0.0119 mm>mm tg rotates about ¢s 200 mm The orientation of the lever arm after it rotates about Geometry. s figure, point A isBshown in Fig. A From the geometry of this figure, C EXAMPLE 2.22–5b. B C 400 mm # Şekildeki sistemde ABC kaldıraç koluna P yükü 400 mm ° = 36.92° 0 mm 2.2 300 mm SOLUTION f = 90° - Ia + 0.05° = 90° - 53.1301° + 0.05° = 36.92° Geometry. The orientation of the lever arm after it rotates about point A is shown in Fig. 2–5b.theorem From thegives geometry of this figure, For triangle ABD the Pythagorean 400 mm LAD = 21300 mm2 -+1 1400 mm2 = 500 mm b = 53.1301° a = tan a 300 mm 2 2 71 400 mm BD D 2 300 mm P C 2 STRAIN A When force P arm ABC in Fig. 2–5a, the tan - 1 a to the rigid b =lever 53.1301° a is = applied 300 mm arm rotates counterclockwise about pin A through an angle of 0.05°. dönmektedir. uygulandığında, (a) kol saat yönünün tersi yönde A noktası etrafında # (a) Determine the normal strain developed in wire BD. Then halatında meydana gelen normal birim deformasyonu belirleyiniz. Ans. Ans. B #25 400 mm A 400 mm 01° (a) 0.05° = 36.92° 2.2 2.2 STRAIN 2.2 2.2 EXAMPLE 2.2 2.2 STRAIN Çözüm - 1 2.2 EXAMPLE EXAMPLE 2.2 EXAMPLE 2.2 = 500 mm 2.2in Fig. STRAIN 71 When force P is applied to the rigid lever arm ABC 2–5a, the Kosinüs teoremi ile halatın dönme sonrasıabout uzamış boyunun hesabı: arm rotates counterclockwise pin A through an angle of 0.05°. When force P is applied to the rigid lever arm ABC in Fig. 2–5a, the force applied to thearm rigidABC leverinarm ABC Fig. 2–5a, the EXAMPLE 2.2 Ptoisthe When force When P is applied rigid lever Fig. 2–5a,in the o triangle AB!D, Determine strain developed in wireof arm rotates rotates counterclockwise counterclockwise about pin pin A through through an angle angle ofBD. 0.05°. 400 mm the normal arm about A an 0.05°. 2.2 STRAIN D arm rotates counterclockwise about pin A through an angle of 0.05°. XAMPLE 2.2 Determine the normal strain developed in wire BD. D When force P is applied to the rigiddeveloped lever arm in ABC Fig. 2–5a, the Determine the normal strain wireinBD. STRAIN 71 71 STRAIN TRAIN S 71 71 71 D D Determine the normal strain developedI in wire BD. 300 mm 2 2 D SOLUTION 2.2 STRAIN P arm rotates counterclockwise about pin A through an angle of 0.05°. 2 71 When force P is applied to the rigid lever arm ABC in Fig. 2–5a, the 300 mm mm P D SOLUTION I 300 P The orientation it rotates Determine the normal strain developed in wire BD. of the lever arm after SOLUTION I Geometry. 300 about mm P EXAMPLE SOLUTION I 2.2 rm rotates about pin A through an angle of 0.05°. 21400 mm2counterclockwise cos 36.92° a 2 300lever mm point A is shown in Fig. 2–5b. From the geometry of this figure, Geometry. The orientation of the arm after it rotates about D Geometry. The of orientation of theafter leverit arm after it rotates about Determine the normal strain developed in wire BD. Geometry. The orientation the lever arm rotates about 300 mm P B C EXAMPLE 2.2 SOLUTION IA point A$is is shown in0.05# Fig. 2–5b. From the geometry of this this figure, When forcepoint P isP applied to the rigid2–5b. leverFrom arm the ABC in Fig. 2–5a, the 2 shown Fig. of figure, LBD " fthe geometry point A is shown in Fig. 2–5b.u in From ofgeometry this figure, 400 mm A B C 400 mm A B arm rotates counterclockwise about pin A through an angle of 0.05°. B 300Bmm C - 1 it rotates about P orientation of the lever arm after A OLUTION Geometry. IWhen force PThe C tanABC a in Fig. 2–5a, b = 53.1301° a =arm is applied to the rigid lever the 400 mm D 400 mm 400 mm A 400 Determine normalinstrain developed in wire BD. of this 300figure, mm point Athe is shown Fig. 2–5b. the geometry 400 mm 400 mm tan apin b = = 53.1301° 53.1301° a = From eometry. arm Therotates orientation of the arm-- 11after itmm rotates about counterclockwise about A through an angle of 0.05°. - 1 lever tan a b a = 2 (a) B¿ A B C tan a b = 53.1301° a = 300 mm mm 400 mm C 300 Then D oint A is shown in# Fig. 2–5b. From the geometry this figure, 300 mm of Determine the normal strain developed in wire BD. 300 mm P 400 mm SOLUTION I (a) A (a) B 2 C ThenAns. a = tan - 1 a 400 mm b = 53.1301° = 0.00116 mm>mm (a) Then Then Geometry. The orientation of the lever arm after it rotates about 300 mm (b) 400 mm SOLUTION I - 1 400 mm f = 90° - a + 0.05° = 90° - 53.1301° + 0.05°P = 36.92° 300 mm b+ = 53.1301° a = tan point A is shown infFig. 2–5b. From the geometry of this figure, (a) = a90° 90° -mm a + 0.05° = lever 90° -arm 53.1301° + rotates 0.05° = =about 36.92° Then 300 a 0.05° 90° 53.1301° 0.05° 36.92° Geometry. orientation of the= after it+ A f = # 90° - afThe += 0.05° = 90° 53.1301° + 0.05° = 36.92° B Fig. 2–5 C For triangle ABD the Pythagorean theorem gives (a) point A is shown in Fig. 2–5b. From the geometry of this figure, 400 mm hen 400 mm 1the For triangle ABD the Pythagorean theorem gives = 36.92° A B f =triangle 90° aABD + -0.05° =theorem 90°b-=53.1301° + 0.05° theorem gives C tan a Pythagorean 53.1301° a -= Forby triangle ABD the Pythagorean btained approximating # For 300 mmLAD gives = 21300 mm22 + 1400 mm22 = 500 mm 400 mm mm - 1 400 + = 90° - a + 0.05° = La90°= 53.1301° 0.05° =53.1301° 36.92° 2–5b. f Here, 22 b+ =1400 (a) =tan 21300 mm2 mm222 = = 500 500 mm mm AD 2 a ABD the Pythagorean theorem gives LAD = 21300 mm2 +2 1400 mm2 ThenFor triangle LAD = 21300 mm2 + 1400 mm2 = 500 300 mm Using this result and applyingmm the law of cosines to triangle AB!D, # 400 mm (a) or triangle ABD the Pythagorean theorem gives 2 2cosines to triangle AB!D, Using this result and applying the law law of =0.05° 21300 mm2 +cosines 1400 mm2 = 500 mm Using and applying the cosines to triangle AB!D, m2 = Using 0.3491Then mm f =result 90° and -Lthis a applying +result = 90° -of53.1301° + of 0.05° = AB!D, 36.92° 400 mm AD this the law to triangle 400 mm Kosinüs teoremi ile: 400 mm D 2LB¿D = 2L2AD 2 + L2AB¿ - 21LAD21LAB¿) cos f LAD = f21300 mm2 + 1400 mm2 = 500 mm =result 90° -Pythagorean + 0.05° = 90° + triangle 0.05° = AB!D, 36.92° 2 2a applying Using this and the law-of53.1301° cosines to For triangle ABD the theorem gives D 2 2 L = 2L + L 21L 21L ) cos f D 400 mm 2AB¿) cos f AB¿ 21500 B¿D = 22LAD AD + LAB¿ AD 2 B¿D D AD mm2 + 1400 mm22 - 21500 mm21400 mm2 cos 36.92° LB¿D = 2LL 21L-AB¿21L ) cos f21LAB¿ a AD + LAB¿ - 21LAD= 2 cosines to triangle 2 Using this result and applying the law of AB!D, 2 2 For triangle ABD the Pythagorean theorem gives = 21500 mm2 + 1400 mm2 21500 mm21400 mm2 cos 36.92° 2+ 2400 mm a 2 2 = 21500 mm2 1400 mm2 21500 mm21400 mm2 cos 36.92° P 300 mm mm mm2 + -1400 mm2 = 500 mm2 mm cos 36.92° 2 2= 21300 a = 300.3491 mm AD =B¿D 21500 mm2 + 1400 mm2 21500 mm21400 D 300 a L = L2L $300 LBDmm u " 0.05# AD + LAB¿ - 21LAD21LAB¿) cos f f P 6 mm>mm 2 Ans. = 300.3491 mm B P $L 2 2 =L21L 300.3491 mm) mm2 " 0.05# 0.05#f P BD = 221300 mm2 = 500AB!D, mm cos 36.92° # 2AB¿ D 300.3491 mm $LBD uu " 2L= + =result L fof2+cosines AD 21500 mm2 + AB¿ 1400 mm2 -1400 21500 mm2 B¿D = Using AD AD21L this and applying thecos law to mm21400 triangle $LBD u " 0.05# a f A 300 mm B 400 f mm Normal Strain. B B 2 2 B¿ P A = 21500 mm2 + 1400 mm2 - 21500 mm21400 mm2 to costriangle 36.92° AB!D, 400 mm =BD 300.3491 mm halatındaki normal birim deformasyon: C au " 0.05#300 A Normal Strain. Using this result and applying the law of cosines $LBD Normal Strain. A mm mm Normal Strain. B¿ 400 f 2 2 LB¿D -) cos LBD 300.3491 mm 300 mm B¿ D 400 mm B C L = 2L + L 21L 21L f 400 mm B¿ C AB¿ B¿D AD AB¿ P = 300.3491 mm PBD AD = = =u "0.00116 Ans. 400 mm 0.05#f mm>mm BD LB¿D - L LBD 300.3491 mm - 300 300 mm mm 300 mm $LC B¿D BD A L (b) 300.3491 mm 2 300.3491 2 -LBD Normal Strain. 2 2 L L mm 300 mm P = = = 0.00116 mm>mm Ans. B D = B¿D 21500 mm2 + 1400 mm2 21500 mm2=cos 36.92°mm>mm Ans. BD BD 2L ) mm21400 cos f0.00116 = = AD-21L 0.00116 a AD= +LLAB¿ - 21L AB¿ B¿ 300 mm PBD = LB¿D =PBD = mm>mm Ans. 300 mm 400 mm (b) (b) BD C LBD L# BD 300 mm 2 300 mm Normal Strain. (b)A 2 300.3491 Fig. 2–5 P = 300.3491 mm Lmm2 mm -- 300 mmmm21400 mm2 cos 36.92° =LB¿D 21500 + 1400 mm2 21500 BD B¿ $LBD 400 mm a u " 0.05# 300 mm f PBD = =SOLUTION II = 0.00116C mm>mm Ans. Fig. 2–5 Fig. 2–5 B LBD - 2mm (b) Fig. 2–5 LB¿D - LBD= Çözüm 300.3491 mm - 300 300 mmmm P 300.3491 $LBD by approximating small, this same result u " 0.05#fA SOLUTION II Since the strain=is 0.00116 = mm>mm Ans.can be obtained BD = Normal Strain. SOLUTION II SOLUTION II LBD Since 300 B B¿(b)Here, the elongation of wire BD as ¢L , shown in Fig. 2–5b. the strain ismm small, this this same result can beBD obtained by approximating Küçük birim deformasyon analizi ile can 400 mmFig. 2–5 Sinceis the strain is small, be by C approximating Since the strain small, this same result same can beresult obtained byobtained approximating A Normal Strain. the elongation of wire BD as ¢L , shown in Fig. 2–5b. Here, LB¿Dthe - elongation L II 300.3491 BD, shown in Fig. 2–5b. Here, SOLUTION of ¢L wiremm BD-as300 ¢Lmm BD Fig.B¿2–5 the of BD wire=BD as 2–5b. Here, BD, shown in Fig. = PBDelongation = 0.00116 mm>mm Ans. 400 mm 0.05° C Since the same can by approximating LBDstrain is small, this 300 mm=result ¢L b1p rad2 uL-AB = mm cbe a obtained d1400 mm2 = 0.3491(b) mm OLUTION II BD mm LB¿D - LBD 300.3491 300 0.05° 180° the ofcinsinden wire BD as , shown in# Fig. 2–5b. Here, 0.05° PBDelongation =uL = 0.00116 BD ¢L b1p rad2 = 0.05° =¢L rad2 d1400 mm2 =mm>mm 0.3491 mm mmAns. ince the strain is=Radyan small,¢L thisBD same result can bemiktarı: obtained by approximating BD AB dönme b1p = uL = ccrad2 aa300 mm2 = 0.3491 mm AB (b) ¢LBD = uLL b1p d1400 mm2 d1400 = 0.3491 mm BD= c a 180° Fig. 2–5 AB 180° he elongation of wire BD as ¢LBD , shown in Fig. 2–5b. Here, 180° Therefore, 0.05° SOLUTION II ¢L b1p rad2 d1400 mm2 = 0.3491 mm Fig. 2–5 BD = uLAB = c a Therefore, Therefore, Since the strain is small, this same 180°result can be obtained by approximating Therefore, 0.05° SOLUTION II ¢L 0.3491 mm BD ¢L b1p = uLAB =ofc wire a d1400 mm2 0.3491 mm the BD asrad2 ¢LBD , shown 2–5b. Here, BDelongation =in =Fig. = obtained PBD = 0.00116 mm>mm Ans. 180° Since the strain is small, ¢L this same result can be ¢LBD 0.3491 mm L 300 mmby approximating BD Therefore, 0.3491 mm BD ¢L 0.3491 mm = = mm>mm P = 0.00116 Ans. BD BD = mm>mmAns. PBD = 0.00116 Ans. the elongation BDLas ¢L=BD=, 300 shown Fig. 2–5b. Here, = of wire =0.05° P# 0.00116 mmin mm>mm BD LBD 300 mm herefore, ¢LBD BD mm = uLABLBD = ¢L c a 300b1p rad2 d1400 mm2 = 0.3491 mm 0.3491 mm BD 180° =0.05° PBD = = 0.00116 mm>mm Ans. LBD mm ¢L = uL = mm c a 300b1p rad2 d1400 mm2 = 0.3491 mm ¢LBD AB 0.3491 BD 180° = PBD = # = 0.00116 mm>mm Ans. Therefore, LBD 300 mm Therefore, ¢L 0.3491 mm BD = PBD = = 0.00116 mm>mm Ans. LBD 300 mm # ¢LBD 0.3491 mm = PBD = = 0.00116 mm>mm Ans. LBD 300 mm #26 A 300 m CHAPTER 2 STRAIN 72 Due a loading, the plate is deformed into the dashed shape shown C H Ato PTER 2 S T72 RAIN C H Aaverage PTER 2 S T R A I Nstrain along the side in Fig. 2–6a. Determine (a) the normal AB, andEXAMPLE (b) the average2.3 shear strain in the plate at A relative to the x and y axes. PLE 2.3 # EXAMPLE 2.3 Due to a loading, the plate is deformed into the dashed shape shown in 2–6a. Determine (a)the thedashed average normal strain along the side Due to a loading, theFig. plate is deformed into shown ythe plate Due to a loading, shape shown C H A Pis T Edeformed R 2 S T R Ainto I N the dashed 72 Due to ayükleme loading, the platedeforme isshape deformed into thedeformasyon dashed shape shown Bir levha altında olmuş AB, and (b) the average shear strain in the plate at Ave relative to the Fig. 2–6a. Determine theFig. average normal along the side in Fig. 2–6a. Determinein(a) the average normal(a) strain along theDetermine side strain in 2–6a. (a) the average normal strain along the side 2 x and y shear axes. AB, andstrain (b) the in the plate at A relative to in the AB, and (b) the average shear in average the plate atAB, A strain relative to the and (b) the average shear strain the plate at A relative to the sonrası hali şekilde kesikli çizgi ile gösterilmiştir. (a) AB x and2.3 y axes. 3 mm x and y 2axes.EXAMPLE x and y axes. kenarı boyuncay ortalama normal birim deformasyonu B y 2 mm y 2 250 mm Due to a loading, the plate is deformed into the dashed shape shown y hesaplayınız. Levhanın noktasında x ve ystrain eksenlerine in Fig. 2–6a. (b) Determine (a)Athe average normal along the side 3 mm AB,ortalama and (b) the average shear strain in the platehesaplayınız. at A relative to the göre kayma birim deformasyonunu 3 mm B x and y axes. B Due to a loading, the plate is deformed into dashed shape shown B to a loading, the plate is average deformed into the dashed shape shown in Due Fig. 2–6a. Determine (a) the normal strain along the side y x strain along the side in and Fig. (b) 2–6a. (a) the average A C normal 300 mm AB, theDetermine average shear strain in the plate2at A relative to the 2 mm mm 250 mm to the AB,yand strain in the plate at A relative x and axes.(b) the average shear (a) 250 mm x and y axes. 250 mm 250 mm 3 mm 3 mm B the Çözüm: y Fig. 2–6 2 mm 2 mm 3 mm B y x A 300 mm x hesaplayalım: C A(a) (a) AB kenarının deformasyon sonrası boyunu A SOLUTION A C 300 mm 300 mm 3 mm (a) after (a)y axis, becomes Part (a). Line AB, coincident with the 3 mm 3 mm line AB¿ B deformation, as shown in Fig. 2–6b. The length of is AB¿ 3 mm B 250 mmFig. 2–6 C 300 mm x C 2 mm x (a) 3 mmFig. 2–6 Fig. 2–6 Fig. 2–6 2 mm 2 2 SOLUTION AB¿ =2 mm 21250 mmB¿- 2 Bmm2 + 13 mm2 =2 mm 248.018 mm x 2 mm mmbecomesCline AB¿ SOLUTION 2 mm Part (a). SOLUTION B¿ Line AB, coincident Awith the y300 axis, SOLUTION 250 mm B¿ 250 mm 250 mm average normal strain for AB(a). is therefore (a) after deformation, as shown Fig.becomes 2–6b. The length of AB¿ is Part Line AB, coincident with the yin axis, line AB¿ B B after The Part (a). Line AB, coincident with the y axis, becomes line AB¿ after Part (a). Line AB, coincident with the y axis, becomes line AB¿ after 50 mm deformation, as shown in as shown 2–6b. of AB¿ is 2–6b. The length of AB¿ is 3 mm Fig. 2–6b. The length in of Fig. is The length AB¿ 250deformation, mm deformation, as shown in Fig. 2 2–6 2 = 248.018 mm 248.018 mm - 250 mm AB¿ - AB AB¿ = 21250 mm 2 mm2 + Fig. 13 mm2 x = 1PAB2avg = AAB B 300 2 21250 2 - x2 mm22 + 13 mm22 = 248.018 mm2 mm mm2=mm mm AB¿# = 21250 mm - 2# AB¿ mm2 + 250 13 Cmm2 = 248.018 mm= 21250 mm - 2 mm2 + 13 mm22 = 248.018 mm A C A AB¿ 300 mm The average normal strain for AB is therefore (a) SOLUTION B¿ -3 (b) A Ans. = - 7.93110 2 mm>mm (a) Aaverage Thefor strain for AB is therefore The normal strain AB is normal therefore Ortalama normal birim deformasyon: Part (a). Line AB, coincident withisthe y axis, becomes line AB¿ after The average normal strain for AB therefore (b) average Fig. 2–6 248.018 mmlength - 250ofmm AB¿ - AB 250 mm (b) Fig. 2–6 deformation, as shown in Fig. 2–6b. The AB¿ is = 1P 2avg = The negative sign strain causes a contraction of AB. AB y indicates the mm AB¿ - 250-mm AB¿ -- 250 ABmm248.018AB 250 mm 248.018 mm AB¿ - AB 248.018 mm - 250 mm AB = 1P =AB2avg = 1P 2 = 2avg 2 AB 250=mm SOLUTION AB avgy AB¿1P=AB21250 mm AB - 2 mm2=2 + 13 mm2 = 248.018 mm AB 250 mm -3 250 mm Ans. = - 7.93110 2 mm>mm SOLUTION Part (b). As noted in Fig. 2–6c, the once 90° angle BAC between the (a).plate Line AB, coincident with ythe axis, becomes line -3 AB¿ after -3 the A Ans. = 7.93110 2 mm>mm -3 sidesPart of the at A changes to due to displacement of B to u¿ Ans. - 7.93110 with 2 mm>mm Part AB,=coincident y axis, becomes line AB¿ after strain mm Ans. = - 7.93110 2 mm(a). # Line The average normal for AB 2ismm>mm therefore as 3shown Fig.g2–6b. Thethe length of AB¿ The negative sign indicates the strain causes a contraction of AB. (b)2–6b. Since then is the angle shown inisthe figure. B¿. deformation, gxy =B p>2 -shown u¿, in xy deformation, as in Fig. The length of is AB¿ B¿ 3 mm 3 mm 2 mm The negative sign indicates the strain contraction of AB. 248.018 Thus,The negative (b) Ortalama kayma birim deformasyonu hesabı sign indicates the strain causes a contraction of causes AB.signa indicates negative strain causes amm contraction of AB. - 250 mm AB¿ the - AB B¿ AB¿ = 21250B mm - 2 mm22 +2 13 mm22 =2 The 248.018 mm B¿ =angle BAC between the 1P 2 = AB avg Part (b). As noted in Fig. 2–6c, the once 90° AB¿ = 21250 mm - 2 mm2 + 13 mm2 = 248.018 mm y AB 250 mm 3 mm gxy -1 sides of the plate at A changes toBAC to the the displacement of B to u¿ due (b). As noted inangle Fig. 2–6c, the once 90° between gxy As Ans. = tan ain Part b = 0.0121 rad Part (b). noted Fig. 2–6c, the once 90° BAC between theangle 250 mm Part (b). As noted in Fig. 2–6c, the once 90° angle BAC between the The average normal strain for AB is therefore gxy 250 mmof-AB 2 mm Since then is the-3angle shown in the figure. B¿. gxy to = u¿ p>2 - to u¿,to gxy for The average normal therefore =displacement -gchanges 7.93110 2 of mm>mm sides the plate at changes due the B toto the xy sides of the plate at Astrain changes to u¿is due to A the displacement of B sides of the plate at A to due displacement of Ans. B to u¿ 250 mm Thus, Since then is the angle shown in the figure. B¿. g = p>2 u¿, g u¿= p>2 AB¿ then is the angle shown in the figure. B¿. Since gxy u¿, g xy xy 250Since mm gxy = p>2 - u¿, then gxy is the angle shown in the figure. - AB xy 248.018 mm - B¿. 3 mm 2 mm = x248.018 mm The 1PAB2avg = Thus, - 250 mm sign indicates the strain causes a contraction of AB. AB¿ - AB negative Thus, u¿ B B¿u¿ AB 250 mmThus, = 3 mm A 1PAB2avg = C AB 250 mm gxy = tan-1 a Ans. b = 0.0121 rad x x 3 mm 250 mm - 2 mm (c) -3 3 mm C -1 x 3 mm -1 Ans. = 7.93110 2 mm>mm A C Part (b). As noted in Fig. 2–6c, the once 90° angle BAC betweenAns. the Ans. b = 0.0121 rad b==tan 0.0121 -1 -3 gxy = tan gxy = tan b = a0.0121 rad gAns. a rad Ans. = -ga7.93110 mm>mm xy 250 mm 2 plate mm (c) xy250 mm2 2 mm 250 mm 2 mm (c) sides of the at A changes to due to the displacement of B to u¿ 250 mm The negative sign indicates the strain causes a contraction of AB. B¿. Since gxy = p>2 - u¿, then gxy is the angle shown in the figure. The negative sign indicates the strain causes a contraction of AB. Thus, u¿ Part (b). As noted in Fig. 2–6c, the once 90° angle x BAC between the 3 mm Part in Fig. 2–6c, once 90° displacement angle BAC between the Cto the # As noted sides of(b). the plate atAA changes to u¿the due of B gxyto= tan-1 a Ans. b = 0.0121 rad ofgthe =plate dueangle to theshown displacement of B to u¿the 250 mm - 2 mm (c) g to is Since then in the figure. B¿.sides p>2at- Au¿,changes xy xy then gxy cinsinden) is the angle shown in the figure. B¿. Since # gxy = p>2 - u¿, (radyan Thus, Thus, x 3 mm x gxy = tan-1 a-1 Ans. b = 0.0121 rad 3 mm 250 mm - 2 mm b = 0.0121 rad g Ans. = tan a xy # 250 mm - 2 mm #27 2.2 2.2 STRAIN 2.2 73 EXAMPLE 2.4 2.4 # EXAMPLE The plate shown in Fig. 2–7a is fixed connected along AB and held in 73 STRAIN 73 STRAIN x y the horizontal guides at its top and bottom, AD and BC. If its right x y plate shown in Fig. 2–7a is fixed connected along AB and held in x D side CD is given a uniform horizontal displacement of 2 mm, A the horizontal guides at its top and bottom, AD and BC. If its right 2 determine (a) the average normal strain along the diagonal AC, and D side CD is givenD a uniform horizontal displacement of 2.4 2 mm, EXAMPLE A A (b) the shear strain at E relative to2the x, y axes. 2 determine (a) the average normal strain along the diagonal AC, and 150 mm E (b) the shear strain at E relative to the x, y axes.The plate shown in Fig. 2–7a is fixed connected along AB and held in 150 mm 2.2150Smm TRAIN 73 STRAIN 73 the horizontal guides at2.2its top and bottom, AD and BC. If its right E SOLUTIONE B C side CD is given a uniform horizontal displacement of 2 mm, Part (a). When the plate is deformed, the determine diagonal AC 150along mm the 2diagonal mm SOLUTION (a) becomes the average normal strain AC, and B C B Fig. 2–7b. The C length of diagonals AC and AC¿ can be found EXAMPLE Part 2.4 AC¿, (a)axes. EXAMPLE 2.4 (b) the shear strain at E relative to the x, y (a). When the2 plate 150 mm 2 mm AC becomes mm mm is deformed, the diagonal AC becomes from the150 Pythagorean theorem. We have # AC¿, Fig. 2–7b. The length of diagonals AC and AC¿ can bey found (a) an be The found (a) x plate shown shown in in Fig. Fig.2–7a 2–7a isis fixed fixed connected connected along along AB AB and and held held in in 76 mm x y The plate A 76 mm from the Pythagorean theorem. We have Şekildeki levha AB kenarı boyunca ankastre mesnetlidir. Ayrıca AD ve BC kenarları boyuncaD¿ the horizontal guides at its top and bottom, AD and BC. If its right 2 and BC. If 2SOLUTION the horizontal guides76at its top and bottom, AD its right AC =horizontal 210.150 m2 + 10.150 m2 0.21213 D 76 mm73 76 mm D¿ 76 mm AD 2.2 STRAIN75 mm side CD CD isis given given uniform displacement of 2= 2 mm, mm, m A a mm D¿ A side a uniform horizontal displacement of düşey hareket engellenerek yatay verilmiştir. CD kenarının yatay olarakthe 2 2 harekete izin 2 Part A (a). When the plate is deformed, diagonal AC becomes u¿ 2 mm determine (a) (a) the the average averageAC normal strain along along the diagonal AC, and mm == 210.150 m2 m2 == 0.21213 2 3 m determine AC¿ 210.150 m22 + + 10.150 10.152 m22AC, 0.21355 normal strain the diagonal and 75 mm AC¿, AC¿ Fig. 2–7b. The length of diagonals AC and can be found 75 mm 2.2 STRAIN 73 (b) the the shear shear strain at atdeplasman E relative relative to the x, x,yy axes. axes. u¿ birim üniform yapması durumunda (a)2AC diyagonali boyunca ortalama normal u¿ (b) strain E to the 2 E¿ 150 mm from the Pythagorean theorem. We have AC¿ = 210.150 m2 + 10.152 m2 = 0.21355 m EXAMPLE 2.4 Therefore the average normal strain along the diagonal is 150 mm 5m E 75 mm E B and held in C. If its right t of 2 mm, onal AC, and yThe E¿ deformasyonuE¿ve (b) E noktasında, x ve y eksenlerine göre kayma birim deformasyonunu 75 mm x mm Therefore the average normalalong strainAB along diagonal isy C¿ shown in75 Fig. 2–7a is fixed connected andthe held in EXAMPLE 2.4 B s The plate SOLUTION AC = B210.150 m22 C + 10.150 m22 = 0.21213 m SOLUTION AC¿ AC 0.21355 m 0.21213 m the horizontal guides at its top and bottom, AD and BC. If its right hesaplayınız. (b) B C C¿ Part (a). When the plate is deformed, the diagonal AC becomes 1P 2 = = B 150 mm 2 mm C¿ AC avg B D 2 (a). When the plate is deformed, the diagonal becomes 150 mm 2 mm xm22 = 0.21355 m y side Part CD is given a uniform horizontal displacement of AC 2 0.21213 mm, The plate shown Fig. 2–7a is fixed connected along AB and held AC m in m AC¿ m2 + 10.152 AC¿ -ACAC 0.21355 m -found 0.21213 A = 210.150 AC¿, AC¿ Fig. 2–7b. Theinlength length of diagonals diagonals and AC¿ can be 13 m AC¿, (a) (b)of Fig.(b) 2–7 2 Çözüm: Fig. 2–7b. The AC and can be found 1P 2 = = (a) determine (a) the average normal along the diagonal thethe horizontal guides atAC itsstrain top and bottom, AD andAC, BC.and If its right avg from Pythagorean theorem. We have AC 0.21213 m = 0.00669 mm>mm Ans. D from the Pythagorean theorem. We have (b) the shear strainisatgiven E relative to the x,horizontal y axes. side CD a uniform displacement Therefore of 2 mm,the averageAnormal strain along the diagonal is Fig. 2–7 Fig. 2–7 (a) AC diyagonal uzunluğunu deformasyon öncesi ve150 sonrası belirleyelim mm mm 76 mm 2 STRAIN A 76mm = 0.00669 mm>mm Ans. 76 mm D¿ determine (a) the2.2 average normal strain73 along the diagonal AC, and Ans. A 76 E D¿ 2 the shear strain 2 Part (b). To find at E relative to the x and y axes, it AC = 210.150 m2 + 10.150 m2 = 0.21213 m 2 (b) the shear at E relative the x,m2 y axes. AC strain = 210.150 m22 + to 10.150 = 0.21213 m 0.21355 m - 0.21213 m 75 mm 150 mm AC¿ - AC is first necessary to find the angle u¿ after deformation, Fig. 2–7b. mmWe SOLUTION 1PAC 2avg =it = u¿E Part (b). To find the shear strain yB75axes, 2 2 at E relative to the x and and y axes, it C u¿ AC 0.21213 m AC¿ 210.150 m2 m22 ++ 10.152 10.152 m2 m22 = 0.21355 m m have AC¿ ==plate 210.150 is first necessary to findthe the diagonal angle =u¿0.21355 after deformation, Fig. 2–7b. We Part (a).WeWhen the is deformed, AC becomes 150 mm 2 mm Fig. 2–7b. # E¿ SOLUTION 0.00669 Ans. have E¿ mm>mm AC¿, Fig. 2–7b. yThe length of diagonals xAC and AC¿ can be found 75 mm= (a) B C Therefore the average normal strain along the diagonal is and held in u¿ 76 mm 75 mm the average normal strain along the diagonal is fromTherefore thePart Pythagorean theorem. We have AC diyagonali boyunca ortalama normal birim deformasyon: (a). When the plate is deformed, the diagonal AC becomes 150 mm 2 mm tan a b = If its right 75mm mm AC and AC¿ can be found C¿ u¿2 Dof diagonals 76 B shear AC¿, Fig. 2–7b. The length (a) 76 mm 76 mm Part (b). To find the strain at E relative to the x and y axes, it C¿ A B of 2 mm, D¿ tan a - bAC = A AC¿ 0.21355 m2-- 0.21213 0.21213 m mis first necessary to find the (b) from the1P Pythagorean theorem. We have 2 75 mm u¿ angle after deformation, Fig. 2–7b. We 2 2 AC¿ AC 0.21355 m p 2 = = (b) al AC, and AC1P=AC210.150 m2 +u¿10.150 m2 = 0.21213 m avg a b190.759°2 = 1.58404 rad AC2avg = AC = =90.759° =0.21213 0.21213 m 75 mm have 76 mm 76 mm AC m A 180° D¿ p u¿Fig. 2–7 2 150=mm Fig. 2–7 2 = 90.759° = 0.21355 am22 =b190.759°2 = Ans. 1.58404 rad AC¿ m22 +u¿mm>mm 10.152 m2 m 404 rad 0.00669 AC = == 0.00669 210.150 m2 + 10.150 0.21213 m E # 210.150 180° mm>mm Ans. u¿ 75 mm 76 mm Applying Eq. 2–3, the shear strain at E is therefore u¿ tan a b = E¿ 2 2 75 mm Therefore the average normal strain along the diagonal is AC¿ = 210.150 m2 + 10.152 m2 = 0.21355 m 2 75 mm (b) Deformasyon sonrası levhanın biçim değişimini çizelim. Deformasyon öncesi x ve y Part (b). To find the shear strain at E relative to the x and y axes, it Applying Eq. 2–3, the shear strain at E is therefore B the shear strain C Part (b). To find at E p relative to the x and y axes, it E¿ C¿ u¿ is first necessary to find the angle after deformation, Fig. 2–7b. We C becomes 150 the mm angle mm B 75Ans. g 2u¿ - deformation, 1.58404 rad =Fig. - 0.0132 rad = after p is first necessary find 2–7b. We Therefore the to average normalxy strain the diagonal =mm 90.759° a b190.759°2 = 1.58404 rad eksenleri açı p # 2 along idi. E =noktası deformasyon AC¿arasındaki -(a)AC 0.21355 m -Deformasyon 0.21213 m issonrası açı ise # u¿ radyan. have n be found (b) 180° have =xy = g - 1.58404 rad = - 0.0132 rad Ans. avg = Ans.1PAC2sonrası C¿ E'AC ne kayacaktır.2 0.21213 m B Fig. 2–7 u¿ The76 negative sign indicates that the angle is greater than 90°. AC¿ AC 0.21355 m 0.21213 m u¿ 76 mm (b) at E is therefore mmmm76 mm Applying Eq. 2–3, the shear strain 1P=AAC 2 76 = mm>mmD¿ = 0.00669 Ans. tanaau¿ tan bb negative ==avg75 mmsign AC 0.21213 m 2 u¿ The indicates that the angle is greater than 90°. n m90°. 2 75 mm NOTE: If the x and y axes were horizontal and vertical at point E,p Fig. 2–7 75 mm 0.00669 Ans. u¿ - 1.58404 rad = - 0.0132 rad = Ans. Part (b). To find the shear=strain at Emm>mm relative to the axes x andwould y axes,not it change xy the then 90° angle these due gto pp u¿ ==the 90.759° = a between b190.759°2 = 1.58404 1.58404and rad vertical 2 NOTE: If the x and y axes were horizontal at point E, almisatfirst point E, u¿ 90.759° = a b190.759°2 = rad u¿ after necessary to deformation, find the angleand = 0 at pointFig. so180° gxydeformation, E. 2–7b. We E¿ angle 180° thethe 90° between axes notychange ehave due toPart the (b). then To find shear strain at E these relative to would the x and axes, it due to the 75 mm = 0 deformation, and so g at point E. The negative xy Applying Eq.2–3, 2–3,the the shear strain at E E therefore u¿isisafter is first necessary toshear find the angle deformation, Fig. 2–7b. We sign indicates that the angle u¿ is greater than 90°. Applying Eq. strain at therefore C¿ B 76 mm have u¿ # tan a b = m NOTE: If the x and y axes were horizontal and vertical at point E, p(b) 2 75 gxymm 1.58404 rad rad == --0.0132 0.0132 rad rad = p -- 1.58404 Ans. thenAns. the 90° angle between these axes would not change due to the gu¿ = xy 276 mm 2 tan a b Fig. = 2–7 p deformation, and so gxy = 0 at point E. 2 75 u¿ = 90.759° = mm a b190.759°2 = 1.58404 rad Ans. 180° u¿ isis greater The negative negative sign sign indicates indicates that that the the angle angle u¿ greater than than 90°. 90°. The p u¿ = 90.759° = a b190.759°2 = 1.58404 rad ndApplying y axes, itEq. 2–3, the shear strain at E is 180° therefore # NOTE: If the x and y axes were horizontal and vertical vertical at at point point E, E, g. 2–7b. We NOTE: If the x and y axes were horizontal and then the 90° angle between these axes would not change due to the pbetween then the 90° Eq. angle these axes not change due to the Applying 2–3, the shear strain at would E is therefore gxyand rad = E. - 0.0132 rad = so Ans. = 00 at deformation, gxy1.58404 at point point = deformation, and so g E. 2 xy # p gxy = - 1.58404 rad = - 0.0132 rad Ans. The negative sign indicates that2the angle u¿ is greater than 90°. 4NOTE: rad The negative indicates that the angle is greater than 90°. If the x andsign y axes were horizontal andu¿vertical at point E, then the 90° angle between these axes would not change due to the 0 aty point deformation, andIfsothe gxyx=and E. horizontal and vertical at point E, NOTE: axes were then the 90° angle between these axes would not change due to the #28

Download
# ders notu