Bölüm 4. Eksenel Yük Saint-Venant Prensibi Bir ucundan ankastre mesnetli diğer ucundan bir P yüküne maruz bir çubuğu gözönüne alalım. Çubuğun deformasyonu öncesi üzerine çizilen karelaj çizgilerinin deformasyon sonrası aldığı şekli incelediğimizde çubuğun her iki ucundaki lokal deformasyondaki düzensizlikler 4.1 görülmektedir. P P SAINT-VENAN P P a b c P a b c P Load distorts lines located near load savg ! 4.1 Lines located away from the load and support remain straight Load distorts lines located P near support savg ! A 121 SAINT-VENANT’S PRINCIPLE # section a–a P 2 section b–b (b) P 2 P A se section c–c Fig. 4–1 (cont.) P A the stress savg ! In the same way, distribution at the support will also even out and become uniform over the cross section located the same distance section c–c ection b–b section c–c (a) # from away the support. (c) (b) The fact that stress and deformation in this manner is referred Fig. 4–1 yeteri kadar uzaklaşıldığında gerilme ve behave deformasyon dağılımın Fig. 4–1 (cont.) Bu lokal düzensizliklerden to as Saint-Venant’s principle, since it was first noticed by the French scientist Barré de Saint-Venant in 1855. Essentially it kesitinin states that the üniformlaştığı gözlenebilir. Bu uzaklaşma mesafesinin, en azından yüklenen çubuk stress and strain produced at points in a body sufficiently removed from en büyük boyutu kadar olması gerektiği Gerilme davranışı the regionanlaşılmıştır. of load application willvebedeformasyonun the same as the bu stress and strain sing this idea, consider the manner in which a rectangular bar will 4 produced by any applied loadings that have the same statically equivalent ess atSaint-Venant thethe support also even rm distribution elastically when bar iswill subjected to a force P applied along its adamının prensibi olarak bilinmektedir. uygulandığı resultant, andBu arebilim applied to the “Yükün body within the samebölgeden region. For er the cross section located the same distance roidal axis, Fig. 4–1a. Here the bar is fixed connected at one end, with example, if two symmetrically applied forces act on P>2 yeteri kadar cisim oluşan gerilme ve birim deformasyon, statikçe the bar, orce applied through a hole uzaktaki at its other end.noktalarında Due the loading, thedistribution Fig.to4–1c, the stress at section c–c will be uniform and eformation in this manner is referred deforms asbehave indicated bybileşkeye the once horizontal and vertical grid lines therefore equivalent to savg = P>A as in Fig. 4–1b.doğan gerilme ve eşdeğer sahip ve aynı bölgede uygulanan herhangi bir yüklemeden ple, since it was first noticed by the French wn on the bar. Notice how the localized deformation that occurs enant in 1855. Essentially it states that deformasyon ile the aynıuniform olacaktır.” ifadesi mevcuttur. Yukarıdaki şekildeki P yüklemesini iki ach end tends birim to even out and become throughout the at points in a body sufficiently removed from section ion willof bethe thebar. same as the stressP/2+P/2 and strainolarak yüklemek, yüklemeden yeteri kadar uzakta aynı etkileri parçaya ayırıp the material remains elastic equivalent then the strains caused by this adings that have the same statically rmation are directly related toregion. the stress meydana getirecektir. to the body within the same For in the bar. As a result, the ss will applied be distributed uniformly the cross-sectional ically forces more on the throughout bar, P>2 act bution at section will be uniform when the sectionc–c is taken farther andand farther from the point where in Fig. = P>A asload is4–1b. applied. For example, consider a profile of the gexternal ation of the stress distribution acting at sections a–a, b–b, and c–c, h of which is shown in Fig. 4–1b. By comparison, the stress tends each a uniform value at section c–c, which is sufficiently removed m the end since the localized deformation caused by P vanishes. minimum distance from the bar’s end where this occurs can be rmined using a mathematical analysis based on the theory of ticity. Notice how the lines on this rubber membrane has been found that this distance should at least be equal to the distort after it is stretched. The localized est dimension of the loaded cross section. Hence, section c–c should distortions at the grips smooth out as stated by Saint-Venant’s principle. ocated at a distance at least equal to the width (not the thickness) of bar.* hen section c–c is so located, the theory of elasticity predicts the maximum stress to be #42 # Lastik membran üzerindeki çizgilerin çekildikten sonra nasıl biçim değiştirdiğine dikkat ediniz. Tutamaklardaki bölgesel bozuklukları prensibinde ifade edildiği Notice howbiçim the lines on thisSaint-Venant rubber membrane gibi giderek düzgünleşmektedir. distort after it is stretched. The localized distortions at the grips smooth out as stated by Saint-Venant’s principle. #43 ed loads at itsloads endsat and variable load distributed of the bar if itload doesdistributed not remain horizontal, or friction forces acting concentrated its aends andweight aexternal variable external th. its This distributed load could, for example, the we wish oncould, the bar’s surface. Here ng length. This distributed load for represent example, represent the to find the relative displacement d bar of if itthe does horizontal, or friction forces (delta) of oneor end ofacting the bar with respect to the other end as caused by ght barnot if itremain does not remain horizontal, friction forces acting d surface. we wish the d the bar’sHere surface. Heretowefind wish torelative find thedisplacement relative this loading. We willdisplacement neglect the localized deformations that occur at elta) endofofone theend bar of with to respect the other as caused by therespect bar with to the other end as causedand by where the cross section suddenly points ofend concentrated loading We will neglect theneglect localized that occurthat at occur loading. We will the deformations localized at changes.deformations From Saint-Venant’s principle, these effects occur within small ncentrated loading and whereand the crossof section suddenly nts of concentrated loading where the section suddenly regions thecross bar’s length and will therefore have only a slight effect on 4.2 Eksenel yüklü bir elemanın elastik deformasyonu mnges. Saint-Venant’s principle, these effects occur within small From Saint-Venant’s principle, these effects occur within small the final result. For the most part, the bar will deform uniformly, so the eions bar’s andlength will therefore onlystress a slight effect on effect oflength the bar’s and will have therefore have only slight on normal will bea uniformly distributed cross section. Değişken kesitli bir çubuk uçlarından kuvvetlereover ve the L uzunluğu boyunca bir yayılı yüke lt. Forresult. the most the bar will uniformly, so the tekil final For part, the most part, thedeform bar will deform soa the Using the methoduniformly, of sections, differential element (or wafer) of length will stress be uniformly distributeddistributed over cross mal will bemaruzdur. uniformly over the cross section. dxthe and cross-sectional area is isolated the bar at the arbitrary Çubuğun birsection. ucunun diğerA(x) ucuna göre from rölatif deplasmanı δ yı hesaplayalım. 4.2 diagram Eof LASTIC Dof EFORMATION OF AN XIALLY Lin OADED EMBER 123 method ofmethod sections,ofa sections, differential element of length Using the a differential element (or wafer) length position(or x.wafer) The free-body this element isAshown Fig.M 4–2b. sectional area A(x) is isolated from the bar at the arbitrary and cross-sectional area A(x)deplasman is isolated from the bar ataxial the arbitrary Çubuğun sonrası alacağı şekil kesikli ile gösterilmiştir. The resultant internal force will beçizgi a function of x since theÇubuğun external sol ucundan e free-body diagram ofdiagram this element iselement shown in Fig. 4–2b. ition x. The free-body of this is shown in Fig. 4–2b. 4.2 E LASTIC D EFORMATION OF AN A XIALLY L OADED the M EMBER 123 distributed loading will cause it to vary along length of the bar. This For the entire length L of the bar, we must integrate this expression to biraxial x be mesafedeki dx elemanı için dδ uzamasını yazalım. Oluşan gerilmelerin orantılılık axial force will a function x since the einternal resultant internal force willload, be aof function of xexternal since the external P(x), will deform the element into the shape indicated by the dashed find d. This yields ading will cause it to vary along the length of the bar. This ributed loading will cause it to vary along thetherefore length ofthe thedisplacement bar. This outline, and of one end of the element with ntire length Lelement of the bar, we integrate this expression to sınırından düşük olduğu varsayımı ilethe (Hooke kanunu geçerlidir): l deform thedeform into themust shape indicated by the dashed d, P(x), will the element into the shape indicated by dashed dd. respect to the other end is The stress and strain in the element are yields L herefore the displacement of one end of the element with line, and therefore the displacement of one P1x2 end of dxthe element P1x2 with dd dand (4–1) other is dd. The and stress strain in= Ethe element 4.2 LASTIC Din EFORMATION 123 pect toend the other end stress is dd. The strain theare element areAXIALLY s = OF AN andLOADED PM =EMBER 0 A1x2E L A1x2 dx L P1x2 P1x2 dd P1x2 dx dd s = P and = P the = stress does Provided not exceed the proportional limit, we can apply d s= = and (4–1) A1x2 dx A1x2 dx ire length L of the must this expression to where ! bar, ! integrate L0 weA1x2E Hooke’s law; i.e., elds does stress not exceed theexceed proportional limit, we can apply vided the stress does not the proportional limit, we can apply d = displacement of one point on the bar relative to the other point s = EP i.e., law; i.e., oke’s L = original length of bar dd P1x2 s =LEP sdx = EP placement of one point onaxial the bar relative the other point P1x2 x from P1x2 =d internal force at the to section, located = Ea b = (4–1)a distance A1x2 dx ginal length ofP1x2 bar P1x2 oneL0endA1x2E dd dd = E a b= E a b P1x2 ernal axial force the section, distance x from as a function A1x2 dx located A1x2 A1x2 = at cross-sectional area dx ofathe bar, expressed of xdx dd = 4 e end A1x2E E = modulus of elasticity P1x2 dx P1x2for dxthe material dd = onbar, dd = relative oss-sectional areapoint of the expressed as to a function x lacement of one the bar the otherofpoint 4 A1x2E A1x2E x dx Constant Load and Cross-Sectional Area. In many cases odulus of elasticity nal length of bar for the material bardxat will have a constant area A; and the material will x axial dx the Pthe P2 rnal force section, locatedcross-sectional a distance x from P(x) P(x) 1 be homogeneous, so E is constant. Furthermore, if a constant external Load and Cross-Sectional Area. In many cases 4.2 E LASTIC D EFORMATION OF AN AXIALLY L OADED MEMBER 123 end P2end, P(x)then P(x) P(x) dd force is cross-sectional appliedP2 at each Fig.the4–3, the internal force P have a constant area A;P(x) and material will s-sectional area of the bar, expressed as a function of x dx L throughout the length of the bar constant. As a result, Eq. 4–1 can neous, so E is constant. Furthermore, if is a also constant external ! 4 d dd eulus entire length L of bar, we must integrate (a) this expression todd (b) of at elasticity forthe the material beLeach integrated to yield pplied end, Fig. 4–3, then the internal force P dx dx L his yields d boyunca dδ entegre edilirse: d Tüm çubuk the length Eq. 4–1 can Fig. (b) 4–2 (a)of the bar is also constant. As a result, (b) a) Load and Cross-Sectional Area. In many cases ed to yield PL Fig. cross-sectional 4–2 Fig. 4–2 ave a constant area L d =the material will (4–2) P1x2 dx A; and AE external eous, so E is constant. d = Furthermore, if a constant (4–1) L0PLA1x2E plied at each end, dFig. P = 4–3, then the internal force(4–2) ! AE he length of the bar is also constant. As a result, Eq. 4–1 can If the bar is subjected to several different axial forces along its length, or to yieldthe cross-sectional Sabit yük veor sabit kesit sözkonusu olduğunda, area modulus of alanı elasticity changes abruptly from The vertical displacement at the top of these displacement ofseveral one of point the relative to the point is subjected different axial forces its other length, or one to region theon bar tobar the next,along the above equation can be applied building columns depends upon the loading the and to the floor attached The vertical displacement at theon top ofroof these ectional orofmodulus of the elasticity changes abruptly from to each segment of bar where these quantities remain constant. The applied original area length bar PL to their midpoint. d = (4–2) building columns depends upon the loading of the bar the next, the above equation can bex applied displacement ofAE one end of the bara with respect to the other is then found nternal axialtoforce at the section, located distance from applied on the roof and to the floor attached ! ment of the bar where these quantities remain constant. The from the algebraic addition of the relative displacements of the ends of one end to their midpoint. nt of oneeach end segment. of the barFor with respect to case, the other is then found this general Bir çubuk uzunluğu boyunca farklı eksenel yüklere maruz ise veya enkesit alanı cross-sectional area of the bar, expressed as a function of x subjected to several different axial forces along its length, or gebraic addition of the relative displacements of the ends of 4 The vertical displacement at the top of these tional area or modulus of elasticity changes abruptly from modulus ofgeneral elasticity for the material nt. For this case, uğruyorsa, PLbe applied building columns depends upon the loading f the bar to the ani next,değişimlere the above equation can d = a (4–3) ent of the bar where these quantities remain constant. AE nt Load and Cross-Sectional Area. In manyThe casesapplied on the roof and to the floor attached to their midpoint. PL of have one end of the bar respect to theA; other then found will a constant area and is the material dcross-sectional =with (4–3)will a AEdisplacements of the ends ebraic addition of the relative of geneous, so E is constant. Furthermore, if a constant external ! t.applied For this general case, at each end, Fig. 4–3, then the internal force P ut the length of the bar is also constant. As a result, Eq. 4–1 can x ated to yield PL d P= a (4–3) P x AE PL L P P d = (4–2) d AE ! L Fig. 4–3 d ar is subjected x to several different axial forces along its length, or Fig. 4–3of elasticity changes abruptly from -sectional area or modulus on of the bar to the next, the above equation Pcan be applied egment of the bar where L these quantities remain constant. The ment of one end of the bar with respect to thed other is then found algebraic addition of the Fig. 4–3relative displacements of the ends of ment. For this general case, The vertical displacement at the top of these building columns depends upon the loading applied on the roof and to the floor attached to their midpoint. #44 Fig. 4–5b. They are PAB = +5 kN, PBC = -3 kN, PCD = -7 kN. This !P Fig. 4–5b. They are PAB = +5 kN, PBC = -3 kN, PCD = -7 kN. This variation in axial load is shown on the axial or normal force diagram for variation in axial load is shown on the axial or normal force diagram for the bar, Fig. 4–5c. Since we now know how the internal force varies the bar, Fig. 4–5c. Since we now know how the internal force varies throughout the bar’s length, end A relative to end D the displacement of end A relative to end D !d the displacement ofthroughout the bar’s length, is determined Positive from sign convention for P and d is determined from d 4–4 124 C HFig. APTER 4 AXIAL LOAD 15 kN2L 1-3ve kN2L 1-7 kN2L 15 kN2L 1-3 kN2LBC 1-7 kN2LCD PL AB BC PLCD AB İşaret Kuralı: Eksenel yük deplasmanlar için pozitif yönler: dA>D = a = + + dA>D = a = + + AE AE AE AE AE AE AEAE Sign Convention. In order to apply Eq. 4–3, we must develop a !P sign convention the force answer and theis displacement the other data arefor substituted andaxial a positive calculated, it of one If the other data are substituted and a positiveIfanswer is calculated, it internal end of the bar with respect to the other end. To do so,elongates), we will consider means that end A will move away from end D (the bar means that end A will move away from end D (the bar elongates), both the force and displacement to be positive if they cause tension whereas a negative result would indicate that end A moves toward !d whereas a negative result would indicate that end A moves toward end D elongation, (the bar shortens). The double subscript notation is used force to and respectively, Fig. 4–4; whereas a negative and end D (the bar shortens). The double subscript notation is used to 1dA>D2; and indicate this relative displacement however, if the displacement displacement will cause compression contraction, respectively. indicate this relative displacement 1dA>D2; however, if the displacement is to For be determined relative tothe a fixed then a single subscript consider bar point, shown in only Fig. 4–5a. The internal axial is to be determined relative to a fixed point, then only a example, single subscript will be used. For example, if D is located at a fixed support, then thesegment, forces “P,” are determined by the method of sections for each will be used. For !P example, if D is located at adisplacement fixed support, then the as simply dA. will be denoted Fig. 4–5b. They are PAB = + 5 kN, PBC = - 3 kN, PCD = - 7 kN. This displacement will be denoted as simply dA. variation in axial load is shown on the axial or normal force diagram for LOAD the bar, Fig. 4–5c. Since we now know how the internal force varies !d throughout the bar’s length, the displacement of end A relative to end D ! 8 kN 4 kN 5 kN from 7 kN Positive sign convention for P and d is determined Sign Convention. In order to apply Eq. 4–3, we must develop a 4 XIAL 8 kN 5 kN 4 kN 7 kN A B sign convention forFig. the4–4 internal axial force and the displacement of one 4 C D LAB LBC LCD end of the bar 15 kN2LAB 1 - 3 kN2LBC 1 - 7 kN2LCD A with respect B to the otherCend. To do so, weDwill PL consider d = = + + (a) both the force and tension acause LAB displacement LBC to be positiveA>D Lif CD they AE AE AE AE and ! elongation, respectively, Fig.(a)4–4; whereas a negative force and displacement will cause compression and contraction, respectively. P (kN) the other data are substituted and a positive answer is calculated, it 5 kNshown in If AB "bar For example, consider Pthe Fig. 4–5a. The internal axial means that end A will move away from end D (the bar elongates), 5 kN A P (kN) forces “P,” are determined by the method of sections for each segment, whereas a negative result would indicate that end A moves toward Fig. 4–5b. They are PAB = +58 kN, kN PBC = -3 kN, PCD = 5-7 kN. This end D (the bar shortens). The double subscript notation is used to variation in axial load is shown on the axial force diagram for PBCor " 3normal kN x indicate this relative displacement 1dA>D2; however, if the displacement 5 kN A 54–5c. SinceBwe now know how the internal the bar, Fig. #3 force varies is to be determined relative to a fixed point, then only a single subscript PBC " 3throughout kN the bar’s length, the displacement of end A relative toxend D #7 will be used. For example, if D is located at a fixed support, then the PCD " 7 kN 7 kN P and d is determined from #3 displacement will be denoted as simply dA. D ! #7 7 kN ! dA>D (c) (b) 15 kN2LAB 1-3 kN2LBC 1-7 kN2LCD PL = a = + +Fig. 4–5 AE AE (c) AE AE 8 kN 5 kN 4–5data are substituted and a positive answer is calculated, it If the Fig. other A means that end A will move away from end D (the bar elongates), LAB whereas a negative result would indicate that end A moves toward end D (the bar shortens). The double subscript notation is used to indicate this relative displacement 1dA>D2; however, if the displacement is to be determined relative to a fixed point, then only a singlePsubscript (kN) PAB will be used. For example, if "D5 kN is located at a fixed support, then the 5 kN A will be denoted as simply d . displacement A 8 kN 5 kN 4 kN B 7 kN C LBC D LCD (a) 5 PBC " 3 kN A x B #3 8 kN 5 kN PCD " 7 kN 4 kN D B A LAB (b) C LBC (a) 7 kN #7 7 kN D LCD (c) Fig. 4–5 P (kN) " 5 kN 8 kN 5 PBC " 3 kN x #3 7 kN #7 (c) Fig. 4–5 #45 PBC " 7 kip 126 126 CHAPTER 4 AXIAL LOAD CHAPTER 4 AXIAL LOAD PCD " 9 kip (b) 126 CHAPTER 4 AXIAL LOAD EXAMPLE 4.1 EXAMPLE 4.1 ! of the external loadings, the 126 ue to the application C H Abar PTER 4 A Xin IAL LOAD The A-36 steel shown Fig. 4–6a is made from two segments in regions AB, BC, and CD will all be different. EXAMPLE Thecross-sectional A-364.1 steel barareas shownofin A Fig. 4–6a is2 made from =two2 segments having and2 A in2. AB = 1 in BD A XIAL LOAD A36ofçeliğinden üretilmiş çubuk iki farklı kesit bölgesine # ained by applyingŞekilde the method sections and the havingthecross-sectional areas of AABA=and 1 intheand ABD = 2 in2. Determine vertical displacement of end displacement force equilibrium as shown in Fig. 4–6b. This ofEXAMPLE The A-36 steel bar shown in Fig. 4–6a is made Determine the vertical displacement of end A and the displacement B relative C.4.1 sahiptir. A ucunun düşey deplasmanını ve Btonoktasının C ye göre göreceli deplasmanını n Fig. 4–6c. having cross-sectional areas of AAB = 1 in2 of B relative to 15 C.kip 15 kip 15 kip 15 kip Determine vertical displacement of end Aa The A-36the steel bar shown in Fig. 4–6a is ma 15 kip 15 kip 15 kip om Athe cover, # Est15=kip2911032 ksi. X I A L inside L O A D back belirleyiniz. of B relative to C. having cross-sectional areas of A = 1 in AB ntion, i.e., internal forces areshown positive A Thetensile A-36 steel bar inand Fig. 4–6a is made from two segments Determine the vertical displacement of end A A 15 kip 15 kip are negative, the vertical displacement of A 15 kip having cross-sectional areas of AAB = 1 in2 and ABD = 2 in2. of B relative to C. upport D is Determine the vertical the displacement 2 ft displacement of end A and A 15 15 kip 4 kip 4 kip kip 4 kip 4 kip 15 kip kip]12 ft2112 in.>ft2 [+7 4 kipkip]11.5 4 kip ft21122in.>ft2 ft of B relative to C. 4 kip 4 kip 4 kip 4 kip The in2 Fig. 4–6a is made fromP two segments +A-36 steel 4 kipbar 43shown kip 2 15 2kip kip 15 kip AB " 15 kip 2 A n22[2911032 kip>inhaving ] 12cross-sectional in22[29110 kip>in ] of15 A areas and = 1 in A = 2 in2. P B AB BD 4 2 ft AB " 15 kip 4 kip 4 kip B vertical displacement of end4A Determine the kipand 4the kip displacement 9 kip]11 ft21124in.>ft2 1.5 ft 8 kip 8 kip of B relative to C. P " 15 kip 8 kip 8 kip 1.5 ft 2 ft 8 kip AB 8 kip in22[2911032 kip>in2] 4 kip 4 kip B4 kip 15 kip 8 kip 8 kip 4 15 kip 15 kip 4 kip 0127 in. Ans. C PBC " 7 kip P " 15 kip 2 ft C 4 kip 4 1 ft sitive, the bar elongates and so the displacement D between points B and C, we 2 ft ft PAB " 15 1kip D obtain, (a) (a) 4 kip B 8 kip4 kip C 8 kip 4 kip 4 kip 4 kip 1.5 ft kip 84 kip 1.5 ft 8 kip ft 41kip 8 kip PBC " 7 kip (b) AB 8 PBC " 7 kip PCD " 9 kip PCD " 9 kip (b) p .5[+7 ft kip]11.5 ft2112 in.>ft2 8 kip SOLUTION PBC " 7 kip DC p The A-36 steel bar shown in Fig. 4–6a is 15made from two segments = +0.00217 in. Ans. P " kip SOLUTION AB 3 2 2 Due to Internal Force. the application of the external loadings, the ! 12 in22[29110 2 kip>in2areas ] 0 of15 AABP=(kip) 1 ft having cross-sectional and 2 in 1 in A = . BD 15 (a) Internal Force. Due to AB, the application of will the external loadings, the (b) internal axialDforces in regions BC, and CD all be different. PBCthe " 7displacement kip 0 P (kip) Determine thethe vertical displacement of end A and from C, since segment elongates. SOLUTION axial forces in regions and of CDsections will alland be different. 1.5 ft Çözüm: kip 8 kipby Theseinternal forces8are obtained applyingAB, theBC, method the 1kip ftB relative to C. of (a) These areforce obtained by applying theForce. method of sections and the (b) of the 15 equation of forces vertical equilibrium as shown in Fig. 4–6b. This Internal Due to the application 0 15 kip 15 kip 15 kip P (kip) equation of vertical force equilibrium as shown ininFig. 4–6b.AB, ThisBC, and CD SOLUTION variation is plotted in Fig. 4–6c. internal axial forces regions PBC " 7 kip PCDis"plotted 9 kip in Fig. 4–6c. variation (b) 3 15 These forces are obtained by applying the metho Internal Due to2 ksi. the application of th Displacement.0 From theP (kip) inside back cover,Force. Est = 29110 2 1 ft 3 SOLUTION Displacement. From the inside back cover, E = 29110 2 ksi. equation of vertical force equilibrium asand show internal axialare forces C 2 st in regions Using the sign convention, i.e., internal tensile forces positive and AB, BC, Using the sign convention, i.e., internal tensile forces are positive and variation is plotted in Fig. 4–6c. forces are obtained compressive forces loadings, are negative, the These vertical displacement of by A applying the met Internal Force. Due to the application of the external the PCD "forces 9 kip are negative, the vertical displacement of A P (kip) compressive (b)and equation of vertical sho relative the fixed D is internal axial AB, CD all support be different. Displacement. From force the equilibrium inside back ascove 2will 4 kip in regions 4 kip 4 kipBC, 4 kip to 3.5 forces to the fixed and support D isUsing variation is plotted in Fig. 4–6c. SOLUTION 7obtained by applying the relative the sign convention, i.e., internal tensile fo These forces are3.5 method of sections the [ + 15 kip]12 ft2112 in.>ft2 [ + 7 kip]11.5 ft2112 in.>ft2 PL PAB " 15 kip 7 =2in dA =asa + [24–6b. + 15 kip]12 ft2112 in.>ft2 [ + 7 kip]11.5 ft2112 in.>ft2 compressive forces are negative, the vertical equation verticalDue force shown Fig. This Displacement. From the inside back co PL InternalofForce. to equilibrium the application of the external loadings, the 3 2 kip>in2] 12+in22[2911032 kip>in2] P (kip) dA AE = a 11 =in 2[29110 2 2 to 2 support 2 Di.e., is internal Using sign tensile variation is plotted in Fig. 4–6c. AB, BC, and internal forces in regions CD will all be different. 11 in 2[2911032 relative kip>in ] thethe 12fixed inconvention, 2[2911032 kip>in ] 3.5 AE !9 axial4.5 8 kip 8 kip 7 [ 9 kip]11 ft2112 in.>ft2 3 compressive forces are negative, the These forces are obtained by applying the method of sections and the 4.5 the inside back cover, E = 29110 2 ksi. !9 From [ + 15 kip]12 ft2112 in.>ft2 vertic [+7 Displacement. PL x (ft) +st - 9 kip]11 ft2112 in.>ft2 = d = + 2[4–6b. 3 2 relative to the fixed support D is equation of vertical force equilibrium as shown in Fig. This A ] a x (ft) 2 3 2 12 in 2[29110 2 kip>in Using the sign convention, i.e., internal tensile forces are positive and 3.5 + (c) AE 11 in 2[29110 2 kip>in ] 12 2 3 7 variation is plotted in 4–6c. the vertical displacement 12in. in 2 kip>in2] PL P(c) " 7Fig. kip BCare [+15 kip]12 ft2112 in.>ft2 [+ compressive forces negative, of 2[29110 A = 4.5 + 0.0127 Ans. !9 3 in. = [ - 9 kip]11 dA = a + Fig. 4–6 ft2112 in.>ft2 0.0127 Displacement. From Est= = +29110 2 ksi. relative to the fixed supportthe D isinside back cover, x (ft)is positive, 11 in22[2911032Ans. kip>in2] 1 Since the result the bar elongates and AE so the + displacement Fig. 4–6 2 3 2 Using the sign[+15 convention, i.e., internal tensile forces are positive and 12 in 2[29110 2 kip>in ] 4.5 !9 Since the result is positive, the bar elongates and so the displacement (c) kip]12 ft2112 in.>ft2 [+7 kip]11.5 ft2112 in.>ft2 at A is upward. PL [-9 kip]11 ft2112 in.>ft2 are negative, vertical displacement of A = forces(b) dAcompressive = a +at PCD "the 9 kip A! is Eq. upward. = ++0.0127 in. x (ft) 2 3 2Applying 3 2 4–2 between B and C, we obtain, !relative AE 11 in support 2[29110 D 2 kip>in ] 12 in22[29110 2 kip>inpoints ] to the fixed is Fig. 4–6 12 in22[2911032 kip>in2] Applying(c)Eq. 4–2 between points B and C, we obtain, SOLUTION Since the result is positive, the bar elongates and [ + 7 kip]11.5 ft2112 in.>ft2 P L BC BC [+15 kip]12ft2112 ft2112in.>ft2 in.>ft2 kip]11.5 ft2112[ +in.>ft2 [-9 kip]11 PL = + 0.0127 in. = + 0.00217 in. Ans. = [+7 the dB>C in.>ft2 PBC=LBC A is2upward. 2 7 kip]11.5 3 at ft2112 nternal Force. dDue to the application of the external loadings, =+ +d ABC E 12 2 kip>in ] Fig. 4–6 A = a 2 3 2 2 3 =in 2[29110 2 = + 0.00217 in. Ans. = 2 3 2 B>C AE 2 3 2 11 in 2[29110 2 kip>in ] 12 in 2[29110 2 kip>in ] kip>in ] different.ABCE Applying Eq. points and C, we Since the result is between positive, the barBelongates a nternal axial forces in regions AB,12BC, and CD will all be in 2[29110 2 kip>in ] 4–2 Here B moves away from C,12 since the segment elongates. at A is upward. These forces are obtained by applying the method of sections and the = + 0.0127 in. Ans. Here B moves away from C, since the segment [ + 7 kip]11.5 ft2112 in.>ft2 [-9 kip]11 ft2112 in.>ft2 PBC Lelongates. BC 6 = = between dB>C Applying Eq. 4–2 points B and C, w equation of vertical force equilibrium as shown3 in Fig. 24–6b. This + 2 Since the result is positive, the bar 2elongates ABC E 12 in22[2911032 kip>in2] 12 in 2[29110 kip>in ] and so the displacement variation is plotted in Fig. 4–6c. [+7 kip]11.5 ft2112 in.>ft2 PBCaway LBC from at A is upward. = + 0.0127 in. B moves C, since the segment e 3 Ans. HeredB>C = = ! Displacement. From the inside back cover, E = 29110 2 ksi. st Applying Eq. 4–2 between points B and C, we obtain, 4–6 ABCE 12 in22[2911032 kip>in2] Using the sign convention, internal tensilethe forces are positive Since thei.e., result is positive, bar elongates andand so the displacement Here B moves away from C, since the segment [+7vertical kip]11.5displacement ft2112 in.>ft2 of A P compressive forces are negative, BC LBC the at A is upward. = +0.00217 in. Ans. = dB>C = 2 3 2 elative to the fixed support is Applying points B and C, AD E 4–2 between 12 in 2[29110 2 kip>in ] we obtain, BCEq. ! [ + 15 kip]12 ft2112 in.>ft2 [ + 7 kip]11.5 ft2112 in.>ft2 Here B moves C,kip]11.5 since the segment elongates. PL [+7 ft2112 in.>ft2 PBCaway LBC from = dA = a + =2 +0.00217 in. Ans. = = d 2 3 2 3 2 in22[29110 2 AE 11 inB>C 2[29110 2 kip>in ] 12 in12 2 kip>in ] ABC E 2[2911032 kip>in ] + Here B moves away from C, since the segment elongates. [ - 9 kip]11 ft2112 in.>ft2 12 in22[2911032 kip>in2] = + 0.0127 in. Ans. Since the result is positive, the bar elongates and so the displacement at A is upward. Applying Eq. 4–2 between points B and C, we obtain, dB>C = PBCLBC = [ + 7 kip]11.5 ft2112 in.>ft2 = + 0.00217 in. Ans. #46 of 10 mm is attached to a rigid collar and passes through the tube. If a tensile load of 80 kN is applied to the rod, determine the displacement EXAMPLE 4.2 EXAMPLE 4.2 4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER of the end C of the rod. Take Est = 200 GPa, Eal = 70 GPa. The assembly in Fig. 4–7aofconsists of an aluminum The assembly shown in shown Fig. 4–7a consists an aluminum tube AB tube AB 4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 2 2 400area mm of 400 having a cross-sectional area steel rod having a diameter 400steel mmrod . Ahaving having a cross-sectional a diameter mmof .A EXAMPLE 10B mm 4.2 isto attached to Aa rigid passesthe through the of 10 mm of is attached a rigid collar andcollar passesand through tube. If a tube. If a tensile 80 kN is the rod, determine the displacement tensile load of 80load kN isofapplied toapplied the rod,todetermine the displacement The assembly shown4.2 in Fig. 4–7a consists 80 ofkN an aluminum tube AB EXAMPLE endrod. C ofTake the rod. Eal80=kN70 GPa. PAB ! 80 kN of the end Cthe of the E GPa, Est =Take 200EGPa, = 70 GPa. ! of st =2C200 having a cross-sectional area of 400 mm a diameter . Aalsteel rod having of 10 mm attached to a rigid and passes the tube. Iftube a AB Theisassembly shown in collar Fig. 4–7a consiststhrough of an aluminum 400 mm 400 mm 2 80 kN tensilehaving load ofa80 kN600 is mm appliedarea to the determine therod displacement cross-sectional of rod, having a diameter 400 mm . A steel A A P ! 80 kN B B (a)TaketoEast rigid of the end of the rod. Ealpasses = 200 GPa,and = 70BCGPa. of 10Cmm is attached collar through the tube.(b) If a kN determine the displacement 80 kNto the80 of 80 kN rod, 4.2 Etensile LASTIC Dload EFORMATION OF is ANapplied EMBER 80 kN 127 kN PAB ! 80 kNPAB ! 80 kN C 80M CAXIALLY LOADED 4–7 mmC of the rod. Take Est = 200Fig. of the400end Eal = 70 GPa. GPa, 4.2 127 4 127 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER A mm 600 mm 400600 mm SOLUTION (a) (a) B Force. The Internal B 127 4 80 kN 80 kNPBC ! 80 kNPBC ! 80 kN (b) (b) 80 kN PAB ! 80 kN C A 4.2 free-body diagram of the tube AB and aluminyum rod segments tüpünden müteşekkildir. 2 olan Şekilde görülen birleşim enkesiti 400 mm –7a consists of aninaluminum tube AB 4–7 80 Fig. Fig. 4–7b, shows that the rod is Fig. subjected tokN a4–7 tension of 80 kN and the 80 kN PAB ! 80 kN C ly shown Fig. 4–7a consists of anaaluminum tube AB A steel rod having a diameter 400 mm2. in 80 kN 600 mm tube is subjected to compression of 80 kN. 10mm çapındaki çelik çubuk rijit bir yakaya bağlıdır ve tüpün içinden geçmektedir. Çubuğa 2 P ! 80 kN ss-sectional area ofthrough steel If rod 400 mmthe . Atube. BC collar and passes a having a diameter (a) (b) ! attached to determine a rigid collar and passes through the tube. If açubuğun SOLUTION SOLUTION to the rod, theçekme displacement Displacement. We will first determine the displacement end C 80 kN yükü uygulandığında C ucunun of deplasmanını 600 mm of 80 kN is applied to the rod, determine the displacement Fig. 4–7 Psegments ! 80rod kN segments Est = 200Internal GPa, EalForce. = 70 GPa. BC The diagram of the tube and rod with respect tofree-body endThe B.Working in units of newtons and meters, we have(b) Internal Force. free-body diagram of the tube and (a) C of the rod. Take Estshows = 200that GPa, GPa. to a tension of 80 kN and the al =is 70 in Fig. 4–7b, theErod subjected # in Fig. 4–7b, shows that the rod is subjected to a tension of 80 kN and the tube is subjected a compression of 80 kN.32 of tube is to subjected 80 kN.m2Fig. 4–7 N]10.6 [+80110 PL to a compression SOLUTION 400 mm Çözüm: dC>B = = +0.003056 m : = 2 9 2 A Internal p10.005 m2 [200110 2 N>m ]andofrod Force. Thefirst free-body diagram the tubedisplacement segments Displacement. We AE will determine the of displacement end C of end C Displacement. We will first determine the 80 kN in Fig. 4–7b, SOLUTION shows that the rod is subjected to a tension of 80 kN and we thehave to respect end80B.Working in units ofPin newtons meters, wemeters, have with to end B.Working units ofand newtons and kN 80 kN C with respect AB ! 80 kN 80 kN Pthe 80 kNrelative to C AB ! positive sign indicates that end C moves to right tubeThe is Internal subjected to a compression of 80 kN. Force. The free-body diagram of the tube and rod segments 3 3 end B, since the bar elongates. 2[+80110 N]10.6 [+80110 N]10.6 m2to a tension of 80 kN4and the in Fig. 4–7b, shows that the rod80m2 is2kN subjected PL PL We =willoffirst determine the displacement of end m C: dDisplacement. = tube =kN. +0.003056 m: = d = = kN +0.003056 80 The displacement end B with respect to the C>B P C>B 600 mm 2 9 2 2 9 2 fixed end A is is subjected to a compression of 80 ! 80 kN BCAE AE m2 [200110 2 [200110 N>m ] 2 N>m P ! 80 kN p10.005 ] meters, we have BC with respect top10.005 end B.Working inm2 units of newtons and (b) (a) ! hesaplayınız. 80 kN 4 (b) = 0.001143 m +m0.003056 m : = 0.00420 = 4.20 mm m = 4.20 mm : 4 4 3 Displacement. We will3 first[-80110 determine the displacement of end C 2 N]10.4 m2 PL Fig. 4–7 The Fig. 4–7 that The positive sign indicates end C moves to moves the right relative torelative to positive sign indicates that end C to the right d = = 2 N]10.6 m2 [+80110 PL B with respect to end B.Working in -6 units newtons and 9 meters, 2 we have [4002mm2110 2 m2of >mm 2 N>m ] dend = bar = 2][70110 +0.003056 m: =theAE end B, since the elongates. B, since bar elongates. C>B 9 2 AE p10.005 m2 [200110 2 N>m ] The displacement of end B with respect to the fixed end A is The displacement of end B with respect to the fixed end A is ! 3 = 0.001143 m : = -0.001143 2 N]10.6 m2 [+80110m PL d = = +0.003056 m: = C>B rce. The free-body diagram ofsegments the tube and rod segments 2 dy diagram of The the tube and rod positive sign indicates that end C movesm2 the 2right relative to 3 392toN>m AE p10.005 m22[-80110 [200110 ]shortens, [-80110 N]10.4 2 N]10.4 m2 PL Here the negative sign indicates that the tube and so B PL shows that the is subjected to a elongates. tension 80 kN and the is subjected toend arod tension and the the bar dB, = of 80 =BkN dthe =right = 2 of -6 B since 22 9 2 2 9 -62][70110 2 2 moves to relative to A. AE AE [400 mm 110 2 m >mm 2 N>m ] cted compression of 80 kN. [400 mmrespect 110 2to mthe >mm ][70110 2 is N>m ] on ofto80a kN. The The displacement of end B with fixed end sign indicates thattoend moves theAright relative Sincepositive both displacements are theCright, thetodisplacement of C to = -0.001143 m = 0.001143 m : = -0.001143 m = 0.001143 m : ! end B, since the bar elongates. ent. We will first determine the displacement of end C relative to the end A is therefore 3 determine the displacement of fixed end C [-80110 2respect N]10.4to m2the fixed end A is PL to end B. Working in units of newtons and meters, we have The displacement of end B with in units of newtons and we= have = sign + 2dBmeters, Here the1Here negative that tube shortens, and so2 B the negative sign indicates that the tube 2d the -6= 0.001143 2 2 m 9shortens, dindicates : = d + + 0.003056 AE C [400 B C>B 2 m >mm ][70110 mm 110 2 N>m ] mand so B 3 moves to the right relative to A. moves to the relative to A. 2 N]10.6 m2rightPL [ + 80110 3 [-801103m 2 N]10.4 m2 2 N]10.6 m2 both 0PL = -0.001143 ==mto 0.001143 + 0.003056 :the = Since displacements to them right, themm displacement of C ==are 0.00420 m 4.20 :: Ans. Since both displacements are right, the displacement of C d = = 2 9 2 B AE = +0.003056 m : ! 2 -6 2 2 9 2 p10.005 m2 [200110 2 N>m ] AE 2 9 2torelative [400 mm 110 2 m >mm ][70110 2 N>m ] relative the fixed end A is therefore to the fixed end A is therefore 2 [200110 2 N>m ] Here the negative sign indicates that the tube shortens, and so B ==right -0.001143 m+to=0.003056 0.001143mm : + 2moves +to e sign indicates that end relative dCCright 1: =moves dB relative +dCto dC>B 0.001143 A. 1: 2the =the dto dC>B =m 0.001143 m + 0.003056 m B + atthe end C moves to the right relative to bar elongates. Since both displacements are to the right, the displacement of C Here the 0.00420 negative sign indicates that the so B m 4.20 mm Ans. andAns. acement of end B with to the fixed end A is= : m 4.20 mm : tube shortens, relative torespect the=fixed end= A0.00420 is= therefore moves with respect to the fixed endtoAthe is right relative to A. 3 + 2 [ -Since are to the the displacement of C 80110both m2dC>B = 0.001143 d2CN]10.4 1: = displacements dB + m +right, 0.003056 m PL = [-80110 = 32 N]10.42m2 relative to the fixed end A is therefore -6 2 2 9 2 AE [400 mm 110 2 m >mm ][70110 2 N>m ] Ans. 2 -6 2 2 2 = 0.00420 m = 4.20 mm : + 9= m 110 2=m ->mm ][70110 ] dm : 0.001143 0.001143 1m : 22 N>m C = dB + dC>B = 0.001143 m + 0.003056 m 43 m = 0.001143 m : egative sign indicates that the tube shortens, B mm : = 0.00420and m =so4.20 Ans. e right relative to A.shortens, and so B ates that the tube h displacements are to the right, the displacement of C A. he fixed end A is therefore re to the right, the displacement of C herefore dC = dB + dC>B = 0.001143 m + 0.003056 m B 4 80 kN Ans. Ans. #47 128 128 128 CCHHAAPPTTEERR 44 AAXXI AI AL L LLOOAADD AXIAL LOAD EXAMPLE 4.3 EXAMPLE EXAMPLE 4.3 4.3 ! CHAPTER 4 90 kN Rigid beam AB rests on the two short posts shown 200Rigid mm beam AB rests on the two short posts shown in Fig. 4–8a. AC is 90 made ofposts steelshown and has a diameter 400 mm 90kN kN Rigid beam AB rests on the two short in Fig. 4–8a. ACof is 20 mm, an 200 mm AB rijit iki kısa direğe oturmaktadır. AChas çelikten yapılmış olup çapı 20mm dir. 200kirişi mm şekildeki made of steel and a diameter of 20 mm, and BD is made of mm aluminum and has a diameter of 40 mm. 1 2 8 C H A P T E R400 4 A X I A L L O A D A made steel andAC hasBisa diameter of 20 mm, and BD is made of Determin 400 short mm AB rests on the two posts shown inofFig. 4–8a. aluminum and has a diameter mm. displacement point FDetermine on AB if a the vertical load of 90 kN is appl A aluminum a diameterofof 40 mm. Determine thealtında, displacement F mm and BDAaluminyumdan yapılmış Bolup çapı 40 dir. has F noktasında 9040 kN düşey yükleme B Rigid beam XIAL LOAD made of steel and has a diameter of 20 mm,ofofand BDonisAB madevertical of load point ofofE 90 isisapplied this point. Take =kN200 GPa, Eover = 70 GPa. pointFFon ABififaavertical load 90 applied this point. st kN alover 128 C H AFPFT E R 4 A X I A L L O A D aluminum andEXAMPLE has a diameter of 40 mm. Determine the displacement Take E = 200 GPa, E = 70 GPa. 300 mm 4.3 Take# Estst = 200 GPa, Ealal = 70 GPa. bu noktanın deplasmanını hesaplayınız. 300 of point F on AB if a vertical load of 90 kN is 300mm mmapplied over this point. 90 kN C Rigid beam AB rests on Dthe twoSOLUTION short posts shown in Fig. 4–8a. AC is Take Est = 200 GPa, E 4.3 = 70 GPa. EXAMPLE C 200 mm D posts shown in Fig. 4–8a. AC is ABalrests on the two short AAXXI AI ALL LLOOAADD Rigid beam C D made of steel and has a diameter of 20Force. mm, and BD is made offorces acting SOLUTION 400 mm SOLUTION Internal The compressive (a)BD is made of madeA of steel and has a diameter of 20 mm, and aluminum and has a diameter of 40 mm. Determine the displacement B 90 kN Internal compressive forces acting the top of post posts are determined the equilibrium of mem Internal Force. The compressive forces acting atfrom the topAC ofeach Rigid beam Force. AB rests The on the two short shown inatFig. 4–8a. iseach (a) (a) aluminum and has diameter of 40 mm. Determine the displacement 200 mm ofpost point Fdetermined on AB ifhas a vertical load of kN is applied over this point. B F a400 from the equilibrium ofof member AB, Fig. 4–8b. These forces are equal to the internal forces in each post are determined the equilibrium member Fig. 4–8b. made ofare steel and afrom diameter of 90 20 mm, and BD isAB, made of mm 90 SOLUTIONof point F on AB if a vertical load is applied over this point. Take EkNforces = 200 GPa, Ealto =the 70 GPa. These are Fig. 4 B of 90 kN These are equal to the internal forcesinineach each post, Fig.4–8c. 4–8c. aluminum and has aequal diameter ofinternal 40 mm. forces Determine thepost, displacement 200 mm stforces A 90 90kN kN Eal = 70 GPa. 400 mm Take E = 200 GPa, 300 mm st Displacement. The displacement of the top of ea 44 of point F on AB if a vertical load of 90 kN is applied over this point. Internal Force. The compressive forces acting at the top of each 200 mm 200AB Fmmrests Rigidbeam beam restson on the twoshort shortposts postsshown shownin inFig. Fig.4–8a. 4–8a.AC ACisis AB the two 400 mm 300 mm Rigid 400 mm Displacement. The displacement of the top of each post is Displacement. The displacement of the top of each post is Take E = 200 GPa, E = 70 GPa. st and al from the equilibrium of20 member 4–8b. C of D of made of steel steel and has diameter of 20 mm, BDFig. made of mpost are determined made and has aa diameter mm, andAB, BD isis made of SOLUTION Post AC: 300 mm ! These forces are equal to the internal forces in each post, Fig. 4–8c. DBB aluminumand andhas hasaadiameter diameterof of40 40mm. mm. Determine the displacement aluminum Determine the displacement SOLUTION Post AC: Post AC: 60 kN kN Internal Force. The30 compressive forces acting at the top of3 each (a) ofCpoint point FkN onAB ABThe ifaavertical vertical load of 90kN kN applied this point. (b) over of FkN on if load of 90 isisapplied point. [-60110 2 N]10.300 m2 PAC AC 60Force. kN post are determined the equilibrium ofLmember Internal compressive forces acting at theover topthis offrom each 60 30 D30kN Çözüm: SOLUTION = AB, Fig. 4–8b. = 3 32dN]10.300 A = Take E = 200 GPa, E = 70 GPa. (b) Take E = 200 GPa, E = 70 GPa. (b) [ 60110 m2 2 9 2 [-60110 2 N]10.300 m2 st al P L st al P L AC AC Displacement. displacement the topInternal of post AAC These forces areis equal to the internal forces inEeach post, Fig. m2 4–8c. AC AC post are The determined from theof equilibrium of each member AB, Fig.= compressive 4–8b. p10.010 -6[200110 2 N>m ] -6 st at the = = 286110 2 m d Force. The forces acting top of each = = = -286110 2 m d 90 kN A 300mm mm A 300 2 9 (a) 60 kNin each 30 kN Estst 4–8c. 4 These forces are equal to the internal forces post, Fig. p10.010 m22[200110 [2001109of N>m2]2] AB, Fig. 4–8b. AAAC E AC p10.010 m2 22N>m 200 mm post are determined from the equilibrium member 400 mm 60 kN 30 kN 60 kN 30 kN mm T post is Displacement. The displacement of =the0.286 top of each Post DAC: These forces are equal to the internal forces in each post, Fig. 4–8c. D SOLUTION90 kN SOLUTION = 0.286 0.286 mm T Displacement. The displacement of the top of each post is = mm T 4 200Force. mm Post AC: at Internal Force. The The acting at the the top top of of each each 3compressive forces Internal compressive acting Post BD:top of each post is 400 mm [-60110 2 N]10.300 forces m2 PAC LAC Displacement. The displacement of the -6 Post AC: post are determined from the equilibrium of member AB, Fig. 4–8b. 60 kN 30 kN post are=determined from the equilibriumPost of member AB, Fig. 4–8b. Post BD: = 286110 2 m dA = BD: 9 2 in each (b) to [ - 6011032 N]10.300 m2 3 PAC LACFig. AAC Est forces These forces areequal equal theinternal internal forces post, Fig.4–8c. 4–8c. p10.010 m2to2[200110 2 N>m 30 kN These are the forces each post, -6 2 N]10.300 m2 [-30110 PBDLBD = - 286110 Post]din AC: = = 2m 3 A 23 3 dB = 9 2 [-60110 2 N]10.300 m2kN = = PACLAC APAC E PAC ! 60 P ! 30 kN N]10.300 m2 ] [ - 30110 m2 [200110 2AN>m PBD BD -6p10.010 22N]10.300 m2 BD BD LstLBD -6 2[7011092 N>m2] 30 kN m2 = == -286110 2 m[-30110 dA! 60=kN (c) BD Eal = p10.020 -6 mm d 102110 = 2 m ! 2 9 2 3 60 kN 30 kN B P ! 60 kN P ! 30 kN Displacement. The displacement of the top of each post is d = -102110 = = 2 m AC BD![200110 Displacement. The displacement of the top of each post is [-60110 A B] P PAC ! 60 E kN PBD 30 kN p10.010 m2 2 N>m N]10.300 L st (b) 2 2[70110 9 9m2 = 0.286 mm TAC AAC p10.0202m2 N>m2]2] ACA (c) BD EE 22N>m (c) BD =alalT p10.020 2m2 [70110 = - 286110-62 m dA = = 0.286 30 kN mm 9 2 mm T = 0.102 A E AC direği: p10.010 m2 [200110 2 N>m ] AC st PostAC: AC: Post = kN 0.286 mm T 0.102mm mmTT 60 30 kN == 0.102 Post BD: 30kN kN 30 Post BD: 3 = 0.286 mm T [ - 60110322N]10.300 N]10.300m2 m2 A diagram showing the centerline displacements LAC [-60110 ACL AC PPAC -6 -6 PostddABD: = = -286110 - 286110 2m m A == = = 2 A diagram showing the centerline displacements A,B, B, andFBy Fon on 2 9 2 the beam ism2shown in Fig. 4–8d. proportion o 3 3 9 2 N>m 2 ] AAC Estst A diagram centerline displacements atatA, and p10.010 m22[200110 [200110 ACE 2 N]10.300 [the - 30110 A p10.010 m2 2 N>m ] PBDisLshowing 2 N]10.300 m2 [-30110 P L BD -6 BD BD the beam shown in Fig. 4–8d. By proportion of the blue shaded Post BD: -6 triangle, the displacement of point F is therefore d= =of-the 102110 =- 102110 =2 m m kN the is shown in Fig. 4–8d. blue 2shaded Bbeam PAC !=60 kN PBD !330 kN 30 d30BkN = 2 By proportion 9 2 2 N]10.300 m2 PBDLBD (c) [-30110 A E p10.020 m2 [70110 2 N>m ] 2 9 2 BD al triangle, the displacement of point F is therefore -6 ABDdE p10.020 2 N>m triangle, ] 0.286 mm displacement F is therefore = the -102110 =TT m2 [70110 2 m of point mm ! B al=== 0.286 D ! 30 kN ABDEal [-3011032 N]10.300 m2 p10.020 m22[7011092 N>m2] PBDLBD 400 mm dB = = 0.102 mm = - 102110-62 m = T PAC ! 60 kN PBD ! 30 kN 2 9 = 0.102 2 mm + 10.184 mm2a d b = 0 A E F p10.020 m2 [70110 2 N>m ] Post BD: BD direği: (c) 400 mm BD al = 0.102 T Postmm BD: 600 mm = 0.102 mm T d = 0.102 mm + 10.184 mm2 a400 mm b = 0.225 mm T Ans. dFF = 0.102 mm + 10.184 mm2a 600 mmb = 0.225 mm T Ans. A showing at A, B, and F on 600 mm = 0.102 mm T -6 the centerline displacements N]10.300m2 m2diagram [ - 301103322N]10.300 LBD [-30110 BDL PPBD BD = -6 the beam is shown in Fig. 4–8d. By proportion of the blue shaded d = 102110 = 2 m PBD!! kN Ashowing centerline displacements and2F Fmon on A3030diagram the displacements at at A,A,B,B,and ddiagram = the 600 mm BB = A showing kN BD Ealcenterline p10.020m2 m222[70110 [70110992triangle, 2N>m N>m22]] =the-102110 0.102 mm BDE A p10.020 displacement of point F is therefore BD al 400 mm B A F theshown beam isinshown in Fig. By 4–8d.proportion By proportion of the blue shaded 600 mm the beam is Fig. 4–8d. of the blue shaded A diagram showing displacements atBA, B, and F on 0.102the mm centerline600 mm 400 mm A F 0.102 mm displacement of F point F is therefore 400 mm A 4–8d. FBy proportion the beam is shown in Fig. of Bthe blue shaded =the0.102 mm triangle, thetriangle, displacement of Tpoint is therefore 0.102 dF ! = 0.102 mm T mm triangle, the displacement of point F isd 400 therefore 0.102 mm 0.184 mm dF = 0.102 mm 0.184 Ans. b 0.102 = 0.225 + 10.184 mm2daF mm mm T (d) mm F 600 mm 400 mm 0.286 mm A diagram showing the centerline displacements at A, B, and F on 0.184 mm (d) AdFdiagram the centerline displacements at A, B,TandAns. F on b = 0.225 mm = 0.102showing mm + 10.184 mm2a 400 By mmproportion 0.286 mm (d) mm the beam beam isis shown Fig. 4–8d. 4–8d. of the the blue blue shaded 400 the Fig. By600 proportion of shaded mm dF = 0.102 Ans. bmm=d 0.225 mm T 0.286 mm + shown 10.184ininmm2a ! Ans.4–8 b = 0.225 mm T Fig. = 0.102 mm + 10.184 mm2a F triangle,the thedisplacement displacementof ofpoint point Fisistherefore therefore 600 Fmm triangle, Fig. 4–8mm 600 600 mm 400 mm F Fig. 4–8 0.102 mm AB çubuğu rijit olduğundan AB boyunda eğilme yoktur.A 600 mm 0.102 mm 400 400mm mm B A F dF == 0.102 Ans. 0.225 mm 0.102mm mm ++ 10.184 10.184 mm2 a400 mm bb == 0.225 0.102 mm 600 mm d Ans. mm TT Amm mm2a F 600mm mm 0.184 0.102 mm F 600 400 mm 0.102 mm B A F dF 0.184 mm ! 0.286 mm d 0.102 mm 0.102 mm A F 0.184 mm A 0.286 mm 0.286 mm (d) 600 mm 0.102 mm 600 mm 400 mm B F 400 mm B F (d)Fig. 4–8 0.184 mm 0.184 mm 0.286Fig. mm 4–8 0.286 mm d dFF (d) (d) 0.102 mm 0.102 mm 0.184 mm 0.286 mm 600dFmm 400 mm (d) dF B 0.102 mm B 0.102 mm Fig. 4–8 (d) Fig. 4–8 Fig. 4–8 Fig. 4–8 #48 EFORMATION OF AN 4.2 EXAMPLE 4.4 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 4.2 EM LASTIC OF AN AXIALLY LOADED MEMBER ! XIALLY LOADED A EMBERDEFORMATION 129 129 129 y member is made from ağırlığına a material ve that a specific weight gsahip and bir malzemeden BirAkonik eleman ɣ özgül E has elastisite modülüne modulus of elasticity E. If it is in the form of a cone having the dimensionsBu shown in Fig. 4–9a,ağırlığında determine asılı how far end is displaced hesaplayınız. üretilmiştir. eleman kendi ikenitsuç deplasmanını r0 duehas toy agravity when it is gsuspended in the vertical position. y eeight fromga and material that specific weight and ity E. If the it is in the form of a cone having the having r0 0 in 4–9a, determine how farrits end is displaced is Fig. displaced SOLUTION n it is suspended in the vertical position. n. Internal Force. The internal axial force varies along the member since it is dependent on the weight W(y) of a segment of the member 4.2Hence, ELASTIC EFORMATIONthe OF AN AXIALLY LOADED MEMBER 129 below any section, Fig. 4–9b. toDcalculate displacement, 4.2 E LASTIC D EFORMATION OF AN AXIALLY LOADED Mits EMBER 1 2 9 we must use Eq. 4–1. At the section located a distance y from free L The internal axial force varies along the member he member end, the radius x of the cone as a function of y is determined by nt on the weight W(y) of a segment of the member the member 4XIALLY 4.2 ELASTIC DEFORMATION OF AN A EXAMPLE proportion; 4.2 ELASTIC DEFORMATION OF AN AXIALLY LOADED MEMBER 129 Fig. 4–9b. Hence,4.4 to calculate i.e., the displacement, splacement, PLE 4.4 4.2 r ELASTIC DLEFORMATION OF AN AXIALLY LOADED MEMBER 129 1. Atits the section located a distanceL y from its xfree r0 rom free 0 y A member is made from a material that has a specific weight g and = ; x = y of the cone as a function of y is determined by ermined by EXAMPLE 4.4y y weight L L ber modulus is made from a material a specific and having 4 of EXAMPLE elasticity E. that If 4.4 ithas is in the form of 4a g cone the r0 s ofdimensions elasticity E. If it in is Fig. in the form of a cone having theis displaced shown 4–9a, determine how far its end EXAMPLE 4.4 The volume of a cone having a base ofAradius x and heightfrom y isr0 a material that has a specific weight xg and member is made r0 Fig.A rit0 is suspended x to gravity ons due shown in 4–9a, determine how far its end displaced when in Ethe vertical position. y member is made from a material that has specific and g EMBER 4.2 LASTIC Dis EFORMATION OF aAN AXIALLY weight LOADED M 129 = it; is suspended x = y in the vertical position. modulus of elasticity E. If it is in the form of a cone having the 2 ravity when pr y L A modulus L 1 of elasticity E. If it is in the form of a cone having the y (a) member is made from a material that2 has a specific weight g and 0 3 in Fig. 4–9a, determine how far pyx dimensions = how2 far y itsshown V =determine r0 its end is displaced dimensions shown in Fig. 4–9a, end is displaced modulus of elasticity E. If it is in the form of a cone having the 3 3L SOLUTION having a base of radius x and height y is due to gravity when it isx suspended in the vertical tne y is r0 position. x to gravity when it is4–9a, suspended in the vertical PLE 4.4 dimensions shown in Fig. determine how far itsposition. end is displaced ONInternal ! dueSince y Force. The internal axial force varies along the member the force at the section becomes gV, it 2 W = topr gravity when is internal suspended in the vertical 1 due (a)position. 0axial (a) 2 3 the l Force. The internal force varies along the member since it is dependent on weight W(y) of a segment of the member y that has a specific weight g SOLUTION = pyx 2 y mber isVmade a= material and 2 Çözüm: gpr 3 from 0 3 3L SOLUTION is dependent onsection, the weight ofHence, a segment ofP1y2 the member below any Fig.yW(y) 4–9b. to calculate the displacement, c P(y) + ©F = 0; = y us of elasticity E. If it is in the form of a cone havingInternal the Force. The internal axial force varies along the member x 3Ly2from any section, Fig. 4–9b. Hence, tosection calculate the displacement, SOLUTION we must use Eq. 4–1. AtForce. the located a axial distance its free the yalong L r y Internal The internal force varies member internal force at the section becomes 0 ions shown in Fig. 4–9a, determine how far its end is displaced since it is dependent on the weight W(y) of a segment of the member t useend, Eq. the 4–1.radius At the located yaxial from L xsection of isthe coneThe asa distance a function of W(y) y its is free determined by since dependent on the weight of a segment of member Internal internal force varies along thethe member 2 it Force. gravity when it is suspended in the vertical position. 4 Displacement. The area of the cross section is also a function of below any section, Fig. 4–9b. Hence, to calculate the displacement, gpr e radius x of the cone a section, function of is determined proportion; i.e., below any Fig. Hence, displacement, it0 yis3as dependent on they4–9b. weight W(y) to of calculate abysegment of the member P(y) the P1y2 =since P(y) 4 position y, Fig. 4–9b. We have 2 we must use Eq. 4–1. At the section located a distance y from its free W(y) ion; i.e., x Fig.At 3L must we the Hence, section located a distance y from its free below any use section, to calculate the xdisplacement, L r0 4–9b. r0 xEq. 4–1. y end, the radius x of the cone as a function of y is determined by 2of y isy determined ION ;Atthe x cone = yas located the xr0of a function weend, Eq. = 4–1. the section a pr distance from its freeby L r0useradius xmust 0 2 2 proportion; y L L he area ofofthe cross section is also a function of 4 i.e., function A1y2 px = of2 yy is determined by = axial ; x y cone end, x= ofLvaries the as the a=function proportion; i.e., al We Force. force along member y theLradius L 4 b. have The internal W(y)y is r r0 i.e., The volume of cone having of radius and height x is dependent onproportion; theaW(y) weight W(y) aofbase a segment ofr xthe member x r0= 0 and y =y L yields = 0 ; xheight x = yx 0 limits y between 2 Applying Eq. 4–1 the of y ume of a cone having a base of radius x and y is = ; x = y pr any section, Fig. 4–9b. Hence, to calculate the displacement, y L L x 2 0 2 r r 2 x 0L pr0 3 1 (a) A1y2 = px = 2 y (b) 2 y = 0y;L = Lfree y 2 t use Eq. 4–1. At the section located a distance yfromx its V =pr20pyx L L= L1 (a) L2 dy LC 1gpr >3L2of 2 ya3 D cone dy having The volume a base of radius x and height y is 2 3 y P1y2 0 3 3L pyx = a dcone y having Vcone = volume of a base= of radius e radius x of theThe as a function by x and height y isx 2= of x y is determined Fig. 4–9 x 3 3L 4 etween the limits of and yields y = 0 y = L ! yields 2 2 y is 2 height 2 y of a force cone having a base of radius x and tion; i.e., W = The 0 A1y2E L0 Since the internal atLthe section becomes gV, volume pr C 1 21pr0>L 2 y D E 0 x (b) pr0 1 y3 V = pyx2 = (a) y force2at(b) the becomes W =L gV, the internal 2 2 section 3 L C 1gpr V2L=1 pyx2 =pr2 2 y3 3 3L r0 0>3L 2 y D rdy P1y2 dy x ggpr (a) 0 0 0 33 2 2= =y P(y) y3 Fig. 4–9 = ; Fig. x4–9 + c ©Fy == 0; P1y2 = V = y pyx dy = 3L gpr 2 2 20 3L 3E3L2 3 2 y P1y2 L 3L Since internal force at xthe sectiony becomes = gV, the A1y2E P(y) 0 0; L= L!0Since 0 L = y 1pr >L 2 y E C W0 = gV,2 the D internal ! ! Wbecomes force at the section x 3L L a cone having Since W internal atythe section becomesof = area gV, 2 force gpr20 3 y ume of a base of the radius xcross and height is also 2 a function The of the section is g Displacement. gL x gpr+0 c3©Fy = 0; P1y2 = y y dy = is also Ans. cof©F ement. The area the a function +4–9b. 0; section P1y2 =gpr2 2 yof position y, Fig. We have y =cross 3L2 P(y) 2 W(y) 3E L 6E 0 3 0 x pr 1 3L (a) 0 3 P(y) 0; 2 = P1y2 = y ! cV©F y, Fig. 4–9b. We+have W(y) y 2 y = y =pyx 2 x 3L 2 pr0 2 3 Displacement. The area yof the cross section is also a function of 3L2 2 gL2 Displacement. The the is also athe function =0 pxarea = of y cross section NOTE: 2A1y2 As aprpartial check notice how4–9b. units ofofthe 2 of this result, Ans. Ans. Displacement. 2 position y, Fig. We have Lthe cross section W(y y A1y2 = px = y The area of is also a function of position y, Fig. 4–9b. We have force at the section becomes W6E= gV, the internal terms, when canceled, W(y) L2We havegive the displacement in units of length as ! x 2 position y, Fig. 4–9b. y W(y) Applying Eq. 4–1 betweengpr the2 limits of y = 0 and y = L 2yields expected. xpx2 = pr0 y2 pr0 2 A1y2 =(b) 3 0 and y ng 4–1the the limits y 0 =ythe y of = the L=yields 2 ial=Eq. check ofbetween this result, notice how unitsA1y2 P(y) 2 2 0;of P1y2 = of px = y yunits L pr 2 2 3 0 22 L L C 1gpr2>3L (b) x Ly P1y2 dy 2of A1y2 led, give as the displacement in3L units length as 2 2 y D 2dy of length 2 3 = 0px = L L 1gpr >3L 2 y dy C D d =dy =0 L Fig. 4–9 P1y2 Eq. 4–1 between the limits of y = 0 and y = L yields x 2 of2 yApplying A1y2E Applying between the 0= Eq. L0 is also 2 y D E=of0 and y = L yields d = The area C 1prlimits Fig. 4–9 cement. of L the cross4–1 section a20>L function x 2 2 2 A1y2E Eq. Applying of y = 0 and y = L yields L L0We L0L4–1 Cbetween (b) 2 3 1pr0>L 2the y D limits E L C 1gpr2>3L n y, Fig. 4–9b. have 2 y D dy P1y2 dy 3 0 L L C 1gpr2>3L22 yW(y) g (b) dy D P1y2 dy 0 L = d = = 2 3 2 y dy y L L g d pr = 2 P1y2 dy = C 1gpr0>3L 2 y D dy L0 A1y2E L0 >L24–9 2 y2 D E C 1pr20Fig. 2 y dy 3E L0 2 = 0 02 A1y2E L0 >L22 y2 D E = C 1pr Fig. 4–9 y 0 3E L0 A1y2 = px d == 2 L L L0 A1y2E L0 L C 1pr20>L22 y2 D E g L gL2 g 2 y dy x = = Ans. L y dy gL 3E L0 ng Eq. 4–1 the 6E limits of y ==g 0 and y = L yields Ans. = between = 3E L0y dy 6E ! (b) 3E 0 2 3 L L C 1gpr2L gL2 >3L 2 y dy 2 D P1y2 dy 0 NOTE: As a partial check of this result, notice how the units of the = gL Ans. 2 = = result,=gL Fig. 4–9 Ans. Asterms, a dpartial check of this notice of theof length as 6E 2how 2 the units 2 when canceled, give the displacement in units A1y2E 6E L0 L0 = C 1pr0>L 2 y D E Ans. whenexpected. canceled, give the displacement 6E in units of length as L d. NOTE: As a partial check of this result, notice how the units of the g #49 NOTE: As a partial check of this result, notice how the units of the y dy = whenthe canceled, NOTE: As a partial check of this result, terms, notice how units of give the the displacement in units of length as 3E L 0 terms, when canceled, give the displacement in units of length as expected. terms, when canceled, give the displacement in units of length as expected. gL2expected. nate Fig. 4–10a, which is subjected to the load P. In Fig. 4–10b, P is replaced 4.3its components, Principle of Superposition 4.3 of by two ofPrinciple P =Superposition P + P . If P causes the rod to deflect 1 2 a large amount, as shown, the moment of the load about its support, The principle of superposition is often to determinestress the stress The of superposition is often usedofused to Pd, principle will not equal the sum of the moments itsdetermine componentthe loads, or displacement at a point in a member when the member is subjected orPd displacement in ad1member the member is subjected Z P1d1 + Pat because Z d2 Z when d. 2d2a, point to a complicated loading. By subdividing the loading into components, to a complicated loading. By subdividing the loading into components, principle of superposition the resultant the the principle of superposition statesstates that that the resultant stressstress or or 4.3 Süperpozisyon Thisprensibi principle will be used throughout this text whenever we assumesumming displacement at the point can be determined by algebraically displacement at the point can be determined by algebraically summing Hooke’sthe lawstress appliesorand also, the bodies that by are considered will be such applied displacement caused load component thebir orEMBER displacement caused by altında each each load component applied 4.4 STATICALLY I NDETERMINATE A XIALLY Lstress OADED M 137 Bu prensip genellikle eleman karmaşık bir yükleme iken bir noktadaki gerilme veya thatseparately theseparately loading will produce deformations that are so small that the to the member. to the member. deplasmanı hesaplamak için kullanılır. Bu prensibe göre bir noktadaki bileşke gerilme change in The position and of the loading will be insignificant and of veya following two conditions must be satisfied the principle of The following twodirection conditions must be satisfied if theif principle deplasman, elemana yük bileşeninin ayrı ayrı yüklenmesiyle oluşan gerilme ve cansuperposition beherbir neglected. superposition is to be applied. is to be applied. Axially deplasmanın cebrik toplamı ile belirlenebilir. Süperpozisyon prensibinin uygulanabilmesi için; 1. loading The loading be linearly related the stress or displacement 1. The mustmust be linearly related thetostress or displacement 1. Yüklemenin hesaplanacak gerilme veya deplasman iletodoğrusal ilişkili olması gereklidir. that is to be determined. For example, the equations s = P>A and that is to be ch is fixed supported at both A determined. For example, the equations s = P>A and 2. Yükleme elemanındgeometrisini önemli derecede değiştirmemelidir. 4.4 S TATICALLY I NDETERMINATE A XIALLY L OADED M EMBER 137 d = PL>AE involve a linear relationship between P and s or d. 4.4 S TATICALLY I NDETERMINATE A XIALLY L OADED M EMBER 137 = PL>AE involve a linear relationship between P and s or d. g. 4–11b, equilibrium requires 4.4 STATICALLY INDETERMINATE AXIALLY LOADED MM EMBER 137 4.4loading STATICALLY INDETERMINATE AXIALLY LOADED EMBER 1 3 7 or or 2. must not significantly change the original geometry 2. The must not significantly change the original geometry LACThe loading configuration of the member. If significant changes do configuration of the member.P1If significant changes do occur,occur, the the Statically Indeterminate Axially = 0 Statically Indeterminate Axially P direction and location of the applied forces and their P andC location of the applied forces and their moment 2moment Loaded Member Ldirection Loaded Member ! " arms will change. For example, consider the slender rod shown Statically Indeterminate Axially Statically Indeterminate Axially arms will change. For example, consider the slender rod shown in in d1 is subjected P which Fig. 4–10a, the P. load P. In4–10b, Fig. 4–10b, P is replaced Fig. 4–10a, which is subjected to thetoload In Fig. P is replaced lly indeterminate, since the Loaded Member Loaded Member Consider shown in Fig. which is fixed supported at both d2 A the rod to deflect onsider thethe barbar shown in Fig. 4–11a which is fixed supported at both LCB d 4–11a A by two of its components, P = P + P . If P causes by two of itsMcomponents, P L= 12 . If P2 causes the o determine the two reactions 4.4 STATICALLY INDETERMINATE AXIALLY L4–11b, OADED EMBER 1OADED 3 71 +MPEMBER 4.4 Fig. SFig. TATICALLY INDETERMINATE AXIALLY 1 3rod 7 to deflect ends. From the free-body diagram, equilibrium requires f of its its ends. From the free-body diagram, 4–11b, equilibrium requires (a) a supported large amount, as shown, the(b) moment ofload the about load about its support, a large amount, as shown, the moment of the its support, der the bar shown in Fig. 4–11a which is fixed at both # er the bar shown in Fig. 4–11a which is fixed supported at both AA 4 Pd, will not equal the sum of the moments of its component Pd,equilibrium will not equal the sum of the moments ofLits component loads,loads, tionFrom needed for solution, it is ends. Fromthe the free-body diagram, Fig. 4–11b, equilibrium requires nds. free-body diagram, Fig. 4–11b, requires Fig. 4–10 AC L NDETERMINATE A XIALLY L OADED M EMBER 1 3 7 Pd Z P d + Pd Z P d + P d d1 Z# dd12 ZZ dd.2 ZACd. 2d2 , because 1 1B 1 1 2 2 ,Pbecause displace. Specifically, an Eğer yük fazla olur çünkü c ©F minate Axially +©F = 0; +Aolursa, F-A -= P# 0= 0 cbar atically Indeterminate Axially = 0; FBF +B F P r displacement is referred to LAC (a) LAC C C aded Member 4.4 Statikçe belirsiz ve eksenel yüklü eleman L ion. In this case, a suitable L F = 0; F + F P = 0 = 0; FB B+ FA A- P = 0 This This principle willused be used throughout this whenever text whenever we assume principle will be throughout this text we assume P e which displacement of one endisatofcalled C P will C This type of problem statically indeterminate, since the Hooke’s law applies and also, the bodies that are considered be such Hooke’s law applies and also, the bodies that are considered will be such is fixed supported both his type of problem is called statically indeterminate, since the A FA Fat A both shown in Fig. 4–11a fixed supported L L A LCB LCB ebar equal to zero, since thewhich endnotissufficient equilibrium equation(s) are to determine the two reactions that the loading will produce deformations that are so small that the that the loading will produce deformations that are so small that the m, Fig. 4–11b, equilibrium requires quilibrium equation(s) are not sufficient to determine the two reactions om theoffree-body diagram, Fig.statically 4–11b, equilibrium requiressince the P Pinsignificant and yon condition becomes problem called indeterminate, bar. ype ofbar. problem isiscalled statically change indeterminate, since the in position and direction of theofloading will be change in position and direction the loading will be insignificant and ntype thethe L LCBCB 4 brium equation(s) arenot notsufficient sufficient todetermine determine two reactions 4.4 4 STATICALLY INDETERMINATE AX Lthe In order establish an additional equation needed for solution, rium equation(s) are the can be ACtwo canneglected. be neglected. at In both order to to establish an additional needed forreactions solution, it isit isLAC A toequation ebar. bar. P = 0 to to necessary displace. Specifically, requires ecessary consider thethe barbar displace. Specifically, an an B B FBconsider + FAhow -how Ppoints =points 0 on on 44 order establish additional equationneeded needed forsolution, solution, itis is C itreferred equation that specifies conditions displacement is to (a) der totoestablish ananadditional equation for quation that specifies thethe conditions forfor displacement is referred to (a) C FA L sary considerhow howpoints points thebar bardisplace. displace. Specifically, B L as atoconsider compatibility condition. case, a an suitable L ononthe Specifically, an B sry a tocompatibility or or kinematic condition. In In this case, a suitable FBthis ACkinematic Statically Indeterminate Axially P ion that specifies the conditions for displacement is referred to (a) of the applied loads by using tically indeterminate, since the P condition would require displacement of one ncompatibility that specifies the conditions for displacement to (a) ompatibility would require thethe displacement of one endend of of problem iscondition called statically indeterminate, since the LCBis referred F F P L FA FAA A Loaded Member compatibility or kinematic condition. In this case, a suitable depends onor the material ent tobar determine the with respect to the toIn be equal to zero, since Cend ompatibility kinematic condition. this case, a since suitable he bar with respect totwo thereactions other end to be equal to zero, thethe endend CB quation(s) are notthe sufficient toother determine the two reactions atibility condition would require the displacement of one end of L avior occurs, d = PL>AE can supports fixed. Hence, compatibility condition becomes ibility condition would require the displacement ofbecomes one end of upports areare fixed. Hence, thethe compatibility condition F FA ar with respect theAsolution, end toP be equal to zero, since Consider the end the bar4shownFAinAFAFig. 4 4–11a which is fixed supported at both nwith segment ACtoistothe +F ,other andend in equation needed for ittois nce the respect other be equal to zero, since establish an additional equation needed for solution, it isthe end LCB rts fixed. Hence, the compatibility condition becomesB of its ends. FromPthe free-body diagram, Fig. 4–11b, equilibrium requires theare bar displace. Specifically, an 4–11c, the above equation can eactions consider how points the barddisplace. Specifically, ts are fixed. Hence, theon compatibility condition becomesan 1 PB1 P = 0= 0 s for displacement is referreddA>B toA>B (a) P P2 P2 specifies the conditions for displacement is referred to (a) ! " ndition. In this case, a suitable 4FB! FA FA " F B ibility dA>B = In 0 this case, a suitable ion, it isor kinematic condition. FB e the displacement of one end of dexpressed 0 in terms of dby B + cby ©F FBF + FF 1 =using A>B (b)one d0; A - P = 0 This equation can be of the applied condition would the = displacement end of ! loads 1 F(c) cally, an This equation can besince expressed of! the applied loads using B in terms A FA A ! require d2 d to be equal to zero, the end F F d2 FAFP P dzero, since a= relationship, depends on material A A espect to the other end to be equal towhich theonend erred to load–displacement relationship, depends thethe material (a) which B 0load–displacement condition becomes (b) sbility equation can be expressed in terms of the applied loads by using Fig. 4–11 (a) (a) F (b) B ixed. Hence, the compatibility condition becomes behavior. For if için linear-elastic behavior occurs, d PL>AE =using PL>AE Çözüm düşey denge denklemi yeterli değil. İlave denklemPgerekiyor. Süreklilik (uygunluk) suitable ehavior. For example, if linear-elastic behavior occurs, d = cancan equation can beexample, expressed in terms applied d–displacement relationship, whichof the depends onloads the bymaterial type of be used. Realizing that the internal force in segment AC isThis + F , and in problem Pis called statically indeterminate, since the end ofRealizing e–displacement used. that the internal force in segment AC ismaterial +F , and in A Fig. 4–10 veya kinematik şart denklemi yazılabilir. relationship, which depends on the A canFig. 4–10 ior. For example, if linear-elastic FA FA behavior occurs, d = PL>AE 0 equilibrium equation(s) are not sufficient to determine the two reactions segment the force-F is -BF 4–11c, above equation the end egment CBCB thethat internal force ,in Fig. 4–11c, the B , Fig. or. For example, ifinternal linear-elastic behavior occurs, d the =above canincancan dthe =using 0 is force ed. Realizing internal segment AC isPL>AE +FFequation , and A>B A on F 1LCB>L 2, so that Bwritten AC # the bar. A be as e written as d. Realizing that theforce internal force in segment AC is +FA , and in FA ent CB the internal can r the reactions become is - FB , Fig. 4–11c, the above equation FB In order to establish an additional FB FFBB FB equation needed for solution, it is t CB as thethe internal force is -F , Fig. 4–11c, the above equation can erms of applied loads by using FB B itten Bu in denklem yük-deplasman ilişkisi cinsinden ifade edilebilir. on candepends be expressed terms ofuygulanan the applied yüklere loads bybağlı using bir necessary to consider how points(c)on(c)the bar displace. Specifically, an P which on the material ten as FB(b) F(b) B P acement relationship, which depends on the material F L F L FALA equation that specifiesFtheF conditions for displacement is referred to AC FBLCB B CB AC can behavior occurs, d = PL>AE B B (c) LAC # = 0= 0 FA - occurs, (b) example, if linear-elastic behavior Fig.Fig. 4–114–11 kinematic condition. In this case, a suitable AE = in P ¢segment ≤ AC is +LFAC , and inLCB AE AEAEd = PL>AE can as a compatibility or Brce A F F (c) (b) A B F Bsegment AC is + F , and in L izing that the internal force in -F can = 0 A compatibility condition would require the displacement of one end of by using Fig. 4–11c, the above equation FAL L Fig. 4–11 AC B AE CB he internal force is - FAE the equation can the bar with respect to the other end to be equal to zero, since the end -P 4–11c, = above 0 B , Fig. material Fig. 4–11 AE # isAE Assuming that is constant, then =B1L FBCB 1L>L >L2FAC 2,supports so that using FBAC AE can that Assuming AEAE constant, then FAF=A F ,Bso that using are fixed. Hence, the compatibility condition becomes CB e direction of the reactions is FB FB the equilibrium equation, the equations for the reactions become , and in he equilibrium equation, the equations for the reactions become (c) (b) ming that AE is constant, then FA = FB1LCB>LAC2, so that using AE sabit ise, # L (c) (b) Btion can ngCBthat constant, then so that using quilibrium equation, the become = 0AE Fis FBequations LCB FA =forFthe B1Lreactions CB>LAC2, Fig. dA>B = 0 ALAC 4–11 AE = 0for the reactions become ilibrium equation, Fig. 4–11 AE the equations AE LCB L LFCB L AC = ¢P ¢B ≤FB ≤andandFBF=B P=¢P ¢AC ≤ ≤ FAF=A P LCB LAC L L This equation can be expressed in terms of the applied loads by using (b)L L (c) # P2¢, so that olacaktır. = AC FB # = P ¢ ≤ and ≤ FA = FB1LF usingve A>L CB L L a load–displacement relationship, which depends on the material L FA = FB1LCB>LAC2,AC L CB at AE is constant, then ns for the reactions F and ≤ Fig. 4–11FB = P ¢ so≤that using behavior. For example, if linear-elastic behavior occurs, d = PL>AE can A = P ¢ become L L m equation, the equations for the reactions become Since both these results positive, direction of the reactions ince both of of these results areare positive, thethe direction of the reactions is is be used. Realizing that the internal force in segment AC is +FA , and in shown correctly on the free-body diagram. hown correctly on the free-body diagram. both of these results are positive, the direction of the reactions is CB the internal force is -FB , Fig. 4–11c, the above equation can segment LAC F P ¢onLresults ≤ free-body using nat correctly the diagram. B oth of=these are positive, theLdirection of the reactions is be written as CB AC FA = P ¢ L ≤ and FB = P ¢ ≤ ecorrectly on the L free-body diagram. L #50 4.4 4.4 4 e, the direction of the reactions is these results are positive, the direction of the reactions is ram. 4.4 FALAC FBLCB = 0 AE AE 4.4 # EXAMPLE 4.5 4.4 STATICALLY INDETERMINATE AXIA 139 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 4.5 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 4.4 4.4 STATICALLY IEXAMPLE NDETERMINATE AXIALLY LOADED MEMBER 1 3 91 3 9 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 139 The steel rod mm. shown in Fig.14–12a fixed 4.4TheASsteel TATICALLY INDETERMINATE AXIALLY OADED MEMBER 3 9 has a diameter of 10 mm. It is 139 STATICALLY INDETERMINATE XIALLY LOADED MEMBER 139aLdiameter rod shown in Fig. 4–12a has It is fixed 4.4of S10 TATICALLY INDETERMINATE AXIALLY LOADED MEMBER0.2 mm Çapı 10 mm olan çelik çubuk A noktasından duvara mesnetlidir. Yükleme B between P! kNöncesi toisthe wall A,ankastre and before it is loaded, there is 20 a gap of 0.2 mm EXAMPLE AMPLE 4.54.5to the wall at A, and before it is loaded, there a gap ofat0.2 mm between A 4.4 S TATICALLY the I NDETERMINATE A XIALLY L OADED M EMBER 1 3 9 wall at and the rod. Determine the reactions at A and B¿ 4.4 Çubuğa STATICALLY INDETERMINATE OADED MEMBER 1 3B¿if9 the wall at B¿ and the rod. the reactions at A and B¿ if the AXIALLY ucutheile B’ arasında 0.2 mmDetermine boşluk mevcuttur. C noktasından 20 LkN yüklendiğinde AB B¿ C The steel rod shown in Fig. 4–12a has a diameter of 10 mm. It is fixed STATICALLY A LOADED Man EMBER 39 rod isXIALLY subjected to axial force of 1P shown. Neglect the =mm20 kN as 800 he steel rod shown inrod Fig.is4–12a has ato4.4 diameter of 10INDETERMINATE mm. fixed 0.2 subjected an axial force of P It= is20 as shown. Neglect kN Pthe ! mm P ! 20 kN20 kN0.2 mm400 mm EXAMPLE 4.5 towall the wall atand A, and before it is loaded, there is aINDETERMINATE gap0.2 ofmm 0.2 mm between 4.4 S TATICALLY A XIALLY Lthe OADED MEMBER 139 size of collar at C. Take = 200 GPa. E the at A, before it is loaded, there is a gap of between st size of the collar at C. Take = 200 GPa. E A ve B’ mesnetlerinde meydana gelecek reaksiyonları hesaplayınız.# of 10 mm. It is fixed st 0.2 mm B¿ B¿ (a) the wall and the rod. Determine the reactions A and if the A B¿ P ! the 20 kN 4.5 EXAMPLE 4.5 eXAMPLE wall at the rod. Determine reactions at Aatand the B¿at B¿ ifB¿ 10 mm. It isand fixed B C C of 0.2 mm between 0.2 mm The steel rod shown in Fig. 4–12a has a diameter of 10 mm. It is fixed B P ! 20 kN rod isasubjected an 10 axial force as shown. Neglect 20 as kNshown. 0.2 mm A LE 4.5 800 mm P ! 20 kN B¿Neglect ads at has diameter of mm. is=Pfixed is tothe antoaxial force ofIt the the Pof 20=kN of 0.2subjected mm between 400 mm 800 mm mm Asteel and if B¿collar PSOLUTION ! 20 kN to the wall at A, and before it is of loaded, there is fixed a of gap ofmm. 0.20.2 mm between 400 mm Cin A SOLUTION size of the at C. Take = 200 GPa. E The rod shown in Fig. 4–12a has a diameter 10 mm. It is B¿ B The steel rod shown Fig. 4–12a has a diameter 10 It is fixed st 4.5 eatof collar C. Take = 200between GPa. Est mm B¿ aded, there athe gap ofwall 0.2 0.2 mm 20 kN(a) A0.2 mm Athe and the B¿ ifisat P ! 20 kN shown. Neglect C the the and rod. Determine theisEquilibrium. reactions at mm A and B¿mm B¿Pif!the (a) Ais0.2 Bthe rod shown inA, Fig. hasitatat ais diameter of800 10 Itof fixed to the wall at and before loaded, there aon gap mm between C B¿ to4–12a the wall A, and before itismm ismm. loaded, there a gap of 0.2 between Equilibrium. As shown free-body diagram, Fig. 4–12b, we 400 As shown on Fig. 4–12b, we 0.2 mmthe free-body diagram, B P ! 20 kN A hown. Neglect the A B¿ mine the reactions at A and ifthe the 800 mm B¿ isassume subjected an axial force of asatshown. Neglect the Pand =reactions 20 kN C atshown A, and before it the isrod loaded, there isB¿ ato gap of 0.2 mm between 800 mm wall at and rod. Determine reactions at A if the B¿ B¿ 400 mm dthe in Fig. 4–12a has a diameter of 10 mm. It is fixed the wall at and the rod. Determine the A and if the B¿ B¿ will that force P is large enough to cause the rod’s end B to (a) B will assume that force P is large enough to cause the rod’s end B to C 0.2 mm C A P ! 20 kN SOLUTION B¿ 400Bmm B of as Neglect the P = 20rod. kN OLUTION size ofthere the collar at C. Take =B¿ GPa. trod and the Determine the reactions atThe AE and if=the B¿and is subjected an axial force of asstof shown. Neglect the P = 20 kN A, before it isto loaded, iswall ato gap of 0.2 mm between 800 mm rod isshown. subjected an axial force as shown. Neglect the P200 20 kN contact the at(a) problem is statically indeterminate since B¿. C 800 mm contact the wall at400 The problem is statically since B¿.mm 800 mm (a)indeterminate 400 mm A B¿B Equilibrium. As shown on free-body diagram, Fig. 4–12b, we 400 mmP ! 20 kN # 200 jected to axial force of Neglect the Pcollar =two kN size ofthe thean collar at C. =the 200 GPa. E and rod. Determine the reactions atshown. Adiagram, and if Fig. the B¿ =quilibrium. 200 GPa. 800 mm As shown on the free-body 4–12b, weare there are unknowns and only one equation of 400 equilibrium. size ofTake the at C.as Take = GPa. E st 20 C st B¿ there two unknowns and only one equation of equilibrium. mm B (a) F will assume that force P is large enough to cause the rod’s end B to (a) FB FA4 A ected at C. Take EstP to an axial force of=isP200 Neglect the end B to (a) = GPa. 20 enough kN as shown. llcollar assume that force large to cause the rod’s 800(a) mm SOLUTION am, Fig. 4–12b, we 3 Çözüm: +B¿. 400 mm contact the wall at The problem is statically indeterminate since 3 (1) : ©F = 0; F F + 20110 2 N = 0 + P ! 20 kN (b) x A B ollar atthe C. Take at = The 200 GPa. EstB¿. ntact wall problem is statically indeterminate : since (1) ©Fx = 0; (a) P !-F m, Fig. 4–12b, we A - FB + 20110 2 N = 0 20 kN the rod’s end to Equilibrium. SOLUTION there are twoB unknowns and onlyAs oneshown equation SOLUTION on of theequilibrium. free-body diagram, Fig. 4–12b, we F 4 ere are two unknowns and only one equation of equilibrium. F A B P kuvvetinin B noktasını B’ ile buluşturacağını varsayarak hesap yapalım. N rod’s end Bsince he to shown ndeterminate 4 FB will assume force Pon large enough to the B no to Amove Equilibrium. As on the diagram, Fig. 4–12b, Pthat !free-body 20 kN force Compatibility. The P causes point Bcause toFwe to4–12b, with B¿,end Equilibrium. As shown the free-body diagram, Fig.rod’s we 3 is + ©F = since Compatibility. The force P causes point B to move to B¿, with no (1) : 0; diagram, -free-body FFig. -P F + 20110 2 N = 0 (b) determinate xshown A B 3 equilibrium. e free-body 4–12b, we um. As on the diagram, Fig. 4–12b, we ! force 20 kN contact the wall at B¿. The istostatically force large cause thethe rod’s end B the toindeterminate further displacement. Therefore condition forsince the (1) :will ©Fassume FAP -is F 20110 N = 0problem assume that P2 to is large enough cause rod’s end B to(b) FP ! 20 kN x = 0; that will B + enough 4 FBcompatibility A further displacement. Therefore the compatibility condition for the quilibrium. me that force Pcause iscontact large enough the rod’s end to there are twoto unknowns only one equation of equilibrium.since contact thetowall at The problem isBstatically indeterminate since B¿.free-body enough the rod’s end toand m. As shown on diagram, Fig. 4–12b, we rod isthe wall atcause The problem isFBBstatically indeterminate B¿. Fthe 4 with AThe P ! 20 kN P ! 20 kN F Compatibility. force P causes point B to move to no B¿, 4 FB (1) F F A (b) indeterminate since rod is A A he wall at two is statically B¿. # problem there are unknowns and only one equation of equilibrium. that force P isThe large enough to cause the rod’s end Btotoequation P ! 20 kN ompatibility. The force Ptwo causes point B toonly move noequilibrium. B¿, with there are unknowns and one of FA blem is statically indeterminate since 3 20 (1) + further displacement. Therefore the compatibility condition for the F 4 (b) F F P ! kN A 4 FB B A two unknowns only one of equilibrium. (1) : ©F =equation 0; -FA -dB>A F 20110 2mN = 0 =+ 0.0002 (b) B at problem is xstatically indeterminate since B¿. Theand FB rther displacement. Therefore the compatibility condition Ffor the P ! 20 kN FB FB 4m 3 ywall one equilibrium. dB>A = 0.0002 +rod 3 A +of©F move toisequation with no B¿, (1) : ©F = 0; F F + 20110 2 N = 0 (1) : = 0; F F + 20110 2 N = 0 (b) x A B (b) FFAB x equation A FA B unknowns and only 4(hiperstatik) (c) do 0; is 3 of equilibrium. # - Fone ….FAof (1) the Statikçe belirsiz.. = - Fno + 20110 2 N The = 0 force This displacement can be expressed FA(1) ove to B¿, with (b)unknown y condition for the A B FA move Compatibility. P causes pointinB terms to to B¿, withFAno FB 4 3 d = 0.0002 m This displacement can be expressed in terms B>A the load–displacement FB FB of the unknown reactions relationship, Eq. applied condition for the (1) + 20110 2AN = 0+force (b) The P3using BPto move tocompatibility no condition B¿, (1) 0;Compatibility. -F - Compatibility. F 20110 2causes = F0point displacement. Therefore the for The force causes point Bwith to move to with nothe B¿,4–2, dB>A =N0.0002 m (b) Bfurther Fig. 4–12Eq. 4–2, applied F F F B B A A using the load–displacement relationship, bility. The force Ptocauses pointAC B and tocompatibility move to B¿, withWorking no reactions segments CB,inFig. 4–12c. inthe units of newtons and (c) further Therefore the condition for rod is further displacement. Therefore the compatibility condition for the Thisdisplacement. displacement can be expressed terms of the unknown FA FA to segments AC and CB, Fig. FA 4–12c. Working Fin splacement. Therefore the compatibility condition (c) meters, we have A units of newtons and This displacement can be expressed inno terms of for the FB¿, Fthe rod isThe ity. force causes point B towith move to no B¿, with B Bunknown rod isload–displacement reactions using the relationship, Eq. 4–2, applied uses point B Pto move to Fwe # FAFAFig. 4–12 FA meters, A have actions using the load–displacement relationship, Eq. 4–2, applied d = 0.0002 m F F B 4–12c. Working BFnewtons Therefore compatibility conditioninB>A for theof LAC FBLCB FA to AC andthe CB, Fig. units and FA FB FB (c) A Fig. 4–12 the compatibility condition for slacement. of segments the unknown dB>Athe =mind0.0002 = m dB>A =Working 0.0002 =m 0.0002 segments AChave and CB, Fig. 4–12c. units of newtons and B>A F F meters, we F L F L F F (c) B B B B AE AE A AC B CB = 0.0002 m ofEq. the4–2, unknown p, appliedd FA FB can be expressed in termskadar ofFA the unknown dB>A = 0.0002 m = FB - (c) Çubuk aradaki boşluk miktarı uzayabilecektir. eters, we have # B>AThis displacement Fig. 4–12 F F (c) (c) AE AE , Eq. applied F L F L A A F 10.4 m2 nits of4–2, newtons and This displacement can be expressed in terms of the unknown A AC B CB This displacement can be expressed in terms of the unknown A using the load–displacement Eq. 4–2,(c)applied F dB>A =expressed 0.0002 m inmFig. = 0.0002 = splacement can bedreactions of F the unknownrelationship, FB 0.0002 m = Fig. 4–12 B>A B Fterms L4–12 A AC BL CB ts of0.0002 newtons and 2 Fig. 9için 2 applied reactions using the load–displacement relationship, Eq. 4–2, Uzamaları AC ve CB bölgeleri ayrı ayrı yazıp cebrik toplamını hesaplayalım. AC bölgesi reactions using the load–displacement relationship, Eq. of 4–2, applied AE AE FAand 10.4 m2 to segments AC and CB, 4–12c. Working in units newtons = m p10.005 m2 [200110 2 N>m ] d = 0.0002 m = Fig. 4–12 using the load–displacement relationship, Eq. 4–2, applied B>A Fig. 4–12 F F B B (c)Fig. 4–12 0.0002 m = AE AE to segments AC and CB, Fig. 4–12c. Working in units of newtons and lacement can be expressed in terms of the unknown to segments AC and CB, Fig. 4–12c. Working in units of newtons and meters, we have 9 2 FA10.4 m2 in units of newtons and nts AC and CB, Fig. 4–12c. Working BLCB p10.005 m22[200110 2 N>mçalışacaktır. ] çekme etkisinde uzamaya çalışırken basınç etkisinde kısalmaya FB(c) 10.8 m2 0.0002 m have = terms meters, ing the we load–displacement relationship, Eq. 4–2, applied CB bölgesi meters, we have unknown F 10.4 m2 2 9 2 A xpressed in of the L e have CB Fig. 4–12 AE F L F L p10.005 m2 [200110 2 N>m ] A AC B CB 0002 m =CB, Fig. 4–12c. 2 9 AC and Working of newtons 9 in units d2F 0.0002 mand =Ap10.005 m2 [200110 2 N>m2] FB10.8 m2 FBL B>A F LAE F p10.005 m22[200110 2 N>m ]FALL= ement relationship, Eq. applied AC CB E AC BLAE CB FA4–2, L AC B CB d = 0.0002 m = have d = 0.0002 m = Fig. 4–12 B>A F 10.8 m2 B>A B d = 0.0002 m = B>A AE AE AE p10.005 m22[20011092 N>m2] 12c. Working in# units of newtons and or - AE AE AE FAF 10.4 m2 10.8 2 m2 9 2 B F L F L m2 [200110 2 N>m ] A AC CB10.4 m2 0.0002 m m2 = -- B Fp10.005 F=A10.4 A 2 2 dB>A = F0.0002 m m2 9 m22] =or 2 3141.59 N # m (2) FAE m2 FB910.8 m2 [200110 2 N>m A10.4 0.0002 m = A10.4 p10.005 m2 [200110 22 N>m=15707950 ] 0.0002 m = =#p10.005 AE Burada # N 2 9 2 0.8 m2 2 9 = or p10.005 m2 2 N>m ] 2 9 [200110 2 p10.005 m2 [200110 2 N>m ] p10.005 m2 [200110 2 N>m ] LAAC Solving AF BLCB Eqs. 1 and 2 yields 10.4 8 m2 F 2 m2F (2) FB10.8 m2 FA10.4 m2 - FB10.8 m2 = 3141.59 N # m 200110 ] -F 10.4 m2 - F 10.8 m2 = 3141.59 N # m m = 92 N>m (2)m2 A B F 10.8 m2 F 10.8 2 9 2 9 AE 2 B kN # …. (2) B Ans. F = 16.0 F = 4.05 kN AE F 10.8 m2 2 9 2 A B p10.005 m2 [200110 2 N>m ] 00110 2 N>m ]F 10.4 m2 - F 10.8 m2 # m p10.005 m2 [200110 2 N>m ] - B= 3141.592N Solving Eqs. 91 and 22 yields - B 2(2) Solving Eqs. 1Aand 2 yields 2 9 [200110 2 92 N>m p10.005 m2 ]2[200110 p10.005 m2 2 N>m ] wall at p10.005 m2 [200110 2 N>m ] Since the answer for is positive, indeed end B contacts the F ve (2)= denklemlerinin F 10.8 m2B ortak çözümünden, Ans. FA = 16.0 kN FB = 4.05 kN or lving 1 and (2) 2(1) yields Ans. 16.0 kNBassumed. FB = 4.05 kN 99or N # mEqs. A originally B¿F-as 2 or 2 9 2 ] p10.005 [200110 2 N>m ] #N>m Ans. kN m2FF = 4.05 Since theatN answer (2)# FAfor= F16.0 N2Since m the # m for FB is(2)positive, indeed end B contacts the wall at B 10.4 m2end -kN F 10.8 m2 =the 3141.59 answer BBcontacts wall Aindeed B is positive, # # (2) F 10.4 m2 F 10.8 m2 = 3141.59 N m # (2) F 10.4 m2 F 10.8 m2 = 3141.59 N m A B A B as originally assumed. B¿ (2) F 10.4 m2 F 10.8 m2 = 3141.59 N m A B as originally B¿the If Findeed wereend a negative nce answerAns. for B contactsquantity, the wall the at problem would be FNOTE: B is positive, Fassumed. 10.8 m2Eqs. 1 Band 2 yields kN BSolving Solving Eqs. 1assumed. and 2statically yields Solving Eqs. determinate, 1 and 2 yieldsso that FB = 0 and FA = 20 kN. as1originally 2 yields Nqs. Fand m2Ans. - FB210.8 m2 =9 3141.592NF#Am= 16.0 kN(2) FB = 4.05 kN A10.4 contacts theIfwall at p10.005 m2 [200110 N>m ] B= =16.0 NOTE: aAns. negative quantity, the problem would be NOTE: were a negative the problem would F 4.05 Fquantity, kNkN F kNbeIf FB wereAns. Ans. FA = B16.0FkN F2BkN = 4.05 A = 16.0 AFkN B = 4.05Ans. ontacts the wall at .OTE: 1statically and 2If yields statically determinate, so that determinate, so that and F F = 0 = 20 kN. Since the answer for FBthe isApositive, indeed end B contacts the wall at FB = 0 and FA = 20 kN. B negative quantity, problem would becontacts FB were answer for is end positive, indeed endwall B the wall at F Since the foraFthe positive, indeed Bthe contacts the at answer foranswer positive, end B wall at FB isSince Bcontacts B isindeed as originally B¿ Ans. FA = 16.0 kN FBFB= =4.05 kN FA = 20 kN. atically determinate, sooriginally that and 0 assumed. assumed. B¿ as assumed. B¿ as originally inally assumed. problem would be (2) the wall at 10.8 m2 3141.59 m end B contacts nswer for F=B is positive,N indeed 0 kN. would roblem be NOTE: If FB were a negative quantity, the problem would be ally assumed. NOTE: If FB were negative quantity, the problem would be NOTE: If FaB negative were a quantity, negative quantity, the problem be If the aproblem would bewould kN.FB were statically determinate, so that FB = 0 and FA = 20 kN. statically determinate, so that FB20=kN. 0 and FA = 20 kN. statically determinate, FA = FBF=A 0=and determinate, so that FBso=that 0 and 20 kN. Ans. would be NFB were FB a=negative 4.05 kNquantity, the problem terminate, so that FB = 0 and FA = 20 kN. # ve, indeed end B contacts the wall at e quantity, the problem would be B = 0 and FA = 20 kN. #51 A P ! 9 kip EXAMPLE 4.6 The aluminum post shown in Fig. 4–13a is reinforced with a brass core. If this assembly supports an axial compressive load of P = 9 kip, applied to the The rigidaluminum cap, determine the average normal in with a brass post shown in Fig. 4–13a is stress reinforced P ! 9 kip 3 the aluminum and the brass. Take and E = 10110 2 ksi core. If this assembly supports an axial compressive load of P = 9 kip, al 140 C H A P T E R 4 A X I A2 Lin.L O A D 1 in. 3 140 C H A P T E R 4 A X IE AL LOAD = 15110 2 ksi. applied to the rigid cap, determine the average normal stress in br 1.5 ft the aluminum and the brass. Take Eal = 1011032 ksi and EXAMPLE 4.6 Ebr = 1511032 ksi. CHAPTER 4 AXIAL LOAD SOLUTION AD 140 1.5 ft 2 in. 1 in. # EXAMPLE 4.6 Equilibrium. The free-body diagram of the with post aisbrass shown in The aluminum post shown in Fig. 4–13a is reinforced P ! 9 kip aluminyum prinç çekirdek ileresultant güçlendirilmiştir. Bu aksama rijit bir başlık The aluminum post shown in Fig. 4–13a is reinforced with a brass Pbrass !core. 9 kolon, kip IfFig. in Fig.2 in. 4–13a is Şekildeki reinforced with a 4–13b. Here the axial force at the base is represented by SOLUTION this assembly supports an axial compressive load of P = 9 kip, 1 in. (a) EXAMPLE 4.6 vasıtasıyla P=9 kip eksenel basınç yükü uygulandığında aluminyum ve prinç (brass) core. If this assembly supports an axial compressive load of 9 kip, in rtsThe an axial compressive load of P = 9 kip, the unknown components carried by the aluminum, and brass, F , 2 in. 1 in. applied to the rigid cap, determineThe the free-body average normal stress al of in aluminum post shown in Fig. 4–13a is reinforced withEquilibrium. a brass diagram the postPis= shown 3 applied to the rigid cap, determine the average normal stress in e core. aluminum post shown in Fig. 4–13a is reinforced with a brass malzemelerdeki ortalama normal gerilmeleri hesaplayınız. The problem is statically indeterminate. Why? . F determine the average normal stress in aluminum brass. Take andbase is represented Eal = axial 10110force 2 ksiat the 1 4 0an axial the C HAPT ER A X I Aof Land LPOFig. A=Dthe br 4 load If this assemblypost supports compressive 94–13b. kip, Here the4–13a resultant by 4 The The aluminum post shown inand Fig. is reinforced with a brass P !shown 9 kip aluminum incompressive Fig. 4–13a is reinforced with a brass 3 3 3 of P the aluminum the brass. Take and E = 10110 2 ksi e.applied Ifbrass. this assembly supports an axial load = 9 kip, Vertical force equilibrium requires Take and E = 10110 2 ksi (a) = 15110 2 ksi. E al al to the rigid cap, determine the average normal stress in br the components carried load by the aluminum, # ve#axial compressive alınacaktır. 1.5 ft core. If this assembly supports an axial compressive of P = 9 kip, Fal , and brass, core. If 2this supports an load of Punknown = 932kip, in. assembly 1 in. 3br = 15110 ksi. Estress plied toaluminum the rigid cap, determine the Take average normal in the applied and the brass. and = 10110 2F ksi 1.5 ftEcal The problem is statically indeterminate. Why? . (1) + ©F = 0; -9 kip + F + F = 0 applied to the rigid cap, determine the average normal stress in to the rigid cap, determine the average normal stress in Çözüm br y al br 3 3140 4 the Cbrass. 2 ksi and EXAMPLE H A P T E R Take 4 AXIE A L4.6 L= O A10110 D = 15110 2 and ksi. Ealuminum al br the force equilibrium the aluminum and brass. Take requires E = 1011032 ksi and SOLUTION aluminum and the brass. Take and E = 101103Vertical 2 ksi the al al 3 = 15110 2 ksi. 3 The rigid cap at the top of the post causes both the 2 ksi. ECompatibility. SOLUTION Ebr = 1511032 ksi. P ! 9 kip1.5 ft br = 15110 Equilibrium. The free-body the post+ F is shown in is reinforced with(1) The aluminum post shown in a brass P ! 9 kip + c ©F =diagram 0; the of -9amount. kip + F4–13a y displace al Fig. br = 0 ….(1) aluminum and # brass to same Therefore, SOLUTION Fig. 4–13b. Here the resultant axial force at the base is represented by Equilibrium. The free-body diagram of the post is shown in ody diagram of the post is shown in core. If this assembly supports an axial compressive load of P = 9 kip, 1 in. EXAMPLE 4.62 in. (a) Statikçe belirsiz.. (hiperstatik) d = d the unknown components carried by the aluminum, and brass, F , LUTION nt axial force at the base is represented by Fig. 4–13b. Here the resultant axial force at the base is represented by Compatibility. The rigid cap at the top of post causes both the al al br Equilibrium. The free-body diagram of the post is shown in applied to the rigid cap, determine the average normal stress in SOLUTION SOLUTION P ! 9 kip (a) brass, 3 The problem is statically indeterminate. Why? . F carried by the aluminum, and F , the unknown components carried by the aluminum, and brass, F , br aluminum and brass to displace the same amount. Therefore, Fig. 4–13b. Here the resultant axial force at the base is represented by the aluminum and the brass. Take and E = 10110 2 ksi al al The aluminum post shown in Fig. 4–13a is reinforced with a brass P ! 9 kip uilibrium. The free-body diagramdiagram of theEquilibrium. postthe is post shown The free-body diagram of the post is shown in al Using the load–displacement Equilibrium. The free-body of is in shown in relationships, 3 Vertical force equilibrium y. 4–13b. indeterminate. Why? The problem is statically indeterminate. Why? . this Fbr the unknown components carried by the aluminum, and brass, FHere ,Ifrepresented =by15110 2 ksi. Erequires core. assembly supports an axial compressive load ofby P = 9 kip, al the resultant axial force at the base is represented by br 2 in. 1 in. Fig. 4–13b. the resultant axial force at the base Fig.Here 4–13b. Here the resultant axial force at the base is 1.5 ft 4 dal =isdrepresented br F L F L m unknown requires al br Vertical force equilibrium requires problem is statically indeterminate. Why? Fbr . The (a) applied to the rigid cap, determine the average normal c ©F components carried by the +aluminum, (1) brass, stress in 0;Fal , andcomponents 9 kipbrass, + carried Fal=+ Fby the unknown the0 aluminum, Fal , and the unknown components carried by the aluminum, Fbrass, y = br = al ,-and Vertical force equilibrium requires A E A E Using the load–displacement relationships, the + caluminum the Take Eal= =0 1011032 ksi and (1) al and alindeterminate. br brbrass.Why? problem isystatically Fbr . The is statically indeterminate. Why? . The+ Fproblem problem is statically indeterminate. Why? . The (1) kip F + F = 0 ©F = 0; -9 kip + F + F br SOLUTION al br al br 3cap at the top of the post causes both the 4 Compatibility.Ebr The rigid = 15110 2(1) ksi. requires A force equilibrium Vertical Vertical force P ! 9equilibrium kip-requires Eal FbrLdiagram of the post is shown in + c ©Fforce 9 kip + requires Fal +1.5Fftbr = Vertical 0 FalalL y = 0;equilibrium Equilibrium. The aluminum and brass to displace the same amount. Therefore, FalThe = Frigid bfree-body aat cap at the top of the post causes both the br a cap = Compatibility. theb top of the post causes both the F A E c P ! 9 kip br (1)base is represented by + ©F = 0; 9 kip + F + F = 0 Abraxial c Fig. 4–13b. Here the resultant at the bralE br (1) + ©F = 0; 9 kip + F + F = 0 A Eamount. (1) ©F = 0; -9 kip + F + F = 0 y al br y al br al br forceTherefore, y al br Compatibility. The Therefore, rigid cap at the top of causes both the lace the same amount. aluminum anddal brass displace the same = dto (a)the postSOLUTION br components carried by the aluminum, F , and brass, the unknown al 3 aluminum and brass to Therefore, The rigid Faldisplace the same amount. A p[12 in.2 - 11top in.2 2alksi E Compatibility. cap at2 the of2d] the=10110 post causes both the al captop at Using thethe top ofload–displacement the postboth causes both the dalCompatibility. = dbr The rigidThe The problem is statically indeterminate. Why? . F mpatibility. at the of post causes the d the relationships, Equilibrium. The free-body diagram of the post is shown in Pcap ! 9rigid kip br al br F = F a b a b Falbrass = Fbr R alR B Therefore, br 3 (b)4 to displace 2equilibrium aluminum and toBdisplace the same amount. Fbr aluminum brass amount. Therefore, A2brbase damount. brsame Vertical force requires minum and brassand to displace the dsame Therefore, al =the p11 in.2 ksi Eisbrrepresented Fig. 4–13b. Here the resultant axial force15110 at the Rijit başlık aluminyum ve prinç malzemelerde aynı kısalmaya sebep by nt relationships, Uygunluk denklemi: F L F L Using the load–displacement relationships, al br (a) the unknown components carried by the aluminum, and brass, F , = d = d al d = d Using the load–displacement br br (1) + c ©F 0; 2F - 92 kip +in.2 Fal2]+ F10110 dalrelationships, = dbr Falal br = (2)032 ksi p[12 - 11 Abry E=Fbral al = br in.2 alEal L FbrL F L F L olacaktır. # problem is statically indeterminate. Why? Fbr . TheA al br F = F B R B R = Using4 the load–displacement al br Using the load–displacement relationships,= 2 relationships, Falrelationships, L F L br(b) p11 in.2 1511032 ksi Solving Eqs. 1 and 2force simultaneously Vertical equilibrium requires Abr Ebr Aal Ayields Ealal rigid ing load–displacement Compatibility. A al the = alEThe brEcap br at the top of the post causes both the P ! 9 kip Fal F= L Fbr a Fb L a b AalEalFalL AbrEbr aluminum brass to displace the same amount. Therefore, FbrL Fbr Aand E br brkip Aal E Fbr+ = al3F (1) + c ©Fy = 0; Falal = 6=kip -9 +kipF = 0E FbrL= A=br 2F (2) al FalL br F = F a b a b AalEal AbrEbrFal = Fbr aal al bdabr=al db = al br s ! 0.955 ksi A E A E A E al al al al br br 2 2 3 A Abrbr FE al br F al brAalEal F A= E Ebr causes both the b a Since b the resultsSolving p[12 in.2 -1 indeed 11 in.2 ]theatstress 10110top 2 ksibr are positive, will be compressive. al br Fbr brbra Eqs. and 2 cap simultaneously yields The rigid Abr A Ebr EFalCompatibility. ARalB the Eal3andof = Fbr Bnormal Rthe post ! 9 kip sal !P0.637 ksi al Thealaverage 2 load–displacement stress in the aluminum brass isTherefore, therefore Using the relationships, 2 2(b) 3 Fin.2 = displace Fbr a b15110 a same b2 ksi brass the amount. 2 Eal b a baluminum and p11 al to in.2 - 11 in.2 ] 10110 2 ksi FalFA al=al Fbr a p[12 11 in.2F ]br =10110 Fbralin.2 = 2E 6-br kip 3 kip32 ksi FF A br 2 2 3 A E = F a b a b br br al br - 11 in.2 ] R B R p[12 in.2 10110 2 ksi 6 kip F = F B R B R F L F L br Abr Ebr R B br (2) 3 2 ksi dp11 dalbr p11 in.22 Fal = F 15110 ksi0.955(b) al == B s3br2 ! R Since Ans. sal =theFal 0.637 al = 2F br in.2 15110 2be ksicompressive. =the br 22are 22 indeed 3ksi stress 2 3 results positive, will Fal 2 2 3 p[12 p11in.2 in.2 - 11sin.2 15110 2 ksi2 ksi p[12 in.2 in.2 -- 11 11 in.2 in.2 ]] Aal10110 Eal 2Aksi p[12 ] ksi 10110 brEbr 0.637 2 al !Eqs. Using the load–displacement relationships, F2alsimultaneously = F Solving and R in B= the R and brass is therefore The normal aluminum Falp[12 = Fbr B 2 - 11 in.2(2) R32B1ksi Rbr Baverageyields Fal = 2Fbr in.2 ] 2 10110 (b) 2 stress 3 (2) F 2F 3 al 15110 br 2 ksi A p11 in.2 in.2 15110 2 ksi Eal (2) Fal2p11 = 2F Fal = Fbr B RB R al br 3 kip 3 F L F L F = F a b a b F = 6 kip F = 3 kip 6 kip al 0.955 ksibr al br p11 in.2 15110 2 ksi Solving neously yields salbr Eqs. Ans. = 1 and F =br 2s simultaneously yields Fbr A= 0.637 Ebrksi Ans. =2F=br (2) Solving Eqs.ksi1 and 2 simultaneouslyFyields al al= sbr ! 0.955 2 2 br p11(2) in.22A al = 2Fbr E A E # ….(2) p[12 in.2 11 in.2 ] al al br br Since the results are positive, indeed the stress will be compressive. 6 kip Fbr = 3 kip (c) (2) Fal ksi = 2Fbr Falyields = 6 kipp[12 F 2= 3 kip FalSolving salF2 !simultaneously 0.637 br therefore 1 andthese 2 simultaneously in.2 -are11shown in.22] in1011032 ksi FbrThe = 3average kip Eqs. Solving Eqs. 1 and yields normal stress in the aluminum and brass is NOTE: Using results, the stress distributions al = 6 kip A E al al s ! 0.955 ksi FFalal==F Ba RB R br e, indeed stress will compressive. b astress b will 3Fbrkip (b)Fig. 4–13c. Since the results are positive, 2 ving Eqs.the 1 and 2 simultaneously yields the Fig.be 4–13 p11 1511032 ksi Ans. 6 kip =br 3 A kip Fbr6 kip results(1) are the stress be Ebrin.2 Falisindeed = Fortak =ksi3çözümünden, kipcompressive.# Fal6 =kip sbr = Fbr indeed = 0.955 ksi be compressive. br br will vepositive, (2) denklemlerinin sal ! 0.637 ssSince in thethe aluminum and brass therefore 2 sal The = average = 0.637 normal stress inin.2 theksi aluminum Ans. and brass is therefore p11 sbr ! 0.955 ksi The average normal and brass is therefore Falare =stress 6positive, kipFin the Fbraluminum = 3the kip p[12positive, in.22 - indeed 11 in.22the ] stress Since are will be (2) = 2Fbr 2 2 Fcompressive. 3 Since the results indeed stressthe willresults be compressive. (c) al 10110 al 6 kip p[12 in.2 11 in.2 ] 2 ksi sal ! 60.637 ksi 6 kip NOTE: Using these results, the stress distributions are shown in kip The average normal stress in the aluminum and brass is therefore Ans. = 0.637 ksi The average normal stress in stress the aluminum and brass is therefore FSolving B R B =yields R al = Fs brEqs. ce indeed will compressive. Ans. 0.637 2 results 2 s (b) the 3 ksi al = 1 and 2 simultaneously Ans. = be 0.637 ksi kip 2 2 in.2the - 11 in.2are ] positive, al = p11 2 ksi 2 Fig.34–13c. 4–132]and brass iss therefore p[12 in.2in.2 - 11 in.22] 15110 p[12 - Fig. 11 in.2 Ans. = = 0.955 ksi 6 kip The average normal stress in in.2 the aluminum br 6 kip 2 =al0.637 ksi = 6 kip Fbr =Ans. 3 kip (2) al =in.2 Ans. 2 sal = = 0.637 ksi sp11 Fal2]=F2F 2 2 br ip p[12 in.2 - 113 in.2 (c) p[12 in.2 11 in.2 ] 6 kip kip s ! 0.955 ksi 3 kip br Ans. = 0.955 ksi NOTE: thethe stress are shown Ans.Ans. sal s = br = = 0.637Using ksi these results, Since are positive, indeed theinstress will be compressive. Ans. = distributions = 0.955 ksi br results 2 2 = 0.955 2 ksi n.22 2ssimultaneously 2 yields p[12p11 in.2in.2 - 11 in.2Fig. ] 4–13c. sal !Solving 0.637 ksi Eqs. 1 and p11 in.2 Fig.# 4–13 3 kipaverage normal stress in the aluminum and brass is therefore The 3 kip (c) sbr = Ans. 0.955 ksi F = 3 kip s = are shown Ans. 2F= = in 0.955 ksi ts, the stress distributions = 6 kip 2 br NOTE: Using theseal results, the stress distributions are shown in p11 in.2 NOTE: Using these3 br results, the stress distributions are shown in 6 kip p11 in.2 kip ksi ! 0.955 Ans. s = = 0.637 ksi al Fig. 4–13c. Fig. 4–13c. sbr =sbr (c) Ans. = Fig. 0.955 ksi 4–13 2 results areinpositive, indeed the stress will compressive. NOTE: Since Usingthe these results, the stress distributions are shown p[12 in.22 11be in.2 ] in NOTE: Usingp11 these are shown in.22results, the sal !stress 0.637 ksidistributions Fig. 4–13c. The average normal stress in the aluminum and brass is therefore Fig. 4–13 Fig. 4–13c. OTE: Using these results, the stress distributions are shown in 3 kip s6brkip Ans. = 0.955ksi ksi Ans. sal = ==0.637 . 4–13c. 2 2 p11 in.2 p[12 in.22 - 11 in.2 ] # (c) Fig. 4–13 (c) Fig. 4–13 NOTE: Using these results, the stress distributions are shown in 3 kip Fig.s4–13c. Ans. = 0.955 ksi br = p11 in.22 NOTE: Using these results, the stress distributions are shown in Fig. 4–13c. #52 4.4 # EXAMPLE 4.7 AXIALLY LOADED MEMBER 141 steelINDETERMINATE bars shown in Fig. 4–14a are pinelemana connectedmafsallarla 4.4 Sçeliğinden TATICALLY AXIALLY LOADED MEMBER 1to 4 1a ÜçThe adetthree A36A-36 imal edilmiş çubuk rijit bir STATICALLY INDETERMINATErigid AXIALLY LOADEDIfMthe EMBER member. applied load141 on the member is 15 kN, determine LY INDETERMINATE 141 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER B bağlanmıştır. 15 kN D F the force developed each bar.oluşacak Bars AB kuvvetleri and EF each have a cross-AB ve EF çubuk alanları 50 yükleme altında herbirinçubukta hesaplayınız. sectional area of 50 mm2, and bar CD has a cross-sectional area of 141 0.5 m ected to a 4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 141 bars shown in Fig. 4–14a are pin connected to a determine B D F napplied connected loadto onathe member is 15 kN, determine ve a4.7 crossB D F E15 E MEMBER kN, determine SOLUTION 4.4 STATICALLY AINDETERMINATECAXIALLY LOADED in bar. Bars AB EF each al each areaEXAMPLE of Band D have a crossF 4.7 2 ach have a crossbar shown CD has cross-sectional area mm steel , andbars 0.5of m diagram Equilibrium. Thepin free-body A-36 inaFig. 4–14a are connected to a of the rigid member is shown sectional area of in Fig. 4–14b. This problem is statically indeterminate since there 0.5 m are threeload A-36onsteel bars shown inkN, Fig.determine 4–14a are pin connected to a er. If theThe applied the member is 15 0.4 m 0.5 m B D F EXAMPLE 4.7 0.2 m 0.2 m three unknowns and only twomember equations. member. If the applied load onhave the 15 kN, determine evelopedrigid in each bar. Bars AB and EF each aavailable cross- is equilibrium 4.4 STATICALLY INDETERMINATE AXIALLY LOADED B D F C E 2 A 4 force developed in each bar. Bars AB area and EF rea of 50the bar CD has a cross-sectional of each have a cross- E mm , and A bars shownCin Fig. 4–14a are pin connected to a 2The three A-36 steel r is shown sectional area of and bar CD has a cross-sectional area of 50 mm , 0.5 m C +Ac ©Fy = 0; rigid FEAIf+the FCapplied + FE -load 15 kN = 0member is 15(1) member. on the kN, determine is shown B D F eree-body there are 0.5 m 30diagram mm2. of the rigid member EXAMPLE 4.7 0.4 m the force developed in each bar. Bars AB and EF each have a cross- 15 kN member is shown roblem is statically0.2indeterminate since there are m 0.2 m ons. 2 0.4 mm2 - FA10.4 d + ©MC = 0; sectional m2 of + 50 15 mm kN10.2 m2bar + FCD = 0 (2) E10.4 area has a cross-sectional area of since there are equilibrium 0.2 ,mand two available equations. A0.2 msteel bars C shown in Fig. E The A-36 4–14a are pin connected to a 4 three Neonly 0.4 2 m (a) 0. 0.2 m 0.230 m mm . equations. A on the4member C E determine member. If the applied load is 15 kN, SOLUTION B D m. The(1) free-body diagram of the rigid member is rigid shown 4 the rigid force developed in each bar. Compatibility. The applied load will cause theishorizontal line Bars ACEAB and EF each have a crossEquilibrium. The free-body diagram of the member shown (1) FAproblem + FC + isFstatically 15 kN = 0 b. This indeterminate since there are E 2 15 kN m CD has a cross-sectional sectional of 50line mm , and0.4bar FA FCarea of FE shown in Fig. is 4–14c to move to the area inclined 0 0 and 0.2 m 0.2are mA¿C¿E¿. The inonly Fig.(1) 4–14b. This problem statically indeterminate since there owns two available equilibrium equations. A C E (2) = 2 15 kN SOLUTION 0.4 m # 30 Emm . be related by similar triangles. displacements of points A, C, and can 0.2 m 0.2 m (2) 10.4 m2 three + 15 kN10.2 m2 and + F(a) 10.4 m2 = 0 unknowns only two available equilibrium equations. 15 kN 4 E Equilibrium. Thethat free-body of the rigid member is shown Thus the compatibility equation relatesdiagram these displacements is 0.4 m2 = 0 (2) Çözüm C 4 (a) in Fig. 4–14b. This problem is statically indeterminate since there are (1) 0; F + F + F 15 kN = 0 A C E l line ACE 0.4 m (a) A C 0.2 m 0.2 m c ©Fwill SOLUTION three and two equations. (1) +load 0; + unknowns FC +ACE 15only kN = 0 available15equilibrium y = cause e applied the horizontal line FA FF FEFE kN The C¿E¿. C A dA=-0Equilibrium. dE (2) dFAC - dEThe Ffree-body line ACE + 15 kN10.2 line m2 +A¿C¿E¿. FE10.4 m2 FE diagram of15the c0;triangles. to -F move tom2the inclined The A10.4 kN rigid member is shown C rizontal 0.4 m = F F F 0.2 m 0.2 m (2) -F d+ ©M = 0; 10.4 m2 + 15 kN10.2 m2 + F 10.4 m2 = 0 ne The A¿C¿E¿. 0.8 m 0.4 m A C E in Fig. 4–14b. This problem is statically indeterminate nts A, C, and E can be related by similar triangles. C A E (a) ments is C + c ©Fy = 0; (1) since there are FA + FC + FE - 15 kN = 0 0.2 m 0.2 m similar triangles. three unknowns and equations. (a) ty equation that relates these displacements is C only two available equilibrium 15 kN 15 kN 1 1 lity. The applied load will cause theChorizontal line ACE placements is dC-F = A10.4 dA + (2) d+ ©M 0; cause m2 d+E15FkN10.2 m2 + C = (b) FC FE10.4 m2FE= 0 Compatibility. The inclined applied load will horizontal line ACE Fig. 4–14c to move to the line The A¿C¿E¿. A 2the 2 c (1) (a) + ©F = 0; F + F + F 15 kN = 0 0.4 m y A C E F F FE shown in Fig. 4–14c to move to the inclined line The A¿C¿E¿. nts of dpoints Edcan be m related by similar triangles. A C A - dA, E C,dand C -0.2 E m 0.2 0.4 m = 15 kN displacements of points A, C, and E can be related by similar triangles. mpatibility that these displacements is applied 0.2 Eq. m 0.2 m we Cthe horizontal line ACE 0.4 m 0.8 mequation 0.4 m relates Using the load–displacement 4–2, have 0.4 m Compatibility. The load will cause 0.4 m relationship, (2) F d + ©M = 0; 10.4 m2 + 15 kN10.2 m2 + F 10.4 m2 = 0 C A E 15 kN Thus the # compatibility equation that relates these displacements is 0.2 m 0.2 m CA # EFC CFA F shown in Fig. 4–14c to move 15 to kNthe inclined line A¿C¿E¿. The dE 1 1 (b) dC = ddA Bilinmeyen +- d dE d -iç dkuvvet displacements of points A, C, and E can be related by similar triangles. 15 kN FCL sayısı1 3, yazılan FAL denge denklemi 1 (b)FELsayısı 2, statikçe E¿ dA ! belirsiz dE sistem. E 2 A 2E = C applied load = d c -Compatibility. d + cThe d displacements 0.4will m cause the C ¿ ACE C dE Thus the compatibility relates is horizontal line 2-(b) 2equation that 2 these d d d A E C E 0.2 m 0.2 m 2 2 dCThe 150shown mm 2E 150 mm 2Est to the inclinedA¿ line0.4dCA¿C¿E¿. 0.8 m 0.4130 m mm 2Est = FA inst Fig. 4–14c to move dA m! dE 0.4 m denklemi: 0.4 m Uygunluk 0.2 m 0.2 m 0.8 m E 0.4 displacements m acement relationship, A Eq. 4–2, we C have of points A, C, and E can be related by similar triangles. (c) 0.4 m 0.4kN m 15 1 1 dE 0.3FA 0.3F e have A+dcompatibility - dE(b) Ethat(3) 0.4FmC = Thus the relates these is dC = dA + d0.4 A - dEE C dC equation Em 15 kN displacements 1 1 Fig. 4–14 dE 0.4 m E¿ = 2 A 2 dA ! dE E C d = d + d 0.2 m 0.2 m d 1 A dE E 0.8 m 0.4 m C ¿ LC (b) F L F 1 E ¿ d d ! d A E 2 2 E A E Est = c A¿d + dcC ! dE d dE C¿ 1–3d2Csimultaneously dA FEL 2 150 mm22E E¿yields A¿ dA Solving ! dE 2 Eqs. d1m Eoad–displacement 150Eq. mm4–2, 2EC¿ dC - dE 15 kN d ! d C dA - dE relationship, we have st st st 1 C E d d 0.4 0.4 dA (c) E =m d = d + d d mm22E(3) C A E A¿ d ! d C relationship, Eq. 4–2, we haveA2 (c) 2 0.8 0.2 m stUsing the load–displacement (b) E0.4mm C m 0.4 E 0.4 m dA C (3) FA = 9.52# kNdE FC = 0.3FA# + 0.3FEFig. 4–14 Ans. A E C (c) FCL FAL FEL 1 1 dE1 d ! dEFig. 4–14 1E¿ = (3)c d +Fig. c4–14the d C =1 3.46AkN C¿ d = d + ddE Ans. F C A Using load–displacement relationship, Eq. 4–2, we have F L F L F L 2 2 2 1 E¿ d ! d 0.4 m 0.4 m C 2E A 2E E A 2 150 mm 2 c 150 mm dCE2 30 mm 2Est 2 E A¿ d d ! d st st = d + c C E dE C¿ d ultaneously yields A A E C 2 2 2 2 150 mm 2Est FE =2 2.02 dC 130 mm 2Est 150 kN mm 2Est A¿ Ans. dE (c) dA dC ! dE (3)the FC = 0.3FA + 0.3FE Ans. FCLUsing FAL FEL Eq. 4–2, 1 load–displacement 1relationship, ! dE (c) wedAhave 0.4Em = c d +4–14c d Fig. C¿ A (3) FC = 0.3FAns. FA = 9.52 kN A + 0.3F 2 E 2 2 2 2 d 130 mm 2E 150 mm 2E 150 mm 2E A¿ st st st Ans. Ans. # Fig. 4–14 dA dC ! dE dE C Ans. FC = 3.46 kN s. 1–3 simultaneously yields FCL FAL FEL 1 1 dA ! (c)dE Ans. = cE d + c (3) d Yazılan 3 denklemi ortak FC = 0.3F Solving 1–3 simultaneously yieldsçözecek olursak, Ans.Eqs. 2 A + 0.3F 2 2 2 2 130 mm 2Est 150 mm 2Est 150 mm 2Est A¿ FE = 2.02 kN Ans. Fig. 4–14 dA dC Ans. FA = 9.52 kN Ans. (3) FC = 0.3FA + 0.3FE Solving Eqs. 1–3 simultaneously yields Ans. FA = 9.52 kN Fi Ans. FC = 3.46 kN Ans. FC = 3.46 kN Ans. Eqs.F1–3=simultaneously Solving yields Ans. 9.52 kN # FE = 2.02 kN A FE = 2.02 kN Ans. Ans. FC = 3.46 kN Ans. FA = 9.52 kN FE = 2.02 kN Ans. Ans. FC = 3.46 kN 4.4 STATICALLY INDETERMINATE AXIALLY LOADED MEMBER 2 ve CD 2 2 mm . çubuk alanı 30 mm dir. 30 mm FE = 2.02 kN Ans. #53 PTER 4 AXIAL LOAD 142 142 142C H A PCTHEAR PC4THE AR A L XLI O 4X I AA A LA D LOAD 4 HAPTER AXIAL LOAD EXAMPLE 4.84.8 4.8 # EXAMPLE EXAMPLE 142 E 4.8 CHAPTER 4 AXIAL LOAD The bolt in4–15a Fig. 4–15a is 2014-T6 made of aluminum 2014-T6 aluminum and is Şekilde aluminyum alaşımdan yapılmış bir bulon, magnezyum alaşımdan The bolt shown in shown Fig. is made of alloy andalloy is The bolt shown in 4–15a Fig. is made of 2014-T6 aluminum alloy and is tightened it compresses a cylindrical tube of Am 1004-T61 tightened so itso compresses a cylindrical tubetube made of made Am 1004-T61 tightened it so compresses a cylindrical made of Am 1004-T61 1 in tir. The bolt shown in Fig. 4–15a is made of 2014-T6 aluminum alloy and is 1 1 of yapılmış bir tüpün içine yerleştirilmiştir. Tüpün yarıçapı 0.5 magnesium alloy.The tube has andış outer it is assumed in., EXAMPLE 4.8 magnesium alloy.The tubetube has an radius of radius is assumed magnesium alloy.The hasouter an outer radius of 2and and itand is assumed in.,2it 2 in., tightened so it compresses a cylindrical tube made of Am 1004-T61 that both the inner radius oftube the tube andradius the radius ofbolt the1are bolt1are 1 in. 3 in. that that bothboth thebulon inner radius of and the ofoldukları the are the inner radius oftube the andradius the ofbolt the 3 in. 3 in. Tüpün 1 the 4 in. 4 in. 4 iç yarıçapı ile yarıçapı 0.25 in alınarak aynı magnesium alloy.The tube has an outer radius of and it is assumed in., The shown in Fig. 4–15a is made 2014-T6 alloy and isto beto be The washers atand the top andof bottom of aluminum the are considered 2 and 1 1 The bolt washers at the top bottom of the are tube considered to be The washers at the top bottom of tube are considered 1 1the tube in. in. 1 in. 2 in. 1that in. 1both in. 4the inner radius of the tube and the radius of the bolt are in. tightened so it compresses a cylindrical tube made of Am 1004-T61 2 4 3 in. 2 4 rigid and have a negligible thickness. Initially the nut is hand tightened 4 rigidBulon and have a negligible thickness. Initially the10.025 nut hand tightened rigid and have a negligible thickness. Initially the isnut hand tightened somunu başlangıç konumundan inisyukarı doğru 142 C H AThe PTER 4 varsayılmıştır. A X Iat A L the L O Atop Dmagnesium washers and bottom of the are considered to be alloy.The tube has an outer radius of and it is assumed in., 1 snugly; then, using a wrench, the nut is further tightened one-half 2 tightened snugly; then,then, usingusing a wrench, the nut further tightened one-half turn. in. snugly; a wrench, the isnut is further one-half turn.turn. 1 4 rigid a negligible thickness. Initially the nut is hand tightened that both the inner radius of the tube and the radius of the bolt are in. sıkıştırıldığında bulonda meydana gelecek gerilmeyi hesaplayınız. 3 in.and have If the bolt has 20 threads per inch, determine the stress in the bolt. 4 If theIfbolt has 20 per inch, determine the stress in the the bolt hasthreads 20 threads per inch, determine the stress inbolt. the bolt. the nutatisthe further tightened one-half turn. The washers top and bottom of the tube are considered to be 1 1snugly; then, using a wrench, in. in. C H A P T E R EXAMPLE 42 A X I A L L O A4If D 4.8 the bolt has 20 threads per inch, determine the thickness. stress in the bolt. the nut is hand tightened rigid and have a negligible Initially SOLUTION SOLUTION SOLUTION (a) (a) (a) snugly; then, using a wrench, the nut is further tightened one-half turn. Equilibrium. The free-body diagram of aaluminum of the bolt and Equilibrium. free-body diagram of section ofsection the bolt and the Çözüm:SOLUTION Equilibrium. diagram of a section the bolt and The bolt shown inThe Fig.free-body 4–15a is made of a2014-T6 alloy and the is the If the bolt has 20The threads per inch, determine the stress inofthe bolt. a)A D 4 4 Fig. 4–15b, is considered in to relate the force in bolt the bolt tube, Fig.tube, 4–15b, considered order to order relate the force in the 4 tube, Fig.so4–15b, is considered in order to relate theofforce in bolt the tightened itiscompresses aincylindrical tube made Am 1004-T61 PLE 4.8 Equilibrium. The free-body diagram of atube, section bolt and therequires that the Fthe . Equilibrium toFthat in the tube, Equilibrium requires Fb magnesium Fttube . tube, in thein Equilibrium requires Ft .of alloy.The has outer radius of 12 in., and it is assumed b to tan b toFthat SOLUTION tube, Fig. 4–15b, is considered in order to relate the force in the bolt F F(a) F tThe bolt shown in Fig. t t 4–15a isc the made of 2014-T6 aluminum alloy and is of the bolt are 14 in. both inner radius ofF the tube and the radius 3 in. c.that c Magnezyum tüpte basınç, aluminyum bulonda çekme etkisi oluşacaktır. (1) (1) (1) + ©F = 0; F = 0 + ©F = 0; F F = 0of the bolt and the shown in Fig. 4–15a is made 2014-T6 aluminum alloy and is + ©F = 0; F F = 0 Equilibrium. The free-body diagram of a section y b t y b t that in the tube, Equilibrium requires Fbofto F y b t t The washers a cylindrical tubeand made of Am 1004-T61 at the top bottom of the tube are considered to be 1 1 tightened so it compresses d so 4it compresses 1004-T61 in. aFbcylindrical tube, Fig.has 4–15b, is considered 1in order to relate the force in the bolt F4 bin. tube made of Am Fb 2 magnesium alloy.The tube an outer radius of and it is assumed in., rigid and have a negligible thickness. Initially the nut is hand tightened c 1 + ©F FCompatibility. Ftalloy =When 0 and Compatibility. the nut is on the bolt, thebolt, tubetube 2 When thetightened nuttightened is (1) tightened on bolt, the the tube Compatibility. When the nut is on the the y =made bin-the um alloy.The hasin anFig. outer radius of0;2 in., and itbistoassumed The bolttube shown 2014-T6 aluminum that tube, Equilibrium F Fthe 1 # 4–15a t . is thatisboth theofinner radius of1shorten the tube and radius ofbolt therequires bolt are in., Fig. snugly; then, using a wrench, the nut is further tightened one-half turn. 3 in. will shorten and the bolt will elongate Fig. 4–15c. Since the d , d , 4 will shorten and the will elongate Fig. 4–15c. Since d , d , will and the bolt will elongate 4–15c. Since the the d , d F t b t b t b tcompresses h the inner radiussoofitthe tube and the radius ofWhen the bolt are in. tightened a cylindrical tube made 1004-T61 4=of 1 1 1 1(1) 1 1 Compatibility. the nut isAm tightened on the bolt, the tube The washers at the bottom ofthreads theone-half tube considered tothe be + ctop ©F 0; Fare Fdetermine =it 0advances 1 If the bolt has 20 per inch, stress in the bolt. 1and nut undergoes one-half turn, it advances a distance of ( )( in.) = y b t nut undergoes one-half turn, it advances a distance of ( )( in.) nut undergoes turn, a distance of ( )( in.) = in. 2 20 Statikçe belirsiz. 2 20 2 20 = magnesium alloy.The tube hastube an outer radius of 2 in., and it is assumed hers at the top and bottomwill ofrigid the considered tothickness. be 4 and have a negligible Initially the nut is hand tightened shorten the bolt will elongate 4–15c. Since theof these dare d1bolt. , Fig. t , and bthe F along the bolt. Thus, the compatibility displacements 0.025 in. along the Thus, the compatibility of these displacements 0.025 in. along bolt. Thus, the compatibility of these displacements 0.025 in. b both thethickness. inner radius of thethe tube and the radius of the bolt are 4 in. 1 1 d havethat a negligible Initially nutone-half is hand tightened Compatibility. the tightened nut of is (tightened on the bolt, the tube snugly; then, using a wrench, nutWhen is further one-half nut undergoes turn, it the advances a distance requires 2 )( 20 in.) =turn. SOLUTION requires requires The washers the topis and bottom of theone-half tube are considered to be (a) hen, using a wrench,atthe nut further tightened turn. will shorten and the of bolt willdisplacements elongate dt , determine db , Fig. 4–15c. Since the If in. thealong bolt has threads per the stress in the bolt. the20 bolt. Thus, theinch, compatibility these 0.025 1 and the and have negligible thickness. Initially is hand tightened Equilibrium. The free-body diagram cbolt. lt has rigid 20 threads per ainch, determine the stress in nut thethe c 21+ dt =itd0.025 1+ 2 nut in. -of 1+ = d0.025 in. -section c 2 one-half undergoes turn, adabdistance ofthe (12)(bolt in.db - dbof t advances requiresthe t = 0.025 20 in.) = snugly; then, using a wrench, nut is further tightened one-half turn. 4 tube, Fig. 4–15b, is considered in order to relate the displacements force in the bolt along the bolt. Thus, the compatibility of these 0.025 in.the SOLUTION Taking moduli of elasticity fromfrom the table on table the Taking the moduli of elasticity the table on inside the backback (a) If the bolt has 20 threads per inch, determine the stress in the bolt. Taking the moduli of elasticity from the on inside theback inside c to that in the tube, Equilibrium requires F F . d 1+ 2 = 0.025 in. d b t t b requires ON cover, and applying Eq. yields cover, and applying Eq. 4–2, Equilibrium. The free-body diagram of a4–2, section of the bolt and the Ft cover, and applying Eq. yields 4–2, yields (1) ©Fthe =in0;order Fbforce -in. Ft in 0b bolt Taking of elasticity the to table the inside ium. SOLUTION The free-body diagram of athe section of the c+2cand y from tube, Fig.moduli 4–15b, is1bolt considered relate the dton + 0.025 -=back dthe F=t13 Fin.2 in.2 13 in.2 t13 F t cover, and applying Eq. 4–2, yields g. 4–15b, is considered in relate force in Fthe bolt F orderFto that the in the requires = = = b to diagram t . Equilibrium 2the the 2 tightened Equilibrium. Theb free-body of tube, aTaking section of the bolt and 2 the moduli of elasticity the2][6.48110 table on32 the back Compatibility. When is tube 2from 2 32 ksi] 3 inside p[10.5 in.2 - 10.25 in.2 ][6.48110 p[10.5 in.2 - nut 10.25 in.2 ksi] Ft the tube, Ft . Uygunluk denklemi: at in Equilibrium requires p[10.5 in.2 10.25 in.2 ][6.48110 2 bolt, ksi] the F in.2-dinF (b) is(b) tube, Fig. 4–15b, considered order to cover, relate the force the t13 (b)+ c ©Fin (1) F = bolt 04–2, applying Eq. willand shorten the yields bolt=willF elongate Fig. 4–15c. Since the d , y = 0; b t , tand b in.2 b13Fin.2 b13F in.2 1 1 2 3 theFtube, requires Fb to that inBulondaki Ft .=Equilibrium b13 (1)in.220.025 =F0; 0 vep[10.5 in.2 - 10.25 ][6.48110 nut undergoes one-half turn, it advances a distance of ( )( in.) = in. 0.025 -2in.ksi] b - Ftuzama tüpteki kısalma miktarlarının toplamı, somunun yukarı doğru ilerleme 0.025 b 2 20 F 13the in.2 -in.2 2 3 2 3 ) Compatibility. When the nut is the tightened ontin. bolt, the tube 2 2 ksi] 3 p10.25 [10.6110 p10.25 in.2 [10.6110 2 =ksi] p10.25 in.2 [10.6110 2 ksi] bolt. Thus, the compatibility of these displacements (1) + c ©Fy = 0; Fb - Ft = 0 0.025 in. along Fb13 in.2 2 d , Fig. 4–15c. willolacaktır. shorten and the elongate Since32the dt , bolt, p[10.5 in.2 - b10.25 in.22][6.48110 ksi] tibility. When the nut is tightened on the the bolt tubewill miktarına eşit 0.025 in. requires (2) (2) (2) 0.78595F =-25 -1251.4414F (b) 11.4414F 2 30.78595F t = 25 t b 0.78595F = - b= 1.4414F [10.6110 ksi] b nut undergoes one-half turn, itin.2 advances a2distance )(20 in.) rten dCompatibility. elongate 4–15c. Since thebolt, db , isFig. t , and the bolt will Fofb13(t2in.2 When the nut tightened onp10.25 the the tube 1 c 2Eqs. 1 and dt these 1+ = 0.025 in. d 0.025 in. - 2of 2and simultaneously, we along the bolt. Thus, compatibility displacements 0.025 Solving Eqs. 1Since simultaneously, we -get ergoeswill one-half itδadvances ain. distance ofSolving (d12)( in.) =the δtüp 0.025 inelongate veya, shortenturn, the= bolt will 4–15c. the dt ,+and 2 get bulon Solving Eqs. 1 2and simultaneously, we3b2get b ,20Fig. (2) 0.78595F 1.4414F t = 25 1 bp10.25 in.2 [10.6110 ksi] 1 requires alongnut the undergoes bolt. Thus, the compatibility these displacements one-half turn, it of advances a distance of moduli (2)(20 in.) =F 11.22 kip kip kip Taking the of=Felasticity from the table on the inside back b =FF b t=F t==F11.22 = 11.22 b t Solving Eqs. 1 and 2 simultaneously, we get 0.78595Ft = 25 - 1.4414Fb (2) Thus, compatibility of these displacements 0.025 in. along the bolt. Final cover, and applying Eq. 4–2, yields d 1+ cFinal 2the Final = 0.025 in. d t in the The The stresses andb tube are therefore stresses inbolt theinbolt and tube are therefore position position The stresses the bolt and tube are therefore requires Fb =Eqs. Ft 1=and 11.22 kip Solving 2 simultaneously, we get dt = 0.025 in. - db position F in.2inside Fb the kip Taking the moduli of elasticity from table11.22 ont13the back Fb 11.22 kip Final F 11.22 kip 57.2 c = db stresses Ans.Ans. sb therefore =sb = = F = 257.2 ksi ksi 1position + 2 in.bolt - dand ddbt = 0.025 =2 b b==Ft = The in applying the tube are b 2 = kip 3 ksi 2 11.22 cover, and Eq. 4–2, yields d Ans. s = = 57.2 0.025 in. A b 0.025 in. b A 10.25 in.2 ][6.48110 2 ksi] in.2 the moduli of elasticity dfrom the table on0.025 thein.inside back p[10.5 p10.25 in.2 b in.2 2 d bp10.25 t t A Final dt p10.25 in.2 (b) b F 11.22 kip Taking the moduli of elasticity from the table on the inside back b The stresses in the bolt and tube are therefore nd applying Eq. 4–2, yields position Ft13=in.2 Fb13 in.2 db Ans. sb = = 57.2 Ft ksi 11.2211.22 kip Ft kip Eq. 4–2, yields F-t3 kip = 11.22 kip 3= 19.1 0.025applying in. 0.025 Ab ksi ksi p10.25 in.2s2t =sF = = 19.1 Initial dt cover, and 2 2b= = in. Initial 11.22 t 2 2 2 Ft13 in.2 p[10.5 in.2 - 10.25 in.2 ][6.48110 2=p10.25 ksi]2in.2 s=A = p[10.5 ksi A in.2 [10.6110 Initial p[10.5 in.2 10.25 in.2 ] 22ksi] t 10.25 in.2 ] 2 = 19.1Ans. t t d s = = 57.2 ksi 2 position position b (b) b A F 13 in.2 2 p[10.5 in.2 10.25 in.2 ] = t t 0.025 3in.position A F 11.22 kip p10.25 in.2 2 2 d b t in.2 t = are F stresses less the reported yield stress for each material, b13than p[10.5Initial in.2 - 10.25 in.2 2tksi] stresses are less than thet reported yield stress for each material, (c) ][6.48110 s (2) 0.78595F =the 25reported - 1.4414F =2 0.025 =These = 19.1 ksi 3These b stress for each material, in. - These 2 stresses are2 less than yield p[10.5 in.22 (c) - 10.25 2 ksi] (c) in.2 ][6.48110 At 1sp[10.5 2 in.2 3 22 in.2 10.25 ] and 1s (see the inside back cover), and and 2 = 60 ksi 2 = ksi F 11.22 kip and 1s (see the inside back cover), 1s 2 = 60 ksi 2 = 22 ksi position Y al Y alp10.25 in.2 Y mgY mg t [10.6110 2 ksi] Fb13 in.2 and 1sY2mg inside 1sEqs. 2tal =1=and 60 = ksi = we 22 ksi =the 19.1 ksi back cover), and Solving 2 analysis simultaneously, get (see Ys Initial Fb13 in.2 Fig. 4–15 2 is valid. therefore this “elastic” is each valid. Fig. 4–15 3 are therefore thisA “elastic” analysis stresses less than the reported yield stress for material, (c) 0.025 in. p[10.5 in.2 - 10.25 in.22(2) ] t - “elastic” 0.025These in. 2position Fig. 4–15 therefore this analysis is valid. 0.78595F = 25 1.4414F p10.25 in.2 [10.6110 2 ksi] t b 2 3 # #and Fbback = Ftcover), = 11.22 kip for each material, 1s (see the inside and 1sY2al p10.25 = 60 ksi =ksi] 22 ksiare in.2 [10.6110 Y2mg2stresses These less than the reported yield stress (c) Final Fig. 4–15 0.78595Ft = therefore this “elastic” analysis is valid. Solving Eqs. 1 and 2 simultaneously, we get (2)ksi in 25 - 1.4414F b the1sbolt therefore and (see the inside back cover), and 1sYThe 2al b=stresses 60 2mgand = tube 22 ksiare 0.78595F 1.4414F Y(2) positiont = 25 (1) ve (2) denklemlerinin ortak çözümünden, Fig. 4–15 F = F = 11.22 kip therefore bthis “elastic” analysis is valid. t Eqs. 1Solving and 2 simultaneously, we get F 11.22 kip b Eqs. 1 and 2 simultaneously, we get Final db Ans. = = 57.2 ksi b =therefore The stresses in the bolt and tube s are 0.025 in. A position Fb = Ft =dt11.22 p10.25 in.22 kip b F = F = 11.22 kip b t # Fb 11.22 kip Ft ksi 11.22 kip Ans. sses inThe thedstresses bolt and intube the are bolttherefore and tube are s therefore = b b = st2 == 57.2 = = 19.1 ksi Initial 0.025 in. Ab p10.25 in.2 2 dt A p[10.5 in.2 - 10.25 in.22] Fb 11.22 kip Fb 11.22 kip t position == 57.2 ksi 2 = 57.2Fksi Ans. 11.22 kipAns. sb = = sb = t These stresses are less than the= reported Ab 2 (c) Ab Initial p10.25 in.2 p10.25 in.2 s = = 19.1 ksi yield stress for each material, # t 2 2 A and 1s 1s 2 = 60 ksi 2 = 22 ksi (see the inside back cover), and p[10.5 in.2 10.25 in.2 ] t Y al Y mg F Ft position 11.22 kip4–15 11.22 kip t Fig. therefore this “elastic” analysis is valid. s = = = 19.1 ksi These stresses are less t st =(c) = = 19.1 ksi than the reported yield stress for each material, At 2 - p[10.5 in.222] - 10.25 in.22] At p[10.5 in.2 10.25 1sY2alin.2 = 60 ksi and 1sY2mg = 22 ksi (see the inside back cover), and stresses arereported less than the reported yield for each material, ressesThese are than the yield stress eachstress material, Fig.less 4–15 therefore thisfor “elastic” analysis is valid. and 1s (see the inside back cover), and 1s 2 = 60 ksi 2 = 22 ksi Y al Y mg 60 ksi and 1sY2mg = 22 ksi (see the inside back cover), and #54 therefore this “elastic” analysis is valid. e this “elastic” analysis is valid.

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