components, namely, ¢Fx, ¢Fy, and ¢Fz, which are taken tangent, tangent, and normal to the area, respectively. As ¢A approaches zero, so do ¢F and its components; however, the quotient of the force and area will, in general, approach a finite limit. This quotient is called stress, and as noted, it describes the intensity of the internal force acting on a specific plane (area) passing through a point. Gerilme: z !Fz z !Fx !F tyz z !Fy x z tyx y sy sx txy !F 1.3 23 STRESS !A 1.3 the force acting normal to ¢A is ).Stress. Since ¢FzThe is normal to of thethe area intensity force acting normal to ¢A is F1 F2 the normal stress, s (sigma). Since ¢Fz is normal to theF1area 0 txz ¢Fz ¢A x (1–4) ¢F (a)z y 1 1 1.3 x (1–4) ! sz = lim ¢A : 0 ¢A ¢A Stress. The intensity of the force acting normal to is n ¢A as shown in Fig. 1–10a, it is 23 STRESS (b) SyTRESS 23 x y (c) Fig. 1–10 s (sigma). normal stress, Since ¢Fz is normal to the area ifthe it force “pushes” on ¢A it ison 1.3 Scalled TRESS ¢A as shown 23 mal or stress “pulls” in Fig. 1–10a, it is Normal Gerilme: o as tensile stress, whereas if it “pushes” on ¢A it is called ve stress. ¢A is ¢F ¢A is calledz rce acting tangentsto (1–4) 1 z = lim he area ¢A : 0 ¢A shear stress ! tress. Thecomponents, intensity of force acting tangent to ¢A is called stress, t (tau). Here we have shear stress components, z Kayma Gerilmesi: ¢A as shown in Fig. 1–10a, it is mal force or stress “pulls” on ¢F x o0 (1–4) as tensile stress, whereas if it “pushes” on ¢A it isscalled z ¢A ¢Fx ve stress. tzx = lim(1–5) ¢A : 0 ¢A ¢Fy (1–5) Tzx 0a, it is The intensity of force¢F 0 ¢A Tzy y tress. acting tangentx to ¢A is called y lim calledt (tau). Heretzy stress, we=have ¢A :shear 0 ¢A stress components, ! Fig. 1–11 z specifies the orientation of the eatthe axes along which each shear ¢F Genel Gerilme x Hali: in this subscript tzxnotation = lim z specifies the orientation of the called ¢A : 0 the ¢A axes along which each Fig. 1–11, and x and y indicate z shear (1–5) . ¢Fy z he body is further by sz tzysectioned = lim t ¢A : 0 ¢A –10b, and the y–z plane, Fig. 1–10c, t zx State of Stress. If sthe body is further sectionedzy by z lement thatFig. represents tyz allel to of thematerial x!z plane, 1–10b, and the y–z plane,tFig. 1–10c, xz (1–5) hosen point in the body. Thiselement at in this subscript notation zstate specifies the orientation txy oftyxthe en “cut out” a cubic volume of material that represents sx sy e components each face Fig. 1–11,acting and xacting and yTon indicate the axes which zxthe chosen of stress around pointalong in the body.each Thisshear state y Tzy y s then characterized x by three components xacting on each face ! ! ment, Fig. 1–12. 1–11 Fig. 1–12 by of the per force area,Fig. the State of unit Stress. If inthe body is further sectioned Gerilmenin Birimi (SI birim sisteminde): h shear he magnitudes both Fig. normal allel to stress the x!zof plane, 1–10b, and per the y–z plane, Fig.in 1–10c, Since represents a and force unit area, the 2 of material that represents units of newtons per square meter n “cut out” a cubic volume element 1 Pa = 1 N/m z nal Standard or SI system, the magnitudes of both normal and 2 Pa = 1specified N>m 2 is small, andof f stress acting around the chosen point in theper body. Thismeter state ss are inrather the basic units newtons square 3 1 kPa = 1000 Pa 2 acting on each face 110 2, as kilosymbolized by k, then characterized by three components ned by s z This unit, called a pascal 11 Pa = 1 N>m 2 is rather small, and tzy 3 2,work -ering 1109Fig. symbolized by such G, are ment, 1–12. 1–10c, tzx 2 prefixes asused kilosymbolized by k, 1 MPa = 1000 kPa110 = 2, 1 N/mm 6 9 alues of stress.* Likewise, in the resents 2, symbolized by M, or giga2, symbolized by G, are used 110 tyz txz ngineers usually express stress in stress.* is state Since stress represents unitLikewise, area, inin the ent larger, more realistic values the ta tyx of per xy force s(ksi), sy x ounds per square inch where ch face nal Standard or SI of system, magnitudes of both normal and nd-Second system units,the engineers usually express stress in 1 z sz Tzx Tzy x y z Fig. 1–11 sz Tzx z T sz zy t t zy x y zx Fig. 1–11 sx txz txy z x tyx tyz sy y sz tzy Fig.tzx1–12 sx txz txy tyz tyx sy y x Fig. 1–12 y ss specified in units ofper newtons er are square inch (psi) orbasic kilopounds squareper inchsquare (ksi), meter where x the his unit, =called d 1kip2 1000 alb.pascal 11 Pa = 1 N>m22 is rather small, and Fig.as 1–12 ering kilo- 11032, symbolized by k, n thework prefixes such 6 2 -3 2, symbolized M, or 110in92, symbolized by G, are used mal and mm , where 1 mm by = 10 m. gigaHowever, nominator of a fraction and therefore it isof stress.* Likewise, in the ntstress larger, more realistic values esmeter is expressed in units of N>mm2, where 1 mm = 10-3 m. However, in 2 m = 1 MPa. d-Second of units, engineers usually express stress itinis all, and aresystem m, prefixes not allowed in the denominator of a fraction and therefore rthe square (psi) 2or= kilopounds equivalent 1 MN>m2 = 1 per MPa.square inch (ksi), where by k, inch1 N>mm d 1kip2 re used = 1000 lb. in the !8 n of m mation 24 CHAPTER 1 STRESS 1.4 Average Normal Stress in an 1 Axially Loaded Bar Eksenel Yüklü Bir Çubukta Ortalama Gerilme In this section we will determine the average stress distribu the cross-sectional area of an axially loaded bar such as the Fig. 1–13a. This bar is prismatic since all cross sections P throughout its length. When the load P is applied to the b 1.4 Average Normal Stress in an centroid of its cross-sectional area, then the bar will defo throughout the central region of its length, as shown Axially Loaded Bar provided the material of the bar is both homogeneous and Region of In this section we will determine the average uniform stress distribution Homogeneous acting on material has the same physical and mechan deformation the cross-sectional area of an axially loaded bar such as the one shown in throughout its volume, and isotropic material has these sa of barsections are Fig. 1–13a. This bar is prismatic since all cross the same all directions. Many engineering materials may be app throughout its length. When the load P is applied to the bar in through the being both homogeneous and isotropic as assumed he centroid of its cross-sectional area, then the bar will deform uniformly throughout the central region of its length, as shown in example, Fig. 1–13b,contains thousands of randomly oriented crystals provided the material of the bar is both homogeneous and isotropic. millimeter of its volume, and since most problems involvin Homogeneous material has the same physical and mechanical properties have a physical size that is very much larger than a sing throughout its volume, and isotropic material has these same properties above assumption regarding its material composition is qu in all directions. Many engineering materials may be approximated as P P ! being both homogeneous and isotropic as assumed here. Steel, for Note that anisotropic materials such as wood have differ (a)thousands of randomly (b) example, contains oriented crystals in cubic directions, and although this is the case, lik ineach different Prizmatik çubuk (tüm enkesitlerin geometrisi aynı) millimeter of its volume, and since most problems involving this material anisotropy is oriented along the bar’s axis, then the bar wi have a physical that merkezinden is very much larger than a single crystal, the P yüküsize ağırlık tesir ediyor above assumption regarding its material composition is quiteuniformly realistic. when subjected to the axial load P. Çubuğun yapıldığı malzeme homojen ve izotropik. Note that anisotropic materials such as wood have different properties in different directions, and although this is the case, like Average wood if the Normal Stress Distribution. If we p Homojen malzeme: Malzemenin kapladığıalso tüm hacim aynı fiziksel ve mekanik özelliklere anisotropy is oriented along the bar’s axis, then the bar will through deform the bar, and separate it into two parts, then equilib uniformlysahip when ise subjected to the axial load P. malzeme homojendir. the resultant normal force at the section to be P, Fig. 1–1 uniform deformation of the it is necessary that th Averageİzotropik Normal malzeme: Stress Distribution. we pass a tüm section Malzemenin Ifözellikleri doğrultuda aynı ise, material, yani kuvvetlerin through the bar, and separate it into two parts, then equilibrium requires be subjected to a constant normal stress distribution, Fig. 1– etkime doğrultusuna özellikler göstermiyorsa the resultant normal force at thegöre section to be P,farklılık Fig. 1–13c. Due to the malzeme izotropiktir. z P uniform deformation of the material, it is necessary that the cross section Ortalama normal gerilme dağılımı be subjected to a constant normal stress distribution, Fig. 1–13d. z P P !F " s!A !F " s!A s P x Internal force s P y Internal force !A !A Cross-sectional area Cross-sectional area y x x External force y 1.4 AVERAGE NORMAL STRESS IN AN AXIALLY LOADE External force P P P ! As a result, each small area ¢A onPthe cross section is subjected to a (c) (d) force ! ¢F = s ¢A, kuvveti and the tesir sum of these forces acting overbirbirlerinden the entire (c) ve Herbir küçük delta A ediyor bu kuvvetler Fig.alanına 1–13 cross-sectional area must be equivalent to the internal resultant force P Fig. 1–13 farklı. Ağırlık merkezine doğru kuvvetlerin büyük beklenir,¢F çünkü eksenel at the bu section. If we daha let ¢A and therefore , then, dış : dAolması : dF recognizing s is constant, we have yük ağırlık merkezinden tesir ediyor. + c FRz = ©Fz; L dF = s dA LA P = sA !9 (d) AVERAGE NORMAL TRESS AN AXIALLY AR25 As a 1.4 result, each1.4 theLINOADED cross is Bsubjected to 25 a ¢A AVERAGE Nsmall ORMAL Sarea TRESS IN AN on ASXIALLY Bsection AR LOADED force ¢F = s ¢A, and the sum of these forces acting over the entire cross-sectional mustis sult, each¢A small the cross section subjected to a to the internal resultant force P ¢A on all area on area the cross section isarea subjected tobe a equivalent theof section. If acting we entire letover ¢A dA and therefore ¢F : dF1 , then, , and theatsum these the:entire s ¢A d= the sum of these forces actingforces over the onalbearea must berecognizing tosthe internal resultant force P ust equivalent toequivalent the internal resultant force P have is constant, we ction. If we let ! and and! ¢F therefore , then, normal gerilme: ¢A : dA ve ¢F : dF let ¢A therefore then, : dA : dF, olursa, ortalama g swe is constant, we have ant, have L + c FRz = ©Fz; ©Fz; dF = dF = s dA s dA LA L LA L P = sA P = sA P s = A! P s = A dF = 1 s dA LA P = sA (1–6) P P s(1–6) = A (1–6) P Here P P the cross-sectional age at point on area tressnormal at anystress point s onany cross-sectional area s ! — the cross-sectional area s! — =theaverage A A normal stress at any point on nal resultant which acts normal force,normal which force, acts through the through centroid the of centroid of P P=isinternal normal force, which acts through the centroid of ross-sectional area. determined using the method of al area. P is determined using theresultant method of s s cross-sectional area. P is determined using the method of ons and the ofthe equilibrium equations ofequations equilibrium the equations of equilibrium -sectional area of thesbar where sand is determined ea of the bar where issections determined A passes = cross-sectional area of of thethebar where s is determined he load P crossoadinternal P passes through the through centroidthe of centroid the crosse uniform stress distribution will produce zero moments about ress distribution will produce zero moments about Since the internal load P passes through the centroid of the crossy axes passing this1–13d. point,To Fig. 1–13d. ng through thisthrough point, Fig. show this,Toweshow this, we section the uniform stress distribution e moment of P about each axis to be equal to the moment of will produce zero moments about f P about each axis to Pbe equal the moment of the andnamely, y axes passing through this point, Fig. 1–13d. To show this, we P distribution about thexaxes, about the axes, namely, each axis to be equal to the moment of Tension the moment of P about Compression !require the stress distributionFig. about the axes, namely, 1–15 y dF =© Mxy; dF0 == ys dA== s ysydA dA= s y dA L LA LA LA LA L AA In other words, the two normal stress components on the element must Normal Kuvvet Diyagramı inxmagnitude but opposite direction. ThisdA is referred as 0= =- s in yx dF = ys = s yto dA R2xdA x; dA 0equal ==-- 1M My; be dF == -=©-M xs dA =© x dF xs s x dA L A L A L A LAL LA L LA LA uniaxial stress. A A Eğer eksenel yüklü bir elemanda farklı yükleme bölgeleri var ise ortalama gerilmelerin The previous analysis applies to members subjected to either tension maksimum olduğu bölgenin tespiti için normal kuvvet değişimin gösteren Normal Kuvvet ationssatisfied, arecompression, indeed satisfied, definition or as shown Fig. As a graphical ndeed since by the 0centroid, =1–15. -of the 1M =since © Mby ; in x centroid, dF = - interpretation, xs dA = - s thex dA R2ydefinition yof x dA = 0. (See Appendix A.) A =E S0.Smagnitude (See Appendix A.) LA P is equivalent A LA s to the volume of the internal resultant force sL 1 T0Rand Diyagramının çizilmesi gerekebilir. 28 C H A P T E R 1 S T R E S S base2. !A under the stress diagram; that is, P = s A 1volume = height * !A ium. Itapparent should These be apparent that only a normal stress exists ould be that a normal stress exists Furthermore, asonly a consequence of the balance of moments, this resultant equations are indeed satisfied, since by definition of the centroid, mall volume of material located each element of element material located at each on point on passes through the of thisat= volume. y dA =centroid 0 and x point dA 0. (See Appendix A.) s 1 1 EXAMPLE 1.6 section of an axially loaded bar. If we consider vertical an axially loaded bar. If we consider vertical ! 1 Although we have developed this analysis for prismatic bars, this m of theFig. element, Fig. 1–14, apply theofequation of lement, 1–14, then apply then the equation assumption can be relaxed somewhat to include bars that ise have amm slight Şekilde verilen çubuğun genişliği 35 mm, kalınlığı 10normal dir.stress Şekilde verilen yükler altında librium, Equilibrium. It should be apparent that only a exists The bar in Fig. 1–16a has a constant width of 35 mm and a thickness taper. ForThe example, it Fig. can be shown, using the morewidth exact analysis of the bar in 1–16a has a constant of 35 mm and a thickness on any oluşacak small volume element ofofmaterial located at maximum each point on normal stress in the bar 10 mm. Determine the average ortalama gerilmeyi hesaplayınız. theory of çubukta elasticity, that for a tapered bar of rectangular cross section, for s1¢A2 = 0maksimum s¿stress s1¢A2 - s¿1¢A2 =-mm. 0s¿1¢A2 s¿ normal of 10 Determine the maximum the bar whenaverage it isbar. subjected to the loadingin shown. the cross section of an axially loaded If we consider vertical which the angle between two adjacent sides is 15°, the average normal s = s¿ 1–14 s =ass¿when Fig. A 1–14 it is subjected to the loading shown. equilibrium element, Fig. 1–14, then apply the of B equation C 4 kN 9 kN D stress, calculated by of s =the P>A, is only 2.2% less than its valueFig. found 12 kN 22 kN force equilibrium, from the theory of elasticity. B C 4 kN 9 kN A D to as a 12 kN mm the Maximum Normals1¢A2 Stress. In our analysis = 0; © FzAverage - s¿1¢A2 = 0 35both 9 kN 22 kN 4 kN internal force P and the cross-sectional area9A along the (a) kNwere constant 4 kN s =normal s¿ stress s = P>A is 35 mm longitudinal as a result the ! axis of the bar, and 12 (a) kN also constant throughout the bar’s length. Occasionally, however, the bar PAB ! 12 kN Çözüm: to several external loads along its axis, or a change in its may be subjected 9 kN cross-sectional area mayyükler occur.altında As a result, the midir? normal stress within the Çubuk verilen dengede 12 kN PBC ! 30 kN 12different kN PAB ! 12and, kN if the maximum bar could be from one section to the next, 9 kN Maksimum hesabıthen niçinitgereklidir? average normal stressortalama is to begerilme determined, becomes important 9 kN to find the location where the ratio P!A is a maximum. To do this it PisCD ! 22 kN necessary to the internal force P at various sections along the 12determine kN PBC ! 30 kN (b) bar. Here it may be helpful to show this variation 9 kN by drawing an axial or normal force diagram. Specifically, this diagram is a plot of the normal P (kN) force P versus its position x along the bar’s length. As a sign convention, PCD ! 22 kN 22 kN P will be positive if it causes tension in the member, and 30 negative if it 22 causes compression. Once the internal loading the bar is (b) throughout s¿ Fig. 1–14 22 kN !10 !A (a) 12 kN 12 kN 12 kN B A 35 mm C 9 kN 9 kN PAB ! 12 kN 4 kN 4 kN 9 kN (a) 35 mm 9 kN 12 kN 12 kN P (kN) 22 kN P (kN) PAB ! 12 kN 9 kN 30 PAB ! 12 kN 22 9 kN PCD ! 22 kN 12 22 kN 9 kN PBC ! 30 kN (b) 9 kN PBC ! 30 kN 12 kN PCD ! 22 kN 22 kN (b) 4 kN (a) PBC ! 30 kN 12 kN 12 kN 22 kN D x (c) 9 kN SOLUTION 22 kN Internal Loading. By inspection, the internal axial forces in regions 22 kN (b) PCD ! 22 kN ! 30 AB, BC, and CD are all constant yet have different magnitudes. Using 22 (b) the method of sections, these loadings are determined in Fig. 1–16b; 12P (kN) and the normal force diagram which represents these results graphically x P (kN) is shown in Fig. 1–16c. The largest loading is in region BC, where (c) 30 PBC = 30 kN. Since the cross-sectional area of the bar is constant, the SOLUTION 22 30 largest average normal stress also occurs within this region of the bar. 12 22 Internal Loading. By inspection, the internal axial forcesxin regions ! 12 Average Normal Stress. Applying Eq. 1–6, we have (c)have different AB, BC, and CD are all constant yet magnitudes. Using PCD ! 22 kN x 10 mm these loadings are determined in Fig. 1–16b; the method of sections, (c) SOLUTION 3011032 N PBC and the normal force diagram which represents these results graphically = = = 85.7 MPa sinBCregions Ans. SOLUTION Loading. ByThe inspection, internal forces 10.035 m210.010 m2 isInternal shown in Fig. 1–16c. largest30the loading is inaxial region BC, whereA kN BC, and are allcross-sectional constant yet have magnitudes. Using Loading. By inspection, axial forces PAB, kN. CD Since the areadifferent ofthe theinternal bar is constant, the in regions BC = 30 Internal NOTE: The stress distribution the method sections, these loadings are determined in Fig. 1–16b; 35of mm AB, BC, and CD are all constant yet have different magnitudes. Using acting on an arbitrary cross section of largest average normal stress also occurs within this region of the bar. 85.7 MPa the bar within region BC is shown in Fig. 1–16d. Graphically the volume and the normal force diagram which represents these results graphically the method of (d) sections, these loadings are determined in Fig. 1–16b; ! the Average Stress. Applying 1–6,represents we “block”) represented by this distribution of stress is equivalent to is shownNormal in Fig.normal 1–16c. force The largest loading is(or inhave region BC, where and diagramEq. which these results graphically AkN; VERAGE Nwhere ORMAL TRESSMPa2135 IN AN Amm2110 XIALLY LOADED Fig. 1–16 the 30region that is, 30 kN =S185.7 mm2. BAR PBC = 30is kN. Since area ofloading theload bar1.4 constant, the shown inthe Fig.cross-sectional 1–16c. The3 largest isisofin BC, 2 N 30110 P largest average normal alsocross-sectional occurs within this region of the bar. BC stress the PBC the1.4 bar is constant, the STRESS IN AN AXIALLY LOADED BAR AVERAGE NORMAL = = area 85.7of MPa sBC= =30 kN. Since Ans. A 10.035 m210.010 m2 largest average normal stress also occurs within this region of the bar. ! Average Normal Stress. Applying Eq. 1–6, we have EXAMPLE 1.7 Normal Average Stress. Eq. 1–6, we section have of NOTE: The stress distribution acting Applying on an arbitrary cross 3 EXAMPLE 1.7in 2 N Graphically the volume 30110 PBCis shown the bar within BC Fig. 1–16d. ! sregion = Ans. BC = 3 = 85.7 MPa 2 N is equivalent to 30110 A by this 10.035 m210.010 m2 PBCdistribution “block”) represented of stress The(or 80-kg lamp is supported by two rods AB and BC asasshown in The 80-kg lamp supported by two rodsAB BC shown in =kN =ağırlığındaki =AB 85.7 MPa sis,BC30 Ans. vasıtasıyla Şekilde verilen 80Aiskg lamba veand BC çubukları asılı bir şekilde the load of 30 kN; that = 185.7 MPa2135 mm2110 mm2. 10.035 m210.010 m2 Fig. 1–17a. If AB has a diameter of 10 mm andsection BC has a diameter of 30 kN Fig. 1–17a. AB has a diameter ofon10 and BC has aofdiameter of NOTE: IfThe stress distribution acting anmm arbitrary cross 8region mm, determine theinaverage normal in each rod. taşınmaktadır. çubuğunun çapı 10stress ve BC çubuğunun çapı 8 mm ise herbir çubuktaki thedetermine bar within BC is AB shown Fig. 1–16d. Graphically the volume 8 mm, the average normal stress inmm each rod. NOTE: The stress distribution acting on an arbitrary cross section of Pa (or “block”) represented by this distribution of stress is equivalent to volume the bar within region BC is shownhesaplayınız. in Fig. 1–16d. Graphically the ortalama normal gerilmeyi the load (or of 30 kN; that is, 30 kN = 185.7 MPa2135 mm2110 mm2. “block”) represented by this distribution of stress is equivalent to y the load of 30 kN; that is,A30 kN = 185.7 MPa2135 mm2110 mm2. C FBA FBC y A C FBA FBC 5 5 60! 60! 5 B 60! 4 B 5 3 4 60! 3 x 4 B x B 80(9.81) " 784.8 N (a) (b) Fig. 1–17 ! SOLUTION (a) Internal Loading. We must first determine the axial force in each rod. A free-body diagram of the lamp is shown in Fig. 1–17b. Applying Fig. 1–17 the equations of force equilibrium, + ©F = 0; : x 80(9.81) " 784.8 N (b) FBC A 45 B - FBA cos 60° = 0 SOLUTION + c ©F = 0; F A 3 B + F sin 60° - 784.8 N = 0 y BC 5 BA Internal Loading. We must determine axial FBC first = 395.2 N, = 632.4 N force in each FBA the rod. A free-body diagram of the lamp is shown in Fig. 1–17b. Applying Byforce Newton’s third law of action, equal but opposite reaction, these the equations of equilibrium, + ©F = 0; : x !11 forces subject the rods to tension throughout their length. Average Normal Stress. Applying Eq. 1–6, 4 FBC A 5 B - FBA cos 60° = 0 FBC 395.2 N 1 3 3 4 8.05 MPa 29 rod. A free-body diagram of the lamp is shown in Fig. 1–17b. Applying the equations of force equilibrium, SOLUTION + ©F = 0; Internal We60° must FBC Loading. : = 0first determine the axial force in each A 45 B - FBA cos x rod. A3 free-body diagram of the lamp is shown in Fig. 1–17b. Applying FBCequations + c ©Fy = 0; the 60° equilibrium, - 784.8 N = 0 A 5 B + FBAofsin force FBC = 395.2 N, FBA = 632.4 N + FBC A 45 B - FBA cos 60° = 0 : ©Fx = 0; 4 AVERAGE NORMAL STRESS IN AN AXIALLY L3OADED BAR 29 By Newton’s third law = of 0;action, FBCequal + c ©F FBAopposite sin 60° -reaction, 784.8 N these = 0 A 5 B + but y forces subject the rods to tension throughout their length. FBC = 395.2 N, FBA = 632.4 N ! Stress. Applying Average Normal Eq. 1–6, 1 8.05 MPa By Newton’s law of FBC third395.2 N action, equal but opposite reaction, these sBC =the rods to tension = 7.86 MPa and BC as shown in =subject forces throughout theirAns. length.8.05 MPa ABC p10.004 m22 BC has a diameter of Average Normal Stress. Applying Eq. 1–6, each rod. FBA 632.4 N = MPa sBA = Ans. FBC 2 = 8.05 395.2 N ABA sp10.005 m2 = = 7.86 MPa Ans. BC = ! 2 ABC p10.004 m2 y acting over a cross C NOTE: The average normal stress distribution FBA 632.4 N FBA is shown in Fig. section of rod AB and this cross BC = = at a point on = 8.05 MPa sBAF1–17c, Ans. ABAas shown p10.005 m221–17d. (d) section, an element of material is stressed in Fig. 32 8.05 MPa 8.05 MPa 632.4 N (c) C H A P T E R 51 3 S T R E S S NOTE: The 4average normal stress distribution acting over a cross B 60! section of rod AB is shown in x Fig. 1–17c, and at a point on this cross section, an element of material is stressed as shown in Fig. 1–17d. 32 CHAPTER 1 632.4 N (d) (c) 1.5 Average Shear Stress STRESS Shear stress has been defined in Section 1.3 as the stress compo ! acts in the plane of the sectioned area. To show how this s 1.5 Average Shear Stress 1 develop, consider the effect of applying a force F to the bar in F A Fig. 1–17 If stress the are considered rigid, and F iscomponent large enough, 1.5supports Shear Stress Shear has Average been defined in Section 1.3 as the stress that it w FC 1 Ortalama Kayma Gerilmesi material ofthe thesectioned bar to deform failhow along planes acts the in the plane of area. Toand show thisthe stress can iden B Shear stress has been defined in Section 1.3 as the stress component that F develop, consider of applying a force to the bar in Fig. 1–20a. AB and CD.the A effect free-body diagram of Fthe unsupported center se D A acts in the plane of the sectioned area. To show how this stress can he axial force in each C If the supports are considered rigid, and F is large enough, it will cause thedevelop, bar, consider Fig. 1–20b, indicates the V = F>2 the effect of applyingthat a force F toshear the bar force in Fig. 1–20a. n Fig. 1–17b. Applying A the material of the bar to deform and fail along the planes identified by C If the supports aresection considered rigid, and F is large enough, it will cause applied at each to hold the segment in equilibrium. The B 1.5 Average Shear Stress AB andthe CD. A free-body diagram of and the fail unsupported center segmentbyof D material of the bar to deform along the planes identified stress distributed overthe each sectioned area that develops t (a) B the shear bar, indicatesdiagram that shear force Vcenter = F>2 must ABFig. and 1–20b, CD. A free-body of the unsupported segment ofbe D force isstress defined by Shear stress has been defined in Section 1.3 as the component applied at each section toindicates holdthat thethat segment in equilibrium. Themust average the bar, Fig. 1–20b, the shear force V = F>2 be e Shear Stress 32 F 80(9.81) C H A"P T784.8 E R 1N S T R E S S (b) N = 0 F applied at eachthis section to hold the segment in equilibrium. Thethis average acts in the show how stress cansectioned shearTostress distributed over each area that develops shear (a) of the sectioned area. ! plane shear stress distributed over each sectioned area that develops this shear 2.4 N (a) develop, consider the effect of applying a force F to the bar in Fig. 1–20a. force is defined by defined in Section 1.3 as the stress component that is defined by Fare considered rigid, and F isforce If the supports he sectioned area. To show Fhow this stress canlarge enough, it will cause pposite reaction, these theapplying materialaof the bar deform along the planes identified by effect of force F totothe bar inand Fig.fail 1–20a. eir length. AB and CD. A free-body diagram of the unsupported center segment of nsidered rigid, and F is large enough, it will cause V the bar, Fig. 1–20b, indicates that the shear force V = F>2 must be tavg = MPa identified by r to deform and fail along the8.05 planes applied atAns. each section to hold the segment in equilibrium. The average A VV V 8.05 7.86 diagram MPa MPa V ody of the unsupported center segment of t = (1–7) t = (1–7) avg avg shear stress distributed AA V over each sectioned area that develops this shear V V ndicates that (b) ! the shear !V = F>2 must be force is defined byV force (b) (b) in equilibrium. The average n8.05 to MPa hold the segment Ans. d over each sectionedFFarea thatF develops this shear on acting over a cross t a point on this cross own in Fig. 1–17d. tavg Here Here Here (d) t tavg avg tavg Here 632.4 N (c)V tavg = average shear stress (1–7)at the section, which is assumed to be the = average shear at section, whichtoisbe assumed to tavg t =avg average shear stress atstress the section, is assumed the A same at each point located on the thewhich section (c) ! ! same at each point located on the section V = same internalat resultant shear force on theon section each point located the determined section from (c) (c) Fig. 1–20 the equations of equilibrium V V = internal resultant shear force on the section determined from V: Kesitte tesir eden kesme kuvveti = (1–7)V = internal resultant shear force on the section determine Fig. 1–20 area at the A = the equations ofsection equilibrium A Fig. 1–20 the equations of equilibrium A: Kesit alanı A = area at the section tavg = area at the section A The=distribution of average shear stress acting over the sections is Ortalama kayma gerilmesinin yönü kesitteki kesme kuvveti ile aynıdır. shown in Fig. 1–20c. Notice that tavg is in the same direction as V, since The which distribution average shear stress acting sectionstois tavg = average shear stress at the section, is assumed tocreate be the the shear stressofmust associated forces all ofover whichthe contribute in Fig. 1–20c. Notice that t is in the same direction as since the internal resultant force V at the section. same at each point located onshown the section The distribution of average shear stress acting V, over the se avg !12 or directto The loading case discussed here is an example of simple the shear stress must create associated forces all of which contribute V = internal resultant shear force on shown the section determined in Fig. 1–20c.from Notice that tavg is in the same direction as shear, resultant since the shear caused the direct action of the applied load F. forceisV at thebysection. the equations of equilibrium the internal the shear stress must create allconnections of which cont This type of shear often occurs in various types forces ofofsimple r stress at the section, which is assumed to beThe theloading case discussed here isassociated an example simple or direct A = area at the section t 1.5 1.5 z t tyz !z z ! Kayma Gerilmesi t Dengesi ytzy Section planetzy z 33 AVERAGE SHEAR STRESS 1.5 1.5 Section plane z AVERAGE SHEAR STRESS AVERAGE SHEAR STRESS 33 1 t Section plane t Section tzy plane Her ne kadar yukarıdaki başlıktan kayma gerilmelerine aitt bir denge denklemi yazılacağı t t !z !x yz t tzy 1.5 t 1 33 AVERAGE SHEAR STRESS anlaşılabilirse de aslında yazılacak denge için olacaktır. !tdenklemi kayma kuvvetleri !z t tyz z t¿yz t¿zy y !y (a) !x (b)t Pure shear tyz !z Fig. 1–21 Fig.!1–21 x y (b) 1–21 (b) ! Fig. t 1 t Pure shear t t t Pure shear (a) (a) t¿yz t y ! Section plane t (a) !y x Fig. 1–21 x t t t¿tzy ! xzy zy !y ! y yz t¿zy !y !x t¿ x t y Pure shear! !x tyz !z t¿ t¿yz tyz !z t¿yz Pure shear (b) t (b) Fig. 1–21 Shear Stress Equilibrium. Figure 1–21a shows a volume element rium. Figure 1–21a a located volume element !y takenshows of material at a point on1–21a the surface a sectioned Pure shear aof Shear Stress Equilibrium. Figure shows volume elementarea t¿zy t Shear Stress Equilibrium. Figureshows 1–21a shows a volume element x Stress Equilibrium. Figure a volume element int located on the surface ofthe asurface sectioned area !of is to aonshear stress andarea equilibrium tzythe material taken at1–21a a point located on surface ofmoment a sectioned area of material which taken at asubjected point located of a. Force sectioned (a) (b) rial taken at a requires point islocated on to thestress surface of aton sectioned area the shear acting this face of the element to be which subjected a shear stress . Force and moment equilibrium zy which is subjected to a shear and stress tmoment equilibrium hear stress Force equilibrium tshear zy. Force and moment zy.gerilmesi s subjected to a stress . Force and moment equilibrium t ! kesit düzleminde tesir eden kayma gerilmesidir. Dengeleyici zy Fig. 1–21 requires the shear stress acting on this face of the element to be accompanied by shear stress acting on three other faces. To show this we kayma requires the shear stress acting on this face of the element to be ss the shear stress acting on this face of the element to be acting on this face of the element to be accompanied byacting shear stress acting faces. show this we accompanied byfirst shearconsider stress on three otheron faces. To yother show this weTo will force equilibrium inthree the direction. Then gerilmelerinin şiddet ve yönleri ise denge denklemleri ile belirlenecektir. aniedwill byfirst shear stress acting on three Tothe show this we Then consider force equilibrium inother theFigure y faces. direction. Then will first consider force equilibrium in y direction. Shear Stress Equilibrium. 1–21a shows a volume element ess acting on three other faces. To show this we force on Then t consider force equilibrium y direction. of material taken atina the point located the surface of a sectioned area quilibrium inis the yforce direction. Then which subjected to a shear force stress tzy. Force and moment equilibrium rce force requires thestress sheararea stress stress actingarea on this face of the element to be stress area accompanied by shear stress acting on three other faces. To show this we stress œ 1¢x force¢y2 equilibrium in¢y the=y0-direction. ©Farea = 0;tzy1¢x ttzy ¢y2 tœ ¢x Then ¢y = 0 yconsider ©Fy = will 0; first ¢x œ zy zy ©Fy = 0; tzy1¢x ¢y2 - tzyœ ¢x ¢y = 0 œ tzy = œ tzy œ tforce zy = tzy ! tzy1¢x ¢y2 - tzy ¢x ¢y = 0tzy = tzy œ œ œ stress area tforce =manner, tzy In a similar In manner, equilibrium inkuvvet the z direction = tyz . bulunur. zy a similar force equilibrium inyields the z!tyzdirection yields tyz = œ tyz . zIn ekseni doğrultusundaki dengesinden a similar manner, force equilibrium in the z direction yields t = t . yz yz Finally, taking moments about the x axis, Finally, taking moments aboutœthe the x axis, Finally, taking moments about x axis, œ œ ©F = 0; t 1¢x ¢y2 t ¢x ¢y = 0 x eksenine göre moment denkleminden: zy y zy ilar manner, in=the yields tyz = tyz . ¢x ¢y2 -force tzyequilibrium ¢x ¢ymoment 0 z direction œ tzy =moment tmoment zy taking moments about the x axis, œ force arm tzy = tzy œ force in the arm force arm In a similarmoment manner, force equilibrium z direction yields tyz = tyz . stress area Finally, taking moments about the x axis, stress area stress area arm œ ©Mx = 0; force -tzyz 1¢x ¢y2 ¢z + tyz1¢x ¢z2 ¢y = 0 equilibrium in the direction moment yields tyz = tyz . stress ©Mxxarea 0; 1¢x¢y2 ¢y2¢z¢z+ +tyzt1¢x ¢z2 tzy =-t == 0; ttzyyzzy 1¢x ¢z2 ¢y¢y = 0= 0 yz1¢x bout the x axis, force arm tzytzy= =tyztyz so that stress 0; -! tzy1¢x ¢y2 ¢z + tarea 1¢x ¢z2 ¢y = 0 yz œ œ so that that tzy = tzy = tyz = tyz = t tzy = tyz moment ©Mx = 0; -tzy1¢x ¢y2 ¢zœ œ+ tyz1¢x ¢z2 œ œ¢y = 0 tzy ==tzy = =tequal = t= t and In other words, all four shear stresses must yztyz yz tzy thave tyz= =tmagnitude zy olarak:or away from tzyeach = tyzother at opposite edges of be directed Sonuç either toward orcethe element, arm In other words, four shear stresses must have equal magnitude andand In other all shear stresses must have equal magnitude œ words, œ 1–21b. This isall referred to the complementary property so that tFig. = tzy =either tyz =toward tfour tasaway yz = or zydirected be from each other at opposite edges of of ! of shear, and the either conditions shown Fig.œ 1–21, material is opposite edges beunder directed toward or in away from the each other at œ sr words, area t = t = t = t = t the element, Fig. 1–21b. This is referred to as the complementary property zy yz zy yz subjected tothe pure shear. element, Fig. 1–21b. This equal is referred to as theand complementary property all four shear stresses must have magnitude of shear, and under the stresses conditions shown in Fig. 1–21, the material is In other words, all four shear must have equal magnitude of shear, andfrom under theother conditions shownedges in Fig. 1–21,and the material is ted either toward or away each at opposite of subjected to toward pure shear. be directed either or away from each other property at opposite edges of subjected to pure shear. ment, Fig. 1–21b. This is referred to as the complementary ¢y2 ¢z +Fig. tyz 1¢x ¢yto=as the 0 yr,1¢x thethe element, 1–21b. This¢z2 isin referred and under conditions shown Fig. 1–21, thecomplementary material is property of shear, and under the conditions shown in Fig. 1–21, the material is ed to pure shear. tzy = tyz 0; 33 AVERAGE SHEAR STRESS area subjected to pure shear. œ œ = tzy = tyz = tyz = t hear stresses must have equal magnitude and or away from each other at opposite edges of !13 36 36 CHAPTER 1 STRESS CHAPTER 1 STRESS S T RCEHS SA P T E R 136 S T R E S S 1 CHAPTER 1 EXAMPLE 1.11 1 EXAMPLE 1.11 1 1.11 AMPLE 1 EXAMPLE 1.11 1 ! STRESS If the wood joint in Fig. 1–23a has a width of 150 mm, determine the If the wood Fig. 1–23a along has a width of 150 mm, determine the average shearjoint stressindeveloped shear planes a–a and b–b. For the wood joint inaaverage Fig. 1–23a hasmm, a width of 150ofthe mm, determine the a–a If the wood jointIfin Fig. 1–23a haseach width of 150 determine shear stress developed along shear planes and b–b. For plane, represent the state stress on an element of the material. If the wood joint 150 in Fig.mm 1–23aolduğuna has a width göre, of 150 mm, Aşağıda verilenaverage şekildeki kalınlığı a-a determine ve b-b the shear elemanların stress developed along shear planes and For average shear stress developed along shear planes and b–b. For each plane, represent thedeveloped state ofa–a stress onb–b. an element theand material. average sheara–a stress along shear planesofa–a b–b. For each plane, represent the state of stress on an element of the material. each plane, represent the state of stress on an element of the material. düzlemlerinde meydana gelen ortalama kaymarepresent gerilmelerini düzlemlerdeki each plane, the statehesaplayınız. of stress on anBu element of the material. gerilme durumlarını malzemeye ait bir elemandaki gerilme hali ile gösteriniz.F a 1 SSTTRREESSSS 6 kN a a a 6 kN 6 kN a 6 kN 6 kN 6 kN b b b mm, determine the 11ig. 1–23a has a width of 150 a a a 6 kN b b a b b F b 6 kN 6 kN 6 kN 6 kN F b 0.1 m 0.125 bm developed along shear planes a–a and b–b. For (b) (b) 0.1 m of0.125 If the the wood joint joint in Fig. Fig. 1–23a has has aa width width 150mmm, mm, determine determine the the wood in 1–23a of 150 the state of stress ! If on an element of the material. 0.1 m 0.125 m 0.1 m 0.125 m (a)0.1 m 0.125 m average shear shear stress stress developed developed along along shear shear planes a–a a–a and and b–b. b–b. For For average planes (a) each plane, represent the state of stress on an element of the material. Fig. 1–23 çizelim: each plane, represent the state of stressçizerek on an element the material. (a) diyagramını Çözüm: cisim kuvvetofdağılımını (a) Serbest (a) F Fig. 1–23 Fig. 1–23 F F F a F 6 kN F 6 kN 6 kN 6 kN F (b) F (b) (b) Fig. 1–23 Fig. 1–23 SOLUTION FF SOLUTION Internal Loadings. Referring to the free-body diagram of the 6 kN SOLUTION aa SOLUTION aa F SOLUTION Internal Loadings. Referring to the free-body diagram of the member, Fig. 1–23b, ! kN 66of kN Internal Loadings. Referring to diagram the free-body of the 6 kN Internal Loadings. Referring to the free-body thediagram member, Fig. 1–23b, 6 kN 6 kN Internal Loadings. Referring to the free-body diagram of the 6 kN F F member, Fig. 1–23b, + member, Fig. 1–23b, Yatay kuvvet dengesini yazalım: kN - F - F = 0 F = 3 kN (b) : ©F x = 0; Fig.61–23b, member, bb + ©Fx = 0; bb 6 kN - F - F = 0 F = 3 kN : +6 ©F + ©Fx = 0; 6= kN - F - FF 3 kN cut across shear planes a–a kN x-=F0;- F Now 0 consider == 30kN6 kN -FofF=segments (b) : : + the equilibrium (b) ©F = 0; F = 0 F = 3 kN : ! x Now consider equilibrium of segments and b–b, shown the in Figs. 1–23c and 1–23d. cut across shear planes a–a 0.1 m m 0.125 0.125 m m 0.1 Now consider thesegments of segments cut across planes a–a Now consider the equilibrium of cut across planes a–ashear and b–b, shown inshear Figs. 1–23c and 1–23d. a-a kesitindeki kuvvet durumu veequilibrium gerilme hali: Now consider the equilibrium of segments cut across shear planes a–a Fig. 1–23 and b–b, shown in : Figs. 1–23c and 1–23d. + ©F and (a) b–b, shown in Figs. 1–23c and 1–23d. (a) 0; shown Vain-Figs. 3 kN = 0 and V1–23d. and b–b, 1–23c x = a = 3 kN 3 kN + ©F = 0; Va = 3 kN : Va - 3 kN = 0 x 3 kN Fig. 1–23 1–23 +V + ©F +Va©F Fig. Va = : ©F 0; = 0: = 0 3 kN 3 kN V : 3 kN =kN30;kN 0 = 0Vb = V3 kN +ax3 ©F x = 3 kN x = 0; a 3 kN : V-a V -b 3=kN x = 0; a = 3 kN 3 kN ta ! 200 kPa Va + ©F : = 0; 3 kN V = 0 Vb = 3 kN x b + +ta ©F ! 200 kPa Va3©F : =V 0;b = 0Average 3 kNV+b- =V = 0Stress. Vb = 3 kN : kNx 3b kN x = 0; Shear : ©F 3 kN - Vb = 0 Vb = 3 kN x = 0; V! a Vtaa ! 200 kPa t ! 200 kPa V (c) a a Referring to the free-body diagram of the Average Shear Stress. SOLUTION 3 SOLUTION Average Shear Stress. b-b kesitindeki kuvvet ve gerilme (c) Average Shear Stress.durumu 3110 2N Vhali: a Shear Average Stress. (c) 1t 2 = = Ans. Internal Loadings. Referring to the free-body diagram of the= 200 kPa a avg 3110 N the Internal Loadings. Referring to the 3free-body diagram32of (c) AV a a =10.1 m210.15 m2 3 = 200 1t 2 = kPa Ans. 3110 member, Fig. 1–23b, 3 a avg2 N V 3= kN a 1–23b, a3 N - F - F = 0 member, FFig. kN 3110 2VN 3110 2Ans. N m2 200 m210.15 = = 200 kPa Aa V=a 10.1 kPa Ans. 1ta2avg = = 1ta2avg = 3 = 200 kPa 1ta2avg = = 3110 2 N Ans. Vbm2 Aa m210.1 m210.15 A3 kN 10.1 m210.15 10.1 m210.15 m2 3 = 160 kPa 1t0b2avg = VF=A Ans. + ©Fx = 0;a a3 kN 3 kN 6 kN F F = = + 2 N 3110 : ©F across = 0; shear 6 Vplanes kN - F 3 kN m210.15 m2 ilibrium of segments cut a–a- F = 0 3 kN : 10.125 3 =AbFb = tb = 160xkPa b 31103V = 160 kPa 1t3110 Ans. N = 2 bN 3 b2avg 2 s. 1–23c and 1–23d. 2N 3110Ans. V A 10.125 m2 ! 1ttbb2=avg160=kPaVb = 1tb2Vavg b b = = = 160 m210.15 kPa Ans. = 160 kPa b 2avgofacross =stress =on elements = 160 kPa a–a and b–b is Ans. 1t Now consider the equilibrium equilibrium of segments segments cut across shear planes a–a a–a The state located on sections Ab of 10.125 m210.15 m2 shear bcut A(d) m210.15 m2 Now planes Vb consider b the10.125 tb = Vb160 kPa Ab 10.125 m210.15 m2 Vb tb = shown 160 kPa in Figs. 1–23c and b–b, and 1–23d. The state of stress on elements located on sections a–a and b–b is shown in Figs. 1–23c and 1–23d, respectively. a-a ve b-b kesitindeki kuvvetleri yatay kuvvet dengesinden hesaplayalım: (d) and shown in Figs. 1–23c and 1–23d. V - 3 kN = 0 3 kN a =b–b, Theon state of stress on elements located on sections a–a and b–b is The state of stress elements located on sections a–a and b–b is (d) shown in Figs. 1–23c and 1–23d, respectively. The state of stress on elements located on sections a–a and b–b is (d) shown in Figs. 1–23c and 1–23d, respectively. shown in==Figs. respectively. ++b ©F V : 0; 1–23c Vaaand kN == = 3 in kNFigs. N - V =©F 3xxkN V shown 1–23c and 1–23d, respectively. V : 0; V -- 331–23d, kN 00 bkN= 0 aa = 3 kN 33 kN ss. Va a 6 kN + ©F = 0; : ! : ©Fxx = 0; + kN -- V Vbb == 00 33 kN kN Vbb == 33 kN V Average Shear Shear Stress. Ortalama gerilmeler: Average Stress. = 200 kPa Ans. m210.15 m2 3 3 3110 2 N Va 3110 2 N = 200 kPa 1taa22avg = Va == Ans. 3 avg 1t = Ans. 3110 2 N A 10.1 m210.15 m2 m2 = 200 kPa = 160 kPa Aaa 10.1 m210.15 Ans. 5 m210.15 m2 N 31103322 N Vbb 3110 V 2 = = = 160 kPa 1t Ans. b avg 2 = = 1t Ans. n elements located bonavgsections and m210.15 b–b is m2 Abb a–a 10.125 m210.15 m2 = 160 kPa Vb A 10.125 V b ! nd 1–23d, respectively. The state state of of stress stress on on elements elements located located on on sections sections a–a a–a and and b–b b–b is is The shown in Figs. 1–23c and 1–23d, respectively. shown in Figs. 1–23c and 1–23d, respectively. 11032 N !14 1.5 EXAMPLE 1.12 AVERAGE SHEAR STRESS 1.5 37 AVERAGE SHEAR STRESS 1 37 The inclined member in Fig. 1–24a is subjected to a compressive force of 600 lb. Determine the average compressive stress along the smooth 1.5 AVERAGE SHEAR S 1.12 ! EXAMPLE 1 areas of contact defined by AB and BC, and the average shear stress along the horizontal plane defined by DB. 1.5 AVERAGE SHEAR STRE The inclined member in Fig. 1–24a is subjected to a compressive force EXAMPLE 1.12 600 lb of 600 lb. Determine the average compressive stress along the smooth 5 4 ahşapshear birleşim areas of contact by AB BC, andverilen the average stress alt yüzeyinde zemin üzerine 1.5 AVERAGE SHEAR STRESSdefined 37 andŞekilde member in Fig. 1–24a is subjected to a compressive force 3 The inclined EXAMPLE 1.12 along the horizontal plane defined by DB. of 600 lb. Determine the averageDiyagonal compressive eleman stress along the smooth 1.5 AVERAGE SHEAR STRESS 37 ankastre mesnetlidir. üzerine 600 lb 1.5 AVERAGE SHEAR STRESS 37 600 lb of contact defined ABisand BC, and average shear stress Theareas inclined member in Fig. by 1–24a subjected tothe a compressive force lb birleşim 5 4 1.5 AVERAGE SHEAR STRESS 37600 kuvvet uygulanmaktadır. ABDB. vestress BC along çizgilerindeki along the horizontal defined by 1.5 AVERAGE SHEAR STRESS of lb.1Determine theplane average compressive the600 smooth 3 5 4 1.5 A VERAGE S HEAR S TRESS 37 areas of contact defined by AB and BC, and the average shear stress 600 lb 1 enkesit alanlarında meydana gelen 3 essive force ortalama gerilmeleri ve along the1 horizontal plane defined by DB. 5 4 A the EXAMPLE 1.12 3 lb sivesmooth force 1 600 600 lb DB hattındaki yatay düzlemdeki ortalama kayma ssive force shear stress 1.12 C 1EXAMPLE in. he smooth 5 4 1 5 4 B heforce smooth ve Fig. 1–24a ishesaplayınız. subjected to a compressive force 3 hear stress 3 D The inclined member ingerilmesini hear stress smooth The inclinedofmember 1–24a subjected to a compressive force the smooth 2in in.Fig. 600 lb. Determine theisaverage compressive stress along A r stress the average compressive stress along theaverage smoothshear stress Cby AB and 3 in.of contact areas defined BC, and the 1.5ofin.600 lb. Determine 1 in. areas of contact defined by AB BC, and the shear stress B and along the horizontal plane defined by average DB. (a) Fig. 1–24 A FAB Çözüm: along the horizontal plane defined by DB. D 600 lb C 2 in. SOLUTION 1 in.600 lb 5 4 600 lb B 3 in. in. A FBC Internal Loadings. The 1.5 free-body diagram of the inclined member 5 4 D 3 5 4 C (a) 600 lb Fig. 1–24 3 is shown in Fig. 1–24b. The compressive forces acting on the areas of 1 in. (b) 2 in. FAB 3 600 lb B 5 4 contact are SOLUTION 3 in. 1.5 in. 360 lb 5600 lb D 4 3 600 lb + ©F = 0; in. : FAB Loadings. - 600 lb A 3 B = 0 FAB =diagram 360 lb of the inclined (a)2 member Fig. x FBC1–24 Internal 5 4 3 5 The free-body 600 lbF 5 4 AB 3 in. 1.5 in. acting on the areas of 5 4 (b) 3 SOLUTION + c ©Fy = 0; is shown FBC in - Fig. 6003 1–24b. lb A 45 B =The 0 compressive FBC = 480forces lb (a) Fig. 1–24 3 contact are V A F FBC Loadings. The Also, from the free-body diagram of theInternal top segment ABD of thefree-body diagram of the inclined member360 lb AB C= 360 lb + ©F = 0; A lb A 3 B = 0 : -SOLUTION 600 600areas lb AB 1 AB in. is shown Fig. 1–24b. The compressive forces(c)acting on the of bottom member, Fig.x 1–24c, the Fshear force acting theF sectioned 5 (b Cinon B 1 in. 5 member 4 4 Loadings. g. 1–24horizontal plane F contact are DB is Internal The free-body diagram of the inclined BC c B + ©Fy = 0; FBC - 600 lb A 5DB = 0 FBC = 480 lb FAB 3 + ©F is shown inx Fig. acting the 2The in. compressive (b) + ©F = 0; Also, from the free-body : = the 0;1–24b. F lb A 35 Bof=forces 0 FAB V=on 360 lbareas of D . 1–24 : AB - 600 V = 360 lb diagram of top segment ABD the x F 2 in. g. 1–24 AB contact are 3 in. (c) 1.5 in. 600 lb FABbottom member, Fig. compressive 1–24c,+ shear force along acting the3lbsectioned FBC average ed member c ©F = 0; F -on 600 = 480 lb A=45 B 0= 0 F FBC 3the in. y stresses Average !Stress. The the + ! ©F –24 1.5 in. (a)FAB BC : = 0; 600 lb = 360 lb A B Fig. 1–24 5 4 x AB F 5 horizontal plane DB is AB areas of V (b) FAB of the horizontal and vertical of the inclined member are free-body diagram Also, from the of the top segment ABD FBCplanes (a) dhemember Fig. 3 4 1–24 FAB lb FBCF SOLUTION d member (c) c + + ©F = 0; F 600 lb = 0 F = 480 A B 360 lb DB yüzeyine etki eden yatay kuvvet: y BC BC AB bottom member, Fig. 1–24c, the shear force acting on the sectioned e areas of :s©Fx= =(b)0; V = Ans. 5 = 360 lb = 360 240 lb psi SOLUTION AB he areas of FBC member V (b) 240 psi F Loadings. Thethe free-body of the inclined member ABD of the horizontal plane DBdiagram is diagram AInternal 11 in.211.5 in.2 Also, from free-body of the top segment BC AB Average Stress. compressive stresses along the 360 The lbThe average areas of (c) FBC Internal free-body diagram of the inclined member (b)Loadings. is shown in Fig. 1–24b. The compressive forces acting on the areas of bottom member, Fig. 1–24c, the shear force acting on the sectioned (b) 360 lb + Fand lb :of©F BC vertical480 horizontal planes thex inclined member are = 0;forces V = 360 of lb is shown in Fig. 1–24b. The compressive acting on the areas (b) Ans. s = = = 160 psi contactlbare Fhorizontal plane BC 160 psi V 360 lbDB is 360 lb ABC 36012 in.211.5 AB in.2 BD of the contact are 3 = The Average Stress. average compressive stresses along the360 lb + Ans. s = = 240 psi AB : ©F = 0; F 600 lb = 0 F = 360 lb + A B (c) x AB AB 240 psi 600 lb 5 : ©F = 0; V = 360 lb A 11 in.211.5 in.2 3 e sectioned are shown AB horizontal + ©F = 0; : FABin! -Fig. 6001–24d. lbxA 5 B = 0and vertical F = planes 360 lb of the inclined member are BD of These the stress! VVdistributions x 4 AB (d) 5= 4 0; BD of the The average shear stress acting on the horizontal plane defined by (c) + c ©F F 600 lb = 0 = 480 A B F 360lb lb F 600 lb 480 lb y BC Average Stress. The average stresses along the ABFBC compressive 5 BC sectioned 4 (c)0; 3 =600 V Ans. s = = 240 psi lb c Ans. s = = 160 psi + ©F = F 600 lb = 0 F = 480 lb A B AB sectioned DB is of the y 160 psi 5 4BC A 240 psi horizontal anddiagram vertical planes of member are 360 lb V Also, BC from the the topthe segment ABD AAB 11inclined in.211.5 in.2 of the 125 in.211.5 in.2BCof BCfree-body (c) V 5 600 lb 4 360 lb (c) ctioned 3 Also, from the free-body diagram of the top segment ABD of the 600 lb F 360 lb member, Fig. the1–24d. shearAB force acting 480 on the tavg =bottom Ans. 801–24c, psiin Fig. These stress distributions are = shown FBC 3 lb sectioned (c)Ans. sAB = = the sectioned = 240 psi 5 in.211.5 4Fig. 1–24c, along the 600 lb 13 in.2 bottom member, the shear force acting on 5 4 240 psi Ans. s = = = 160 psi horizontal planeacting DB ison the horizontal (d) 11 in.211.5 by in.2 BC AABplane The average shear stress ABC defined 12 in.211.5 in.2 5 4 3 horizontal plane3DB is uniformly distributed over the sectioned area in along This the stress is shown 80 psi + DB is FBClb are shown 480 lb 3 : ©Fx = 0; These stress distributions V = 360 along Fig. the1–24e. in Fig. 1–24d. 360 lb Ans. s = = = 160 psi + ©F = 0; (e) Ans. lb lb BC V360 = average 360 x ( A 12 in.211.5 in.2horizontal 240: psi ng the The stress acting onAns. the BC = =shear 80compressive psi Averagetavg Stress. The average stresses along theplane defined by 13 in.211.5 in.2 Ans. DBstress is planes Stress. average compressive stresses along the These distributions are shown in Fig. 1–24d. horizontalThe and vertical of the inclined member are 240Average psi Ans. is shown uniformly distributed over the lb sectioned area in (d) 240This psi stress and Ans. 360 lbhorizontal horizontal vertical planes of the inclined member are The average shear stress acting on the plane defined80bypsi F 360 160 psi AB Ans. tavg = Ans. = 80 Ans. psi s = = = 240 psi (e) F 240 psiFig. 1–24e. AB 360 lb DB AB is 13 in.211.5 in.2 240 psi Ans. AAB 11 in.211.5 in.2 Ans. sAB = = 240 psi 160 psi = Ans. 360distributed lb 240 psi A 11 in.211.5 in.2 160 psi This stress is shown uniformly over the sectioned area in AB (d) tavg Ans. = 80 psi FBC 480= lb defined Ans. by 13 in.211.5 in.2 Ans. s = = = 160 psi Fig. 1–24e. 160 psi (e FBC BC 480 lb 12 in.211.5 in.2 160 psi (d) ! Ans. s360 = This! =stress A = 160 psidistributed defined by BC lb isBC shown uniformly over the sectioned area in 160 psi (d) 80 p A 12 in.211.5 in.2 defined by BC These stress distributions are shown in Fig. 1–24d. Ans. Fig. 1–24e.gerilmesi: (e) (d) distributions 360 lb ined by DBThese yüzeyindeki ortalama kayma (d) stress are shown in Fig. 1–24d. The360 average shear stress acting on the horizontal plane defined by lb (d) Ans. The average shear ned area in DB islb stress acting on the horizontal plane defined by 80360 psi Ans. 360 lb DB is 360 lb (e) Ans. in ed area 360 lb 80 psi t Ans. = = 80 psi avg lb 360 ed area in 80 psi 13 in.211.5 in.2 tavg = Ans. = 80 psi (e) 13 in.211.5 in.2 distributed over the sectioned area in area in (e) 80 psi This stress ! ! is shown uniformly 80 psi (e) is This stress shown uniformly distributed over the sectioned area in 80 psi Fig. 1–24e. (e) Fig. 1–24e. (e) !15

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